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I
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^i^^irnpht^n ,md f,^. t,,,,, ^gy
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LONDON :
PRINTED BY MILLS, JOWETT AND MILLS,
(late bensley)
BOLT-COURT, FLEET-STREET.
• » •
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TO THE RIGHT HONOURABLE
THE (latb) lords COMMISSIONERS
FOR EXECUTING THE OFFICE OF LORD HIGH ADMIRAL
OP THE
UNITED KINGDOMS OF GREAT BRITAIN AND IRELAND^
THE RIGHT HONOURABLE ROBERT, LORD VISCOUNT MELVILLE, K.T.
VICE-ADMIRAL SIR WILLIAM JOHNSTONE HOPE, G.C.B.
VICE-ADMIRAL THE RIGHT HON. SIR GEORGE COCKBURM, G.C.B.
SIR GEORGE CLERK, BART.^-and
W1LUAM ROBERT KEITH DOUGLAS, ESa
My Lords,
In brin^g to a close the following Treatise, I feel that
it cannot with so much propriety be inscribed tb any other
department in the State, as to that which has so successfully
presided over, and so long and judiciously directed the Naval
operations of Great Britain.
It will ever be to me, my Lords, a cause of the most sin-
cere gratitude, that to the condescension of your Lordships,
in accepting the Dedication of my mathematical labours, I
am principally indebted for the encouragement and support
which I have received, in presentiDg the result of those
labours to the Royal Naval Service of His Majesty^ and to
the Merchant Service, in general, of the British Empire.
I have the honour to be,
My Lords,
With the utmost deference.
Your Lordships* most humble,
And most obedient Servant,
THOMAS KERIGAN.
PtfrUmouihf
Dtcember, 1827.
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LIST OP SUBSCRIBERS. XXI
Capt Arthur Fanshawe, R.N.
Capt. B. M. Festiog, R.N., Fareham, Hants.
Capt. Peter Fisher, R.N.
Capt. Oshome Foley, R.N.
Capt. Foster, R.N., F.R.S.
Lieut. Edmd. H. Fitzmaurice, Scout Revenue Cutter.
Mr. Thomas Fairweather, Purser, H.M.S. Wolf.
J. M. French, esq., Royal Exchange, London.
Rear Admiral John Giffard.
Capt. Sir James A. Gordon, K.C.B., R.N., Resident Commissioner^ Plymouth
Hospital.
Capt. Henry Garrett, Resident Commissioner, Haslar Hospital.
Capt. Robert Gambier, R.N.
Capt. George C. Gambier, R.N.
Lieut R. F. Gambier, H.M.S. Asia.
Capt. J. G. Garland, R.N., Poole.
George Garland , esq., Poole.
Capt Charles Gordon, H.M.S. Cadmus.
Capt. Thomas S. Griffinhoofe, H.M.S. Primrose.
Lieut. 0. G. Sutton Gunning, H.M.S. Wellesley.
Mr. Jas. Geary^ R.N., Portsmouth.
Joseph Grout, esq. Stamford Hill, Middlesex.
•Vice Admiral Peter Halkett, Uplands, Fareham, Hants.
Rear Admiral G. E. Hamond, C.B., Yarmouth, Isle of Wight
Rear Admiral Sir Tliomas M. Hardy, hart., K.C.B.
Capt H. C. Harrison, R.N., Southampton.
Capt. Henry Haynes, R.N.
Capt John Hayes, C.B., R.N., Shallots, Hants.
Capt William Hendry, R.N., Kingston Crescent
Capt. P. Heywood, R.N., Highgate, 2 copies.
Capt. T. Huskissou, R.N., Paymaster of the Royal Navy.
Lieut. George Hales, R.N.
Lieut Frederic Hutton, H.M.S. Dispatch.
Lieut Charles Hopkins, (b) R.N.
George Hall, esq., Chichester.
Mr. Harrison, Bookseller, Portsmouth, 12 copies.
Mr. Harvey, Royal Naval College, Portsmouth Dock^yard.
Mr. T. S. Herring, Daniel Street, Portsea.
Edward James Hopkins, M.D., Queen-square, St. James's Park*
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XXll LIST OF SUBS^CBIBERS.
Capt, the Hon. C. L. Irby, H.M.S. Ariadne*
Lieut. R. Ingram, H.M.S. Gloucester.
Admiral Jones, 10, Curzon Street, May Fair.
Capt. Theobald Jones, R.N., Bamibttry Row^ Islington.
Mr. Jeringham, H.M.S. Galatea.
The Rev. J. Kirkby, Sheemess Dockyard.
Capt. Abraham Lowe, R.N.
Lieut. Gower Lowe, H.M.S. Valorous.
Alexander Lumsdale, esq.. Master Attendant, Plymouth Dock-yard.
Mr. H. Lawrence, R.N., Kingston, near Portsea.
Mr. Thomas Lock, Weymouth, 2 copies.
Capt. The Hon. J. A. Maude, H.M.S. Glasgow.
Capt. Jas. Mangles, R.N.
Capt. Joseph Maynard, R.N.
Capt. W. J. Mingaye, H.M.S. Hyperion*
Capt. Andrew Mitchell, R.N.
Capt. John Molesworth, R.N. Clapham.
Capt. C. R. Moorsom, R.N.
Capt. William Mudge, R.N.
Lieut S. Meredith, H.M. Cutter Vigilant
Capt. C Morton, R.N., Lower Eaton Street, Grosvenor Place.
J. M'Crea, esq.. Surgeon, R.N., Bamsbury Row, Cloudesley Square.
John M' Arthur, esq.^ Hinton Lodge^ Homdean, Hants.
Mr. George Miller, R.N., Portsmouth.
Lieut Thomas M'Gowan, R.N.
Admiral the Right Hon. Earl Northesk, Commander in Chief, Plymouth.
Capt. the Right Hon. Lord Napier, R.N.
Lieut. H. Nurse, R.N., Pinner, Middlesex.
Mr. Joseph Nalder, Guildhall, London.
Rear Admiral Sir E. W. C. R. Owen, K.C.B. and M.P.
Rear Admiral R. D. Oliver, Dublin.
Rear Admiral the Right Hon. Lord James 0*Bryen.
Capt. Hayes O'Grady, R.N.
Capt W. F. W. Owen, H.M.S. Eden.
B. E. O'Meara, esq., Montague Square.
Mr. Joseph Oakey, R.N.
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LIST OV SUBSCRIBBRS* Xxiii
Vice Admiral C. W. Patenon, Coaham, Hants.
Capt. Lord William Paget, WiUiam and Mary Yacht.
Cape William E. Parry, F.R.S., R.N., Hydrographer td tlie Admiralty.
Capt Charles G. R, Phillott, R.N.
Capt. W. H. Pierson, R.N., Havant.
Capt H. Prescott, C.B., R.N., FarDham, Surrey.
lient J, T. Paulson, R.N.
Mr. J. B. Paddon, H.M.S. Galatea.
George Peel, esq., George Yard, Lombard Street
Mr. Joseph Pym, Bartholomew Close.
Capt J. C. Ross, R.N.
Capt Edwin Rkhards, R.N.
Lieut Harry B. feichaitls, R.N.
Lieut Curtis Reid, R.N., Southampton.
Lieut Beoj. Roberts, H.M.S. Wolf.
Mr. PercETal Roberts, H.M.S. Wolf.
Lieut Edward Rogier, R.N.
Mr. Rolhtnd, H.M.S. Galatea.
The Right Hon. Earl Spencer, K.G., &c. &c.
Admiral the Hon. Sir R. Stopford, K.C.6., Commander in Chief, Portsmouth.
Thomas Asherton Smith, esq., M.P.^ Penton Lodge, Andorer.
Capt. W. Sanders, R.N., Kingston^ Portsea.
Capt Thomas Sanders, H.M.S. Maidstone.
Capt. G. R. Sartorius, H.M.S. Pyramus.
Capt G. F. Seymour, C.B., R.N., Hampton Court
Capt. Charles Shaw, R.N.
Capt Henry Shifiher, R.N., Sompting Abbotts, Shoreham.
Capt Houston Stewart, R.N.
Capt Charles Stnmgways, R.N.
Capt C. B. Strong, R.N., King's Terrace, PorUmouth.
Capt H. E. P. Sturt, R.N.
Lieut Archibald Sinclair, R.N.
Lieut. M. A. Slater, R.N.
Lieut Thomas Spark, H.M. Revenue Cutter Fancy.
Lieut John Steane, R.N., Ryde.
Lieut W. B. Stocker, R.N., Poole.
Lieut George F. Stow, H.M.S. Espoir.
The Rev. T. Surridge, H.M.S. Ocean.
The Rev. J. E. Surridge, M.A., R.N.
Mr. George Starr, R.N.
Mr. George Saulez, Alton, Hants.
Mr. W. D. Snooke, Professor of Mathematics, Ryde, Isle of Wight.
Mr. W. Selby, Portsmouth.
c2
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XXiv LIST OF SUBSCRIBERS.'
Capt. N. Thompson, H.M.S. Rerenge.
Capt. John Tancock, R.N.
Capt. John Jervis Tucker, R.N., Trematon Castle, Plymouth.
Lieut. John Thompson (5), R.N., North Potherton.
Mr. Thomas P. Thompson, H.M.S. Pyramus.
llie Rey. John Taylor, H.M.S. Ramiliies.
Mr. S. Tuck, R.N., Kingston Cross, Portsea.
Mr. Joseph Tizard, jun., Weymouth, 2 copies.
The Hon. G. Vernon, Ryde, Isle of Wight.
Capt. A. E. T. Vidal, R.N.
Commodore J. C. White, R.N.
Capt. James Wemyss, R.N. and M.P., Wemyss.
Thomas P. Williams, esq., M.P., Berkeley Square.
lieut. H. Walker, R.N., Cosham, Hants.
Lieut. William Wilson, H.M.S. Challenger.
Lieut. Joseph C. Woolnough, Com. H.M. Cutter Surly.
Lieut. J. L. Wynn, R.N.
Edward D. Warrington, esq., Charles Square, Hoxton.
Thomas S. Whitney, esq., Newpass, Rathone, Ireland.
Mr. Thomas Woore, H.M.S. Alligator.
The Right Hon. Lord Viscount Yarhorough, 2 copies.
Captain Thomas Young, R.N., Fareham, Hants.
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CONTENTS.
X
DESCRIPTION OF THE, TABLES.
Table. Page.
I. To conyert longitude, or d^;rees into time, and conversely 1
II. Depreflsion of the horizon -. ., 3
III. Dip of the horizon at different distances from the obsenrer 6
IV. Augmentation of the moon's semi-diameter 8
V. Contraction of the semi-diameters of the sun and moon • • 11
VI. Parallax of the planets in altitude 12
VII. Parallax of the sun in altitude 13
VIII. Mean astronomical refraction. • w 13
IX. Correction of the mean astronomical refraction 15
X. To find the latitude by the north polar star 17
XI. Correction of the latitude deduced from the preceding table. • . • • • 20
XII. JVIean right ascension of the sun 21
XIII. Equations to equal altitudes of the sun, part First 22
Xiy. Equations to equal altitudes of the sun, part Second 22
XV. To reduce the sun's longitude, right ascension, and declination ;
and, also the equation of time, as given in the Nautical Almanac,
to any given time under a known meridian 25
XVI. To reduce the moon's longitude, latitude, right ascension, declin-
ation, semi-diameter, and horizontal parallax, as given in the
Nautical Almanac, to any given time under a known meridian 30
XVII. Equation of the second difference of the moon's place 33
XVIIL Correction of the moon*8 apparent altitude 38
XIX. To reduce the true altitudes of the sun, moon, stars, and planets,
to their apparent altitudes 40
XX. Auxiliary angles • 42
XXI. Correction of the auxiliary angle when the moon's distance from a
planet is observed 45
XXII. Error arising from a deviation of one minute in the parallelism of
the surfaces of the central mirror of the circular instrument of
reflection 46
XXIII. Error arising from an inclination of the line of collimation to the
plane of the sextant, or to that of the circular instrument of re-
flection • 47
XXIV. I^rithmicdiference .,-,-j,.,<3oOgl^
XXvi CONTENTS.
Table. Page.
XXV. Correction of the logarithmic difference for the sun's, or star's appa-
rent altitude 61
XXVI. Coirection of the logarithmic difference for a planet's apparent
altitude 52
XXVIL Natural Tersed sines, and natural sines 53
XXVIII. Logarithms of nifmb^rs ^ , 62
XXIX. Proportional logarithms 75
XXX. . Logarithmic half elapsed time • 84
XXXI. Logarithmic middle time 86
XXXII. Logarithmic rising 87
XXXIII. To reduce points. Qf the compass to degrees, and conversely 89
XXXIV. Logarithmic sines, tangents, and secants to every point and quar-
ter point of the compass 89
XXXV. Logarithmic secants to every second in the semi-circle ••...... 90
XXXVI. Logarithmic sines to every second in the semicircle 93
XXXVII. Logarithmic tangents to every second in the semicircle 97
XXXVIII. To reduce the time of the moon's passage over the meridian of
Greenwich ta the time of her passage over any other meridian 100
XXXIX. Correction to be applied to the time of the moon's reduced transit
in finding the time of high water at any given place 102
XL. Reduction of the moon's horizontal parallax on account of tfie
spheroidal figure of the earth 104
XLL Reduction of terrestrial latitude on account of the sphenoidal
figure of the earth 105
XLII. A genera] traverse table, or difference of latitude and departure 106
XLIIL Meridional parts , 113
XLIV. The mean right ascensions, and declinations of the principal fixed
stars 114
XLV. Acceleration of the fixed stars, or to reduce sidereal time into
mean solar time 117
XLVI. To reduce mean solar time into sidereal time 119
XLVII. Time from noon when the sun's centre is in the prime vertical ;
being the instant at which the altitude of that object should be
observed, in order to ascertain the apparent time with the great-
est accuracy 119
' XLVIII. Altitude of a celestial object (when its centre is in the prime ver-
tical), most proper for determining the apparent time with the
greatest accuracy 120
XLIX. Amplitudes of a celestial object reckoned from the true east or
west point of the horizon 122
L. To find the times of the rising and setting of a celestial object. . . . 123
LI. For computing the meridional altitude of a celestial object, the
latitude and the declination being of the same name 138
LII. For computing the meridional altitude of a celestial object, the
latitude and the declination being of contrary names 138
LIU. The miles and parts of a mile in a degree of longitude at every
degree of laatttde.,.; •. 144
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LIV. IVfifXtflioiial miles lor congtnictiag MeroaWr*a chltfW 145
LV. Tq find the disUace of terre^triftl objocto at tea 147
LYI. To leducse the Freach oentMiioal d^yiiioa of the cirde into the
Eng^ah aexafesiiaal diwioB; or, to reduce Fiench degrees
into EngUah degrees, and odhversely • . . • ^ 150
hVlU A g«^eral table for gaagiogt or finding the content of all circular
headed casks 1 52
LVIII. Latitudes and longitttdes of the prinoipal aea-poits, islands, capes,
shoals, &c in the luiawn world ; with tbe time of high water,
at the full and ch^Age of the moon, at all plaees where it is
known ••.... , 154
Alphabeticai refeienoe to. the pieoeding table 155
Form of a treni^t table •.. 155
Miscellaneous niunbeca with their corresponding logarithms .... 155
A table showing the tn^e time and degree at which the hour and
minute hands of a well-regulated watch, or clock, should
exactly meet, or be in coi\|unction, &c. in every revolution. ... 155
A concise system of decimal arithmetic • 156
Solution of Problems in Plane, and Spherical Trigono-
METRT 168
Plane trigonometry, solution of right aiigled triangles 171
solution of oblique angled triangles. • ' 177
Sphencal trigonometry, solution of right angled triangles . . 1 81 1 82
solution of quadrantal triangles 193
solution of oblique angled triangles . . . • 197
Nayisation.., •• ..••.. 211
Solution of problems relative to the difference of latiUide and dif-
ference of longitude • 214
Solution of problems in parallel sailing 217
middle latitude sailing , 221
Mercator's sailing 236
oblique sailing ^ 255
windward sailing 262
current sailing 266
Solution of pioUems relative to the errors of the log line and the
half minute glass 272
Solution of a very useful problem in great circle sailing 276
To find the time of high water at any known place • • •• 103
. To make out a day's work at sea by inspection •••• 249
SoLUTiOK or Problsms jk Nautical Astronomy »... 296
L To convert longitttde, or parts of the equator into time 296
IL To conTjert time into longitude or parts of the equator • 296
HI. Given the time under any known meridian, to find the corres*
pondiof time it Gifsnwicb •,•••.•««•.«*• ••«««.f«ft,t 397
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XXviii ' CON TBNTS •
Problem. Page.
IV. GiveD the time at Greenwich, to find the corresponding time
under a known meridian. • 297
V* To reduce the sun's longitude, right ascension, declination, and,
also, the equation of time as given in the Nautical Almanac,
to any other meridian, *and to any given time under that
meridian ..••••• • • • • * 2^98
VI. To reduce the mpon*s longitude, latitude, right ascension, declin-
ation, semi-diameter, and horizontal parallax, as given in the
Nautical Almanac, to any other meridian, and to any given
time under that meridian ••••• 302
VII. To reduce the right ascension and declination of a planet, as
given in the. Nautical Almanac, to any given time under a
known meridian • ; 307
VIII. To compute the apparent time of the moon's transit over the me-
ridian of Greenwich '. •••••• 309
IX, Given the apparent time of the moon's transit over the meridian
of Greenwich, to find the apparent time of her transit over any
other meridian • • • • • • 312
X. To compute the apparent time of a planet's transit over the meri* -
dian of Greenwich •• •• .••• 313
XI. Given the apparent time of a planet's transit over the meridian of
Greenwich, to find the apparent time of its transit over any
other meridian •••.»«••.. • 315
^^ XII. - To find the apparent time of a star's transit over the meridian of
any known place •••.••••'-....••••.. 317
XIII. To find what stars will be on, ov nearest to the meridian at any
given time ..•••• ••••••^•••••«« 319
XIV. Given the observed altitude of the lower or upper limb of the
sun, to find the true altitude of its centre , 320
XV. Given the observed altitude of the lower or upper limb of the
moon, to find the true altitude*of her centre 323
XVI. Given the observed central altitixde of a planet, to find its true
altitude 325
XVII. Given the obsen'ed altitude of a fixed star, to find its true
altitude 327
SoLUTiOK Of Problems relative to the Latitude.^ •••• 328
I. Given the sun's meridian altitude, to find the latitude of the
place of observation ...• 328
II. Given the moon's meridian altitude, to find the latitude of the
place of observation . . • • • .••••••••.. 330
III. Given the meridian altitude of a planet, to find th* latitude of
the place of observation • t 333
IV. Given the meridian altitude of a fixed stav, to find the latitude of
the place of observation • • • • 335
V. Given the meridian altitude of a celestial object observed below
the pole, to find the latitude of the place of observation. . • . • . 336
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CONTENTS. XXix
Problem. • Page.
VI. ' ' Given the altitttde of the north polar star^ taken at any hour of
the night ; to find the latitude of the plape of dbeerration .... 337
VII. Oiren the latitude by account, the sun's declination, and two
observed altitudes of its lower or upper limb ; the elapsed dme^
and the course and distance run between the observations ; to
find the latitude of the ship at the time of observation of the
greatest altitude ..» « 341
VIII. Given the altitudes of two known fixed stars observed at the same
instant, at any time of the night ; to find the latitude of the
place of observation, independent of the latitude by account^
the longitude, or the apparent time of observation 347
IX. - Given the latitude by account, the altitude of the sun's lower or
upper limb, observed within certain limits of noon, the
apparent time of observation, and the longitude ; to find the
true latitude of the place of observation • 354
X. Given the latitude by account, the altitude of the moon's lower or
upper limb, observed within certain limits of the meridian, the
apparent time of observation, and the longitude' ; to find the
latitude of the place of observation « ••••.. 358
XI. Given the latitude by account, the altitude of a planet's centre
observed within certain hmits of the meridian, the apparent
time of observation, and the longitude ; to find the true latitude
of the place of observation 362
XII. Given the latitude by account, the altitude of a fixed star observed
within certain limits of the meridian, the apparent time of
observation, and the longitude ; to find the true latitude of the
place of observation •••••••.••••«.. 365
To find the latitude by an altitude taken near the meridian below
thepole 368 369
Captain Wilfiam Fitzwilliam Owen*^B general Problem for finding
the latitude r 371
Xin. Given the interval of time between the rising or setting of the
sun's upper and lower limbs ; to find the latitude • . . 373
SOLUTIOV or PROBI.EHS RELATIVE TO THE APPARENT TiME 375
I. To find the error of a watch by equal altitudes of the sun 377
II. To find the error of a watch by equal altitudes of a fixed star .. 380
^'UI. Given the latitude of a place, and the altitude and declination of
the sun ; to find the apparent time of observation, and, thence,
the error of the watch. Method I ••... 383
Method II. Of computing the horary distance of a celestial
object from the meridian •••••• 388
Method III. Of computing the horary distance of a celestial
object from the meridian. • 390
Method IV. Of computing the horary distance of a celestial
. object from the »«ddiaA.t,« 392
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Problem. • J'oQI^-
)V. Gifen the Mtude ami lovyitiide of ^ pkce^ tl^ %Ktit«4e» n^l
iu8ceiiai(ui, and dec^ifAtiaa of a known fixed star, and the sun's
right ascension ; to find the apparent time •.•.•..«....••... 394
V. Qiven the latitude and longitude of a place, and the a].titude of a
planet ; to find the apparent time of obserT^tioa. . . ^ 397
Vf , Given th^ latitude and longitude of a place, the estimated time
at that place, and the altitude of the moon*s limb ; to find the
apparent time of observation •••«••..•• ^« i^« ., .^ • .^ ..«•• . 400
SoLDTIOir pF PaOI^LVMS E^LATIVE TO riKDIVG THE AX'TJ^TUDES OF
THE Heavenly Bqdies.v ••«. .,., , ^••,«, , ^..^ 403
L Given the latitude and longitude of a place, and the apparent
dpie at that place ; to find the true a^d thet apparent altitude
of the sun's centre • • * 404
II. Given the latitude and longitude of a place, and the apparent
time at that place ; to find the true and the apparent altitude
of a fixed star •«•••... , 406
III. Given the latitude and longitude of a place, and the apparent
time at that place ; to find the true and the apparent altitude
of a planet 408
ly. Given the latitude and longitude of a place, apd the apparent time
at that place ;• to find the true and the apparent altitude of the
moon's centre • « • • 410
Solution of PaoBLEMs relative to the Lqvg^tude ••• 413
I. To convert apparent Ume into mean time • • 415
II. To convert mean time at Greenwich into apparent time 416
III. Given the ktitude of a place, and the observed altitude of the
sun*s limb ; to find the longitude of that place by a chrono-
meter or time-keeper 417
IV. Given the latitude of a place, and the. observed altitude of a
known fixed star ; to find the longitude of that place bj a
chronometer or time-keeper , 420
V. Given the latitude of a place, and the i>b8erved altitude of a
planet ; to find the longitude of the place of observation by a
chronometer or time-keeper 423
VI. Given the latitude of a place, and the observed altttode of the
moon's limb ; to find the longitude of the place of observation
by a chronometer or time-keeper. 426
VII. To find the longitude of a ship or place by celestial observation,
commonly called a LuKAR Observation ..••• • 431
Method I. Of reducing the apparent to the true central
distance 433
Method 11. Of reducing the apparent to the true central
distance 436
Method IIL Of reducing the apparent to the true central
fliitaooe ^••«,,.«o*ftff • ft 439
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Method IV. Of reduciag the appiprent to the tnie central
difitancse. •• 439
Method V. Of reducing the apparent to the true central
distance • ^ 441
Method . VI. Of reducing the apparent to the tr4ie central
distance : ; 442
Method VII. Of reducing the apparent ta the true central
distance 443
Method VIII. Of reducing the apparent to the true central
.distance • 445
Method IX. Of reducing the apparent to the true central
distance 447
Method X. Of reducing the apparent to the true central
distance .• 448
Method XI. Of reducing the apparent to the trae central
distance • 450
Method XII. Of reducing the apparent to the true central
distance • 451
Method XIII. Of. reduciBg the apparent to the true central
distance • • 453
Vill. Given the apparent time, and the true central distance between
the moon and sun, a fixed star, or planet ; to determine the
longitude of the place of observation • 454
IX. Given the latitude of a place, its longitude by account, the
' observed distance between the moon and sun, a fixed star, or
a planet, and the observed altitudes of these objects ; to find
the true longitude of the place of observation ••••«.•,•••••• 456
X. Given the observed dbtance between the moon and sun, a fixed
star, or planet, the apparent time, with the latitude and longi-
tude by account ; to find the true longitude of the place of
observation 470
Xi. To find the longitude of a place by the eclipses of Jupiter's
.satellites 478
XII. To find the longitude of a place by the eclipses of the moon .... 481
SOLITTIOV OF PrOBLBI^S RELATIVE TO TH* VaRIATIOK OP THE
Compass • 483
I. (Hven the latitude of a place, and the sun's magnetic amplitude ;
to find the variation of the compass • 484
II. Given the latitude of a place, the sun's altitude, and his magnetic
azimuth ; to find the variation of the compass 487
A new method of computing the truo azimuth of a celestial
object \ 490
III. To find the variation of the compass by observations of a circum-
polar star 492
IV, To^find ibe varit^tioQ of the compass hj the maj;netic bearing of
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XXxii CONTENTS.
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a fixed star or planet, taken at the time of its transit over the
meridian of any known place 494
V. Giyen the true course between two places^ and the variation of
the compass ; U> find the magnetic or compav coarse ••.••• 496
VI. ' Given the magnetic course^ or -that steered by compa»» and the
variationrof the compass; to find*the true course.* •'••••••••• 497
Description of an improved azimuth compass card • • • • 497
\'
Solution of PaoBLEMs relative to the Risivg akd Setting
OF THE Celestial Bodies ..•••••. • ••••«.•• 500
I. Given the latitude of a place, and the height of the eye above
the level of the horizon ; to find the apparent times of the sun's
rising and setting • •••:.. •• 500
II. Given the latitude of a place, and the height of the eye above ,
the level of the horizon ; to find the apparent times of rising
and setting of a fixed star .•••••• • 504
III. Given the latitude of .a place, and the height of tbe eye above the
level of the horizon ; to find tbe apparent times of a planet's
rising and setting, • •• 506
IV. Given the latitude of a place, and tbe height of the eye above
the level of the horizon ; to find the apparent times of the
moon's rising jind setting •••••• • • 5l 1
V. Given the latitude and longitude of a place, and the day of the
month ; to find the times of the beginning and end of twilight,
and the length of its dilation .•..•••...••« 516
VI. Given tbe latitude of a place ; to .find the time of the shortest
twilight, and the length of its duration • •••••• 519
VII. Given the httitude of a place between 48° 32' and 66'>32' (the
limits of regular twilight)'; to find when real night or darkness
ceases, and when it commences .•••••••••••• {^20
VII L Given the latitude, and the sun's declination ; to find the interval
of time between the rising or setting of the upper and lower
limbs of that luminary ., , , , 620
Solution of Problems in Gnomonics oa Dialling 522
L Given the latitude of a place ; to find the angles which the hour-
lines make with the substyle, or meridian line of a horizontal
•sun-dial *, 523
II. To find the angles on the plane of an erect direct south dial for
any proposed north latitude, or on that of an erect direct north
dial for any proposed south latitude 526
Solution of Problems relative to the Mensuration of
Heights ani^" Distances 528
I. To find the height of an accessible object 529
II. Given the angle of elevation, and the height of an object ; to find
the observer's horizontal distance from that object , . 530
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CONTENTS. XXXIU
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III. To find the height of an ioaccestdble object 531
IV. To find the distance of an inacceesible object, which can neither
be receded from nor approached, -in its vertical line of
direction . • • • 532
V. To find the distai\ce between two inaccessible objects •. . • • 534
VI. Given the distances between three objects, and the angular dis-
tances between those objects taken at any point in the same
hoii^ntal plane ; to find the distance between that point and
each of the objects •.... 53^
VIL Given, the distances between three objects, and the angular dis- '
* tances between those objects taken at any point within the
' triangle formed by the right lines connecting them; to
find, the distance between that point and each of -the. objects. • 539
VIII. Given the distances between three objects situated in a straight
line, and the angular distances between those objects taken at
any point in the same horizoikal plane ; to find the distance
between that point and each of the objects, • 541
IX. Given the height oi, the eye, to find the distance of the visible
horizon ••••.-••••••• ••. 544
X» Given the measured length of a base line, to find the allowance
for the curvature or spherical figiure of the earth 545
XL Given a base line measured on any elevated tevel, to find its true
measure at the suriace of the sea • 547
XII. To find the height and distance of a hill or mountain 549
XIII. To find the height of a mountain, by means of two barometers
and thermometers •••••.••••••••• 550
XIV. To find the distance of an object by observing the interval of '
time between seeing the flash and hearing the report of a gun
or of a thunder cloud • • ••••.. 552
XV. Given three bearings of a ship sailing upon a direct course, and
the intervals of time between those bearings ; to find the course
steered by that ship, and the time of her nearest distance firom
the' observer. . • ••••••••• ••••••..•••••• 553
SOLUTIOH OV PkOBLEMS IN PfiJiCTICAL GuNNERY 557
I. . Given the diameter of an iron ball, to find its weight ..•..••• 557
II. Given the weight of an iron ball, to find its diameter 558
III. Given the diameter of a leaden ball, to find its weight 558
IV. Given the weight of a leaden ball, to find its diameter •••..... 559
V. Given the internal and external diameters of an iron shell, to
find its weight 560
VI. To find how much powder will fill a shell • • 561
111. To find the size of a shell to contain a given weight of powder. • 562
VIII. To find how much powder will fill a rectangular box 562
IX. To find the size of a cubical box to contain a given weight of
powder «;••«••• t •..• • • • , 563
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X3tkiv
Problem,
X.
XL*
XII*
xni.
XIV.
XV.
XVI.
XVII.
XVIII.
XIX.
XX.
XXI:
xxn.
XXIII.
XXIV.
XXV.-
XXVI.
XXVII.
XXVIIL
XXIX.
XXX.
XXXI.
XXXII.
XXXIIL
CONTENTS,
Pistge.
Te find 'how much powder will fill a cylinder , 564
To find What length of a cylinder •will be filled hf a given
weight of powder , 565
To'find the number of balls or shells in a triangular pile ..•••. 566
To find the number of balls or shells ifl a square pile 567
To find the number of balls or shells in a rectangular pile. . . . ; . 567
To find the number of balls or shells in an incomplete triangular
pile ; 4 568
To find the number x)f balls or shdk in an incomplete square pile 569
To find the number of balls or bhells in an incomplete rectkngnlar
pile .; ;i i..., 570
To find the velocity of any shot or shell. .;.••• '. 57 1
To fitid the terminal velocity of a shot or shell ; that is, the
greatest velocity it can acquire in descending through the air
by its bwn weight 572
To find the height from which a body must fall, in vacuo, in
order to acquire a given velocity. .' 573
Concise Tables for determining the greatest horizontsd range of
a shot or shell, when projected in the air with a given velo-
city ; with the elevation of the piece to produce that range 574
To find the greatest range of a shot or shell, and the elevation of
the piece to produce that range .'...•;••.•....•••• 575
Given the range at one elevation, to find the range at another
elevation .:... 576
Given the elevation for one range, to find the elevation for another
• range ..••••••••,••. 4 • • • 577
Given the charge for one rtinge, to find the ch&rge for anothet
range •••.k« 578
Given the range for one charge, to 6nd the range for another
chal-ge » ' i . . . ^ . . . 579
Given the range and the elevation, to find the impetus . . • . . .' 579
Given the devation and the range, to find the time of fiight • • • • 580
Given the range and the elevation, to find the greatest altitude
of the shell 581
Given the inclination of the plane, the elevation of the piece, and
the impetus ; to find the radge ....<••••• 58^
Given the inclinatiob of the plane, the eletation of the piece, and
the range ; to find the impetus. .••...••«. * • 583
Given the weight of a ball, the charge of powder with which it
is fired, and the known yelocity of that ball ; to find the velo*
city of a shell, when projected with a given ch&rge of powder 584
A Table, showing the velocities of the different-siccd shells,
when projected with a given charge of powder • • • < • 585
Givbn the elevation and the rangej to find the impetus, velocity,
and charge of powder ••,•••••.•.*.•••••..•• 685
Given the inclination of the plane, the elevation of the j^ece, anit
the range I tofiad th« charge of powdtri «»«• # •b«i»« ••••«• 586
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CONTENDS. xxxr
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XXXIV. Given tiiB inclinatioii of the plane, the eleration of the piece, and
the impetus ; to find the time of flight ••••. 688
XXXV. Given the impetus and the elevation, to find the horisontal range 589
XXXVI. Given the impetus and the elevation, to find the time of flight on
the horizontal range 590
XXXVIL Given the time of flight of a shelly to find the length of the fuse 591
•
Solution of Paobliems ik the Mensuratiow of Planeb, &c .. 592
I. Given the hase, and the perpendicular height of a plane triangle ;
to find its area • ••••••• i •••.'•••. • 592
IL Given two sides, and the conti^oed angle of a plane triangle ; to
find its area * 592
III. Given the three sides of a plane triangle ; tor find its area • • • • 593
IV. Given the diameter of.a circle; to find its circumference, and
conversely • ••..•...•••.•• 594
V. Given the diameter, or the circumference of the earth ; to find
the whole area of its surface • •• •»..•• 594
VL To find the length of any arc of a4^irele • ^ •..•••••• 595
Solution OF Paoblems ik Gauging.. • 596
I. To reduce the old standard wine measure into the Imperial
standard measure. »«••• «•••••*••••••»«•••• 597
I}, To reduce t)ie Imperial standard measure into the old stand*
ard wine measure •••^•••••. ••••••••••••«*••••• 597
III. To reduce the old standard ale measure into the Imperial
standard measure ••••••»• • .«... 598
rV. To reduce the Imperial standard measure into the old standard
ale measure • ••••••••••••• 598
V. Given the dimensions of a circular headed cask; to find its con-
tents in ale and in wine, gallons, and, also, id gallons agreeably
to the Imperial standard measure *«••••• • 599
VI. To find the ullage of a cask lying in a horisontal position 60 1
VII. To find the ullage of a cask standing in a vertical position • . . • 604
A general Table for converting ale or wine measure into the im-
perial standard measure^ and conversely • • « • • 606
Solution of Miscellaneous Problems • 607
h To find the weight of a cable » 607
II. To find the circumference of a circle »•!.••••• 608
III. To find the area, or superficial content of a circle • • • • • « 609
IV. Given the area of a circle, to find its diameter « «•.•••« 609
V. To find the side of a square equal in area to a given circle 610
VI. To find the side o( a square inscribed in a given circle 610
VIL Tofindtheareaof andlipsis ..•• • 611
VIII. To find the diameter of a circle equal in area to a given ellipsis. • 61 1
IX. To find the circumference of an ellipsis. ••••• • 612
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XXXvi CONtENTS,
Problem. Page.
X. To find the solid content of a sphere or globe 612
XL To find the height from which a person could see the one third of
the earth's surface • • . • Q13
XII. To find the distance of the sun from the earth . . .^ • • 614
' XIII. To find the measure of the 8un*8 diameter in English miles ..... 614
XIV. To find the ratio of the magnitudes of the earth and sun 615
XV. To find the rate at which the inhabitants under the equator are
carried in consequence of the earth's diurnal motion round its
axis ••••. « • 615
XVI. To find the rate at which the inhabitants under any given parallel
of latitude are carried, in consequence of the earth's diurnal
motion round its axis .« 616
XVII. To find the length of the tropical or solar year 616
XVIII. To find the rate at which the earth moves in the ecliptic • 617
XIX. To find the measure of the moon's diameter in English miles . . 617
XX. To find the ratio of the magnitudes of the earth and moon 618
XXI. To find how much lai^er the earth appears to the lunar inha-
bitants than the moon fippears to the terrestrial inhabitants • • 618
XXII. To find the rate at which the moon revolves round her orbit* • • • 619
XXIII. To find the true distance of a planet from the sun ••••• •• 619
XXIV. To find the comparative heat and light which the difiierent planets
receive from the sun ., •• 620
XXV. ' ' To find the measure of a planet's diameter in English miles • • • • 621
XXVI. To find the time that the sun takes to turn round its axis •••••• 622
XXVII. To find the length of a penduhrai for vibrating seconds 623
XXVIIL To find the length of a pendulum for vibrating half seconds • • • • 623
A compendium of Practical Navigation, &c. &c. &c 624
To make out a day's work at sea by. calculation 633
Of the.Iog book . . . , ' * ; 639
. Of the measure of a knot on the log line ; and of the true figure
of the earth •• 649
The true method of finding the index error of a sextant, &c. so as
to guard against the error arising from the elasticity or spring
of the bar, &c , 653
Of taking altitudes by means of an artificial horizon 655
A new and correct method of finding the longitude of places on
shore , 661
Solution OF Useful Astronomical Problems , 672
I. To find the latitude and longitude of a celestial object ........ 672
II. To find the right ascension and declination of a celestial object. • 677
III. To compute the lunar distances, as given in the Nautical Almanac 68 1
Appendix, showing the direct application of logarithms to the
doctrine of compound interest • • • • 687
Description and use of the general victualling table 717
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TO
HIS ROYAL HIGHNESS WILLIAM HENRY, DUKE OF CLARENCE
AND ST. ANDRE\<^S, K.Q., &c. &c. &c.
«
LORD HIGH ADMIRAL
UNITED KINGDOMS OF GREAT BRITAIN AND IRELA"ND,
MAY IT PLEASE YOUR ROYAL HIGHNESS;
This Treatise, which I am graciously permitted to lay-
before your Royal Highness, is the result of long study
apd labour ; the chief aim of which, .has been, to can-
tribute, in some measure, to the benefit of the Naval Service
of His Majesty. To this end, I have sought to combine
simplicity, perspicuity, and conciseness, in trigonometrical
calculations, in a greater degree than has hitherto been
attempted by the writers of nautical works; and to comprise,
in one book, a compendium of all the sciences that may
be useful or interesting to the practical navigator.
Tliat my humble attempt has met with your Royal High-
nesses approbation and high sanction, I shall ever esteem to
be the most honourable circumstance of my life ; that it has
been deemed worthy of the honour of your Royal Highnesses
patronage, I cannot but feel as the greatest mark of the
condescension of your Royal Highness.
I have the honour to subscribe myself.
With the most profound respect,
Your Royal Highnesses
Most obedient, most devoted.
And most grateful Servant,
THOMAS KERIGAN.
PcrUmmUhf
JDcctmbetg 1827.
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PREFACE.
Although the importance and general utility of: the subjects treated of in
this work are sufficient to recommend it to public attention, without the
aid of prefatory matter^ yet, since there is an extensive variety of nautical
publications now extant, I think it right to say nomethmg relative to what
I have done, were it for no other purpose than that of satisfying the reader
that the present work is widely different from any former treatise on nau-
tical and mathematical subjects. The jToUowing observations will develope
my motives for commencing so laborious an undertaking.
In perusing the various, nautical publications which have appeared for
many years past, I observed that they* all fell considerably short of the
objects at which they professed to sum ;— some, by being too much con-
tracted, and others by not including all the necessary tables, or by
being generally defective : and that, therefore, a great deal remained to
be done, particularly in the tabular parts^ beyond what had yet been
brought before the public.
Of the nautical works that came under my notice, some have proved, on
examination, to be so inaccurately executed, as to be entirely unfit for
the consultation of any person not sufficiently skilled in the mathematics
to detect their numerous errors.. Many of the works in question are ex-
tremely incomplete, through their want of particular tables, and their logar-
ithms not being extended to a sufficient number of decimal places: such
as those by Mendoza Rios, where the decimals are only continued to Jive
places of figures, and where the logarithmic tangents are entirely wanting;
for, although the addition of a logarithmic sine and a logarithmic secant
will always produce a logarithmic tangent, yet there are few mariners so
far acquainted with the peculiar properties of the trigonometrical canon^
as to be aUe to find by Rios' tables the arch corresponding to a given
b2
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viii PREFACE.
logarithmic tangent.* Hence, when the course and the distance between
two places are to be deduced from their respectiye • latitudes and longi-
tudes, by logarithmical computation, the mariner is invariably obliged
to have recourse to some other work for the necessary table of logarithmic
tangents. Besides, since none of the nautical works now in use exhibit
the principles upon Which the tables contained therein have been con-
structed, the mariner is left without the means of examining such tables,
or of satisfying himself as to their accuracy; though it is to them that he
is obliged to make continual reference, and on their correctness that the
safety of the ship and stores, with the.liv^s of 41 o^. board, so materially
depend. . . * •
Notwithstanding that Mr. Taylor's Logarithmical Tables are the most
extensive, the best arranged, and by far the most useful for astronomical
purposes, of any that have ever appeared in print^ — yet, since they, do not
contain the necessary navigation tables, they are but of little use, if of
any, to the practical qavigator : and, since the same objection is applicable
to the very excellent system of tables published by the learned Dr. Hutton^
these are, also, ill adapted to nautical purposes, and but rarely consulted
by mariners.
Being thus convinced that thei'e was something either deficient or
very defective in all the works that had hitherto been published on this
subject^ I was ultimately led to the conclusion that a general and com-
plete ^t qf Nautical Tables was still a desideratum to mariners : with
this conviction on my mind, I was at length induced to undertake the
laborious task of drawing up the following work ; in the prosecution of
which I found it necessary to exercise the most determined perseverance
and industry, in order to surmount the fatigue and anxiety attendant on
such. a long series of difficult calculations.
These points premised, \t .remains to present to the reader a familiar
$s\d comparative view of the nature of* this work, and of the improve-
ments that have been made in the tables immediately connected with' the
elements of narigation and nautical astronomy : confining the attention to
those that possess the greatest claims to originality, or in which the most
useful improvements have bee^ made.
Table VI. contains th.e parallaa^es of the . planets in altitude ; and
will be found particularly useful in deducing the apparent time from the
altitudes of the planets, and, also, in prbblems relating to the longitude.
The hint respecting this was originally taken from the Copenhagen edition
of "The Distances of tlie Planets from the Moon's Centre, for the Year
1823 ^" but this design has. been considerably enlarged and improved
upon.
• See Remarki page 98 ^ with dia^am apd calculations^ page 99.
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PRBFACB. IX
Table VIH. is so arranged that the mean astronomical refraction may
be taken out at first sight, without subjecting the mariner to the necessity
of making proportion for the odd minutes of altitude. This improvement
wOI have a tendency to facilitate nautical calculations.
Table X. — The arrangement of this table is an improvement of that
originally given by the author^ in his treatise called ^^ The Young
Navigator's Guide to the Sidereal and Planetary Parts of Nautioaf Astro-*
nomy/' By this improved table, the correction of the polar star's altitude
may be readily taken out, at sight, to the nearest second of a degree, by
means of five columns of proportional parts; and^ to render the table
permanent for at least half a century, the aimual variation of that star's
correction has been carefiilly determined lo the hundredth part of a second.
By means of this table, and that which immediately folfows (Table XL),
the latitude may be very correctly inferred at any hour of the night, in the
northern hemisphere, to every degree of accuracy desirable for nautical
purposes.
Tables XHI. and XIV. contain the equations to equal altitudes of the
sun : these have been computed on a new prinpiple, so as to adapt them to
proportional logarithms, by means of which they are rendered infinitely
more simple than those given under the same denomination in other
treatises on nautical subjects ; they will be found strictly correct, and, from
their simplicity, a hope may be entertained that the truly correct and ex-
cellent method of finding the error of a watch or chronometer by equal
altitudes of the sun, will be brought into more general use«
Tables XV. and XVI., whfch are entirely new, contain correct equations.
for readily reducing the longitudes, right ascensions, declinations, &c. &c.,
of the sun and moon, as given in the Nafttical Almanac, to any given
meridian, and to any given time under that meridian,
TmUe XVII. contains the equation corresponding to the mean second
difference of the moon's place in longitude, latitude,, right ascension, or
decUnation ; this table, besides being newly*arranged, will be found more
exteasive than those under a similar denomination, usually met with in
bocdca on navigation.
. Table XVI IL is so arranged as to exhibit the true correction of the
moon's apparent altitude corresponding to every second of horizontal
parallax, and to every miiiute of altitude from the horizon to the zenith :
and will prove very serviceable in dl problems where the moon^s altitude
forms one of the u^ments either given or required.
Table XIX. is fijUy adapted to the reduction of the true Utitudef of the
hesranly bodies, obtained by calculation, to their apparent central alti-
tudes : the lednctions of altitude may be very readily taken out to the de-
cimal part of a second. This table will be found of considerable utility in
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X PRBFACB.
deducing the longitude from the lunar observations^ when the distance
only has been observed.
Table XX. is new; and by its means the operation of reducing
the apparent central distance between the moon and sun, a fixed star, or
planet, to the true central distance, is very much abridged, as will appear
evident by referring to Method I., vol. i., page 433, where the true central
distance is found by the simple addition of five natural versed sines.
T^ble XXL, which is also new, contains the correction of the auxiliary
angle when the moon'si distance from a planet Ls observed : this will be of
great use in finding the longitude by the moon's central .distance from a
planet. ' .
Table XXIV. — ^The form of this table is entirely origin^ 5 and though
It is comprised in nine pages, yet it is so arranged that the logarithmic
difference may be obtained, strictly correct, to the nearest minute of the
moon's apparent altitude, and to every second of her horizontal parallax*
This table will be found of almost general use in the problem for finding
the longitude by the lunar observations. - •
Table XXVL, which is original, contains the correction of the logarithmic
difference when the moon^s distance from a planet is observed : this table
will be found of great use in computing the lunar observations whenever
the moon's distance from the planets appears in the Nautical Almanac; an
improvement which, from the advertisement* prefixed to the late Alma-
nacs, may be shortly expected to take place.
Table XXVIL, Natural Versed Sines, &c.— Tlie numbers corresponding
to the first 90 degrees of this table are expressed by the arithmetical com-
plements of those contained in the Table of Natural Co-sines published
by the author in ^^ The Young Navigator's Guide,". &c. ; the arithmetical
complement of the natural co-«ine of an arch being the natural versed sine
of the same arch. The numbers contained ii) the remaining 90 degrees of
this table are expressed by the natural sines, frotn the abovementipned
work, augmented by the radius.
This table is so arranged as: to render it general for every arch contained
in the whole semi-oircle, and conversely, whether that arch or its corre-
lative be expressed as tf. natural versed sine, natural versed sine supplement,
natural co-versed sine, natural sine, or natural co-sine.
Table XXVIII. is- an extension of that published by. the author in
^^ The Young Navigator's Guide/' &c. : it is arra,nged in a familiar
maimer, and, thoi^gh concise, contains, all the numbers that can be use-
fully employed in the elements of navigation ; ' for^ by means of nine co-
lumns of proportional parts, the logarithmic value of any natural number
under 1839999 may be obtained nearly at sight, and conversely.
Tables XXX., XXXI., and XXXII., have been carefully drawn up, and
proportional parts adapted to them, by means of which the logarithmic
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FRBFXCB. XI
half-elapaed time, middle time, and logarithmic rising may be very readily
taken out at the iiret sight, and conversely.
Table XXXV., Logarithmic Secants.— The arrangement of this table is
original, as well as its leng^ : the numbers contained therein are expres-
sed, by the arithmetical complements of those contained ih the table of
logarithmic co-siries published by the author iii '^ The Young Navigator's
Guide," &c.
This table is so drawn up as to be property adapted to every arch
expressed in degrees, minutes, and seconds, in the whole semi-cirole,
whether that arch or its correlative be considered as. a secant or a co-secant;
and by means mf proportional parts, the absolute value of any arch, and
conversely, may be readily obtained at sight.
Table XXXVI., -Logarithmic Sines. — ^This table is rendered general for
every degree, minute, and second, in the whole semicircle. The Table of
Logarithmic Tangents, which immediately fellows, is ako rendered gene-
ral to the same extent; and by means of proportional parts, the true yabie
of any arch, and conversely, may be instantly obtuned, without the trouble
of either multiplying or dividing t this improvement, to the practieal navi-
gator, must be an object of great importance, in* redueing the labour
attendant on computations in Nautical Astronomy.
Table JCXXVIIL has. been newly computed to the nearest second of
time, so that the mariner may be readily enabled to reduce the time of the
moon'«6 passage over the meridian of Greenwich to that of her passage over
any other meridian. This table will be found very useful in determining
the apparent time of the moon's, rising or setting, and also in ascertaining
the time of high water at any given place by means of Table XXXIX.
Table XLU.— This general' Traverse Table, so useful in practical navi-
gation, is arranged in a very different manner froih the Traverse Tables given
in the generality ef nautical book»; and although comprised in 38 pages,
is more comprehensive than the two combined tables of 6 T pages usually
found in those books, under the head ^'Difference of Latitude ^and Depar-
ture." Ill this table, every page exhibits all the angles that a ship's cour^
can possibly make with the meridian, expressed both in points and de^
grees ; which does away with the necessity of consulting two tables in find-
ing the difference of latitude and departure corresponding to any given
course and distance.
Table XLIV. contains the mean right ^censions and declinations of the
principal fixed stars. The eighth column of this table, which is origi-
nalj and is intended to facilitate the method of finding the latitude by the
altitudes of two fixed stars observed at any hour of the night, contains
the true spherical distance between the stars therein contained and
those preceding or abreast of them on the same horizontal line. The
mnth or last colunm of the page contains tha annual variation of that
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Xll PRBFAGB.
dhtancBf expressed in seconds and decimal p&rts of a second. Gieat
pains have . been taken^ in order to, find the absolute value of the an-
nual variation of the true spherical distance between the fixed stars ; and
the author ti^ists.that he has so far succeeded as to render this part of the
table permanent foi; aloHg period of years wbseqiient to 1824* •
Tables XLV. and XL VI *j which are adapted to the reduction of sidereal
time into mean solar time, and conversely, have been newly constructed :
these will be found considerably more extensive and uniform, than those ge-
nerally given under the same denomination.
Tables. LL.aiid LIL are entirely new: .these will be found exceedingly
useful in finding the latitude by the altitude of a celestial -object observed
at certain intervals from the meridian.; and since they are adapted to pro*
portional logarithms, the operation of finding, the latitude thereby becomes ,
extremely simple^* and yet far more accurate than tKat resulting from don*
ble altitudes,, even after repeating a trouUesome operation, and then
applying correctkmto correctiofu
Table LIV«— This table will be of service to Masters iki the Royal Navy,
to. officers employed in maritime surveys, and to all others who may be
desir6us of constructing charts agreeably to Merpator's principles of
projection. • .
Table LVL will be found essentially useful in reducing the. Erench
centesimal division of the circle into the English sexagesimal division,
and conversely ; and since most of the mpdem French works on astronomy
are now adapted to the centesimal principle, this table will be found of aik-
sistance in consulting those works j^nor will it be of less advantage to the
French navigator, in enabling him readily to consult the works of Jthe BngHsh
astronomers, where the degrees, &c., are expressed agreeably to the original
or sexagesimal principle.
Table LVIL is new; and although -it m^y not immediately affect the
interest of the mariner, yet it qannot fail to be usefiil to officers in charge
of His Majesty's Victualling Stores, in consequence of the late Act of Par-*
liament for the establishment of a new generd standard or imperial gallon
measure throughout the United Kingdoms^ — See Practical Gauging,
page 596 to 606.
Table LVIII. contains the latitudes and longitudes of all the principid
sea-ports, islands^ capes, shoals, rocks, &c. &c.) in the known worid ; tl^se
are so arranged as to exhibit to the navigator the whole line of coast along
which he may have occasion to sail, or on which he may chaiiee to be
employed, agreeably to the manner in which it unfolds to his view on a
Mercator's chart ; a mode of arrangement much better adapted to nautical
purposes than the alphabetical. But since the table is not intended' for
general geographical purposes, the positions of places inland, which do
not immediately concern the mariner, have, with a. few exceptions, beeii
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PkfS»ACA. xifi
pmpoeely omitted. The time 6f high water, at the full and change of
the moon/ is giveu at all places where it is known; which will be found
considerably more convenient than referring for it to a separate table.
The series of latitudes and longitudes that have; been established, astro-
nomically and chronometricdly, by Captatn William F1t£william Owen,
of Hia Majesty's ship Eden, during his recent and extensive survey along
the coasts of Africa, Arabia, Madagascar, Brazil, &c., follow as an
Appendix to the last^mentioned table. These series are published by the
express permission of Captain Owen ; And from his general kno}vIedge as
a navigator^ hydrographer, and practical astronomer, there is every rea«
SOD to believe that the geographical positions have been determined with
astronomical exactness.
A general Victualling Table forms ah addition to the Appendix ; And as
this exhibits the foil allowance of sea provisions (calculated agreeably to
the n6w Victualling Scale), from one man to any given number of men,
it will lie found useful l;o the Pursers of the Royal Navy, to Lieutenants
serving as Commanders and Pursers, and to the gentlemen who are
officially employed in the auditing of the Naval Victualling Accounts. '
The smt's declination is not^ given in this work ; nor is it necessary that
it should be, since it is contiuned,' in the most ample manner, in the Nctu^
tieal Almanac; a work which is so truly valuable to mariners that few
now go to sea without it ; the judicious never ^ilL
Hairing thus taken a survey of the principal part of the Tables, I must
briefly norice their cbsimpfion and U$e ;->-these will be found at the com-*
menocanent' of the first Volume. The principles and methods of their
computation are here fully detailed ; and the reader is. furnished with the
means, in the most simpkr,foTmttlffi> of examining any part of the Tables ;
which ia far more satisfactory than trusting to the* author^ mere word for
their entire accuracy \ though, I flatter myself with the hope t)>at, in this
extensive mass of flgarea, very few errors Will be found ;-^at all events,
none of pruicipU.
My origbal plan bad been to clofte the work with the description and use
of the Tables^ but being apprehensive that a series of Tables alone, how-
ever well arranged, or clearly illustrated, would not be sufficient to ensure
genehd aoceptation, I was induced to show their direct application to the
different elements connected with the sciences of navigation and nautical
astronomy, as well as to other subjects- of a highly interesting nature, such
as the art of gunnery, &c< &c. In this part of the work, since my design
did not extend beyond an ample illustration of the various mathematical
purposes to Which these tables may be applied, I have restricted myself to
the practical parts of the sciences on Which I have Had occasion to touch ;
because those are the points which most concern themariner,*and the com-
inerei&l intereets of this maritime nation. Nevertheless^ wherever it has
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appeared necessary to notice the elementary parts of the sciences, refer*
ence has been made to relative problems iq *^ llie Young Navigator's
Guide," where, it is hoped, the reader will find his inquiries fully satisfied.
The various sciences touched upon commence with a concise system of
decimal arithmetic, and complete courses of plane and spherical trigono-
metry. In the latter, the solution of the quadrantal triangles vdll be found
much simplified.'
Thepractical parts of Navigation begin with . parallel sailing ; but, with
the view of preventing the work from swelling to an unnecessary size, the
cases of plane sailing, usually met with iti other nautical books, have been
omitted in this ; as these are, in effect, no more than a mere repetition of
the cases of right angled plane trigonometry under a different denoniination.
Middle latitude sailing will be found exceedingly simplified by means of a
series of familiar analogies or proportiions : and in Mercator*s sailing a se-
ries of rational proportions is given ; which, it ishoped, may tend'to in-
duce mariners to substitute the rules of reason for the rules of rote j and
thus do away with the mistaken system of getting canone by hearty a
system which has too long prevailed in the Royal Navy.
The two very useful sailmgs, pblique and windward, which have been
hitherto little noticed by mariners, are also rendered so simple, parliculariy
the latter, that it is to be hoped they will, ere long, be brought into general
use.
In current sailing (Example 3,) the true principles of steering a vessd in
a current, or tideway are familiarly illustrated. This problem cannot fail of
behig interesting to every person who is at all curious in the art of !»•«
vigation.
The solution of a problem in great ^circle sailing is. given, which will be
found essentially useful to ships. bound from the Cape of Good Hope to
New South Wales*: comprising a table which exhibits, at -sight, all the
scientific particulars attendant on the true spherical track between those
two places ; by which it will be seen that a saving of 585 miles may be
effected by sailing near the arc of a greatxircle a»laid down in that table $
which saving ought to be an object ofvery high consideration to^all sliips
bound from the Cape of Good Hope to. Van Diemen*s Land, <v to Us
Majesty's colony at New South Wales with either troops or convicts ; .be*
cause the length of the voyage on the old track, or that deduced from the
common principles of navigation, generally occasions a great scarcity of
fresh .water, and this, eventually, adds distress to the many privations vnder
which those on board usually labour. In the same problem, there is a table
showing the true spherical route from Port Jackson, in New South Wales,
to Valparaiso, on the coast of Chili : in this route there is a saving of 745
miles when compared with that resulting from Mercator*s sailing ; and this
must be of considerable importance to the captain of a ship sailing between
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PRBVACB. Xy
these places, who is desirous of making his port in the shortest space of
time ; particularly since* few shijps can carry a liberal allowance of fresh
water to serve during a passage which measures very nearly one fourth of
the earth's circumference.
The introductory problems to Nautical Astronomy will be found rang^
in the most natural order ; all of which, except those relating to the alti-
tudes of the objects, are concisely solved by proportional logarithms :
the greater part of these will appear entirely . new to the . navigator.
The Vlth problem relating to the latitude exhibits the method of finding
the latitude by an altitude of the uorth polar star taken at any hour of
the night, which will be found very useful in all parts of the northern
hemisphese. — ^The Vlllth problem shows the method of finding the latitude
by the altitudes of two stars taken at any time of the night, agreeably to
the computed spherical distance between them contained in Table XLIV ;
this method of ascertaining the. latitude is general; it will be found very
correct, «nd far less troubletome than that by double altitudes which im-
mediately precedes it.— Problems IX, X, XI, and XII, contain new and
accurate methods of deducing the latitude from the altitudes of the celestial
bodies observed at given intervals from the meridian : the operation con-
sists of very little more than the common addition of three proportional
Icgarithms, and yet the latitude resulting from it will always be as cor-
rect as that deduced from the object's • meridional altitudes, provided the
watch shows apparent time at the place of observation, and the altitudes
be taken within the limits prescribed. These problems will be found
highly advantageous to the practical navigator ; because, in the event of
the sun's, or* other celestial object's meridional altitude being neglected
to be taken, or of it's being obscured by -clouds at the time of transit, he
is^ thus, provided with the most sale and jready means of determining his
latitude with as much certainty as if the altitude of the object had been
observed actually upon the meridian eitlier above or below the poIe« See
remark, page 368.
A. most ingenious problem in this part, of the work, for determining the
latitude, which for neatness and general utility stands unrivalled, ha» been
communicated to the author by the scientific Captain W. F. W. Owen.
In the methods of computing the altitudes of the heavcf^ly bodies, the
solutions to the several problems are rendered exceedingly concise and
explicit.
The Ilird, IVth, Vth> and Vlth problems relating to the longitude con-
tain the methods of finding the longitude by a chronometer and the res-
pective altitudes of the sun, stars, planets, and the moon i the three last
of which will be found considerably elucidated.
The lunar observations commence with the Vllth problem on the lon-
gitude. In this problem thirteen methods are given for reducing the ap«
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jtfi PREFACB*
parent central distance between the moon and enn^ a fixed star, or planet,
to the true central distance ; several of which are'entirely original, and all
of them. adapted to solve this interesting and important problem in the
moqt simple and expeditious manner.
• In the series of problems relative to finding the variation of the compass
byamplitudes, azimuths, transits of the fixed stars and planets, and by ob«
servations of the circumpolar stars, Problem II exhibits a n«tom^fAod for
computing, the true azimuth of a celestial object : and Problems V and VI,
contain the methods of reducing or correcting the true and the magnetic
courses, between 'two places, tigreeably to any given variation of the. com-
pass.—-An improved azimuth compass card is<described in this part of the
work, which may be applied to the determination of the longitude by the
lunar observations : — See the last two paragraphs in page 499. ' *
The series of problems for finding the apparent times of the rising or
setting of the celestial bodies, and of the beginning or tHe end of twi-'
light ;^and that for determining the interval 'of time between tiie rising
or setting of the sun's upper and lower limbs, it is-hoped will prove ac«
cep table to the lovers of the science of Nautical Astronomy ; ^likewise the
art of Dialling, which, although it may appear foreign or irrelevant to the
pursuits of the mariner, cannot fail to be interesting as a. branch of
science. * It is here treated of in a fieimiliar manner. * . ' -■
The IVth Problem in the mensuration of heights and distances, exhibits
the method' whereby the officers on board two ships* .of war can readily
ascertain their absolute distance firom any fort or garrison which tfiey may
be directed to cannonade ;-^afler which follow several' problems that will
be fpund exceedingly useifol on many military occasions.-^See remark at
page 53S, and also at page 543. Problem XL showing the method of re*-
ducing a base line, measured on any elevated horizontal plane, .to its true
level at the surface of the sea;, and Problem XIII. exhibiting a new rule
for Ending the height of e mountain, or other eminence, by means* of two
barometers and two thermometers, may be of considerable use to engi-
neers, or to others employed in conducting surveys* A problem is also
given for finding the direct course steered by- a ship seen at a distance;
and being a subject highly interesting to all nautical persons, it is reduced
to every desirahle degree of simplicity both by geometry and trigonometry.
All the problems in Practical Gunnery are readily solved by logarithms x
it contains three very concise tables whicji considerably facilitate the oper-
ation for finding the greatest range of a shot or shell, and the elevation of
the piece to produce that range. A small table is also given, which will
be found extremely useful in problems relating to shells, when it is requirecl
that they should strike an object at a given distance. — The rules and oper-
ations for computing the time of flight of a shell in Problems XXVII>
XXXIV, and XXXVI, wiU be found very simple and concise.
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PRSFACB. XVn
Although the art of gunnery may^ in some measure, be eonsidered as
not being immediately connected with that of navigation; yet it is a
subject with which all naval officers ought to have some acquaintance ;
since it very frequently happens, in time of war, that they are called upon
to go on shore with a party of men for this purpose of working the great
guns of the besieging batteries in co-operation with his Majesty'a Land
Forces.: — and since this truly interesting art is here, for the first time, un-
veiled of its mystic dressy and reduced to a state of simplicity, every
officer may make himself thoroughly acquainted with it in a very little
time, without any other assistance than that afforded in this treatise.
The problems on the mensuration of planes may be found useful on
many occasions ; particularly to persons employed in carrying on surveys
on shore.
Practical Gauging contains a few interesting problems ; the last of which
will be found essentially useful to such persoQS as may have occasion to
purchase wine, or spirits on his Majesty's account in foreign countries;
because it enables them to ascertain, in a very few minutes, the absolute
number of gallons ' contained in any given quantity of foreign liquor,*
agreeably to the newly established standard or Imperial gallon measure.
The compendinm of Practicial Navigation, given in this volume, exhibit-
ing the direct manner of making out a day^s work at sea, is intended for
the benefit of such persons, as may be unacquainted with the elements of
geometry and trigonometry : and includes the true method of finding the
index error of a sextant or quadrant so as to guard against the error arising
from the elasticity or spring of the index bar, with the method of appl}ing
the corrections to altitudes taken on shore by means of an artificial
horizon.
A new and correct method of finding the longitude of a place on shore
by means of the moon's altitude (observed in an artificial horizon,) and the
apparent time of observation, follows the above compendium; and will be
found of considerable utility in settling the geographical positions of places
inland or along the sea coast. An Appendix, which concludes the first
volume, contains everything relating to the doctrine of compound interest;
and developes the extraordinary powers pf logarithipical numbers in a more
striking point of view than any other department of science to which
they have been applied.
I have thus given a brief account of the more original parts of the sub-
jects comprised in this work,' the completion of which has cost me several
years of incessant labour ; during which time I had to contend with as
many infirmities, vexations, and disappointments as generally fall to the lot
of persons doomed to drudge through the toils of life : but stimulated by
the hope of ultimately succeeding in rendering myself useful to the Naval
Service of his Majesty, and to the nautical world in general, I have been
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XYIU PRBFACB.
SO far enabled* to bear up against the vicissitudes of health and fortune^
as to bring my long and arduous task to a close.
How far I have succeeded in my endeavour to supply the desideratum
which has l)een hitherto felt by navigators, it is not for me, but for a gene-
rous British public to determine : to their decision I submit my labours,
tinder the conviction tl)at, whatever may be the defects in its execution,
they will do justice to my motives, in this attempt to lessen the existing
obstructions' in the way of attaining a practical knowledge of the elements
of Navigation and Nautical Astronomy*
, . TOOMAS KERIGAN.
PwUmmiik, December l«r.,
182r.
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LIST OF SUBSCRIBERS.
His Royal Higfaneas the Duke of Clarei^pe and St Andrews, Lord High Admiral
of the United Kingdoms of Great Britain and Ireland, dec. drc. Sec*
The Right 'Honourable the (late) Lords CommisaionerB of the Admiralty, One
Hundred Guineas for 10 copies.
The Elder Brethren of the Honourable Trinity Corporation^ One Hundred Pounds
for 5. copies.
The Court of Directors of the Honourable East India Company, One HuAdred
Guineas for 10 copies.
The Honourable the Commissioners of His Majesty's Nafy, 5 copies.
The Honourable the Commissibners for Victualline His Mi^esty's Navy^ 6 copies.
The Right Honourable and Honourable the Directors of Greenwich Hosftital,
The Committee of Lloyd's, Ten, Guineas for 2 copies.
The Royal Naval Club, New Bond Street. * ' *
The British library, St. Helier*s,^ Jeney.
Capt. R. Anderson, R.N.
Capt» F, W. Austen, C.B., R.N., Gosport
Lieutenant J. W. Aldridge, R.N., North Street, Bristol.
Liettt. H. T; Austin, R.N:, Chatham.
Mr. Herbert Allen, H.M.S. Heron.
ISenry Adcock, esq., Polygoq, Somen' Town.
Vice Admual the Hon. Sir Henry Blackwood^ bart. K.C.B., Commander in
Chief at the Nore.
Commodore 'C. Bulten^C. B. .
Capt.H.W. Bayfield, R.N.
Cape. A. B. Branch, R.N.
Capl. J. W. Beechey, H.M.S. Blossom.
Capt. Edward Brace, C.B., It.N.
Capt. R. L. Baynes, H. M. S. Alacrity.
Lieut. A. B. Becher, R.N., Hydrographical Office, Admirarty*
Lieut. Philip Bisson, R.N., St. Heller's, Jersey.
The Honourable Frederic Byug.
lient. Jacob Bucknor, R.N.
Robert Bried, esq., Surgeon, R.N., Spencer Street, ClerkenwelL
Mr. Wm. H. Brown, Purser, H.M.S. Musquito.
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XX ^ LIST OF SUBSCRIBERS.
Mr. W. P. Browne, R.N., Plymouth.
Mr. John Browning, R.N., Ann's Hill Place, Gosport.
Thovias Best, esq. '
Alexander P. Bond, esq., Edgewortbstown, Ireland.
Mr. James Bradlej, Hanoyer Street, Portsea.
Admiral Sir Isaac Coffin, hart., Titley Court, Hereford.
Yice Admiral Sir Edward Codnngton, ^.C.B., Commander in Chief, Mediter-
ranean, 6 copies.
Capt. Janes Campbell* H.M^. Slaney.
Capt. Henry D. Chads, C.B., R.N.
Capt E. Chetharo, C.B., R.N., Gosport.
Capt D. C. Clavering, H.M.S. Redwing.
Capt Benj. Clement, R'.N., Chawton, Hants/
•Capt. Augustus W. J. CliflFord,C.B., H.M.S. Undaunted. .
Capt Charles Crole, R.N.
Capt E. Curzon, H.M.S. Asia.
Lieut Edward St. L. Cannon, Q.M.^. Wolf.
Mr. Jfunes Cannon, H.M.S. Thetis: *
Lieut. W..J. Cole, Royal George Yacht*
Lieut P. E. Cqjlins, H.N.
Lieut Edward Corbet, R.N.
Mr. Champronier, H.M.S. Eden'.
Mr. Thos. Cox, Purser, H.M.S. Pyramus.
Simon Cock, esq. New Bank Buildings, London.
William Curtis, esq., Portland Place.
The Rev. Colin Campbell, Widdington Rectory, Bishop's Stortford, Ettex.
Mr. Comerford, Bookseller, Portsmouth, 6 copies.
Mr. Crew, Bookseller, High Street, Portsmouth, 6 copies,
Capt Nevinson IJeCourcy,- R.N*., Stoketon House, Plymouth.
C^pU Manley Hall Dixon, R.N., Stoke, near Deyonport.
Capt George Shepherd Dyer, R.^. •
Lieut. Henry M. Denham, Linnet Surveying Vessel.
The Rev. E. Davies, H.M.S. Pyramus.
— Douthwaite, esq., Commander of the Circassian India Ship.
Admiral tlie Right Hon. Lord Viscount Exmotttbi G.C.B,
Capt. R. Evans, R.N.
Lieut. The Hon. Wm. Edwardes, H.M.S. Asia.
Lieut John Evans, (a) R.N.
Lieut. Thos. Eyton, R.N.
Lieut. W. W. Eyton, H.M.S, Wolf. .
The Rev. J. M. Edwards, H.M.S. Otdatea.
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THE
DESCRIPTION AND USE
OF TBB
TABLES;
WITH TB«
PRINCIPLES UPON WHICH THEY HAVE BEEN COMPUTED.
Table I.
To convert longitude^ or Degrees^ into Time, and conversely.
THIS Table consists of six compartments, each of which is divided
iiito two columns : the left-hand column of each compartment contains
the longitude, expressed either in degrees, minutes, or seconds ; and the
right-hand column the corresponding time, either • in hours, minutes,
seconds^ or thirds. The proper signs, for degrees and time, are placed at
the top and bottom of their respective columns in each compartment, with
the view of simplifying the use of the l^ble :— hence it will appear evident
that if the longitude T)e expressed in degrees, the corresponding time will
be either in hours or minutes ; if it be expressed in minutes, the corre-
sponding time will be either in minutes or seconds ; and if it be expressed
in seconds, the corresponding time will be expressed either in seconds or
thirds. The converse of this takes place in converting time into longitude.
The extreme simplicity of the Table dispenses with the formality of a
rule in showing its use, as will obviously appear by attending to the^folbw-
iog examples.
Example 1.
Required the time corresponding to 47?47'47? of longitude ?
47 degrees, time answering to which in the Table is 3* 8? 0! Of
• 47 minutes, answering to which is • • « .0. 3. 8. 0
• .47 seconds, answering to which is • ' • . 0. 0. 3. 8
Urn. 47?47'47^ the time corresponding to which is . 3M1?U!8!
B
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.• . jDESCHIPTION A^ USB OF THB TABLES*
Example 2.-
Required the longitude corresponding to the given time 8*52T28! ?
8 hours; longitude answering to which in the Table is • 120?0'0^
• 52 minutes, answering to which is .•..•• 13.0.0
• . 28 seconds^ answering -to which is . * . ^ 0.0.7
'rime8t52?28!, the longitude corresponding to which is . 133?0'7^
Besides the use of this Table in the redactioh of longitude into time,
and the contrary, it will also be found very convenient in problems relating
to the Moon, where it becomes necessary to turn the right ascension of
that object into time.
Example.
The right ascension of the Moon is 3S5?44C48r; required the corre-
sponding time ?
355 degrees, time answering to which
intheTableis . . . . • 23*40? 0! Of
• 44 minutes, answering to which is * 0. 2. 56. 0
« . 48 secs.^ answering to which ifl 0. 0. 3,12
Right ascen^on 355?44U8Tj tlie time corresponding to
which is • • • . 23*42T59!12f
Since the Earth makes one complete revolution on its axis in the space
<Jf 24 hours, it is evident that every part of , the equator will describe
a great circle of 360 degrees in that time, and, consequently, pass
the plane of any given iheridian once in every 24 hours ; whence it is
manifest that any given number of degrees of the equator will bear the
same proportion to the great circle of 360 degrees that the corresponding
time does to 24 hours; and that any given portion of tdme will be in the
same ratio to 24 hours that its corresponding number of degrees is to 360.
Now since 24 hburs are correspondent or equal to 360 degrees, 1 hour
must, therefore, be equal to 15 degrees; 1 minute of time equal to 15
minutes of a degree; 1 second of time to 15 seconds of a degree, and so
on. And as 1 minute of time is thus evidently equal to 15 minutes or one
fourth of a degree, it is very clear that.4 minutes of time are exactly equal
to 1 degree ; wherefore since d^ees and time are similarly divided, we
have the following general rule for converting longitude into time, and
vice t)0f«a» ,
Multiply the given degrees by 4, and the product will be the corre*
sponding time :-*<*observing that seconds multiplied by 4 produce thirds ;
minutes, so multiplied, produce seconds, and degrees minutes j which,
divided by 60, will give hours, llie convene <tf thb is evident :--thw.
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BBSCAimON AND USB OF THB.TABUUU
reduce the hotm to minutes; then these minutes^ divided by 4, trill give
d^eea ; the seconds^ ao divided, will give minute, and the thirdsi if any^
seeonds. Hence the prificiples upon which the Table has been copiputed.
The following eumples are given for the puipoae of illustrating the above
rale.
Example !•
Required the time corresponding
toS6?44:32f?
Given degrees = 36?44'32r
Multiplied by 4
Corresponding time 2^26r58!8!
JExampZe2.
Required the degrees correspond-
ing to 3 * 45^48 1 20 f ?
Given times3M5?48f20f
60
Divide by 4)225.48.20
Corresponding degs. 56?27-5T
Tablb II.
Depression of the Horizon.
The depression or dip of the horizon is the angle contuned between a
horizontal line passing through the eye of an observer, and a line joining
his eye and the visible horizon.
This Table contains the measure of that angle, which is a correction
expressed in minutes and seconds answering to the height of the observer's
eye above t^e horizon ; and which being subtracted from the observed
central altitude of a celestial object, when the fore observation is used, or
added thereto in the back observation, will show its apparent central altitude.
The corrections in this Table were deduced from the following considera-
tions, and agreeably to the principles established in the annexed diagram.
Let the small circle
A B C 6 represent the
terrestrial globe, and eO
the height of the ob-
server's eye above ita
surface ; then HOQ,
drawn parallel to a tan-
gent line to the surface
at e, will be the true or
sensible horizon of the
observer at O; and O P,
touching the surface at
T, the apparent horizon.
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• Let S be an object whose altitude is to be taken by a fore observation,
by bringing its image in contact with the apparent horizon at P ; then the
angle SOP will be the apparent altitude, which is evidently greater than
the true altitude S O H by the arc P H, expressed by the angle of horizontal
depression PO H. But if the altitude of the object S is to be taken by a
back observation, then, the observer's back being necessarily turned to the
object, his apparent horizon will be in the direction O F, and his whole
horizontal plane represented by the line D O F ; in which case his back
horizon O D, td which he brings the object S, will be as much elevated above
the plane of the true horizon HOQ as the apparent horizon OF will be
depressed below it ; because, when two strdght lines intersect each other, the
opposite angles will be equal. (Euclid, Book I., Prop. 15.) In this case it is
evident that the arc or apparent altitude S D is too little; and that it must
be augmented by the arc D H = the angle of horizontal depression FO Q,
in order to obtain the true altitude S H. Hence it is manifest that altitudes
taken l>y the fore observation must be diminished by the angle of horizontal
depression^ and that in back observations the altitudes must be increased by
the value of that angle.
The absolute value of the horizontal depression may be established in
the following manner :*-From where the apparent horizon O P becomes a
tangent to the earth's surface at T (the point of contact where the sky and
water seem to meet) let a straight line be drawn to the centre E, and it
will be perpendicular to OP (Euclid, Book III., Prop. 18) : hence it is
obvious that the triangle E TO is right-angled at T. Now, because O T is
a straight line making angles from the point O upon the same side of the
straight line O E, the two angles EOT and TO H are together equal to the
angle EO H (Euclid, Book I., Prop. 13) ; but the angle EO H is a right
angle; therefore the- angle of depression TOH is the complement of the
angle EOT, or what the latter wants of being a right angle : but the angle
T E O is also the complement of the angle E OT (Euclid, Book I., Prop. 32);
therefore the angle T E O is equal to the angle of horizontal depression ;
for magnitudes which coincide with one another, and which exactly fill
up the same space, are equal to one another* Then, in the right-angled
rectilineal triangle ETO, there are given the perpendicular TE, »= the
earth's semidiameter, and the hypothenuse E O, = the sum of the earth's
semidiameter and the height of the observer's eye, to find the angle T E O
s= the angle of horizontal depression TO H : — hence the proportion will be,
as the hypothenuse EO is to radius, so is the perpendicular T E to the
cosine of the angle T E O, which angle has been demonstrated to be equal
to the angle of horizontal depression HOP. But because very small arcs
cannot be strictly determined by cosines, on account of the differences being
so very trivial at the beginning of the quadrant as to run several seconds
without producing any sensible alteration, and there b^ing no rule for showing
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OSSCRIFTIONAND USB OF THB TABLS8. 5
why one second should 1>e preferred tp another in a choice of so many, the
following method is therefore given as the most eligible for computing the
true value of the horizontal depression, and which is deduced from the
86th Prop, of the third Book of Euclid.
Because the apparent horizon OP touches the earth's surface at T, the
square of the line O T 4s equal to the rectangle contained under the two
lines CO and 6 O. Now as the earth's diameter is known to be 41804400
English feet, and admitting the height of the observer's eye eO to be 290
feet above the plane of the horizon; then, by the proposition, the square
rootof CO, 41804690 x eO,290= the line OT, 110105.75 feet; the
distance of the visible horizon from the eye of the observer independent
of terrestrial refraction.
Then, in the right-angled rectilineal triangle ET O, there are given the
perpendicular ET = 20902200 feet, the earth's semidiameter, and the base
0T= 110105.75, to find the angle TEO. Hence,
As the perpendicular TB ^ 20902200 feet, log. arith. compt.^ 2. 679808
Is to the radius . . . . 90?0'0r log. sine . . .10.000000
SoisthebaseOTs . • 110105.75 feet, logi 5:041810
TotheangleTEO=s . 18f7^ = log. tang. . . . 7,721618
But it has been shown that the angle TEO, thus found, is equal to the
angle HOP; therefore the true value of the angle of horizontal depression
HOP, is 18 '7^ Now, according to Dr. Maskelyne, the horizontal de-
pression is affected by terrestrial refraction, in the proportion of about one-
tenth of the whole angle ; wherefore, if from the , angle of horizontal
depression 18^7^ we take away the one- tenth, viz. 1C49'', the allowance
for terrestrial refraction, there will remain 16 ^8^ for the true horizontal
depression, answering to 290 feet above the level of the sea. The prin-
ciples being thus clearly established, it is easy to deduce many simple for-
mufae therefrom, for the more ready computation of the horizontal de-
pression ; of which the following will serve as an example.
Tq the proportional log. of the height of the eye in feet, (estimated as
seconds,) add the constant log. .4236, and half the sum will be the propor-
tional log. of an arc ; which being diminished by one-tenth, for terrestrial
refraction, will leave the true angle of horizontal depression.
Example.
Let the height of the eye above the level of the sea be 290 feet, required
tlie.depresaion of the hori^n conesponding thereto ^
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6 DEiCRIFTlOH AKD USB OF THB TABLS8.
Height of the eye !290 feet, esteemed; as Bec8.s=4:50r, proporJog.ssl. 5710
Constaatlog. . • . . .j4236
Sum=5 1.9946
Arc= 187^ Proportional log, .9973
Deduct one-tenth =5 1.49, for terrestrial refraction.
True horizontal depression 16 ' 18^, the same as by the direct
method.
In using ihe Table, it may hot be unnecessary to remark that it is to be
entered with the height of the eye above the level of the eea, in the column
marked Height, tfc. ; opposite to which, in the following column, stands
the corresponding correction ; which is to be subtracted from the observed
altitude of a celestial object when taken by the fore observation ; but to be
added thereto when the back observation is used, as before stated. Thus
the dip, answering to 20feet above the level of the sea, is 4' 17^
Tabu III«
Dip of the Hariztm at different Dittances from the Obeerver.
If a ship be nearer to the land than to the visible horizon when uncon*
fined, uiid that an observer on board brings the image of a celestial object
in contact with the line of separation betwixt the sea and land, the dip of
the horizon Mali then be considerably greater than that given in the preced-
ing Table, and will increase as the distance of the ship from the land
diminishes : in this case the ship's distance from the land is to be estimated,
with which and the height of tiie eye above the level of the sea, the angle
of depression is to be taken from the present TaUe* Thus, let the distance
of a ship from the land be 1 mile, and the height of thie eye above the
sea 30 feet; with these elements entear the TaUe, and in the angle of
meeting under the latter and opposite to the former will be found 17 -
which, therefore, is the correction to be applied by subtraction to (he
observed altitude of a celestial object when the fore observation is used,
aud vice versa.
The corrections in this Table were computed after the following manner }
viz.,—
Let the estimated distance of the ship from the land represent the base
of a right-angled triangle, and the height of the eye above the level of
the sea its perpendicular; then the dip of the horizon will be expeessed
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AUCftfPTIOll AND tf 8fi OF THll TABUM. 7
by the measure of tke angle opposite to the perpendicular : hence^ since the
base and perpendicular of that triangle are known> we have the following
general
/Zu/e.— As the base or ship^s distance from the land, is to the radius, so
is the perpendicular, or height of the eye above the level oiP the sea to the
tangent of its opposite angle, which being diminished by one-tenth, on
account of terrestrial refiraetion, will leave the correct horizontal dip, as in
the subjoined example.
Let the distance of a ship from the land be 1 mile, and the height of
the eye above the level of the sea 25 feet, required the corresponding
horizontal dip
As distance 1 mile, or 5280 feet, Logarithm Ar. Comp.ss 6. 277^66
Is to radius • . . 90?, Logarithmic Sine . . 10.600000
So is height of the eye 25 feet. Logarithm .... L 397940
To Angle 16'. 17^=Log. Tang. = 7.676306
Deduct one-tenth for terrestrial
refraction ««•••• L37
True borisootal dip » f • • 14'40r,orl5:nearlyasintheTable«
iS^arlc-^AlQiough a skilful mariner can always estimate the. distance
of a ship from the shore horizon to a sufficient degree of accuracy for
taking ouMhe horizontal dip from the Table, yet since some may be de-
sirous of oDtaining the value of that dip independently of the ship's dis-
tance from the land, and consequently of the Table^ the following rale is
ghren for their guidance in such cases :—
Let two observers^ the one being as near the mast head as possible, and
the other on deck immediately under, take the sun's altitude at the same
instant. Then to the arithmetical complement of the logarithm of the
difference of the * heights, add the logarithm of their sum, and the loga-
rithmic sine of the difference of the observed altitudes ; the sum, rejecting
10 from the index, will be the log. sine of an arch ; half the sum of which
and the difference of the observed altitudes will be the horizontal dip cor-
responding to the greatest altitude, and half their difference will be that
contqponding to the least altitude^
Example.
Admit the height cyf an observer's eye at the main'-topmast head of a ship
elose in with the land, to be 96 feet, that of another (immediately under)
on dedt 24 feet; the altitude of the sun's lower limb found by the forme?
to be 89^37 ^ and by the latter^ taken at the same instant, 39?21 ' ; required
the dip of the shore horizon corresponding to each attitude 1
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8 ]>B$CEIPTION AND USE OF THE TABLES^
Height of mast head observer 96 feet.
He^ht of deck observer . 24 do. .
Difference of heights , • 72 do,^ Log.Ar.Comp.sS. 142667
Sum of ditto .... 120 do. Logarithm . 2.079181
Difference of altitudes • 16' Log. sine • 7- 667845
Arch s 26U0? Log. siae 7. 889693
Sum =42'40r,|=2i:20^=diptothegreatestheight
Diff. = 10. 40, 1= 5. 20=dip to the least height.
^0^6.— When the dip answering to an obstructed horizon is thus care-
fully determined, the ship's distance from the land may be ascertained to
the greatest degree of accuracy by the following rule : viz. As the Log.
tangent of the horizontal dip of the shore horizon is to the logarithm of
the height of the eye at which that dip was determined, so is radius to the
true distance.
Thus, in the above example where the horizontal dip has been deter-
mined to the corresponding height of the eye and difference of altitudes.
As horizontal dip = 5 ^26f Log. tang. ar. compt.=:2. 809275
Is to the height of the eye 24 feet^ Logarithm • . • 1 . 3802 1 1
So is radius. ... 90? Logarithmic sine . 10.000000
To true distance • • 15469.8 feet . Logarithm=4.f69486
The same result will be obtuned by using the greatest dip and its cor-
responding height ; and since the operation is so yery simple^ it cannot
fell of being extremely useful in determining a ship's true distance from
the shore.
Table IV.
jfugfiieniation of the Moon's Semidiameter.
Since it is the pr(q)erty of an object to increase its apparent diameter in
proportion to the rate in which its distance from the eye of an observer is
diminished ; and, since the moon is nearer to an observer, on the earthy
when she is in the zenith than when in the horizon, by the earth's semi-
diameter; she must, therefore, increase her semidiameter by a certain,
quantity as she increases her altitude from the horizon to the zenith. This
increase is called the augmentation of the ^loon^s semidiameter^ and d^«
p^nds upon the follovidng pnn^^jplest
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1NI5CRIPTI0N AND USE OF THB TABLB8»
Let the. circle A BCD
represent the earth ; A E
its semidiameter^ and M
the moon in the hori-
zon. Let A represent the
place of an observer on the
earth's surface; BDM
his rational horizon^ and
H A O, drawn parallel
thereto^ his sensible hori-
zon extended to the moon's
orbit ; join A M^ then
A ME is the angle under
which the earth's semi-
diameter A E is seen from
the moon. M, which is
equtl to the angle M A O^ the moon's horizontal parallax ; because the
straight line A M which falls upon the two parallel straight lines E M and
AO makes the alternate angles equal to one another. (Euclid, BookL Prop.
29.) Let the moon's horizontal parallax be assumed at 57 '30^, which is
about the parallax she has at her mean distance from the earth j then in the
right angled triangle A E M, there are given the angle A M E=57'30^, the
moon's horizontal parallax, and tjie side AE=3958. 75 miles, the earth's
semidiameter ; to find the hypothenuse AM=the moon's distance from
the observer at A : hence by trigonometry,
As the angle at the moon, A M E=a57'30r Log. sine ar. comp. 1. 776626
Is to the earth's semidiameter=A E=3958. 75 miles. Log. . 3. 597558
So is radius ....... 90? . . . Log. sine lO.OOOOOO
To moon's horizontal distance A M=236692.35 miles, Log. . 5. 374 184
Now, because the moon is nearer to the observer at A, by a complete
semidiameter of the earth when in the zenith Z, than she is when in the
horizon M, as appears very evident by the projection ; and, because the
earth's semidiameter A E thus bears a sensible ratio to the moon's distance ;
it hence follows that the moon's semidiameter will be apparently increased
when in the zenith, by a small quantity called its augmentation; and
which may be very clearly illustrated as follows, viz.
Let the arc Z O M represent a quarter of the moon's orbit ; Z her place
in the zenith^ and Z S her semidiameter : join E Z, A S, and E S ; then the
angles Z E S and Z A S will represent the angles under which the moon's
semidiameter is seen from the centre and surface of the earth; their diffe*
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10 MSCEfPTtOK AND USB OT TRB TABLB8.
rence^ viz., the angle A S E is, therefore, the augmentation of the moon's
semidiameter, which may be easily computed ; thus---»
In the oblique angled triangle A S E, there are given the side A E
=3958. 75 miles/ the earth's semidiameter ; the side A S,r=A M — * AEsa
23273.3. 6 miles, the moon's distance when in the zenith from the observer
at A ; and the angle AE.S=15' SCT, the moon's mean semidiameter; to
find the angle ABE=:the greatest augmentation corresponding to the
given horizontal parallax and horizontal semidiameter : therefore,
As moon's zenith distance = AZ=232733. 6 miles. Log. ar. co. 4. 633141
Is to moon's semidiameter A E S = 15 ' 30r Log. sine 7- 654056
So is earth's semidiameter E A = 3958. 75 miles. Log. • . 3. 597558
To augment, of semidian^ ASE=:0'16r Log. sine 5.884755
Now, having thus found the augmentation of the Moon-'s semidia-
meter, when in the zenith, answering to the assumed horizontal parallax
and horizontal semidiameter ; the increase of semidiameter at any given
altitude, from the horizon to the zenith, may be computed in the following
manner.
Let S A be produced to F. and draw E P parallel to Z S ; then will E P
represent the greatest augmentation to the radius E Z. Let the moon
be in any other part of her orbit, as at }) with an altitude of 45 degrees ;
joinDE, and])F, and makeDG=])E; then will EG (the pleasure of
the angle E]>G to the radius E}) ,) be the augmentation corresponding to
the given altitude. Then, in the right angled triangle BGP, right
angled at G, there are given the angle EFG=43 degrees, the moon*s
apparent altitude, and the side E F=?16 seconds, the augmentation of
semidiameter when in the zenith ; to find the side E G, which expresses
the augmentation of semidiameter at the given altitude. And, since the
angles expressing the augmentations are so very small, the measure of each
may be substituted for its sine, vehich will simplify the calculation] iho».
As radius • 90?0'0'' Log. sine ar. comp. 0.000000
Is to moon's greatest augment, of semidiam»=sE F 16^, Log. as 1. 204120
So is moon's given apparent alt. =5 ^ E FG, 45? Log; sine a: 9. 849485
To the augmentation, or side . E G = 1 P p 3 1 . Log. s 1 . 053605
iffaicby therefixre^ is the augmentation of the moon's semidiameter cor-*
responding to the given apparent altitude of 45 degrees ; horizontal semi*
diameter 15 ^SOl! and horizontal parallax 57 ' 30?
Explanation of the Table.
His Table contains the augmentation of the moon's semidiameter (de-
termined after the above manner,) to every third degree of altitude : the
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DXSC&IPTION AND ITSS OF TUB TABLBB, 11
augmentatioti is expressed in seconds, and is to be taken out by entering
the TaUe mth the moon's horizontal scmidiameter at the top, as given in
the Nautical Almanac, and the apparent altitude in the left-hand column ;
in the angle of meeting will be found a correction, which being applied
by addition to the moon's horizontid scmidiameter will give t)}e true semi-«
diameter, corresponding to the ^ven altitude. Thus the augmentation
answering to moon's apparent altitude 30 degrees, and horizontal semi-
diauiete^ 16C30? is 0 seconds; and that corresponding to altitude 60?
and aemidiameter 16* is 14 seconds.
Tablb V.
Contraction of the semidiameters oftlie Sun and Moon^
Since all parts of the horizontal semidiameter of the sun or moon are
equally elevated above the horizon, all those parts must be equally affected
by refraction, and thereby cause the horizontal semidiameter to remain
invariable. Bu^when the semidiameter is inclined to the plane of the
horixon^ the lower extremity will be sO much more affected by refraction
than the upper, as to suffer a sensible contraction, and *thus cause the
aemidiameter, so inclined, to be something less than' the horizontal semi*
diameter given in the Nautical Almanac, Hence it is manifest that the
aemidiameter of a celestial object, measured in any other manner than
that parallel to the plane of the horizon will be always less than the true
aemidiameter by a certain quantity : — ^this quantity, called the contraction
of semidiameter is contained in the present Table i the arguments of
which are, the apparent altitude of the object in the left-hand column, and
at the top the angle comprehended between the measured diameter and
Aat parallel to the plane of the horizon ; iii the angle of meeting will be
foBod a correction, which being subtracted from the horizontal semi*
diameter in the Nauticar Almanac, will leave the true semi-'diameter.
Thus, let the sun's or moon's apparent altitude be 5 degrees, and the
inclini^on of its semidiameter 72 degrees ; now, in the angle of meeting,
of these arguments, stands 23 seconds • which, therefore, is the contraction
of aemidiameter, and which is to be applied by siiblraetioti to the semi-
diaftictef given in the Nautical Almanac.
To compute the contraction of Semidiameter.
Bule. — Find by Table VIII. the refraction corresponding to the object's
apparent- central altitude, and also the refraction answering to that altitude
augmented by the semidiameter} (whichi for this purpose, may be estiamted
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at 16 minutes,) and their difference will be the contraction of the vertical
semidiaroeter. Now, having thus found the contraction corresponding to
the vertical semidiameter, that answering to a semidiameter which forms
any given angle with the plane of the horizon^ will be found by multiply-
ing the vertical contraction by the square of the angle of inclination.
Example.
Let the sun's or moon's apparent central altitude be 3? and the indi-
nation of its semidiameter to the plane of the horizon 72?; reqjiired the
contraction of the semidiameter ?
Apparent central altitude . 3? 0' Refractions 14.' 36f
Do. augmented by semidiam. = 3 ? 1 6 ' Ditto • = 13 . 46.
Contraction of the vertical semidiameter . . . 0 ' 50rLog.= 1 . 698970
Inclination of semidiameter , =72? twice the log. sine • =19.956412
Required contraction of semidiameter • . . .45'. 22 Log.=l. 655382
And so. on of the rest.— It is to be remarked, however, that the correc-
tion arising from the contraction of the semidianieter.of » celestial object
is very seldom attended to in practice at sea.
Table VL
Parallax of the Planets %7i AUitude.
The arguments of this Table are the apparent altitude of a planet in
the left or right-hand margin, and its horizontal parallax at the top ;
under the latter, and opposite the former, stands the corresponding parallax
in altitude ; which is always to be applied by addition to the planets ap-
parent altitude. Hence, if the apparent altitude of a planet be 30 degrees,
and its horizontal parallax 27 seconds, the corresponding parallax in
altitude will be 23 seconds; additive to the apparent altitude.
Tlie parallaxei ofJUUude m this Table were computed by the follomng
J&ifc.— To the proportional logarithm of the planet's horizontal parallax
add the log. secant of its apparent altitude, and the sum, abating 10 in
the index, will be the proportional logarithm of the parallax in altitude.
Example.
If the horizontal parallax of a planet be 23 seconds, and itsapparei^t
i^ltitude 30 de^ees } required the parallax in altitude ? .
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BBSCftlPTION AND USE OF TMB TABLES. 13
HoricoBtal parallax of the planet=23 Seconds^ proportional log.= 2. 6717
Apparent altitude of ditto . . 30 Degrees^ log. secant . .10. 0625
Parallax in altitude . . • « 20 Seconds^ proportional log. 2. 7342
Table VII.
Parallax of the Sun in Altitude*
lie difference between the places of the sun^ as seen from the surface
and centre of the earth at the same instant, is called his parallax in al-
ticnde^ which may be computed in the following manner.
To the log. cosine of the sun's apparent altitude, add the constant log.
0.945124, (the log. of the sun's mean horizontal parallax estimated at
8'. 813,) and the sum, rejecting 10 from the index, will be the log. of
the parallax in altitude; as thus,
Given the sun's apparent altitude 20 degrees ; required the correspond-
ing parallax in altitude ?
Sun's apparent altitude 20 degrees, log. cosine ; . 9. 9729S6
Constant log 0.945124
Parall. corresponding to the given altitude S"". 282 Log. 0. 9181 10
This Table, which contains the correction fo^ parallax, is to be entered
with the sun*8 apparent altitude in the left-hand column ; opposite to which,
in the adjoining column, stands the corresponding parallax in altitude ;—
thus, the parallax answering to 10? apparent altitude is 9 seconds ; that
answering to 40? apparent altitude is 7 seconds, &c. &c. — And since the
parallax of a celestial object causes it to appear something lower in the
heavens, than it really is ; this correction for parallax, therefore, becomes
always additive to the sun's apparent altitude.
Table VIII.
Mean Astronomical Reaction.
Since the density of the atmosphere increases in proportion to its prox-
imity to the 'earth's surface, it therefore causes the ray of light issuing from
a celestial object to describe a curve, in its passage to the horizon ; the
convex side of which is directed to that part of the heavens to which a
tangent to that curve at the extremity of it which meets the earth, would
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14 PBSCEIPTIOK AND D«B OF THB TABUS.
be directed. Hence it is, that the celesUal objects are apparently more
elevated in the heavens than they are in reality ; and this apparent increase
of elevation or altitude is called the refraction of the heavenly bodies ; the
effects of which are greatest at the horizon, but gradually diminish aa
the altitude increases, so as to entirely vani&h at the zenith.
In this Table the refraction is -computed to every minute in the first
8 degrees of apparent altitude ; consequently this part of the Table is to
be entered with the degreea of apparent altitude at the top or bottom,
and the minutes in the left-hand coluixm : in the angle of meeting, stands
the refraction.
In the rest of the Table the apparent altitude is given in the vertical
columns, opposite to which in the adjoining columns will be found the
corresponding refraction. Thus, the refraction answering to3?27« appa-
rent altitude, is 13'14f; tiiat corresponding to 9?46' is 5'52r; that
corresponding to 17^55 'is 2'54?, and so on. The refraction is always to
be applied by subtraction to the apparent altitude of a celestial object, on
account of its causing such object to appear under too great an angle of
altitude. The refractions in this Table are adapted to a medium state of
the atmosphere ; that is, when the Barometer stands at 29. 6 inches, and
the Thermometer at 50 degrees ; and were computed by the following ge^
neral rule, the horizontal refraction being assumed at 33 minutes of a
degree.
To the constant log. 9. 999279 (the log. cosine of 6 times the horizontal
refraction) add the log. cosine of the apparent altitude ; and the sum,
abating 10 in the index, will be the log. cosine of an arch. Now, one-
sixth the difference between this arch and the given apparent altitude will
be the mean astronomical refraction answering to that altitude.
Example.
Let the apparent altitude of a celestial object be 45?, required the cor-
responding refraction ?
Constant log 9. 999279
Given apparent altitude 45?0'0T Log. cosine 9.849485
Arch 45?5U2r Log. cosine 9.848764
Difference 0^5 U2r ^ 6 = 0^57^ ; which, therefore,
is the mean astronomical refraction answering to the given apparent alti-
tude.
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DBSCEIPTION AND T78B OP THB TABLB8« 15
Tabib IX.
Corrictknofihe Mean J$tr(m(nnical Refract
Since the refraction of the heavenly- bodies depends on the density and
temperature of the atmosphere, which are ever subject to numberless varia-
tions ; and since the corrections contained in the foregoing Table are
adapted to a medium state of the atmosphere, or when the barometer stands
at 29. 6 inches, and the thermometer at 50 degrees : it hence follows, that
when the density and temperature of the atmosphere differ from those
quantities, the amount of refraction will also differ, in some measure, from
thut <»ntained in the said foregoing Table. To reduce, therefore, the
corrections in that Table to other states of the atmosphere, the present
Table has been computed ; the arguments of which are, the apparent ald^
tude in the left or right hand margin, the height of the thermometer at the
top, and that of the barometer at the bottom of the Table; the correspond-
ing corrections will be found in the angle of meeting of those arguments
respectively, and are to be applied, agreeably to their signs, to the mean
refraction taken from Table VIII, in the following manner :—
Let the apparent altitude of a celestial object be 5 degrees; the height
of the barometer 29. 15 inches, and that of the thermometer 48 degrees;
required the true atmospheric refraetion I
Apparent altitude 5 degrees, — mean refraction in Table VIII = . . 9'54T
Opposite to 5 degrees, and over 29. 15, in Table IX, stands • . — 0. 9
Opposite to 5 degrees, and under 48 degrees, in ditto . . • • + 0. 3
True atmospheric refraction, as required 9 '.48?
The correction of the meitn astronomical refraction, may be computed by
the following rule, viz.
As the n^an height of the barometer, 29.6 inches, is to its observed
hright, so is the mean refraction to the corrected refraction ; now, the
diflereuce between this and the mean refraction will be the correction for
barometer, which will be afEurmatlve oc negative, aoe<mling as it is greater
or less than the latter.— And,
As 350 degrees* increased by t|ie observed height of Fahrenheit's ther-
mometer, are to 400 degrees f, so is the mean refraction to the corrected
refraction ; the difference between which, and the mean refraction, will be
the correction for thermometer; which will be affirmative or negative, ac-
cording as it is greater or less than the latter.
* Seven times 50 degrees, the mean temperature of the atmosphere.
t Eifht timet J^Odegrcss, the mean temperature of the atmosphere.
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16 DB8CRIPTI0K AND USB OF THS TABLBS.
Bsample I.
Let the apparent altitude be 1 degree^ the mean refraction 24^29^, the
height of the barometer 28. 56 inches, and that of the thermometer
32 degrees; required the respective corrections for barometer and ther-
mometer ?
As mean height of barometer • • 29. 60. Log. an co. • • 8. 528708
Is to observed height of ditto • •28.56. Log 1.455758
Soismean refraction 24 ;29r= . )469r Log 3.167022
To corrected refraction • . • • 1417^ Log 3.151488
Correction for barometer ... • — 52^, which is negative^ because the
corrected refraction is the least.
And
As 350?+ 32?=. ..... 382? Log. ar. co. . . 7.417937
Isto ...:...... 4()0? Log 2.602060
So 18 mean refraction 24:29^'= . 1469r Log 3.167022
To corrected refraction . . . 1538r Log 3.187019
Correction for thermometer . . +69^=1 '9?^ which is affirmative,
because the corrected refracUon is the greatest.
Example 2.
Let the apparent altitude be 7 degrees, the mean refraction 7-201"^ the
height of the barometer 29. 75 inches, and that of the thermometer 72 de-
grees ; required the respective corrections for barjometer and thermometer ?
As mean height of barometer . . 29. 60. Log. ar. co. . . 8.628708
Is to observed height of ditto .. 29.75. Log. . . . . 1.473487
Soismeanrefraction7'20^ = . . 440f Log 2.643453
To corrected refraction . • . . 442f Log 2. 645648
Correction for barometer . • . • +' 2^, which is affirmative.
And
As 350? + 72?= 422? Log. ar. co. . . 7.374688
Is to 400? Log 2. 602060
So is mean refraction 7'20r = . . 440'/ Log 2. 643453
-
To corrected refraction . . . .417^ Log 2.620201
Correction for thermometer . • • — 231", which is negative.
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DBSCRIPTION AND V8E OF THB TABLKS* 17
Table X.
To find the LaHiude by an JUitude of the North Polar Star.
^ The correction of altitude, contained in the third column of this Table^
expresses the difference of altitude between the north polar star, and the
north celestial pole, in its apparent revolution -round its orbit, as seen from
the equator : the correction of altitude is particularly adapted to the be-
ginning of the yesf 1824 *, but by. meana of its annual variation, which is
determined for the sake of accuracy to the hundredth part of a second,
it may be readily reduced to any subsequent period, (with a sufficient de-
gree of exactness for all nautical purposes,) for upwards of half a century,
as will be seen presently.
The Table consists of five compartments ; the left and right hand ones
of which, are each divided into two columns containing the right ascension
of the meridian: the second compartment, which forms the third column
in the Table, contains the correction of the polar star's altitude : the third
compartment consists of five small columns, in which are contained the
proportional parts corresponding to the intermediate minutes of right
ascension of the meridian; by means of which the correction of aldtude, at
any given time, may be accurately taken out at the first sight : the fourth
compartment contains the annual variation of the polar star's correction,
which enables the mariner to reduce the tabular correction of altitude to
any future period : for, the product of the annual variation, by the number
of years and parts of a year elapsed between the beginning of 1824, and
any given subsequent time, being applied to the correction of the polar
star's altitude by addition or subtraction, according to the prefixed sign^
mil give the true correction at such subsequent given time*
JSxample I.
Required the correction of the polar star's altitude in January 1834, th6
right ascension of the meridian being 6 hours and 22 minutes ?
Correction of altitude answering to 6t20C, is • • • • ; 0?16; 9?,
Proportional part to 2 minutes of right ascension « « • • 0. 50
G>rfection of polar star's altitude in January 1824s: • « • 0. 15. 19
Annual variation of correction i •+ 2^.90
Number of years after 1824 • • • . . 10
Product ^ . . +29''.0=s • • • + 0.29
Correction of the polar star's altitude in Jan. 1834, as required 0? 15 M8?,
c
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18 DESCRIPTION AND USB OV THE TABLES.
* Example 2.
Required the eorrection of the polar star's dtitude in January 1854,
the right ascension of the meridian being 5 hours and 13 minutes i
Correction ofaltitude answering to 5 MOrifl 0944124?
Proportional part to 3 minute of right ascension • • • * ^^* ^
Correction of polar star's altitude in Jan, 1824 • • • . 0. 43. 16
Annual variation of cortection . . —3'^. 34
Numberofyears after 1824 ^ . 30
Product -100".20= . . : - 1.40
Correction of the polar star's altitude in Jan. 1 854, as required, 0?4 1 1 d6f
t^hich diflfers but 8 seconds from the true Result by spherical trigonometry,
as will be shown hereafter ; and which evidently demonstrates that the
column of annual variation may be safely employed in reducing the correc-
tion of altitude to any future period, for a long series of years, since the
error in the space of thirty years only amounts to 8 seconds of a degree,
which becomes insensible in determining the latitude at sea.
The corrections of altitude contained in the present Table were com*
ptited iri conformity, with the following principles :—
Since to an observer placed at the equator, the poles of the world
will appear to be posited in the horizon, the polar star will, to such
observer, apparentiy revolve round the north celestial pole in its diurnal
motion round its orbit. In this' apparent revolution round the celestial
pole, the star's meridional or greatest altitude above the horizon will be
always equal to its distance from that pole ; which will ever take place,
when the right ascension of the meridian is equal to the right ascension of
the star. In six hours ctfter this, the star will be seen in the horizon, west
of the pole ; in six hours more it will be depressed beneath the horizon (on
the meridian below the pole), the angle of depression being equal to its
p<rfar distance ; in six hours after, it will be seen in the horison east of the
pole ; and in $ix hours jnore, it will be seen again on the meridian above the
pole t allowance being made, in each case, for its daily acceleration.
Now, since the north celestial pole represents a fixed point in the hea<*
Tens, and that the star apparently moves round |t tn an uniform manner,
making determinable angles with this meridian j it is, therefore, easy to
compute what altitude the star will have, as seen from the equator, in
every part of its orbit ; for, in this computation^ we have a spherical trian-
gle to work in, whose three sides are expressed by the complement of the
latitude, the complement of the polar starts altitude, and the complement
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DSSCHIPTIOIf AND tTSB OF THB TABLES. 19
of its declination $ ip which there are given two sides and the included
angle to find the third side; via,, the star's co-declination or polar distance
and the complement of the latitude, with the comprehended angle, eqpal
to the star's distance from the meridian, to find the star's co-altitude ; the
difference between which and 90 degrees will be the correction of altitude,
or the difference of altitude between the polar star and the north celestial
pole, as seen from the equator.
In. January 1854, the mean right ascension of the north polar star
will be 1!5T23!, and its polar distance l?28^5''j now, admitting the
right ascension of the meridian to be 5*13T, the correction of thp polar
star's altitude, as seen frpm the equator, is required ?
Right asc. of the merid. 5*13" Oi
Right asc. of the pol. star 1 . 5 . 23
P. star's disu from mend. 4 1 7*37 ? = 61 ?54 i 15?
Half dd. do, in degrees . . . . 30.67 • 7i Twice (he
log.sinesl9.4S2469
Star's polar distance ^ • , . . 1 . 28 . 5 Log. sine 8* 408572
Complement of the latitude , . , 90. 0. 0 Log* sine 10.000000
I ■ Sum 37.881041
DMT. between polar dist. and eo-lat. 88?3l ^55? Half sum 18. 915520}
Half do , . . . 44? 16'. 574^ Log. sine 0.843849
Arph« 6?48:35KW,»ngt9t 071671*
Log, sine of this arch , ^ • 9. 068670
Half the polar star's co-altitude , ,44?39:i6r Log. sine 9.846850*
Pokr star's co-altitude 89?16^32C'
will I ■ ■■ >■ l»
Cor. of polar star's alt. in Jan. 1 854 =» 0?4 1 ^ 28^ Now, by comparing
this result with that shown in Example 2 (page 18), it will be seen
that the correction of altitude, deduced directly from the Table, may be
reduced to any period subsequent to 1624, without its being affected by
any error of sufficient magnitude tp e^cjaoger the interest of the mariner
in any respect whatever.
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20 DSSCUIPnON AND USE OF THB TABIDS.
ffotej^Vor further information on this subject, the reader is referred to
the author's Treatise on the Sidereal and Planetary Parts of Nautical
Astronomy, page 144 to 156.
Tablb XL
Correction of (he Latitude deduced froth tlie preceding Table.
Although the latitude deduced from Table X. will be always sufficiently
correct for most naufacal purposeSi, yet, «ince observation has shown that it
will be something less than the truth in places distant from the equa-
tor, the present Table has been computed ; which contains the number
of minutes and seconds that the latitude, so deduced, will be less than what
would result from actual observation at fevery tenth or fifth degree from
the equatoY, to within five degrees of the north pole of the world.
The elements of this Table are, the approximate latitude, deduced from
Table X., at top, and the right ascension of the meridian in the left or
right-hand column ; in the angle of meeting will be found the correspond-
ing correction, which is always to be applied by addition to the approx-
imate latitude. Hence, if the approximate latitude be 50 degrees, and
the right ascension of the meridian GMQ?, th^ corresponding correction
willbel^SSr additive.
Hernark. — Since the corrections of altitude in Table X. have been com-
puted on the assumption that the motions oi the polar star were witnessed
from the equator, they ought, therefore, to show what altitude that star
ivill have at any given time, in north latitude, when applied to such latitude
with a contrary sign to that expressed in the Table ; this, however, is not
the case; because when the altitude of the polar star is computed by sphe-
riciJ trigonometry, or otherwise, it will alw&ys prove to be something less
than that immediately deduced from Table X. : it is this difference, then,
thftt becomes the correction of latitude in Table XI., and which is very
•easily determined, as may be seen in the following
Example.
Let the right ascension of the meridian in. January 1824 be 6 MO?, and
the latitude 60 degrees north ; required the true altitude of the polar star,
and thence the correction of latitude ? .
Latitude or elevation of the pole • • . • ' 60?0^ OC north.
Correction in Table X., answ. to 6M0r, is + 0. 7. 41
Altitude of polar star, per Table X. = . ; 60?7'4K
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BSSCJtJPTfON ANJD USB OF THB TABLB|k 31
Now, to compute the true altitude of the polar star, on spherical prin-
ciples, at the given time and place, we may either proceed as in last
example, or, more readily, as follows : —
Right ascension of the mend. 6t 40? 0!
Star's right ascension • • 0.58. 1
Star's dist. from the meridian 6 1 4 1 ?59 ! . . • -Log. .rising 5 • 96448 1
Star's polar distance . . l?37'48r . . . Log. sine 8.454006
Complement of the latitude 30. 0. 0 • . • Lfog.'sine 9.698970
Difference 28 . 22 . 1 2 Nat cos. 879897 ■
Natural number .... 013106 Log.=4. 117457
Sur's true altitude . . 60?5n6r Nat. sine 866791
SUr's alt. per Tab. as above 60 . 7. 41
Difference 0?2' 25'^; which, therefore, is the correction
of latitude.
JVote.— The correction of latitude, thus found, differs 4 seconds from that
given m Table XL : this difference is owing to the star's apparent polar
distance having been -taken, inadvertently, from .the Nautical Almanac of
1824, instead of its mean polar distance ; but since this can only lead to a
trifling difference, onii not to any erroTy it wa^ not, therefore, deemed
necessary to alter or recompute the Table.
Tablb XII.
Mean Right Ascension qfthe Sun.
This Table may be used for the purpose of finding the approximate time
of transit of a fixed star, when a Nautical Almanac is not at hand ; it may
also be employed in finding the right ascension of the meridian, or mid-
heaven, when the latitude is to be determined by an altitude of the north
polar star : for, if to the sun's right ascension, as given in this Table, the
apparent time be added, the sum. (rejecting 24 hours if necessary) will be
the right ascension of the meridian, sufficiently near the truth for deter-
lipning the latitud^^
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88 9S8CltIPnO|f ANB USB OF TUB TABLW*
Table XIII.
Equations id equal Altitudes. — first part.
The arguments of this Table are^ the interval between the observations
at top or bottom, and the latitude in either of the side columns ; in the
angle of meeting stands the corresponding equatipn, expressed in seconds
and thirds.: hence the equation to interval 6 hours 40 minutes and latitude
60 degrees, is 15 seconds and 33 thirds.
The equations in this Table were computed by the following rule, viz. :—
To the log. cb-tangent of the latitude, add the ' log. sine of half the
interval in degrees ; the proportional log. of the whole intierval in ' time
(esteemed as minutes and seconds), and the constant log. 8. 8239;* the
sum of these four logs., rejecting 29 from the index, will be the propor-
tional log. of the corresponding equation ii\ minutes and seconds, which
are to be considered as seconds and thirds.
Example.
Let the latitude be 50 degrees, and the interval between the observed
equal altitudes of the sun 4 hours ; required the corresponding equation ?
• Latitude 50?0C0^ Log. co-tang. 9. 9238
Half int. = 2 hours, in deg;8.=: 30. 0. 0 Log. sine . 9. 6990
Whole interval 4 hours, esteemed as 4 min.^ propor* log. 1 . 6532
Constant log 8.8239
Required equation * . . . U'^IS'^ Plropor.log. 1.0999
The equations in the abovementioned Table were computed by Mrs. T.
Kerigan.
Table XIV.
Equations to equal Jltitudes.^^VART sbcond.
In this Table, the interval between the observations is m^irked at top or
bottom, and the sun*s declination in the left or right-hand margin ; under
or over the former, and opposite to the latter, stands, the corresponding
equation, expressed in seconds and thirds : thus, the equation answering to
6 hours 40 minutes, and declination 18?30', is 2 seconds and 48 thirds.
The equations contained in this Table were computed as follows, viz. :— •
To the log. co-tangent of the declination, add the log. tang, of half the
interval in degrees ; the proportional log. of the whole interval in time
(esteemed as minutes and seconds), and the constant log. 8. 8239}t the
* t The viUunetical compleitLent of 12 hours considered u i
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DB8C&IPTi<^ AND USB OF THfi TABU& 28
sam of these four logs., rejecting 29 from the index^ will be the propor-
tional log. of the corre^onding equation in minutes and secondfly which
are to be considered as seconds and thirds. .
Example. .
Let the sun*s declination be 18?30', and the interval between the
obserred equal altitudes of the sun 4 hours ; required the corresponding
equation ?
Sun's declination .... 1893G' Log. co-tang. 10. 4755
Half interval = 2 ho. in degs.=30 . 0 . Log. tang. . 9. 7614
Whole interval 4 ho. esteemed as 4 min. Prop. log. 1 . 6532
Constant log 8.8239
Required equation =5 • . . 3729r Pirop. log. . 1.7140
The equations in the abovementioned Table were^ also> computed by
Mrs. T. Kerigan.
To find the Equation of Equal Altitudes by Tables XIU. and XIV.
Rule.
Enter Table XIIL, with the latitude in the side column and the interval
between the observations at top $ and find the corresponding equation^ to
which j^refix the sign + if the sun be receding from the elevated poIe> bul
the sign — if it be advancing towards that pde.
Enter Table XIV., with the declination in the side eolumn, and the
inlerral between tiie observations at top, and take out tiie corresponding
equation, to which prefix the sign + when the sun's declination is increoi^
big, but the sign — iwheh it is decreasbig. •
liofw, if those two equations are of the same signs ; that is, both afSniH
ative or both n^ative, let their sum be taken) but if contrary signs, namely^
one affirmative and the otiier negative, their difierence is to be taken : then.
To tlie proportional log. of this sum or difference, considered as minutes
and seconds, add the proportional log. of the (bdly variation of the sun's
declination ; and the sum, rejecting 1 from the index, will be the propor*
tional log. of the true equation of equal altitudes in minutes and seconds,
which are to be esteemed as seconds wd thirds, and which will be always
of the same name with tiie greater equation.
Example 1.
In latitude 49? south, the interval between equal altitudes of the sun
was 7*20? j the suns declination 18? north, increasing, and the variation
of decUnation 15 '12? ; required the true equation of equal altitudes ?
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24 DBSCRIPTIOK AND USB O^ THB TABLB8*
Opposite lat. 49? under 7^20? Tab. XIII. 8tand8+ 15^27^
Opposite dec. 18? under 7 ' 20r Tab. XIV. stands + 2. 30
Sum 17^57^^ Pro. log.l. 0012
Variation of declination . . 15 '. I2r Pto. log.l . 0734
True equation, as required + 1 5 ri Or Pro. log. 1 . 0746
Example 2.
In ladtude 50?north, the interval between equal altitudes of the sun was
5^207; the sun*s declination IS^SO^north^ increasing, and the daily vari-
ation 6( declination 14^34?; required the true equation of equal altitudes ?
Op.lat. 50? under5*20?Tab.XIII.8tands-l4r5br
Op.dec.l8?30'.under5.20 Tab. XIV. stands + 3.11
Diflference .... -Iir39r Pro. log. = 1.1889
Variation of declination 1 4 ' 34 ? Pro. log. = 1 . 09 1 9
True equation, as required^ 9^26^ Pro. log. = 1 . 2808
Memarktr^ln north latitude the sun recedes from the elevated pole from
the summer to the winter solstice ; that m^ from the 21st June to the 21st
December $ but advances towards that pok from the winter to the summer
solstice; viz., from the 21st December to the 2l8t June. The converse of
this takes place in soutli latitude: thus, from the 2l8t June to the 21st
December, the sun advances towards the south elevated pole ; but recedes
from that pole the rest of the year, viz., from the 2Ut December to the
21st June.
Here it may be necessary to observe, that in taking out the equations
from Tables XIII. and XIV., allowance is to be made for the excess of the
givaui above the next less tabuhir arguments, as in the following examples;
Example U
Required the equation from Table XIIL, answering to latitude 50?48^9
and interval between the ob^iervations 5 hours 10 minutes ?
Equation to latitude 50?, and interval 4 UOr = ^ . 14r33r
Tabulardiff.tol?oflat,== + 3irjnow,^^^' = + 0,24^
Tab. diff.to40f of inter. = + I7'J'jnow,l^J??- ^ + 0. 12|
Equation^ as required' ••«•,*,••• .15^10?
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DJUCRIPnoN AND USB. OF THE TABtM. 25
jtxample 2,
Required the equation from Table XIV., answering to sun's decUnatloa
20?47 ' , and interval between the observations 5 hours 10 minutes ?
Equation to declination 20?30; and intenral 4?40Tz3 3r44r '
TabdardiflF. to 30^ declination ==+ 6^5. now, 5liiJ2- =5 +0. 3i
3u.
Tabular diff. to 401 interval = - lOT; now, ^9L^f^^^ 0. 7J
Equation, as required . • • • 3^40^^
JVo/tf.— Should the latitude exceed the limits of Table XIII., which is
only extended so far as to comprehend the ordinary bounds of navigation,
viz., to 60 degrees, the first part of the equation, in this case, must be
determined by the rule under which that Table was computed, as in
page 22.
Table XV.
To reduce the Sim*8 Longitadej Right Jscemion, and Declination ; and
also the Equation of Time, as given in the NaiUical Almanac j to any
givefi Meridian, and to any given Time wider that Meridian.
This Table b so arranged, that the proportional part corresponding to arty
given time, or longitude, and to any variation of the sun's right ascension,
declination, &c. &e., may be taken out to the greatest degree b( accuracy,
— even to the two hundred and sixteen tliousandth part of a second, \f
necessary.
. Precepts.
In the general use of this Table it will be advisable to abide by tiie solar
day ; and hence, to estimate the time from noon to noon, or from 0 to 24
hours, after the manner of astronomers, without paying any attention to
either the nautical or the civil division of time at midnight. And to guard
against falling into error, in applying the tabular proportional part to the
sun's right ascension, declination, &c. &c., it will be best to reduce the
apparent time at ship or place, to Greenwich time ; as thus :
Turn die longitude into time (by Table I.), and add it^to the given time
at ship or place, if it be ti;e»/; but subtract it {{east; and the sum,^ or
^foeocei w31 be. the c9rr^pQnd}ng time at Greenwich^
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9§ DBSCRIpTION AND USB 09 Tfll^ TA^LM
From page II. df the month in the Nautical Almanac, take out the
sun's right ascension, declination, &c. &c., for the noons immediately
preceding and folhmng the Greemoich time, and .find their difference^
which will. express the variation of those elements in 24 hours ; then,
Enter the Table. with the variation, thus found, at top, and the Ghreen-
wich time in the left-hand column ; under the former and .opposite the
latter will be found the corresponding equation, or proportional part* And^
since the Greenwich time may be estimated in hours, minutes, or seconds,
and the variation of right' ascension, &c. &c. &c., either in minutes or
seconds s the sum of the several proportional parts making up the whole
of such time and variation will, therefore, express the required proportional
part* The proportional part, so obtained, is always to be applied by
addition to the sun's longitude and right ascension at the preceding noon ;
but it is to be applied by additioUy or subtraction, to the sun's declination
and the equation of time at that noon, according as they are increasing or
decreasing. — See the following examples : —
Example I.
Required the sun's right ascension and declination, and also the equation
of time May 6th,l 824, at 5 * 1 OT, in longitude 64 ?45 ^ west of the meridian
of Greenwich?
Apparent time at ship or place 5^ 10?
Longitude 64?45^ west, in time =: . . • + 4. 19
Greenwich time •••••...• 9^29?
To find tlie &ia'8 Right Ascension :~
Sun's right ascension at noon, May 6th, 1824, per Nautical
Almanac, . « 2^53:3l!42f
Variation iq 24* =3^52'/
Pro. part to 9* 0?and 31 Or= V, 7^30r OV
Do, to 0. 29 and 3. 0 = 0. 3. 37. 30
Do. to 9. 0 and 0.50 = 0.18.45. 0
Do. to 0.29 and 0.50 = 0. 1. 0.25
bo. to 9. 0 and 0. 2 = 0. 0.45. 0
Do. to 0. 29 and 0. 2 = 0. 0. 2. 25
Pro. part to 9*29?and 3^52? is 1.81.40.20±: +r31*40r
Sun's right ascenrion, as required • • • , # . « • • 2^55? 3-22>
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To find the Sun's Declination :-—
San*8 declination at noon^ May 6th^ 1824, "pet Nautical
Almanac^ .*......•,.•....• 16?36'5?
north, increasing, and var. in 24 ho.= 16^38T
Pro. part to 9* OTand 16^ 0^= 6^ 01 OT 0"."
Do. to 0.29 and 16. 0 = 0.19.20. 0
Do. to 9, 0 and 0.30 = O.ll.lS.H)
Do, to 0.29 and 0.30 = 0. 0.36.15
Do. to 0. 0 and 0. 8 = 0. 3. 0. 0
Do. to 0. 29 and 0. 8 = 0. 0. 9. 40
Pro. part 'to 9*29rand IG^SSris 6.34.20.55 « + 6^34r
Sim's declination, as required » • • . • • • • • 16?42'39C
To find the Equation of Time : —
Equation of time at noon, May 6thy 1624^ p«r NtuUeal
Almanac, • • . 3T36! 6f^
increasing, and variatioti in 24 hours aa 4?d6T
Pro. part to 0* 0?and 4r OT « l*30r OV
Do. to 0.20 and 4. 0 » (K 4.50
Do. to 9. 0 and 0.30 » O.lhlS
Do. to 0«29 and 0.30 s 0. 0.36
Pro. part to 9t29? is 4r30r = K46.41 = + 1^47^
* 'tm '
Equation of time, as required i • 3T37'53f
Example 2. '
Required th^ sun's right ascension and declination, and alio the eqtiation
of time, August 2d, 1824, at 19^22?, in longitude 98?45t east of the
meridian of Greenwich ?
Apparent time at ship or place . >. . . 19^22?
Longitude 98?45: east, in time a • . .-6.35
Greenwich time •..«••••• 12M7*
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^9 BBSCRIPTIOK AND USB OP THB TABIJ&I.
To find the Sun's Right Ascension :—
Sun's right ascension at noon^ August 2d^ 1824^ per Nautical
Almanac, • 8*50rO!48!
Variation in 24 hours =3^521^
Pro. part to 12t Or and 3^ Or = l^SOr OT 0''/
• Do. to .0.47 and 3. 0 = 0. 5.52.30
Do.^ to 12. 0 and 0.50 = 0.25. 0. 0
Do. ^ to 0.47 and 0.50 ,= 0. 1.37.55
Do. to 12. 0 and 0. 2 =0. I. 0. 0
Do. to 0. 47 and 0. 2 z= 0. 0. 3. 55
Pro. part to 12*47^ and 3^52^ is 2. 8.34.20= +2^3r34r
Sun's right ascension, as required • 8^52^4! 22 f
To* find the Sun's Declination :—
Sun's declination at noon, August 2d, 1824, per Nautical
Almanac, w . . 17?44Mir
north, decreasing, and var. in 24t =: 15^ 36?
Pro. part to 12* OTand 15C 07 = 7'30r OT OV
Do. to 0.47 «nd 15. 0 = 0.29.22.30
Do. to 12, 0 and 0.30 =: 0.15. 0. 0
Do. to 0. 47 and 0. 30 =0. 0. 58. 45
Do. to 12. 0 and 0. 6 =0. 3. 0. 0
Do. to 0.47 and 0. 6 =: 0. 0.11.45
Pro. part to 12*47rand 15:36r is a. 18.33. 0= - 8^9?
gun's declination, as required 17?36^22r
To find the Equation of Time :— -
Equation of time at noon, August 2d, 1824, per Nautical
Almaniie, Sr54!24f
decreasing, and variation in 24 hours =: 4^30^
Pro. part to 12* OTand 4? Or = 2r Or OV
Do. to 0.47 and 4. 0 == 0. 7.50
Do. to 12. 0 and 0.30 s 0.15. 0
Do. to 0.47 and 0.30 = 0, 0.58
Pro. part to 12t47^and 4r30r is 2.23.48 = - 2r24r
Equation of titpc; as required .'••••,•,• 5752! 0!
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BBSCRIFTION AND USB OP THE TABLES. 29
Hemark.^^hould the proportional part corresponding to the daily
variation of the sun's longitude and any given time be required, it may be
taken from the first page of the Table, by esteeming the seconds of varia-
tion, in that page, as minutes, and then raising the signs of the correspond-
ing proportional parts one grade higher ihan what are marked at the top
of the said page : the seconds of variation will, of course, be taken out after
the usual manner. Thus,
Suppose that the daily variation of the sun's longitude be 57- 40T, and
the Greenwich time 9 hours 50 minutes, to find the corresponding equationy
or proportional part.
Vto. part to 9* OTand 50C Or = l8M5r Or OV
Do. to 9. 0 and 7. 0 = 2.37.30. 0
Do. to 0.50 and 50. 0 s 1.44.10. 0
Do. to 0.50 and 7. 0 s 0.14.35. 0
Do. to 9. 0 and 0.40 = 0.15. 0. 0
Do. to 0.50 and 0.40 '= 0. 1.23.20
Pro. part to.9t50rand 57'40r is 23.37.38.20= 23J38f +
Kote. — It is easy to perceive that the foregoing operations might have
been much contracted, by taking out two or more of the proportional parts
at once ; but, lest doing so should appear anywise ambiguous to such as
are not well acquainted with the method of taking out tabular numbers, it
was deemed prudent to arrange the said operations according to their
present extended form, so as to render them perfectly intelligible to every
capacity.
The present Table was computed agreeably to the established principles
of the rule of proportion ; viz.. As one day, or 24 hours, is to the variation
of the sun's right ascension, declination, &c. &c., in that time, so is any
other portion of time to the corresponding proportional part of such
variAtion.
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30 BESCniPTlON AND TJSB OF THK TABUSS.
Tablb XVI,
To reduce the Moon's Longitude, Latitude, Right Jscension, Declination,
Semidiameter, and Horizontql Parallax, as given in the Tiautical
Almanac, to any given Meridian, and to any given Time under that
Meridian.
This Table is arranged in a manner so nearly similar to the preceding,
that any explanation of its use may he considered almost unnecessary ; the
only difference being, that the proportional parts are computed to wiatioa
in 12 hours, instead of 24. By means of the present Table, the proper*
tional part corresponding to any variation of the moon's longitude, latitude,
right ascension, &c. &c. &c., may be easily obtained, to the greatest degree
of accuracy, as follows ; viz.
Turn the longitude of the ship or place into time (by Table I.), and add
it to the apparent time at such ship or plaise, if it be west; but subtract it
if east : and the sum^ or difference, will be the corresponding time at
Qreenwich.
Take from pages V., VI., and VII. of the month, in the Nautical
Almanac, the moon's . longitude, latitude, right ascension, declination,
semidiameter, and horizontal parallax, (or any one of tliese elements,
a<icording to circumstances,) for the noon and midnight imme^Kfttely
preceding and following the Greenwich time, and find their difference ;
which difference will express the variation of those elements in 12 hours.
Enter the Table with the variation, thus found, at top, and the Green-
wich time in the left-hand column ; in the angle of meeting will be found
the corresponding equation, or proportional part, which is always to be
added to the moon's longitude and right adcension et the preceding noon
or midnight, but to be applied by addition, or subtraction^ to the moon's
latitude, declination^ semidlanieter, and horizontal parallax, according as
they are increasing or decreasing. And, since the Greenwich time and the
variation in 12 hours will be very seldom found to correspond exactly; it is
the sum, therefinre, of the several equations making up those terms^ that
will, in general, express the lEequired proportional part*
* JExample.
Required the moon's longitude, latitude, right ascension, decliiMtaon,
semidiameter, and horizontal parallax, August 2d, 1824, at 3 MO", in
longitude 60?30', west of the meridian of Greenwich ?
Apparent time at ship or place . , . . . 3*10?
Longitude 60? 30^ west, in time = . ,- . 4. 2
Greenwich time ••••••,•• 7n-2?
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DBSC&fFTtOK AND USE OF THB TABIS9. 81
To find the Moon's JLiongitude :—
oon's longitude at noon, August 2d, 1824, per Nautical
Almanac,
.......; 7M7?16(27?
Variation i
in 12* =6° snsgr
PropoT. -part
to 7* or and 6? 0'. 07 = 3»30^ 0? Or
Do.
to 0. 12 and 6. 0. 0 =z 0. 6. 0. 0
Do.
to 7. 0 and 0.30. 0 = 0.17'.30. 0
Do.
to 0.12 and 0.30. 0 =0. 0.30. 0
•
Do.
to 7. 0 and 0. 1. 0 =0. 0.35. 0
Do.
to 0.12 and 0. 1. 0 =s 0. 0. 1. 0
Do.
to 7. 0 and 0. 0.50 = 0. 0.29.10
Do.
to 0. 12 and 0. 0.50 = 0. 0. 0.50
Do.
to 7. 6 and 0. 0. 9 = 0. 0. 5. 15
Do.
to 0. 12 and 0. 0. 9 = 0. 0. 0. $
Proper, part
to 7^2? and 6?3i:597 U 3.55.11.24=
tude, as required 7
;+S?55MI*
Moon's lengi
!21?li:38'
To find the Moon's Latitude ^-««-
Moon's latitude at noon^ August 2d, 1824, per Nautical
Almanac, 4?6'59:
south, decreasing, and var* in 12 hours s 23' 35 Y
Proportional part to 7? 0? and 20^ O: = U'AOl OT
Do. to 0.12 and 20. 0 = 0.20. 0
Do. to 7. 0 and 3. 0 = 1.45. 0
Do. to 0. 12 and 3. 0 = 0. 3. 0
Do. to 7. 0 and 0.30 = 0.17.30
Do. to 0. 12 ^d 0.30 =z 0. 0.30
Do. to 7. P and 6. 5 df 0. 2.55
Do. to 0. 12 and 0. 5 = 0. 0. 5
Proportional part to 7*12? and 23:35r is 14. 9. 0 =s- W. 9t
Moon's latitude, as required . « . • 3?52^50r
Note. — In consequence of the unequal motion of the moon in 12 hours,
(when her place is to be determined with astronomical precision,) the
proportional part of the Tariation of her longitude and latitude, found as
above, mtist be corrected by the equatioa of second difference contained in
Table XVIL ; and the same may be obsenred of her right ascension and
declination.
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82 BJBSCRIPTION AND USE OF THB TABLB8#
Tq find the Moon's Right Ascension :—
Moon's right ascension at noon, August 2d, 1824, per
Nautical Almanac, 223?33^36^
Var.inl2* ='6?51M9r
Propor. part to 7i or aAd 6? Of Or = 3?30f 01 OT
Do. to 0. 12 and 6. 0. 0 = 0. 6. 0. 0
Do. to 7. 0 and 0.50. 0 =: 0.29.10. 0
Do.. to 0.12 and 0.50. 0 = 0. 0.50. 0
Do. to 7. 0 and 0. 1. 0- = 0. 0.35. 0
Do. to 0.12 and 0. 1. 0 = 0. 0. .1. 0
Do. to 7. 0 and 0. 0.40 rs 0. 0.23.20
Do. to 0.12 and 0. 0.40 =0. 0. 0.40
Do. to 7. 0 and 0. 0. 9 =s 0. 0. 5. 15
Do.' to 0.12 and 0. 0. 9 = 0. 0. 0. 9
Propor. part to 7*12? and 6?5i:49ris 4. 7. 5. 24= +4? 7^ 51
Moon's right ascension, as required ••••.«. 227?40'4lr*
To find the Moon's Declination :—
Moon's declination at noon, August 2d, 1824, per Nautical
Almanac,-. '. 20?57f H
8outh, increasing, and var. in 12 ho.= I ^23' 431
•or. part to 7- 0'
and 1? 0'. or
:=
35^ or or
Do.
to 0.12
and 1. 0. 0
=
1. 0, 0
Do.
to 7. 0
and 0. 20. 0
^z
11.40. 0
Do.
to 0.12
and 0.20. 0
zz
0.20. 0
Do.
to 7. 0
and 0. 3. 0
ZZ
1.45. 0
Do.
to 0. 12
and 0. 3. 0
=
0. 3. 0
Do.
to 7. 0
and 0. 0.40
""
0.23.20
Do.
to Oi.12
andO. 0.40
™*
0. 0.40
Do.
to 7. 0
and 0. 0. 3
^
0. 1.45
Do.
to 0.12
and 0. 0. 3
=
0. 0. 3
Propon pirt to ■7M2? and 'l ?23'43r is 50. 13. 48 = + 50; 14r
Moon's declination^ as required 21?47;2lr*
* When ftecuracy is required, the moon's rif^ht aicension and declination must be cor-
rected by the equation of second dlfferencCi on account of the Irref^arities of her motioa
in 12 hours.
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DBSCRIPTiON AND USB OF THB TAJBLB6. S3
To find the Moon's Semidtameter :— «
Moon's semidiameter at noon, August 2d, 1824, per NouUcal
Almanac, # • . . • 15'33^
decreasing, and var. in 12 hours = %".
Proportional part to 7* 0? and 6r = SrSOT
Po. to 0.12 and 6 s= 0. 6
Proportional part to 7*12? and 6r is 3.36 =s - 4
Moon's semidiameter, as required ••«;•••• 15^291
To find the Moon'9 Horizontal Parallax :—
Mood's horizontal parallax at noon, August 2d, 1824, per
Nautical Almanac, • • • • 57*6!
decreasing, and var. in 12 hours = 231"
Proportional part to 7* OT and 20r = lir40r
Do. to 0.12 and 20 = 0.20
Do. to 7. 0 and 3 = 1.45
Do. to 0. 12 and 3 = 0. 3
»
Proportional part to 7*12? and 23r is 13.48 = - 14?
Moon's horizontal parallax, as required ^ t « • • « 56'52T
Remarks. — 1. It is evident that, in the above operations, the greater
part of the figures might have been dispensed with, by taking out two or
more of tlie proportional parts at once ; but since they were merely intended
to simplify and render familiar the use of the Table, the whole of the pro-
portional parts have been put down at length.
2. This Table was computed according to the rule of proportion; viz.:-^ ,
As 12 hours are to- the variation of the. moon's longitude, latitude, right
ascension, &c. &c. &c., in that interval, so is any other" given portion of
time to the corresponding proportional part of such variation.
Tabls XVII.
Squation of Secmd Difference.
Since the moon's longitude and latitude, and also her right ascension
and declination, require to be strictly determined on various astronomical
pceasioi^; porUcularl^ the twQ latter when the apparetU time is to be
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S4 DESCRIPTION AND tJSfi 6F THE TABLES.
inferred from the true altiiude of that object; and since the reduction
of these elements^ to a given instant, 'cannot be performed by even propor-
tion, on account of the great inequalities to which the lunar motions ««
subject }— a correctio/r, therefore, resulting from these .inequalities, must be
applied to the proportional paYt of the moon's longitude or latitude, right
ascension or declination, answering to a given period after noon ^f mid-
night, as deduced from the preceding Table or otherwise, in 4»rder to have
it.truly accurate. This correction is contained in the present Table, the
arguments of which are, — ^the mean second difference of the moon's place
at top ; and the apparent or Greenwich time past noon, or midnight, in
the left or right-hand column ; in the angle of meeting stands the corre-
sponding equation or correction.
The Table is divided into two parts ! the upper ' part ia adapted to the
mean second difference of the moon's place in seconds of a degree, and in
which the equations are expressed in . seconds and decimal . parts* of a
second ; the lower part is adapted to minutes of mean second difference ;
the equations being expressed in minutes, and seconds, and decimal parts
of a second.
In using this Table, should the mean second difference of the moon's
place exceed its limits, the sum of the equations corresponding to the
several terms which make up the mean second difference,- in both parts of
the Table, is in such case to be taken. The manner of applying the
equation of second difference to the proportional part of the moon's
'motion sn latitude, longitude, right ascension, or' declination, as deduced
from the preceding Table, or obtained by even proportion, will be seen in
the solution to the following
Problem* *
7b reduce the JU9m*s LatUude, Longiiude, Eight McentUm^ <»td J>e^&a-
tion, a given w the NauHcal jUmame, to any given Time under a
hwm Meridian.
Rule.
Turn the longitude into time, (by Table I.) and apply it to the apparent
time at ship or place by addition in west, or subtraction in east longitude j
and the sum, or difference, will be the corresponding time at Greenwich.
Take from the Nautical Almanac the two longitudes, latitudes, right
ascensions, and declinations immediately preceding and following the
Greenwich time, and find the difference between each pair successively j
find also the second difference, and let its mean be taken.
Find the proportiood part of the middle Jbrst difliereiice, (the tariation
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BiiscRipnoN akd ttsb of fnA fABLSs. 65
of the moon's motion in 1^ hours^) by Table XVL^ answering to the
Greenwich lime^
With the mean second difference, found as above, and the Greenwich
time, enter Table XVIL, and take out the corresponding equation. Now,
thia equation being • addtd to the proportional part of the moon's motion
if the first first difference is greater than the third first difference^ but
fubtracted if it be less, the sum or difference will be the correct propor-
tional part of the mood's motion in 1 2 hours.
The correct proportional p&rt, thus found. Is always to be added to th4
moon's longitude and right ascensioa at the noon or midnight preceding
the Greenwich time ^ but to te applied by addition or subtraction to her
latitude and declination, accoitling.as they may be increasing or de-
creasing.
^ Example.
Required the moon's correct longitude, latitude, right ascension, and
declination, August 2(1, 1^24, at 3^ 10? apparent time, in longitude 60^30.,
west of the meridian of Greenwich }
Appajrent time at ship or place • • • • . SMO?
Longitude 60?30^ west, in time a « • • 4. 2
Greenwich time • « • V 7M2?
To find the Moon's comet Longitude :-^
First Second Mean
Diff. Diff. 2dDiff.
Moon's long. Aug. l8t,atmidnt. 7* 10^38 M9r 1^^007/00//!
Do. 2 atnoon 7.17.16.27 [Ai JLi^'^^'^U'oji^
Do. i atmidnt.7.23.48.26 j«-31.59 jg j^ j^*?*-
Do. 3 atnoon 8. 0.15/ 9 l6.26.43
Pfopor. part from Table XVL,ans.to7M2?and6?3l^59ris3?65nK24r
Bq. fromTab. XVIL, corres. to 7* 12? and 5 ^ Or =: 36^ 0
and0.20 =2.4
andO. 7J= 0 .9
Eq.ofmeaiisecoriddiff.an6.to7*12Tand5^27|ris39 .3= + 39n8r
Correct proportional part of the moon's motion in longitude 3?55 '60^42?
Idoon's longitude at noon, August 2d, 1824 . . . . 7 • 17. 16. 27. 0
Moon's correct longitude, at the given time • • . . 7 ' 2 1 ? 1 2 : 1 7 742r
j)2
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To find the Moon's correct Latitude :— •
First Second Mean
Diff. Diflf. 2dDiff.
Moon's lat. Aug. Ist, at midnt. 4?27 '37^ S* lon^^fi'^i
Do. 2 atnoon 4. 6.59 I f2^57^lo/^^.
Do, 2 at midnt. 3. 43. 24 }23.35 \^ gj J2.44.
Do. 3 atnodn 3.17.18 }26. 6 ^
Pro.partfromTableXVI.,an8.to7*12rand23^35r is0?14^ 9r
Bq.fromTab.XVIL,cor.to7*12r and2^ Or= 14^4
and 0.40 =4.8
andO. 4=0.5
Eq.ofmean8ec.diff.,ans.to7*l2rand2U4fis 19 .7= — 19^^.7
Correct proportional part of the moon's motion in lat. 0?13C49^.3
Moon's latitude at noon^ August 2d, 1824 • • • . '4. 6.59 .0 S.
Moon's correct latitude at the given time • • . « 3?53' 9^". 7south.
To find the Moon's correct Right Ascension:—
First Second Mean
' • Diff. Diff. 2d Diff.
Moon's R. A. Aug. 1st, at midnt 216?44 ;43r « ^\, .<. , .„«.
Do. 2 atnoon 223.33.36 ° [2;56ri„,„q,
Do.. 2 atmidnt.230.25.25. 6-51.49 I2 gJ
Do. 3 atnoon 237' 1^.22 r6.53.57 ' '
Propor.partfromTableXVI.,an8.to7M2?and6?51M9ri8 4? 7' 5r24r
Eq. from Table XVII., ans. to 7 M 2? and 2 C Or = 14".4
andO.30 = 3 .6
andO. 2 = 0.2
^.ofmeansec. diff.,ans.to7M2?and 2!32ris 18 .2 = - 18M2r
Correct prbpor. part of the moon's motion in right ascension 4? 6^47^12^
Moon's right ascension at noon, August 2d, 1824 . . 223.33.36. 0
Moon's correct right ascension at the given time . . • 227?40'23ri2*r
To find the Moon's correct Declination :
First . Second Mean
DiflF. Diflf. 2dDiflF.
Moon's dec, Aug. 1st, at midnt* 19? 15 C49r 8.1,0^1/10.^1
Do. 2 atnoon 20.57. 7 L ! h7-35r),-..ft^
Do. 2 at midnt. 22. 20. 50 U.23.43 {^g gj jl/'^^-
Do. 3 atnoon 23.26.12 ' }l. 5.22 *
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JttSCRlPnON AND ITSB OF THB TABLES. 87
fto, part fir. Tab. XVI., ans. to 7 * 1 2? and 1 923 M3r is 0?50r 13r48r
Eq.fr.Tal>.XVII.cor.to7M2?andl5^ 0^=1 M8^0
and 2. 0 =0.14 .4
and 0.50 =0. 6 .0
and 0. 8 =0. 1 .0
Eq.ofmn8ec.diff.,an8.to:r*12?andl7'Mrw2. 9 .4= + 2^ 9r24r
Correct prop, part of the moon's motion in declination 0?52 ' 23T 12T
Moon's declination at noon^ August 2d^ 1824 . • 20.57. 7. OS.
Moon^s correct declination at the given time . • 21 ?49 ' 30? 1 2T south.
Note. — It frequently happens that the three ^r«t differences first increase
and then decrease^ or tnce versa, first decrease and then increase ; in this
case half the difference of the two second differences is to be esteemed as
the mean second difference of the moon's place : as thus^
Rrst Second Mean
Diff. Diff. 2dDiff.
Mn'sdec.Aug.l8th,1824,atmidt.24?23C26rN.>,Q,rt,..>
Do. 19 atnoon 24.41.47 T [l4'26fi-,o«^
Do. 19 atmidt.24.37.52 / 3-55 {33 jo r*^^*
Do. 20 atnoon 24.10.39 }27. 13 ^
Here the two second differences are 14 '26?, and 23' 18f respectively;
therefore half their difference/viz., 8f52r h- 2 = 4C26r is the mean
Second difference. Now, if the Greenwich time be 5 MO? past noon of
the 19th, the corresponding equation in Table XVII. will be 33f sub-'
traciive^ because the first^W difference is less than the third Jirst differ^
ence; had it been greater, the equation would be addiiive*
Remark. — ^When the i^parent time is to be inferred from the true '
altitude of the moon's ccfntre, the right ascension and declination of that
object ought, in^neral, to be corrected by the equation of second differ-
ence ; because an inattention to that correction may produce an error of
about 2| minutes in the right ascension, and about 4 minutes in the de-
elination ; which, of course, will affect the accuracy of the apparent time.—
See the author's Treatise on the Sidereal and Planetary Parts of Nautical
Astronomy, pages 171 and 172,
Tlie equation of second difference, contained in the present Table, was
computed by the following .
Rule.
To tha constant log. 7« 540607 add the log. of the mean second differ-
mce reduced to seconds ; the log. of the time from noon, and the log. of
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98 IffiSCRIfTfOlif 4ND U8H 09 THl TABU9,
the diiferenc^ of ttial time to 12 hours (both expressed !n houro and
decimal parts of an hour) : the sum, rejecting IQ from the index, will
be the log. of the equation of second difference in seconds of a degree*
Example.
Let the mean second difference of the m<>on'8 plac6 be 8 mii^utesi and
the apparent time past noon or midnight 3 t20T; required the correspond-
ing equation } '
Mean second difference, 8 minutes = 480 seconds. Log. =z 2. 681241
Apparent time past noon or midnight == 3* . 333 Log. == Q. $3SS835
Difference of do. to 1 2 hours 8 * . 666 Log. = 0. 9378 1 9
Constant log. (ar. co. of log. of 288 = 24 x 12) . . =7. 640607
Required equation • 48''. 14 Log. == 1 . 68250S
Table XVIII.
Correction qf the Mgon^s Jpparent Jltitude.
By the correction of the mqon's apparent altitude is ^ant^ the diBfer-
ence between, the parallax of that object, at any given altitude, and the
refraction corresponding to that altitude.
This correction was computed by the following rule ; viz.
To the log. secant of the moon's apparent altitude, add the proportional
log. of her horizontal parallax; and the sum, abating 10 in the index, wilt
be the proportional log. of the parallax in altitude ; which, being diminished
by the refraction, will leave the correction of the moon's apparent altitude.
Let the moon's apparent altitude be 25?40^, and her horizontal parallax
59 minutes ; required the correction of the apparent altitude ?
Moon's iq>parent altitude . , , . 25?40^ Log, secant =: 10.0451
Moon's horizontal parallax • • • * 0. $9 Propor, log. =: 0* 4844
Moon^ parallax in altitude ; . . 53'IH scProporJog.s: 0.5295
Refraction ans. to app. ait. in Tab. VIIL 1.58
Correction of the moon's appar. altitude 51^131^
The correction^ thus compnt^d, is arranged* in the present Tahle, where
it is given to every tenth minute of apparent altitude^ and to each minute
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]»4CKIFnON AND US^ OF THS TABLES* 89
of horizontal parallax. The proportional part for the excess of the g^ven
above ttie next less tabular altitude^ is contained in the right-hand column
of each page ; and that answering to the seconds of parallax is given in
the intermediate part of the Table^
This correction is to be taken out of the Table in the following manner ;
viz.
BntCK the Table with the moon's apparent altitude in the left-hand
eoluDiii, or the altitude iiext iess if there be any odd minutes ; opposite to
whichy and wider the minutes of the moon's horizontal parallax^ will be
found the approximate correction. Enter the compartment of the ^' Pro-«
poitiona] parta tP seconds of parallax/' abreast of the approximate correct
tioDy with the tenths of seconds, of the moon'« horizontal parallax in the
vertical column, and the units at t(ie top ; in the angle of meeting will ba
feuttd the proportioiial pari for second^ which add to the approximate,
correction. Then,
Enter the last or right-band column of the page, abreast of the approxi*
mate coivection or nearly so, and find the proportional part corresponding^
to the odd minutes of altitude. Now, this being added to or subtracted
from tlfte approximate correction, according to its signy will leave the trua
eocraelfam of the moonV apparent altitude, . And since the apparent alti-
tude of a eelestial oltfect is depressed by parallax and raised by refraction,
and the lunar parallax being always greater than the refraction to the
aaiqe allitude, it hence ioUaws that the correction, thus deduced^ i» always
to liQ iiPpUdi by {^Mlieu, (o th^ moon's apparent altitude,
Example U
Let Iha bbooii*s apparent aiti^de be &?38C, aad her bori«(mtal parallax
57^46? I requiied dM» corresponding c<weetion }
Correction to alL 8?3a^ and horiz. parallax 57 'Or is 50n4r
P^opor. part to 46 seconds of horiz. parallax • • + 0. 46
Do. to 8 min. of alt, (8* x Of. 5=4 -'. 0) = + 0. 4
Correction of the moon's apparent altitude^ as required 5 1 C it
Let the moon's apparent altkude be S3?I0C, and her horizontal parallax
59^34T ; required the corresponding correction ?
Correction to alt 33? 10^ and horiz. parallax 59^ Or is 47'56r
Propor. part to 34 seconds of horiz. parallax • • • + 28
Do. to 6 minutes of altitude — 3
Correction of the moon's apparent altitude, as req^irfd 48 ' 2 K.
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40 bESCRlPf ION AND USS OF THE TABLES.
•Table XIX.
To reduce the Tme Jllitudes qf the Sun, Moon, Stars, and PUmefs, *
to their apparent Jltitudes.
This Table is particularly useful in that method of finding the longitude
by lun{ir observations, where the distance only is given, and where, of
bourse, the altitudes of the objects must be obtained by computation.
' The Table consists of two pages, each page, b^ing divided into two*
parts : the left-hand part contains four columns ; the first of which com-
prehends the true altitude of the sun or star; the second the reduction of
the sun's true altitude ; the third the reduction of a stair's true altitude f
ind the fourth the common difference of those reductions to 1 minute of
altitude for sun or star.
The other part of the Table is appropriated to the moon ; in which the
true altitude of that object is given in the column marked " Moon's true
altitude," and her horizontal paralla^f at top or bottom ; the two last or
right hand columns of each page contain the difference to 1 minute of
idtitude, and 1 second of parallax respectively ; by means of Which the
reduction may be easily taken oiit to minutes of altitude and seconds of
horizontal parallax.
The first part of the Table is to be entered with the sun's or star's true
altitude (or the altitude next less when there are any odd minutes, as there
generally will be,) in the left-hand column ; abreast of which, in the proper
column, will be found the approximate reduction ; from which let the pro-
duct of the difference to 1 minute by the excess of the odd minutes above
the tabular altitude, be subtracted, and the remainder will be the true
* reduction of altitude for sun or stftr.
Example 1. .
Let the true altitude of the sun's centre be 8? 15 f j required the reduc-
tion to appar^it altitude ?'
Correction corresponding to altitude 8 de^ees . , . . . . 6^5^
Cor. for min. of alt. ; viz. diff. tol min. of alt.=0''. 70 x 15 C =t lO''. 5 =— M)
Required reduction = ••••••«•••••,• 6'5r
Example 2.
Given the true altitude of a star 19?45; ; the reduction to apfmrent
altitude is required ?
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MMCKIPTIOV AKD tTSB OV THE TABLES. 41
CSorrection corresponding to altitude 19 degrees • • ^ / # 2M4f
Cor.formin.of alt;viz. diff«to 1 inin.ofalt«=:0'^. 15x4S'z:6''.75=--7
Required reduction =r
The reduction of the moon's true altitude is to be taken from the second
part of the Table, by entering that part with the true altitude in the proper
column (or the altitude next les9 when there are any odd minutes) and the
horizontal parallax at top or bottom ; in the angle of meeting will be found
a correction ; to which apply the product of the' difference to 1 minute by
the excess of the odd minutes above the tabular altitude by Mubtradian, and
the product of the difference to . 1 second by the odd seconds of parallax by
addition : and the true reduction will be obtained^ as may be seen in the
following
Sstample, - -
Let the true altitude of the moon's centre be 29? 13 ' , and her horizontal
paralUx 58(37? ; required the corresponding reduction to apparent alti-
tude ?
Correc. corres. to alt 29 degs. ^ and horiz. parallax 58;= . • 49 ' 22 ?
Cor. for min.of alt.; viz., diff.to 1 min.of alt.=: 0''.41 x 13^=5*'. 3=^5
Gor.for sees, of par.; viz.^ diff. tolsec. of par.=0''. 90 x 37''=:33'', 3= +33
Required reduction //.'.'.'. . . . 49' 50?
2ZefifarJlr.->-The reduction of the sun's true attitude* is obtained by increas-
ing that altitude by the difference between the refraction and parallax cor-
responding thereto : then, the difference between the refraction and paral-
tax answering to that a;ugmented altftuJe, witf be*the reduction of the true
altitude.
MxampU.
«
Let the true altitude of thd sun's centre be 5 degrees ; required the
reduction to apparent altitude }
Sun's true altitude . . . . 5? OC OT
ftefract.Tab. VlIL=9^54q ..„ . ., .,,
Faral. table VII. 0. 9 J^*- + »-^*-
Augmented altitu<le • • 5? 9' 451, refrac. ans. to which is 9^38?
and parallax • 0. ^
Bequir^ reduction =: •••••*•••••••• tt'291
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Ti|e cprr^ctiQn for reduciing a itar'a tni^ alUMe to iU upparenl^ la
obtained ip the same manner, omitting what relates to p^n^llax. Thv^s if
the true altitude of a star be 8 degrees^ and the corresponding refraction
9"29% (heir $uni/ viz.^ 8^6;29r will, be the augmented altitude; diQ
refraction 'answering to this is 6^24^, which, therefore, is the reduction of
the true to the apparent altitude of the star*
The correction for reducing the true altitude of the moon to the appa-
rent, is found by diminishing the true altitude by the difference between
the parallax and refraction answering thereto ; then the difference between
the parallax and refraction corresponding to the altitude so diminished,
will be the reduction of the true to the apparent altitude. As thus s — •
Let the true altitude of the moon^s centre be 10 degrees, and her hori*
Bontal parallax 57 minutes ) required the reduction to apparent altitude ?
Moon's true altitude • • . • , 10? 0^ 0^ Log. secant 10.0066
Do. horizontal parallax , . . • 57-0^ Propor.log. 0.4994
l^iraHaK in altitude! 56^ 8r Propor.log. 0.506Q
Refrac. to altitude 16?, Table VIIL =: 5. 15
jOiffcrence between parallax and refrac. s 50'. 53^
Dfanhiished altitude 9? 9f 7^ Log. secant 10.0056
Horizontal parallax 57-0 Proper, lo^. 0*4994
Parallax ip altitude *.,..« 56 U6? Propor^log* 0,5050
Refrac, to dimini8he4 alt. Table VIII. 5.42
Difference « • • • • ^ • , ^ 50^34^ j whiph^ therefore^ U
the required reduction.
AuxiKary Angles.
Since the solution of the Problem for finding the longitude i^t sea, hy
celestial observation, is very conaideraMy abridged by the ii^trodu^pn <^
an auxiliary angle into the openition;| the true central dbtance being hence
f ^adily determine to the nearest second of a degree by the simple additi(H;i
of fiv^ natur^ yer9ed ^iiies ; this Table has, therefore, been computed ', and
to render it as convenient as possible, it is extended to every tenth minute
ff ih$ m.oon's apparent altitude| and to each minute of ly^ I^wmqiIM
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p«fall»} ivilh proportion^} parts adapted to the iatenufdif^U minvtei <if
altitude, and to the seconds of horizontal parallax*
This Table was calculated in the following manner :-r^
To the moon's apparent altilude apply the correction from Table XVIIL,
^^d the sum will be hH^r true al|i|«de. j fron\ thf Ipg. tsosiM of whioh (^
indax being augmented by 10) subtract the log. cosine of her apparent
altitude, and Ibe remainder will be a log.^ which, being diminished by the
canstaiH Iqg. .900910/ will give the logarithmip cosine of the auxiliary
Iiet tha mopn.'s apparent altitude be 4 degrees, and her horizontal paral-
lax 55 minutes ; required the corresponding' auxiliary angle ?
Moon's apparent altitude . 4? 0^ Of Log. cosine • 9.998941
Correction fromTable XVIII. + 43.. 2
Moon-s true altitude ; . 4?43' 2r Log. cosine • 9.99atS27
tl W MlllliM
Log. . . , 9.999)586
CotMtantli^, 0.S0091Q
ij I nwinyn
AiixiHar7angk,a8requbed60M:2ir»L<i(.eMiM .'^.Omii
The correction of the auxiliary angle for the sun's or star's appartot
altitude, given at the bottom of each page of tbe Talile, was oompated by
the following rule — ^viz. • .
FVom the log. cosine of the sun's or star's true altitude subtract dia logk
coane. of the apparent altitude^ and And the diffsfeuc^ between the
remainder and the constant log. . OOOlSO.f Now thia ^iftience, heiag
subtracted from the lo^. cosine of 60' degrees, will leave the log. cosine of
an arch]} the diiference between which and 60 degrees Drill be the eovrec-
tion of the auxiliary angle depending on the apparent altiUide of the sun
or^tar.
I^t thf^ sun's or star^ apparent altitude be 3 degrees; required the
^rrection of the auxiliary angle ?
* TUs u the lof. se^uit, 1ms ndiutt of (^ decrees dinuaithed by • 000 120t the differf nee
^etweea tbelfif . oMines of a star's true and apparent altitude betwixt 30 and 90 degrees.
t This is the difference betweeo the lo^. cosines of a star's true and apparent altiluda,
hetvcco 3# end 9(^degrecs«
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44 JhtitAtVnoV ASB tSSJt OP <rHB- TABI JM.
Sun's Apptirent altitude . . . '. 3? 0' Or Log. cosine 9. 999404
Refract. Table VHI. WMI) ,.-, i^'o-r*
Pardlax.TableVH. 9 ( ^ffe^'x^^-l^^/r
< II ■ li
8un'^ true altitude . . • ^ . « 2?45;33f Log. cosine 9. 999497
Remainder 0.0Q0093
Const, log. 0.000120
Difference 0.000027
60? 0'. 01 Log. cosine 9. 69S970
Arch • . . . . \. • . * . 60. 0. 8 Log. cosine 9. 698943
Difference .•••••... 0? 0^ 8T ; which^ thereforCi is the
required correction of the auxiliary angle.
In this Table the auxiliary angle is giveu to every tenth minute of the
moon's apparent altitude (as has been before observed) froift the horizon
to the zenith^ and to each minute of horizontal parallax. The proportional
part for the excess of the given, above the next less tabular altitude is con-
tained in the right-hand column of each page ; and that answering to the
Seconds of )>arallax is given in the intermediate part of the Table. The
correction depending on the. sun's or star's apparent altitude is placed at
the bottom' of the Table in each page.
As the size of the paper would not admit of the complete insertion of
the auxiliary angle, except in the first vertical column of each page under
or over 53 C ; therefore, in the eight following columns, it is only the excess
of the auxiliary angle above 60 degrees that is given : hence, in taking out
the auxiliary angle from those columns^ it is always to be prefixed with 60
degrees.
The auxiliary ang^e is to he taken out of the Table, as thus :—
.. Enter the Table with the moon's apparent altitude in the left-hand
colump of the page, or the altitude next less if there be any odd minutes^
opposite to which and under the minutes of the moon's horizontal parallax
at top, will be found the approximate auxiliary angle.
, Enter the cpmpartment of the ^'Proportional parts to seconds of paral-
lax," abreast of the approximate aiixiltary angle, with the tenths of seconds
of the moon's horizontal parallax in the vertical column, and the units a£
the top ; in the angle of meeting will be found a correction, which place
under the approximate auxiliary angle ; then enter the last or right-hand
column of the page abreast of where the approximate auxiliary angle was
found, or nearly so, and find the proportional part corresponding to th«
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PESCRIPTIOM AND U8B OF THIt TJkBUm^ 4$
odd minutes of altitude, which place under the formen To these three
let the correction, at the bottom of the Tfible, answering to the sun's or
star's apparent altitude, be applied, and the eum will be the correct
auxiliary angle.
JBxilmple.
Let the moon's apparent altitude be 2o?37^9 the sun's apparent altitude
58?20'9 and the moon's horizontal parallax 59U7^; required Uie cor*
responding auxiliary angle ?
Aux. angle aiis. to moon's wp. alt, 25^30% and hor. par. 59' is 60?13'47 v
Proportional parts to 47 seconds of horizontaf parallax is • 12 ^
Proportional part to 7 minutes of altitude is 4
ConrectioQ corresponding to sun's ajpp. alt. (5 8? 20^) is | ^ . 4
Auxiliary angle^ as required • • • • 60fl4' 7^
Table XXI.
^rr^ciioii of the Auxiliary Angle vohen the Maon^s DUtance from a
Planet if observed*
The argi*ments of this Table are, a planet's apparent altitude in the left
or right-hand column^ and its horizontal parallax at top ; in th^ angle of
meeting stands the correction, which is always to be applied by addition to.
the auxiliary angle deduced from the preceding Table : hence, if the appa-<
rent altitude of a planet be 26 degrees, and its horizontal parallax 23
seeoncb^ the correction of the auxiliary angle will be 6 seconds, additive.
This Table was calculated by a modification of the rule (page 43)
for computing the correction of the auxiliary angle, answering to the sun's
or star's apparent altitude ; lis thus :— *-
To the logarithmic secant of the planet's apparent altittde, add the
logarithmic cosine of its true altitude, and the constant logarithm.
9. 698850;* and the sum (abating 20 in the index) will be tlie logarithmic
cosine of On arch ; the difference between which and 60 degrees will be the
required correction.
I ■■ . — * ■ III. 1 - '
* Tbif is tlve \og» cosine of 60 dcffrecs diminished by . 000120, the diflference between the
las* cosines of the true and SpfKUrent altitude of #.fi;^e4 9^ between 30 and M deems^
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A6 ]>ft6clitM6» kM mic 6t tHS tA)StJI^«
ReampU*
Let the apparent altitude of a -planet be 30 degrees^ and Its horisOQtel
parallax* 23 seconds^ required the correction of the auxiliary angle ?
Planet's apparent altitude . . . 30? 0^ Or Log. secant 10. 062469
ParalUx,TabkVL0.20 }
■ Const log. 9.698850
True altitude of the planet . . • 29?58'.42f Log. cosine 9. 937626
ktchtA •••..•.. 60? 0' 7^= W cosine 9.698945
60. 0. 0
Difference i i \ . • • • . 0? 0' 7^ J which 1§ the requited
correction.
Table XXIL
Error ariring from a Deviatimi o/ one Minute in the ParaUelism of
the Surfaces of the Central Mirror qf the Grcular Jnstrument of
B^fiecHofi.
This Tahle contiuns the error of observation arising from a deviation of
ona nutate in the parallelistn of Ihe surfaces of the central mirror of the
reflecting eircley the axi^ of the telescope being supposed to make an ahgle
•f 80 degrees iftdth the hori^eon mirfor } it i^ very useful in finding the
verifieatioti of the parallelism of the surfaces of ihe central tnirror in the
^fleeting circle, or of the index glass in ttie sextant $ as thus f-^
Let the instrument be carefully adjusted, and then take four or live
observations of the angular distance between two well^d^ed objects,
whose distance is not less Ihan 100 degrees } the sum of these, divided by
their number;^ will be the mean observation. Then,
Take out ik central mirror, and turn it b6 that the (fdge which Was
before uppefmoM may ndw be downwards, or next the plane 6f the instru-
ment ; rectify its position, and take an equal ndmber of observation!! of th^
angular distance between the same two objects, and find their mean, as
before : now, half the difference between the mean of these and that of the
former, will be the error of the mirror answering t6 the observed Migles
If the first mean exceeds the second, the error is subtractive | otherwise
additive i the mirror being i& its first or natutal position. Hemse^ if the
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i>Js8CRirridi^ AM t^BB ol^ tHft tabus. 47
mean of the first set of obsetvations be I15?9^40?, atid that of the second
114?59'20r, half their difference, viz., ICSOf 4- 2 =2 40r, will be the
error of the obsenred angle, and is subtractive; because* the first mean
angular distance, or that taken with the mirror in its natural position, \i
greater than the second, ot that taken With the mirror inverted.
Having thus determined the error of the observed angle, that aiisW^ring
to any giveii angle may be readily computed by means of the present TVtble,
as follows :^— -
Enter the left-hand column of the Table with the angular distance, by
which the error of the central mirror was determined; and take out the
corresponding number from the adjoining column, or that marked ^Ob-
servation to the right ;^^ in the same manner take out the number answer-
ing to the given angle ; then,
To the arithmedcal complement of the proportional log. of the Jirsi
number, add the proportional log. of the second, and the proportional log.
of the observed error; the sum of these three logs., rejecting 10 from the
tndex^ urill be the proportional log. <rf the error answeriDg to such given
angle.
jfirample.
Hanng found the error snAng from h Ahhtt of parallelism in the centra!
mirror, at an angle of 115 degrees, to be 40 seconds subtractive; required
the error corresponding to an angle of S5 degrees }
Obs. ang.1 15 deg. opp. to which is 3 '23? Arith. cdMp. prop, log.^8. 2741
Given ang. 85 deg. opp*. to which is 1 H5 ? Propor. log. . • =: 2. 1584
Observed error of central mirror 0.40 Propor.log. . • =: 2.4313
■ • —
Required error = • • . • - O'lSr = Propor. log. • =: 2.8638
Tabui XXIII.
JError qfObiervatim ariringfrom an IneUnaHon o/tke Line qfColUmoh
Hon to the Plane qfthe Sextant, or to tliat qfthe circular Imtrument
of Refleetion.
If the line of sight is not parallel to the plane of the instrument, the
tngle measured' by such . instrument will always be greater than the true
angle. This Table contains the error arising from that cause, adapted to
the most probable limits of. the inclination of the line of coUimation, and
to any angle under 120 degrees : hence the arguments of the Table are^
tbe oibserved angle in the left*hand column^ and the inclination of the liiie
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48 ;dbscription and usb of thk tablbs.
of collimation at top; oppoSte the former, and under the latter, will be
fbund the corresponding correction.
Hius, if the observed angle be 60 degrees, and the inclination of the
)ine of collimation 30 minutes, the corresponding error will be 13 seconds.
The error or correction taken from this Table is always to be applied by
m^btracHon to the observed angle.
Tlie corrections in this Table were computed by the following
nule.
To ' the log. sine- of half the observed angle, add the log. cosine of the
inclination of the line of collimation ; and the sum, rejecting 10 in the
index, will be the log.'sine of an arch. Now, the difference between twice
(his arch and the. observed angle, ivill be the error of the line of collimation.
Example^
Let the observed angle be €0 degrees, and the inclination -ot the line of
collimation 1?30^ ; required the corresponding correction ?
Obs. angle 80 degs. and 80? -^ 2 = 40? Log. sine 9. 808068
Inclinatofline ofcoUim, , . \ . . 1?30' Log. cosine 9.999851
Arch ~ 39?59: 11 - Log. sine 9.807919
Twice the arch = 79?58' 21
bitference 0? 1^58?, which, therefore, is the require4
error.
Table XXIV.
LogdrUhmic Difference.
This Table contains the logarithmic difference, adapted to every tenth
minute of the moon's apparent altitude from the horizon to the zenith, and
to each minute of horizontal parallax. The proportional part for the
excess of the given above the next less tabular altitude, is contained in the
right-hand compartment of each page, and that answering to the second^
of parallax is given in the intermediate part of the Table.
As the size of the paper would not admit of the complete insertion of
the logarithmic difference, except in the first vertical column of each page^
under or over 53^, therefore m the eight following columns it is only the
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BBaCRIFTION AND ttSB OF TRB TA9I3S« 49
foHr last figures of the logarithmic difference that Bie given : hence, in
taking out the numbers from these columns, they are always to be prefixed
by the characteristic, and the two leading figures in the first column. The
logarithmic difference is to 6e taken out in the rollpwing manner.
Bnter the Table with the moon's apparent altitude in the If ft-hand
column of the page, or the. altitude next lets if there be any odd minutes^
opposite to which^ and under the minutes of the moon's horizontal paral-*
lax, at top, will be found a number, which call the appraximate logarUhmic
^Sfference.
Enter the compattment of tha ^ Proportional parts to seconds of paral^
lax," abreast of the approximate logarithmic difference, with the tenths of
seconds of the moon's horizontal parallax in the vertical column, and
the units at the top, and take out the corresponding correction. Enter the
right-hand compartment of the page,* abreast of where the. approximate
logarithmic di&renoe was found, or nearly so, with the odd minutes of
altitude, and take out the corresponding correction, wliich place under the
former. £nter Table XXVV or XXVI., with the sun's, star's, or planet's
apparent altitude, apd take out the corresponding correction, which also
place under the former. Now, the suin of these three corrections -being ,
taken from the. approximate logarithmic difference, will leave.tho correct
I<^;arithmic difference, /
ExampU 1.
Let the moon's apparent altitude be 19?25C^ her horizontal paralW
60' 38^, and the. siin's apparent altitude 33 degrees; required the loga-
rithmic difference 7
Log. difference to app. alt. 19?20', and hor. par. 60' is 9. 997669
Propor. part to 38 seconds of parallax is . 28*1
Propon part to 5 minutes ci akiti|de is, « .« 1 1 >sum = — 49
Cor. from Tab. XXV. ans. to sun's apparent alt. is 1 0 J
Logarithmic differedce, as required .»,;«; 4 9.997620
* lo taking out the correctioa correspondiDg to the odd minatei of altitude in this com-
Tartaient, attention is to be paid to the moon's horizontal parallax : thus, if the paratlax
he between &3' and h^^ tlie correction is to be taken out of the first column, or that adjoin-
ing the minutes of altitude ; if it be between 56' and 59', the correction is to be taken out of
the second, or middle column ; and if it be between 59^ and 62'^ the correction is to be
taken out of the tUfd^ or last' column.
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80 MsomiPTioK Am> utB or ran mbum*
Example 2.
Let the moon's apparent altitude W 63^87', fcerhorirontal parallax
58U3r, the apparent altitude of a planet 85? 10', and its horicontal
parallax 237 } required the logarithmic diflerenee ? '
Log. difference to i^par. alt. 63?30C, and hor. par. 58t, is 9.993682
Propor. part to 4a? of parallax is • • • » 88^
Ph>por. part to 7 • of altitude is • « • • » 7 ?9ttm =s-»l 18
Cor. from Tab. XXVI. ans. to planet's q>par. alt; 28 j
Iiogarithmic difference, as required •««•••• 9.993504
JBemarfc.-*The logarithmic difference waa oonpnled by the f
2bcb.
To the. logarithmic. secant of the moon^s apparent altitude, add the
^ logarithmic cosine of her true altitude, and the constant log. .000120;*
the sum of these three logs., abating 10 in the index, will be the loga-
rithmic difference.
Exafnple.
Let the moon's apparent altitude be 19?20t, and her borisontal parattaa^
60 minutes ; required the logarithmic difference ?
Moon's apparent altitude . . 19?20' Or Log. secant 10.025208
Correction from Table XVIIL 5S« 56 . Conataat log. 0. 000120
Moon's true altitiide • • « • 90.18.5S Lof.eosbe 9.972341
Logarithmic difference, as re<}uired • , • • , • « 9.997669
* The <^lference between the Io|*. coiines of the true and sppsrent tlUtode of s star
betwixt 3Q sad 90 degrees.
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SmCfUFTION AND ITM OF fHS TABUS* 51
Table XXV.
CcrrecAon of th^ Logarithmic Difference.
lliis Table is divided into two parts : the first, or left-hand part, con-
tains the correction of the logarithmic difference when the moon's distance
from the sun is observed j^and the second, or right-hand part, the correc-
tion of that log. when the moon's distance from a star is observed. Thus^
if the son's apparent altitude be 35 degrees, the corresponding correction
will be 11 ; if a star's apparent altitude be 20 degrees, the corresponding
correction will be 1; and so on. These corrections are always to be
applied by mdiiriaction to the logarithmic difference deduced from the
preceding Table.
The corrections contained in this Table were obtained in the following
naiiner, viz.
To the log. secant of- the apparent altitude, add the log. cosine of the
true altitude; aiidthesum, rejecting 10. from the index, will be a log. ;
whieh being mibtracted from the comtaat log. .000120^* will leave the
tabttlar correctioR,
Easamfle 1.
Let the sun's apparent altitude be 35 degrees j required the tabular cor<«
rection?
Given apparent altitude ss 85? 0' Or Log. secant =: 10.086635
Rcfrac. TaWe VIII. V.2V.\..^ \ aa
Parallax Table Vll. 7 /^^^•=-"*-*^ •
Siitiytrae akituda • • .. . 84?58:46? Log. cosine • 9.913474
Sum • . • 6.000109
Constant log. 0.000120
Tabular oonection^ as required .«••••••«• 0.000011
BxampU2.
Let the apparent ^titude of a star be 10 degrees ; required the tabular
correction?
• aBtNi9te»iisce50.
k2
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52 BBSCRIPTION AND USB OF THE TABUS*
Star's apparent altitude 10? 0 ' 0^ Log- secant =10. 006649
Refraction Table VIIL -.5.15
Star's true altitude • • 9.54.45 Log. cosine = 9.093467
Sum= . . 0.000116
Constant log. 0.000120
Tabular oorrcctien, IB required 0.000004
Tablb XXVI.
Correction dfthe Logarithmic Difference iohen the MomttB jlKstance
from a Planet is observed.
The arguments of this Table are^ the apparent altitude of a planet in
the left or right-hand marginal column, and it9 horizontal parallax at top;
in the angle of meeting stands the corresponding correction, which is to be
applied by subtractum to the logarithmic difference deduced from Table
XXIV., when the moon's distance from .a planet is observed. Hence, if
the apparent altitude of a planet be 20 degrees, and its horizontal parallax
21 seconds, the corresponding correction will be 16 subtractive, and so on.
This Table was . computed by the rule in page 51, under which the
correction corresponding to the sun's apparent altitude in Table XXV.
was obtained, as thiis :^-
Let the apparent altitude of a planet be 23 degrees, and its horizontal
parallax 21 seconds; required the correction of the logarkhmie difference ?
Planet's apparent altitude ... 23? 0' 0^ Log. secant 10.035974
Refrac Table VIIL = 2^ 14^ \ ,.^ _ • - .
Parallax Table VL == 20r/ '""^ • "*
Planet's true altitude • . . .. 22?58: 6? Log. cosine 9.964128
Sum . • 0.000102
Constant log. 0. 000120
Correction of the logarithmic difference, as-required . . . 0.000018
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PEfCAIFTION AND VSB OF THB TAfiLBSt
53
Tablb XXVII.
Natural Verged Sines, and Natural Sines^
Since the methods of computing the true altitudes of the. heavenly
bodies^ the apparent time at ship or place^ and the true central distance
between the moon and sun, or a fixed star, are considerably facilitated by
the ^plication-of natural versed sines, or natural sines, this Table is giv^;
which, with the view of rendering it generally useful and convenient, is
extended to every tenth second of the semicircle, with proportional parts
corresponding to the intermediate seconds; so that either the natural
versed sine, natural versed sine supplement, natural co-versed sine, natural
sine or natural cosine of any arch,, may be readily taken out at sight.
The numbers expressed in this Table may.be obtained in the following
manner : —
Let ABC represent a qua*
drant, or the fourth part of a
circle; and let the radius CB
= unity or 1, be divided into
an indefinite number of decimal
parts: as thus, 1.0000000000,
&c. Make B D = the radius
CB; and since the radius of a
circle is equal to the chord of
60 d^rees, the arc BD is equal
to 60 degrfies i draw DM, the
sine of the arc BD, and, at
right-angles thereto, the. cosine
]>B: biaect the arcBD in F,
and draw FN and FG at rights
angles to each other; then will
the former represent th^ sine,
and the latter the cosine pf the
arc BF = 30 degrees: bisect
BF in H; then HOwill express
the sine, and H I the cosine pf
the arc BH ss ,15 dctgrees*
Proceeding in thi^ manner, after 12 bisections, we come to an arc of
0?0' 52^44^3 V45^r, the cosine of which approxin^ates so very clcsely to
the radius C B, that they may be considered as being of equal value. Now,
the absolute measure of this arc may be obtained by numerical calculation,
us foUdws^ vias.
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54 DtSCklPTION AKD tfSB OV THX TABLBS*
Because the chord line B D is the side pf a hei^^n, inscribed In a circle^
it is the subtense of 60 degrees, and, consequently, equal to the radius C B
(corollary to Prop, 15, Book IV., of EucUd) ; wherefore half the radius B S
= BM, will bp the sine of 30 degrees = FN,, which, therefore, is
. 5000000000. Now, having found the sine of 80 degrees, its cosine may
be obtained by Euclid, Book L, Prop. 47 : for in the right-angled triangle
FNC, thehypothenuse.FC is given = the radius, or !• 0000000000,
and the perpendicular F N = half the raditis, or • 6000000000>
to find the base C N = the cosine G F ; therefore
V FC X FC-FN X FN = CN .8660254037, or its equal OF;
hence the sine of 30 degrees is . 5000000000, bnd its cosine . 8660254037-
Again,
In the triangle FNB, the perpendicular FN is given = • 5000000000>
and the base CB-CN=:NB=. 1339745963, to find the hypothenuse
B F : but half the side of a polygon, in9cribed in a circle, is equal to the
sine of half the circumscribing arc ; therefijrc its half, BT = H O, will be
the sine of the arc of 15 degrees : hence '/FN x FN + NB x NB
= BF .51763809025 the half of which, viz., . 2588190451, is therefore
equal to BT, or to its equal HO, the sine of 15 degrees, and from which
its cosine HI maybe easily obtained; for, in the triangle COH| the
hypothenuse C H is given = ' 1^ 0000000000, and the perpendicular H O
= . 2568190451, to find the base C O z= the cosine H L Now^
^/CH X CH- HO x HO = CO .9659258263 = the cMitine HI;
hence the sine of 15 degrees is . 2588190451, and its oosiiie » 965925826S.
Thus proceeding, the sine of the 12th bisection, viz^ 52r44rS^MS'?^9
will be found = . 0002556684; And becaoae smidl Arc^ are very neariy
as their corresponding sines, the measure of 1 minute may be easily deduced
from the sine of the small arc, or 12th bisection determined as above; for^
As die arc df 52f44r3''f45^'^ is to.an are of 1 minute, so is the sine of
the former to the sine of the latter : that is, as 52r44r8V45^^ ; K :t
.0002556634 : i 0002908882; which, therefore, is the sine of 1 minute^
the cosine of whicb is . 9999999577 ; but this approximates so very dosely
to the radius, that it may be esteemed as being aetually equal to H m all
calculations ; and hence, that the cosine of 1 ^ is 1 . 0000000000*
'Now, having thus found the sine and cod&e of one tnifiute, the sihes of
every minute in the guadrant may be obtained by the following rule ; viz.
As radius is to twice the cosine of 1 minute, so is the sine of a mean are
to the sum of the sines of the two equidistant extremes ; from which let
either extreme be subtracted, and the remainder vrill be the sine of the
other extreme : as thus^
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lUMeitapnoH amo va <a nm tabu»« 5S
Tb^Snd' tlu Siae ^the Arc afTMxmiet,
As radius =1:2:! .0002908882 to .0005817764, and .0005817764
- .0000000000 = .0005817764; which, thiurtfore, is the sine of the
arc of 2 nunutes.
lb fi»A th« <9iM ({f 3 Mtmfev. .
Aamdin = 1:2:: .0005817764 to .001 1635528, and .0011635528
* . 0002908882 = . 0008726646 } which, theiefore, is the sine of an are
of S minwUa.
As nuiiuB = 1:3::. 0008726646 to . 0017453292, and . 0017453292
- .0005817764 = ^0011635528; which, therefore, is the sine of the
arc of 4 minutes.
To JmA the Sine ofi Minutes.
As radins =; 1 : 2:: .0011635528 to .0023271056, and .0023271056
- • 008726646 s . 0014544407 i which, therefore, is the sine of the arc
of 5 minutes.
In this manner, the sines may. be found to 60 degrees; from which, to.
the end of the quaidrant, they may he obtained by addition only; for the
dne of an arc greater than 60 degrees; is equal to the sine of an arc as
inuch less than 60, augmented by the sine of the excess of the grven are
above 60 dej^ees : thus.
An the sines being fomid to 60 degrees; required the sine of 61 degrees) '
Aifalioiiw^Sine of 699s . 8571673, and sim of 1? r= . 0174524 ; their
rnns •8746197s«lueh, timsfore, iatbesineof 61 d^ees^ as required*
Agdn,
All the sines being found to 60 degrees ; required the sine of 62
degrees?
iSbfefioii.— Siaeof 58? = . 8480461, and sine 3? sr . 0348995 ; their
= . 8829476 ; wUeb, therefore, is the sine of 62 dqirees, |» required.
Now, liie nstBBal sines beiiig thus found, the natural versed rines^ natu^
nd tangents, and natnral aecants, may be readily deduced therefrom^
agreeahiy to die tviacifiks of aimifair triangles, as demonatrated in Euclid^
Book VI^ Prop. 4. I^
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56 2>B8CAIPTION AND USK OF TU£ TABUS.
To find the NaUtral Versed Sine of 30 Degrees = N B, in tJie Diagram.
Since the versed aine of aif arc is represented by that part of the diame-
ter which is contained between the sine and the arc ; therefore N B is the
versed sine of the arc B F^ which is the arc of 30 degrees ; and since the
versed sines are measiired upon the diameter^ from -the extremity B to C
continued to the other extremity, the natural versed sines under 90 degrees *
are expressed by the difference betweeil the radius and the cosine, and
those above 90 degrees by the sum of the radius and the sine : henc^, the
radius C B 1, 0000000 - the cosiqe FG, or its equal N C . 8660254 =
^B p 13S9746; whicb^ therefore, U the natural versed sine of 30 degrees.
To find the Natural Tangent qfQO degrees zzBQ,in the Diagram.
As the cosine C M is to the sine D M, so is the radius CB to the tangent
BQ: that is,
As CM. 5000000 : DM .8660254 :: CB UOOOOOOO : BQ =
1 • 7320508 ] which is the natural tangent of 60 degrees.
To find the Natural, Secant of 60 Degrees == C Q, in the Diagram.
As the cosine CM is to the radius C D^ so is the radius C B to the secant
C Q : that is.
As CM .6000000 : CD KOOOOOOO :: CB 1.0000000 : CQ =:
2. 0000000 ; which is the natural secant of 60 degrees; Hence, the man-
ner of computing the natural co^^tAngent A P, the natural co-secant C P,
and the natural co-versed sine E A, will be obvious.* The versed sine sup-
plement of an arc is represented by the difference between the versed sine
of that arc and the diameter or twice the r^ius: thus, the versed. nne
supplement of the arc B F is expressed by the difference between twice the
radius C B, and the versed sine N B ; viz., twice C B = 2. 0000000 - N B
• 1339746 = 1. 8660254 ; which, therefore, is the natural versed sine sup-
plement of the arc B F or the arc of 30 degrees, and so of any other.
Now, the natural sines, versed sines, tangents, and secants, found as
above, being principally decimal numbers, on account of the radius being
assumed at unity or 1 3 therefore, in order to render these numbers all
affirmative, they are to be multiplied by ten thousand millions respectively;
and then the conmnon logs, corresponding thereto will be the logarithmic
sines, versed sines, tangents, and secants^ which are generally given in tho
different mathematical Tables under these denQQuiliations,
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DBSCJIIPnON AND 0SS OF THJt .TABLES* 57
Of m Table.
In this Tablei the natural verted sines are given to every tenth second of
the semicircle ; the corresponding arcs being arranged at the top, in nume-
rical order, from 0 to 1 80 degrees. The natural versed sines supplement
are given to the same extent; but their corresponding arcs are placed at
the bottom of the Table^ and numbered, from the right hand towards the
left, or contrary to the order of the versed sines. Hie natural co-versed
sines begin at the end of the first quadrant, or of the 90tb degree of the
verted sines ; the arcs, corresponding to which are given at the bottom of
the page and numbered, like the versed sines supplement, towards the left
hand from 0 to 90 degrees, and then continued at top of the page from 90
to 180 degrees, towards the right hand, until they terminate at the 90th
degree of the versed sines, where they first began. The natural sines begin
where the co*versed sines end ;- viz., at the end of the first quadrant^ or 90th
degree of the versed sines, with which they increase by equal increments ;
the arcs corresponding to those are placed at the top of the page to every
tenth second o( the quadrant, the 90th degree of which terminates with
the 180th of the versed sines. The natural cosines b^gin with the versed
sines supplement 5 the ares corresponding to which are given at the bottom
of the page, being numbered, like the latter, contrary to the order of the
versed sines and natural sineSf to every tenth second from 0 to 90 degrees,
or to the end of the first quadrant of the versed shies, thus ending whera
the co-versed sines begin.
Note. — ^In the general use of this Table, it is to be remarked, that the
naturaVversed sine supplement, natural conversed sine Under 90 degrees, or
natural cosine, of a given degree, is found in the same page witK the next
less degree in the column marked 0? at top, it being the first number in
that column ; that answering to a given degree and minute is found on the
same line with the next less minute in the column marked 60^^ at the bottom
of the page ; and that corresponding to an arch expressed in degrees,
minutes, and seconds, is obtained by deducting the proportional part, at
bottom of the page, from the natural versed sine supplement, natural co-
versed sine under 90 degrees, or natural cosine of the given 4icgree, minute,
and less tenth second*
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58 PBSCRIFTIOM AND 08B OW THB TABLM*
* ♦ :
PROBLEMS- TO ILLUSTRATE THE USE OP THE TABLE.
Probibic.
To find the Natural Vened Sine^ Naiwral Versed Sine Si^^pkmeni,
Naiumd CSo'Vened Sme^ Naharal Sine^ and Noiwrul Coeine, e/ an^
gwen Jrch, evpreeeed in D^eeSy MmUes, and Seeondim
AuLB.
Enter the Table^ and ftnd the natural Yetaed aine, versed sine aopplement^
co-^rersed aine, naitaral ame, or natural cosine, anawermg to the ghren
d^^ee^ minute, and next leaa tenth aecond ; to which add the proportional
part anawering to the odd aeeonds, found at the bottom of the page, if a
natural versed sine, eo-versed sine ^bove 90?^ or nolural sine be wanted;
but subtract the proportional parl^ if a versed sine supplement, co-versed
dne under 90?, or nolifrol come, be required t and the sum, or remainder^
will be the natural versed sine, natural sine, natural versed sine supplenieat,
oo^versed sine, or naiwral come, of the given arch.
MxampU 1«
Required the. natural versed sine, versed sine supplement, co-versed
sine, natural me, and natural cosuie, answering to 42? 12'36T?
To find the Natural Versed Sine :-p
Natural versed sine to 42?12^90r ^ ..... . S5929S
Proportional part to • 6? » • • • • Add 20
/KfcnArch 42?12:s<{r Natural versed sine B.85931S
To find the Versed Sine Supplement :— -
Versed sine supplement to 42?12'30r « 1.740707
Proportional part to • « 6r ... Subtract 20
Given arch . • 42? 12^36? Versed sine sup. =; 1.740687
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DBsoBirriow and uib of thb tablbs« 59
To find the Co-veraed Sine t«—
Convened sine to . » 42?12'30r ...... S28173
Proportional part to^ ^ 6T « • Subtract 21
Giveq.arch 42?12:36r Co^vers. sine = 328151
To find the Natural Sine :—f .
Natural sine to » , 42n2:30r • • • . . . 671828
pR>porUonal part to ^ 6T • • • Add 21
Given arch. 42?12:3Qr Nat. sine ::? 671849
1V> find the Natural GMine i'^
Natural cosine to • 42?12:80r .\ 740707
Proportional part to 6^ • . Subtract 20
Given arch 42? 12:36r Nat. cosine s 74068/
Example 2.
Required the natural versed sine^ versed sine supplement, co-versed
one, Mliiral iiniy and natural came, answering to 109?53!45??
To find the.NaUual Vetted Sine :—
Natural versed sine to 109?53:40r = / . ... 1.340288
nopoiiioDal pan to &<r is # • • Add 23
■ ■ ' '■ ^
Given arch. . 109?53'45? Nat. versed sine = 1.340311 '■
To find the Versed Bine Supplement :-«
Ycisdlsfaiesup.to 109t53!40? 659712
Phiportiond pan to 5' i . Subtract 23
GIvenarch 109^53:45? Vers, sine sup, s 659689
To find Ae Co-versed Sine :•«
Co-vened sine to lOQfSSUO? ; 059679
Fn^rtionalpartto - S? • « . .Add 8
Oina nvh .MOtSSta? Gs-verftd idne ss 059687
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,00 DfiSCMVTlOS AND VSB OF THB TA«UI9»
To find the Natural Sine :—
Natural 8i«c to • 70?6nOr Sup. to l09?63:50-r 940305
Proportional part to 5^ , • • • Add 8
SuppHement 70^6 1 15r to given arch^ nat. sine =: 940313
To find the Natural Coeine :—
Natural cosine to 70?6ClOr Sup. to 109?53:50r . 340334
Proportional part to ' Sr ; • . Subtract 23
Supplement 70?6 '. 1$ ^to given arcb^ nat. co8ine=34031 1
'JSfmariir.— -Since the natural rines and natural comes are hot extended
beyond 90 degrees^ therefore^ when the given arch exceeds that quantity,
its supplement, or what it wants of 180 degrees, is to be taken, as in the
above examyte. And when the given arch is expressed in degrees and
minutes, the corresponding versed sine supplement, co-versed sine under
90 degrees, and natural cosine, are to be taken out ogreeably to the not«
in page 57> which see.
Probuoc IL
To find the Arch corresponding to a gwen Natural Verged &nef Versed
Sine Supplemeiay Conversed Sine, Natural Sine, and Natural Cosine* .
R0LB.
Enter the Table, and find the arch answering to the next less natural
versed sine, or natural sine, hnt to the next greater versed sine supfrie-
ment, co-versed sine, or natural cosine; the difference between which and
that given, being found in the bottom of the page, will give a number of
seconds, which, being added to the arch found as above, will give the
required arch.
Example 1.
Re<)uired the arch answering to the patUral versed 9me 363985 ?
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DSSC&IPTION AKD tSB OF THB TABLES. 6l'
&6*/ion.— The next less natural versed sine is 363959, corresponding to
which is 50?30' lO^j the difference between 363959 and the given natural
versed sine, is 26; corresponding to which, at the bottom of the Table, is
7^ which, being added to the above-found arch, gives 50?30' 17% the
required arch.
Note. — ^The arch corresponding to a given natural sine is obtained
preciiely in the same manner.
Example 2.
Required the arch corresponding to the natural versed sine supplement
1,464138?
SbZttJion^— The next greater natural versed sine supplement Is 1 . 464 1 55 ;
corresponding to which is 62?20'40r; the difference between 1.464155
and the given natural versed sine supplement, is 17 ; answering to which,
at the bottom of the Table, is 4f , which, being added to- the above-found
arch, gives 62t20U4^, the required arch.
No^e.— The arch corresponding to a given cci-versed sine, t>r natural
cosine, is obtained in a similar manner.
Bemarkh
The logarithmic versed sine of an arch may be found by taking out th^
eommon logarithm of the product of the natural versed sine of such arch
by 10000000000; as thus:
Required the logarithmic versed sine of 78?30U5f ?
The natural versed sine of 78?30U5r is . 800846, which, being multir
plied by 10000000000, gives 8008460000 ; the common log. of this is
9. 903549 ; which, therefore, is the logarithmic versed sine of the given
arch, as required.
Remark 2.
The Table of Logarithmic Rising may be readily deduced from the
natural versed sines ; as thus :
Reduce the meridian distance to degrees, by Table I., and find the
natural versed sine corresponding thereto ; now, let this be esteemed tfi
an integral number, and its corresponding common log. will be the loga«
rithmic rising.
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62 mtscRiPTioN Aim VOL OF m TABUBS.
Required the logarithmic rising answering to 4^50*45 ? ?
4!50r45! s 72?4r.l5r, the natural versed sine of which is 702417 ;
the common log. of this is 5. 846595, whicb^ therefore, is the logarithmie
rising required.
Tabu XXVIII.
LogarUhnu of Iiumber$k
Logarithms are a series of numbers invented, and 'first published in
1614, by Lord Napier, Baron of Merchiston in Scotland, tot (he purpose
of facilitating troublesome calculations in plane and spherical trigonometry.
These numbers are so contrived, and adapted to other numbers, that the
sums and di£ferences of the former shall correspond to, and show, the pro*
ducts and quotients of the. latter.
Logarithms may be defined to be the numerical exponents of ration, or
a series of numbers in arithmetical progression, answering to another
series of numbers in geometrical progmston } as^
Thusi
0, 1.
2.
9.
4.
5.
6.
7.
8. tud-ortog.
1.2.
.4.
8.
16.
32.
64.
Or,
128.
S56.ge<Kprag.
0. 1.
2.
3.
4.
5.
6.
7.
8. ind.orlog.
1. s.
9.
27.
81.
243.
729.
Or,
2187.
6561. geo.pro.
0. 1.
2.
3.
4.
5.
6.
7.
8. ind.orl<^.
1. 10. 100. 1000. 10000. 100000. 1000000. 10000000. 100000000 ge. pro.
Whence it is evident, that the same indices serve equally for any
geometrical series ; and, consequently, there may be an endless variety of
systems of logarithms to the same common number, by only changing the
second term 2. 3, or 10. &c. of the geometrical series of whole mmi«
hers*
In these series it is obvious, that if any two indices be added together.
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]>BtGBIPTIOM Aim VtM OV TU TABLW. M
thdr turn iriU be the index of that number whidi !■ eqilal to the product
of the two terms^ in the geometrical progression to whieh those indicee
belong : thus^ the indices 2. and $• being added together^ make 8; and
the corresponding tenns 4. and 64. to those indices <in the first series),
being multiplied together^ produce 256, which is the number correspohding
totheindexS. *
It is also obvious^ that if anyone index be subtracted from another, the
differenee will be the index of that number which is equal to the quotient
of the two corresponding terms : thus, the index 8. mimto the index 3 s=
5 ; and the terms corresponding to these indices are 256 and 8^ the quo-
tient of which, viz., 32, is the number corresponding to the index 5, in the
first series.
And, if the logarithm of any number be multiplied by the index of its
power, the product will be equal to the logarithm of that power ; thus, the
mdex, or l<«arithm of 16, in the first series, is 4 ; now, if this be multiplied
by 2, the product will be 8, which is the logarithm of 256, or the square
of 16.
Again,— if the logarithm of any number be divided by the Index of its
foot, the quotient will be equal to the logarithm of that foott thus, the
index or logarithm of 256 is 8; now, 8 divided by 2 gives 4 ; which is
the logarithm of 16^ or the square root, of 256, according to the first
series.
The logarithms most convenient for practice are such as are adapted to a
geometrical series increasing in a tenfold ratio, as in. the last of the fore-
going series ; being those which are generally found in most mathematical
woika, and which are usually tailed oommon kgarUhn^ in Older to distin^
goish them from ether species of logarithms.
In this system of log^thms, the index or logarithm of 1, is 0 ; that of
10, is 1 ; that of 100, is 2; that of 1000, is 3} that of 10000, is 4, &c.
&e.; whence it is manifest, that the logarithms of the intermediate num-
bers between 1 and 10, must be 0, and some fractional parts ; that of a
number between 10 and 100, must be 1, and some fractional parts ; and so
on for any other number : those fractional parts may be computed by the
followii^
Attle.— To the geometrical series h 10. 100. 1000. 10000. &c., apply the
arithmetical series 0. 1. 2. 3. 4. &c., as logarithms. Find a geometrical
mean between 1 and 10, or between 10 and 100, or any other two adja-
cent terms of the series between which the proposed numbet lies. Between
the mean thus found and the nearest extreme, find another geometrical mean
in the same manner, and so on till you arrive at the number whose loga-
rithm is sought, fmd as many aritiimetieal means, according to the order
in which the geomelrical ones, were founds and they will be the logarithms
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64 DBSCHTPnON AK]> t78« OF TUB TABLEd.
of the said geometrical means ; the last pf which will be the logarithm of
the proposed number.
Esample.
To compute the Log« of 2 to eight Places of Dechnals :-—
Here the proposed numbl|r lies between 1 and 10.
Firsts The log. of 1 is 0, and the log. of 10 is 1 ;
therefore 0+l-ft-2=:.5isthe arithmetical mean,
and V 1 X 10 = 3. 1622777 is the geometrical mean :
hetice the log. of 3. 1622777 is . 5.
Second, The log. of 1 is 0, and the log. of 3. 1 622777 is . 5 ;
therefore 0 + 5 -^ 2 s . 25 is the arithmetical m^an,
and V 1 X 3. 1622777 = 1. 7782794 the geometrical mean :
hence the log. of 1. 7782794 is . 25.
Third, The log. of 1 . 7782794 is . 25, and the log. of 3. 1622777 is . 5 ;
therefore .25 + .5^2 = . 375 is the arithmetical mean^
and n/ 1.7782794 x 3.1622777= 2.3713741 the geo. mean:
hence the log. of 2. 37 13741 is . 375.
Fourth, The log. of 1 . 7782794 is . 25, and the log. of 2. 371374 1 is . 375 ;
therefore . 25 + . 375 -f- 2 = . 3125 is the arithmetical mean,
and V 1.7782794 x 2.3713741 = 2.0535252 the geo. mean :
hence the log. of 2. 0535252^ is . 3 125.
Fifth, Thelog. of 1. 7782794 is . 25, and the log. of 2. 053925218 .3125;
: therefore . 25 + . 3125 -f- 2 = . 28125 is the arith. mean,
and V 1.7782794 x 2.0535252 = 1 . 9109530 the geo. mean :
hence the log. of 1 . 9 109530 is . 281 25.
Sixth, Thelog.ofL9109530is.28125,&thelog.of2. 053525218. 3125|
therefore . 28125 + . 3125 -ft- 2 = . 296875 is the arith. mfean^
and V 1.9109530 x 2.0535252 = 1.0809568 the geo. mean:
hence the log. of 1. 9809568 is . 296875.
Seventh, Thelog.ofl. 9809568 is. 296875, & the log; of 2. 0535 252 is .3125;
therefore . 296875 + . 3125 -f- 2 t=. 3046875 is the arith. mean^
and V 1.9809568 x 2.0535252 = 2. 0169146 the geo. mean:
hence the log. of 2. 0169146 is . 3046875.
Eighth, Thelog.of2. 0169146 is. 3046875, &log.of 1 . 98095 68 is 296875 ;
therefore . 3046875 +'. 296875-4- 2=. 30078125 is the ar. mean,
and V 2.0169146 x 1.9809568 = 1. 9988548 the geo. mean s
hence the log. of 1. 9988548 is . 30078125/ .
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DESCRIPTION AND USB OP THB TAfiLBS.
65
Proceeding in this manner, it will be found, after 25 eKtractions, that
the log. of L 9999999 is . 30103000; and since 1. 9999999 may be con-
sidered as being essentialjy equal to 2 in all the practical purposes to
^ich it can be applied, therefore the log. of 2 is . 30103000.
If the log. of 3 be determined, in the same manner, it will be found that
the twenty-fifth arithmetical mean will be . 47712125, and the geometrical
mean 2. 9999999 ; and since this may be considered as being in every re*
spect equal to 3, therefore the log. of 3 is . 47712125.
Now, from the logs, of 2 and 3, thus found, and the log. of 10, which
18 giTen^l, a great many oth^r logarithms may be readily raised; because
the sum of the logs, of any two numbers gives the log. of their product ; and
the difference of their logs, the log. of the quotient ; the lo^. of any num-
ber, being multiplied by 2, will give the log. of the square of that number;
or, multiplied by 3, will give the log, of its cube; as in the following
examples :—
Example 1.
To find the Log. of 4 :—
To the log. of 2 = . 30103000
Add the log. of 2 =x . 30103000
Sum is the log. of 4 ». 60206000
Example 2.
To find the Log. of 5 :—
From the log. of 10=1. 00000000
Take the log. of 2 = . 30103000
Rem. is the log. of 5 s . 69897000
Example 3.
To find the Log. of 6 :—
Totbelog. of 3= .47712125
Add the log. of 2 = . 30103000
Sum is the log. of 6 = . 77815 125
Example 4.
To find the Log. of 8 :—
To the log. of 4 =3 . 60206000
Add the log. of 2 = . 30103000
Sum is the log. of 8 = . 90309000
Example 5.
To find the Log. of 9 :—
To the log. of 3 = . 47712125
Add the log. of 3 =s . 477 1 21 25
Sum is the log. of 9 = • 95424250
Example 6.
To find the Log. of 1 5 : —
To the log. of 5 = . 69897000
Add the log. of 3 a . 47712125
Sumisthelog.of 15r=L 17609125
Example 7.
To find the Log. of 81 s th^
square of 9 :—
Log.of9= • . . .95424250
Multiply by • « • 2
Pro. is the log. of 81 = 1 . 90848500
Example 8.
To find the Log. of 729 :;=: the
cube of 9:
Log. of 9 « , . . .95424250
Multiply by . . . 3
Pro.is thelog.of 729*52. 86272750
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66 BBSCRIPTtOK AND USB OF THB TABLBS.
Siuce the oddnumbere 7. 11. 13. 17. 19. 23. 29. &c. cannot be exactly
deduced from the multiplication or division of any two numbers, the logs,
of those must be computed agreeably to the rule by which the logs, of 2
and 3 were obtained ; after which, the labour attending the construction of
a table of logarithms will be greatly diminished, because the principal part
of the numbers may then be very readily found by addition, subtraction,
-and composition*
Of the Table.
This Table, which is particularly adapted to the reduction of the appa-
rent to the troe central distance,. by certain concise methods of computa-
tion, to be treated of in the Lunar Observations, is divided into two parts :
ibBjirst of which contains the decimal parts of the logs., to six places of
figures, of all the natural numbers from unity, or 1^ to 999999 ; and the
iecondj the logs, to the same extent, of all the natural numbers from
1000000 to 1839999 ; — and although the logs, apparently commence at
the natural number 100, yet the logs, of all the natural numbers under that
are also given : thus, the log. of 1, or 10, is the same as that of 100 5 the
k^, of 2, or 20, is the same as that of 200; this log. of 3, or 30, is equal
to that of 300; that of 11, to 110; that of 17> to 170; that of 99, to 990;
and BO on : using, however, a different index. And as the indices are not
affixed to the logs., they must therefore be supplied by the computer : these
indices are always to be considered as being one less than the number of
integer figures in the corresponding natural number. Hence the index to
the log. of any natural number, from 1 to 9 inclusive, is 0 ; the index to
the log. of any number firom 10 to 99 inclusive, is 1 ; that to the log. of any
number from 100 to 999, is 2 ; that to the log. of any number from 1000
to 9999, is 3; &c. &c. &c. The second part of the Table will be found
very useful in computing the lunar observations, by certain methods to be
given hereafter, when the apparetit distance exceeds 90 degrees, or when
it becomes necessary to take out the log. of tt natural number connsting
of seven placer of •figures, and conversely.
In the left-hand column of the Table, and in the uppw or lower hori*
zontal row, are given the natural numbers, proceeding in regular succes-
sion ; and, in the ten adjacent vertical columns, their corresponding loga-
rithms.
As the size of the paper would not admit of the ample insertion of the
logs., except in the first column, therefore only the four last figures of each
log. are given in the nine following columns ; the two preceding %ures
belonging to which will be found in the first column under 0 at top, or
over 0 at bottom ; and where these two preceding figures change, in the
body of the Table, large dots are introduced instead of O's, to catch the
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eye and to incUcate that from thence, through the rest of the line, the
said two preceding figures are to be taken from the next lower line in the
column under or over 0 : those dots are to be accounted as ciphers in tak^
ing out the logarithms.
The log. of any natural number consisting of four figures, or under, and
conversely, is found directly by the Table ; but because the log. of a natural
number consisting of five or more places of figures, and the converse, is
frequently required in th^ reduction of the apparent to the true central
distance, and also in many other astrc^nomical calculations j proportional
parts are, therefore, adapted to the Table, and arranged in the nine small
columns on the right-hand side of each page; by means of which the loga-
rithms of all the natural liumbers, not consisting of more than seven places
of figures, and vice versGy may be found to a sufficient degree of accuracy
for all nautical purposes, as may be seen in the following problems.
Probum L
Gioen a Natural Number consisimg^ of Jive^ siXf or seven Placei qf
Figures, to find the corresponding Logarithm,
RULB.
Look for the three first figures of the giren natural number in the left-hand
column 3 opposite to which, and under the fourth figure, in the horizontal
column at top, will be found the log. to the four first figures of the given
natural number : on the same line witfi this, and under the fifth figure of
the natural number at top, in the proportional parts, will be found a numberi
which, being added to the above, yn\l give the log. to five places of figures
of the given natural number; on the same line of proportional parts, and
under the sixth figure of the natural number at top, will be found a number,
which, being divided 1^ 10, aqd the quotient added to the last found log.,
will pye the log. to six places of figures of the given natural number. In
the same manner, the log, may be taken out to seven places of figures ;
observing, that the number in the proportional parts, corresponding to the
seventh figure of the natural number, is to be divided by 100.
JVb^e.— In dividing by 10 or 100, we have only to strike off the right-
hand, or two right-hand figures.
Example.
Rw|aii«d ihdlog. oonesponding to the gifen natural number 1378078?
y2
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68 DESCRIPTION AND ITSB OF THB TABLBS.
Log. corresponding to 1378 (four first figures) is ... . 139249
5th fig. of the nat. num. .9 ans. to which in the pro. parts is 284
6th fig. of . do. . . 7 ans. to which in the pro. parts is
221, which, divided by 10,
gives 22.1 22
7th fig, of do, ... 8 ans. to which in the pro. parts is
252, which, divided by 100,
gives 2. 52 . • • • • 2
Given liatural number 1378978 Corresponding log. =s . • 6. 139557*
' Problem II.
To find the Natural Number to five, six, or ieoen Places of Figures,
corresponding to a given Logarithm.
Rule.
Find the next less log. answering to the given one in the column under
0 ; continue the sight along the horizontal line, and a log,, either the same
as that given, or somewhat near it, wiQ be found ; then, the three first
figures of the corresponding natural number will be found in the left-^hand
column, and the fourth figure, above the log., at the top of the Table. Should
the given log. be found exactly, let one, two, or three ciphers be annexed
to the natural number found as above^ according to the number of figures
wanted, and it will be the natural number required. But, if the log. cannot
be exactly found (which in general will be the case), find the difference
between the given log. and the next less log. in the Table : with this
difference, enter the proportional pai'ts, on the same horizontal line in
which the next less log. was found, and find the next less proportional
part; answering to. which, at the top or bottom, will be found the fifth
figure of the required natural number : find the difference between the
above-found difference and the aforesaid next less proportional part;
which being multiplied by 10, and the product foimd in the same line of
proportional parts, the number corresponding thereto, at top or bottom,
will be the sixth figure of the required natural number, Now, the differ-
ence between the above product and its next less proportional part, being
multiplied by 10, also, and the product found in the same line of propor-
tional parts, the number answering thereto at top or bottom will be the
seventh figure of the required natural number.
* The iadtx 6 is prefixed, because the given natural number consists of seven places of
fi^ires.
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DBSCRIPITON AND USE OP THB TABLBS. 69
Example.
Required the natural number correspondingto the given log, 6, 1 19558?
Given log 6,119558
1316 s= fourfirst figs, ofthe required nat. num.
answering to next less log. • . .1 19256
Difference ; 302
. 9 = fifth fig. ofthe required nat. num. ans.
to the pro. part next less ... 297
Difference . 5 x lQ=50product,
. . l=sixth fig. of the required' nat. num.
ans. to pro. part next less •••••• 33
Difference 17x10=170
• . . 5=:seventh fig. of the required nat. num. ans. to the
nearest pro. part ••#••165
1316915 which is the natural number corresponding to the given log.
6. 119558^ as required.
Note. — ^From the above Problems^ the manner of using the second part
of the Table will appear obvious.
Remarks.
1. The whole of the operation is inserted at length, for the purpose of
illustrating, more clearly, the use ofthe Table; but in practice, the logs,
may, in most cases, be taken out at sight, and cbnversely ; particularly
from the second part, where the natural numbers are given to five places of
figures, froni lOQOOOO to 1839999.
2. In taking out the log. of a decimal, fraction, or any number less than
unity, if the first decimal place be a significant figure, the index of its log.
is to be accounted as 9 ; but if the first significant figure of the decimal
stands in the second, third, or fourth place, &c«, the index of the corre-
sponding log. is to be taken as 8, 7, or 6, &c. The converse of this, — that
is, finding the significant decimals corresponding to a given log., will
appear obvious.
3. Hie arithmetical complement of a log, is what that log. wants of the
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radius of the Table ; viz.^ of 10. 000000 : this is most easily found, by begin-
ning at the left hand, and subtracting each figure from 9, except the last
significant one, which is to be taken from '10; as thus:
The arithoiedeal complement of tiie log. 4.97^8Sd is 5.627147 j and
80 on.
Problbm III.
To petform MuUvficaii(m ijf Lagariihm*
tlCLB.
To the log. of the multiplicand, add the log. of the multiplier, or add
the logs, of the foctors together, and the sum will be the log. of the' pro-
duct ; the naturaT number corresponding to which will be the product
required.
Example h
Multiply 436 by 19.7.
436 Log. = . . 2.639486
19.7 Log. =: . . 1.294466
Prod.=:8589. 18 Log.=:3. 933952
Example 2.
Multiply 437. 8 by 14.07, and
ako by 0. 239.
437.8 Log.r:
14.07 Log.=:
0.239 Log. =
• 2.641276
. 1.148294
• 9.378398
Example 3.
What is the product of 0.049^
9. 875, and 0.753?
0.049 Log.= . . 8.690196
9.875 Log.= . . 0.994537
0.753 Log.= . . 9.876795
Prod.=0.3642 Log. = 9. 561528
Example 4.
What is the product of 0. 0567
and 0.00339?
Pro.=1472.204Log. = 3. 167968
0. 0S67 Logf. >6
0. 00BS9 Logi a
« 6.768588
. 7.580200
Pro.«0. 0001628Log.«6. 883783
Abte.-«R6speeting the index of a decimal fraction^ akid eonmtiely^ see
Remaik 2,page69,
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PBSCftlPnON AND USB OV TBI TABUUU
71
Paoblbm IV.
To perfima I)ms%on by Logarithms.
RULB*
From the log. of the diyidend^ subtract the log. of the divisor^ and the
remainder will be the. log. of the quotient; the natund 'number corre-'
sponding to which will be the quotient required.
JEsQxnplti 1*.
Divide 1497 by 98. . , „
1497 Log. = . . 3.175222
93 Log. = . •' 1:988483'
Quo.= 16.0968 Log.= 1.2067S9
JEjr(impfe2.
Divide 469. 76 by 0.937.
469.76 Log. s , 2.671876
0.937 Log. =. . 9.971740
Qao.sS01.84S Log.s2. 700136
EsampteS.
, Divide49.73by0.06S2.
49.73 Log. = 1.696618
0.0632 Log. = 8.800717
Qtto.=:786.869=Log.=2. 895901
Example 4.
Divide 0. 00815 by 0. 000275.
0.00815 Log. =1 • 7.911158
0.000275 Log. =: • 6.439333
Quo.=29. 6368 Log. = 1. 471825
Problbm V.
To petform Proportkm^or ike RdeofThree^ or Golden Hub^ by
. ' LogftnihiM.*
RULB.
To the aridlmetical coippleDient of the log. of th^. first term, add the
logs, of the gjscond and third terms { and the sum will be the log, of the,
fourth term, or answer. *
Example U
If a ship sails 19$ miles in 2i hours^ how many miles will she run, at
the same rate, in 24 hours ?
As 2.25 hours, arith. comp. log. =: . . • 9.647817
Is to 19. 5 miles, log. . 1.290035
So is 24 hours, bg. « ; 1.380211
To 208 mfles, log. = ....... 2.318063
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72 DBSCRIPTION AMD USB OF TH« TABLBS.
Example 2,
If the interest of 1001. for 365 days be 41 lOs., what will be the interest
ofl78I.15«. for213daya?
. 5 100 Arith. conp. of log.
■^1365 Do. dp.
T ♦ J 178.
I» »°l213.
8.000000
. . 7.437707
• •
178.75 Log 2.252246
Log 2.328380
So 18 4.5 Log.' ...... 0.653213
To 4.69403 Log 0.671546
Example 3.
A man of war^ sailing at the rate of 9 knots an hour, descried a ship,
distant 26 miles, sailing at the rate of 6^ knots, to which she gave chase :
after two hours' chase, the breeze freshened, and increased the man of war's
rate of sailing to 1 1^ knots, and that of the chase to 8^. In what time did
the man of war come up with the chase ?
Sb&iHon.— Since the man of war gained, atthe commencement, 2^ miles
an hour on the chase, therefore, at the end of the first two hours, the dis-
tance between them was reduced to 21 miles ; during the rest of the chase,
the hourly gain of the man of war was 2} miles.
Hence, As the hourly gun 2. 75 Ar. comp. log. 9. 560667
Is to .... 1 hour,. Log. • • 0. 000000
So is distance . 21 miles, Log. . . 1.322219
To .... 7.6363 Log. = 0.882886,
or 7 hours and 38 minutes from the time the breeze freshened.
Problem VI.
To perform InvobMon by Logarithms.
Rule.
Multiply the log. of the given number by the index of the power to
which it is to b^ raised, and the procjuct will be the log. of the required
power. • • . - ,
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DBSCRIPTION AND USB OF THB TABLB8*
73
Mxample I.
Required the square of 346 ?
346 Log. . . . 2.539076
Ind. of the power= 2
Answer 119716 Log.=5. 078152
Example 2.
Required the cube of 754 ?
754 JLog 2.877371
Ind. of the .power= 3
Ans. 428661064 Log.=8.632113
Pboblbm VIL
To perform Evolution by Lc>garUhms,
Rulb.
Divide the log. of the given number by the index of the power^ and the
quotient will be the log. of the root.
Example 1.
Required the square root of
76176?
76176 Log. . . y4. 881818
Ans. 276 Log. = 2.440909
Example 2.
Required the cube root of
21952000 ?
21952000 Log. . f 7. 341475
Ans. 280 Log. =: 2.447158^
PaoBLBM Vin.
To find ike Tonnage of a Ship by Logarithms.
RuiB.
To the log. of the length of the keel^ reduced to tonnagCy add the log. of
the breadth of the beam, the log. of half the breadth of the beam, and the
constant log. 8.026872*; the sum, rejecting 10 from the index, will be
the log. of the required tonnage. .
Example.
Let the length of a ship's keel, reduced to tonnage^ be 120. 5 feet, and
the breadth of the beam 35. 75 feet; required the ship's tonnage ?
* This is the uithmetical complemeat of the logt of 94 ; the common divisor for fiiiding
the fopno^e of sl)ipt»
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74 J>K8CRIPTtON AVD USB OF TBB TABUS.
Length of the ked for tonnage . 1 20. 5 feet Log. 2. 080987
Breadth of the beam .... 35.75 feet Log. 1.553276
Half ditto 17.875 feet Log. 1.252246
Constantlog. • .« 1 . . 8.026872
Required tonnage 819.18 . . . Log. 2.913381
Problem IX.
Given the Measured Length of a Knot, the Number of Seconds run by
the Glass, and ihe Distance sailed per Log, to find the true Distance
by Logarithms. *
RULB.
Tq the* arithmetical complement of the log. of the oomber of aeoonda
run by the glass, add the log. of the measured length of a knot, the log. of
the distance sailed, and the constant log. 9. 795880* ; the sum of these
four logs., rejecting 20 from the index, will be the log. of the true distance.
Example 1.
The distaoce sailed by the log is ISO mil&ij the measured length of a knot
is 43 feet, and the time by the glass 32 seconds ; required the true distance ?
32 seconds, arith. comp. log. . . • 8. 494850
43 feet, log. s 1.633469
180 mUes, log. = 2.255273
Constant log. as 9.795880
True distance = 151. 2 miles. Log. s 2. 179472
Example 2.
The distance sailed by the log is 210 miles ; the measured length of a knot
19 5 1 feety and the time by the glass 27 seconds ; required the true distance ?
27 seconds, arith. comp. log. th . . 6. 568636
51 feet, log 1.707570
210 miles, log .... 2.322219
Constant log. 9. 795880
True distance = 247. 9 miles. Log. = 2.394305
• This It the ram of Om arlUimtClcsl compkoMnt of Hie hog. ol 46 (the gtnenl teagtii
of a knot) and the log.of 30 secondi, the true measure of the half-miDttls flaas;
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DUCaiPTIoy AND USB OF THB TABLBO. 79
Tabu XXIX.
Prapcrtional Logarithmic
This Table cpntuns the proportional log. conesponding to all portions
of time under three hours^ and to every second under three degrees. It
^vas originally eomputed by Or. Maskelyne^ and particularly adapted to the
operation for finding the apparent time at Greenwich answering to a given
distance between the ftioon and sutr^ or a fix^ star; but it is now applied
to many other important purposes^ as will be seen hereafter.
Proportional Logarithms may be computed by the following
RULB.'
From the common log. of 3 hours, reduced to seconds^ subtract the
common log. of the given time in seconds ; and the remainder will be the
proportional log. corresponding thereto.
Example.
R^red the proportional log. corresponding to 0^40^26! ?
3 home reduced to seconds s lOSOOr Log. :£: <. 033424
40r26! given tilne^ in sees, s 2426? Log. a 3.384891
Proportional log. corresponding to the given time s 0* 6485* 33
As hours and degrees are similarly divided, therefore, in the general use
ot this Table, the hours and parts of an hour, may be considered as
degrees and parts of a degree, and conversely. And to render the use of
it more extensive, one minute ihay be Esteemed as being either one degree,
•r one second^ and vice versa.
Since proportion is performed by adding together the arithmetical com-
plement of the proportional logarithm of the first term, and the pr<^or-
tional logarithms of the second and third terms, rejecting 10 from the
index, the present Table is of great use in reducing the altitudes of
the moon and sun, or a fixed star, to the mean time apd distance, when
fhe obeerradons are made by one person, as will appear evident by the
following
• Example.
Let the lirsl aWlude of the moon*s lower limb be 27?25t20'', and the
corresponding tim6 per watch 21M2r8!^ and the last altitude 25?24;20ry
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76 DBSCRIPTION AND USE OF TAB TABLBS.
and its corresponding time 21^55T57- ; }t is required to reduce the first
altitude to what it should be at 21M9T33!, the time at which the mean
lunar distance was taken ?
l8ttime2lM2r 8! Isttime21?42r 8! l8tait,27?25C20r 27?25:20r
Last do. 21. 55*57 Mean do. 21. 49. 33 Last do. 25. 24. 20
Ditt • 0.18.49 Diff. , 0. 7.25 Diff . 2. 1. 0
As 13?49!, arithmetical comp. frop. log. = 8. 8851
Is to 7*^25 ! proportional log. • • • • =1.3851
So is 2? 1' proportional log =0.1725
To prop. log. of reduction of Moon's alt. . :± 0. 4427 = — 1? 4 '57*
Moon's alt. reduced to mean time of observation • • = 26?20^23T
And in the same manner' may the altitude of the sun^ or a fixed star^ be
reduced to the time of taking the mean lunar distance.
Remark. — ^Although this Table is only extended to 3 hours or 3 degrees^
yet by taking such terms as exceed those quantities one grade lower^ that
is, the hours^ or degrees, to be esteemed as mintites, and the minutes as
seconds, the proportion nlay be worked as above : hence it is evident that
the Table may be very conveniently applied, to the reduction of the sun's^
moon's, or a planet's right ascension and declination to any given time afiter
noon or midnight ; and, also, to the equation of time 3-— for the illustration
of which the following Problems are given.
Problbm L
To reduce the Sun's Longitude^ Itight Ascension and Decimation; and,
also, the Equation of Tlrne, as given in the Nautical Ahnanac, to anjf
. given Meridian, o^td to any given time under that Meridian.
RULB.
To the apparent time at ship, or place, (to be always reckoned from the
preceding noon *,) add the longitude, in time, if it be west, but subtract it
if east ; and the sum, or difference, will be the Greenwich time.
From page II. of the month in the Nautical Almanac, take out the sun's
• See precepte to Tftblc XV.— page 25,
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longitude^ right ascension, declination, or equation of time for the noons
immediately preceding and following the Greenwich time, and find their
difference 5 Uien,
To the proportional log. of this difference, add the proportional log. of
the Greenwich time (reckoning the hours as minutes, and the minutes as
seconds,) and the constant log. 9. 1249* ; the sum of these three logs., re-
jecting 10 from the index, will be the proportional log. of a correction
which is always to be added .to the sun's longitude and right ascension at
the noon precedkig the Greenwich time ; but to be applied by addition or
subtraction to the sun's declination and the equation of time^ at that noon,
according as they may be increasing or decreasing,
Esample 1.
Required the sun's longitude, right ascension and declination, and also
the equation of time. May 6th, 1824, at 5 MO?, in longitude 64?4S^ west
qf the 4aiieridian of Greenwich ?
Apparent time at ship or place^ = 5 MO?
Longitude 64?45C west, in time, = . . * • . .+4. 19
Greenwich time, = 9*29?
To find the Sun's Longitude.
Diff. in 24 hours = 57 '59? prop, log = .4920
Greenwich time = 9*29? prop. log. ...;... =1.2783
Constant log. . =9. 1249
Correction of sun's long = . + 22:55? p. log. = 0.8952
Sun's long, at noon. May 6, 1824 . = 1 ! 15?51 ^ 13?
Sun's long, as required . . . . = 1M6?14'. 8?
To find the Sun's Right Ascension.
Diff. in 24 hours = 3^52? prop. log. ; =1.6679
Greenwich time = 9*29? prop, log * =1.2783
Constant log = 9. 1249
Correction of sun's right asc. . . . = + 1 ^ 32? p. log. = 2. 07 1 1
Sun's right asc. at noon, May 6, 1824, = 2*53?31 ' . 7
Sun's right asc. as reqtured • . v • =^ 2*55? 3*. 7
t The aridunetical €oniplemeiit of the proporUonsl lo^* of 24 hours esteemed as minutes.
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To find the' Sun'^ Declination*
Diff. in 24 hours = 16^38r prop, log , . = 1,0343
Greenwich time = 9*29T prop. log. *.• =1.2783
Constant log. • , . . . . . , == 9. 1249
Correetion of sim'a dee ss+ 6^34? p. log. -a 1.4375
Sun's dec. at noon^ May 6^ 1824, . . sl6?S6f 51 north.
rfi ■ ■■■■III I 111^
Sun's dec. as required • , . • • bb16?41I^39? north.
To find the Equation of Time.
Diff. in 24 hours ss 4^.5 prop, log, • , • • » « 7=3.3829
Greenwich time cs 9i297prop.log. »,..•« ;= 1,2783
Constant log. »«,•«»»;= 9. 1249
Correction of. the equation of time « :s'+ l'« 8 p, logf s: 3* 7861
Equation of time^ May 6,1824 . . s STSd'.l
Equation of time as required . . • - z 8?37 ' • 9
JRanorfc.— Since the daily difference of the equation of time is ex-
pressed, in the Nautical Almaiiao, in secoAds and tenths of a second ; i^
therefore, these tenths be multiplied by 6, the daily differenoe will be re-
duced to seconds and thirds :-^NoW) if those seconds and thirds be esteem-
ed as minutes and seconds, the operation of reducing the equation of time
will become infinitely more simple ; because the necessity of making pro-
portion for the tenths, as above, will then be done away with 2— remember-
ing, however, that the minutes and seconds corresponding to the sum of the
three logs, are to be considered as seconds and thirds.
Example 2.
Required the sun's longitude, right ascension, and declination, and also
the equation of time, August 2d, 1824, at 19^22?, in longitude 98 ?45^ east
of the meridian of Greenwich ?
Apparent time at ship, or place, 19^22'
Longitude 98 ?45'. east, in time, 3= • • « » — 6.35
'm
Greenwich time^ • « • • » • • • • « 12t47?
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To' find the Sun's Longitude.
Dlff. in24hours=: 57'28rprop.log = .4959
Greenwich time =: 12M7* prop, log. ....... =1.1486
Constant log . = 9. 1249
Correction of sun's longitude, . • = + 30^37^ p. log. =5 0. 7694
Sh&'s long, at noon, Aug. 2, 1824, =4! 10? Si 8r
Son's long, as required . • • «. ^4;iO?3SM5;
To find the Sun's Right Ascension.
Diff. in 24 hours = 3!52r prop. log ; s U6679
Graenwich time s= 12^7? prop. log. • « s= 1.1486
Constant log =9. 1249
CorreGtionof sun's right asc. • • =:+ 2C 4? p. log. • =: 1.9414
Sim's right ase. at noon, Aug. 2, 1 824,:^ 8 1 507 0. 8
Son's right ase* as required « • • =:8i52? 4'.8
To find the Sun*s Declination.
Diff. in 24 hours = 15'36r prop. log. ....... =1.0621
Greentdch time = 12^47" prop. log. . . . . • • . =1. 1486
Constant log. ........ =9.1249
Correction of sun's dec. . . . = - 8^9^ p. log. = 1. 8356
Su'sdec. at noon, Aug. «, 1824^; . = 17? 44Mlf north.
Sun's dec. as required, .... = 17? 36'22r north.
To find the Equation of Time.
Diff. in 24 hours = 4r30r prop, log =1.6021
Greenwich time = 12^47* prop. log. =1.1486
Constant log • . = 9. 1249
>j ■ ■ ■ . ■ —
Correction of the equation of time . = — 2^24^ p. log. = 1. 8756
Bqn.of time, at noon, Aug. 2, 1824, = 5? 54!24f
Eqiitof tiflM as required^ • • • s 57 52! 0!
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Problem II.
To reduce the Moon's Longitude, Latitude, Right Ascension, Declination,
• Semi-diameter and Horizontal Parallax, as given in the Nautical Alma-
nac, to any given Meridian, and to. any given time under that
Meridian,
Rule.
To the apparent time at ship, or place, (reckoned from the preceding
noon or midnight,) add the longitude, in time, if it be west, but subtract it
if east, i^d the sum, or difference, will be the Greenwich time past that
noon or midnight, according as it may be.
Take from page V. VI. ox VII. of the month in the Nautical Alma-
nac, the moon's longitude, latitude, right ascension, declination, semidi-
ameter, or horizontal parallax for the noon and midnight immediately
preceding Bud following the Greenwich time, and find their difference;
then.
To the proportional log. of this difference, add the proportional log. of
the Greenwich time past the preceding noon or midnight, (reckoning the
hours as minutes, and the minutes as seconds,) and the constant logarithm
8. 823.9 * ; the sum of these three logsl, abating 10 in the index, will be
the proportional log. of a correction which is always to be added to the
moon's longitude or right ascension at the noon or midnight preceding the
Greenwich time ; but to be applied by addition or subtraction to her lati-
tude, declination, semidiameter or horizontal parallax, at that noon or. mid-,
night, according as it may be increasing or decreasing.
Nof«.— Since the difference of the moon's longitude and right ascension,
in 12 hours, will always exceed the limits of the Table, and also the differ-
ence of her declination in that interval, at times ; if, therefore, the one half
or one third of , such difference be taken, and the correction, resulting there-
from, multiplied by 2 or 3, the required correction will be obtained.
Example.
Required the moon's longitude, latitude, right ascension, declination^
semidiameter and horizontal parallax, Aug. 2d, 1824, at 3! 10? past ttoon>
in longitude 60?30' west of the meridian of Greenwich ?
Apparent time at ship, or place, • . . , r: 3M0?-
Longitude 60?30^ west, in time,. • . , =: + 4. 2.
Greenwich time, past noon, Aug. 2, 1824, = 7*12T
f The arithmeticitl complement of the proportional log. of 12 hou^s esteemed as mintU^,
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To find the Moon's Longitude.
Diff. in 12 hours = 6?31 ^59? -*-3=29l0^39|-r, prop, log, = , 1391
Greenwich time = 7M2? = prop. log. =1.3979
Constant log. =8. 8239
^ '"
One third the corr. of the moon's long. =3 1 ? 1 8 ( 25 f p, tog, as 0. 3609
Multiply by , , , , . 3.
Corr. of moon's long. . , . . . + 3?55'15r
Moon's long, at noon, Aug. 2, 1824 . =7: 17?16^ 27r
Moon's lon^, as required • • . , '7!21?ir.42r
To find the Moon's Latitude.
DiflF.inl2l]iOurs=:23'.35rprop. log. .....,: a .8827
Greenwich time = 7H2+ prop. log. ......... =; 1.3979
tlonnant tog^ .....«•• ss 8. 8239
Correction of moon's lat - 14^ 9^ p. log. a 1. 104S
Moon's lat. at noon, Aiig. 2, 1824, . = 4? 6/.59r south.
Moon'3 lat. as required . ... • • • 8?52'50? south,
To find the Moon'i Kigfat Ascension :-^
Dirf. in 12 hours =s ^?5r.48r ^3«2? 17^ 16ir, prop. log. ss . 1 177
Greenwich time, :s . . . i . .7*12^ prop. log. =1.3979
Constant log. . .=8.8239
One third the corr. of the moon'« rtasc. cs 1?22'22? p. tog, =0.3395
Multiply by ..... 3.
Corr. of moon's right asc -f-'4? 7- 6^
NIoon's rt. asc. at noon, Aug. 2, 1824, s= 223. 33. 36.
Moon's right asci as wquirccl . . . 227?40:421'
o
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To find' the Moon's Declination :—
biff, in 12hoifr8=:l?23M3r, prop. log. ...... = -3325
Greenwich time = 7*12? prop, log, =1.8979
• • • Constant log. \ . • =8.82^
(Donrectlbn of moon's dec, • . . . • + 50M4r p. Jog. = 0,6J43
Moon's dec. Aug. 2, 1824, at noon, . := 20967^ 7"- south.
Moon's declination as re<pired • . . 2t?47'%l' south.
^ofe.— The cpnectiofti or proportional part of the moon's jnptioq, found
as above, must be corrected by the e*quatioh of second difference contained
in Table XVII., as expluned in pages 33 and 34.
To find the Moon's Semidiameter :—
0iff, in* 12 hours r: ' 6'/ prop. log. • • • , •
Greenwich time =:7* 12? prop. log. • » • • •
G>n8taiit log. • . • . • • •
Corf, of the moon's semidiametec • • • • — 4f
Moon's semidiameter at noon, Aug. 2, 1824,2 13-^38
• • !!!? o« 2553
. . = 1.3979
. .. = 8.8239
p. log, 3 3.4771
Moon's ^mkBameter as required • • • • • 15C 29^
T« fiBd tlie Moon's Horisontal ?wMmX :—
Diff. in 14 houra r= 28f prop, log/ • .• . •. • • •5:2.6717
Greenwich time zsl 7Marprop.log • « •^1«3979
Constantlog. .«««'..» . = ^8.8239
Conr. of moon's horiz. para} « ; -^ 14? p. k>g« s 2. 8935
Moon's horiz. paraL at noon, Aug. 2^ 1824, s57t 6T
Moon's horiz, paraL as required • • • • 56'52^ /•
iVbfo.— ^The moon's semidiameter, thus found, must be augmented by
the correction contained in Table IV.,. as jexflained ia pagc» 10«
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PROBLBIf III.
To reduce the Eight Ascension and Declination qf a Planet, as given in
the Nautical Jlmanacj to any ^ven time under a knoum Meridian^
RULB,
Turn the longitude iiitd tlmey^ and add it lo the apparent time at ship e?
place if it be west^'but subtract it if east ; and the sum^ or difference^ will be
the corresponding time at Greenwich.
From page IV, of the month in the Nautical Almanac^ Cake oot the pla»
net's right ascension and declination for the nearest days preceding and
following the Chreenwich time^ and find the difference ; find, also, the
difference between the Greenwich time and die nearest preceding day ;
then, —
To the proportional log. of this difference, add the proportional log. of
the difference ef right ascension, or declination, and the constant leg.
9.d031 *; the sum^of these three legs., rejecting 10 from the index, will be
the proportional log. of a correction, which being applied by addition, oi
subtraction, to the right ascension, or declination, (on the nearest day pre-
ceding' the Greenwich time,) according as it* may be increasing or decreas-
ing, the sum, or difference, will be the correct right ascension or declin-*
ation at the given time and place. -
* Example.
Required the right ascension of tiie planet Venus, July 3, 1S24, at
10^20? apparent time^ at a place 75?SO^ west of the meridian of Green-
wich ? ^ .
Apparent time at given place, -=:•••«• 10^20?
Longitude 75 ?80C west, in time = • • • • + 5. 2.
Oieenwich tfane •••. = •• 8 days, 15! 22?
To find tiie Right Ascension :—
R. A. ofVenus, July 1 = 6* 8? . . . '. . 1*. 0* 0?
Ditto • • • • 7 = 6. 40. Gr. time.= . 3-. 15. 22
I
Diflference. = 0*a2r Diff. , =.2-15*22? s aS?22?j
which are to be Esteemed as nunutes and seconds t — Whence, *
* The arithmetical complement of the proportional log. of 144 hour} (6 day^} citeemed
ummates; and, hence taken as 2 houfs and 24 miautes.
02
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64 DESCRIPTION AND USB OF THB TABLES.
Diff. bet G. time and nearest preceding day 63*22? prop, log, = . 4534
DiflF. of right ascension in 6 days . • . •0*32* prop. log. = . 7501
Constant log. • •- • 9»903l
Correction of right ascension * . . . + 14C 5^ p. log,:i:l. 1066
Planet's R. A. on July 1,1824 = .. • . .6* 8? 0!
R. A. ofVenus, as required • • #- • . 6i22T 5!
To find the Declinatioiv:—
Dec.ofVenus,July 1 =:23?36C N. . . ... H 0* Or .
Ditto. . . . . 7 =23.32. N.Gr. time = 3.15.22
Difference . . 0? 4^ Diff. =2^115*22.?= 63*22?}
which are to be esteemed as minutes and seconds ; hence, .
Diff; bet. G. time and nearest' day preceding 63*22? prop. log. = .4534
Diff. of declination in 6 days = . . . . 0? 4^ prop. log. 1. 6532
Constant log 9.9031
Correction of declination = • .^ — n46^p. log, =2.0097
Ranet'sdec.on July 1, 1824, . . = 23.36. 0. north. •
Dec. of Venus, as required . . • • 23?34i: 14? north.
Tablb XXX.
LogQinihmic Half -elapsed Time. •
This Table is useful in finding the latitude by two altitudes of the sun ;
aiid also in otlier astronomical calculations, as will be shown hereafter.
The Table is extended to every fifth second of time under 6 hours, with
proportional parts, adapted to the intermediate seconds, in the right hand
margin of each page ; by means of which, the logarithmic half-elapsed time
answering to any given period, and converisely, m.ay be readily obtained at
Bight. .
As the size of the page would not admit of the indices being prefixed to
the logs, except in the first column, under Of, therefore where the indices
change in the other columns, a bar is placed over the 9, or left hand figure
of the log., as th\is, 9, to catch the ey^, and to indicate that from thence,
through the rest of the line, the index is tp be taken from the next lower
line in the first column,, or that marked 0! at' top and bottom. It is to
be observed, however, that the indices are only susceptible of change when
the half-elapsed time is under 23 minutes.
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■
The logarithmic half-elapsed time corresponding to any given period, is
to be taken out by entering the Table with the hours and fifths of seconds
at the top, or next less fifth if there be. any odd seconds, and the minutes
in the left-hand column ; in the angle of meeting will be found a number,
which being diminished by the proportional part answering to the odd se*-
conds^ in the right hand margin, will give the required logarithm,
Exatnple*
Required tlie logarithmic half-elapsed time answering to 2t47*28! ? '
2*47?25! answering to which is . . . , , 0. 17572
Odd seconds « • . 3. pro. part answering to which is • • -* II
Given time = 2M7^28? corresponding log. hf.-elapsed time . 0. I756I
In the converse of this, that is, in finding the time corresponding to a
given fog. j-^if the given log. can be exactly found, the corresponding
hours, minutes, and seconds, will be the time required :— but if it cannot
be exactly found (which in general will be the case), take out the hours,
minutes, and seconds answering to the next greater log.; 'the difference
between which, and .that given, being found in the column of 'proportional
parts, abreast of where the next greater log. was found, or nearly so, will
give a certain number of seconds, which being added to the hours, minutes
and seconds, found as above, will give the required time.
Example.
Required th^ time corresponding to the logarithmic half-elapsed time
0.14964?
Sohaion. — ^Th^ next greater log. is 0. 14973, corresponding to which is
3^0*25! ; the difference between this log. and that given ifl 9 ; ans\vering
to which in the column of proportional parts is. 3 seconds, which being
added to the above found time gives 3*0T28t for that required.
Amarfc.— The numbers in this. Table are expressed by the logarithmic
^o-secants* adapted to given intervals-of time, the index being diminished
by radius, as thus :
Let^the hal^elapsed time be 3t20T45'. ; to compute tlje corresponding
logarithm.
Given time =:.3^20T45! in degrees = 50?11U51' ; log. co-secant less
radius = 0. 1 14557 ; which, therefoie, is the required log. ; and since it is
not necessary that this number should be extended beyond five decimal
places, the sixth, or right hand figure, may be struck off; observing,
however, io increase the fifth figure by unity or 1, when the right hand
figure, so struck off, amounts to 5 or upwards : — hence, the tabular number
corresponding to 3!20T45 ^ is 0. 1 1456 } mid so of others.
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JLojrari^Amic Middle Time.
Thid Table is^ also^ useful in finding the latitude by two 'altitudes of the
sun ; for which purpose It is extended to every fifth second under 6 hours^
with proportional parts for the intermediate seconds^ in the right-hand
margin of each page ; by means erf which the logarithmic middle time an-
swering to a|iy given period^ and conversely^ may 'be readily taken out at
sight. ' .
As the indices are' only prefixed to the logs, in the first column^ there-
fore where those change in the other columns a bar is placed over the cy«
pher^ as thus, 0, to catch the eye, and to indicate that from thence through
the rest of the line, the index is to be taken from the next lower line, in
the first column.
The Idgarithmie middle time answering to any ghreii period is to be
taiken out by entering the Table with the hours and fifths bf seeoiids at die
top, or the nextUssJifih second (when there are atiy odd seconds, as there
generally will be), ahd the mihutes in the left-hand eolutnn yvx the angle tff
meeting will be found a. number, which being aikgmMbd by the propor-
tional part answering to the odd ^econds^ in the compartment alireast of the
angle ef meeting, will give the log. required. '
Example.
Required the logarithmic middle time answering to • 3 M 7*23? ?
3M7':^20! ahswering to which is .. . ^ • . 6. 18099
Odd seconds . • 3. pro. part answering to which is • • + 8
Given tune s 3M7"23! corresponding log. middle time • .. ^. 18107
The time corresponding to a given logarithmical number, is founcl by
taking out the hours, minutes, and seconds, dtiswering to the iiext Use ta-
bular number ; the differenee between which and that given, being fostld in
the. compartment of proportional paits, abreast of the saidnejrt 1^ tabular
number, will give a certain number of seconds, which being added to the
hours, minutes^ and seeonds found as aboire, will be the time required*
. Example.
Required the time eorrespondltig to the bg« middle time^ 6» 01787 }
Solution.— The next Ic^« tabular log. is 6; 01757, answering to which is
2?5?30? J the difference between this log. and that given is 10, answering
to which in the colunm of propQrtiond parts is 2 seconds, which being
added to the time founds aa above, gives 2!5?32!, for that required.
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•
Hemarlc.^^The logarithmic middle time may be readily eomputed by the
following rule ; via :— ^
To the logarithmic sine of the given time expressed in degrees^ add the
constant log. Q. 30lQ30| and the sum, abating 10 in the index, will be the
required logarithm. ' . ••
Let the middle time be. 4^ 10T25 ' , Tequircd the corresponding log* }
.Given tim^ = 4^ 10^25'., in degs. = 62?36^ l^r log. sine » 0. 948339
Constatttlqg« » • » 6.301030
Logarithmic middle lime, as required ..;••••= 6. 249369 ;
and since it is not necessary that this log. should be extended t^yond five
plaoei; of decimals^ the sixths or rjght-hand figure may, therefore, be struck
off; observing, however, to increase the fifth figure by unity or I, when the
right-hand figure, so struck off, amounts io 5 or upwards ; hence the tabu-
lar number corresponding to 4* 10T25 !,, is 6. 24937^ and so on.
Table XXKIU
Logarithmic Rising.
This Table, vrith the two preceding, is particularly useful in finding the
latitude by two altitades of tfie sUn ) it is also of oon^derable ute in mtfny
other asti'onomical calculations, such as 'h computiBg the apparent time
from the altitude of a celestial object; determining the altitude of a celes-
tial obyeel from the i^parent time, &c. &ew— -The arrangement of the pre-*
sent Table is so yery uniform with the. preceding, that it is not deemed
necessary to enter into* its description any farther than by observing that
the indices are only prefixed to the logs, in the first column :— thfit where
those change in the other columns, large dots are introduced instead of O's
to catch the eye, for the purpose of indicating that '^rom thence through
the rest of the line, the index is to be taken from the next lower line in the
first column ; and that, in the general use of. the Table^ these dots are to
be accounted as cyphers. • •
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Required the logarithmic rising pmsweringto 1M3?27' ?
1*43725!, answering to which is • . . • • 5.00040
Odd seconds • • 2, pro. paiit answering to which \s • * • 28
Given time = 1M3T27 ', conesponding logarithmic rising 5. 00068 .
The converse of this, that is, finding the time corresponding to a given
log. will appear obvious ; thus, .
Let the given logarithmic rising be 5.^9088, to find the corresponding
thne. *. •
The next less tabular log. is 5. 66071, answering to which i»3*57"30' ;
the difference b^tV^een this log. and that given is^ 17> answering to which
in tKe column of proportional parts,abreastof the tabular log., is 3 se-
conds ; now, this being added to the tifnefouiid, as above, gives 3^57*33' ;
which, therefore, is the time corresponding, to the given logarithmic
rising. #
Nbfe.— The numbers in this Table were computed, by the following
rule, vt« : — * * ^
To twice the logarithmic sine of half the meridian distance, in degrees,
add the constant log. 6. 301080, and the sum, rejecting 20 from the index^
will be the logarithmic riaiifig. *
Example.
Required the logarithmic-rising answering to 4^ 10^45'. ?
(Kven meridian distance =: 4^ 10T45 ?, in degrees = 62?41 i 15f
Halfthe meridian distance, in degrees • • • • =: 3 1 . 20. 37^9 twice the
logarithmic sine ••••••*•••••.• • 19.432293
Constant log, . ••...*... 6.301030
Logarithmic rising answering to the given meridian distance r: 5, 733323
Hie numbers in the present Table may be also computed by means of tlie
natural versed sines contained in Table XXVIL, as thus ;
Reduce the meridian distance to degrees, and find the natural versed
sine corresponding thereto; the common log, of Which will be the loga-
rithmic rising.
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BifiCRIPTION AND TTSB OF THB TABUTS. 89
Example.
Required the logarithmic rising answering to 4^22r30^ or 65?37^30^ ?
Meridian distance in degrees = eS^S/'SO"^, natural-versed sine = 587293^
log. :s= 5. 768855-5 which^ therefore, is the logarithmic rising answering
to the given meridian distance*
In this method of computing the logarithmic rising, the natural versed.
9ine is to be conceived as being multiplied by 1000000, the radius of the
Table^ and thus reduced to a whole number.
Table XXXIIL
To reduce Pointf qfthe Compass to Degrees, and conversely.
This Table is divided into six columns ; the two first and two last of
which contain the names of* the several points and quarter points of the
compass ; the third column contains the corresponding number of points
and quarter points' reckoned from the meridian; and the fourth column the
degrees and parts of a degree answering thereto.-7-The manner of using
this Table is obvious ) and so is the method by which it was computed :— -
for since the whole compass card is divided into 32 points, and the whole
circle into 360 degrees ; it is evident. that any given number of points will
be to their corresponding degrees in the ratio of 32 to- 360 ; and vice versa,
that, any given nXimber of degrees will be to their corresponding points as
360 is to 32 : — H^nce, to find the degrees corresponding to one point.—
As82f : 360? :: If: 11?15:; so that one point contains 11 degrees and
15 minutes ) — two points, 22 degrees 30 minutes, &c. &c.
Tabus XXXIV.
Ijogarithmic Sines, Tangents, and Secants, to every Point and Quarter
Point of the Compass*
In this Table the points and quarter points ate contuned in the left and
right hand maq;inal columns, apd the log. sines, tangents, and secants,
correisponding thereto^ in the intermediate columns.
If the course be given in points, it will be found more convenient to take
the log. sine, tangent, or secant of it from this Table, than to reduce those
points to degrees, and then find the corresponding log. sine, &c. &c. in
either of the following Tables.— The mariner of using this Table must
appear obvious at first sight.
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TAButJOCXV.
Logarithmic Secants.
In the first 10 degrees of this Table, the logarithmic secants are giren to
arery tet^h second, with proportional parts, answering to the intermediate
seconds, iti the right hand marginal colniiin. — ^Thence to 88 degrees, the
log. secants are given to every ^ft second, with proportional parts, adapted
to the intermediate seconds, in the right hand column of each page :— ^tmd
because the numbers in<;rease rapidly between 60 and 88 degrees, produc-
ing very considerable diflferences between any t^o adjacent logs. ; there-
fore betwixt those' limits, there are two pages. allotted to a degree ; every
page being divided into two parts of 15 minutes eacb^ so that ub portion
whatever of the proportional parts might be lost, and that the whole might
have room to be fully inserted. — In the twq last degrees, viz. from 88 to
90, the log. secants are gTven to every second.— The Table is so arranged
as to be extended to every second in the -semicircle, or from 0 to' 180 de-
grees ; as thus t the arcs corresponding to the log.' secants are giv^n in
regular succession at top from 0 t& 90 degrees, and then continued at bot-
tom, reckoning towards the left hand, from 90 to 180 degrees 2-*the arcs
correspdnding to the eo-secants are placed at t^e bottom of the Tabled in
numerical order, from the right hand- towards the left (li^e the secants in
the second quadrant), from 0 to 90 degrees, and then continued at top^
agreeably to the order of the secants in the first quadrant, from 90 degrees
to the end of the semicircle.— Tins mode of arrangement, besides doing
away with the necessity of finding the supplement of an arch when it*
exceeds 90 degrees, possesses the peculiar advantage of ^a^ling the
reader to take out the log. secant, or co- secant of any arch whatever, and
conversely, at sight, as will appear evident by the followhig problems.
No/e.— The log. co-secant of a giyen degree, or secant of a degree above
90, will be found in the same page with the next less degree in the first
column under 0^ at top, it being the first number in that column i and the
log. co-secant of a given degree and minute, or secant of a degree and
minute above 90, will be found on the aame line vrith the neati less minnto
in the eohunn macked 60jf at bottom of th^ Tia)le.
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Problbm i. •
To find ihe LogarUhndc Secant^ and Ca-secatU of any given Jrchy
expresied in IhgreUy ffSnuleSf and Seconds*
•
If the gir^n ftrch bfe cotxiprieed iVlthiti the limits of the two last degrees
qf the first quadrtirit^ that is, between 88 and 90 degrees, the Table will
direotljr exhibit hs corresponding log. sceant or co*secant ; — but when it
falk without those, limits, then find the log. secant, or co-secant, in the
^iigfe of meethig nrade bjr the giren degree and il^t less fifth or tenth
second at top, and iht iiiinutef in the lefthand coliitnn ; to which, Aid
the propoftiotial ptirt eorrespcaiding to th^ odd seconds from the right hand
column abreast of the angle of meeting, if a secant be wanted, ot«a co-ie^
eoKt above 90 iegtet9 ; but subthict that part when a co-secant i^ required,
or a secant fibove 90 degrees ; and the sum, or difiierence, will be the \og.
secant or co-secant answering to the given arch.
Example U
Reqmred the logarithmic secant^ and co-secant» correspoEding to
23?14:23r?
To.find the Log« Secant :—
S3?14;3(K^iin8«lo.whiehis « i . 10.086747
Odd sccoiMb 3 pnqpor. part to which is + 8
Gitenarch =r 88?I4!2af Corres.log. secant =: 10.036750
To iind the Log. Co-secant :-^
23?l4^20r,ans.towhichis . . 10.403881
Odd seconds 3 propor. part to which is — 15
lii n.
Girenarcbs 23?14C23r Corres. log: co-secant = 10.403866
Be^dred th* leg. smmt, and co^eaot, eoirespoiKliog to Ud?83M7? }
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To find the Log. Secant :—
113?23M5r,ans. towhichis . . 10.401121
Odd seconds 2 propor. part to which is -r ' 10. .
Given arch =; 1 13?23U7? Corres. log. secant r: 10. 401 11 1
To find the Log. Co-secant :— *
113?23M5r,au8.*to which IS • , 10.037260
Odd seconds 2 propor. part to which is + 2
% •
Given arch = 113?23<477 Corres.log. cb-8ccant=:l 0.087262
Note. — In that part of the Table which lies between 10 and 80 degrees,
the size of the page would not admit of the indices - being prefixed to ahy
other logs, than those contained in the first jrolumn of each page ; nor, in-
deed, is it necessary that they should be, tince they are uniformly the same
as those contained in the said first coliimn; viz., 10 for each log. secant or
co-secant*
PbpBLEM n.
To find the Arch corresponding to a gwen Logarithmic Secant or
Co-secant:
If ihe given log. secant, or co-sectuit, exceeds the secant of 88 degrees,
viz., 1 1. 457181, its corresponding arch will be found at 8rst sight in the
Table ; but if it be under tliat number, find the arch answering to the next
less secant, or next greater co-secant; tfie difierence b^twe'en which and
that given, being found in the column of proportional parts, abreast of the
tabular log., will give a certain number of seconds, which, being added to
the above-found arch, will give that required.
Example I. ^ .
Required the arch corresponding to the given log. secant 10. 235421 ?
Sblufion.— The next less ^cmt, in the Table, is 10.235412, com-
sponding to which is 54? 26^25 ^ ; the difierence between this log. secan^
and that given, is 9 ; answering to which, in the column of proportional
jparts abreast of the tabular log., is 3? ; which, beii\g added to the al^ve-
found arch, gives 54?26U8r for that required.
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Example 2.
Required the arch corresponding to the given I<^. co-secant 10« 5621 14 ?
So&iiion.— The next greater co-secant, in the Table, is 10.562129,
corresponding to which is 15945 '25r; the difference between this log.
co-secant and that given, is 15 ; answering to which, in the column of pro-
portional parts abreast^ of the tabular log., is 21 *, which, being added to
the above-found arch, gives 15?45'.27? for that required.
Remark. — ^llie log. secant of any arch is expressed by the difference
1)etween twice the radius and the log. co-sine of that arch ; and the co-
secant of an arch, by the difference between twice th^ radius and the log.
sine of such arch. Hence, to find the log. secant of 50M0'. — ^The log.
co-sine of 50?40^ is 9, 8019^4, which, being taken from twice the radius,
viz., 20. 000000, leavea 10. 198026 for the log. secant: from this, the
manner of computing the co-secant will be obvious.
Tablb XXXVI.
ZfOgarithmic Sines.
Of all the Logarithmic Tables in this work, this is, by far, the most
generally useful, particularly in the sciences of Navigation and Nautical
Astronomy ; and^ therefore, much pains have been taken in reducing it to
that state of simplicity which appears to be best adapted to its direct
application to the many other purposes for which it is intended, besides
those above-mentioned.
In this Table, the log., sines of the two first degrees of the quadrant are
given to every second. The next eight degrees, viz^ from 2 to 10, have
their corresponding log. sinesf to every fifth second, with proportional parts
answering to the intermediate seconds in the adjacent right-hand column ;
and because the log.sines increase rapidly in those degrees, two pages are
allotted to a degree ; every page being divided into two parts, and each
part containing 15 minutes of a degree : so that no portion whatever of
the proportional parts might be lost, and that the whole might have room
to be fully inserted. In the following seventy degrees, that is, from 10 to
80, the log. sines ar^ alsp given to every fifth second, wjth proportional
parts corresponding to the intermediate seconds in the right-hand column
of each page. In this part pf the Table, each page contains a degree; and,
for want of sufficient room, the indices ar^ only prefixed to the logs, ex-
pressed in the first column.
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From 80 to 90 degrees^ the . log. sines are only given to every tenth
second, because of the small increments by which the sines increase
towards the end of the first quadrant ; the pr opordooal parts for thf{ inter-
mediate seconds are giveain the right-hand column of each page^ as in the
preceding part of the Table.
The Table is so arranged, as to be extended to every second in the
Beraicirde, t)r from 0 to 180 degrees ; as thus : the arcs corresponding to
the log. sines are given in regular succession at top, from 0 to 90 d^prees^
and then continued, at bottom, reckoning towards the left hand, from 90
to 180 degrees. The arcs corresponding to the co-siines are given at At-
torn of the Table, and ranged in numerical order towards the left hand^
from 0 to 90 degrees, (according to the order of the sines between 90 and
180 degrees,) and then continued at top, from 90 degrees to the end of the
semicircle, agreeably to the order of the sines in the first quadrant. This
mode* of arrangement does away^th the necessity of finding the siipple-
ment of an arch* when it Exceeds 90 degrees, and possesses the peculiar
advantageof enabling the navigator to take out the log. sine or co-sine of
any- arch, and conversely, at sight, as will appear obvious bjr the following
Problems.
JNbto.— -The log. co-sine of a given degree is found in the same page with
the next less degree in the column marked O'' at top, it being the firtt
number in that column ; and the co-sine of a given degree uid minute is
found on the same line with the nest le$$ minute in tba ooluiyiii mAAed
60' at bottom of the page. •- ' ^
Problem I.
To find the Logarithmic Sine, and Co-^ine ofemy pven Arch, espreseed
in Degrees, MmUeSy and Sewnds.
RULB.
If the given arch be comprised within the limits of the two first degrees
of the quadrant, the Table will directly exhibit its corresponding log. sine
or co-sine; but when it exceeds those limits, then find the log. sine, or co-
sine, in the an^le of meeting made by the given degree and next less fifth
or tenth second at top, and the minutes in the left-hand column ; to which
add the proportional part corresponding to the odd seconds in the right-
hand column abreast of the angle of meeting, if a sine be wanted, at a
co-sine above 90 degrees; But subtract that part when* a co-sine is required^
or a sine above 90 degrees: and the sum, or difference, will be (he log. sine,
or co-sine, answering to the given arch.
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Example I.
Required the log. sine, and co-sine, corresponding to 23?I4'23f ?
To find the Log. Sine :-^
23?14^20r,an8. towhichis . . . 9.596119.
Odd seconds 3 propor. part to which is + \5
Given arch = 23?14C23r Corresponding* log. sine 9.596134
To find the. Log. G)-8iqe : —
23?14^20r,an8.towhiehi8 • • . 9.96S253
Oddseeonds S propor. pwt to whieh !• «- 8
-•m
Given^ch == 23?14C23r Corresponding log; op-siiie 9, 968250
Example 2.
Required the log. sine^ and co-sine^ conespondipg U> 113?23'47? ?
To find the Log, Sine s—
I13?SSU57,en8.towhiehi8 . . 9.962740
Odd iseeonda 2 propor. part to which is ^ 3 «
limn ■ '■ ^»
Given arch s 113?23M7r Conmpondiog log. sine 9.062798
To find the Log. Cocaine :«*-
ll$?23'45r,ans.towhichi6 • , 9,598879
Odd seconds 2 propor. part to which is + 10
Given arch = 1 13?23 :47? Corresponding log. co-s. 9. 598889
Problem U.
Tojind the Arch corfeipMding to a given LogaHthmic Smef, or Co^fmf^
RULB.
If the given log. sine, or co-sine^ be less than the sine of 2 degrees, viz.,
8. 54281 9» its corresponding arch will be found at first sight in the Table 3
but if it exceeds that number^.find the arch an9wering to the next less sine,
or next greater co^s|ne ; the difference between which and that given, being
found ia the column of proportional parts abreast of the tabular log., will
give e certain number of seooodaj wUcb, being added to the above-found
arcbj will give that required.
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Abf^.— Since the arcs corresponding to the sines between 90 and 180
degrees are found at the bottom of the Table^ and those corresponding to
the c6-sines between the same limits at its top ; if, therefore, it be required
to find the arch above 90 degrees answering to a given log. sine, or co-sine,
the first term is to be taken out as if it w^e a cosine tinder 90 degrees^ and
the other term as if it were a sine und[er 90 degrees.
Example I.
Required the arch corresponding to the given log. sine 9. 437886 ?
Sohdionj^^The next less log. sine in.the Table is 9. 437871, correspond-
ing to which is 15?54^25^ ; the difference between this and that given, is
15 ; answering to* which, in the column of proportional parts, abreast of
the tabular log., is 2'; which, being added to the above-found arch, gives
15?54'. 27T for that required.
Example 2.
Required the arch corresponding to the given log. co-fsine 9. 764570 i
Sbltifion.— The next greater co-sine in the Table is 9. 764588, corre-
sponding to which is 54? 26 '25?; the diffel^nce between this and that
given, is 9 J answering to which, in the column of proportional parts,
abreast of the tabular log., is 8^3 which, being added to the above-found
arch^ gives 54?26C28r for Aat required.
JSem'arfc.— The log. ^ines are deduced directly from the natural sines ; as
thus: —
Multiply the ftatural sine by 10000000000; find the' common log. of
the product, and it will be the log. sine.
Example 1.
Require the log. sine of S9?30' ?
Sblutioii.— The natural sine of 39?30^ is . 636078, which, being multi-
plied by 10000000000, gives 6360780000.000000, the common log. of
which is 9. 8035 1 1 5 which, therefore, is the log. sine of 39 degrees and 30
minutes, as required.
Example 2.
Required the log. co-sine of 68 degrees ? \
Solutim.rr-The natural co-sine of 68 degrees is . 374607, which, being
multiplied by 10000000000, gives 3746070000. 000000, the common log.
of which is 9, 573575 j which, therefore, ia the log. co-sine of 68 degrees^
as required. •
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Table XXXVII.
LogarUhmic Tangents.
Tliis Table is arranged in a manner so very nearly similar to that of the
I<^. sines, that it is not deemed necessary to enter into its description any
fiEUther than by observing, that it is computed to every second in the two
first and two last degrees of the quadrant, or semicircle, and to every fifth
second in the intermediate degrees. The log. tangent, or co- tangent, of a
given arch, and conversely, is to be found by the rules for the log. sines in
pages 94 and 95.
Example 1.
Required the log. tangent, and co-tangent, correspondbg to31?10M7??
Tf find the Log. Tangent :—
31?10M5r,an8.towhichis . • • 9.781846
Odd seconds 2 propor. part to i^hich is + 10
Given arch = 31?10^47^ Corre8pondinglogitang.=:9. 781855
To find the Log. Co-tangent :-—
31M0:45r,ans.towhichis . . 10.218155
Odd seconds 2 propor. part to which is — 10
Givenarchzi 31? 10^47^ Corrcs. log. co-tang. = 10.218145
Example Z
Requiied the log. tangent, and co*tangent, corresponding tol39? 1 1 '53??
To find the Log. Tangent :—
i39?lll50r,ans.towhichis . . 9.936142
Odd seconds 3 propor. part to which is — 13
CHvenarch=: 139911C53? Corres.log.tang. = . 9.936129
To find the Log. Co-tangent :-?*
139?in50r, ans. to which is • . 10. 063858
Odd seconds 8 propor. part to which is + 13
Givenai€b=: 139?lH53r Corres. log. co-tang. =: 10.063871
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Eicampk 3.
Required the arch corresjtonding to the given log. tang, 10. 155436 ?
Sohition.'^Tht next leu log. tangent in the Table U 10. 165428, cone-
sponding to which is 55?2'25f ; the difference between this log. tlWgent
^d that given^ is 13; answering to which, in the column of proportional
parts abreast of the tabular log., is 3r ; which, being added to the above-
found archi gives 55 ? i2 ^ 28 T for that required.
Example 4.
Required the arch corresponding to the given log. co-tang. 9. 792048 ?
Solutim.-^The next greater Ipg. co-tangent in the Table is 9. 792057,
corresponding to which is 58? 13' ISI' ; the difference between this log. co-
tangent and that given, is 9 ; answering to which, in the column of propor-
tional parts abreast of the tabular log., is 2^ ; which, being added to the
above-found arch, gives 58? 13U7 • for that required.
Remark.
The arch corresponding to a given log. tangent may be found hj means
of a Table of log. sines, in the following manner } vb^.
Find the natural number corresponding to twice the given log. tangent,
rejecting the index, to which add the radius, and find the common log. of
the sum;, now, half this log. will be the log. secant, less radius, of the
required arch ; and which^ being subtracted from the given log. tangent,
will leave the log. sine corresponding to that arch.
EsfOimple.
Let the given log. tangent be 10. 064 158 ; nquivad theiuidi Aomapond-
ing thereto by a table of log. sines ?
Given log; tang. . 084153 x 2 = . 168306, Nat num. =: 1. 473349
to which add the radius = 1. 000000
Sum= 2.473349, the
common log. of which is 0. 393286 ; the half of this is 0. 196643, the
secant, less radius of the required arch.
Given log. tangent = . • • . 10.084153
Corresponding log. sine = ...,.»••. 9.887510«
answering to which is 50?31' ; and which, therefore, is the
required arc)i eomespooding to the given log* tmgipu
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Tlie arch corresponding to a given log. tangent may also be found in the
following manner^ which, it is presumed, will prove both interesting
and instructive to the student in this department of science.
Find the natural tangent, that is^ the natural number corresponding to'
the given log. (angent, to the square of which add the square of the radius ;
extract the square root of the sum, and it will be the natural secanf corre-
sponding to the required arch ; then, say, as the natural secant, thus found,
is to the natural tangent, so is the radius to the natural sine : now, the
degrees, &c, an9Wering to this in the Table of Natural SineSy will be the
arch reqii|^ed^ or that corresponding to the given log, tangent*
Esample^
Let the givei^ log. tangent be 10; 084153 $ it is required to (n4 the arch
corresponding thereto by a Table of Natural Sines ?
Soluium. — Given log. tangent = .084153; the natural number corre-
sponding to this is 1.213816; which, therefore, is the natural tangent
answering to the given log. tangent.
F-
In the anne3^e4 diagnmn, let B C represent the
natural tangebt =: 1. 213816, and AB the radius
== 1. 000000. Now, since the base and perpendi-
cular of the right«angled triangle ABC are
known, the hypothenuse or secant AC may be
determined by Buclid, Book L, Prqs. 47. Hence
^/BC«= 1.213816* + AB>= 1.000000* =
A C = 1. 572689, the natural secant corresponding
to the given log. tangent. Having thus found the natural secant A C, the
natural sine DE may be found agreeably to the principles of similar tri-'
angles, as demonstrated in Euclid, Book VI., Prop. 4 ; for, as the natural
secant A C is to the natural tangent B C, so is the radius AD =: AB to
the natural sine D E : hence,
AsAC 1.572689 :BC 1.213816 :: AP 1.000000 : DE- 7718JO,
the corresponding natural sine ; now, the arch answering to this, in the
Table of Natural Sines, is 50?31C ; which, therefore, is the arch corre-
sponding to the given log. tangent, as required.
JWo/e^^^The Table of log. Qingents may be very readily deduced from
Tables XXXV. and XXXVI., as thus :— To the log. secant of any given
aich, add its log. sine; and the sum, abating 10 in the index, will be the
Ii^. tangent of that arch; the difference between which and twice the
radiiia, will be its co*tangent.
h2
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Example.
Required the log. taDgent, and co-tangent, of 25?27 -35^ ?
Log. secant of the given arch 25?27'35r = 10.044366
Log. sine of ditto 9.633344
Log. tangent corres. to the given arch =: • 9.677710
Log« co-tangent corres. to ditto • ; • • 10.322290
The Table of log. tangents may also be computed in th€ following man-
ner ; viz..
From the log. sine of the given arch, the index being increased by 10,
subtract its log. co-sine, and the remainder will be the log. tangent of that
arch; the difference between which and twice the radius, will be its log.
co-tangent.
Example.
Required the log. tangent, and co-tangent, of 32?39U0r ?
Log. sine of the given arch 32?39 U07 = . . 9. 732 1 28
Log. co-sine of ditto 9. 925249
Log. tangent corres. to the given arch r= . « 9.806879
Log. co*tangent corresponding to ditto • • 10. 193121
Table XXXVIII.
7b reduce the Time of the Moon^s Passage over the Meridian qf
Greenwich, to the Time of her Passage over any other Meridian.
The daily retardation of the moon's passage over the meridian, given at
the top of the Table, signifies the difference between two successive trans-
its of that object over the same meridian, diminished by 24 hours ; as thus:
the moon's passage over the meridian of Greenwich, July 22d, 1824, is
21 *7'^^ and that on the following day 22*9?; the interval of time between
these two traxisits is 25^2?, in which interval it is evident that the moon is
1^2? later in coming to the meridian; and which, therefore, is the daily
retardation of her passage over the meridian.
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DBSCRIPTION AND USB OP THB TABLES. 101
This Table contains the proportional part corresponding to that retard-
ation and any given interval of time or longitude ; in computing which, it is
easy to perceive that the proportion was.
As 24 hours, augmented by the daily retardation of the moon's transit
over the meridian, are to the said daily retardation of transit, so is any
given interval of time, or longitude, to the corresponding proportional part
of such retardation. The operation was performed by proportional logs.,
as in the following
Example.
Let the daily retardation of the moon's transit over the meridian be 1 ?2?;
required the proportional part corresponding thereto, and 9? 40? of time, or
145 degrees of longitude ?
As 24 hours + 1*2? (daily retard.)=: 25^ 2? Ar. comp. pro. log. 9. 1432
Is to daily retardation of transit z: . 1. 2 Propor.log. • • 0.4629
So is given interval of time =: • • • 9.40 Propor.log. • • 1.2700
To corresponding proportional part =: 23T57 • = Pro. log. =: 0. 876 1;
and in this manner were all the mimbers in the Table obtained.
The corrections or proportional parts contained in this Table are ex-
pressed in minuted and seconds, and are extended to every twentieth
minute of time, or fifth degree of longitude : these are to be taken out and
applied to the time of the moon s transit, as given in the Nautical Almanac,
in the following manner :—
Pind, in page VL of the month in the Nautical Almanac, the difference
between the moon's transit on the given day (reckoned astronomically)
and that on the day followmg, if the longitude be west ; but on the day
preceding, if it be east. With this difference enter the Table at the top,
and the given time in the left-hand, or the longitude in the right-hand
column ; in the angle of meeting will be found a correction, which, being
applied by addUion to the time of transit on the given day, if the longitude
be west, but by subtraciumy i( east, the sum, or difference, will be the
reduced time of transit.
Exodtvple 1.
Required the apparent time of the moon's passage over a qimdian 80
d^ees west of Greenwich^ July 22^1824 ?
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Mn'0pafi.t>veriner«ofGreenw.ongtv.dayis21t7" • • • 2H 7* 0!
Ditto on the dsy following = 22. 9
Retardation of moon's transit = . • . l*2!';an8.towhich
and80deg8.i8+I3. 13
Apparent time of the moon's transit over the given meridian s: 21 t20T13!
Example 2.
Required the apparent time of the moon's passage over a meridian 120
degrees east of Greenwich, August SOth, 1824 ?
Mn'8pa6.overmer^ofGreenw.ongiv.dayi820^54r • « • 20^14? Ot
Dittd on the day precedit^zz 1 9. 54
Retardation of the moon's transit = • • 1 * 0?; ans. to which
andl20deg8,is-19.12
Apparent time of the moon's tratlsit over the given meridian r: SOt 34*^48!
Table XXXlX.
Correcthn to be appUed eo the Time of the Mom'e I\xmit mjm^g
the Time of High Water.
Since the moon is the principal agent in riusing llie tides, it might be
expected that the time of high water would take place at the moment of
her passage over the meridian; but observation has shown that this is
not the case, and that the tide does tlot cease flowing for some tii^e allir :
for, since the attractive influence of the moon is only diminished, imd dot
entirely destroyed, in passing the meridian of any place, die ascepding im*
pulse previously communicated to the waters at that place tnust| therefore^
Continue to act for some time after the moon's meridional' paBStige. The
ascending impulse, thus imparted to the waters, ought to cause the time of
the highest tide to be about 80 minutes after the moon's passage over the
meridian ; but owing to the disturbing force of the sun, the actual time ef
high water difiers, at times, very considerably from that period.
The effect of the moon in raising the tides exceeds that of the sun in
the ratio of about 2^ to 1 ]; but this effect is far from being uniform : for,
since the moon's distance from the earth bears a very sensible proportion
to the diameter of this planet, and since fthe is constantly chatiging Aat
d}9timce, (being sometimes nectreri aad At otiw tiom mere remete m everjf
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AS^CRIPTION AND USB OP THB TABLBS. 103
limation,) it fa evidfiht that die must attract the waters of the oceati with
^ry imeqtial forces: but the sun's distance ft'om the earth being so very
immense^ that, compared with It^ the diameter of this planet is rendered
nearly inseaeible^ his attraction is consequently more uniform^ and there-
fore is affeets the different parts of the ocean with nearly an equal force.
By the conibined eff^t of these two forces, the tides eome on sooner
when the tnoon is in her Jirst and third quarters, and later when in the
second and jbiHA qualrtera, than they would do if raised by the sole lunar
agieuey : it is, therefore, the mean quantity of this acceleration and retard-
adon tiiat is contained in the present Table, the arguments of which are,
the apparent times of the moon's reduced transit } answering to which, iil
the dijoitiing colttiiin, stands a correction^ which, being applied to the ap-
parent time of the moon's passage over the meridian of any given place by
addition or sobtraetioti^ according to its title, the sum, or dilierence, will be
the corrected time of transiti Now, to the corrected time of transit, thus
found, let the time of high water on full and change days, at any given
place in Table LVL, be applied by addition, and the sum will be the time
of high water at that place, reckoning from the nooti of the given day:
should the mm exceed 12^24?, or 24?48?, subtract one of those quantities
from it, and the remainder will be the time of high water very near the
truth*
Example h
Required the time of high water at Cape Florida, America, March 7th,
1824 ; the longitude being^ 80?5 ' west, and the time of high water on full
and change days 7t30??
Moon's transit over the meridian of Greenwich, per Nautical
Almanac, March 7th, 1824, is 5* 2* Of
Correction from Table XXXVIII,, answering to retardation of
tranttt58r,andlongitude80?5^ west = ..... -f 12.23
Moon's transit reduced to the meridian of Cape Florida • * 5 ^ 14?23 !
Cotrectaoii aoswerii^ to reduced transit (5* 14T23') in Table
XXXDC^fa ,...'•.•*•..•.•- 1. 5. 0
Corrected time of transit 4^9^23!
Time of high water at Ci^ Florida on full and change days ?• 30. 0
I^&eofhigfa water at Cq>« Florida on the given days « • U?39r23!
Example 2.
Reqnhred the time of high water in Queen Charlotte's Sound, Nev^
Zeakmd, April IStfa, 1824; tiie longitude being 174?56: easti and t)l«
time of high wi^t^r on Mmi change da^ 9^0? ?
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Moon's transit over the meridian of Greenwich^ per Nautical
Almanac, April 13th, 1824, is ........ .10*27^0*
Correction from Table XXXVIIL, answering to retardation of
transit 50T, and longitude 174?56' west = .... ^ 23.29
Moon's transit reduced to the meridian of Queen Charlotte's
Sound 10! 3T31I
Correction answering to reduced time of transit (lOtSTSl!)
in Table XXXIX., is + 23. 0
Corrected time of transit 10i26?31!
Time of high water at given place on full and change days • 9. 0. 0
Trnie of high water at Queen Charlotte's Sound, past noon of
the given day . • . . 19?26T8l!
Subtract 12.24. 0
Hme of high water at given place, as required 7 • 2T31 !
Table XL.
Beduction qf tlie Moon's Horizontal Parallax on account of the
Spheroidal Figure qfthe Earth.
Since the moon's equatorial horizontal parallax, given in the Naiftical
Almanac, is determined on spherical principles, a correction becomes
necessary to be applied thereto, in places distant from the equator, in
order to reduce it to the spheroidal principles, on the assumption that the
polar axis of the earth is to its equatorial in the ratio of 299 to 300 ; and,
when very great accuracy is required, this correction ought to be attended
to, since it may produce an error of seven or eight seconds in the computed
lunar distance. The correction, thus depending on the spheroidal figure
of the earth, is contained in this Table} the arguments of which are, the
moon's horizontal parallax at the top, and the latitude in the left-band
column } in the angle of meeting will be found a correction, expressed in
seconds, which being subtracted from the horizontal parallax given in th^
Nautical Almanac, will leave the horizontal parallax agreeably to the sphe-
roidal hypothesis.
Thus, if the moon's horizontal parallax, in the Nautical Almanac, be
57-58^, and the latitude 51?48^ ; the corresponding correction will be 7
seconds subtractive. Hence the moon's horizontal parallax ou the sphe*-
roidal hypothesis, in the given latitude, is 57 *5K,
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DSSCUPTtON AMD VSB OF THE TABLBS« 105
JtewuurJc^^The correctioiii contained in this Table may be computed by
the following
Rule.
To the logarithm of the moon's equatorial horizontal parallax, reduced to
seconds, add twice the log. sine of the latitude, and the constant log*
7. 522879 ;* the sum, rejecting the tens from the index, will be the loga*.
rithm of the corresponding reduction of parallax.
Example.
Let the moon's horizontal parallax be 57 '58?, and the latitude 51?48^ ;
required the reduction of parallax agreeably to the spheroidal hypothesis ?
= 3478r . Log. = 3.541330
Twice the log. sine =19. 790688
Constant log. . . 7.522879
Moon's equatorial horiz. par. 57^58r = 3478? . 'Log. = 3.541330
Latitude; 51?48' Twice the log. sine =19. 790688
Reduction of horizontal parallax = . . 7^ 159 Log.=0. 854897
Table XLL
Jteduetion of LaiUude on account of the Spheroidal Figure of the Earth.
Since the figure of the earth is that of an oblate spheroid, the latitude of
a place, as deduced directly from celestial observation, agreeably to the
q>herical hypothesis, must be greater than the true latitude expressed by
the angle, at the earth's centre, contained between the equatorial radius
and a line joining the centre of the earth and the place of observation.
This excess, which is extended to every second degree of latitude from the
equator to the poles, is contained in the present Table ; and which, being
iubtracted from the latitude of any given place, will reduce that latitude
to what it would be on the spheroidal hypothesis : thus, if the latitude be
50 degrees, the corrjespoiiding reduction will be 1 1 M2?, subtractive ; which,
therefore, gives 49?48^ 18? for the reduced or spheroidal latitude.
Remark. — ^The corrections contained in this Table may be computed by
the following rule ; viz..
To the constant log. . 003003,t add the log. co-'tangent of the latitude.
* The arithmetical complemeDt of the log. of the earth'i ellipticity assumed at ^^
t The excess of the spherical above the elliptic arch in the paraUel of 45 dei^rees from the
equator, is 1K887» or 1^53^' (Robertson's Navi^tion, Book VIII., Article 134) i hence
45« - ir 53^ a 44* 48^ 7", tiis lo(. co-t«iigeat of which, lejeclins the index, U . 003003.
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106 DBtCRimOK AND VtB. OF THB TABUIti
and tte turn will be the log. e o-tangent of ui arch ; the difference between
which and the given latitude will be the required reduction.
Example.
Let it be required to reduce the spherical latitude 50?48f to what it
would be if determined on the spheroidal principles ; and^ hence^ to find
the reduction of that latitude.
Latitude 50?48' Or Log. co-tang.=:9.911467
Constant log.= .003003
Reduced or spheroidal latitude = 50^36^21? Log. co-tang.f:9. 914470
Reduction of latitude^ as required 0?llC39r
Tablb XLIL
A General Traverse Tabk ; or Difference of Latitude and Depaartwre.
This Table, so exceedingly useful in the art of navigation, is drawn up in
a manner quite different from those that are given, under the same deno-
mination, in the generality of nautical books i and, although it occupies
but 88 pages, yet it is more extensive dian the two combined TaUea tff
61 pages, which are contained in those books. In this Traverse Table,
every page exhibits all the angles that a ship^s course can make with the
meridian, expressed both in points and degrees; which does away with tfatf
necessity of consulting two Tables in finding the diflerence of latitude and
the departure corresponding to any given course and distance. If the
course be under 4 points, or 45 degrees, it will be found In the left-hand
compartment of each page ; but that a&ote 4 points, or 45 degrees, in the
right-hand compartment of the page. The distance is given, in numerical
order, at the top and bottom of the page, from unity, or 1, to 304 miles $
which Qlomprehends all the probable limits of a ship's run in 24 hours j and,
by this arrangement, the mariner is spared the trouble of turning over
and consulting twenty-three additional pages. Although the manner of
using this Table m\ut appear obvious at first sight, yet since its mode of
arrangement differs so very considerably from the Tables with which the
reader may have been hitherto acquainted^ the following Problems are
given for its illustration.
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MioRimoif AND vnm op thi tabus. 107
Problbm L
Ghen fftv C(mr$e and Distance fotkil, or betwem two Places, to find the
Difference qf Latitude and the Depdarturei
RULB.
Enter the Table with the course in the left or right-hand column^ and
the distance at the top or bottom | opposi^ to the former^ and under or
over the latter, will be found the corresponding difference of latitude and
departure i these are to be taken out as marked at the top of the respective
columns if the course be tmcfer 4 points or 45 degrees, but as marked at
the bottom if the course be more than either of those quantities.
Note* — If the distance exceed the limits of the Table, an aliquot part
thereof may be taken, as a half, third, fourth, &c. ; then the difference of
latitude and departure corresponding to this and the given course, being
multiplied by 2, 3, 4, &c., (that is, the figure by which such aliquot part
was found,) the product will be the difference of latitude and departure
answering tb the given course and distance.
Example 1.
A ship sails S.S.W. | W. 176 miles; required the difference of latitude
and the departure i
Opposite ^ points and under 176 milesi stand 155. 2 and 88. 0 1 heiie«
the dlffefenee of latitude is 155^ 2^ and the departure SS. 0 miles.
Example 2.
A ship sails N« 57? B. 236 miles ; required the difference of hititude and
the departure ?
OffOMe to 57?5 and under 236 miles^ stand 128. 5 and 197^ 9 : hence
Use diilbi«nce of hititude is 126. 5^ and the departure 197* 9 miles.
Example 3.
'the course between two places is E. b. S. ^ S^ and the distance 540
miles ; required the difference of latitude and the departure ?
Distance divided by 2, gites 270 miles } under or over which, and oppo«
Hie to 8i points, stand . . . 91.0 and 254.2
Multiply by 2 2
IVoducU =: 182. 0 and 508. 4 1 hence (he differeuce
pf Mtode is 182. 0| and the departure 508* 4 milest
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Example 4.
TTie course between two places is N. 61 W. anil the distancel 176 miles;
required the difference of latitude and the departure ?
Distance 1 176 divided by 4, gives 294 miles ; under or over wluch^ and
opposite to 61?^ stand • • • 142.5 and 257*1
Multiply by 4 4
Product = 570. 0 and 1028. 4 : hence the <tifier-
ence of latitude is 570. 0^ and the departure 1028. 4 miles.
Problem IL
Given the Difference ofLat^ude and the Departure, to find the Omree
and Distance.
Rule.
With the given difference of latitude and departure^ enter the Table and
find, in the proper columns abreast of each other, the tabular difference of
latitude and departure either corresponding or nearest to those g^ven } then
the course will be found on the same horizontal line therewith in the left or
right-hand column, and the distance at the top or bottom of the compart-
ment where the tabular numbers were so found.
Note.-^If the difference of latitude be greater than the departure, the
course will be less than 4 points^ or 45 degrees ; and, therefore, it b to be
taken from the left-hand column : but when the difference of latitude is
less than the departure, the course will be more than 4 points or 45 degrees^
and, consequently, it must be 'taken from the right-hand column.
Note, also, that when the difference of latitude and the departure, or
either of them, exceed the limits of the Table, aliquot parts are to betaken,
as a half, third, fourth, &c., with which find the course and distance as
before $ then the^distance, thus found, being multiplied by 2, 3, 4, &c., the
product will be the tu/ioie distance corresponding to the given difierence
of latitude and departure. The course is never to be multiplied, because
the angle will be the same whether determined agreeably to the whole dif-
ference of latitude and the departure, or according to their corresponding
aliquot parts*
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BBSCRIFTION AND USB OP THB TABLES* 109
Exatnple I.
If the difference of latitude made by a ship in 24 hours be 177* 4 miles
north, and the departure 102. 6 miles east, required tlie course and distance
made good?
SohUim. — The tabular difference of latitude and departure, nearest
corresponding to those g^Yen, are 177. 5 and 102. 5 respectiYcly : these are
found in the compartment under or over 205, and opposite to 30 degrees ;
hence the course made good is N. 30 E., and the distance 205 miles.
trample 2.
The difference of latitude made by a ship in 24 hours, is 98.5 miles
south, and the departure 140. iS miles west; requfared the course and dis-
tance made good ?
SohUmi. — The tabular difference of latitude and departure, nearest to
those giTcn, are 98. 7 and 140. 9 respectively : these are found in the com-
partment under or over 1 72, and opposite to 55 degrees ; hence the course
made good is S. 55? W., and the distance 172 miles.
Example 3.
The difference of latitude is 700 miles south, and the departure 928 miles
west; required the course and distance ?
Solution, — Since the difference of latitude and the departure exceed the
limits of thcTable, take therefore aify aliquot part of them, as one fourth, and
they will be 175 and 232 respectively : now, the tabular numbers, answering
nearest to those, are 175. 1 and 232. 4 ; these are found in the compartment
under or over 29.1, and opposite to 53 degrees : hence the course is S. S3?
W«, and tiie distance^91 X 4 = 1 lfi4 miles, as required.
Remarks — ^Whenever it becomes necessary to take aliquot parts of the
difference of latitude, the same must be taken of th^ departure, whether it
falls without the. limits of the Table or not; and, vice versGy whenever it
becomes necessary to take aliquot parts of the departure, the same must
be taken of the difference of latitude.
And, in all cases where the tabular numbers differ considerably from
those given, proportion must be made for that differeuce.
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IIQ DiacRifnov Aim va ot tbb vabui.
Problem .III.
Given the proper Difference of Latitude between two Places, the Meridimal
Difference of Latitude, and the Departure, to find the Course, Distance,
and Difference of Longitude.
With tht proper diftrenci pf latitude and the cbparture^ find t]i« couna
and di^tanoct by Problem IL; tben^ with the course thus found and the
meridional difference of latitude, (in a latitude column,) take out the cor-
responding departure, and it will be the differenpe of longitude required ;
as thus : run the eye along the horizontal line answering to the course,
from where the ptopex difference of latitude was found, {ahjoays to the
right hand,) and find, in* a latitude column, the tabular difference of lati-
tude answering nearest to the given meridionid difference of latitude |
abreast of which, in the departure column, will be found tl^fs difference of
longitude.
The proper difference of latitude between two places, is 142 miles
north, the departure 107 miles west, and the meridional difference of
latitude 169 miles ; required the course, distance, and difference of Ion-
^tudei
S6Itt(ia»K— The tabular difference of latitude and departure answering
xu^est to those giyei), are 142. 2 and 107. 3 respectively i these are (bond
in the compartment under or over 17S, and opposite to 37 degr^ies: heppfi
the ppur^e is N. 37? W., a^d tb^ distance 178 miles. Now, with the
cpurse 37 degreos, and the meridional difference of latitude 169 mleu,
the difference of longitude is foundt as thus : from where the proper dif-
ference of latitude was fpundj run the eye along the horizpntat liu^ answer-
ing to 37 degrees, {ahoays towards the right hand,) and the tabular differ-
ence of latit|ide answering nearest to the given meridional diffisrence of
latitude will be found in the compartment under or over 212, viz. 169. S; ^
correspanding to which, in the departure column, is 127«63 and which^
therefore, is the difference of bnptude, as required.
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Paqbdsm IV,
Qioen the proper Differefice qf LalUwUy the Meridkmal Diffisrence qf
Latitude, and the Difference of Longitude, to find the Course and
Distance.
Rule.
^th the meridional difference of Ulitttde and the difference of longi-
tude, esteenied as difference of latitude and departure, find the course by
Problem II. } then with the course^ thus found, and the proper difference
of latitude^ the dbtance is to be obtained, as thus : run the eye {akoays to
the ^fi hand) f^png the horizontal line answering to the comae^ from wl^re
the meridional differei|ce of latitude was found| mi ieekj in th^ proper
cohimn, the differeiic^ of latitude answering nearest to that given ; over or
under which, at the top or bottom of the column, will be foup4 (he
required distance.
Note. — ^When the meridional difference of latitude exceeds the differ-
ence of longitude, the course is ^ be taken from the lefk-hand column i
but otherwise from the right.
ExampJe.
Hie proper differoice of latitude between two places is 78 miles south,
the meridional difference of latitude 107 miles south, and the difference of
longitude 119 miles east; required the course and distance ?
Mmtioeu-^Th^ tirinilar diflbrence of latitude md departmie^ answering
nearest (o th^ meridional difference of latitiidn md the dtiEerenee of longi-
tudiK, if« 107. 1 and U8»0 respeittmiy i these are foi|nd in the comparts
ment under or pver )60, and opposite to 48 degrees s heiice the eoune i«
S. 48? EL Now, the eye being run along tlie horizontal line fmsweriog to
48, [towards the ^ft hand,) the nearest tabular difference of latitude,
answering to the proper difference of latitude, will be found in the com-
partmant imder or over U7 s hence the distance is 117 miles.
Probum V.
Gioen the middle Latitude^ and the Meridian DietwmcfDepoflme^
to find the Difference €f J^n^ptude^
Rule.
Bntat tha Tablt with the middle latitude, taken as « eeimi^ and the
departure in a latitude eolumnj run the eye along the horizontal Una
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112 DSSCRIPTION AND USB OF THB TABLES.
answering to that course (towards the right hand or the left, acoording as
the first tabular difference of ItOitude which meets the eye therein is greater
or less than the given departure), and find a difference of latitude that
either agrees with, or comes nearest to, the given departure ; then .the
distance Corresponding to this, at the top or bottom of Uie column, will be
the difference of longitude.
Example.
The middle latitude between two places is 20? north, and the meridian
distance or departure 140 miles; required the difference of longitude?
Solution. — The middle latitude, 20 degrees, taken as a course, and the
departure 140, as difference of latitude, will be found to correspond in the
compartment under or over 149 : hence the difference of longitude is 149
miles, as required.
Problbm VI,
Given the middle Latitude, the Difference of Latitude, and the Difference
of Longitude between two Places, to find the Coureeand Distance,
RULB.
Enter the Table with the difference of longitude, esteemed as distance,
at the top or bottom of the page, and the middle latitude, taken as a course,
in the left or right-hand column ; answering to which, in the difference of
latitude column, will be found the departure. Now, with this departure
and the given difference of latitude, the course and distance are to be found
by Problem II.
Example.
The middle latitu4.e is 26 degrees north, the difference of latitude 200
miles north, and the difference of longitude 208 miles east; required the
course and distance ?
Solution. — In the compartment under or over 208 miles (the given longi*
tude), and opposite to 26 degrees (the middle latitude taken as a course),
stands 186.9 in the difference of latitude column, which, therefore, is the
departure. Now, the tabular numbers answering nearest to the given
difference of latitude and the departure, thus found, are 200. 4 and 186. 9
respectively ; these are found in the compartment under or over 274, and
opposite to 43 degrees: hence the course is N. 43? E., and the distance 274
miles.
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113
-Bemorfc.— The numbers in the general Traverse Table were computed
agreeably to the following rule; via..
As radius is to the distance^ so is the co-sine of the coufse to the differ-
ence of latitude ; and so is the sine of the course to the departure.
Bsample.
Given the course 35 degrees, and the distance 147 miles j to compute
the difference of latitude and the departure.
To find the Difference of latitude.
As radius • . •
Is to distance . .
So is the course •
s 90? log. sine . . s 10.000000
. 147 miles . . log. = 2.167317
= 35? log. co-sine . = 9.913365
To difference of lat. = 120.4, miles . • log. = 2.080682
As radius . .
Is to distance .
So is the course
To departure
To find the Departure,
. . s 90? log. sine . . s 10.000000
. . 147 miles . . log. =: 2. 167317
, . s= 35?, log. sine . . = 9-758591
= 84. 3 miles . • log. = 1.925908
Table XLIII.
Meridimal Parts*
This Table contains the meridional parts answering to each degree and
minute of latitude from the equator to the poles ; the arguments of which
are, the degrees at the top, and the minutes in the left or right hand mar-
ginal columns ; under the former, and opposite to the latter, in any given
latitode, will be found the meridional parts corresponding the^to, and
conversely. Thus, if the latitude be 50?48C , the corresponding meridional
parts will be 3549. 8 miles.
Remark. — The Table of meridional parts may be computed by the fol-
lowing rule ; viz..
Find the logarithmic co-tang^t less radius of half the complement of
any latitude, and let it be esteemed as an integral number ; now, from the
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114 DB8CRIPTION AND USB OF THB TABUW.
common logarithm of this, subtract the constant log. 2.101510*, and
the remainder will be the log. of the meridional Iparts answering to that
latitude. .„ , ,
Example i.
Required the meridional parts corresponding to latitude 50?48^ ?
Given lat. = 50M8- complement = 39? 12^^.2 = 19936., the half
complement; hence, aaojao ^\. % f
Halfcomp.«19^36: log. co.tangent7««radit« =« ^ *^^**\^t,2f'''
whichis «-65 71^^
Constant log ^'^Q^^^"
Meridional parts correspoudinj? to given lat, 3549. 78«log.=:3. 550202
Example 2.
Iteqtured the mcridiotial parts corresponding to latitude 89^30^ ?
Given lat. = 89?80^ } comp. = 0?30^ -^2 ^ 0?15^ the half comple-
meut; hence, k^aida \. t
Half comp. = 0*ll5^ log. co-tangent less radius = 2.360180, the log.
ofv^hichis . • 6.372945
Constantlog 2.101510
Meridional parts corresponding to given lat 18682. 49=log.=:4. 271435
Table XLIV.
T%e Mean Right Jscensions and Declinations of the principal ficed
Stars.
This Table contains the viean right ascensions and declinations of the
principalfixed stars adapted to the beginning of the year 1824.— The stars
are arranged in the Table according to the order of right ascension in
ivhich they respectively come to the meridian; the annual variation, in
right ascension and declination, is given in seconds and dedmal parts of a
second} that of the former being expressed in time, and that of the latter
motion.
The stars marked ft h«ve been taken horn the Nautical Almaiiae for
the year 1824.— The stars that have asterisks prefixed to tiiem are tfaoee
from which the moon's distance is computed in the Nautical Almanac for
the purpose of finding the longitude at sea.
* TliemttSQTe affile arc of 1 minute (pa^ 54,) is .00029088821; which beinf multiplied
by 10000000000, (the radius of the Tables) produces 290.8882000000 ; and, this being:
multiplied by the modulus of the common ^iguflthms, tia., Adl29448190y ghres
126.331140109823580 1« the coanitt kg. of which is 2.l01dl0»«s abov«.
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ABSCRIPTIOK Airs nSS OF THB TABLB9. 115
The places, of the etars, as given in this Table, may be reduced to any
future period by multiplying (he annual variation by the number of yean
and parts of a year elapsed between the beginning of 1824, and such fiiture
period : the product of right ascension is to be added to the right ascen*
sions of alt the stars, except fi and Z, in Ursa Minor, from whose right as-
censions it is to be subtracted :. but the product of declination is to be ap-
plied, according to the sign prefixed to the annual variation in the Table,
to the declinations of all the stars without any exception ;— thus,
To find the right ascension and the declination of a Arietis^ Jan. 1st, 1884.
R. A. of « Arietis, perTab. 1*57^16!, and its dec. . . 22?37'83r N.
Annual var. . +3*'. 35 Ann. var.+ 17'«40.
Number of years Num. of yrs.
after 1824= 10 after 1824 = 10
Product. +33-.5 +0^38^ 5 Prod.+ 174''.Os= + 8(54f
Rt. asc. of a Arietis, as req. 1^57*49'. 5, and its declination 22?40^27^ N.
Should the places of the stars be required for any period antecedent to
1824, it is evident that the product^ of right ascension ^nd deplination iwt^
be applied in a contrary manner.
The eighth column of this Table contains the true spherical distance and
the approximate bearing between the stars therein contained and those prer
ceding, or abreast of them op the same horizontal line ; and the pinth^
or last column of the page, the annual variatiop of that distance expressed
in seconds and decimal parts of a second.— By means of the last column,
the tabular distance may be reduced very readily to any future period, by
multiplying the years and parts of a year between any such period and the
epoch of the Table, by the annual variation of distance ; the product being
applied by addition or subtraction to the tabular distance, according as the
sign may be affirmative or negative, the sum or difference will be the dis-
tance reduced to that period.
flsample.
Required the distance between a Arietis and Aldebaraii, Jan. 1st, 1844 ?
TdHilar dist. between the two given stars == . . . 35?32'7^
Annual var. of distance . — 0-. 02
Number of years after 1824 « 20
Product. . f-.0<',40= -0^.40
True spherical distance between the two given stars, as
required • 85?32;6^60.
x2
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Ifemari,— The trae spherical distance between any two stars, whose
right ascensions and declinations are known, may be computed by the fol-
lowing rule } viz..
To twice the log. sine of half the difference of right ascension, in jdegr ees
add the log. sines of the polar distances of the objects ; from half the sum
of these three logs, subtract the log. sine of half the difference of the polar
distances, and the remainder will be the log. tangent of an arch ; the log.
sine of which being subtracted from the half sum of the three logs., will
leave the log. sine of half the true distance between the two given stars.
Example.
Let it be required to compute the true spherical distance between a
Arietis and Aldebaran, January 1, 1844.
R. A. of a Arietis red. to 1844 =: U5Sr23!, and its dec. =22?43^2ir N.
R. A. of Aldebaran red. to 1844 = 4. 26. 58. 6, and its dec. = 16. 1 1. 28 N.
Difference of light ascension = 2 ! 28'?35 ' . 6 =
37?8^54r^2=18?34:27.r
":Sil!^r!'"^*-'r'*:}i8?34c27ri;«.n?i:}i9. 0063060
«.poi«ai...»f.Arua.=: {67.16.39 {^ } 9.9649129
N.p«i«d»it.«fAia,ur«= {73. 48. 32 {|ff} 9. 9824236
Sum . . 38.9536425
Diff. of Polar dists.6?3i:53r Half=19.4768212| . . . 19.4768212.5
Half diff. of ditto 3?15r56jr Log S. 8.7556177i
Arch 79?14:27^ 5826 log. tang. . 10. 7212035 Log. S. 9. 9922976. 3
Half the req. dist. .... 17?46'.3'^. 4424 . Log. S. 9. 4845236. 2
True spher. dist. between
the two given stars , . 35?32^6''. 8848 on Jan. 1, 1844.
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BESCRIPTION AND USB OF THE TABLB8. 117
Now, by comparing this computed distimce with*that directly deduced
from the Table, as in the preceding example, it will be seen that the differ-
ence amounts to very little more than the fifth part of a second in twenty,
years ; which evidently demonstrates that the tabular distances may be re-
duced to any subsequent period,, for a considerable series of years, with all
the accuracy that may be necessary for the common purpose9 of navi-
gation.
Note* — ^The tabular distances will be found particularly useful in deter-
mining the latitude, at sea, by the altitudes of two stars, as will be shown
hereafter.
Tablb XLV,
Acceleration of the Fixed Stare ; or to reduce Sidereal to Mean Solar
Time.
Observation has shown that the interval between any two consecutive
transits of a fixed star over the same meridian is only 23t56T4' . 09, whilst
that of the sun is 24 hours : — the former is called a sidereal day, and the
latter a solar day ; the difference between those intervals is 3*55 '. 91, and
which difference is called the acceleration of the fixed stars.
This acceleration is occasioned by the earth's annual motion round its
orbit : and since that motion is from west to east at the mean rate of 5 9 ^ 8 '^ . 3
of a degree each day ; if, tlierefore, the sun and a fixed star be observed on
any day to pass the meridian of a given place at the same instant, it will be
found the next day when the star returns to the same meridian, that the
sun will be nearly a degree short of it ; that is, the star will have gained
3" 56 ! . 55 sidereal time, on the sun, or 3t 55' • 9 1 in mean solar time ; and
which amounts .to one sidereal day in the course of a year: — for
3r55'.91 X 365'. 5 U8^ 48*.= 23t56"r4!:— hence in 365 days as mea-
sured by the transits of the sun over the same meridian, there are 366 days
as measured by those of a fixed star.
Now, because of the earth's equable or uniform motion on its axis, any
given meridian will revolve from any particular star to the same star again
in every diurnal revolution of the earth, without the least perceptible differ-
ence of time shewn by a watch, or clock, that goes well : — and this pre-
sents us with an easy and infallible method of ascertaining the error and
the rate of a watch or clock :*«to do which we have only to observe the
instant of the. disappearance of any bright star, during several success ive
nights^ behind some fixed, object, as a chimney or pomer of a bouse at a
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116 BJIftCIlIPTlbN AWB USB OF THB TABLSd.
litdii dlsUtice^ lh6 position o( thi^ ejnei bleing fix^d dt 8otil6 pftrtittulBr ftpot,
auch as at ft smftll hole in a window-dhutter nearly in the plane of the meri-
. dian 5 then if the observed timed of disappearance correspond with the ac-
celeration contftihed iti the second coliimn of the first compartment of the
jpresent Table, it will be an undoubted proof that the wateh is well regik-
lated :-^hence, if the wiitch be exactly true, the disappeiirance of the samfc
star will be 3*? 5 6'. earlier every night; that is, it will disappear 3^56'
sooner the first night ; 7*52! sooner the second night; 1 1*48'. sooner the
third night, and so on, as in the Table. — Should the watch, or clock de-
viate from those times, it must b^ corrected acbordingly ; and sinc6 the dis^
appearance of a star is instantaneous, we may thus determine the rate df a
watch to at least half a second.
The first compartment of this Table consists of two columns ; the first of
which contains the sidereal days, or the interval between two successive
transits of a fixed star over the same meridian, and the second the accele-
ration of the stats expressed in mean solar time $ which is extended to 80
days, so as to afford ample opportunities for the due regulation of clocks or
watches.— The five following compartments consist of two columns each,
and are particularly adapted to the reduction of sidereal dmfe into mean
solar time :— the correction expressed in the column marked dcceleralfen,
&c. being subtracted from^ its corresponding sidereal time, will reduce it to
mean solar time ; as thus.
Required the meah solar time corresponding to 14^40^55*. iid^eal
time }
Given sidereal time := 14^40^55'.
Corresponding to 14 hours is • , 2* 17 ' • 61 ")
Do. 40 minutes, .0. 6 .55 > Sura as — 2*^24 .81
Do. 55 seconds • . 0. 0.15 J -:— ^ ^.^....^
Mean solar time as required ...,.••• 14t38r30'.69
J{6mdr)lr.-^Thi6TaMe was computed in the fddlowing manner; vit.^
Since the earth performs its revolution round its orbit^ that is, roftnd
the sun, in a solar year; therefore as 366f5*48'?48'. ; 360?:: 1*. : 69;8''.a|
which, therefore, is the earth's daily advance in its orbit : but winle the
earth is going through this daily portion of its orbit, it turns once rowid
on its axis, from West to east, and thereby describes an arc of 360?59' 8""^ 8
in a mean solar day, and an arc of 360? in a sidereal day.
Httice, as 360?59:8''.3 : 24*::3609 : 23*56?4'.09,.the length of a
sidereal day in mean solar time; and which, therefore^ evidently anticipaites
3" 55 ' . 9 1 upon the solar day as before-mentioned, Now,
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nSORIPnON AND U8B OV THB TABUI8« 119
Ab one aidtteal day, is to 3*55' * 91, so is any given portion of sidereal
time to its corresponding portion of mean solar time i-^-^and hence, the me*
thod hy which the Table was computed.
Tabus XLVJ.
Ta reduce Mean Solar Time into Sidereal Time.
Since this Table is merely the converge of the preoeding, it is presumed
that it does not require any explanation farther than by observing, that the
correction is to be applied by addition to the corresponding mean solar
time, in order to reduce it into sidereal tifife ; as thus.
Required the sidereal time corresponding to 20M 5*33! mean solar
time?
Given mean solar time = • 20M5"3dl
Corresponding to 20 hours is 3t 17' . 13^
Do. 15 minutes 0. 2 .46V Sum = . + 3rl9;.68
Do. 33 seconds 0, 0 . 09) —
Sidereal time as required .•..••.. 20?18T52%68
Tabijs XLVIL
Time from Noon tohen the Sun*s Centre is in the Prime Vertical; being
the instant at which the Altitude of that Object should be observed in
crier ioaseertain the apparent Time with the greatest Accuracy.
•
Since the change of altitude of a celestial object is quickest when that
object is in the prime vertical, the most proper time for obser^ng an alti-
tude from wiuch the apparent time is to be faiferred, h therefiMPe when the
object is due east or west; because then the apparent tine is not likely to
be affected by the unavoidable errors of observation, nor by the inaccuracy
of the assmned latitude. — This Table contains the lyparent time when a
celestial object is iu the above position.«-The declination is marked at top
and bottom, and the latitude in the left and right hand marginal columns z
hence, if the latitude be 5{) degrees, and the declination 10 degrees, both
being of the same name, the object will be due east or west at 5 ! 26*? from
its time of transit or meridional jMssage.
£emarfc,<— TUs Tabic vsfl computed by the felk^wing rule j vii^
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120 DBSCRIFTION AND USB OF THB TABLES.
To the log. co-tangent of the latitude, add the log. tangent of the decli-
nation ; and the sum, abating 10 in the index, will be the log. co-sine of the
hour angle^ or the object's distance from the meridian when its true bearing
is either east or west.
Esamfle.
Let the latitude be 50 degrees, north or south, and the sun's declination
10 degrees, north or south ; requiredthe apparent time when that object
will bear due east or west ?
Given latitude = 50? log. co-tangent = 9.923814
Declination of the sun = . . . 10? log. tangent ^ 9. 246319
Hour angle = • . . 81 929'30'r =lpg. co-sine = 9. 170133
In time ss . . . 5^25?58! j which, therefore, is the ap-
parent time when the sun bears due east or west.
Note. — ^During one half of the year, or while the sun is on the other side
of the equator, with respect to the observer, that object is not due east or
west while above the horizon ; in this case, therefore, the observations for
determining the apparent time must be made while the sun is near to the
horizon ; the altitude, however, should not be under 3 or 4 degrees, on ac-
count of the uncertiunty of the effects of the atmospheric refraction on low
altitudes.
Table XLVIIL
Altitude of a Celestial Object {when its centre is in the Prime Vertical,)
most proper for determining the apparent T^me with the greatest
Accuracy.
This Table is nearly similar to the preceding; the only difference being
that that Table shows the apparent time when a celestial object bears due
east or west, and this Table the true altitude of the object when in that po-
sition ; being the altitude most proper to be observed in order to ascertain
the apparent time with the greatest accuracy : — thus, if the latitude be 50
degrees, and the declination 10 degrees, both being of the same name, the
altitude of the object will be 13?61 , when it bears due east or west from
the observer; which, therefore, is the altitude most proper to be observed^
for the reasons assigned in the explanation to Table XL VII.
7s^o<e.f-This Table was computed by the following rule ; viz.^
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]>BSCRIPTION AND U8IS OF THE TABLBS. . 121
If the declination be less than the latitude ; from the log. sine of the
former (the index being increased by 10). subtract the log. sine of the lat-
ter, and the remainder will be the log. sme of the altitude of the object
when its centre is in the prime vertical :-— But, if the latitude be less than
the declination, a contrary operation is to be used ; viz., from the log. sine of
the latitude, the index being increased by 10, subtract the log. sine of the
declination, and the remainder will be the log. sine of the altitude of the
object when its centre is in the prime vertical, or when it bears due east or
ivcsti
Example h
Let the latitude be 50?, and the declination of a celestial object 10"?,
both being of the same name ; required the altitude of that object when iu
centre is in the prime vertical.
Declination of the object s= 10? log. sine s 9. 239670
Latitude 50. log. sine = 9. 884254
Altitude required . . • I3?6'6r log. sine =s 9.355416
Example 2.
Let the latitude be 3?, and the declination of a celestial object 14?, both
being of the same name ; required the altitude of that object when its
in the prime vertical.
Latitude 3? log. sine = 8.718800
Declination of the object = 14 log. sine = 9.383675
Altituderequired . 12?29'S8r log. sine = 9.335125
^ole.— Altitudes under 3 or 4 degrees should not be made use of in
computing the apparent time, on account of the uncertainty of the atmos-
pheric refraction near the horizon.
And since the Table only shows the altitude of a celestial object most
favourable for observation when the latitude and declination are of the same
name; therefore during that half of the year in which the sun is on the
other side of the equator, with respect to the observer, ' and in wliich he
does not come to the prime vertical while above the horizon, the altitude
is to be taken whenever it appears to have exceeded the limits ascribed to
the Oncertainty of the atmospheric refraction in page 120.
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182 raSCEIPTION AKB VME OP TUB TAAUS*
Tablb XLDC.
AmptUudes of a Celestial (^ect, reckoned from the true East, or West
Point of the Horizcn.
^1^e arguments of this Table are, the declination of a celestial object at
top or bottom, and the latitude in the left, or right hand column ; in the
angle of meeting will be found the amplitude : proportion, however, is to
be made for the excess of the minutes above the next less tabular argu-
ments.
Example 1,
Let the latitude be 50?48' north, and the sun's declination 10?25C
north ; required the sun's true amplitude at its setting ?
True amplitude corresp. to lat. 50?, and dec. 10?, =W. 15? 40C N.
Tab. diff. to 1^ of lat. =21 '. ; now ^^^ = +17, nearly;
T.difr.tol?ofdec.=l?36:,or96^inow^^^^' = +40
Sun's tme amplttude at required • • « • • as W. 16. S7« N.
Example 2.
Let the latitude be S4?24' north, and the sun's declination 16? 48'
south ; required the sun*s true amplitude at the time of its rising ?
True amplitnde corresponding to latitude S4? N. and
declination 16?30r S. = E. 20? 2' S.
Tab.di£tDl?of]aU ^ 15?} Mwl^^^iH^a. • + 6
Tab. diff. to 30i of dec. = 37^} now ^^iili' m. . +22, aetriy.
Sun's tone amplitude as required s ...««& 20?30' S,
Remark.
This Table was computed agreeably to the following rule ; viz..
To the log. sec^t of the latitude, add the log. sine of the declination,
and the sum, abating 10 in the index, will be the h^. sine of the true am-
plitude.
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J^SICftlPnON AKD Va Of TBS TAftUIS* 12^
Example*
hbt the latitude be 50?48'.^ and the dedinatioa of a celeetial olject
10^25' ; required the true amplitude of that object ?
Latitude . . . • • 4 . ^ 50?48C log. secant 10. 190263
Deelination 10. 2S log. sine 9.2S72U
True amplitude asrequired 16?37'22r log. sine . • 9. 456474
Table L.
Tx>JM the Timet ttfUm Ritbig ond Setting tfa Cde&tM Olgeet.
HiU Table contains the semidiurnal arch, or the time of half the continu-
ance of a celestial object above the horizon when its declination is of the
same name with the latitude of the place of observation; or the. time of
Kalf its continuance below the horizon when its declination and the lati-
tude are of different denominations. — The semudiurnal arch espreesee the
time thai a celestial object takes in ascending from the eastern horizon to
the fNeridian; or of its descending from tke meridian to the western
hcriton.
As the Table is only extended to 2d| degrees of declination^ being the
greatest declination of the sun, and to no more than 60 d^ees of latitude ;
therefore, when the declination of any other celestial object and the lati*
tude of the place of observation exceed those limits, the semi-diurnal arch
is 16 be computed by the following rule ; viz..
To the log. tangent of the latitude, add the log. tangent of the decHna-
tion^ and the sum, rejecting lO in the index, will be the log. sine of an
vdi; whieh being converted into time, and added to 6 hoyrs when the
latitude and declination are of Ihe same name ; or subtracted from 6 hours
when these elements are of coiitrary names 3 the sum, or difference, will be
the eemi-diumal arch.
JSTmrnpIe 1.
Let the latitude be 61 degrees, north, and the declination of a celestial
olgectN|^5? 10^, north ; required the corresponding semi-cUumal arch ?
Latitude % « . . . 61? 0^ north, log. tangent 10. 256248
Declination. • . .25.10 north, log. tangent 9.671963
Afch= . . . 57?57f2ir = log. sine. . .9.928211
Arch eeov. mtotime 3^51^49! + 6i s 9tSlU9'., the semrdittr*
Hal arch, as r^^yArefk
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I
124 PBSCRIPTION AKD USB OF THE TABLBS.
Example 2.
Let the latitude be 20?40', south, and the declination of a celestial ob-
ject 80?29', north 5 required the corresponding semi-diurnal arch ?
Latitude. . • . • 20?40: south, log. tangent . .9.576576
Declination ... 30. 29 north, log. tangent . . 9. 769860
Arch = I2?49:45r = log. sine ... 9. 346486
Arch conv. into time 0*51-?19f j and 6t -0?51'?19? = 5*8?4l!,
the semi-diurnal arch.
The present Table has been computed agreeably to the first example ;
but as in most nautical computations, it is not absolutely necessary that
the semi-diurnal arch should be determined to a greater degree of accu-
racy than the nearest minute; the seconds have, therefore, been reject^^
and the nearest minute retained accordingly.
Since the Table for finding the time of the rising or setting of a celes-
tial object (commonly called a Table of semi-diurnal and semi^nocturnal
arcs,) is scarcely applied to any other purpose, by the generality of nau-
tical persons, than that of merely finding the approximate time of the rising
or setting of the sun ; the following problems are, therefore, given for the
purpose of illustrating and simplifying the Aise of this Table ; and of show-
ing how it may be employed in determining the apparent times of the rising
and setting of all the celestial objects whose declinations come within its
limits.
Problbm L
Given the Latitude and the Sun\8 DecJination, to find the Thne of Us
RiHng or Setting.
RULB.
Let the sun's declination, as given in the Nautical Almanac, be reduced
to the meridian of the given place by Table XV., or by Problem L, page
76 5 then,
Enter the Table with this reduced declination at top, or bottotti, and
the latitude in either of the side columns ; under or over the former, and
opposite to the latter, will be found the approximate time of the sun's set-
ting when the latitude and declinatioti are of the same name ; or that of
its rising when they are of contrary names.— The time of setting being
taken from 12 hours will leave the time of rising, and vice versa, the time
of rismg being taken from 12 hours will leav^ that of setting*
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BBSCRIPTION AND USB OF THB TABLES. 125
Note. — ^Proportion must be made, as usual, for the excess of the minutes
of latitude and declination above the next less tabular arguments.
Example 1.
Required the approximate times of the sun's rising and setting July 13,
1824, in latitude 50?48', north, and longitude 120 degrees west ?
Sun's declination July 13th. per Nautical
Almanac, is ,..•..... 21?49C5ir north.
Correction from Table XV., answering to
' var. of dec. StSS^, and long. 120? W. - 2^59^ •
Sun's dec. reduced to given meridian • • 21?46'52r; or 21?47', N. *
Time, in Table L., ans. to lat. 50?, north, and
dec. 21 ?30^, north =...., 7*52?
4' v4fi'
Tabular diflference to 1? of lat. = 4C] now ^ ^^, =+3
3' X 17'
Tab. diflference to 30' of dec. = 3'; now ' a^ • =:+ 2, nearly.
Approximate time of the sun's setting 7^57"
Approximate time of the sun's rising . .. .. 4*3?
Nofe.— Twice the time of the sun's setting will give the length of the
day ; and twice the time of its rising will give the length of the night.
Example 2.
Required the approximate times of the sun's rising and setting October
1st, 1824, in latitude 40?30' nortli, and longitude 105 degrees east ?
Sun's declination October 1st. per Nautical
Almanac, is 3?16C 6T south.
Correction from Table XV., answering to
var. of dec. 23:20r, and long. 105? E. - 6:48r
Sun's dec. reduced to the given meridian 3? 9' 18^, or 3?9' south.
Time in Table L., ans. to lat. 40? north, and
dec. 3? south, is 6* 10?
Tab.diff. to iroflat. = Oi ; now qqT = 0
Tab.diff. to 1? of dec.= 3: 5 now ^^ = 0
Approximate time of the sun's rising €M0?
Approximate time of the sun's setting 5*50?
* Tiie nearcjit minute of declination is sufBcieiitly exact for the purpose of finding the
approuinate thnet of the risUlg and setting of a celestial object.
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126 DBSCaiPTION AVD VHB Of THB TABLIU.
Remark.
Since the times of the sun's rising and settings found as above, will differ
a few minutes from the observed^ or apparent times in consequence of no
notice having been taken of the combined effects of the horizontal refracr
tion and the height of the observer's eye above the level of the sea, by
which the time of rising of i^ celestial object is accelerated, and that of its
setting retarded ; nor of the horizontal par^lax which iiffects these tivnes
in a contrary manner; a correction, (I)^refore, must be applied to tl^e ap-
proximate times of rising find setting, in order to reduce them to the appa-
rent tijDes.-^Tbis correction may be ooioputed by tha Mlowing role; by
which the apparent times of the son's rising and setting will be always
found to within a few seconds of the truth*
Rule.^^To the approximate times of rising and setting, let the longitude,
in time, be applied by addition or subtraction, ac^prding as it is west or
east, and the corresponding times at Greenwich will be obtained : to these
times, respectively, let the sun's declination be reduced by Table XV*, or by
Problem I., page 76 5 then.
Find the sum and the difference of the natural sine of the latitude, and
the noturtd co-pine of the declination (rejecting the two right band figpres
from each term), and take out the common log. answering thereto, reject-,
ing also the two right hand figures from each : — now, to half the sum of
these two logs, add the proportional log. of the sum of the horizontal re-
fraction and the dip of the horizon diminished by the sun's horizontal
parallax, and tha constant log. 1 . 1 76 1 * ; the sum of these three logs., abat-
ing 4 in the index, will be the proportional log. of a correction ; which
being subtracted from the approximate time of rising, and added to that
of setting, the apparent times of die sun's rising and setting will be
obtiiined.
Tht»,«^Let it be required to- reduce the approximate tipies of the
sun's rising and setting, as found in the last Example, to the respective ap-
parent times ; the horizontal refraction being 33^ ; the dip of the horizon
5^ 15?, and the sun's horizontal parallax 9 seconds*
The sun's declination reduced to the .approximate time of ridupg, ia
3?3'37^ and to that of setting 3514:^58? south.
* This is the proportional log. of 12 hours esteemed as mfaiQtcs.
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SMCKIFTION ANS VSM OF THX TABLBS. 127
LaUtudfe . ; 40?30: Or nat. tine . , :? 6494
DecUuation , 3? S:37' nat. co-tine . = 9986
Sum 1648a log. =4.2170
Difference 3492 hg. = 3.5431
Sum 7. 7601
■^^
Half-rams . . . 3.8800|
33^+5n5r-9r=38^6r,prop.log. . . . 0.6743
Constant log. 1.1761
Correction ~ 8^2K|nopJof.Bl.7a04|
Approximate time of rinng = . . . 6M0? 0!
Appwent time of aiQi'a riaing s: . .6^6739:
Latitade . . 40?30: 0? nat. aine . . = 6494
Declination . 3?14'58f nat. co-aine . == 9984
Sum ..,..,. 16478 loy. = 4.2169
Diffnence 3490 log. s 3.^428
Sum, . • . p . 7.7597
Half sum = . • , 3.8798^
33^ +5M5r-9r=38^6r, prop, log. = . 0. 6743
Constant log. . . . . ^ , • • • • 1. 17<Sl
Correction ....... -f St21f prop. log. =: 1.7S02|
Approximate time of setting =: . . 5*50? Of
Apparent time of son's setting =: .5^53721?
Note. — In this method of reducing the approximate to the appaiwH
time of rising or setting, it is inunatcriai whether the latitude and declina-
tion be of the same, or of contrary names :— nor b it of any consequence
whether t)te dedtnation ht reduced to the approximate tfmea of rising and
setting or not, since the declination at noon will be fdways sufteientiy
uact to determJM the correction within two iv three seconds of Um» trutliu
on aoooimt of its natural co-sine h^xtg only required to four places of
wiUan^air evkkot by Arfefriag to tfae ftbovt Man
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128 DESCRIPTION AND USS OP THB TABLES.
although there is a difference of 11 '2K between the reduced declinations
at the approximate times of rising and setting ; yet this difference has no
sensible effect on the correction corresponding to those times.
Problem II.
Given the LatUude of a Place and the Decimation of a fixed Star, to find
the Times of its Rising and Setting.
Rule.
Let the right ascension and declination of the star, as given in Table
XLiV^ be reduced to the given day ; then^ from the right ascension of the
star, increased by 24 hours if necessary, subtract that of the sun, at noon
of the given day ; and the remainder will be the approximate time of the
star's transit, or passage over the meridian 3 from which, let the correction
answering thereto and the daily variation of the sun's declination (Table
XV.,) be subtracted, and the apparent time of the star's' transit will be ob-
tained.
If much accuracy be required, and the place of observation be mider a
meridian different from that of Green\^nch, a correction depending on the
longitude and variation of the sun's right ascension (Table XV.,) must be
applied to the time of transit : — this correction is subtractive in west, and
additive in east longitude ; the time being always reckoned from the pre^
chding noon : now.
Enter Table L., with the declination at top or bottom, and the latitude
in the side column ; and in the angle of meeting will be found the sesri-
diurnal arch, or the time of half the star's continuance above the horizon,
when the latitude and declination are of the same name; but if these ele*
ments are of different names, the time, so found, is to be subtracted from
12 hours, in order to obtain the half continuance above the horizon:
then this half continuance * being applied by subtraction and addition to
the apparent time of transit, will give the approximate times of the star's
rising and setting. . <
Example 1.
At what times will the star a Arietis rise and set January 1st, 1824^ in
latitude 50';48^ north?
* In strictness the semi-diurnal arch, or half continuance above the horizon ought to be
corrected by subtracting therefrom the proportional part (Table XV;,) corjrespondiiif to it
and the rariation of the sun's right ascension for the given day.
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DBSCftlPTION AND US]^ OF THB TA^LBS. 129
Star's dec. on given ^ay is 22?37'33r, or 22?38C north,
and its right ascension 1?57"16!
Sun's right ascension at noon of the given day is • • • 18. 43. 58
Approximate time of star's transit 7* 13. 18
Correction from Tab. XV., ans.* to 7 * Wrl 8 !, and 4 ^ 24^,
the v&r. of the sun's right ascension . — 1.20
Apparent time of star's transit, or passage over the meridian 7 * 1 1 T58 !
Time, in Tab. L. ans. to lat. 50? N., and dec.
22?30'. N.= ..;..... 7^58?
Tabular diff. to V. of lat. = 5:^ now ^^— =+ 4
Tab. diflf. to 30^ of dec. = 4'.-, now ^'3^.^'=+ ^
Semi-diurnal Arch, or time of half the star's
continuance above the horizon • • • =s 8^ 3?. •8?3?0'.
Approx. time of star's tis^ng, past noon of Dec. 31st, 1823 23^ 8T58!
Approx. time of star's setting, past tioon of the given day • 15M4T58!
Example 2.
At what times will the star Sirius rise and set January 1st, 1824,jn lat.
40?30C north, and long. 120 degrees, west of the meridian of Greenwich ?
Star's dec. on given day is 16^28^53: or 16?29^ south,
and its right ascension 6*37T23!
Sun's right ascension at noon of the given day is . • . 18. 43. 58
Approximate time of the star's transit • • • • . • 11.53.25
Corr. from.Table XV., ans. to ll*53r25 f, and 4^24%
the var. of the sun's right ascension <^ 2.11
Corr. from ditto, ans. to long. 120? west, and 4'24T the
var. of the sun's right ascension . ' — 1.28
Appar. time of star's transit over the given meridian . . • 1 1 1 49T46'
Time, in Table L., an^. to lat. 40? north, and
declination 16? S. = ....... . 6*56?
Tab. diff. to 1? of lat. = 2^ ; now ^^^^ = + 1
Tab. diff. to 30i of dec. = 2^ ; now y* s= + 2, nearly.
Semi-nocturnal arch 6*59?,
which being subtracted from 12* leaves 5* 1? 0'.
Approximate time of the star.'s rising • 6*48?46'.
Approximate time of the star's setting •••.... 16*50746!
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130 DBSCElipTION AND X78B OF THB TABLB8.
iZemarft.— 'The approximate times of the rising and setting of a fixed
star may be readily reduced to the respective apparent times by the rule
given for those of the siin^ in page 126 3 omittingi however, the first part,
or that which relates to the reduction of declination : and, since the fixed
stars have no sensible parallax, the words '^horizontal parallax'' are, also,
to be omitted } thus : —
To reduce the approximate times of rising and settings as found in the
last example, to the respective apparent times, the dip of the horizoh being
assumed at 6'30r
Latofplace of observ. 40930' Nat. sine = 6494
Declin. of the star =: 16. 29 Natco-sine=:9589
Sum r: • « • 16083 Log. = . 4. 2064
Difference =: . 3095 Log.=: » 3:4907
Sum= • 7.6971
Half sum =3.84851
Horiz. iPefrac.=:33^ +dip of horiz.=6^30^=39:30r Prop. log.= 0. 6587
Constant log. = ....'• 1.1761
Corrections. ......... SM4r Prop. log. =: 1.6833 J
Now, this correction being subtracted from the approximate time of
rising, and added to that of setting, shows the former to be 6!45*2!, and
the latter 16*54?S0:
Pboblbm III.
Given the Latitude qf a Place, and the DecUnatim qf a Planet, to find
the Times of its Rising and Setting.
Rule. .
Take, from page IV« of the month in the Nautical AlmatiaC, the times
of the i^anet's transits for the days nearest preceding and following the
given day, and find their difference; then say, as 6 days arc to this differ-
ence, so is the interval between the given day and the nearest preceding
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DBSCJtIPTfON AVD USB OF THB TABUS. 131
davj to a correction ; which, being applied by addition or subtraction to
the time of transit on the nearest preceding day, according as it i^ increa^
ing or decreasing, the sum or difference will be the ajJproxiipate time of
transit. Find the interval between the tiiAes of transit on the days nearest
preceding and following the given day ; and then say, as the interval
between the times of transit is to the difference of transit in that interval,
so is the longitude, in time, to a correction ; which, being added to the ap-
proximate time of transit if the longitude be west and the transit increasing,
or subtracted if decreasing, the sum or difference will be the apparent time of
the planet's transit over the meridian of the given place ; but if the longi-
tude be east, a contrary process is to be observed : that is^ the correction b
to be subtracted from the approximate time of transit if the transit be
increasing, but to be added thereto if it be decreasing.
.To the apparent time of transit, thus found, apply the longitude, in time,
by addition or. subtraction, according as it is west or east ; and the sum or
difference will be the corresponding time at Greenwich. To this time, let
the planet's declination be reduced by Problem IIL, page 83; or -as
thus:—
Take, from the Nautical Almanac, the planet's declination for the days
nearest preceding and following the Greenwich time, and find the differ-
ence ; find, also, the difference between the Greenwich time and the nearest
preceding day :.then say, as 6 days are to the difference of declination, so
is the difference between the Greenwich time and the nearest preceding
day, to a correction ; which, being applied to the declination on the nearest
preceding day, by addition or subtraction, according as it may be increas-
ing or decreasing, the sum or difference will be the planet's correct decli«*
nation at the time of its transit over the given meridian. Now,
With the planet's declination and the latitude of the given place,- enter
Table L., and find the corresponding semidiurnal arch^ by Problem II., page
128; and, thence, the approximate times of rising and setting, in the same
manner as if it were a fixed star that was under consideration.
Example 1.
At what times will the planet Jupiter rise and set, January 4th, 1824, in
latitude 36? north, and longitude 135? west of the meridian of Greenwich 7
* In ftrictaesa the semidiurnal arch ought to be corrected by adding thereto^ or sub*
tracting therefrom, the proportional part corresponding to it and the daily yariation of
transit, according as the transit may be increasing or decreasing.
k2
-^
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132 DKSCRTWION AND USE OF THB TABLES,
Timcofpreced.trans. Jan.l,is ll*38?nearestprec.dayl8t,traa8.ll*38T 0!
Time of follow, trans. Jan.7> is 1 1 . 8 given day 4th
• , , ,
As6fi8toO!30T,8ois 3f to - ISr 0?
Approximate time of transit on the given day = •• • • 11*23?0!
Time of preceding transit = • • I'll ^SS?
Time of following transit =. • • 7*11« 8
Interval between the times of trans.=5 f 23 1 30?
As interval between times of trans.=5 f 23*30* ! diff. of
transits 30" :: longitude in time =: 9* to . . . — 1.53
Apparent time |)f transit over given merid. Jan. 4th, 1824 = 11*21? 7'
Longitude 135 degrees west, in time zz 9. 0. 0
Corresponding time at Greenwich =: . . 20*21? 7*
Plahet'sdec.Jan.lis=23?17'N.;near.prec.lf 0* 0?0! dec.23«17.' O^N.
Ditto 7 is=23. 20 N.^Gr. tim.=4. 20- 21 . 7
As efistoO? 3: BO is . . 3f20?2ir7' to + 1.55
Jupiter's dec. reduced to his app. time of transit over the
given meridian = 23?18'55rN,
Time, in Table L., ans. to lat. 36? north, and dec. 23?'N. = 7* 12? 0!
Tabular diflference to 30: of dec. = 2',; now^^'^/^' = . +1.16
Semidiur. arch, or time of half planet's contin. above the hor. = 7^ 13? 16!
Apparent time of Jupiter's transit over the given meridian = 11. 21. 7
Approximate time of Jupiter's rising at the given meridian =: 4t 7?51 !
Approximate time of Jupiter's setting at ditto = «... 18*34?23f
Example 2.
At what times will the planet Mars rise and set, January 16th, 1824,
in latitude 40? north, and longitude 140? east of the meridian of Green-
wich ?
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BESCRIPTION AND USE OF THE TABLES* 133
Timeof prcccd.trans.lSth, i8l6*54?3 near.prec.dayl3th,tran8.= 16?54? 0!
Tiineoffollow,trans.l9th,j8l6.34; given day 16th
As 6f is to 0? 20?, 80 is . . 3f to- . - 10? 0!
Approximate time of transit on the given day =:•••• 16M4? 0!
Interval between the times of transit = 5"? 23*40?
Aa interval between times of transit == 5*23?40! ; diff, of
trans. = 20? :: long, in time = 9*20?, to . . . • + 1. 18
Apparent time of trans, over given merid. Jan. 16th, 1824 =: 16*45?18!
Longitude of the given merid. = HO? east, in time =: • • 9. 20. 0
Corresponding time at Greenwich = 7* 25? 18'
Dcc.ofMars,Jan.l3,i8 0?37'S.;near.prec.l3f 0* 0? 0:dec.0'?37' O^S.
Ditto 19,isl.ll S.;Gr.time=16. 7.25.18
As 6f i8to0?34^8ois . 3f 7*25?18'. to + 18.45
Dec. of Mars reduced to his apparent time of transit over
the given meridian = «•«..• 0?55U5^S.
Semidiurnal arch in Table L.,* answering to lat. 40? N. and
dcc.0?55M5rS.,ia6*2?48!;sub. from 12* leaves • . . 5?S7?12.'
Apparent time of the planet's transit over the given meridian:=16. 45. 18
Approximate time of rising of the planet Mars = . • • . 10^48? 6 .'
Approximate time of setting of ditto = . • . • 22M2?30!
IZemarfc.— The approximate times of a planet's rising and setting may
be reduced to the respective apparent times, by the rule in page 126, for
reducing those of the sun ; omitting, however, the first part, or that which
relates to the reduction of declination, and reading planet's instead of sun's
horizontal parallax : this, it is presumed, does not require to be illustrated
by an example. .
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134 DBSCRIITION AND USB OF THB TABLES.
Paoblem IV.
Given the Latitude of a Place, and the Moon's DediMiion^ to find Hie
Times of her Rising and Setting.
Rule.
Take; from page VI. of the month in the Nautical Almanac^ the moon's
transit, or passage over the meridian of Greenwich, on the given day, and
also her declination. Let the time of transit be reduced to the meridian of
the given place by Table XXXVIII.j to which apply the longitude, in time^
by addition or subtraction, according as it is west or east; and the sum, or
difference, will be the corresponding time at Greenwich : to this time, let
the declination be reduced by Table XVL, or by Problem II.j page 80;«—
then.
With this reduced declination, and the latitude of the given place, find
the moon's semidiurnal arch, or the time of half her continuance above the
horizon, by IVoblem IL, page 128, and, thence, the approximate times of
rising and setting, in the same manner precisely as if it were a fixed star
that was under consideration : call these the estimated times of rising and
setting.
To the estimated limes of rising and setting, thus found, let the longitucfe^
in , time, be applied by addition or subtraction, according as it is west or
east ; and the sum, or difference, will be the corresponding times at Green-
wich.
To these times respectively, let the moon's declinati<Hi be reduced by
Table XVL, or by Problem 11^ page 80; with which, and the Utitttde,
find the moon's semidiurnal arch at each of the estifnated times.
- To the respective semidiurnal arches, thus found, apply the corrections
corresponding thereto, and the retardation of the moon's transit (Table
XXXVI JL) by addition, and the correct semidiurnal arches will be ob*
tained.
Now, the semidiurnal arch answering to the estimated time of rising,
being subtracted from the moon's reduced transit, will leave die approx-
imate time of her rising at the given place ; and that corresponding to the
estimated time of setting, being added to the moon's reduced transit, will
give the approximate time of her setting at the sud place.
Example 1.
Required the times of the moon's rising and setting, Jan. 17th, 1824, in
latitude 51?29' north, and longitude 78?45C west of the meridian pf
Greenwich ?
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MSCftlFTtON.AND VSft OF THB TABLM. 13S
Moon's transit over mend, of Greenwich on the given day 'is 13*34? 0'
Corr. fr. Tab. XXXVIIL^ans. to retard. 53 % and long. 75? west + 10.39
App. time of moon's transit reduced to the given meridian • 13M4T39!
Longitude 78?45^ west, in thne zz ....... . 5. 15. 0
Corresponding time at Greenwich 18t59?39!
Moon's dec. red. to Gr. time, by Table XVI., is 10?25f SO^N.
Semidiurnal arch, in Table L., answering to lat. 51?29'N.,
and declination 1 0? 25 ^N., is ......;,. 6t54? Of
Moon's reduced transit ••••••• 13.44.39
JBrtJmaled time of the moon's rising . • « • .- • • 6t50r39'.
J&tima<€d. time of the moon's setting • 20^38739!
To find the ap(Mroximale Time of Rising :•«-
Estimated time of rising 6^50739!
Loi^tude 78?45^ west, in time = , « 5. 15. 0
Greenwich time past noon of the given day 12t 5?39!
Moon's dec. reduced to Greenwich time, is 12? 10^ 53^N.
Time, inTableL,ans.tolat.51?29'N.anddec. 12?ll'N.,is 7* 3r 0!
ConrectSon,Table.XXXVIU.,An^tQ53:aiid7^3r = » . -h 15. 0
Moon's correct semidiurnal arch at rising . • 7^8? 0!
Moon's reduced transit .>• \ •«•••.•• 13. 44. 39
Approximate time of moon's rising '• 6t26T39!
•To find Utie approximate Time of Setting :*--
EfflMMKed time of setting 20^38r39!
Longitude 78M5< west^ in time s: 5. 15. 0
Greenwichtimepastnoonof the 18th • 1^53^39!
Moon's dec. reduced *to Green^vich time, is 8?41 ' 1 KN. .
Time, in Table L., answ. to lat. 51?29:N. and dec. 8?41 <N., is 6M4r 0!
Correction,TableXXXVlII.,ans. to53C and6*44r = ; . +14.0
Moon's correct sepiidiurpal.arch at setting 6^58T 0'
Mood's reduced transit 13.44.39
Approximate time of moon's setUng ••«•...« 20M2T39:
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136 PB6CR1PTI0N AND USB OF THB TABLBS.
• ' Example 2.
Required the approximate times of the moon's rising and setting, Janu-
ary 20th, 18^24^ in latitude 40^30' north, and longitude 80 degrees east of
the meridian of Greenwich ?
Moon's transit over the merid. of Greenwich on the given day is 16S 6T 0!
Cor. fr. Tab. XXXVIIL, ans. to retard. 49: and long. 80^ east — 10. 32
Moon's transit reduced to the given meridian • • •' . . 1S^55?28!
Longitude 80 degrees east, in time ;= 5.20. 0
Greenwich time 10^35T281
Moon's dec. red. to Green, time, by Table XVI., is 5?55t40rS.
Seminoctumal arch, in Table L., angering to lat. 40?30^N.
and dec. 5?56^S. = 6*20r, subtracted from 12^, leaves 5M0r 0!
^Moon's reduced transit • • • • . . 15.55.28
£9timafed time of the moon's rising 10M5T28!
£^ma<ed time of the moon's setting 21t35?28'
To find the approximate Time of Rising :—
JStffimaf^ time of rising 10M5r28!
Longitude 80 degrees, east, in time == 5. 20. 0
Greenwich time = • . . • 4t55?28!
Moon's dee. reduced to this time, is 4?30'49TS.
Time, in Table L., answering to lat. 40?30'N., and dec.
4?31 CS. is 6t I5r, which, subtracted fwm 12\ leaves . 5U5? 0!
Corr. Table XXXVIIL, answering to 49^ and 5M5r ..+11.0
Moon's correct semidiurnal arch at rising .... . • • 5t56T 0!
Moon's reduced transit 15.55.28
Approximate time of moon's rising 9^59T28!
To find the approximate Time of Setting : —
JSsHmafcd time of setting 21*35?28!
Longitude 80 degs. east, in time = 5.20. 0
Greenwich time = 16*15T28!
Moon's dec. reduced to this time, is 7° 17^52^8,
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]>B8CRtFriON AND VSi OF THS TABLES. 137
Time, in Table L., answering to lat. 40?30^N. and dec.
7?18'S,, is 6i25T, which, subtracled from 12t, leaves . 5*35? 0!
Corr. Table XXXVIIL, ans. to 49C and 5*35? ...*.+ 11. 0
Moon's correct semidiurnal arch at setting 5*46T 0!
Moon's reduced transit • • • •* • • • 15.55.28
Approximate time of moon's setting 21*41?28!
Remark. — ^The approximate times of the moon's rising and setting may
be reduced to the respective apparent times by the following rule ; viz.,
Find the sum and the difference of the natural sine of the latitude and
the natural co-sine of the declination at the estimated times of rising and
setting (rejecting the two right-hand figurcjs from each term), and find the
common log. answering thereto^ rejecting also the two right-hand figures
from each. Now, to half the sum of these two logs, add the constant log.
1. 1761,* and the proportional log. of the difference between the horizontal
parallax and tlie sum of the horizontal refraction and dip of the horizon :
the sum of these three logs., abating 4 in the index, will be the propor-
tional log. oC a correction, which, being added to the approximate time of
rising and gubtracled from that of setting, the respective apparent times of
rising and setting will be obtained : thus,
Let it be required to jeduce the approximate times of rising and settings
as found in the last example, to the respective apparent times, the dip of
the horizon being 4 ^ 50^
Note* — The moon's horizontal parallax computed to the reduced eiti-
fnJated time of rising, is 59 '6?, and that at the reduced time of setting
58U0?
Latitude 40?30C Nat. sine 6494
Declination 4.31 Nat.co-sine9969
Sum ...... 1B463 Log. ..... 4.2165
Difference .... 8475 Log. ... . 3.5410
Sum • ... • 7.7575
Half sum • . 3. 8787i
.59^6f-37'50^ (33C + 4^50r) =: 2ia6^ Prop. log. . . 0.9276
Constant log. 1.1761
Correction + 1^52'/ Prop. log. . . 1.9824^
Approximate time of rising = 9* 59^28 f
Apparent time of moon's rising = 10 1 1 r20'.
^ lliis is the proportional to^* of 12 hours esteemed as minutes.
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ISA OBSCftlPTIOK AND tT6B O^ THB tA&LBft«
Latitude 40?30f Nat. sine 6494
Declination 7^18 Nat. co-sine 9919
Sum •...•• . 16413 Log 4.2152
Difference . • . ; 3425 Log 3.5347
.Sum . • . • 7.7499
Half sum • . 3.8749^
58!40r-37'50r(33: + 4r50r)=:20C50r Prop. log. . . 0.9365
Constant log 1.1761
Correction ~ 1^5K Prop. log. • . 1.98751
Approximale tiqae of setting =: . 2M41'r28!
> ■ •
Apparent time of moon's setting = 21 t39T37 '•
Nofe.— The direct method of solving thb and the three preceding Pro*
blems, by spherical trigonometry, is given in some of the subsequent pages
of this work.
Tables LI. akd LU.
For computing the Meridianal JltUude of a Celestial ObjecL
Since it frequently happens, at sea, that the meridional altitude of thft
sun, or other celestial object, cannot be taken, in consequence of the iater*
position of clouds at the time of its coming to the meridian ; and since it
is of the utmost importance to the mariner to be provided at all tines, with
the means of determining the meridional altitude of the heavenly bodies,
for the purpose of ascertaining the exact position of his ship .with respect to
latitude, these Tables have therefore been carefully computed j by means of
which the meridional altitude of the suq, or any other celestial object whose
declination does n6t exceed 28 degrees, may be very readily obtained to a
sufficient degree of accuracy for all nautical purposes, provided the altitude
be observed within certain intervals of noon, or time of transit, to be
governed by the meridional zenith distance of the object: thus, /or thesun^
the number of minutes ai)d parts of a minute contained in the interval
between the time of observation and nopn, must not exceed the number of
degrees and parts of a degree contained in the object's meridional zenith
distance at the place of observation. And since the'meridional zenith dis-
tance of a celestial object is expressed by the difference between the lati-
tude and the declination when they are of the same name, or by their sum
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DiSCBIPTION AND USB OF THB TABtBS. 139
when of contrary names ; therefore the extent of the interval from noon
(within which the altitude should he observed) maybe determined by
means of the diff(ftrence between the latitude and the declinadon when they
are both north or both south, or by their sum when one is north and the
other south. Thus, if the latitude be 40 degrees, and the declination 8
degrees, both of the same name, the interval between the time of taking
the altitude and noon mast not exceed 32 minutes ; but if they be of differ-
ent names, the altitude may be taken at any time within 48 minutes before
or after noon : if the latitude be 60 degrees, and the declination 10 degrees,
both of tbe same name, the interval between the time of observation and
noon ought not to exceed 50 minutes ; but if one be north and the other
south, the interval may be extended,- if necessary, to 70 minutes before or
after noon, and so on.
The limits within which the altitudes of the other celestial objects should
be observed, may be determined in tbe same manner; taking care, how-
ever, to estimate the interval from the time of transit or passage over thi
meridian, instead of from noon.
Now, if the' altitude of the sun or other celestial object be observed at
am/ time within the limits thus prescribed, and the time of observation
be caieAilIy noted by a well-regulated watch, the meridional altitude of
such object may then be readily determined, to every desirable degree of
accuracy, by the following rule; viz.,
Enter Table LI. or Lll., according as the latitude and the declination
are of the same or of contrary names, and with the latitude in the side
cohunn, and tfie declination (reduced to the meridian of the place of 6b-
serration) at the top or bottom ; take out the corresponding correction in
seconds and thirds, which are to be esteemed as mimUei aitd seconds ; —
then.
To the proportional log. of this correction,* add twice the proportional
log. of the interval between the time of observation and noon, or time of
transit, and the constant log. 7* 3730 ; and the sum urill be the proportional
log. of a correction, which, being added to the true altitude deduced from
observation, will give the correct meridional altitude of the object.
Note ] • — In taking out the numbers from Tables LI. and LIL, proportion
must be made for the excess of the given latitude and declination above
the next less tabular arguments.
* When the object either ooracf to, or wlthio, one decree of the zenith, the anple of
meetiiic nade by the latitude and declination will fall ivithin the aip^ag double lines
which mn through the body of Table LI., and (hroogfi the upper left-hand corner of Table
UI : ia IliU ease, since the intenral between the time of obserfation and noout or meri*
dional paisa^, most not exceed one minute, tbe corresponding number wiU be tbe correc-
tion of tltitude direct, independently of any calculation whatever.
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140 DESCRIPTION AND VSB OF TUB TABLEB.
2.-^The interval between the time of observation and noon may be
always known by means of a chronometer, or any well-regulated watch ;
making proper allowance, however, for the time comprehended under the
change of longitude since the last observation for determining the error of
such watch or chronometer.
Example I.
In latitude 45? north, at 34740! before noon, the sun's true altitude was
found to be 54?12'49'/, when his declination was 10? north; required the
meridional altitude ?
Corr. in Table LL, ans. to lat. 45? and dec. 10?, is 2^23*^. 1 j
the propor. log. of which is » K8778
Interval between time of obs. and noon, 34740!, twice prop. log.= 1.4308
Constant log. « . « 7*2730
Correction of altitude . . 0?47' lOr Prop. log. =: . , 0. 5816
True alt. at time of observ. 54. 12. 49
Sun's meridional altitude 54?59^59!'; which is but one second less
than the truth.
Example 2»
In latitude 48? north, at 1^5748! past noon, the sun's true altitude was
found to be 20?25'5^', w;hen his declination was 20 degs* south} required
the meridional altitude ?
Corr. in Table LIL, answering to lat. 48?N. and dec. 20?S., is
1^19*^.9, the propor log. of which is V. ... , . . . 2.1308
Intend between time of obs. and noon 1*5748!, twice prop. log.=0. 8740
Constant log. ....«««.,••».««* 7- 2730
Correction of altitude • • 1?34!57^ Prop. log. = . 0.2778
True alt. at time of obseiy* 20. 25. 5
Sun's meridionalaltitude . 22? OC 2''} which is but two seconds more
than the truth.
Example 3.
At sea, March 22d, 1824, in latitude 51? 16! north, at 50732'. past
noon, the sun's true altitude was found to be 38?20!567 ; required the
meridional altitude, the declination being 0?43!5 17 north?
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DESCRIPTION AND USB OF THE TABLES. 141
Corr. in Table LI., answering to lat. 5 1 ? 16Und dec. 0?43 '51?
is 1 ^35*'. 6,* the propor. log. of which is 2, 0530
Interval between time of obs. and noon 50T32', twice prop. log.=s 1. 1034
Constant log :....• 7-2730
Correction of altitude • . 1? 6^58r Prop. log. s . . 0. 4294
True alt. at time of observ. 38. 20. 56
Sun's meridional altitude . 39?27'541'; which differs but three seconds
from the truth.
Example 4.
At sea, December 2l8t, 1824, in latitude 60? 22^ north, at 10*36f 10!
A.M., or 1^23T5p' before noon, the sun's true altitude was found to be
4?26C38?; required his meridional altitude, the declination being
23?27'45r south? •
Corr. in Table LII., ans, to lat. 60922C and dec. 23?27U5?,
is OrSS*'. 8,t the propor. log. of which is 2.3026
Interval between time of obs. and noon 1*23?50!, twice prop.log.ssO. 6638
Constant log. 7. 2730
Correction of altitude . . 1 ?43'43r Prop. log. = , • . 0. 2394
True alt. at time of observ. 4 . 26. 38
Sun's meridional altitude . 6?10'2K; which differs but six seconds
from the truth.
After this manner may the meridional altitude of the moon, a planet, or
a fixed star be obtained, when the declination does not exceed the limits of
the Table.
Remarks, ifc^
From the above examples it is manifest, that by means of the present
Tables the meridional altitude of a celestial object may be readily inferred
♦ Corr.tolat.50Oanddec.0o- l"38'^^8
Diff. to 2"* lat. = 6"'.8;now.6«/.8x76'+120' » - 4 .3
Diff. to l** dec. « .1'". 5 ; now, 1'". 5 x 44'+60' = + 1.1
Corr.toUt.50o32'anddec,0<'43'51''» ..... r35'''.6
t Corr. to lat; 60O and dec. 23** « 0*54'". 6
Diff. to 2o lat. » 3'". 5 ; now 3'".5 x 22'+120' « . . -0.6
Diff.tol«dcc.-0"^5JnowO'".5x28'-^ 60^= . . - 0 .2
Corr. to lat 60<»22' and dec. 23*^7'45'' a .... 0"53'".8
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142 rasCAIFTtON AND USB OF THB TABLB9.
from its true altitude oI)8enred at a knowa interval from noon (within the
limits before prescribed), with all the accuracy to be desired in nautical
operations ; and that it is immaterial whether the observation is made before
or after noon, of time of tjrai\sit, provided the tiine be but correctly known;
and, since most sea-going ships are furnished with chronometers, there can
be but very little diiRculty in ascertaining the apparent time to within a
few seconds of the truth.
It is to be observed, however, that the nearer to noon or time of transit
the observation is made, the less susceptible will it be of being affeeted by
any error in the time indicated by the watch : thus, in example 4, where the
interval or time from noon is 1? 23?50?, an error of one minute in that
interval would produce an error of 2^ minutes in the sun^s meridional
altitude; but if the observation had been made within a quarter of an hoiir
of noon, an error offive minuiee in the time would scarcely affect the meri-
dional altitude to the value of 2 minutes : hence it is evident, that although
the observation may be safely made at any time from noon to the fiili extent
of the interval, when dependance can be placed on the time shown by tho
watch, yet when there is any reason to doubt the truth of that time, it wiH
be advisable to take the altitude as near to noon, or the time of transit, as
circumstances may render convenient.
In all narrow seas trending in an easterly or westerly direction, where
the meridional altitude of a celestial object is of the greatest consideration,
such as in the British Channel, the mariner vrill do well to avail himself of
this certain method for its actual determination ; particularly during th»
winter months, when the sun is so very frequently obscured by clouds at the
time of its coming to the meridian.
These Tables were computed by the following rule; viz.,
To the constant log. 0. 978604,* add the log. co-sines of the latitude and
the declination ; the sum, rejecting 20 from the index, will be the log. of a
natural number, which, being subtracted from the natural co«sine of the
difference between the latitude and the declination, when they are of the
same name, or from that of their sum if of contrary names, will leave the
natural co-sine of an arch ; now, the difference between this arch, and the
•difference or sum of the latitude and the declination, according as they are
of the same or of contrary names, will be the change ofaltitude in one
minute from noon.
Example 1.
Let the latitude be 18 degrees, and the declination of a celestial object 2
degrees, both of the same name; required the variation or change ofaltitude
in one minute from noon ?
* This it the log. versed sine, or loff. rblDf, of one miaute of tine.
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Constant log. = , . 0.978604
Latitude =: • 13 degrees. Log. co-sine • 9. 988724
Dfcltnatioii =s 2 degrees. Log. co-«s{ne • 9. 999735
Difference zs . 11 degrees. Nat. co-sine=981627
Nat^umberc 9. 269:=:Log« 0. 967063
Arch = . . 11? 0:i0r =:Nat,co.s.=:981617.731
Difference =! • 0? 0^ \0f ; which^ therefore^ is the change of altitude in
one minute from noon.
Let the latitude be 40 degrees^ and the declination of a celestial object
8 degrees/ of a contrary name to that of the latitude j required the Tariation
or change of altitude in one minute from noon ?
Ckmstant log. =: 0.978604
Latitude = 40 degrees. Log. co<^ne . 9.884254
Declination =: 8 degrees. Log. co-sine . 9. 995753
Sum =: • • 48 degrees. Nat c6-sine=:6691Sl
Nat. num. = 7. 221 Log.=0. 85861 1
Arch =3 • . 48? 0^ 2r=rNat.co-sine=669123.779
Di&renoe =: 0? 0' 2? ; which^ therefore^ is the change of altitude in
one minute from noout
It is to be observed^ however^ that, with the view of introducing every
possible degree of accuracy into the present Tables, the natural and log.
co-sinesy &c., employed in their construction, have had their orcspective
numbers extended to seven places of decimals.
Nofe.«^The difference between the meridional altitude of a celestial
object and its altitude at a given interval. from noon, is found, by actual
observation, to be very nearly proportional to the square of that interval,
under certain limitations, as pointed out in page 138 ; and hence the rule,
in page 139^ for computing the meridional altitude of a celestial object.
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144 description akd use of thb tables.
Table LIIF,
7%6 Miles and Parts of a Mile in a Degree of Longitude at every
Degree of Latitude.
This Table consists of seven compartments : the first column in each
compartment contains the degrees of latitude, and the second column the
miles and parts of a mile in a degree of longitude corresponding thereto.
In taking out the numbers from this Table, proportion is to be made, as
usual, for the minutes of latitude ; this proportion is subtractive from the
miles, &c., answering to the given degree of latitude.
Example*
Required the number of miles contained in a degree of longitude in
latitude 37 ?48'. ?
Miles in a degree of longitude, in latitude 37 degrees s . • • 47* 92
Difference to 1 degree of latitude = . 64 ; now ^^. r^ = — -51
Miles in a degree of long, in latitude 37 degs. 48 min., as required^s 47. 41
Remarks* — Since the difference of longitude between two places on the
earth is measured by an arch of the equator intercepted between the meri-
dians of those places ; and since the meridians gradually approach each
other from the equator to the poles, where they meet, it hence follows that
the number of miles contained in a degree of longitude will decrease in
proportion to the increase of the latitude ; the ratio of decrease being as
radius to the co-sine of the latitude. Now, since a degree of longitude at
the equator contains 60 miles, we have the following rule for computing
the present Table ; viz..
As radius is to the co-sine of the latitude of any given parallel, so is the
measure of a degree of longitude at the equator to the measure of a degree
in the given parallel of latitude.
Example.
Required the number of miles contained in a degree of longitude in the
parallel of latitude 37 degrees ?
As radius . . 90 degrees Log. sine = . . 10.000000
. Is to latitude = 37 degrees Log. co*sine . . 9. 902349
Sow . . .60 miles Log. = . . . 1.778151
To .... 47. 92 miles Log. = . . . 1.680500;
Hence the measure of a degree of longitude in the given parallel of lati-
tude, is 47. 92 miles.
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DSSCRIPTION AND USB OF THB TABLES* l45
Table LIV-
Proportional Miles for constructing Marine or Sea Charts.
In this Table the parallels of latitude are ranged in the upper horizontal
column, beginning at 0?, and numbered 10?, 20?, 30?, &c., to 89? ; the
horizontal column immedi^ltely under the parallels of latitude contains the
number of miles of longitude corresponding to each parallel's distance
from the equator ; under which, in the horizontal column marked " Differ-
ence of the Parallels, &c.," stands the number of miles of longitude con-
tained between the parallel under -which it is placed and that immediately
preceding it.
The left-hand vertical column contains the intermediate or odd degrees
of latitude, from 0? to 10? ; opposite to which, and under the respective
parallels of latitude, will be found the number of miles of longitude corre-
sponding to each degree of latitude in those parallels : these are intended
to facilitate, and render more accurate, the subdivision of the different
parallels of latitude into degrees and minutes.
To rnake a Chart of the. World, in which the Parallels of Latitude and
Longitude are to consist of 10 Degrees each.
Draw a straight, or meridian, line along the right hand, or east margin x>f
the paper intended to receive the projection; bisect that line, and from the
point of bisection draw a straight line perpendicular to the former, which
continue to the left-hand or west margin of the paper,, and it will represent
the equator.
Fi'om any diagonal scale of convenient size take 600 miles in the com^
passes (the number of miles of the equator contained in 10 degrees of
longiitude), and lay it off from the point of bisection along the equator^
and it will graduate it into 36 equal parts of 10 degrees each; through
which let straight lines be drawn at right angles to the equator, and parallel
to that drawn along the right-ht^d margin, and they will represent the
meridians or parallels of longitude. Take, from the same scale^ 60 miles
in the compasses, and it will subdivide each of those 36 divisions, or paral-
lels of longitude, into ten equal parts consisting of one degree each ; and
then will the equator be divided into 360 degrees of 60 miles each.
On the meridian lines drawn along the right and left-hand margins of
the paper, let the parallels of latitude be laid down, as thus : — ^For the first
parallel, or 10 dc^ees from the equator, take 603. 1 miles in the compasses
(found in the horizontal column immediately under the parallels of latitude^
and marked^* Ditto in miles of the Equator^ &c/')j place one foot on the
L
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146 BBSCRIPTION AND USB OF THB TABLBS.
equator, and where the other falla upon the right and left-hand marginal
lines, when turned northward and southward, there make points ; through
which let straight lines be drawn parallel to the equator, and they will
represent the parallels of latitude at 10 degrees north and south of the
equator: in the same manner, for 20 degrees, lay oflF 1225. 1 miles; for
30 degrees^ 1888. 4 miles ; for 40 degrees, 2622. 6 miles, and so on.
But since the common compasses are generally too small for taking off
such high numbers^ it will be found more convenient to lay down the paral-
lels of latitude by the numbers contained in the third horizontal column,
or that marked *^ Difference of the Parallels, &c.'' Thus, for 10 degrees,
take 603. 1 miles in the compasses ; place one foot on the equator^ and
with the otlier make points north and south thereof on the east and west
marginal lines, through which let straight lines be drawn, and they will
represent the parallels of latitude at 10 degrees north and south of the
equator. From these parallels respectively, lay off 622, 0 miles, by pUcing
one foot of the compasses on the respective parallels and the other on the
east and west marginal lines ; through the points thus made by the com*
passes draw straight lines, and they will represent the parallek of latitude-
at 20 degrees north and south of the equator. From the parallels, thus
obtidned, lay off 663. 3 miles, and the parallel of 30 degrees will be deter-
mined: thence lay off 734. 2 miles, and it will show the parallel of 40
degrees ; and so on for the succeeding parallels.
The numbers for subdividing those parallek will be found in the vertical
columns under each respectively, and are to be i4)plied as follows; thus, to
graduate the parallel betwcfen 50 and 60 degrees ; take 94. 3 miles in the
compasses, and lay it off from 50 degrees towards 60 degrees, and it will
give the parallel of 5 1 degrees ; from which lay off 96. 4 miles, and it will
show the parallel of 52 degrees; from this lay off 98.6 miles, and the
parallel of 53 degrees will be obtained ; and so on of the rest. In the
same manner let tiie other parallels of latitude be subdivided ; then let the
parallels of latitude be numbered along the east and west marginal columns^
from the equator towards the poles, according to the number of degrees
contained in that arc of the meridian which is intercepted between th.em
and the equator^ as 10?, 20?, 30?, 40?, &c. &c.; and let the parallels
of longitude be numbered at the top and bottom, and also along the
equator ; these are to be reckoned east and west of the first meridian^ as
10?, 20?, 30?, 40?, &c., to 180?, both ways ; and since the first ineridian is
entirely arbitrary, it may be assumed as passing through any. particular
place on the earth, such as Greenwich Observatory : then will the chart be
ready for receiving the latitudes and longitudes oiF all the4>rincipat places
on the earthy and which are to be placed thereon by the following rule i
via..
Lay a ruler Qver the giv^a loogitud^ found at the top wd botloia of the
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chart, and with a pair of compasses teke the latitude from the east or west
marginal columns ; which being applied to the edge of the ruler, placing
one foot on tlie equator or on the parallel that the latitude was counted
fiom, the other foot turned north or south according to the name of the
latitude, will point out or fall upon the true position of the given latitude
and longitude.
From what has been thus laid down, the maimer of constructing a chart
for any particular place or coast must appear obvious,
iVp/e.— Since this Table is merely an extract from the Table of Meridi-
onal parts, the reader is referred to page 1 13 for the method of computing
the different numbers contained therein.
Tablb.lv.
Tb Jind the Distance of Terrestrial Ohfects at Sea.
If an observer be elevated to any height above the level of the earth or
sea, he can not only discern the distant surrounding objects much plainer
than he could when standing on its surface, but also discover objects
which are still more remote by increasing his elevation. Now, although
the great irregularity of the surface of the land cannot be subjected to any
definite rule for determining the distance at which objects may be seen
from different elevations; yet, at sea, where there is generally an uniform
curvature of the water, on account of the spherical figure of the earth, the
distance at which o|pjects may be seen on its sur&ce may be readily
obtained by means of die present Table ; in which the distance answering
to the height of the eye, or to that of a given remote object, is expressed
in nautical miles and hundredth parts of a mile ; allowance having been
niade for terrestrial refraction, in the ratio of the one-twelfth of the inter-
cepted arch.
No^e.— The distance between two .objects whose heights are given, is
found by pdding together the tabular distances corresponding to those
heights. And, when the given height exceeds the limits of the Table, an
aliquot part thereof is to be taken ; as one fourth, one ninth, or one six-
teenth, &c.; then, the distance corresponding thereto in the Table, being
multiplied by the square root of such aliquot part, viz«, by 2, 3, or 4, &c.,
according as it may be, will give the required distance.
l2
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148 DESCRIPTION AND ^SE OF THE TABLES.
Example I.
The look-out man at the mast-head of a man-of-war, at an elevation of
160 feet above the level of the sea, saw the top of a light-house in the
horizon whose height was known to be 290 feet; required the ship's dis-
tance therefrom ?
The distance answering to 160 feet is . . 14., 57 miles.
Ditto . to'290 feet is . • 19. 62 do.
Required distance = 34. 19 miles;
which, therefore, is the ship's distance from the light-house.
Example 2.
The Peak of Teneriffe is about 15300 feet above the level of the sea; at
what distance can it be seen by an observer at the mast-head of a ship,
supposing his eye to be 170 feet above the level of the water ?
One ninth of 15300 is 1700, answering to.which is 47*50 miles; this
being multiplied by 3 (the square root of one ninth) gives 142. 50 miles.
Distance ans. to 170 feet (height of the eye) is . 15.03 do.
Required distance = 157* 53 miles.
JRetnark 1. — Since the distances given in this Table are expressed in
nautical miles, whereof 60 are contained in one degree, and there being
69. 1 English miles in the same portion of the sphere ; if, therefore, the
distance be required in English miles, it is to be found as follows ; viz..
As 60, is to 69.1; so is the tabular distance to the corresponding distance
in English miles ; which may be reduced to a logarithmic expression^ aa
thus ; —
To the log. of the given tabular distance, add the constant logarithm
0. 061327,* and the sum will be the log. of the given distance in English
miles.
Example.
Let it be required to reduce 157. 53 nautical miles into English miles ?
Given distance in nautical miles = 157*53, log. = 2. 197364
Constant log ,...,... 0.061327
Distance reduced to English miles 181.42 = Log. s 2.258691
• Thelog.of69.1 « 1.839478, less the log. of 60 » 1. 778 IM is 0.061327; which, there-
fore, is the constant logarithm.
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DBSCRIPTION AND V8E OF THE TABLES. 149
The converse of this (that is^ to reduce English miles into nautical miles^)
must appear obvious.
Remark 2. — ^This Table was computed by the following rule 3 viz.,
To the earth's diameter in feet, add the height of the eye above the
level of the sea, and multiply the sum by that height; then, the square root
of the product being divided by 6080 (the number of feet in a nautical
mile), will give the distance at which an object may be seen in the visible
horizon, independent of terrestrial refraction. This rule may be adapted
to logarithms, as Uius : —
Let the earth's diameter in feet be augmented by the height of the eye ;
then, to the log. thereof add the log. of the height of the eye ; from half
the sum of these two logs, subtract the constant log. 3. 783904,* and the
remainder will be the log. of the distance in nautical miles, which is to be
increased by a twelfth part, of itself, on account of the terrestrial refrac-
tion.
Example.
At what distance can an object be seen, in the visible horizon, by an
observer whose eye is elevated 290 feet above the level of the sea ?
Diameter of the earth in feet = 41804400
Height of the eye ..... 290 Log.= 2.462898
Sum= 41804690 Log. = 7.621225
Sum . 10.083623
Halfsum5.0418ni
Constant log. = . . 3.783904
Distance uncorrected by refraction 1 8. 1 1 =Log. =: 1 . 257907J
Add one* 1 2th part on ace. of refr^c. 1.51
Distance as required ss • . • 19. 62 nautical miles.
^o^^.— rFor the principles of this rule, see how the distance of the visible
horizon, expressed by the line O T, i&..determined in page 5.
* This IS the lop of 6080, the number of feetiu a nautical mile.
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Table LVI.
To reduce the French Centesimal Division of the Circle info the EtigUsh
Sexagesimal Division; or, to reduce French Degrees, Ifc, into English
Degrees, jfc, and conversely.
This Table is intended to facilitate the reduction of French degrees of
the circle into English degrees, and conversely. The TVble is divided into
two parts : the first or upper part exhibits the number of English degrees
and parts of a degree contained in any given number of French degrees
and parts of a degree ; and the second or lower part exhibits the number
of French degrees, &c., contained in any given number of English
degrees, &c.
Note, — In the general use of this Table, when any given number of
French degrees exceeds the limits of the Arst part^ take out for 100 degrees
first, and then for as many more as will make up the given number;. and,
when any given number of English degrees exceeds tlie limits of the second
part, take out for 90 degrees first, and then for as many more as will make '
up the given number.
Ejtample 1.
If the distance between the moon and a fixed star^ aeoording to the
French division of the circle, be 128?93'9.6'/, required the distance agree-
ably to the English division of the circle ?
100 French degrees are equal to . . 90? 0' 01 English.
28 Ditto are equal to .. 25. 12. 0 do.
93 French miniates are equal to . . 0. 50. 13 . 20 do.
96 French seconds are equal to . • 0. 0.31 .10 do.
Distance reduced to English degs., as required 116*? 9^4', 30
Example 2.
If the distance between the moon und sun, according to the English
division of the circle, be 116?53'47'', required the distance agreeably to
the French division of the circle ?
90 English degrees are equal to . . 100? 0' Or French.
26 Ditto are equal to . . 28. 88. 88 . 89 do.
53 English minutes are equal to . • 0. 98. 14 . 81 do.
47 English seconds are equal to . . 0. 1.45 .06 do.
Distance reduced to French degs, as required = 129?88C48''. 76
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DBSCRlpnON AND USB OP TUB TABLBS, 151
Remark 1. — ^ITiis Table was computed in conformity with the following
considerations and principles ; viz.^
The French writers on trigonometry have recently adopted the cente-
simal division of the circle^ as originally proposed by our excellent coun-
tryman Mr. Henry Briggs, about the year 1600. In this division, the
circle is divided into 400 equal parts or degrees, and the quadrant into 100
equal parts or degrees; each degree being divided into 100 equal parts or
minutes, and each minute into 100 equal parts or seconds : these degrees,
&c. &c., are written in the usual maimer and with the customary signs, as
thu8 5l28?93:96^
Hence, the French degree is evidently less than the English, in the ratio
of lUO to 90 ; a French minute is less than an English minute, in the ratio
of 100? X lOOUo 90? X 60 f 5 and a French second is less than an Eng-
lish second, in the ratio of 100? X 100'. x lOOr to 90? x 60^ x 60'/ :
now, the converse of this being obvious, we have the following general rule
for converting French degrees into English, and the contrary.
As 100, the number of degrees in the French quadrant, is to 90, the
number of degrees in the English quadrant ; so is any given number of
French degrees to the corresponding number of English degrees.
As 10000, the number of minutes in the French quadrant, is to 5400,
the number of minutes in the English quadrant; so is any given number of
French minutes to the corresponding number of English minutes. And,
As 1000000, the number of seconds in the French quadrant, is to 324000,
the number of seconds in the English quadrant; so is any given number of
French seconds to the corresponding number of English seconds.
English degrees, minutes, and seconds^ are reduced into French by a
cronverse proportion; viz.^
As 90, is to 100 ; so is any ^ven number of English degrees to the corre-
sponding number of French degrees.
As 5400, is to 10000 ; so is any given number of English minutes to the
corresponding number of French minutes. And^
As 324000, is to 1000000 > so is any given number of English seconds to
the corresponding number of French seconds.
Remark 2. — ^French degrees and parts of a degree may be turned into
English, independently of the Table^ by the following rule ; viz..
Let the French degrees be esteemed as a whole number, to which annex
the minutes and seconds as decimals ; then one-tenth of this mixed num-
ber^ deducted from itself, will give the corresponding English degrees^ &€.
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152 BB8CBIPTI0N AND USB OF THE TABUS,
• Example.
The latitude of Paris^ acc6rding to the French division of the quadrant,
is 54?26'36^ north; required the latitude agreeably to the English divi-
sion of the quadrant ?
Given latitude = 54?26:36r=54^ 2636
Deduct one-tenth
. . 5 .42636
English degrees, &c. »
^ .48». 83724
60
•50'. 23440
60
14'. 06400
Hence, die latitude of Paris, reduced to the English division of the qua-
drant, is 48?50: 14r north.
Remark 3. — English degrees and parts of a degree may be turned into
French, independently of the Table, as thus : —
Reduce the English minutes and seconds to the decimal of a degree, and
annex it to the given degrees ; then one-ninth of this mixed number, being
added to itself, will give the corresponding French degrees, &c.
Example.
The latitude of the Royal Observatory at Greenwich is 51°28M0r
north, agreeably to the English division of the quadrant ; required the
latitude according to the French division of the quadrant ?
Given latitude= 51?28U0r = 51°. 4777777, &c.
Add one -ninth 5 .7197530, &c.
French degrees, &c 57^ 1975307 = 57? 19'.75''. 307
Hence, the latitude of Greenwich Observatory, according to the French
division of the quadrant, is 57? 19'75''. 307 N. '
Tabus LVII.
A general Table for Gatigwg, or finding the Content of all Orcular-
headed Casks.
Although this Table may not directly affect the interest of the mariner;
yet, since it cannot fail of being exceedingly useful to officers in charge of
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DBSCAIFTION AKD USB OF THB TABLES. 153
His Majesty's victualling stores (such as Pursers of the Royal Navy^ Lieu-
tenants commanding gun-brigs^ &c. &c.), it has therefore been deemed
advisable to give it a place in this work^ particularly since it may be found
interesting to those whom it immediately concerns.
This Table is divided into two parts : the first part consists of five com-
partments^ and each 'compartment of three columns; the first of which
contains the quotient of the head diameter of a cask divided by the bung
diameter ; the second the corresponding log. adapted to ale gallons ; and
the third the log. for wine gallons. The second part of the Table contains
the bung diameter and its corresponding Ic^rithm. -
The use of this Table will be exemplified in the following
Problem.
Given the Ditnensums of a Cask, to find its Contents in Ale and Wine
GaUons.
Rule.
Divide the head diameter by the bung diameter to two places of deci-
mals in the quotient ; then add together the log. for ale or wine gallons,
corresponding to this quotient, in the first part of the Table ; the log. cor-
responding to the bung diameter, in the second part of the Table, and the
common log. of the length of the cask ; the sum of these three logs., reject-
ing 10 in the index, will be the log. of the true content of the cask, in ale
or wine gallons, according as the content may be required.
Example.
Let the bung diameter of a cask be 25 inches, the head diameter 19. 5
inches, audits length 31 inches; required the contents in ale and wine
gallons ?
25) 19. 50(. 78, quotient of the head diameter divided by the bung
175 diameter.
200
200
. 78 ■= quotient, log. for ale gallons = • . . 7. 862671
25 inches, bung diameter, corresponding log. = 2. 795880
31 inches, length of the cask, common log. = 1. 491362
Content in ale gallons ^ 44. 66 common log, ss 1. 649913
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154 mUCRIPTION AND USB OP THS TA8LB8.
• 78 ts quotient) log. for wine gallons as • • 7* 449340
25 inches, bung diameter, corresponding log. ss 2. 795880
Si inches, length of the caak, common log. ss 1 . 49136%
Content in wine gallons s= 54. 52 common log.s 1. 736582
jRtfmofJk.^— Should the bung diameter liot come within the limits of the
second part of the Table ; that is, should it be under 10 or above 50
inches, the^ twice the common log. corresponding thereto will express the
Ictg, of the said bung diameter, with which proceed as before : hefice, the
rule becomes universal for all circular-headed casks, be the me ever so
great or ever so trivial.
This subject will be revived in a subsequent page of the present work.
Tablb LVIII.
Latitudes and Longitudes of the prindpal Sea-Ports, Islands, Capes,
^c. ^.c, with the Time of High fVaier at the Full and Change of the
Moon at all Places where it is known.
In drawing up this Table, the greatest pains have been taken to render
it not only the most accurate, but also the most, extensive of any now
extant* Perfect accuracy, however, is not to be expected in a Table
which principally depends on the observations made, at different periods^
by the navigators of most civilized nations j because^ in those periods, or
at the time when a very considerable portion of the latitudes and lon^tudea
were established, the nautical instruments and tables employed in their
determination were far from being in that highly-improved state in which
they are found at present : besides, it is a fact well known to the generality
of nautical persons, that if two or more navigators be directed to ascertain
the position of any particular place, they will, in most ease^ difier four or
five miles in the latitude, and perhaps thrice as many in the longitude.
In constructing all the other Tables in this work, there were fixed data
to work uponj with certain means of detecting and exterminating errors ;
but, in this, tiiere were no determinate means of ensuring the desired
degree of accuracy, except in those positions where chancn or profes-
sional duties happened, from time to time, to conduct the author.
Hence, although every possible degree of attention has been paid in con-
sulting the most approved works of the present day, and in colUting this
with the- best modem Tables; yet the mariner must not expect to find
it perfectly free from blemishes ; though, doubtless, he will find it con«
siderably less sq than any with which he may have been hitherto
acquainted.
Since this- Table is aol inteaded for general geog^phkal purposes, the
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DESCRIPTION AND USE OF THE TABLES. 155
positions of places inland, which do not concern the navigator, have, with
one or two exceptions, been ptirposely omitted : hence, the latitudes and
longitudes are limited to maritime places. These are so arranged as to
exhibit to the mariner the whole line of coast along which he may chance
to sail, or on which he may be employed, agreeably to the manner in
which it unfolds to his view on a Mercator's chart. This mode of arrange-
ment is evidently much better adapted to nautical purposes than the
alphabetical mode.
With the view of keeping up the- identity of the Table with the line
of coast laid down on partiQular charts, a^/eip positions have been inserted
a second time. This, it Is presumed, if not conducive to good, will not,
at least, be productive of any evil, since the repetition is so very trivial as
not to embrace, in the whole, more than ten or twelve positions.
The time of high water, at the full and change of the moon, is given '
at all places where it is known. This, it is hoped, will be found not a
little coDvenienti since it does away with the necessity of consulting a
separate Table far that particular purpose*
In order to render this Table. still more complete, an alphabetical
reference has been annexed, which will very essentially contribute towarcU
assisting the mariner in readily finding outmost of th« principal coasts
an4 islands contiuned in that Tablet
The page which immediately follows the alphabetical reference to Table
LVIII. contains' the form of a Transit Table, and the next page a variety
of numbers with their corresponding logarithms, &c., which may, perhaps,
be found useful on many occasions. At the foot of these numbers there
is a small Table, showing the absolute time at which the hour and minute
hands of a well-regulated watch or clock should exactly be in conjunction,
and also in opposition, in every revolution.
Having thus completed the Description and Use of the Tables eontained
in this work, it now remains to show their application to the different
elements connected with the sciences of navigation and nautical astronomy.
In doing this, since the author's design carries him no farther than that
of giving an ample iUustratkm of the various purposes to which they
may be apfdied} the reader must nbt, therefore, expect to find the
dementary part of the seiences treated of. Hence, in this part of the work,
tha anthor will endeavour to confine himself to such Problems and subject
matters as may appear to he most interesting and useful to nautical per*
sons, without entering into particulars or the minutiae of the sciences,
and duis sw€Uing the work to an unnecessary size^; — a thing which he
most anxiously wishes to avoid*
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156
A CONCISE SYSTEM
OF
DECIMAL ARITHMETIC.
Although^ from what has been 6aid in the last paragraph^ it may appear
somewhat irregular, and even contrary to the general tenor of this work,
to introduce any subject therein that does not come immediately under the
cognizance of logarithms ; yet, since the reader may be desirous of having
some little acquaintance with the nature of decimal fractions previously to
his entering on the logarithmical computations, the following concise
system is given for that purpose.-*-It haa been deemed advisable to
touch upon this subject for two cogent reasons ;**first, because a short ac-
count of decimals may be acceptable to the mariner whose early entrance
on a sea life prevents him from going through a regidar course of scholas-
tic education on shore ; and, second, that he may have directly under his
view all tliat is essentially necessary to be known in the practically useful
branches of science, without being under the necessity of consulting any
other author for the purpose of assisting him in the comprehension of the
different subjects contained in this work.
DECIMAL FRACTIONS.
A decimal fraction signifies the artificial manner of setting down and ex-
pressing natural vulgar fractions as if they were whole number8.--'A decimal
fraction has always for its denominator an uait (1,) with as many ciphers
annexed to it as there are places in the numerator ; and it is generally ex-
pressed by setting down the numerator only, with a point before it, on the
left hand ;— thus, ^% is . 5 ; ^-Vir is . 75 j -rgg^ is • 025 ; ^uVolny w .00114,
&c. &c. :--hence the numerator must always consist of as many figures as
there are ciphers in the denominator.
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BBCIMAL FRACTIONS.
157
A mixed number is made up of a whole number and a decimal fraction,
the one being separated from the other by a point i thus 5. 75 is the same
Ciphers on the right hand of decimals do not increase their value; for
• 5 . 50 . 500 • 5000, 8ic,, are decimal fractions of the same value, each being
equal to Z^, or i; — But when ciphers are placed on the left hand of a deci-
mal they decrease its value in a tenfold proportion ; — thus, . 5 is -^ or 5
tenths; but .05 is only -j^ or 5 hundredths; .005 is only -j-^jf or 5
.thousandths, and so on : — ^hence it is evident that in decimals as well as
in whole numbers, the value of the place of the figure increases towards
the left hand, and decreases towards the right, each being in the same
tenfold proportion.
ADDITION OP DECIMALS. .
Addition of decimals is performed in the same way as addition of whole
numbers, obsemng to place the numbers right ; that is, all the decimal
points under each other, units under units, tenths under tenths, hundredths
under hundredths. Sic, ; taking care to point off from the total or sum as
many places for decimals as there are in . the line containing the greatest
number of decimal places.
Example 1.
Add together 41.37; 3.7625
137.03; 409, and .3976.
41.37
3.762
137-03
409;
.3976
591.5596, the sum.
Example 2.
Add together 3.
268;
208
276 J
4.7845, and 1.07.
3.268
208.1
276.
• 4.7845
1.07
493. 2225, the sum.
SUBTRACTION OP DECIMALS.
#
Subtraction of decimals is likewise performed the same way as in whole
numbers; observing to place the numbers right; that is, the decimal
points under each other, units under units, tenths under tenths, hundredths
under hundredths, &c. &c.
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158
DBCIMAL FRACTIONf.
Example 1.
Pfom • • . . 489.7265
Take . . . • 98.283
Remains •
34 K 4435
Example 2.
Prom - . . . 179.087
Take .... 54.932468
Remains . . . 124.104532
MULTIPLICATION OF DECIMALS.
Multiplication of decimals is also performed the same way as in whole
numbers ; observing to cut off as many decimal places in the product as there
are decimal places in both factors ; that is^ in the multiplicand and muU
tiplier.
Example \, Exampk2.
MulUply . . . . 2.4362
By. .... . .275
Product ss
121810
170534
48724
0.6699550
Multiply
By . .
Product ss
. 376.09
. 13.48
112827
150436
112827
37609
5050. 8887
Note. — If a decimal fraction be multiplied by a decimal fraction the pro-
duct will be less than either the multiplicand or the multiplier. — ^And if any
number eiAer whole or mixed^ be multiplied by a decimal fraction^ the pro-
duct will be always less than the multiplicand, as in example 1 } — ^hence if
a decimal fraction be multiplied by itself, its value will decreaee in the pro-
portion of its multiple :-^thus.
Multiply ... .25
By .... .25
Product
125
50
.0625
Multiply ... .75
By 75
Product
375
525
.5625
DIVISION OF DECIMALS.
Division of decimals is performed in the same manner as in whole
numbers } observing to point off as many decimal places in the quo-
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DECIMAL FRACTIONS.
159
tient as the decimal places in the dividend exceed those in the divisor :-<-
But if there be not as many figures in the quotient as there are in that
excess^ the deficiency must be supplied by prefixing ciphers, with a
point before them; — for the decimal places in the divisor and quotient
taken together^ must be always equal to those in the dividend.— When
there happens to be a remainder after* the division ; or when the decimal
places in the divisor are more than those in the dividend^ then ciphers
may be annexed to the latter^ and the quotient carried on as far as may h€
necessary.
Example 1.
Divide .6699550 by .375
Dividend. Quotieni^
IKotfor. 275 \. 6699550 /2.4362
; 550 V
EjttunplB 2.
Divide 5050.8887 by 13.43
Div. 13. 43 \ 5050. 8887 /376. 09
;4029
10218
9401
\
.8178
8058
.12087
120S7
Nofe.— -If a decimal fraction be divided by a decimal firaction^ the quo-
tient will be greater than either the divisor or dividend^ as in Example 1.
Andy if any whole, or mixed number be divided by a decimal fraction, the
quotient will be greater than the dividend ; but if a decimal fraction be
divided by a whole, or mixed number, the quotient will be less than the
dividend, — ^If a decimal fraction be divided by itself, its value will increase
in the proportion of its division, or of the decrease of the parts into which
the decimal is divided ; because, in this case,, the quotient will be a natural
number: — thus, . 25 divided by .25, quotes 1.— -And, .5625, divided by
. 5625, quotes 1 also. Hence it is manifest that the dividing of a decimal
fraction by itsdf increases its value.
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160
DECIMAL FRACTIONS.
REDUCTION OF DECIMALS.
Case L
To reduce a Vulgar Fraction to a Decimal Fraction of equal value.
RULB.
Annex a cipher or ciphers to the numerator ; then divide by the denomi*
nator, as in whole numbers, and the quotient will be the required decimal.
Examples,
Reduce | to a decimal fraction.
4)100
Required dec. s • 25
Reduce f to a decimal fraction.
4)300
Required dec.= . 75
Examples,
Reduce i to a decimal (ra^tion.
2)10
Required dec. = « 5
Reduce { to a decimal ifiraction.
8)5000
Req.dec. == , .625
Casb II.
To reduce Numbers of different Denominations^ such as Degrees^ Time,
Cdny Measure, ^c. into Decimals.
Rule.
Reduce the given degrees, time, coin, measure^ &c. into the lowest de-
nomination mentioned, for a dividend, aimex ciphers thereto, and then
divide by the integer, reduced also into the lowest denomination mentioned ;
the quotient will be the required decimal fraction.
Examples.
Reduce 30 minutes to the deci-
mal of a degree.
The given number being in the
lowest denomination required, an-
nex a cipher and divide by 60, the
number of minutes in a degree;
the quotient will be the required
decimal ; — thus,
60\ 800 / . 5, the Answer.
>/300V
Examples.
Reduce 49^30? to the decimal of
a degree.
The given number being reduced
to the lowest denomination men*
tioned, gives 2970'' ; to this annex .
ciphers, and divide by 3600, .the se-
conds in a degree ; the quotient will
be the required decimal : — thus>
3600\ 2970. 000 / . 825, Answer «
; 28800 \
..9000
7200
18000
18000 '
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I>SC1MAL FRACTIONS.
161
Reduce 15?50f to the decimal
of an hour.
The given terms being reduced
to the lowest denomination give
950 seconds ; annex ciphers and
divide by 3600^ the seconds in an
hour ; as thus,
3600\ 950. 0000/. 2639 nearly
Moo V Ans.
23000
21600
.14000
10800
.32000
32400
Reduce 4? 10T50! to the decimal
of a day.
The given time being reduced to
the lowest denomination mentioned
is 15050 seconds; annex ciphers
and divide by 86400, the seconds
in a day, or 24 hours ; — thus,
86400\ 15050. 00000 /. 17419
) 86400 A nearly Ans.
641000
604800
.382000
345600
. 164000
86400
776000
Reduce 3 '. 4 f to the decimal of a
pound sterling.
The given sum being reduced to
the ioiyest denomination mentioned
gives 40 pence, annex ciphers and
divide by 240, the pence in a
pound sterling ; as thus,
240\40.0000 /. 1666 Answer.
;24W V
1600
1440
.1600
1440
.1600
1440
.160
Reduce 45 minutes to the deci-
mal of an hour.
The given number being in the
lowest denomination mentioned,
annex ciphers and divide by 60,
the minutes in an hour ; as thus,
60\ 45. 00 \ . 75 which is the
)420 J ' Ans.
• 300
300
Reduce 100 fathoms and 2 feet
to the decimal of a nautical mile.
The given measure being reduced
to the lowest denomination men-
tioned is 602 feet ; annex ciphers
and divide by 6080, the number of
feet in a sea mile ; as thus,
6080\ 602. 00000/. 09901 Ans.
; 54720 V
.54800
54720
...8000
6080
1920
' Reduce 3 qrs. 21 lb. to the deci-
mal of a hundred weight.
The given weight being reduced
to the lowest denomination men-
tioned is 105 lbs. annex ciphers, and
divide by 112, the number of
pounds in a himdred weight y as
thus,
112\ 105. 0000/. 9375 Ans.
; 1008 V
.420
336
.840
784
.560
560
M
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Casb III.
lb find the vahke of any Decimal FYaction in the known parts of an
Integer ; such as Degrees, Time, Coin, Weighty Measure^ ic^
Rule.
Multiply the given deeinial by the number of parts Contained in the next
inferior denomination ; and, from the right hand of the product, point oiF
so many figures as the given decimal consists of. — Multiply those figures
so pointed off by the' number of parts contained in the next inferior deno*
mination, and from tRe 'result cut off the decimal places as. before:-—
proceed in this manner till the ieavt known, or required parts of the integer
are brought out ;-^hen, the several denominations on the left hand of the
decimal points, will express the value of the given decimal fraction.
Example 1.
Requbed the vahte of • 825 of a
degree.
Given decimal • 825
Multiply by 60 minutes.
49'. 500
Multiply by 60 seconds.
30-. 000
Hence, the reqmred value is 49 ^ 30'/
Example 3.
Required the vahie of . 166666
of a pound sterling.
Given decimal s • 166666
Multiply by 20 shffl.
S-. 333320
Multiply by 12pence
3*. 999840
Henee^ the reqiured vahie is 3;4f
very nearly.
Example 2.
Reqttred the value of • 2639 of
an hour.
Given deeimal ss • 2639
Multiply by 60 min.
15\8840
Multiply by 60 seconds.
50*. 040
Hence, the required value is 15?
50*. 040.
Example 4.
Beqiiirad Um nlue of . 09901 of
a naatieal or m» mik.
Given deeimal = .09901
Multiply by 60eO,tlMft.
ia a aea mjU, •• ■• -
792080
594060
601.98080
Hence, the required value, is 602
feet veiy nearly.
DBCIMAL VRACnONg. IflB
THE RULE OF PROPORTION IN DECIMALS.
Prepare the terms by reducing the fractional parts to the highest deno-
mination mentioned ; .then state the question and proceed as in the com-
mon Rule of Three Direct 5 — thus, place the numbers in such order that the
first and third may be of the same kind, and the second the same as the
number required :— bring the first and third terms into the same name^
and the second into the highest denomination mentioned. — Then,
Multiply the second and third terms together ; divide the product by the
first term, and the quotient will be the answer in the same denomination
as the second number ;^— observing, however, to point off the decimal
places ; the value of which is to be found by the Rule to Case III.^
page 162.
iVbte.— In the rule of proportion there are always three numbers given
to find a fourth proportional ; two of these are of supposition and one of
demand; the latter must ever -be the third term in the statement of the
question ; and, as this is interrogatory, it may, therefore, be known by the
words — ^What will ? What cost ? How many ? How far ? How much ?,
&c, — ^The jirst term must always be of the same name as jthe third j the
fourth, or term sought, will be of the same kind and denomination as the
second term in the proportion.
Example U
If a degree of longitude, measured on the surface of the earth under the
equator, be 69. 092 English miles ; how many miles are contained in die
earth's circumference under the same parallel, it being divided into 360
degrees ?
As ... 1? : 69-. 092 :: 360?
360
4145520
207276
Answer ... 24873. 120 English miles.
Example 2.
The earth turns round upon its axis in 23t56T ; at what rate per hour
are the inhabitants carried from west to east by this rotation under the
M 2
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.164 ABCIMAL FRACTIONS,
equator where the earth's circumference measures 24873. 12 English miles ;
and at what rate per hour are the inhabitants of London carried in the
same direction, where a degree of longitude measures 42. 99 miles.
First. — For the InhaUtantt at the Etputtor,
23 hours 56 minutes are equal to 23. 9333 hours. — ^Now,
Aa 23*. 9333 : 24873-. 12:: V. : 1039 miles.
24873. 1200
239333
..939820
717999
2218210
2153997
..64213
Second.— i'br the Inhabitants of London.
360 degrees multiplied by 42. 99 miles, give 15476. 4 miles ; — ^And,
As23\9333 : 15476"'. 4 :: 1* : 646 miles.
15476.4000
1435998
> 1116420
957332
. 1590880
1435998
. 154882
Hence, the inhabitants under the equator are carried at the rate of 1039
miles every hour, and those of London 646 miles per hour, by the earth's
motion round its axis.
Example 3.
If a ship sails at the rate of 11^ knots per hour ; in what time would
she circumnavigate the globe, the circumference of which b 24873. 12
miles?
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PROPORTION^ AND PROPRRTtBS OF NUMBERS. 165
11 J knots are equal to 1 1. 25 mile8.^Now,
As 11-'. 25 : 1! :: 24873-M2 : 2210.9 hours.
2250
.2373
2250
.1231
1125
.10620
10125
..495
Hence^ the required time is 2210. 9 hours ; or 92 days^ 2 hours^ and 54
minutes.
PROPORTION, AND PROPERTIES OF NUMBERS.
If three quantities be proportional, the product or rectangle of the two
extremes will be equal to the square of the mean.
If four quantities be proportiotial, the product of the two extremes will
be equal to the rectangle or product of the two means. — ^Thus,
Let 2.4. 8. 16 be the four quantities ; then, the rectangle of the ex-
tremes, vi2. 16 X 2, is equal to the rectangle of the means, viz. 4x8,
or 32.
If the product of any two quantities be equal to the product of two
others, the four quantities may be turned into a proportion by making the
terms of one product the meanSf and the terms of the other product the
extremes.— Thus,
Let the terms of two products be 10 and 6, and 15 and 4, each of which
is equal to 60 ; then, As 10: 4:: 15 : 6. ^4: 6:: 10: 15. As 6: 15::
4:10, &c. &c.
If four quantities be proportional, they "shall also be proportional when
taken inversely and alternately.
If four quantities be proportional, the sum, or difference, of the first and
second will be to the second, as the sum, or difference of the third and
fourth is to the fourth. — ^Thus, let 2.4.8.16 be the four proportioufil
quantities; then
A«2 + 4:4::S+l6:l6} or,a84-2:4::i6-8: 16.
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16C PROPBRTIBS OF NUHBBR8.
If from the sum of any two quantities either quantity be taken, the re^
mainder will be the other quantity.
If the difference of any two quantities be added to the less, the sum will
be the greater quantity ; or if subtracted from the greater, the remainder
will be the less quantity.
If half the difference of any two quantities be added to half their sum,
the total will give the greater quantity 3 or if subtracted, the remainder will
be the less quantity.
If the product of any two quantities be divided by either quantity, the
quotient will be the other quantity.
If the quotient of any two quantities be multiplied by the less, the pro-
duct will be the greater quantity.
The rectangle or product of the sum and difference of any two quan-
tities, is equal to the difference of their Squares. — ^Thus,
Let 4 and 10 be the two quantities j then 4+ 10 = 14 ; 10-4 = 6, and
14x6=84.— Now, lOxlOelOO; 4x4 = 16, and 100-16 = 84.
The difference of the squares of the sum and difference of any two quan-
tities, is equal to four times the rectangle of those quantities. — ^Thus,
Let 10 and 6 be the two quantities; then 10+6 = 16x16 = 256;—
10-6 = 4x4= 16.— Now, 256-16 = 240; and 10x6x4 = 240.
The sum of the squares of the sum and difference of any two quantities,
is equal to twice the sum of their squares. — ^Thus,
10+6= 16x16 =256; and 10-6=4x4=16} then 256+16*272.
Again, 10x10=100; 6x6 = 36, and 100 + 36 = 136x2^272.
If the sum and difference of any two numbers be added together, the
total will be twice the greater number. — ^Thus,
10+6 = 16; and 10.-6 « 4; then 16+4 « 20; and 10x2 = 20,
If the difference of any two numbers be subtracted from their sum, the
remainder will be twice the less number. — ^Thus,
10-6 = 4; and 10+6 =16; then 16-4s 12;-^and6x2 = 12.
The square of the sum of any two numbers is equal to the sum of their
squares, together with twice their rectangle. — Thus,
10+6=16; and 16x16 = 256. Again, 10x10= 100; 6x6 = 36,
and 100+36= 136; then, 10x6x2 = 120; and 120+136 = 256.
The sum, or difference, of any two number9 will measure the sum, or
difference, of the cubes of the same numbers ; that is, the sum will mea-
sure the sum, and the difference the difference.
The difference of any two numbers will measure the difference of the
squares of those numbers.
The sum of any two numbers differing by an unit (1,) is equal to the dif-
ference of the squares of those numbers. — ^Thus,
9+8=17; and9x9 = 81; 8x8=64; now, 81-64 = 17.
If the sum of any two numbers be multiplied by each aunber respect-
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.fLANB TftlGONOMXTBT* 167
ively, the sum of the two rectangles will be equal to the square of the sum
of those numbers.
Thus, 10+6=16; now, 16x10=160; 16x6 = 96; and 160+96
= 256.
Again, 10+6= 16; and 16x 16 = 256.
The square of the sum of any two numbers is equal to four times the
square of half their sum. — ^Thus,
10+6 s 16 ; and 16 X 16 = 256 ; then 10+6 = 16-4*2 = 8, and 8 k 8
x4=2S6.
The sum of the squares of any two numbers is equal to the square of
their difference, together with twice the rectangle of those numbers.—
Thus,
10x10= 100; 6x6x3 36; and 100+36= 136.— Again^
10^6 = 4; and 4x4= 16; 10x6x2 = 120; and 120+16=136.
The numbers 3, 4 and 5, or their multiples 6, 8 and 10, &c. &c., will
express the three sides of a right angled plane triangle.
The sum of any two square numbers whatever, their difference, and
twice the product of their roots, will also express the three sides of a right
angled plane triangle. — ^Thus,
Let 9 and 49 be the two square numbers : — then 9+49 = 58 ;. 49—9 =i
40.— Now, the root of 9 is S, and that of 49 is 7 ;— then 7 x 3 x 2 = 42 i
henee the three sides of the right angled plane triangle will be 58, 40,
and 42.
The sum of the squares of the base and perpendicular of a right angled
plane triangle, is equal to the square of the hypothenuse.
The diference of the squares of the hypothenuse and one leg of a right
angled plane triangle, i» equal .to the square of the other leg.
The rectangle or product of the sum and difference of the hypothenuse
and one leg of a right angled plane triangle, is equal to the square of the
other leg.
The cube of any number divided by 6 will leave the same remainder as
the number itself when divided by 6.—- The difference between any number
and its cube will divide by 6, and leave no remainder.
Any even square number will divide by 4, and leave no remainder; but
an uneven square number divided by 4 will leave 1 for a remainder.
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168 FI.ANB TRIGONOMETRY.
PLANE TRIGONOMETRY.
The Resolution of the different Problems, or Cases^ in Plane Trigonometry,
by Logarithms^
Although it is not the author's intention (as has been already observed^)
to enter into the elementary parts of tlie sciences on which he may have
occasion to touch in elucidating a few of the many important purposres to
which these Tables may be applied 3 yet, since this work may, probably,
fall into the hands of persons not very conversant with trigonometrical
subjects, he therefore thinks it right briefly to set forth such definitions, &c.
as appear to be indispensably necessary towards giving such persons some
little insight into this particular department of science*
Plane Trigonometry is that branch of the mathematics which teaches
how to find the measures of the unknown sides and angles of plane trian-
gles from some that are already known. — It is divided into two parts ; right
angled and oblique angled : — in the former case one of the angles is aright
ungle, or 90? ; in the latter they are all oblique.
Every plane triangle consists of six parts ; viz., three sides and three an*
gles ; any three of which being given (except the three angles), the other
three may be readily found by logarithmical calculation.
In every triangle the greatest sideis opposite to the greatest angle ; and,
vice versa, the greatest angle opposite to the greatest side. — But, equal
sides are subtended by equal angles, and conversely.
The three angles 6f every plane triangle are, together, equal to two right
angles, or 180 degrees.
If one angle of a plane triangle be obtuse, or more than 90?, the other two
are acute, or each less than that quantity : and if one angle be right, or 90?,
the other two taken together, make 90? : — ^hence, if one of the angles of a
right angled triangle be known, the other is found by subtracting the known
one from 90? . — If one angle of any plane triangle be known, the sum of the
other two is found by subtracting that which is given from 180? ; and if two
of the angles be known, the third is found by subtracting their sum
from 180?
The complement of an angle is what it wants of 90? 5 and the supple-^
ment of an angle is what it wants of ISO?
In every right angled triangle, the side subtending the right angle is
called the hypotlieniise ; the lower or horizontal side i& called the 6(we, and
that which stands upright, the peTpendkuIar*
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PtANB TRIGOKOMBTRY.
169
If the hypothenuse be assumed equal to the radius^ the sides^ that is, the
base and the perpendicular, will be the sines of their opposite angles. And,
if either of the sides be considered as the radius, the other side will be the
tangent of its opposite angle, and the hypothenuse the secant of the smne
angle»
Thus. — Let A B C be a right angled plane triangle ; if the hypothenuse
A C be made radius, the side B C will be the sine of the angle A, and
AB the sine of the angle C— If the side AB be made radius, BC will
be the tangent, and A C the secant, of the angle A :— -And, if B C be the
radius, A B will be the tangent, and A C the secant of the angle C.
For, if we make the hypothenuse A C radius
(Fig. l.)> and upon A, as a centre, describe the
arch C D to meet A B produced to D ; then
it is evident that B C is the sine of the arch D .C,
which is the measure of the angle B A C ; and
that A B is the co-sine of the same arch : — and
if the arch A E be described about the centre
C, to meet C B produced to E, then will A B be
the sine of the arch A E, or the sine of the angle
A C B, and B C its co-sine.
Again, with the extent A B as a radius (Fig.-
2.), describe the circle B D ; then B C is the
tangent of the arch B D, which is evidently the
measure of the angle BAC; and AC is the
secant of the same arch, or angle.
Lastly, with CB a(s a radius (Fig. 3.), describe
the arch B D ; then A B is the tangent of the
arch B D, the measure of the angle A C B, and
A C the secant of the same arch or angle.
In the computation of right angled triangles, any side, whether given or
Tcquired, may be made radius to find a side ; but a given side must be
made radius to find an angle : thus,
To find a Side;—
Call any one of the sides of the triangle radius, and write upon it the
word rodtitf .•^^-observe whether the other sides become sines^ tangents, or
secants, and write these words on them accordingly, as in the three pre*
ceding figures : then say, as the name of the given side, is to the given
side} 80 is thename of the side required, to the side required*
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170 PLAVB miQovoumr.
Andj to find an Angle :-r-
Call one of the gweti rides the radius^ and write upon it theirord radius t
observe whether the other sides become sines^ tangents, or secants, and
write these words on then) accordingly, as in the three foregoing figures ;
then say, as the side made radius, is to radius ; so is the other gicen ride
to its name : that is^ to the sine, tangent, or secant by it represented.
Now, since in plane trigonometry the sides of a triangle may be eonsi-
dered, i^thout much impropriety, as being in a direct ratio to the sines of
their opposite angles, and conversely ; the proportion may, therefore, be
stated agreeably to the established principles of the Rule of Three Direct,
by saying
As the name of a given angle, is to its opposite given side $ so is the
name of any other given angle to its opposite side.-— And, as a given side,
is to the name of its opposite given angle ; so is any other given side to the
name of its opposite angle.
The proportion, thus stated, is to be worked by logarithms^ m the fol*
lowing manner ; viz.,
To the arithmetical complement of the first term, add the logs* of the se-
cond and third terms, and the sum (rejecting 20, or 10 from the index,
according as the required term may be a side or an angle,) will be the
logarithm of the required, or fourth 'term.
Rem(n'ks,r^l. The arithmetical complement of a logarithm is what
that logarithm wants of the radius of the Table ^ viz.^ what it is short of
10. 000000 ; and the arithmetical complement of a log. sine, tangent, or
secant, is what such logarithmic sine, &c. &c. wants of twice the radius
of the Tables, viz., 20. 000000.
2. The arithmetical complement of a log. is most readily found by be-
ginning at the left hand and subtracting each figure from 9 except the last
significant one, which is to be taken from 10, as thus ; — ^If the given log;
be 2. 376843, its arithmetical complement will be ?• 623157 : — ^if a given
log. sine be 9. 476284^ its arithmetical complement will be 10. 523716^ and
soon,
3. The arithmetical complement of the log. sine of an arch, is the log.
co-secant of that arch ;-^the arithmetical complement of the log. tangent of
an arch, is the log. co- tangent of that arch; and conversely^ in both
cases* • , :
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PLANS TRIOONOMITBT. t71
Solution of Right-angled Plane Triangles, by Logarithms.
PROBLBItf h
Given the Jnglei and the Hypothenuse, to find the Base and the
Perpendicular.
' ' Example.
Let the hypothcniise A C, of the annexed trian-
gle ABC, be 246.5, and theangle A 58?7M8r |
required the base A fi, and the perpendicular B C ?
Note. — Since there is no more intendedf in this
placcy than merely to show the use efthe Tables;
the geometrical construction of the diagrams is,
therefore, purposely omitted.
By making the hypothennse A C radius ; B C 'becomes the sine of the
angle A9 and A B the co-sine of the same angle, — Hence,
To find the Perpendicular B C :—
As radius =: • . • . 90? =: Log.slne = • • 10. 000000
btokypotbenu0eAC=:246.5 Log, =s , . . « 2.391817
So ia the angle A= 53?7M8r Log. sine zz . . 9. 903090
To the perpendicular B C = 1 97- 2 = Log. = ... 2. 294907
To find the Base A B :—
As radius = . . . ^ 90? = Log. sine = . . 10. 000000
Is to hypotbenuse A C = 246. 5 Log. = ... 2. 391817
So is the angle A = 53?7 ' 481 Log. co-sine =: . 9. 778153
Tfc the base AB = 147.9 =■ . Log. = ... 9. 169970 -
Making the base A B radius ; B C becomes the tangent of the angle A^
and A C the secant c^the same angle. — Henee^
To find tfie Perpendicular B C :—
As the angle A = • 53? 7 • 48? Log. secant Ar. comp.a 9. 778 153
Is to hypothennse A C = 246. 5 Log. =z . • . . . 2.391817
So is the angle A = 53?7'48r Log. tangent =: . . 10. 124937
TotbepcriKsiMtteuUrBC^ 179.2 =Wp.;5 » .- • ?.294907 .
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172 PtANB TRIGONOBfBTRT.
To find the Base A B:—
As the angle A = 53?7'48r Log. secant Ar. compt. = 9. 778153
Is to hypothenuse A C = 246. 5 Log. =:«..•» 2. 391817
So is radius = 90"? Log. sine =: . • • • 10. 000000
To the base AB 147.9=: Log. = • , 4 • « 2.169970
The perpendicular B C being made radius ; the base AB becomes the tan-
gent of the angle C, or co-tangent of the angle A, and the hypothenuse A C
the secant of the angle C^ or co-secant of the angle A.*— -Hence,
To find the Perpendicular B C :
As the angle A == 53?7'48? Log. co-secant .Ar. compt. =: 9. 903090
Is to hypothenuse A C == 246. 5 Log • « 2. 391817
So is radius = 90"? « • • Log. sine • • • • • 10.000000
To the perpendicular B C = 197. 2 = Log. = . . • • 2. 294907
To find the Base A B:—
As the angle A it 53?7'48? Log. co-secant Ar. compt. = 9. 903090
Is to hypothenuse AC zz 246. 5 Log. = 2. 891817
So is the angle A =: 53?7 • 48r Log. co-Ungent . • 9. 875063
Tothe base AB=: 147.9= Log. = 2.169970
Problem II.
Given the Afigles wdOne Side, to find the Hypothetme and the other
&de.'
Example.
Let the base A B of the annexed triangle ABC,
be 300.5, and the angle A 40? 54 MO'/ 5 required
the hypothenuse A C, and the perpendicular B C ?
J CCS
Tlie hypothenuse- A C being made radius; the perpendicular B C mil be
the «ine of the angle A^ and the base A B the co-sine of the same a&gle«
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PLANB TRIGONOMETRY* l??
To find the HypotheDUse A C : —
Aa the angle A = 4O?54U0r Log.co-sine Ar. compt. = 10. 121635
Is to the base AB=: 300.5 Log. = 2.477845
So is radius = 90? Log. sine = . • . . 10. 000000
To the hypothcnuse A C = 397- 6 = Log. = . . • 2. 599480
To find the Perpendicular B C :—
As the angle A = 40?54M0r Log. co-sine Ar. compt. = 10. 121635
Is to the base AB = 300.6 Log. = 2.477845
So is the angle A =: 40?54M0r Log. sine = * , . • 9. 816167
To the perpendicular B C = 260. 4 = Log. = . . . . 2.41 5647
The base A B being made radius ; the perpendicular B C will be the tan-
gent of the angle A, and the hypothenuse A C the secant thereof.— Hence,
To find the Hypothenuse AC: —
As radius = 90? Log. sine = • . • . 10.000000
Is tothe base AB = 300.5 Log. = 2.477845
So is the angle A =: 40?54M0^' Log. secant .... 10. 121635
To the hypothenuse A C = 397. 6 = Log. = .... 2. 599480
To find the Perpendicular B C :—
As radius = 90*? = Log. sine = 10.000000
Is to the base A B = 300. 5 Log • . 2.477845
So is the angle A =r 40?54M0r Log. tangent .... 9. 937802
To the perpendicular B C = 260. 4 r: Log. = ... 2. 415647
Th^ perpendicular B C being made radius ; the base A B will be the tan-
gent of the angle C, or co-tangent of the angle A, and the hypothenuse the
secant of the angle C, or co-secant of A.— Hence,.
To find the Hypothenuse AC : —
As the angle A dt 40? 54 :40r Log. co-tang. Ar. compt. = 9. 937802
Is to the base A B = 300. 5 Log. = 2. 477845
So is the angle A = 40?54M0r Log. co-secant = .. 10. 183833
To the hypothenuse A C = 397. 6 = Log. = • . . • 2. 599480
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PLANB TRIGONOMSTET*
To find the Perpendicular B C :—
As the angle A = 40?54U0r Log. co-tang. Ar. compt. =: 9. 937802
Is to the basi A B ^ 300. 5 Log. = 2. 477845
So is radius =: 90? Log. sine = 10.000000
To the perpcildicular BC = 260. 4 =: Log. =;
2.415647
pROfiLBM IIL
Oicen tlie Hypothenuse and One Side, to find the Angles and the
Other Side.
Example.
Let the hypothenuse A C, of the annexed tri-
angle ABC, be 330. 4, and the base A B 280.3 ;
required the angles A and C^ and the perpendi-
cular B C ?
2Sa.j
By making the hypothenuse A C radios ; the perpendicular B C becomes
the sine of the angle A, and the base A B the co-sine of the same angle, —
Hence^
To find the Angle A:—
As the hypothenuse A C == 330. 4 Log. Ar. compt. = 7. 480960
Is to radius =: 90? Log. sine = 10. OOOOOO
So is the base AB = 280.3 Log. = 2.447623
To the angle A =31?57'56r Log. co-sine =1 . • . 9.92858»
To find the Perpendicular B C.
As radius = 90? Log. sine = , * • 10. OOOOOO
Is to hypothenuse A C =: 330. 4 Log. s= 2. 519040
SoistheangleA=:31?57^56rLog.aines . . . 9.723791
To the perpendicular B C = 174. 9 = Log. = • . . 2. 24283 1
The base A B being made radius ; . the perpendicular B C becomes the
tangent of the angle A> and the hypothenuse A C the secant of that angle.
— Hence^
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VUaa TBIQOMOlIBTaT.
17S
To find the Angle A : —
As the base A B = 280. 3 Log. Ar. compt. = . . • 7. 552377
Is to the radius = 90? Log. sine = 10. 000000
So is the hypothenuse A C >=s 330. 4= Log. :&:... 2.5 19040
TotlieangleA = 31^57?56rLog. secanta . .
• - TofindthePerpeiKiicidarBC:^
As nMlius s= 90? .Lg^. sine »••••.••*
Is to the base AB = 280. 3 Log. =
So is the angle A ^ 31?57 -56^ Log. tangent a .
1*0 the perpendicwlaF B C a 174» 9 a Log. » »
10.071417
10.000000
2.447623
9. 795208
2.242831
jBpmari.^The perpendicular B C may be found independently of the
angles by the following rule (deduced from Euclid, Book I. Prop. 47, and
Book U. Prop. 5^9 viz..
To the log. of 'the sum of the hypothenuse and giren side, add the log.
0f Atk diffevenee ; then, half the sum of these two li>gs. will be the log. of
the required side : — as thus ;
Hypothenuse
Base . . ,
AC =
AB =
rBC =
330.4
280.3
Difference . ,
610.7 Log. .
50. i Log. .
Sum .
174.9 = Log.
2.785828
1.699S38
4.485666
Perpendicula
2.242833
PUOBLBM IV.
Giioen the Base and the Perpendicular^ to find the Jngles and the
Hypothenuse,
Example.
Let the base AB, of the annexed triangle ABC,
be 262. 5, and the perpendicular B C 210. 4 ; re*
quired the aogleSi and the hypothenuse A C 2
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176
FLANB TRIGONOMSTRT..
By making the base A B radius ; the perpendicular B C becomes the
tangent of the angle A, and the hypothenuse A C the secant thereof.
—Hence,
To find the Angle A : —
As the base A B = 262. 5 Log, Ar. compt. = .... 7. 58087 1
Is to radius =: 90? Log. sine ae 10. 000000
So is the perpendicular B C = 210. 4 Log. s • . . • 2. 323046
To the angle A s38?42M7i: Log. tangents « . « 9.903917
To find the Hypothenuse A C :—
As radius s 90^. Log. sine s . . . . ^ . . . 10.000000 '
Is to the base AB = 262.5 Log. = 2.419129
Sobth«angleA=x.38?42M7^ Log. secant = . . . 10.107745
To theliypothenuse A C = 336. 4 = Log. = . • . 2. 526874
The perpendicular B C being made radius ; the base A B will be the tan-
gent of the angle C, or co-tangent of the angle A, and the hypothenuse A C
will be the secant of C, or the co-secant of the angle A. — ^Hence,
To find the Angle A :—
As the perpendicular B C = 210. 4 Log. Ar. compt. s= • 7* 676954
Is to radius = 90? Log. sinfe = . . . . . . . .• 10.000000
Soistheba8eABs=262.5Log. = 2.419129
To the angle A = 38?42:47? = Log. co-tangent = . 10. 096083
To find the Hypothenuse A C :—
As radius = 90? Log. sine = lO.OOOOOO
Is to the perpendicular B C =: 210. 4 Log. = . . . 2. 323046
So is the angle A = 38?42U7^ Log. co-secant = . . 10. 203828
To the hypothenuse A C =; 336. 4 = Log. = . . . 2. 526874
The angle A subtracted from 90? leaves the angle C ; thus 90? — 38?
42C47^ = 51?17'. W^ the measure of the angle C.
Bcmarfc.— The hypothenuse A C may be found independently of the an-
gles by the following rule, deduced principally from Euclid, Book I*
Prop. 47 } Book U. Prop. 5 -, and Book VL Prop. 8^ viz..
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TLAKB TRtGONOMBTRY.
177
From twice the log. of the base subtract the log. of the perpendicular^
and add the corresponding natural number to the perpendicular ; then^ to
the log. of this sum add the log. of the perpendicular^ and half the dum of
these two logs, will be the log. of the hypothenuse. As thus : —
Base A B = . . • 262. 5 twice the log. == 4. 838258
PerpendicutarBC= 210.4 Log. . . = 2.323046 • . 2.323046
Natural number zz 327* 5 Log. •
Sum • • • . • 537. 9 Log. s •
Hypothenuse A C =: 336. 4 Log. = .
= 2.515212
. . . 2.730702
Sum = 5.053748
. .. 2.526874
Solution of Oblique-angled Plane Triangles by Logarithms.
Problem L
Given ike Angles and One Side qfan Oblique^angkd Plane TYiangle, to
find the other Sides.
RULB.
' As the Log, sine of any given angle^ is to its opposite given side ; so is
the log. sine of aiiy other given abgle to its opposite side*
Example.
Let the side A B, of the triangle ABC,
be 300.2, the angle A 39?39^^0r the
angle C 90?33C26r and, hence, the angle
B 49?47'14-
BC.
to find the sides A C and
Ci^fi. z
TofindtheSide AC:— .
As the angle C =: 90?33'.26: Log. sine ar. compt. = 10. 000021
IstothesideBC = 300.2Log 2.477411
So is the angle B =.49?47 ' Ur Log. sine .... 9. 882895
TothesideAC =; 229,3=; Log. r= ...... 2.360327
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178 PLAMS TRlGOMOMBTRt.
To find the Side BCr-
As the angle C = 90*33 '. 26f Log. wn* ar. compt. = U). 000021
Is tothe8ideBC = 300.2 Log. = 2.477411
So is the angle A = 39?39; 20r tog. dne = . ... . 9. 804937
To the8ideBC= 191.6 = Log. = 2.282369
Note.— When a log. sine, or log. co-sme, is the first term in the propor-
tion, the arithmetical complement thereof may be taken directly from the
Table of secants Iqr using a log. co-secant in the foimer case, and a log.
secant in the latter.
Probi^bm II*
Givea two Sidet and an Angle apposUe tooneqf ikem, tojind the other
Angles and the third Side.
Ruui.
As any given side of a triangle is to the log. sine of its opposite given
angle, so is any other given side to the log. sine of the angle opposite
thereto.
The angles being thos foond, the tlurd side-is to be computed by the
preceding Problem.
Example. *
Let the side A B, of the triangle A B C, be 4^. 7$
the side AC 684. 5, and the angle B 100?7C36? ; re*
quired the angles A and C, and the side B C ?
^^f7
TofiDdtfaeane^C:— »
As the side AC s . 684. 5 Log. ar. comp. 7* 164626
IstotheangleA= 100?7^35? Log.sine=: 9.993181
SoisthesideABs 436.7 Log. = . . 2.640183
TQtheaDgleCs38?54:22; Log«tt . . 9.797990
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PIAMV TmiOONOMXTEY. 179
To find the side B C :—
As the angle B = . 100?7'35'/ Log. sine ar. comp, = 10. 006819
Is to the side A C = 684. 5 Log. = . . . • 2. 835374
So is the angle A == 40'?58:3^ Iiog. sine = . . . 9. 816659
To the side B C = 455. 9 ;= Log. = .... 2. 658852
JVbte.-.The angle A = 100?7J35r + the angle C = 38^54^22^ =
1399l^57r;andl80? - 139?1^57^ = the angle A = 40?58C3r
Remark. — ^An angle found by this rule is ambiguous when the given
side opposite to the given angle is less than the other given side ; that is^
the angle opposite to the greater side may be either acute or obtuse : for
trigonometry only g^ves the sine of an angle, which sine may either repre-
sent the measure of the angle itself, or of its supplement to 180 degrees.
But when the given side opposite to the given angle is greater than the
other given side, then the angle opposite to that (other given) side is always
acute, as in the above example.
Problsm IIL
Given two Sides and the inchded Angle, to find the other Angles amd the
Hwrd Side,
Rule.
Find the sum and difference of the two given sides j subtract the given
angle from 180? ; take half the remainder, and it will be half the sum of
the unknown angles i then say, '
As the sum of the sides is to their difference ; so is the log. tangent of
half the sum' of the unknown angles, to the log. tangent of half their
difierence.
Now, half the difference of the angles, thus found, added to half their
sum^ gives the greater angle, or that which- is opposite to the greater side;
and being subtracted, leaves the angle opposite to the less side.
llie angles being thus determined^ the third side is to be computed by
Problem I., page 177*
C
Example.
Let the side A B, of the triangle AB C, be
210. 3, the side B C 160. 2, and the angle B
110?!' 20^; required the angles A and C, and a
the side AC? 2^0. J
N 2
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180
PLAKB TRIGONOMSTRT.
180? - the angle B 110?i:20r = 69?58M0r h- 2 = 34?59^20r =
half the sum of the angles A and C«
Side AB =:
Side BC =
210.3
160.2
As sum = 370. 5 LiOg ar. comp. =: 7* 431212
Is to difference = . . . 50.1 Log. = • . 1.699838
So is i sum of angles = 34?59C 20i: Log. tang. =: 9. 845048
Tojdiflfer. of angles = 5?24^24r Log. tang. = 8.976098
Angle C = . .
Angle A = • .
. 40?28M4^
. 29?34:56-
As the angle A =:
Is to the side B C =
So is the angle B =
To find the side AC:
. 29^34^56^ Log. sine lar. comp. =: 10.306561
. 160.2 Log.= .... 2.204663
. 110?lC20r Log. sine = i . • 9.972925
To the side A C =: . 304. 9 = Log. =
2.484149
Problbm IV.
Gken tlie three Sides of a Plane TViangle, to find the Angles.
IluLB.
Add the three sides' togiether, and take half their sum ) the difference,
between which and the side opposite to the required angle call the remain^
der; then,
To the arithmetical complements of the logs, of the other two sides, add
the logs, of the half sum and of the remainder: half the sum of these four
logs, will be the log. co-sine of' an archj which, being doubled, will give
the required angle.
Now, one angle being thus found, either of the other two angles miiy'be
computed by ftoblem II., page 178.
Example. •
Let the side A B, of the triangle ABC,
be 260. 1, the side AC 190. 5, and the side
B C 140. 4 J required the angles A, B, and C? ^ .
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8PHBRICAL TRlGONOIiETRY. 181
The side A B = 260. 1 '
V BC = 140.4 Log. ar. comp. . • 7.852633.
AC = 190.5 Log. ar. comp, . . 7.720105
• « ■ ■■■ «
. Sum z= . • . 591.0
Half sum = . . 295.5 Log. = . • . . 2.470558
Remainder =: • 35.4 Log. == .... 1.549003
Sum = 19.592299
Arch = . . 5H17'22r Log.co^sine = . . 9.796149§
Angle C = . l02?34U4r
To find the angle B :-^
As the side AB = 260.1 Log. ar. comp. = 7.584860
Is to the angle C =102^34 '.44? (jog. sine = . . 9.989448
. 'SoisthesidcrAC =2 190.5 Log. =: ... 2.279895
To the angle B= 45?37'45r Log. = . . . 9.854203
Now, angle C 102?34M4? + angle B 45?37M5r = 146n2^29r;
and 180? - 148?12^29r = 3I?47-'3ir = the angle A.
THE RESOLUTION OP THE DIFFERENT PROBLEMS, OR
CASES, IN SPHERICAL TRIGONOMETRY, BY LOGARITHMS.
Spherical Trigonometry is that branch of the mathematics which shows
bow to find the measures of the unknown sides and angles of spherical
triangles from some that are already known. It is divided into three
parts ; viz., right-angled, quadrantal, and oblique«> angled.
A right-angled spherical triangle has one right angle; the sides in-
cluding the right atigle are called legs, and tt\at opposite thereto the hypo-
thenose.
A quadrantal spherical triangle has one side equal to 90?, or the fourth
part of a circle.
An oblique-angled spherical triangle has neither a side nor an angle equal
to 90?
. A spherical triangle is formed by the intersection of three great circles
on the surface of the sphere.
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182 8PUBRICAI. TRIGONOMBTRY.
The three angles of a spherical triangle are always more than two, but
less than six, right angles.
The three sides of a spherical triangle are always less than two semi-
eircles, or 360?
Any two sides of a spherical triangle, taken together, are greater than the
third.
The greater side subtends the greater angle ; the lesser side the lesser
angle, and conversely.
Equal sides subtend equal angles, and, vice versay equal angles are sub-
tended by equal sides.
The two sides or two angles of a spherical triangle, when compared
together, are said to be alike, or of the same affection, when both are less
or both greater than 90? ; but wh^n one is greater and the other less than
90?, they are said to be unlike, or of different affections.
Every side of a right-angled spherical triangle exceeding 90?, is greater
than the hypothenus^ ; but every side less than that quantity, is less than
the hypothenuse.
The hypothenuse is less than a quadrant, if the legs be of the same
affection ; but greater than a quadrant, if they be of different affections:
The hypothenuse is, also, less or greater than a quadraQt, according as
the adjacent angles are of the same or of different affections.
When the hypothenuse and one leg, or its opposite angle, aie of the
same or of different affections, the other side, or its opp6site a^gle, will
be, accordingly, less or greater than a quadrant.
The legs and their opposite angles are always of the same affection.
The sides of a spherical triangle may be changed into angles, and con-
versely.
Every spherical triangle consists of six parts : viz», three rides and three
angles i of which, if any three be given, the remaining three may be readily
computed 3 but in right-angled spherical triangles, it is sufficient that two
only be given^ because the right angle is always known*
SOLUTION OF RIGHT-ANGLED SPHERICAL TRUNGLES, BY
LOGARITHMS, AGREfiABLY TO LORD NAPIER'S RULES*
In every right-angled spherical triangle there are five . circular partly
exclusive of the right angle, which is not taken into consideration, lliese
five parts consist of the two legs, or sides ; the complement of the hypfh*
thewuae; and the cimplemenU ^f the two angles. T^iey are caUed circular
parts^ because each of them is measured by the arc of a great circlet
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IPHB&JOAL TRIOONOMBTRY* 188
Hiree of these circular parts, besides the radius, enter into every propor-
tion ; two of which are given, and the third required. One is called the
Tttiddleparty and the other two the extremes conjunct or disjunct.
The middle part, and also the extremes conjunct or disjuncty may be
determined by the foUowIng rules.
Mule I.-— When the three circular parts under consideration are joined
together^ or follow each other in successive order, the middle one is termed
Uie middk part, and the other two the extremes conjunct, because they
are directly conjoined thereto.
IhJe 2.-^When the three circular parts do not join, or follow each
other in successive order, that which stands alone, or disjoined from the other
two, is termed the middle part, and the other two the extremes disjunct,
because they are separated or disjoined therefrom by the intervention of a
tfide^ or an angle not concerned in the proportion*
Ab/«,— In determining the middte part, it is to be observed,^ that the
right angle does not separate yr disjoin the legs : therefore, when these
^e under consideration, they are always' to follow each other in succession.
These things befng premised, the required parts are to be competed by
the two following equations } viz.,
1st. — The product of radius and the sine of the middle part, is equal to
the product of the tangents of the extremes conjunct >
2d/-»TA^ product if radius and the sine qfthe middle part, is equal to
the product of the co^sHmes of the extremes disjunct.
Since these equations are adapted to the complements of the hypothec
nvse and angles, and since the sine or thf tangent of the complement of
an arch is represented directly by the co-sine or co-tangent of that arch,-—
therefore, to save the trouble of finding the complements,, let a co-sine or
co-tangent be used instead of a sine or tangent, and a sine instead o^ a
co-sine, &c. &c., wheft the angles or the hypothenuse are in question.
Now, the middle part being determined by the rules 1 or 2, as above,
according as the extremes are conjunct or disjunct, the terms under con-
sideration are then to be reduced to a proportion, as thus :— Put the
unknown or required term last, that with which it is connected^r^/, and
the remaining two in the middle, in any order; this being done, the equa*
tion will then be ready for a direct solution by logarithmical numbers*
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1 84 sraBRlCAL TRICONOMBTIIT*
Problem L
Given the Hypothevme and one Leg, to find the Angles and the
other Leg.
Example.
Let the hypothenuse A B, of the spherical triangle ABC,
be 64^20M5r, and the leg AC 51 ? 10^57 ; required the'
migles A and B, and the leg B C ?
To find the angle A :— •
Here the hypothenuse A B, the given leg A C, and the required angle A^
are the three circular parts which enter the proportion ; and since the angle
A evidently connects the hypothenuse and the given leg^ it is ther^ore
the middle parf, and the other two the eSDtremes conjuncty according to
rule 1, page 183 ; therefore, by equation 1, page 183,
Radius x co-sine of angle A = tangent of A C x co-tangent of A'B.
Now, since radius is connected with the required term, it is ta be the
first term in the proportion. Henc^,
As radius = , . • . . 90? 0^ 0? Liog. sine ar.comp.=:l 0.000000
Is to the leg A C = « « 5U 10. 13 Log. tangent = ^ 10. 094280
So is the hypothenuse AB ^ 64. 20. 45 Log. co-tangent = 9. 681497
To the angle A =: ^ . * 63*2 li 50: Log. co-sine = ^ 9.775777
2Vb^e.-*-The angle A Is acute, because the hypothenuse and the given
leg are both of the same affection.
To find the angle B :— .
The three circular parts which enter the proportion, in this case, are the
hypothenuse AB, the given leg AC, and' the required angle B; and since
the leg A C is disjoined from the other two parts by the angle A, it is
therefore the tniddfe party and the' other two the extremes disjunct, ac-
cording to rule 2, page 183 ; therefore, by equation 2, page 183,
Radios X sine leg AC =^ sine hyp. AB x sine of angle B.
Now, since the hypothenuse is connected with the required term, it is to
stand first in the proportion. Hence,
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SPHBBI€4L TRIGONOMBTRY. 185
As the hypothenuse A B = 64?20' 45 r Log. sine ar. comp. = 1 0. 04507 1
Is to radius == ... 90. 0. 0 Log. sine = . . 10.000000
SoistheIegAC= . . 51.10.15 Log. sine =r . . 9.891548
To the angle B= . . 59? 47. 34 r Log. sine = . . 9.93661^.
^o<€.— The angle B is acute, because the hypothenuse and the giren leg
are of the same affection.
To find the leg B C :—
In this case the three circular parts which enter the proportion, are the
hypothenuse and the two legs ; and since the hypothenuse is disjoined from
the legs by the angles A and B, it is the middle part, and the other two are
the extremes digunct ; therefore,
Radius x co-sine hyp. A B = co*sine leg A C x co-sine leg B C.
Now,* the leg A C, being connected with the required term, is, therefore,
to stand first in the proportion. . Hence,
As the leg AC =: . . 51?10'15r Log. co-sine ar.comp.= 10. 202732
Is ta radius = . . . 90. 0. 0 ' Log. sine = . . . 10.000000
So is hypothenuse A B = 64. 20. 45 Log. co-sine =r .. . 9. 836426
To the leg B C = . . 46? 19^ 52^/ ^og. co-sine = . . 9. 839158
Note.-^The leg B C is acute, because the hypothenuse and the given leg
are of the same affection.
Problem II.
Given the Hypothenuse and one Angle, to find the other Angle and
the tfvo Legs.
Example*
Let the hypothenuse AB, of the spherical triangle
ABC, be jS6?44^35r, and the angle A 61*59^55^5 ^;
required the angle B and the legs A C and B C ? B
To find the angle B ;—
Here the three circular parts are connected or joined together; there-
fore the hypothenuse A B is the middle part^ and the angles A and B
extremes co^Junct (rule 1, page 183} j therefore, by equation 1, page 183,
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186 mnuLiCAL tiigoiioiibtey.
Radius x co-nne byp< A B ±: co-tangeht angle A x co-tangeikt angle B«
Now^ the angle A, being connected with the required part, ie therefore
to stand first in the proportion. Hence^
As the angle A =: . 61?59C55r Log. co-tang. ar. comp. =: 10. 27430O
Is to radius = . . 90. 0. 0 Log. sine = .... lO.OOOOOO
Sols the hyp. ABs: 66.44.35 Log. co-sine . s . . 9.596438
To the angle B = . 53^24^ 12r Log. co-tangent =: . . 9. 870738
Ao^e.— The angle B is acute^ because the hypothcnuse and the given
angle are of the same affection.
To find the leg A Cl-
in this case^ the three circular parts are joined together ; therefore the
angle A Is the nmddle part^ and the bypothenuse A B and required leg A C
are the esr(remef cofytmd; therefore,
Radius x co-sine of angle A := co-tangent AB x tangent AC*.
And since the bypothenuse is connected with the required part^ it is
therefore to be the first term in the proportion. Hence,
As the hyp. AB s 66?44:35r Log. eo-tang. ar. comp. = 10.366756
Is to radius = . . 90. 0. 0 Log. sines .... lO.OOOOOO
So is the angle. A s 61.59.55. Log. cp-sitte . . « . 9.671629
To the leg AC 8 . 47?3U42r Log. tangent s . . » la 038885
Note,, — ^The leg A C is acute, because the bypothenuse and the given
angle are of the same affection. ,
To find the leg BC:—
In this case the leg B C is the middle part, because it stands alone, or
is disjoined firoift the other two circular parts concerned, by the angle B :
hence the hypothenuse' AB and the given angle A are extremes disjunct,
according to rule 2, page 183 ; therefore, by equation 2, page 183,
Radius x sine of leg B C == sine of hyp. A B x sine of angle A.
And since radius is connected with the required part, it is to be the first
term in the proportion. Hence, '
As radius =s ... 90? Of Or Log. sinear. ebmp. s= lO.OOOOOO
IstohypothenuseABs 66.44.35 Log. sine s . '. • 9.963194
So is the angle A <= . 61 . 59. 55 Log. sine = ... 9. 945929
To the leg BC a . 54n2:45r Log. sine a . . . 9.909123
No(«.— -The leg BC is acute, because the bypothenuse and the given
angle are of the same affection.
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SPHJiRtCAL. TAtGOKOMBTKY* 187
Problem IIL
Given a Leg and its opposite Angle, to find the other Angle, the other
Legj mid the Hypothenuse.
Example.
Let the leg AC, of the spherical triangle ABC, be
56?30M0^ and the angle B 70?23:35r; required the angle
A, the leg B C, and the hypothenuse A B ?
. To find the angle A :*-•
Here the three circular parts which enter the proportion, are the given
angle B, the given leg A C, and the required angle A ; and since the angle
B is disjoined from the other two parts by the intervention of the
hypothenuse ' AB, itT is the middle part, and tlie other two are the
extremes disjunct, according to rule 2, page 183 ; therefore, by equation 2,
page 183,
Radius X co-stne of the angle B s sine of the* angle A x co^sine of
the leg AC.
And since A C is connected with the required part, it is to be the first
tern in tfie proportion. Hence,
As the leg A C s 56?30(.40r Log. co-sine ar. comp. a 10. 258238
la to radiuses . SO. 0. 0 Log. sine = .... 10.000000
8obtlieamteBaB70«2d.85 Log«co«sine • . . • 9.525778
To the angle A= { i32!32!37 ' } ^** ^^^^ = . . . . 9. 784016
^oto.— The angle A is ambiguous, since it cannot be determined, from
the parts given, whether it is acute or obtuse.
TofindtheiegBC:—
Hie three circular parts concerned in this case, are the legs A C and B C,
and the given angle A ; and since the right angle never separates the legs,
B C is the middle part, and A C and the angle B are the extremes conjunct,
by rule 1, page 183) therefore, by equation !» page 183,
Radius X sine, of the leg B s tangent leg AC x co-tangent angle B*
Now^ since radius is connected with the required term^ it is to stand first
io the pmportaen. Hence,
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As radius == • . 90? 0^ 0? Log. sine ar. coup. = 10.000000
Is to the leg A C = 56. 30. 40 Log. tangent = . . 10. 179400
So is the angle B s 70. 23. 35 Log. co-tangent ^ . 9. 55 17 19
To the leg B C = { iJy.' 25,' 27 ' } ^8- "^« = • • • 9- 731 1 19
No^e.— «The leg B C is ambiguous^ since it cannot be determined^ from
the parts given, whether it is acute or obtuse.
To find the hypothenuse A B :—
Here the given leg A C is the middle part, because it is disjoined from
the other two circular parts concerned^ by the intervention of the angle A:
hence the angle B and the hypothenuse AB are e^emes dujjunct ; diere-
fore,
"Radius x sine of leg A C = sine of hyp. AB x sine of imgle B.
And since the angle B is connected with the required term> it is to stand
first in the proportion. Hence,
As the angle B = • . 70?23 '33 r Log. sine ar. comp, = 10. 02594 1
Is to the leg A C = ^ . 56. 30. 40 Log. sipe s * • • 9. 92 1 162
So is radius = . • . *90. 0. 0 Log. sine a • • . 10.000000
To the hyp.AB = Ij^^'J^'^* jLog.si^^ • , . 9.947103
Noto.— The hypothenuse AB is amlnguous; that is, it may be either
acute or obtuse, from Jtl^e parts given.
Paoblbm IV.
Given a Leg and its adjacent Angle, to find the other ybigle, the other
Leg, a^id the Hypothenuse.
Example. Pv^^
Let the leg AC, of the' spherical triangle ABC, be / ^. >^
68?29M5r, and the angfe A 74?45a5r 5 required the . * ' '^ ^'^
angle B, the leg B C, and the hypothenuse AB ?
SPHERICAL TRIGONOMBTRT. ' 189
To find the Angle B :—
Here the circular parts concerned are, the leg A C, the given angle A,
and the required angle B; and since the angle B is disjoined from the
other two parts by the hypothenuse A B, it is the middle part, and the
other two are the estremei disjunct, by rule 2, page 183; therefore/ by
equation 2, page 183,
Radius X co-sine angle B = sine of angle A x co-sine leg A C.
Now, since radius is connected with the cfquired term, it is to stand
first in the proportion. Hence,
Asradms= . . 90? 0' Or Log, sine ar. comp. = lO.OOOOO*
. Is to the angle A = 74. 45. 15 Log. sine = . . . 9. 984440
So is the.leg A C » 68. 29. 45 Log. co-sine = • • 9. 564156
Tothe-angleB= 69?17'.17^ Log. co-sine = . . 9.548596
JNb^.— The angle B is «cute, or of the same affection with its opposite
given leg A C.
' To fin4 the Leg B C :—
III .this case, since the right af^gle never separates the legs, the three
^circular parts are joined together : hence the leg A C is the middle part,
and the leg B C and .the jangle A arQ the extremes conjunct, according to
rule 1, page 183 ; therefore, by equation 1, page 183,
Radius x sine of leg A C = co-tangent, angle A X tangent of leg B C.
And fiince the angle A is connected with the required part, it is to be the
first teem in the proportion. Hence,
As the angle A =s 74?45' 15? Log. co-tang. bx. comp. =. 10. 564549
Id to radius = . . 90. 0. 0 Log. sine = .... 10.000000
So is the leg A C = 68. 29. 45 Log. sin^ = . . . . 9. 968666
TothelegBC« 73 ?40^20ir Log. tangent =s . . . 10.533215
Note* — ^Thei^leg B C is acute, or of the same affection with its opposite
given angle A. .
To find the Hypothenuse A B :—
In this case, since the three circular parts which enter the proportion
^e joined together^ the given angle A is the middle part, and the leg A C
and the hypothenuse A B are the extremes conjunct : therefore,
Radius X co-sine of angle A = Ungent of leg A€ x co-tangent.hypo-
then'useAB,
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190 $?BBaiCAL TaiGQVOMnftT*
Now, the leg AC, being connected with the required part, Is therefore
to be the first term in the proportion. Hence,
As the leg A C = . . 68?29'45r Log, tang, ar, comp. = 9. 595490
Is to radius = ... 90. 0, 0 Log. sine = . . • 10.000000
So is the angle A = . 74. 45. 15 Log. co-sine = . • 9. 419891
To the hypothenuse A Cs 84? 5i 6r Log. co-tangent a 9. 015381
Note.'^^The hypothentise v acute, because the given leg and angle are
of the same affection.
. Peoblbm V.
Gwen the two Legiy to find the Aagle$ and the H^fothemue.
. A-
Example.
Let the 1^ A C, of the spherical triangle A B C, be
70?10:20r, and the legBC 76?38M0r; required the
angles A and Bi and the faypoth^nuse AB ?
To find the Angle A :—
tiere, since the right angle woer 9eparat€9 the legs, the l^g A C is the
middle part, and the leg B C and the jequired angle A are the e&itemeg
conjunct, agreeably to rule 1, page 183 ; therefore, by equation 1, page 183,
Radius X sine leg A C s tangent leg B C x co-tangent angle A.
Now, since the leg B C is connected with die required part, it is to be
the .first term in the proportion. Hence,
AsthelegBC » 76?88140r Log. tangent ar. comp. « 9.375506
Is to radius =s • 90. 0. 0^ Log. sine s= • • . . . 10. 000000
So is the leg A C «70. 10. 20 Log. sine =....! 9. 973459
To the angle A == 77?24'37^ Log. co-tangent = . . 9.348965
JVb/e.— The angle A is acute^ or of the same affection with its oppoaite
given legBC,
To find the Angle B :—
In this case the leg BC is the middle part, and th^ leg AC and the
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8PMRICAL TRI430NOMBfF&T. 191
required angle B are the extremes conjunct, according to rule 1, page 183 ;
therefore, by equation 1, page 183,
Radius x sine of the leg B C s= tangent of leg A C x co-tangent
angle B.
And since the leg A C is connected with the required part, it is to be the
first term in the proportion* Hence^
As the leg A C = 70? 10^ 20r Log. tangent an comp. = 9. 556990
Is to radius = • . 90. 0. 0 Log. sine == • • • • 10.000000
So is the leg B C =: 76. 38^40 Log. sine = . « . . 9. 988093
To the angle B =: 70?40! Si': Log. co-tangent =: . • 9. 545083
No/e.— The angle B Is aeute^ or of the same a&ction with its opposite
pvnlegAC.
To find the Hypothenus^ A B :-^
Here the hypotheniise AB is the middle part, because it Is disjoined
from the legs by the angles A and B: hence AC and BC are exiremes
digjuncty agreeably to rule 2, pagq 183 ; therefore, by equation 2, page 183>
Radius X co-sine hypotbanuae A9 =: co-sine kg AC x co-sitie leg BC.
4Qd radius^ being connected with the middle par^ b therefore to be th»
fijst term in the proportion. Hra^e^
As radius = . . 90? Ot 0? Log. sine ar. comp. i=: • . 10.000000
Is to the leg AC =: 70. 10. 20 Log. co-sine = . . • 9. 530448
So b the leg B C = 76. 38. 40 Log. co-sine == ... 9. 363599
Tothehyp.AB = 85?30^22? Log. co-sine ;= • . . 8.894047
Hoiee-^Tht hypothenuse AB b acute^ because the given legs AC and
B C are of the same affection.
Peoblsm VI.
Gwenthe tu>o Jngles, to find the Hypothenuse and ilie two Legs.
ExamplSm
Let the angle A, of the qpherical triangle ABC, be
50? 10r20% and the angle B 64?20^25f ; r«|mred Ibe lege
ACandBC, and the hypothenuse A B ? H
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192 SPHERICAL TRIGONOMBTRY.
To find the Hypothenuse AB :— •
Here, because the three circular parts are joined together^ the hypothe-
nuse AB is the middle part, and the angles A and B are the extremes
coigunct, agreeably to rule 1, page 183 5 therefore, by equation 1, page 183^
Radius x co-sine hypothenuse A B 1= co-tangent angle A x co-tangent
angle B.
Now, since radius is connected with the required part, it is to be the
^ first term in the proportion, Hence^ .
As radius = . . • 90? O: Or Log, sine ar.' comp. *= 10.000000
Is to the angle A s= 50. 10. 20 Log. co-tangent = . 9. 921 161
So is the angle B = 64.20.25 Log. co-tangent = . 9.681605
To the hyp. AB = 66*?22'.52r Log. co-sine = . . 9.602766
Note,, — The hypothenuse A B is acute, because the given angles A and
C are of the same aiSfection.
To find the leg AC:—
f
Here, since the angle B is disjoined by the hypothenuse A B from the
dther two circular parts concerned, it is the middle part ^ and the angle A
and the required leg A C are the extremes disjtincty agreeably to rule' 2,
page 183 5 therefore, by equation 2, page 183,
Radius x corsine angle B = sine of angle A X po-sine of leg AC.
And because the angle A is connected with the required part, i< Is to
stand first in the proportion. . Hence, . •
As the angle A = Stt? 10' 20^ Log. sine ar. comp. = . 10. 1 14654
Is to radius = . 90. 0. 0 Log. sine == .... lO.OOOOOO
So is the angle B = 64. 20. 25 Log. co-sine .... 9. 6365 14
TothelegAC = S5M0'38r Log. co-sine =: . . . 9.751168
^ofe.-^The leg A C is acute, or .of (he same affection with its opposite
given angle B.
To find the Leg B C :—
In this case the angle A is -the middle part, because it is disjoined from
the other two circular parts by the hypothenuse A B i hence .the angle B
and the required leg B C are extremes disjwictj therefore,.
Radius x co-sine of angle A = sine of angle B x co-sine of leg B C.
And as the angle B is connected with the required part, it is to be the
first term in the proportion. Hence, * . *
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SPHSmCAL TRt GONOIIBTRY. 1 93
As the angle B = 64?20^45r Log. sine ar. comp* = . 10. 045091
Is to radius = . 90. 0. 0 Log, sine = .... 10.000000
So is the angle A = 50. 10. 20 Log. co-sine 9. 806507
TothelegBCzz 44?43niir Log. co-sine = 9.851598
Note. — ^The leg B C is acute^ or of the same affection with its opposite
given angle A. -
SOLUTION OP QUADRANTAL SPHERICAL TRIANGLES,
BY LOGARITHMS.
Problem I.
Given a Quadrantal Side, its opposite jingle^ and an adjacent Angle, to
find the renuwiing Angle and the other two Sides,
' IZemarlr.— Since the sides of a spherical ' triangle may be turned into
angles, and, vice versa, the angles into sides, all the cases of quadrantal
spherical triangles may^ be re<K>lved agreeably to the principles of right-
angled spherical triangles ; as thus : let the quadrantal side be esteemed
the radius ; the supplement of the angle subtending that side, the hypo-
thenuse ; and the other angles legs, or the legs angles, as the case may be.
Then the middle part, and the extremes conjunct or disjunct, being esta-
blished, the required parts are to be computed, and the affections of the
angles and sides determined, in the same manner precisely as if it were a
right-angled spherical trijangle that was under consideration.
Example.
Let AB, in the spherical triangle ABC, be the qua-
drantal side = 90?, the angle C 120?19^30^, and the
angle A 47?d0^ 20r ; required the sides A C and B C^ and
the angle B?
Solution. — Let the supplement of the angle C (59M0'30^), subtending
the quadrantal side A B, represent the hypothenuse a 6 of the dotted
spherical triangle abc. Let the given angle A 47^30' 20T represent the
leg i6 c of the said dotted triangle, and the required angle B the leg a c.
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ThePi in the rigfat-uigled spherical triangle a be, given the ^
hypothennac ab 59°40C30r, and the leg i5c 47?30120^, to |/\
find the 1^ a c = the angle B in the quadrantal triangle ; the ^/ \
angle « ■= the leg B C, and the angle i = the leg A C, of the ^ <^ \ ^
said quadranul triangle. '^'Si'^^-
To find the Leg ac == the Angle B in the Quadrantal Triangle :~
Here the hypothenuse ab is the middle party and the legs &c and ac
are the extremes diejunct -, therefore^
Radius x co-sine hyp. abzz co-sine leg & c x co-sine legac.
Now» lince the leg 6 c is connected with the required part, it is to be the
first term in the proportion. Hence^
As the leg 6 c = 47?30' 20r Log. co-sine ar. comp. = . 10. 170363
Is to radius =: • 90. 0. 0 Log. sine =: • • . . 10. 000000
So u the hyp. a 6= 59. 40. 30 Log. co-sine = , , . . 9. 703209
Tothelegac ;;; 41?97'54? Log. co-sine r: . . • . 9.873572
Noie.'^^Tht leg a c is acute, because thfc hypothenuse and the given leg
are of the same dfcction : hence the angle B (in the quadrantal triangle),
represented by the leg ac, is also acute » 4l?37'54'
To find the Angle a = th^ Leg B C -in tb^ Quadrantal Triangle ^^
Here the leg b c is the middle party and the hypothenuse a b and angle a
are the extremes di^unct ; therefore,
Radius x sine of leg be^ sine of hypothenuse ab x sine of the
angle a.
And since the hypothenuse is connected with the required part, it is to
be the first term in the proportion. Hence,
As the hyp. a c = . 59?40f SOr Log. slne.ar. comp. = 10. 068901
Is to radius = . . 90. 0. 0 Log. sine = ... 10.000000
So is the leg J c = „ 47. 80. 20 Log. sine = . . . 9. 867670
To the angle a = • 58940C26r Log. sine = ... 9.931571
JNb^e.— The angle a is acute, becduse the hypothenuse and the gi?en
leg are of the same afiection : hence the leg BC (of the qu^dn^ntal
Uiangle)^ represented by th« an^le a, is also acut« ^ 53?40 C 26r
SPHBEICAt TRIOONOMBTEY* 195
To find the angle b = the Leg A C in the Quadrantal Triangle :~
In this case the angle b is the middle part, and the hypothenuse ab and
the leg b c are the exir^es conjunct ; therefore,
Radius x co-sine of the angle b = co-tangent hypothenuse ab x tan-
gent of leg b c.
And radius^ being connected with the required part| is, therefore, to
stand first in the proportion. Hence,
As radius = , . . . 90? 0' OT Log« sine ar. comp. = 10. 000000
Is to the hyp. a i = ,59. 40. 30 Log. co-tangent =: 9. 7671 10
So is the leg ic= . . 47.30,20 Log. tangent ;;: • 10.038032
To the angle A - , . 50?19'.19f Log«co*Mne =: . t 9.805142
iVbto.-— llie angle b is acute, because the hypothenuse and the given leg
are of the same affliction. Hence, tke leg A C (of the quadrantal triangle),
repreMnted by the angle £, is also acute s 50? 19' 19?.
Problem II.
Qxom the QuadraaiUal &de and the other two Sides, to find the three
Angles.
Example.
Let AB, in the spherical triangle ABC, be
the quadrantal side = 90? ; the side A C,
U5?19M$f; and the side BC, I17?39:35r:
required the angles A, B, and C ?
jSb2ti/ton.— Let the angle c, iii the dotted
spherical triangle a i c, be radius, and represent
the side AB = 90? of the quadrantal triangle
ABC. Let the angle a, of the dotted .triangle,
represent the side B C of the quadrantal triangle
r= I17?39'45r, and let the angle £ represent
the side A C of the said quadrantal triangle ==
115?19M5r« Then, in the right-angled sphe-
rical triangle a be, right-angled at c, given the angle a =: 117?39^35^,
and the angle £ = 115?19'45^, to find the hypothenuse a i, the leg ac,
and the leg i c; the first of which represents tht supplement of the angle
o2
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196 sraBRlCAL TRIGOKOBfBTRT.
C opposite to the quadrantal side AB^ in the triangle ABC; the second
represents the an^le B ; and the third the angle A, in the said quadrantal
triangk»
To find the Hypothenuse a i = the Supplement of the Angle C^
subtending the Quadrantal Side A B :— >
Here the hypothenuse a£ is the middle part, and the given angles a and
h are the extremes conjunct ; therefore,
Radius X co-sine hypothenuse ab = co-tangent of angle a x co-tan-
gent of angle 6.-^Now, since radius is connected with the required part, it
is to be the first term in the proportion. — Hence,
As radius = . . . 90? OC Or Log. sine ar. coropt. = 10.000000
Is to the angle a =* 117.39.35 Log. co-tangent = 9.719427
So is the angle 6 = 115. 19. 45 Log. co-tangent = ; 9. 675 156
To the hypo. ai= 75?38nir Log. co-sine = . . . 9.394583
JNb<e.— The hypothenuse ai is acute because the given angles are of the
same affection :— but since it only represents the supplement of the angle C ;
therefore the angle C is obtuse, or 104'?21'49r.
To find th^e Leg ac = the Angle B in the Quadrantal Triangle.
The angle b, in this case, is the middle part, . and the angle a and leg a c
extremes disjunctj^Thevefore, radius x co-sine of angle b = sin^s of angle
a X co-sine of leg ac.
And the angle a being connected with the required part, is, therefore, to
be the first term in the proportion. — Hence,
As the.angle a = 117''^l9^35r Log. sme ar. compt. = 10.052703
Is to radius =. . 90. 0. 0 Log. sine = . . . 10.000000
So is the angle is 115.19.45 Log. eo-sine = . . 9. 63 1 259
To the side a c S3 118952^57^ Log. co-sine = . . 9.683962
Note. — The side ac is obtuse, or of the same affection with its opposite
angle b : — ^and since a c represents the angle B ; therefore the angle B, in
the quadrantal triangle, is obtuse, or 1 18?52'57r.
To find the Leg £ c = the Angle A in the Quadrantal Triangle.
In this case the angle a is the middle part^ and the angle b and leg i c
extremes dw/tmct.— ^Therefore, radius x co-sine of the angle a = sine of
the angle b X' co-sine of the leg b c«
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SPHERICAL TRIGONOMETEY. 197
And since the angle b is connected with the required part, it is to be the
first term in the proportion.— Hence,
As theangle b = 115?19M5r Log. sine ar. compt. = 10.043896
Is to radius = . . 90. 0. 0 Log. sine s . . • . 10. 000000
So is the angle a = 11 7. 39. 35 Log. co-sine = . . . 9. 666723
To the leg i c := 120^54^ \2t Log. co-sine = . . . 9. 710619
Vote. — The leg i c is obtuse, or of the same affection with its opposite
angle a: — and' since the leg b c represents the angle A, in the quadrantal
triangle; therefore the angle A is obtuse, or 120?54^ 12?.
jRtfmorJIr.— From the ample solutions of the two preceding Problems, it
must appear obvious, that all the cases of quadrantal spherical triangles
may be easily resolved by the equations for right-angled spherical triangles.
And if the analogies of those two Problems be well understood, all the ap-
parent difficulty attending the trigonometrical solution of quadrantal trian-
gles will entirely vanish.
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRUNGLES
BY LOGARITHMS.
The most natural, and, perhaps, th^ easiest method of solving the four
first ProblemSy or cases of oblique-angled spherical triangles, is by means
of a perpendicular let fall from an angle to its opposite side, continued if
necessary ; and thus reducing the oblique into two right-angled spherical
triangles.— The perpendicular, however, should be let fall in such a manner
that two of the given parts in the oblique triangle may remain known in
one of the right-angled triangles : — ^Then, the other parts may be readily
computed by means of Lord Napier's analogies, as given in the equations 1
and 2, page 183. — ^But, since the solution of oblique-angled spherical trian-
gles without a perpendicular is possessed of many advantages in astrono-
mical calculations ; and, besides, since the author's object is to establish
the use of the Tables contained in this work by a variety of rules and for-
mulae which, it is hoped, may not be found quite uninstructive to persons
but slightly informed on trigonometrical subjects ; the different cases of ob-
lique triangles will, therefore^ be resolved independently of a perpendicular,
agreeably to the propositions generally used in such cases.
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108 SPHXRICAL TRIGONOMirrET.
Problem I.
ISiven Ttjoo Sides of an Oblique-angled Spherical Triangle, and an jingle
opposite to one of them; to find the remaining Angles and the Third
S&te.
RUUE.
l.^Tbfind on angle opposite to one of the gioen sides.
As the log. sine of the side opposite to the given angle, is to the log. sine
of the given angle ; so is the log. sine of the other given side^ to the
log. sine of its opposite angle.
Now, to know whether the angle thus found is determinate ; that ie,'
whether it is ambiguous, acute, or obtuse, proceed in the following man-
ner, viz.<^To the angle so found, and its supplement, add the given angle^
or tl)at used in the prop6rtion. — ^Then, if each of these sums be of the same
affeciim with respect to 180? as the stun of the two given sides^ or those
used in the proportion, the angle is amAigtiottf ; that is, it may be either
acute or obtuse ; and, therefore, indeterminate.*^But, if those sums are of
different affections with respect to the sum of the sides, the angle is deter-
minate, and, therefore, not ambiguous : — ^In this case that value of the
angle is to be taken, whether acute or obtuse, which, when added to the
given angle, produces a quantity of the same affection with the sum of the
two sides.
2.-^Tofind the angle contained between the two given sides.
Find half the difference, and half the sum of the two given sides :--find^
also, half the difference of their opposite angles. Then say,
As the log* sine of half the difference of the sides, is to the log. sine of
half their sum ; so is the log, tangent of half the difference of their oppo«
site angles, to the log. co-tangent of half the angle contained between the
two given sides ; the double of which will be the angle sought.
3.— To find the third side.
Since the sides are proportional to the sines of their opposite angles |
therefore the third side may be found by the converse of the first part of
the rule ; as thus :
As the log. sine of a given angle opposite to a given side, is to the log,
sine of that side ; so is the log. sine of the given angle opposite to the re-
quired side, to the log. sine of the required side«
Note. — ^When the angle comes out ambiguous, or indeterminate, in the
first proportion ; the contained angle and the third side, found by the other
proportions, ^^11 also be ambiguous.
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8P»fia(CAt tniQoKoittrHY. 199
Example.
In the oblique-angled spherical triangle A B C^ let
the side A B be 74?59^50r, the side B C 68?10:30r
and the angle A 63?58'.32f ; required the angles B
and C^ and the side AC?
To find the Angle C :—
As the side BC = 68?10f30r Log. sine ar. compt. 2= 10.032301
Is to the angle A a 63.58.32 Log. sine . • . = 9.953570
SoisthesideABs 74.59.50 Log. sine . . , s 9.98493^
To the angle C« 69? 13^37^ Log. sine . . , = 9.970809
To determine whether the Angle C is Ambiguous, Acute, or Obtuse : —
Angle Ctt69? 13^37: Sap. » 1 10^46^ 23? SideBC» 68?]0C30r
AngleA S3 63.58.32 AngUA » 68.58.33 SideAB» 74.59.60
Sum = 133?12'. 9r Sum = 174?44;55r Sums I43?l0r20r
Here, since .the three sums are of the same affection with respect to 180?
the angle C is ambiguous ; therefore it may be either 69? 13 '37? or the
supplement thereof; viz., 110? 461 23r.
To find the Angle B '.-^
Astheside AB^thesideBC-H2=:3?24U0?Log.S. ar.compt. 11.325483
IsittheS.AB+tbeS.BC-H2=='71.35.10 Log.siness . . 9.977174
SoistheaDg.C-theang.A-t-2a=2.37.32^ Log. tangent = 8.661426
To half the angle B m • * • 5^?49C 22? Log. co-tangent 9. 864089
Angle B =5 . • « , • • 107.38.44} Which is amb^ous because
the angle C came out indeterminate.
TofindtheSide AC:—
As the angle A s • « 63?58'.32? Log. sine ar« compt. 10.046430
Is to the side BC s: .68. 10.30. Log. sine » • • . » 9.967699
So is the angle B s . 107. 38. 44. Log. sine s «... 9. 979070
To the side A C=: , 100? 6'47T Log. sine = . . , , 9.993199
The side A C is also ambiguom because the angle C came out indeter-
minatet
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200 SPHB^ICAL TRIGONOMBTRY, .
Peoblem II.
Gioen Ttoo Angles of an Oblique Angled Spherical Triangle, and a Sde
opposite to one of them ; tojind tlw remaining Angle and the other Two
Sides.
Rule.
\.— Tojind a side opposite to one of the given angles.
A« the log. sine of the angle opposite to the given side, is to the log. sine
of the given side : so is the log. sine of the other given angle, to the log;
sine of its opposite side.
Now, to know whether the side thus found is ambiguous, aeute, or ob-
tuse, proceed as follows 3 viz..
To the side so found, and its supplement, add the given side, or that
used in the proportion.— Then, if each of these sums be of the same affec-
ttofi with respect to 180? as the sum of the two given angles, or those used
in the proportion, the side is ambiguous ; that is, it may be either aCute^
or obtuse ; and, therefore, indeterminate.
But, if those sums are of different affectiom with respect to the sum of
the angles, the side is norambiguous : in this case that value of the side is
to be taken, whether acute or obtuse, which, when added to the given
side, produces a quantity of the same affection with the sum of the angles.
2.— Tojind the side contained between the two given angles.
Find half the difference, and half thel sum of the two given angles :'-
find, also, half the difference of their opposite sides. — ^Then say.
As the log. sine of half the difference of the angles, is to the log. sine of
half their sum j so is the log. tangent of half the difference of their oppo^
site sides, to the log. tangent of half the side contained between the two
given angles ; the double of which will be the side sought
3. — Tojind the third, or remaining angle.
As the log. sine of a given side opposite to a given angle, is to the |og«
sme of that angle ;• so is the log. sine of the side opporfte to the required '
angle, to the log. sine of the required angle.
Note.^-^When the side comes out ambiguous, or indeterminate, in the
first proportion ; the contained side and the third angle, found by the other .
poportions, will also be ambiguous.
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8PHBE1CAX. TRIGONOMSTRY. 201
* Example.
Let the angle A, of the spherical triangle ABC, / ^^o^X.
be 130?40M3r, the angle C 41?39U0r, and the / % *i\
side B C 1 15° 10' 25^ ; required the angle B, and B ^ — Z5??""^\^ v\
the sides A B and AC? ""^-'^^
To find the Side A B :—
As the angle A s 130?40U3^ L<«. sine ar. coropt. = 10. 120114
Is to the side B C = 1 15. 10. 25 Log. sine ='«,.. 9. 956660
So is the angle C =: 41.39. 40 Log. sine = . . » . 9. 822641
To the side A B = 52?29^ 28r Log. sine = • . • . 9. 899415
To determine whether the Side A B is Ambiguous, Acute, or Obtuse : —
Side AB 52?29:28r Supplement =: 127^30^32^ Angle A 130?40U3?
SideBC115.10.25 Side B C = 115.10.25 Angle C 41.39.40
Sum = 167?39C53r Sum = . . 242?50:57? Sum = I72?20:23r
Here, since.the two first sums, viz. A Band B C, and. the supplement of
A B and B C, are of different affections with respect to 180?, the side A B is
not ambiguous; — and since the sum of the acute value of AB added to
B C is of the same affection .with the sum of the angles } therefore the side
ABisacuter=52?29C28r.
TofindtheSide ACi-
Astheang. A— theang. C-^2r=44?30'31i^ Log.S. ar.compt. 10. 154271
Is to angle A + angle C -h 2=S6. 10. 1 1| Log. sine =: . . 9. 999029
Sois the S.BC-S. A B-h2^31. 10.28^ Log. tangents: . 9.784614
TohalftheBideAC= • . 40^55^ 6 ^ Log. tangent • 9.937914
Side AC = 81?50n2r; which is acute, because the
side A B came out determinate, and that its acute value applied to B C is
of the same affection with the sum of the angles.
To find the Angle B :—
Afl the side BCs . . 115^10' 25 V Log. S. ar.compt. . . 10.043340
Is to the angle As . 130. 40. 43 Log. sine = . . . .' 9. 879886
So is the side A C ss • 81.50.12 Log. sine == . . /. 9.995577
TotbeaiigleB» . . 56? 2Ml^rLog.8iae s . . . . 9.918803
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202 SPttfiAICAl tHlOONOMiTRt.
Note.'^The aiigle B is acute like its opposite side A C, because the side
A B is not ambiguous ; and that its acute value applied to the side B C is of
the same affection with the sum of thd angles.
PaofiLBM IIL
Owen T\oo Sides of an ObUque-aiigled Spherical Ti-iangle, and the Angle
contained between them ; to find the other T\oo Jnglee and the Third
Side.
RUJLB.
1. — To find the other two atiglea.
As the log. co^sine of half the sum of the ^two given sides^ is to the log.
co-sine of half their difference $ so is the log. co-tangent of half the con-
tained anglc^ to the log. tangent of half the sum of the other two angles*
Half the sum of the angles thus found, will be of the same affeetbn with
half the sum of the sides^— -Again : As the Icfg. sine of half the sum of tli«
two given sidets^ is to the log. eine of half their diffemiee } so is the log. co-
tangent of half the contained angle, to the 1<^. tangent t>f half the differ*
ence of the other two angles.— Half the difference Of the angles, thua foiiiid^
will always be acute.
Now, hair the sum of the two angles, added to half their difference, will
give the greater angle ; and half .the difference of the angles subtracted from
half their sum will leave the lesser angle.
2.-'Tofind the third side.
.The angles being known, the third or remuningside is to be computed
by Rule 3, Problem L, page. 198.
'Example.
Let the side A C, of the spherical triangle
ABC, be 78?45n6:, the side A B 69^55! 55 r,
and the contained angle I26?30l20r; required
the angles B and C, and the side B C ?
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8l>fflBftTCAL TIttGOKOBflSTRlr. ^3
To find the Angle B s-.
As the side AC+AB-H2a74?20^55r Log. co-sine ar.comp.= IO. 568984
IstothesideAC-AB^2= 4.24.40 Log. co-sine = . . . 9.998712
So is the angle A -»- 2 == 63. 15, 10 Log. co-tang. = • . . 9. 702414
To i the sum of the an. = 61"? 46^ 8^ Log. tangent = • • .10. 2701 10
Half diff. of the angles = 2. 18. 19, as below
Sum= ... 64? 4127^ = Angle B.
To find the Angle C :—
As theside AC + AB -^ 2 = 74?20'.55'; Log. sine ar. compt.= 10. 016409
Istotb6 8ideAC-^AB-^-2s 4.24.40 Log. sine « • « . 8.885996
So isthe angle A -H 2 = 63. 15. 10 Log. co-<tangent ss . 9. 702414
To half the diff. of the ang.s 2<^ 18' 19? Log. Ungent = . . 8i 604819
Half sum of the angles s ^61. 46. 8> as abov^
Difference = . . 59^27 '49? = Angle C.
Note. — The half sum of the angles came out acute, because the half sum
of the sides is acute ^ the half difference of the angles ii alway$ acute.
To find the Side BC:—
As the angle B s= 64? 4^27? Log. sine ar. compt. = 10. 046066
Is to the side A C s 78.45. 15 Log. sine ss . . » . 9. 991580
So is the angle A = 1 26. 30. 20 Log. sine = . . \ . 9. 905 1 48
TothesideBCs 118?46: ir Log. sines . . . . 9.942794
Remark l.-^The side B C may be found directly, independently of the
ati^ea B and C, by the following- general Rule.
To twice the log. sine of half tip contained angle, add the log. sines of
the two containing sides; from half the sum of these three logs, subtract
the log. nne of half the difference of the sides, and the remainder will be
the log. tangent of an arch : the log. sine of which being subtracted from
the half sum of the three logs, will leave the log. sine of half the required
aide*
Example.
Let the side A C, of a spherical triangle^ be 62? 10^25?, the sido A B
50?14C45^ and the included angle A 123?li:40r; required the aide
BC?
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204 SPHERICAL TRIGONOMETRY.
Half ang. A= 61^35^50? r:f.7*!^'»:}= 19. 888596
Side A C = 62. 10, 25 Log. sine = 9. 946632
SideAB= 50.14.45 Log. sine =9.885811
Sura = 39. 721039
DiflF.ofSides ll?55M0r Half = 19. 86051 9i, ...19.8605194
Half ditto = 5?57^50r Log.sine= 9.016622
Arch = . . 81 ?50;52r Log. tang. = 10. 8438971 Log. S.=9. 9955881
iSideBC = 47?^ 6^50r =Log.8ine =....... 9.864931
Side B C s 94? 13 ^ 40r, as required.
Remark 2.«— The side B C may be also computed by the following ge«
neral rule, viz.
To twice the log. si^e of half the contuned angle, add the log. sines of
the two containing sides, and the sum (rejecting 30 from the index,) will be
the log. of a natural number.— -Now, the sum of twiot this natural number
and the natural versed sine of the difference of the containing sides, will be
the natural versed sine of the third side.
Thus, to find the side B C in the above example.
Half included ang. A=6-l.?35 '. 50? twice the log. sine= . .19. 888596
Side AC == . . . 62.10.25 Log. sine == .. ... . 9.946632
SideAB= . . '. 50.14.45 Log. sine = -9.885811
Natural number = . 526065 »: Log. 9. 721039
Twice the nat. numb.=: ....... 1052130
Diff. of the given sides ll?55UO: N.V.S.i= 021591
SideBC= . . . 94? 13^ 40r N.V.S. 1073721; the same as above.
Note. — In taking out the natural number corresponding to the sum of
the three logs. : if the index be 9, the natural number is to be takien out to
six places of figures ; if 8, to five places of figures ; if 7j to four places of
figures, &c.
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SPHERICAL TRIQONOMBTRT. 205
Problem IV,
Given Two jingles of a Spherical Triangle, and the Side comprehended
between them ; to find the remaining jingle and the other T\vo Sides.
Rule.
l.'^Tofihd the other two sides.
As the log. co-sine of half the dum of the two given angles, is to the log.
co-Bine of half their difference ; so is the log. tangent of half the compre-
hended side, to the log. tangent of half the sum of the other two sides.
Half the sum of the sides, thus found, will be of the same affection with
the half sum of the angles.
Again. — ^As the log. sine of half the sum of the two given angles, is to the
log. sine of half their difference ; so is the log. tangent of half the compre-
hended side, to the log. tangent of half the difference of the other two
sides.
Half thetlifference of the angles, thus found, will always be acute.
Now, half the sum of the two sides, added to half their difference, will
give the greater side; and half the difference of the two sides, subtracted
from half their sum^ will leave' the lesser side.
2.^Tofind the remaining angle. .
The sides and two angles being known, the remaining or third angle is
to be computed by Rule 3, Problem II., page 200.
Example.
x.
Let the angle A, of the spherical triangle ABC,,
be 63?50'.25r ; the angle C 58?40'. 15r, and the
comprehended side A C 87?30MO'r ; required the
sides A B and B C, and the remaining angle B ?
To find the Side B C :—
As the angle A+angleC^2=61 <" 15 '. 20^ L. co-sine ar.com. = 10. 317942
Is to the ang. A— ang.CH-2= 2. 35. 5 Log co-sine = . .9. 999557
So is the side A C -h 2 s:: 43. 45. 20 Log. tangent := • . 9. 981 129
To half the sum of (he sides=63?18^28^ Log. tangent = . 10. 298628
Halfdifferenceof the sides = 2.49. 10, as in the next operation.
. ■ —— .» .
Sum = 66? 7 ' 38? s= the side B C.
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206 SPHBRICAl. TRIGONOMBTRT*
To find the Side A Bi-
as the angle A4-angleC^2=61 ? 15 ^ 20r L. sine ar. compt. = 10. 0571 13
Is to angle A — angle C-h2=: 2.35. 5 Log, sine = . . 8. 654144
SoisthesideAC -t- 2= 43.45.20 Log tangent , . 9.981129
Tohalfthediff.oftheside8=2949n0? Log tangent = . 8.. 692386
Half sum of the sides = . 63. 18.28, as in the last operation.
Difference ^ 60V29: 18r » the tide A B«
No/e.~The half sum of the sides joaxae out acute because the half sum of
the angles is acute ; the half difference of the sides must be akoayi acute.
To find the Angle B :—
As the Side B C s 66? J'SSI Log. sine ar. compt. s 10. 038842
Is to the angle A a 63. 50. 25 Log. sine s .... 9. 953068
SoisthesideAC = 87.30.40 Log. sine ==: . . . . 9.999590
To the angle B ss . 78^42^ 3r Log. sine » . « . .9.991500
Bemark 1.— The angle B may be found directly by the following gene-
ral rule.
To twice the log, co-sine of half the given side^ comprehended between
the two given angles^ add the log. sities of tho^e angles : from half the sum
of these three logs, subtract the log. sine of half the difference of the ui-
gles, and the remainder will be the log. tangent of an arch.— -Now, the log.
sine of this arch being subtracted from the half sum of the three logs, will
leave the log. sine of half the required angle. ^
Thus^ to find the angle B in the above example.
Half side A C =5 43?45 '. 20^ C1iS&!^' }=« 19. 717432
Angle A = 63. 50. 25 Log sine = 9. 953068
Angle C s 58. 40. 15 Log. sine ^^ 9. 931557
Sum= 39.602057
Diff.oftheang.=5?10U6r Half = 19. 8ai028i . . 19.801028J
Halfdiff.ofdo.=:2. 35. 5 Log. sine = 8.654144
Arch » 85?55 : 17^ . U>g. tangents: 1 1 . 146884^1 Lg.S.s9. 998899
Halfthe required angle = 39?^inrLdg.sine = . . . 9.802129^
Hence, the angle B is = 78?42:2r} which differs K from the angle
found as above.
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8PBBAICAL TRIGOVOIiBTilY. 807
Remark 2.— The angle B may be also very readily computed by the fol-
lowing general Rule 5 viz.^
To twice the log. co-nne of half the given side^ comprehended between
the two given aisles, add the log. sines of those angles^ and the sum (reject-*
ing 30 from the inde^c)^ will be the log. of a natural number.-— Now, the sum
of twice this natural number and the natural versed sine of the difference of
the angles^ will be the natural versed sine of the required angle.
Thus^ to find the angle B in the last example.
Half the given side A Cs 43?45 ', 20r twice the log. co-sine s 19. 717432
Angle A « . . . • 68. 50. 25 Log. sine s .... 9. 953068
Angle Cs k . . • 58.40. 15 Log. sine = . . . . 9.931557
Natural number =:,... S99998=:Log. 9. 602057
Twice the natural number =: . 799996
Diff.oftheang.= 5?10nOrnat versed sine = 004067
Angle B = 78?42' 2", nat. versed sine = 804063 j the same as by
the former Rule.
Pboblim V.
Gftoen the Three Sides of a Spherical JHangle, tojind the Angles.
RULB,
Add the three sides together and take half their sum ; find the difference
between this half sum and the side opppsite to the required anglcj which
eall the remainder ; then.
To the log. co-secahts, less radius^ of the other two sides, add the log.
aihes of the half sum and the remuqder :-<i>-half the'sum of these four logs,
will be the log. co^sii)^ of an arch, which being doubled will be the required
angle.
. Chie angle being thus found, the remaining amgies may be computed by
Role 3^ Problem IL, page 200.
Example.
In the spherical triangle ABC, let the side A B ^7
be70?llM5r, the side AC 81?59'.55r, and the ^ "^ '
aideBC 120? IOC SOT; required ^the angles A, B, ^
and C?
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20S
SPHBRICAL TRIGONOMETRT.
SideBCs
Side AC =
SideABs:
Sum
To find the Angle A :—
120nOC50^
81 . 59, 55 Log. co-secant| less radias=0. 004248
70. 11.45 Log. co-secant, less radiusssO. 026477
272.22.30
Halfsums 136.11. 15' Log. sines. .... 9.840295
Remainders 16. 0.25 Log. sine = ..... 9.440522
Sum = . . . 19.311542
.Arch=. ; 63? 5' 8r = Log. cp-sine = . . 9.655771
AngIeAs= 126?10n6r
As the side B C = .
Is to the angle A =
So is the side A C =
To the angle B = .
To find the Angle B:—'
. 120?10:50r Log. co-secants 10.063262
. 126. le. 16 Log. sine s . . 9.907012
. 81 . 59, 55 Log. sine = . . 9. 995752
67?37^52r Log, sines
9. 966026
As the side B C s .
Is to the angle A =
So is the side A B s
To the angle C s . .
To find the Angle C .—
. 120?10:5ar Log. co-secant s 10.063262
. 126.10.16 Log.«ine= . . 9.907O12
. 70.11.45 Log. sine = . . 9.973523
. 6l928:31.r Log.sine =. . "^.943797
-JR^mr^.^-The required angle of a spherical triangle (when the three-
sides are given), may be also found by the following general Rule ; viz,.
Add the three sides together and take half their sum : find the difierence
between this half sum an.d each of the sides containing the required an^le,
and note the remainders. — ^Then,
To the log. co-secants, lesis radius, of those sides, add the log. sines of the
two remainders z^-half the sum of these four logs, will be the log. sine of
half the required angle.
' Thus, to find the angle A in the last example.
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SPHERICAL TRlGONbMETRT.' 209
SideBC = . . . 120?10f50r
Side A C = . . 8 1 . 59. 5a Log. co-secant, less radiu8=0. 004248 .
Side A B = . . 70. 11. 45 Log. co-secant, less radiu8=0. 026477
Sumss . . , . 272.22.30 *
Half sum , . 136°lin5r
Remainder, by AC = 54. 11.20 Log. sine - . . . ; .9.908994
Remainder,byAB = 65.59.30 Log. sine = •* . • . •9.960702
Sum = . . 19.900421
Half thie angle A = 63? 5^ 8? Log. sine ^ . . • . 9.950210J
Which beirtg doubled, shows the angle A to be 128? 10n6r j the same
as by the former rule.
Probum VI.
Owen the Three Angles of a Spherical THangUy to find the Sides.
Add die three angles together and take half their sum ; find the differ-
ence between the half sum and the angle opposite to the required side,
which call the remainder. — ^Tlien, %
To the log« do-secants, less radius, of the other two angles, add the log.
CO- sines of the hsilf sum', dnd the remainder; half the sum of these four
logs, will be the log. sine of half the required side;
One side being thus found, the remaining sides may be computed by
Ruled. Problem L, page 198.
E:tample0
In the spherical triangle A B C, let the angle A
be I25?16^25?; the angle £84? 20 ^50r, and the
angle C 72?40: 15r ;. required the sides B C, A B, /jv
«ndAC?
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210 SPHERICAL TOIOONOMKTRY.
To find the side B C :—
Angle A = . . 125?16r25r
Angle B =■ . . 84. 20. 50 Log. co-secant, less' fadius s= 0. 002117
Angle C= , . 72.40. 15 Log. co-secant, less radius = 0.020174
Sum= .. . . 282.17.30
Half sum = . . 141? 8M5r Log. co-sine = ' ' ' ' 9-891395
Remainder 3= . 15. 52. 20 Log. co-sine = . . . . 9.983118
Sam s . .19. 896804
HalfthesideBC*. . 62?37a3? Log. sine =» ... .9.948402
The double of which gives 125?14'.26'r, for thewhole Vide B C.
To find the Side A B :—
As the angle A = . . 125? i6C25f Lpg» coisecant = . .10. 088095
l8tothesideBC= • 125. 14. 26 Log. sine = . . . . 9.912083
So is the angle C a . 72.40,15 Log. sine » . • • . 9.979826
TothesideAB = . . 72 944 U6r Log. sine = . ,.,9.980004
To find the Side A C :—
As the angle A = 126? 16^ 25r Log. co-sectot s ... 10. 088093
Is to the side B C = 125. 14. 126 Log. sine =..... 9. 912083
So is the angle B = 84.20.50 Log. sine = • . , . , 9.997883
To the side AC = 84?35f 25r Log. sine » .... 9. 998061
IZ^arA;.—- The required side of a spherical triangle (when the three an-
gles are given,) may he also found by the (pUowing general rule ; viz., •
Add the three angles together and take half their siim ; find the differ-
ence between the half sum and each of the angles comprehending the re-
quired side, and note the remainders. — ^Then to the log. co-secants less ra-
dius, of those angles, add the log. co-sines of the two remainders : half the
sum of these four bgs. will be the log. co-sine of half tlm required side.
Thus, to find |)&e side B C in the last example.
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MAVIOATION. 211
Angle Ass. ; 125?16*25r
Angle B a . 84. 20. 50 Log. co-secant^ less radius s 0. 002 11 7
Anjie C =s . 72. 40. 15 Log. co-secant, less radius =s 0.020174
Sum = 282. 17.30
Half sum ss . • 141? 8'.45r
Remainder byB = 56. 47. 55 Log. co-sine = .... .9.738450
Remainder by Cs 68. 28. SO Log. co-sine a 9.564556
Sums 19.325297
Half Side B C a . . . . 62?S7< I3r Log. co*sine a . 9. 662648|
Which being doubled gives S3 U5?14C26f^ for the tide BCj the same
as by the foriper rule.
THE HESOLUTION OP PROBLEMS IN NAVIGATION BY LOG-
ARITHMS) AND, ALSO, BY THE GENERAL TRAVERSE
TABLE.
Lest the mariner should feel some degree of disappointment in not find-
ing a regular course of navigation in this work : the author thinks it
right to remind Kmj that his present intention carrieta him no farther than
merely to show the proper application of the Tables to some of the inost
useAil paris of the sciences on which he may touch : — it being completely at
variance with the plan of this work, to enter into such parts of the sciences
as' could reasonably be dispen^d with, without ^itirely losing sight of their
principles. — Hence it is, that the cases of plane sailing, usually met with in
books on narigation, will not be noticed in this.— However, since it is not
improbable that this voluine may fall into the hands of persons not very
deeply versed in nautical matteis ; it therefore may not be deemed unne*
eeseary to give a few introductory definitions, &c. for their immediate guid-
ance, previously to entering upon the essentially useful parts of the sailings.
Navigation is the art of conducting a ship, through the wide and path-
leas ocean, from one part of the world to another. — Or, it is the method of
finding the latitude and longitude of a ship's place at sea ; and of thence
determining her course and distance from that place, to any other given
place.
p 2
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212 .NAVIGATION.
The Equator is a great circle circumscribing the eartl?, every point of
which is equally distant from the poles ; thus dividing th^ globe into two
equal parts, called heniispheres : that towards the North Pole is called the
northern hemisphere, and the other, tlie southern hemisphere.— Tlie equa-
tor, like all other great rircles, is divided into 360 equal parts, called de-
grees ; each degree into 60 equal parts, called minutes j each minute into
60 equal parts, called seconds, and so on.
The Meridian o{ fiXiy pUce on the earth is a great circle passing through
.that place and the poles, and cutting the equator at right angles.— Every
point on the surface of the sphere may be conceived to haye a meridian
line passing through it 5 —hence there may be as many pieridiahs -as there
are points in the equator.— Since the First Meridian is merely an imagin-
ary circle passing through any remarkable place and the poles of the
world ; therefore it is entirely arbitrary.— Hence .it is that the. British
reckon their Jvrst meridian to be that which passes through the Royal Ob-
servatory at Greenwich :' the French esteem their first meridian to be. that
which passes through the Royal Observatory at Paris ; the Spaniards that
which passes througK Cadiz, &c. &c. &c.
Every meridian line may be said, with respect ta the place through
which it passes, to divide the surface of the sphere into two equal parts^
called the pastern and westei^n hemispheres.
The Latitude of any place oh the earth is that portiofi of its meridian
which is intercepted between the equator and the given place ; and is named
north or south, according as the giveii place is in the northern or southern
hemisphere. — As the latitude begins at the equator, where it is nothing,
and is reckoned thence to the poles, where it terminates ; therefore the
greatest latitude any place can have, is 90 degrees.
The Difference of Latitude between two places on the earth is an arc of
the meridian intercepted between their corresponding parallels of latitude ;
showing how far one of them is to the northward or southward of the
other :— The dlflference of latitude between two places can never exceed
180 degrees.
The Longitude of any place on the earth is that arc or portioir of the
equator which is contained between the first meridian and the meridian of
the given place ; and is denominated east, or west, according as it may be
situated with respect to the first meridian. — ^As the longitude is reckoned
both ways from the first meridian (east and west) till it meets at the same
meridian on the opposite part of the equator ', therefore .the longitude of
any place can never exceed 180 degrees*
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^AVIGATIO^. 218
The difference of Longitude between two places on the earth is an arc of
the equator intercepted between the meridians of those places ; shovHng
how far one of them is to the eastward or westward of the other : — ^The
difference of longitude between two places can never exceed 180 degrees.
When the latitudes of two places on* the earth are both north or both
south ; or their longitudes both east or both West, they are said to be of the
same name. — But, when, one latitude is north and the other south ; or one
longitude east and the othef west ; then they are said to be of different
names.
The Horizon is that great circle which is equally distant from the zenith
and n^dir, and divides the* visible from the invisible hemisphere ; this is
called the rational horizon.«-The sensible horizon is that which terminates
the view of a spectator in any part of the world.
Tie Manner's Compass is an artificial representation of the horizon : — it
is divided into 32 equal parts, callett points ; each point consisting of
U?15C. — Hence the whole 'Compass card contains 360 degrees; for
ll?15f multiplied by 32 points = 360 degrees.
A Rhumb line is a right line, or rather curve, drawn from the centre of
the compass to the horizon, and obtains its name from the point of the ho-
rizon it falU in with. — Hence there may be as many rhimib-lines as there
are joints in the horizon.
The Course steered by a ship is tlie angle contained between the meri-
dian of the place sailed from, and the rhumb-line on which she sails ; and
is either estimated in points or degrees. .
The Distance is the number of miles intercepted between any two
places, reckoned on the rhumb line of the course $ or it is the absolute
length that a ship has sailed in a given time.
The D^parfKre.is the distance of the ship from the meridian, of the place
s^led from, reckoned on the paraltel of latitude at which she arrives ; and
is named ea«t ot west) according as the course is in the eastern or western
hemisphere. . ^
If a ship's course be due north or south, she sails on a meridian, and
therefore giakes no departure : — hence the distance Bailed will be equal to
the difference of latitude.
If a ship's coufse be due east or west, she sails either on the equator, or
on some parallel of latitude ; in this case since she makes no difference of
latitude^ the xlistance sailed willj therefore, be equal to th^ departure.
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914 . NAVIQATIOir.
«
Whea the course is 4 points^ or 45 degrees, the difference of latitude
and departure are equal.
When, the course is less than 4 points^ or 45 degrees, the difference of
latitude exceeds the departure ; but when it is iiiore than 4 points, or 45
degrees, the departure exceeds the difference of latitude.
Note. — Since the distance sailed, the difference of latitude, and the de-
pai^ture form the sides of a right angled plane Jriangle | in which the hypo-
thenuse is represented by the distance ; the perpendicular, by the differ-
ence of latitude ; the base, by the- departure ; the angle opposite to the
base, by the course; and the angle opposite to the perpendicular, by the.
complement of the course } therefore any two of these five parts being
given, the remaining three may be readily found by the analogies for right
angled plane trigonometry.
%.
These being premised, we will now proceed to the following Introduciory
Problems.
Problbm L
Gioen the LatUwies qf Jloo Places on the Eairth, to find the difference of
RuuB.
When the latitudes are of the same name ; that is, both north, or both
south, their difference will be the difference- of latitude ; but when one is
north and tbo other south, th^ir sum will express the difference of la-
titude,
Note.— The same Rule is to be observed in finding the meridicmal differ-
ence of latitude between two places.
"Exomple L,
Required the difference of la-,
titude between Portsmouth and
Cape Trafalgar?
Lat. of Portsmouth =50?47fN.
Lat. of C. Trafalgar = 86. 10 N.
Diff. of Lat. = • . 14?37t
Pitto in Miles =♦ , 877
Exarnple 2.
Required the difference, of la«
titude between Portsmouth and
James Town, Stw Helena ?
Lat. of Portsmouth = «0?47' N.
Lat.ofJariiesTown=s 15.55 S«
Diff.ofLat.se .\ 461*42;
PittpiDMil^ss , ,2809
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DIFF£RBNCB OP LATlttTDB AND LONGITUDE.
215
Note.^^lu finding the diiFerence of latitude, or the difierence of longi*
tude between tviro i^aceii (when any of the sailings are under consideration),
it will be sufficiently exact to take out the latitudes and longitudes from
Table LVIIL to the nearest minute of a degree^ as above.
Problbm IL
Oven the Latitude left and the difference of Latitude, to find the
Latitude tn.
RUtB.
When the latitude left and the difference of latitude are of the same name
their sunr will be the latitude ; but when they are of contrary denomin-
ations^ their difference will be the latitude required : — ^This latitude will
always be. of the same name with the greater quantity.
Example 1.
A ship^ from a place in latitude
30?45: north sajled 497 miles in
a northerly direction ; required the
latitude at which she arrived ?
Latitude left » . . 30N5: N«
Diff.ofLiit3iB497m«.or8.17 N.
Lat. arrived at =
sa? 2: N.
Example 2.
A ship from a place in latitude
2?50' norths sails 530 miles in a
southerly direction; required the
latitude come to )
Latitude left m ^ . . 2?50' N.
'Diff.oflata530ms/or 8.50 8.
Lat. come to =
6? 0^ S.
Peoblbm III.
Gioen the Longitudes of Tioo Places on the Earth, to find the difference
^ qf Longitude.
Ru£b.
When the longitudes are of the same name : that is^ both east, of both
west, their difference will express the' difference of longitude > but when
one is east and the other Wesf, their sum wHl be the difference of longitude.
If the sum of the longitudes exceed ISO?, subtract it from 360% and the
remaioder will be the difference of longitude*
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316
NAVIGATION*
Example I.
Required the difference of lon-^
gitude between Portsmouth and
Fayal, one- of the western islands ?
Long, of Port8mouth= 19 6^ W.
Long, of Fayal, HorU,28. 43 W.
Diff. of long, =. . 27^37^
Ditto in miles = • • 1657
JSxample 2.
Required the difference of lon^
gitude between Canton and Point
Venus^ in the island of Otaheite ?
Long, of Canton = 1 13? 3 C E,
Long, of PointVenus= 149. 36 W.
Sum = « • •
Diff. of Long. = •
Ditto in miles =
262939C
97921 C
. 5841
Paobleat IV.
Given the Jjmg%tude\left and the difference of Longitude^ to find the
Longitude in.
RULE;
When the longitude left and the difference of longitude are of thie same
name^ their sum will be the longitude in; should that sum exceed 180?,
subtract it from 360? ; and the remainder ytill.be the longitude in, of a
contrary name to the longitude left^-^Buiy when the. longitude left and the
difference of longitude are of contrary names, their difference will be the
longitude in, of the same name with the greater quantity.
Example 1.
A ship from a pla<;e in longitude
50?40' west, sails westward till
her difference of longitude iir 4H)
miles ; required the longitude in ?
Long, left = . • . 50940^ W;
Diff.oflong.=41pm8.or6..50 W. ,
Longitude in si • • 57?30C W.
Example 2.
Let the long, left be 174945.'
west, and* the difference of longi*
tude 13? 17 f west; riequired the
'longitude in?
Longitude left = . Hr4?45^ W.
Diff. of Long. = . 13.17 W.
Sum:
Longitude in i
188. 2
171958CE.
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PARAtLBt SAILIKC.
21?
Example 3.
Let the longitude left be 4 1 ^ 37 •
east, and the difference of longi*
tude ll?20' west; required the
longitude come to ?
Longitude left = •
Diff. oflong. = .
Longitude in = •
41?37r E.
11.20W.
30^.17f E.
Example 4.
Let the longitude left be 5?40f
east,' and the difference of longi*
tude 10? 17' west; required the
longitude in ?
Longitude left =
Diff. oflong. = 4
Long, in = * .
5?40^ E.
10. 17 W.
4?37IW.
Remarks. — ^If a ship be in north latitude sailing northerly, or in south
latitude sailing southerly, she hicreases her latitude, and therefore the dif«-
ference of latitude must be added to the latitude left, in order to find the
latitude in r—rbut, in north latitude sailing southerly, or in south latitude^
northerly, she decreases htr latitude ; therefore the difference of latitude
subtracted from the latitude left will give the latitude in :— should the dif-
ference of latitude be the greatest, the latitude left is to be taken from it ;
in this case the ship will be on the opposite side of the equator with res-
pect to the latitude sailed from. — Again,
If a ship be in east longitude sailing easterly, or in west longitude sailing
westerly, she increases her longitude ; therefore the difference of longitude
added to the longitude left will give the longitude in ; should the sum ex-
ceed 180?, the ship will be on tlie opposite side of the^^t meridian with
respect to the longitude sfliiled from. — But, in east longitude sailing west*
erly, or in west longitude sailing easterly, she decreases her longitude, and
therefore the difference of longitude is to be sttbtracted from the longitude
left;, in order to find the longitude in;— should the difference of longitude
be the greatest, the longitude left is to be taken from it ; in this case the
ship will, also, b^ on the opposite side of the first meridian with respect tq
the longitude sailed from. * These remarks will appear evident on a com*
parison with the above Examples;
SOLUTION OF PROBLEMS IN PARALLEL SAILING.
Parallel iSailing is the method of finding the distance between two places
situate under the same parallel of latitude ; or pf finding the difference .of
longitude corresponding to the meridional distance, when a ship sails due
east or west.
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^8 NAVIGATlOlf.
Problem I.
Given the Difference ofLongiiude between two Places, both tii the same
PardUel qf Latiixidey to find their Distance^
Ruus.
As radius, is to the co-sine of the latitude; so is the difference of long!-*
fude, to the distance.
Example*
Required the distance between Portsmouth, in longitude 1?6' W,^ and
Qreen Island, Newfoundland, in longitude 55?35' W.j their oomtnon lati-
tude being 50?47 C N. ?
Long, of Portsmouth=i 1? 6^W.
Long.of Green l8land=55 . 35 W,
Diff. of long. = . 54?29I =3269 miles.
Sohi&on*
In the right-angled triangle A B C, where the hypothenuse A C represenU
the difference of longitude between the two giv^ places, the angle A the
latitude of the parallel of those places, and the base AB their meridional
distance: given the side AC = 3269 miles, and the angle A = 50?47C,
to find the side A B.' Hence, by right-angled plane trigonometry, problem
1., page 171,
«
As radius ...... 90? 0' Or Log, co-secant s . 10.000000
I»to the diff. of long. A C '= 3269 miles Log. = . . 3. 514415
So is the lat.=the angle A=50?47 ' Or Log. «o-sine =r . , .9. 800892
Tothemerid. dist. AB = 2066. 8 miles Log. = . , S.315S07
To find the Meridional Distance by Jn^pecHon in the general Traverse
Table:—
Note.'^ThiB case mny be solved by Problem L, page 107, as thus r
.To latitude 50? as a bourse, and one-eleventh of the difference of longi-
tude (via. 297. 2) aa a distance, the corresponding difference of latitude is
190. 9; and to latitude 5 1 ?^ and distance 297. 2, the difference of l^titud^
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PAHALLU SAILING. 219
is 186. 9 : hence the change of meridional distance (represented by diiFer-
ence of latitude,) to 1? or 60' of latitude, is 4^ Now, 4! x 47^ -i- 60^
= 3 ' . 1 ; this being subtracted from the first difiTerence of latitude, because
it is decreasing, gives 187. 8; and 187-8 multiplied by U, the aliquot
part, gives 2065. 8 for the meridional distance ; which comes within one
pile of the result by calculation.
PaofiLSM II.
Given the Distance between two Places, both in the same Parallel of
JLij^iude, tojind the Difference of Ixngitude between those Places.
RULB.
As the co-sine of the latitude, is to radius ; so is the distance, to the
difiTerence qf longitude.
Example*
A ship from Cape Clear, in latitude 5 1 ?25 ' N. and longitude 9?29^ W.,
sailed due west 1040 miles 5 required the longitude at which she then
arrived ?
JSobaim j--ln the right angled triangle ABC, let the bypothenuse A O
represent the difference of longitude ; the angle A, the latitude of the parallel
on which the ship sidled ; and the base A B, the q
neridiooal distance: then, in this jangle, there
are '^vmn^ the. angle A as 51?25;, and the base
A B = 1040 miles^ to find the 9ide 'A C. Hence, ^
by right angled plane trigonometry. Problem II .^
page 172,
As radius » . . . . ^ 90? 0^ 0? Log. co-secant a. 10.000000
Is to the merid. dist AB = 1040 miles. Log. « . 3.017033
So is the lat s die angle A = 5 1 ?25 ' Or Log. secant = . 10. 205057
To the difference of long. A C = 1667. 6 miles. Log. = • 3. 222090
LongitndeofCape Clear ss 9?29Cwese.
IMfftrence of longitude ;i 667. 6 miles, or . .^ . 48 west^
I^ongitade 4^ which (be shig arrived ^ •
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"SiZO KAVI6ATIOK*
To find th^ Difference of Longitude by Inspection in the general
Traverse Table :—
Note,— Tliis case falls under Problem V., page 111: bence^
To latitude 51? as a course^ and one^eighth of the meridional distance
=: ] 30, in a difference of latitude column, the corresponding distance is
207 ; and to latitude 52?, and difference of latitude 130, the diatanee is
211: hence, the difference of distance to 1? of latitude, is 4 niiles. Now,
4'. X 25^ -H 60^ = 1'. 6,^ which being added to the first distance,
because it is increasing, gives 208. 6 ; this being multiplied by 8 (the
aliquot part), gives 1668. 8 fojr the difference of longitudf.
. PROBf^M III,
Given the Difference of Longitude, and the Distance between two Places,
in the same Parallel of Z^ititudey to find the Z,atitude of that
Parallel
Rule,
As the difference of longitude, is to the distance; so is radius,' to the
po-sine of the latitude*
Example,
A ship, from a place in longitude 16^30' W., sailed due east 456 miles^
and then by observation was found to be in the longitude of 4?15t W,;
required the latitudie of tlie parallel on which she sailed ? •
liong. sailed from = 16?30^ W.
Long, come to 33 . 4. 15 W.
Diff.oflong, =s « 12?15C ^ 735 rnilea*
Mr.M4i/. 4S6
/ &)IttHon.~In the right angled triangle ABC, let the hypothenuse A C
represent the difference of longitude \ the angle A, the latitude of the
parallel ; and the base A B, the meridional distance : then, there are giv^n,
the side AC = 735 miles, and the leg AB = 456 miles, to find the angle
A. Hence, by right angled jrtane trigonometry, Problem III., page 174,
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MIBDLB XATITth>fi SAlLtNG* 221
As the diC of longitude A C =: 735 miles. Log. ar. comp. = 7. 1337 13
Is to radius = . • . . 90? 01 Or Log. sine = . 10.000000
50 is the merid. distance AB = 456 miles. Log. = •' . • 2. 658965
Tolat.ofparall.= ang,A=51?39a4: Log. co- sine = 9-792678
To find the Latitude of the Parallel by Inspectim in the general
Traverse Table :•»-
Enter the Table with one-third the difference of longitude =: 245 as a
distance, and one-third tlie meridional distance == 152, in a difference of
latitude column ; and the latitude corresponding to them will be found to
lie between 51? and 52?. Now, to latitude 5 1 ?, and distance 245, the*
corresponding difference of latitude is 154-.2, which exceeds half the
meridional distance by 2' . 2 ; and, to latitude 52?, and distance 245, the
difference of latitude is 150. 8, 'which is 1 ' . 8 less than half the meridional
distance. . Hence, 1 ' . 8 + 2' . 2*=: 4' is the change of meridional distance
to 1? of latitude; And, as 4^ : 2\ 2 :: 60'. : 38: 3 this, being added to
5 1 ?9 gives 5 1 ?38 ' for the required latitude*
- SOLUTION OF PROBLEMS IN MIDDLE LATITUDE.
SAILING.
Middle Latitude Sailing is the method of solving the several cases, or
problems, in Mercator's-saiiing, by principles compounded of plane and
parallel sailing. This method is founded on the supposition \that the
meridional distance, at that point which is a middle parallel between the
latitude left and tlie latitude bound to, is equal to the departure which the
ship makes in sailing from one parallel of latitude to the other. *
. This method of sailing, though not quite accurate, is, nevertheless, suffi-
ciently so for a migle day's ruriy particularly in low latitudes, or when the
ship's course is not more than two or three points from a parallel. But,
in high latitudes, or places considerably distant from the equator, it fails
of the. desired: aecuracy : in such places, therefore, the mariner should
never employ it in the determination of a ship's place, when he wishes to
draw correct nautical conclusions from his operations.
With the intention of avoiding prolixity and unnecessary repetition, in
resolving the different problems in this method of sailing, we will here
briefly give a general view of the -principles en which the solutions of those
problems are founded } as thus :— * * .
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222 NAVIGATION,
In the annexed diagram, let the triangle ABC
be a figure in plane sailing, in which A C repre-
sents the distance, A B the difference of latitude,
B C the departure, and the angle A the course.
Again, let D B C be a figure in parallel sailing,
in which D C represents the difference of longi.
tude, BC the' meridional distance, and the angle
C the middle latitude. Hence, the parts con-
cerned form two connected right angled triangles,
tn which the departure or meridional distance B C
is a side common to both.
Now, in one of these triangles, there will be always two terms given, aad
in the other one term, at least, to find the required terms. The required
parts in that triangle which has two terms given, may be readily found hf
the analogies for right angled plane trigonometry, page 171 to 177; tad^
hence, the unknown terms in the other triangle.
When the departure B C is not under consideration, the two connected
triangles may be considered as one oblique angled triangle, and resolved as
such. In this case, if the course, distance, middle latitude, and difference
of longitude, are the terms in question, any three of them being given^^ the
fourth may be found by one direct proportion. Thus, in the oblique angled
triangle A CD, the side AC is the distance 3 the angle A, the course ; the
angle B C D, the middle latitude ; and, consequently, the angle D its com-
plement, and the side D C the difference of longitude. Now, if any three
of these be known, the fourth may be found by one of the following analo-
gies; via.,
1., As co-sine middle latitude = C : sine of course s A :: distance =
AC : difference of longitude =: PC
2. As sine of course = A ! co-sine middle latitude s C :: difference of
longitude = D C : distance = 'A C.
3. As distance = A C : difference of longitude = D C : : co-sine of
middle latitude = C : sine of course = A.
4. As difference of longitude = D C *. distance = A C : : sine of course
= A : co-sine of middle latitude = C.
Again, if the course, middle latitude, difference of latitude^ and diffeitnoe
of loBgitude^ be the terms under consideration, the resulting analogies
will be,
5. As difference of latitude = AB ! difference of longitude = DC ::
co-sine of middle latitude = C : tangent of course =3 A*
6. As difference^of longitude = DC : difference of latitude = AB :;
tangent of course = A *• co-sine of middle latitude = C»
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MIDDLB LATrrUBH SAILING.
228
7f As co-sine of middle latitude = C : tangent of course =8 A ::
difference of latitude = A B : difference of longitude ss A C.
8. As tangent of course = A : co-sine of middle latitude =s C : : differ-
ence of longitude = D C '.difference of latitude = A B.
In these four analogies, it is evident that the course must be a tangent,
because the difference of latitude AB is concerned.
Note.— nSince the sine complement pf the middle latitudes the angle D, is
expressed directly by the co-sine of the angle BC D, therefore, with the view
of abridging the preceding analogies, the co-sine of the middle latitude has
beeu used instead of its sine complement; and, in the operations which
follow, the same term will be invariably employed.
IZemarfc.— *The middle latitude between two places is found by taking
half the sum of the two latitudes, when they are both of the same namci or
half their difference if of contrary names.
ProIbjlbm I. '
Gwen the LcOUfides and LmgUudes o/two Places, tofiid the Cwree
and Distance between them. .
Exajnpk.
Required the course sjid distance from Oporto,
in latitude. 4 1?9' N. and longitude &?37' W.
to Porto Santo, in latitude 33 ?3: N. and longi-
tude 16? 17^ W.? . ,
Latitude of Oporto 41? 9' N. Longitude s=
Lat. of Porto Santo 33. 3 N. Longitude =
. 8?37^W.
. 16. 17 W.
Diff. of latitude = 8? 6ts486 miles. Diff. of long. = 7?40f =460ms.
Sum of latitudes = 74?12C h-2 = 37?6: = the middle latitude.
To find the Course = Angle A :—
Here, since the departure is not in question, the parts concerned come
imder the 5th analogy in page ^22 : hence,
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224
NAVlCATIOlf.
As the diff. of latitude = 486 miles, Log. ar. coinp. = 7. 313364
Istothediff. ofIong.=: 460 miles. Log. = . • . 2.662758
So is the mid. latitude = 37'?6: Log. co-sine = . 9, 901776
To the course =* . • 87?3C Log. tangent = 9.877898
To find the Distance = A C :—
The course being thus found, the distance may be determined by trigo-
nometry. Problem II., page 172 : hence,
As radius = . . . 90^QC Log. co-secant = 10.000000
Istothediff.oflat. =s 486 miles. Log. = . . • 2.686636
So is the course = . 37°3' Log. secant — • 10. 097937
To the distance = • 608. 9 miles. Log. =£
2. 784573
Hence, the true course from Oporto to Porto Santo is S. 3/^3' W., or
S.W. i S. nearly, and the distance 609 miles*
To find the Course and Distance by Inspection in the general
Traverse Table :—
To middle latitude = 37? as a course^ and one-fourth the difference of
longitude ^ 115, as a distance, the corresponding difference of latitude is
91. 8 = the meridional distance. . Now, one-fourth the difference of lati-
tude = 121.5, and the meridional distance 91. 8 in a departure column,
are found to agree nearest at 37?, under distance 152. Hence, the course
is S. 37? W., and the distance 152 X 4 = 608 miles.
Problem IL
Gvoen the Latitude and Longitude of the Place sailed from, the Course,
and Distance; to find, the Latitude and Longitude of the Place
come to.
Example.
A ship from Corvo, in latitude 39?41 ' N., and
longitude 31?3: W.,swledN.,E. ^E., 590 miles;
required the latitude and longitude come to ?
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MIDDLB LATITUDE SAILING. 225
To find the Difference of Latitude = AB ;—
Here the course s= A, and the distance = A C, being given, the differ-
ence of latitude^ AB may be found by trigonometry. Problem I., page
171 ; as thus :
As radius = . . . 90?0! Log, co-secant = 10.000000
Is to the distance =s 590 miles. Log. = . . • 2. 770852 .
So IS the course = . 4| points, Log. co-sine = • 9. 802359
Tothediff.oflat. = 374.3mile8, Log. = . . • 2.573211
Utitudelcfk= 39941:N 39?41^N.
Diff. of lat. = 374. 3 miles, or = 6. 14 N. Half = 3. 7 N.
Latitude come to s . • . . 45?55:N. Mid.lat.s 42948^
To find the Difference of Longitude s C D :—
Here, since the^departure is not concerned, the parts in question come
under the 1st analogy in page 222 : hence.
As the mid. lat^ s • . 42?48f . Log. secant » 10.134464
Is to the course =s • • 4^ points. Log. sine = • 9. 888185
So is the distance = . 590 miles. Log. = • . . 2. 770852
To the diff. of longitude s 621. 6 miles. Log. = . . . . 2. 793501
Longitude left s 31?8fW.
Diff. of longitude s 621 . 6 miles, or « 10. 22 E.
Longitude come to = ...... 20?41^W.
To find the Difference of Latitude and Difference of Longitude by
Inspection :—
Under or over one-fifth of the given distance =: 1 18, and opposite to the
course ^ 4| points, is difference of latitude 74. 9, and departure 91. 2.
Tabular difference of latitude 74. 9 x 5 = 374. 5, the whole difference of
latitude; whence the latitude in, is 45? 55' N., and the middle latitude
42?48C. Now« to middle latitude 42?, and departure 91. 2, in a latitude
column, the corresponding distance is 123 miles; and to middle latitude
43?, and departure 91. 2, the distance is 125 miles : hence, the difference
of distance to 1? of latitude, is 2 miles; and 2^ x 48' -i- 60^ = r.6,
which, added to 123, gives 124. 6 ; this, being multiplied by 5 (the aliquot
part), gives 623 miles ^ the diffierence of longitude, or 10?23' E.
Q
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226
KAVIOATIOK.
Problem III.
«
Gioen both LaAtudi^ and the Coune; to find the JXttanee and the
L(mgitude tit.
JEyompfe.
A ship from Brava, in latitude 14t46' N*, and
longitude 24?46^ W.^ sailed S.E. b. S., until, by
obserration, she was found to be in latitude
10?30C N.j required the distance sailed and her
present longitude ?
Lat.ofBrava=:14?46^N. . . . 14?46^N.
Lat.byob8. = 10.30 N. . . . 10.30 N.
Diflf.oflat-= 4n6i=256m.Sum=25?16r
Middle latitudes 12?381
To find thfc Distance = A C r—
With the course s A, and the difference of latitude ss AB, the distance
is found by trigonometry, Problem II., page 172 } as thus :
A&rddius es .... 90?0^ Log. co*s6cant = . 10. 000000
Is to the diif. of latitude = 256 miles . Log. = • • • 2, 408240
So is the course =v • • • 3 points, Log. secant » » 10. 080154
To the distance == •
307. 9 miles. Log, =
2.488394
To find the Difference of Longitude = C D :— •
Here, since the departure b not in question, the parts concerned fall
under the 7th analogy, page 222 : hence.
As the middle latitude s 12?38:
Is to the course :s • . 8 points.
So is the diff. of lat ^ 256 miles,
Log. secant 9 10.010644
Log. tangent a 9.824898
Log. « . 4 2.408240
To the diff. of long. = 175. 2 miles. Log. = . . 2.243777
Longitude of Brava, the place staled from = 24?46^ W.
Difference of longitude « 175. 2 miles, or = 2. 55 E.
Longitude of the ship <
21?51^W*
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MIBBLB LATnum SAILING. 227
To find the Distance sailed, and the Difference of Longitude^ by
Inspection :—
To the course 3 points, and half the difference of latitude sss 128^ -the
distance is 154, and the departure 85. 5. Now, 154 X 2 =s 308 miles, is
the required distance. Again, to middle latitude 12?, and departure 85. 5,
in a latitude column, the corresponding distance is 87; and to latitude 13%
and departure 85. 5, the distance is 88 : hence, to middle latitude 12'?38C,
and departure 85. 5, the distance is S7i ; the double of which es 175 miles^
is the difference of longitude, as required.
Probuem IV.
Given the Latitude and Longitude qftbe Place sailed fiom, the Couree,
and the Departure; to find the Distance sailed^ and the Latitude and
Longitude of the Place cahne to.
D
Example.
A ship from Cape Finisterre, in latitude
42?54^ N.^ and longitude 9?16(W., sailed
N.W. b. W., till her departure was 468 miles;
required the distance sailed, and the latitude and
longitude come to ?
To find the Distance » AC:~
With the course = A. and tiie departure B C, the distance may be found
by trigonometry. Problem IL, page 172 ; Jis thus :
As radius = 90?0i Log. co-secant = 10.000000
Is to the departure = • . 468 miles. Log. = ... 2. 670246
So is the course = . . • 5 points. Log. co-secant » 10.080154
To the distance s . . . 562. 9 miles. Log. = ... 2. 750400
To find the Difference of Latitude = AB :-^
With the course = A, and the departure B C, the distance is found by
trigonometry. Problem IL, page 172 ; as thus :
a 2
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As the course =: . . . » 5 points. Log. co-tangent = 9. 824893
Is to the departure = • • 468 miles. Log. =3^ . • • 2. 670246
So is radius = 90?0: Log. sine = . •10.000000
To the diff. of latitudes . 312.7 Log. = • . . 2.495139
Latitude of Cape Finisterre =r 42? 54: N 42?54^ N.
DiflF. of lat. s= 313 miles, or = 5. 13 N, Halfdiflf.of lat.= 2.36 N.
Latitude of the ship =i . • 48? 7^ N. Middle lat. s= 45?30:
To find the DiiFerenee of Longitude s C D :-^
With the middle latitude = B CD, and the departure B C, the differ-
ence of longitude is found by trigonometry. Problem II., page 172. —
As radius = . • . • 90? 0' Log. co-secant = 10.000000
Is to the departure = . . 468 miles. Log. = . . 2. 670246
So is the mid. lat = . . 45?30: Log. secant = 10. 154338
Tothe diff.oflong. = . 667. 7 miles, Log. = . . . 2.824584
Longitude of Cape Finisterre = • • • 9?16^ W.
Difference of long. = 667. 7 miles, or = 1 1. 8 W.
Longitude of the ship = 20. 24 W.
To find the Distance, Difference of Latitude, and Difference of Longitude,
by Inspection : —
To course S points, and one-fourth of the departure =117, the distance
is 141, and the difference of latitude 78. 3. Now, 141 x 4 = 564 miles,
thedistance, and 78.3 x 4 = 313.2, or 5? 13', the difference of lati-
tude ; whence the latitude in, is 48?7*N., and the middle latitude 45.?30'.
Again, to middle latitude 45?, and one-fourth the departure = 117, in a
latitude column, the distance is 166; and to middle latitude 46?, and
departure 117, the distance is 168 : hence, to middle latitude 45?30^, and
departure 117, the difference of longitude is 167 x 4 = 668 miles; nearly
the same as by calculation.
UIDDLB LATITt7J>B 8AILIN0.
229
Problsm v.
Given both LatUudes and the Distance ; to find the Course and Differ-
ence of Longitude.
Example.
A ship from St Agne«, Scilly, in latitude 49e54C
N., and longitude 6? 19^ W., sailed 320 miles between the
south and west, and then, by observation, was found to
be in latitude 45?8C N.; required the course, and the
longitude come to ?
Latitude of St Agnes = 49? 54 ' N.
Latitude of tlie ship = 45* 8 N.
49?54C N.
45. 8 N.
Difference of latitude = 4? 46 f = 286 miles. Sum = 95? 2".
Middle latitude = 47?3i:
To find the Course = A :—
With the distance A C, and the difference of latitude s AB, the course
may be found by trigonometry, Problem IIL, page 174 ; as thus :
As the distance = . •
Is to radius = . . .
So is the diff. of lat. = •
320 miles, Log. ar^ comp. == 7« 494850
90? 0^ or Log. sine = . . 10.000000
286 miles. Log. = . . . 2.456366
To the course s . . • . 26?39' 6? Log. co-sine = • 9.951216
To find the Difference of Longitude = C D :—
With the course, middle latitude, and distance, the difference of longi*-
tude is found by the Ut analogy, page 222 ; as thus :
As middle latitude =s
Is to the course = .
' So is the distance =
To the diff, of long. =s
47?31' 01 Log. secant = 10.170455
26.39. 6 Log. sine = . 9.651825
320 miles. Log. = . . 2.505150
212,5 Log.
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230
KAVlGATtOV.
Longitude of St. Agnes = . . . 6?19^ W.
Diff. of long. = 212. 6 miles, or = 3. 33 W.
Lons^tttde of the ship =s .... 9?S2: W.
The course is S. 26?39C W., or S.S.W. i W., nearly.
To find the Course and Difference of Longitude by Inspection : —
To half the distance = 160^ and half the difference of latitude = 143,
the course nearest agreeing is 27» and the departure 72. 6. Now, to
middle latitude 47? as a course, and departure 72. 6, in a latitude column,
the distance is 106 ; and to middle latitude 48?, and departure 72. 6, the
distance is 108 c hence, the differetice of distance to I? of latitude, is 3
miles; therefore, 3^ x31'-i-60 = r.5, which, added to 105, makes
106.5 : this, being multiplied by 2, gives 213 miles = the difference of
longitude.
PaOBJLBBC VI.
Given one Latitude, Distance, and Departure ; to find the other LatUnde,
the Course, and the Difference qf Longitude.
Example.
A ship from Cape B^joli, Minorca, in latitude
40?3: N., and longitude 3?52' B., sailed 280
miles between the north and east, upon a direct
course, and made 186 miles of departure; re- .
quired the course, and die latitude and longitude
come to ?
To find the Course =a A :—
The distance ss A C, and the departure B C, being given, the course
may be found by trigonometry, Problem III., page 174; as thus :
As the distance s .
Is to radius s= • •
So is the departure =
To the course =? • •
280 miles. Log. ar« comp. ss 7. 552842
90? OC or Log. sines . lO.OOOOOO
186 miles, Log. = . . 2.269513
41?87^39r Log.^ne
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MIDDLB LATITITDB SAILING. 33)
To find the DiflFerence of Latitude = AB :—
The course = A^ and the distance, being thus known, the difference of
latitude may be computed by trigonometry, Problem III., page 174««^
As radius = ... 90? 0^ 01 Log. co-secant = 10. 000000
Is to the distance ac . 280 miles, Log. >» . . . 2..447158
So is the courses. . 41?37'39? Log. co-sine = • 9.873599
To the diff. of lat. = . 209. 3 miles, Log. = ... 2. 320757
Latitudeof Cape Bajoli = . . 40? 3^ N 40? 3' N.
Diff. of lat. = 209. 3 miles, or s 3. 29 N. Half =: . 1. 44i N.
Latitude come to = .... 43?32^ N. Middle latitude ss 41 ?47i«N.
To find the Difference of Longitude = *C D : —
The middle latitude = angle BCD, and the departure B C, being given,
the difference of lon^tude may be found by trigonometry, Prblblem II.,
page 172 i as thus:
Aa radius s .... 90? 0^ Log. co-secant s 10.000000
la to the departure s . 186 miles, Log. s • » . 1.269513
So is the mid. lat. s: . 4 1 ?47i ' Log. secant = . 10. 1 275 10
To the diff. of long. = . 249. 5 miles. Log. = ... 1. 397023
Longitude of Cape Bajoli == 3?52^ E.
Diff. of long. = 249. 5 miles, or s • • . 4. 9 E.
Longitude come to ........ 8. 1 E.
The course is N. 41 ?38^ E., or NJB. i N., nearly.
To find the Course, Difference of Latitude, and Difference of Longitude,
by^ Inspection !^
The cBstanee 280, and depatture 186, are found to agree between 41 ?
and 42?, ^d the corresponding difference of latitude 208. 1 : whence the
middle latitude is 41?46'. Now, to middle latitude 41?, and departure
186, m a latitude column, the corresponding distance is 247 j and to lati-
tude 42?, and departure 186, the distance is 250 : hence, the difference of
distance to I? of latitude, is 3 miles; and 3' x 46 h- 60' = 2'.3, which,
added to 247» gives 249. 3= the difference of longitude, as required ; which
pearly, igr^ Witi\ th^ result, by calculatioo.
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232 NAVIGATION.
- Paoblbm VII.
Gwen both Latitudes and Departures to find the Course, Distance, and
Difference of Jj)ngitud£.
Example.
A ship from Cape Agulhas, in latitude 34?55 C
S., and longitude 20? 18' £.^ sailed upon a direct
course between the south and east, till she was
found, by observation, to be in latitude 40?47*
S., and. to have made 436 miks of easting;
required the course, distance, and longitude at
which the ship arrived ?
Latitude of Cape Agulhas = 34?55'S. ....... 34?55^S.
Latitude of the ship .= • 40.47 S . 40.47 S.
Diflf. of latitude = 5?52' =352 miles. Sum = 75.42
Middle latitude =37?5 1 :
To find the Course = Angle A : —
Here, the difference of latitude = A B, and the departure B C, being
given, the course is jfound by trigonometry. Problem IV., page 175 ; as
thus :
As the diiF. of lat. =? 352 miles. Log. ar. comp. = * 7*453457
Is to radius = . . . 90?0^0r Log. sines . . 10.000000
So is the departure = • 436 miles. Log. = . • . 2. 639486
To the courses . . 5l?5:5r Log. tangent = 10.092943
To find the Distance =. A C :—
With the course, thus found, and the difference of latitude A B, the
distance may be computed by trigonometry. Problem IV., page 175 :
hence,
As radius = . . . 90?0<0r Log. co-secant = 10. 000000
Is to the diff. of lat. s 352 miles, Log. = i . . 2. 546543
So is the course = . 5 1 ?5 ' 5 r Log. secant = ,10. 201922
To the distance = 560. 4 miles, Log« s » • . 2. 748465
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MIDDI.B LATITUDB SAILING. 233
Hence^ the course is S. 51?5' E., or S.E. i B.^ nearly, and the distance
560. 4 miles.
To find the Difference of Longitude = C D :—
With the middle latitude = B CD, and the departure B C, the differ-
ence of longitude is found by trigonometry, Problem IV., p^ 175] as
thus :
As radius s • ,. . 90? 0^ Log. co- secant = 10.000000
Is to the departure = 436 miles. Log. = • . . 2. 639486
So is the middle lat. =: 37?5 H Log. secant =s . 10. 102582
To the diff. of long.=s552. 2 miles. Log. = . . , 2. 742068
Longitude of Cape Agulhas = « . « . 20? 1 8 ' E.
Diff. of longitude = 552. 2 miles, or » . 9. 12 £.
Longitude at which the ship airived =: . 29.30 E.
To find the Course, Distance, and Difference of Longitude, by
Inspection : —
Half the difference of latitude = 176, and half the departure = 218,
are found to agree nearest at 51? under or over distance 280 : hence, 280
X 2 = 560 miles, is the distance. Again, to middle latitude 37? as a
course, and departure 218, in a latitude column, the corresponding distance
is 273 ; and to latitude 38? and departure 218, the distance is 277 : hence,
.the change of .distance to 1? of latitude, is 4 miles. Now, 4^ x 51.^ -h
60 = 3' . 4, which, added to 273, gives 276. 4 ; and this, being multiplied
by 2, gives 552. 8 miles ; which very nearly corresponds with the result by
calculation.
- . - . ■ ■ ■■■.., ■■ .- — — ^ ■ ■
Problem VIII.
6w€U one LatUude, Departure, and Difference of Longitude ; to find
the other Latitude, Cour$e, and Distance.
Example.
A ship from the Snares, New Zealand, in lati-
tude 48?3C S., and longitude 166?20: E., sailed
upon a direct course between the south and west,
till she was found byjobservation to be in longitude
151?27- E., and to have made 546 miles of
departure; required the latitude come to, the
course steered, and the distance sailed ?
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284 NAVIQATfON.
Longitude of the Snares = . . 166?20fE.
Long, of the ship by observation = 151. 27 E.
Difference of longitude s • • • 1 4 ? 53 ' :s 893 miles.
To find the Middle Latitude s the Angle BCD:—
With the departure = B C^ and the difference of longitude s= C D^ the
angle of the middle latitude may be found by trigonometry, Problem IIL,
page 174; as thus:
As the diff. of long. = 893 miles. Log. ar. comp. = 7* 049148
Is to radius s: . 90? O: Or Log. sines . 10.000000
So is the departure = 546 miles, Log. = . « 2. 737193
To the mid. lat. = 52? 1 8 '. 28r Log. co-sine sa 9. 78634 1
Twice mid. lat = 104?37' Of nearly.
Lat.ofthe Snares=48. 3. OS.
Latitude come to =56?34f O^S.
Diff. of latitude » 8?3l! 0?s 511 miles.
To find the Course =c the Angle A :—
With the difference of latitude A B, and the departure B C, the eourat
may be found by trigonometry, Problem IV., page 175 ; as thus :
As the diff. of lat. = 511 miles. Log. ar. comp. = 7* 291579
Is to radius s • . 90? 0' Or Log.sine^ . 10.000000
So is the departure = 546 miles, Log. = . • 2. 73/193
To the course as . . 46?5aM8r Log. tangent at 10.026772 .
To find the Distance ss A C :~
Wiih the angle of the course, thus found, and the difference of lati^de
A B, the distance may be computed by trigonometry. Problem IV., page
175 : hence.
As radius « . . 90? 0^ Or Log. co-secant s 10. 000000
Is to diff. of lat = 511 miles. Log. &r . . . 2.708421
So is the course = 46?53M8r Log.secant = . 10. 165378
To the distance s • 747. Smiles, Log. =: • . • 2.873799
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MIDDLB LATITODJI SAILING. 23$
Hence, the course is S, 46? 54^ W., or S.W, J W, nearly^ and the dis-
tance 747* 8 miles.
To find the Latitude come to, Course, and Distance, by Inspection in
the general Traverse Table :—
One^fourth of the difference of longitude = 223^, taken as distance, and
one-fourth of the departure = 136. 5, in a latitude column, will be found to
agree between 52? and 53?. Now, to latitude 52?, and distance 223, the
difference of latitude is 137.3, which is 0\ 8 more than 136.5; and to
latitude 53?, and distance 223, the difference of latitude is 134. 2, being
2'. 3 less than 136. 5 : hence, the difference of meridional distance to 1?
oflatitudeisO'.S + 2'.3 =^3M : therefore, as 3M : 0'. 8 :: 60^:
16 C, which, added to 52? (proportion being made for the quarter of a mile
in the distance), gives the middle latitude ss 52? 18§^ : hence, the latitude
come to is 56?34^ S., and the difference of latitude 511 miles. Again, to
one-fourth of the difference of latitude = 127* 75, and one-fourth of the
departure « 136. 5, the course is 47 ?> and the distance 187*; which, mufti-
plied by 4, gives 748 miles s the whole distance.
Problem IX«
Gken the Distance, Differenae of Longitude, and Middle ZatUude;
to find the Course and both Latitudes.
Eaample.
A ship, in north latitude, toiled 500 miles upon
a direct course between the south and west, until
her difference of longitude was 440 miles; required
the course steered, the latitude sailed from, and
the latitude come to; allowing the middle latitude
to be 43 ?45' north?
To find the Angle of the Course =s A :.^
The course may be found by the 3d analogy, page 222, as thus :
As the distance = ... 500 miles, Log. ar. comp. = 7. 301030
Is to the diff. of longitude 5= 440 miles. Log. = ^ . . 2. 643453
So is the middle latitude =5 43?45' OT Log. co«sine b 9. 858756
* the courses . . S.39?28:i4r W. Log, sine = 9. 803239
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To find the Difference of LaUtude r: A B :—
The difference of latitude may be found by the 8th analogy, page 222j ai
thus:
As the course = . • . 39 ? 28 '14^ Log. co-tangent = 10.084350
Is to the middle latitude = 43.45. 0 Log. co-sine =: • • 9.858756
Soisthediff.oflong. =: • 440 miles, Log. = . . . . 2.643453
To the diff. of latitude = 386 miles, Log. =: .... 2. 586559
Middle latitude =: . . .... 43?45^ N.
Half the diff. of lat. =: 193 miles, or =: . . 3. 13 S.
Latitude of the place sailed from =
Latitude of the place come to = .
46?58: N.
40.32 N.
SOLUTION OF PROBLEMS IN MERCATOR'S SAILING.
Mercator's Sailing is the method of finding, on a plane surfiice, the
motion of a ship upon any assigned point of the compass, which shall be
true in latitude, longitude, and distance sailed.
Mariners, generally speaking, solve all the practical cases in Mercator's
Sailing by stated rules, called canons, which they early commit to memory,
and, ever after, employ in the determination of a ship's place at sea.
Those mnonSf certainly,- hold good in most cases ; but . since they are
destructive of the best principles of science, inasmuch as that they have
a direct tendency to remove from the mind every trace of the elements of
trigonometry, the very doctrine from which they were originally deduced,
and on which the whole art of navigation is founded, the following observa-
tions and consequent analogies are, therefore, submitted to the attention of
naval people, under the hope that they will serve as an inducement to the
substitution of the rules of reason for the rules of rote » and thus do away
with the necessity of getting- canons by heart.
In the annexed diagram, let the triangle ABC
be a figure in plane sailing, in which the angle A
represents the course, A C the distance, A B the
difference of latitude, and B C the departure. If
A B be produced to D, until it is made equal to
the meridional difference of latitude, and D E be
drawn at right angles thereto,, and parallel to B C ;
then the triangle A D E will be a figure in Merca-
tor's sailing, in which the angle A represents the
course, the side A D the meridional difference of _
latitude, and the side D E the difference of longi- ^ ^
tud^. Now, since the two triangles ABC and A DE are right angled.
c/
/ ^
1
B
/
D
mhrcator's sailing. 287
and that the angle A is common to both; therefore they are equi-
angular : and because they are equi-angular^ they are also similar.; there-*
fore the sides containing the equal angles of the one are proportional to
the sides containing the equal angles of the other, — Euclid, Book VI.^
Prop. 4.
Now, from the relative properties of those two triangles, all the analogies
for the solution of the different cases in Mercator's sailing may be readily
deduced agreeably to the established principles of right angled trigono*
metry, as given in page I7I9 and thence to 177 ; as thus : —
First, in the triangle ABC, if the distance AC be made radius, the
analogies will be,
1. As radius I distance A C 1 1 sine of the course A '. departure EC ;
and I * co^sine of the course A I difference of latitude A B.
2. As sine of the course A *. departure B C ! I radius *. distance A C ;
and 1 ! co-sine of the course A I difference, of latitude AB.
3. As co-sine of the course A I difference of latitude A B * ; radius I
distance A C ; and : : sine of the course A '. departure B C.
4. As the distance A C : radius : : departure B C ". sine of the course A ;
and 1 1 difference of latitude A B '• co-sine of the course A.
Agaun, by making the difference of latitude A B radius, the analogies
will be, .
5. As the difference of latitude AB : radius : ! departure BC : tangent
of the course A; and ! *. distance A C I secant of the course A.
6. As radius : difference of latitude A B : : tangent of the course A I
departure.B C ; and 1 1 secant of the course A '. distance A C.
And by making the departure B C radius, it will be,
7. As the departure B C '. radius , ; difference of latitude AB t co- tan-
gent of the course A ; and 1 1 distance A C *. co-secant of the course A.
8. As radius I departure B C 1 1 co-tangent of the course A : difference
of latitude A B ; and ', I co-secant of the course A '. distance A C.
Now, in the triangle ADE, if the meridional difference of latitude A D
be mrde radius, the analogies will be,
9. As the meridional difference of latitude AD '. radius ! ; difference of
longitude D E '. tangent of the course A.
10. As radius I meridional difference of latitude AD.*; tangent of the'
course A • difference of longitude D E.
And by making the difference of longitude D £ radius, it will be^
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238 MAVIGATIOlf.
1 1. As the difference of longitude D E : radius : ; meridional difference
of latitude D £ I co-tangent of the course A.
12. As radius : difference of longitude D E ! I co«tangent of the course
A I meridional difference of latitude A D.
Finally, since the triangles ABC and A D E are equi-angular and simi-*
lar^ we have,
13. As the difference of latitude A B 'departure BC :: meridional
difference of latitude AD I difference of longitude D E.
Hie meridional difference of latitude is found by means of Table XLIII.,
by the same rules as those for the difference of latitude given at page 214;
as thus : — If the two given latitudes be of the same name, the difference of
their corresponding meridional parts will be the meridional differ^ce of
latitude ; but if the latitudes be of contrary names, the sum of these' parts
will be the meridional difference of latitude.
PaoBtEM L
Given the LaHtudee and Longitudes of two Places ; to find the Course
and Distance between them.
2Hff.lmv. nS, p
.£rampfe.
Required the course and distance between
Cape Bajoli, in latitude 40? 3' N., and longi-
tude 3?52; -E., and Cape Side, in latitude
43?2' N., and longitude 5?58: E,?
Lat. of C. Bajoli 40? 3'N. Merid.pts. 2626.6. Longitude 3°52^E.
Lat. of C. Sicie 43. 2 N. Merid.pts. 2865.8. Longitude 5. 58 E.
Diff. of latitude 2?59t Merid.diff.lat. 239.2. Diff. long.2i 61
s 179 miles. s 1 26 miles.
To find the Course = Angle A :—
This comes under the 9th analogy,. in page 237 : hence.
As the merid. diff. of lat. =: 239. 2 miles. Log. ar. comp. =: 7* 621239
Istotadjusz: . . . . 90? 0'. OIT Log, sines . lO.OOOOOO
So is the diff. of long. = . 126 miles, Log. =: • • 2.100371
To tlie course =: • . . 27?46:42r Log. Ungent = 9.721610
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MSRCATOBfs SAILING. 289
To find the Distance = A C :—
This comes under the 6th analogy^ in page 237 : hence.
As radius = ... 90? 0' Or Log. co-secant = . . 10.000000
Is to the diff. of lat s 179 miles. Log. ...... 2. 25285S
So is the course s . 27?46M2r Log. secant b . • . 10.0^3176
To the distance =. 202. 3 miles, Log. = 2.306029
Hence, the true course from Cape Bajoli to Cape Sicie is N. 27?47^ E.,
or N.N £. 4 B. nearly, and the distance 202. 3 miles.
To find the Course and Distance, by Inspection in the general
Traverse Table :— .
The meridional difference of latitude 239. 2, and the difference of longi-
tude 126, as departure, are found to agree nearest at 28?, which, there-
fore, is the course. Now, to course 28?, and difference of latitude 179,
the corresponding distance is 203 miles ; which nearly agrees with the
result by calculation.
Paoblbm IL
Giioen the Latitude and Longitude of the Place eaiied from, the Couth
and Dittance; to find the jLatUude and Lmgitude of the Place come
to.
JHff. limni
Example.
A ship from Cape Ortegal, in latitude 43?47'
N., and longitude 7?49: W., sailed N.W. | N.
560 miles ; required the latitude and longitude
come to ?
A
To find the Difference of Latitude = A B :—
This comes under the 1st analogy, page 237 : hence.
As radius = . . . . 90? 0' . Log. co-secant :c . 10.000000
Is to the distance = . 560 miles, Log. = .... 2.748188
Sp is the courses . . 3i points. Log. co-sine = . . 9.888185
To the diff. of latitude = 432. 9 miles. Log. = • . . . . 2. 636373
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240 NAVIGATION.
To find the Difference of Longitude = D E : —
This comes under the 10th analogy^ in page 237 : hence^
As radius = . • • 90?0: Log. co- secant s . . lO.OOOOOO
l8toiherid.diff.oflat.=641 miles,Log. s: ..... 2.806858
So is the course = • 3^ points^ Log. tangent = . . • 9.914173
Tothediff*oflbng.=526.1 miles, Log. = 2.721029
Lat.of C. Ortegal 43^47 'N. Mer. pts 2927. 8. Long.of C.Ortegal 7^9^ W.
Diff.lat.=433m.or7. 13 N. Diff.long.=526m.or8. 46 W.
Latitudecometo=51? O'N. Mer. pts 3568. 8 Long, come to = 16^35 ^W.
Merid. diff. of lat. = 641 . 0
To find the Difference of Latitude and Difference of Longitude, by
Inspection :—
To course 3| points,, and half the distance = 280, the difference of
latitude is 216.4; the double of which, or 432. 8, is the difference of
latitude: hence, the latitude come to is 51?0' N., and the meridional
difference of latitude 641.
Now, to course Sf points, and one-third of the meridional difference of
latitude = 213. Jy the corresponding departure is 175. 4, proportion being
made for the excess of the given, above the tabular difference of latitude ;
then 175. 4x3 = 526 miles ; which, therefore, is the difference of longi-
tude.
Problem III.
Gwen tlie Latitude and Longitude ofilie Place miUd from, ilie Course,
and tlie Departure, to find the Distance sailed, and the Latitude and
Longitude of the Place come to.
Example.
A ship from Wreck Hill, Bermudas, in latitude
32?15' N., and longitude 64?47^ W., sailed
S.W. i W., and made 340 miles of departure ;
required the distance sailed, and the latitude and
longitude come to ? Y." ^^^ ^^
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mbrcator's sailing. 241
To find the Distance = A C :—
This comes under the 8th analogy, page 237 : hence.
As radius = • . 90?0' • Log.' co-secant s 10.000000
Is to the departure s 340 miles, Log. = . • • 2. 53 1479
So is the course = 4^ points, Lojg. co*secant == 10. 11 1815
To the distance = 439. 8 miles. Log. = . . . 2. 643294
To find the Difference of Latitude == A B :-—
This comes under the 8th analogy, page 237 : hence.
As radius = . 90? . Log. co-secant = . lO.QOOOOO
Is to the departure 340 miles. Log. =: • . « • 2. 53 1 479
So is the course = 4^ points, Log. co- tangent = • 9. 914173
To the diff. of lat.=279 miles. Log. = . . . . 2. 445652
Lat. of Wreck Hill, Bermudas, 329 15 1 N. Merid. parte = 2046. 1
Diff. of latitude = 279 miles, or 4. 39 S.
Latitude come to = . . . 27?36' N. Merid. parts =s 1724.0
Meridional difference of latitudes. • ; 322.1
To find the Difference of Longitude D E :—
This comes under the 10th analogy, page 237 : hence,
As radius = . . 90?0' Log. co-siecant = 10.000000
Is to merid. diff. of ]at.=322. 1 miles. Log. = . 2. 507991
So is the coarse = 4| points. Log. tangent = .10. 085827
To the diff. of long. =s 392. 5 miles. Log. s . . 2. 593818
Longitude of Wreck Hill, Bermudas, = . . 64?47' W.
Difference of longitude = 392^ 5 miles, or = 6.32 W*
Longitude come to = . ... ^ ... 71^19^ W.
The distance sailed b 440 miles, very nearly.
To find the Distance sailed, and the Latitude and Longitude come to,
by Inspection : —
To the course 4i points, and half the departure = 170, the corresponding
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Z4as
NAVIGATION.
difference of ladtude is 139.6, under distance 220 j twice the latter, or
440 miles, is, therefore, the distance sailed ; and twice 139. 6 = 279. 2
miles, or 4?39^, is the difference of latitude: whence the latitude in, is
27?36' N., and the meridional difference of latitude 322. 1. Now, to
course 4| points, and half the meridional difference of latitude a 161 miles,
in a latitude column, the corresponding departure is 196. 3; the double of
which, or 392. 6 miles, is the difference of longitude : hlBnce^ the lou^tude
cometoi8 7ni9iC W.
Problem IV.
Oiven both Latitudes and the Course; to find the Distance and the
JLongiittde m.
Example.
A ship from the east end of Martha's
Vineyard, in latitude 4l?2H N., and lon-
gitude 70?24C W., sailed S.E. i S., and, by
observation, was found to be in latitude
32?21 ' N. ; required the distance sailed^ and
the longitude at which she arrived ?
Lat. of the east end of
Martha's Vineyard = 41?21( N.
Lat. in, by observation =32.21 N.
Ih/f l^nq.
Mend. part^!= 2729.5
Merid.part9= 2053.2
Difference of latitude = 9? OC =540 miles. Merid.diff.oflat=676.3
To find the Distance = ACs~
This comes under the 6th analogy, page 237 ; therefore,
As radius = . . 90?0C Log. co-secant = . . 10. 000000
Is to the diff. of lat. 540 miles. Log. = . . 2.733394
So is the course = 2\ pts. Log. secant = . . . 10. 1 1 1815
To the distance = 698. 6 miles, Log. = .... 2. 844209
To find the Difference of Longitude = D E ; —
This comes under the 10th analogy, pag« 237 \ therefore.
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850 NAVIGATION,
To find the Course and Distance made good ^—
To the whole difference of latitude and departure, so found, find the cor-
responding course and distance by Problem II, page 108, and thus the
course and distance made good will be obtaindl.
To find the Latitude in, by Account, or Dead Reckoning :—
If the difference of latitude and the latitude of the place from which the
ship's departure was taken> or the yesterday *s latitude, be of the same name^
their sum will be the latitude in, by account; but if Uiey are of contrary
names, their difference will be the latitude in, of the same nam* with the
greater quantity.
To find the Difference of Longitude; —
With the course made good, and the meridional difference of latitude, in
a latitude column, find the corresponding departure, by Problem III. page
UO^ and it will be the difference of longitude.
Or,— W^tli the middle latitude as a course, and the departure, in a lati-
tude column, find the corresponding distance, by Problem V., page 111,
and it ^11 be the difference of longitude.
To find the Longitude in, by Account, or Dead Reckoning :—
If the difference of longitude and the longitude of the place from which
the ship's departure was taken, or the yesterday's longitude, be of the same
name, their sum wiD be the longitude in, by account, when it does not ex*
eeed 180? i othenvise, it is to be taken from 360?, and the remainder will
be the longitude in, of a contrary name to thatleft :«— but, if the difference
of longitude and the longitude left are of contrary names, thw difference
will be the longitude in, of the same name with the greater quantity.
To find the Bearing and Distance from the Ship tothe Port to which
she is bound :— .
By Mercator's Sailing.
With the meridional difference of latitude^ in a latitude colunm, and the
difference of longitude, as departure, find the course, by Problem IV. page
111; then, with the course, thus found, and the difference of latitude, the
distance is to be obtained by the same ProbIem.**Or,
By Middle Latitude Sailing,
With the middle latitude between the ship and the proposed place, as a
course, and the difference of longitude, as distance, find the corresponding
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MERCATOH'fl SAILING.
251
meridional distance^ or departure, by Problem VI. page 112; then^ with this
departure, and the difference of latitude, the course and distance are to be
obtained by the same Problem.
Note, — ^The true bearing or course, thus found, may be reduced to the
magnetic, or compass bearing, if necessary, by allowing the value of the
variation to the right hand if westerly ; and to the left hand if easterly ;
being the converse of redtlcing the course steered by compass, to the true
course.
And| this rule .comprises the substance of that nautical operation which
is generally termed a day's work at sea.
Example 1«
A ship from Cape Espiehell, in latitude 38° 35^ norths and" longitude
9? 13' west, bound for Porto Santo, in latitude 33?3^ north, and longi*
tude 16?17-'we8t, by reason of contrary winds was obliged to sail upon
the following compass courses | vis.^W. by S. 56 miles ) N. W. by W,
110 miles; W. N. W. 95 miles ; S. by E. i E. 50 miles ; S. by W. J W.
103 miles, and 8. 8. W. 1 16 miles ; the variation was 2 points westerly on
the three first courses, and If- point on the three last : required the course,
and distance made good, the latitude and longitude at whieh the ship ar-
rived ; with the direct course, and diatance from thenge to her intended
port?
Tkavbrsb Tabls.
•
Corrected
Course!.
Dis-
tances.
Difierence of Latitude.
Deparl
ture.
N. 1 S.
G.
W.
S.W.byW.
W.byN.
West.
S.E.fS.
South.
S.iW.
56
110
95
50
103
116
n
21.5
99
99
»
31.1
40.2
103.0
115.9
»»
>»
>»
29.8
- »
46.6
107.9
95.0
M
>»
5.7
21.5
Diff.Latss
290.2
21.5
29.8
Departure:;^
255.2
29.8
268.7
225.4
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252 NAVIGATION.
To find the G>nr8e and Distance made good :«-
Half the difference of latitude = 134. 35, and half the departure =
112. 7; are found to agree nearest abreast of 40? under distance 175; —
now, 175 X 2 = 350 miles. — Hence, the course made good is S. 40? W.
or, S. W. i S, nearly, and the distance 350 miles.
To find the Latitude and Longitude come to, by Account :—
Lat.ofC.E8pichell= 38925 C N, Mer.pts. 2500. 1. Long.=9M3nV,
Uiff.oflat.=269m9., or 4.29 S. . . . • . Diff. Iong.=s4.40 W.
Latitude come to = 33?56: N. Mer. pts. 2166. 7* Long.= 13?53f W.
Merid.diff.oflat. . / s 333.4
.To find the Difference of Longitude made good :-.
To' the cpurse made good = 40? and half the meridional difference of
latitude = 166. 7 the corresponding departure is 140. 1, which, multiplied
by 2, gives the difference of longitude 280. 2 miles = 4?40.t west. — Or,
^ith the middle latitude = 36? 10^ and half the departure = 112. 7^
in a latitude column, the corresponding -distance is 139.3 (proportion
being made for the 10 minutes of latitude) ; hence, 139.3 X 2 = 278.6
miles, the difference of longitude $ being about a mile- and a half less than
the result by Mercator's sailing.
To find the Course and Distance from the Ship to her intended Port :—
Lat. of the ship 33?56^ N. M. pts. 2166. 7. Longitude 13?53^ W.
Lat. Porto Santo 33. 3 N. M. pts. 2103. 1. Longitude 16. 17 W.
' . ' ' • — — - — [miles.
Diff. of Lat. = 0?53^s53msJd.diff.L.63.6.Diff.LoBg. 2?24'.=:144
By Mercator's Sailing.
The meridional difference of latitude = 63.<$ in a latitude column, and
the difference of longitude s 144, in a departure column, are fotmd to
agree nearest abreast of 66? the course.— Now, to course 66? and differ-
ence of latitude 53, the corresponding distance is 190 miles.— Or,
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BfSRCATOR's SAILING. 253
With the middle latitude = 33^301 as a course, and the difference of
longitude a= 144 as distance, the corresponding difference of latitude is
120. 1.— Now, witli 120, 1 in a departure column, and the difference of
latitude = .53, in its proper column, the eorresponding course is 66? and
the distance 131 miles^
Hence,— The course made goad is S. 40? W. or S. W. | S, nearly.
The distance made good is 350 miles.
The latitude by account is 33*56 '. north.
The long, by account is 13. 53 west
And,
Porto Santo bears from the ship S. 66? W. or W. S. W. nearly.
Distance 130 miles as required.
Note. — If the latitude and longitude bf the ship, or either of them, have
been deduced from celestial observations, they are to be made use of,
instead of those by account, in determining the course and distance between
the ship find the place to which she is bound.— See the compendium of
Practical Navigation near the end of this Volume.
Example 2.
A ship frpm Port Royal, Jamaica, in latitude 17?58' north, and longi-
tude 7&?5S C west, got under weigh for Hayti, St. Domingo, in latitude
18?30' north, and longitude 69?49^ west, and* sailed upon the following
courses, viz. ; S. 40 miles ^ S. £!. by S. 97 miles; N. by E. 72 miles;
S.E.^S. lOStoiles; N/byE.^E. 114 miles; S. E. 126 miles; N.N.E.
86 miles ; and then by observation was found to be in latitude 16?55 '. N.,
and longitude 72?30' W.;— the lee- way on each of those courses was a
quarter of a point (the wind being between E. S. E. i S. and E. by N.'^ N.),
and the variation of the compass half a point easterly : — required the true
course and distance made good ; the latitude and longitude at which the
ship arrived by account, with the direct course and distance between her true
place by observation and the port to which she is bound ?
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254
NAVIOATIOM.
Travbiub Tablb. 1
Corrected
• Courses.
Dis-
tance's.
• Difference of Latitude.
Departure.
. N. 1 S.
E. 1 W.
S.iW.
S. S.E.iE.
N.byE.jE.
S.S.E.|E.
N. by E. } E.
S.E.byS.iE.
N.N.E.JE.
40
97
72
108
114
126
86
99
69.8
107^3
99
77.7
39.6
.87.7
99
92.6
99
101.2
99
41.5
17.5
65.5
38.4
75.1
36.8
5.9
»
»»
»
254.8
Diff. Lat. =
321.1
254. 8
264.8
5.9
5.9
Departure
66.3
258.9 =
To find the Course and Distance made good :^— '
Half the difference of Littitude = 33. 15, and half the departure =
129.45, are found toagree nearest between 75? and 76?, und[er distance
134 ; and by making proportion for the difference between the ^ven' and
the tabular numbers^ the true course will be found :» 75?38t ; and the
distance 134, x 2= 168 miles.^Hence the course made good is S. 75?88^
E. or E. by S, i S. nearly ; and the distance 268 miles.
To find the Latitude and Longitude come to, by Account :-—
Lat.of Port Royal=17?58: N.Mer.pts, 1096. 1 Long. = .76?53^ W.
Diff. Lat. 66.3, or 1. 6 S. Diff. Long. =4. 30 E.
Ut.cometobyac.=:16?52f N. Men pts. 1026.9 Long, by Ace. 72? 23 C W.
Merid. diff. of Lat. = 69. 2
To find the Difference of Longitude made good : —
To the course made good = 75f38rand the meridional difference of
latitude = 69.2, the corresponding departure is 270.3, proportion being
made for the 38^ in the course beyond 75? — Hence, the difference of lon-
gitude.is 270.3, or 4?30: east.— Or, with the middle latitude = 17?25^
as a course, and half the departure made good ss 129.45 in a latitude
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OBLiaUB SAILING. 255
colmniiy the corresponding distance, at top or bottom^ !• 1S5 } which^
multiplied by 2, gires the dMhrence of longitude ts 270 miles.
To find the Course and Distance from the Ship to her intended Port : — .
Lat. of ship by ob. 16?55 ^ N. Mer. pts. 1030. 1 Long, by ob. 72?30^ W.
Lat. ofHayti= 18.30 N. Mer. pte. 1129.8 Lg.ofHayti69.49 W.
Diff. of Lat =r 1 fSS ', M.D.L. i= 99. 7 Diff. of Long. 2MI C
= 95 miles* ' a 161 miles.
The meridional difference of latitude as 99. 7, and difference of longi-
tude = 161^ in a departure column, are found to agree nearest between
58? and 59? under distances 188 aiid 194 ; and by making proportion for
the difference between the given and the tabular numbers, the true course
will be found » 58? 14 ^— Now, to course 58? 14^ add difference of lati-
tude 95, the corresponding distance is 180 miles.*— Or, with the middle la-
titude s 17?42i^ as a course, and the difference of longitude =s 161 as a
distance, the corresponding difference of latitude is 153.4:-— now, with
153. 4, in a departure column, and the difference of latitude sr 95, fai its
proper column, the course, nearest agriseing, is 58 degrees, and the dis-
tance 181 miles.— Hence,
The Courte made good is S. 75?38: E. or E. by S. i S. neariy.
Distance made good a 268 miles.
Latitude cojne to by account » 16? 521 N.
Latitude by observation s • . 16?55' N.
Long, come to by account » • 72?231 W.
Long, by observation =: . . 72?30C W.
Hayti bears from the ship N. 58? 14 C E. or N. E. by E. ^ E. nearly.
Distance 180 miles, as required.
Abte.— This example and the preceding exhibit all the particulars
attendant on making out a day*$ work at sea. — See mofe of this in the com-
pendium of Practical Navigation near the end of this Volume.
SOLUTION OP CASES IN OBLIQUE SAILING.
Oblique sailing is the application of oblique angled plane trigonometry
to the solution of certaim cases at sea : such as in coasting along shore ;
approaching, or leaving the land ; surveying coasts and harbours, &c., where
it becomes necessary to determine the distance of particular places from
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the ship^ and from each other.— And^ alao, when it is required to settle the
position of any place^ cap^ or head-land from a ship, by obsenrations taken
on board.
Example 1«
A ship being about to take her departure from Madeira, set the lizard
Point, which bore, by azimuth compass, N. W. by N. ; and after sailing
S. W. 20 miles, it was again set and found to bear N. i E. ; required the
ship's distance from the Lizard at both stations.
SoJuiion. — ^In the annexed diagram
let the point C represent the Lizard,
and the points A and B the stations or
places of the ship, whence the bearings
of the point C were taken.— Now, the
difference between the bearing A C =3
N. W. by N. and the ship's course
A B =^ S. W. is 9 points, which is the
value of the angle BAG, measured by
the arc a i : — The difference between
N. W. by N. and N.iE. is 3i points
ss the angle A C B, measured by the
arc 6 d ; and the difference between N. | E. and N. £. (the opposite
point to S. W.) is 3^ points = the angle ABC, measured by the
arc d e.— Then, in the oblique angled triangle ABC, given the an-
gles and the side A B, to find the sides A C and B C = the distance of
the ship from the Lizard at the respective stations.-^Hence, by oblique
angled trigonometry. Problem L, page 177*
To find the Distance A C :--»
As the angle C = • • . 3} pts. Log. co-secant =
Is to the distance A B= 20 ms.. Log. = . • .
So is the angle B = 3i pts. Log. sine =:
To the dist. A C=: 17- 74 ms.. Log. n . . .
To find the Distance B C :—
As the angle C = . . 3} pts. Log. co-secant
Is to the distance A B =: 20 ms,. Log. = • » • •
So is the angle A =: 9 pts. Log. sine r: •
To the dist, B C =29. 21 miles^ Log. =: . .
10. 172916
1.301030
9.775027
1.248973
= 10.172916
. 1.301030
. 9.991574
. 2.465520
OBUQ0B SAILING. 257
Hence^ the distance of the ship from the Lizard ajt the first station is I?}
miles } and at the second station 29i miles nearly.
* Example 2. '
Two ships sail from the same port^ one N. W. by W. 80 miles^ and the
other S;W. 68 mil^s; required the bearing and distance* of those ships
from each other ?
Solution.-^ln the annexed diagram let the
side A C represent the course steered by one
of the ships, and the side AB the course
steered by the other ship 3 and let the side.
BC represent the relative bearing and distance
of the. ships from each other.— Now, the
difference between the bearing A = N. W.
by W. and the bearing A B = .S. W. is 7
points = the angle BAG, measured by the ^
arca£. — Hence, in the oblique angled triangle ABC, given the side
AC 80 miles; the. side AB 68 miles,* and the included angle A=:7
points; to find the other angles, and the pideBC— Therefore, by ob-
lique angled trigonometry, Problem IfL, page 179,
To find the Angles B and C :—
As the sum of A B and A C = ' 148 miles, Log. ar. compt. =7. 829738
Is to difference of A B and A C := 12 miles, Lpg. = . . . 1 . 079 1 81
So is i sum of angles B and C=; 50?37/30r Log. tangent = ,10. 085827
To i diff. of angles B and C = 5?38:32r Log. tangent = 8, 994746
Angle B= . . . 56?16C 2^
Angle C± , . 44^58^58r
To find the Side B C =: the Distance between the Ships :-«
As the angle B = 56? 16.' 21 Log. cosecant = . . 10. 080066
Is to the side AC = 80 miles, . Log. = . ...... .1. 903090
So is the angle A= 7 points Log. sine = .... .9. 991574
To distance BC:=: 94.34 miles, Log. =: • • . . . 1.974730
s
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2SS NAVIGATION.
To find the relative Bearing* of the Ship« :—
From the angle B =: 56? 16^ 2'^ subtract the course finom A to B = 45?,
and the remainder = 1 1 ? 16' 2^' is the bearing of C frpm B == N. 1 1? 16'
W. or N. by W. nearly,— And frpm the course A C = 56? ISC subtract
the angle C = 44?58C58^ and the- remainder = 1 l?16C2r ia the course
from C to B = S! 1 1? 16' B. or S. by E. nearly.
Exampls 3.
Coasting along shore two head-lands were observed ; the first bore, by
azimuth compass, N.N.E., the second N.W.:— after swlingW. by S.
16 miles, the first bore N. E. i E. and the second N. by E. J E. j required
the relative bearing and distance of those head-lands from each other ?
SolutionJ-^ln the diagram ABDC,
let the side A B represent the course
steered by the ship; AC the bear-
ing of the first head-land, and AD the
bearing of the second head-land from
the place of the ship at A; and, let
B C represent the bearing of the first
head-land, and B D the bearing of the
second head-land from the shipf's
place at B.-^— Now, in the triangle
ABD, the angles and the side A B ^ ••...^--•-*
are giveit, to find the side A*D. — ^
Thuff, the difference between N. W, and W. by S. is 5 points ±: the an-
gle BAD| measured by the arc. a e;— the difference between N. byE.
i E. and E. by N. (the opposite point to W. by S. the ship's course^)
is 5^ points = the angle DBA, measured by the arc c d, and the differ-
ence between N. by E. ^ K and N. W* is 5^ points = the angle A D B,
measured by the arcce; and the side AB = 16 miles; to find the
side AD. — Hence, by oblique angled trigonometry. Problem L, page 177,
As the angle A D B =: 5| pts. Log; co-secaat =: 10. 054570
Is to the side AB= 15 miles. Log. = . . . .1.176091
So is. the angle AB D = 5^ pts. Log, sine = . . 9. 945430
To the side AD. t= 15 miles. Log. = . . 1.176091
Nofe.— The side A D might be determined independently of calculation,
aa thus j the angles B and D are equal, for each ia meaanred by aa ctre of
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OBL^aU^ 9AIUN6. S59
5| poiQta ; and since equal sKigles are subtended by eqnal ni^es^ therefore
the side A D is equal to the side A B =: 15 miles.
' Ag^ih.-^Inthe triangle A BC^ the angles and the aide A B aia given,
to find the side AC; thus^ the difference between N. N. B. and W.by S.
ia 11 points =: the angle BAG, measured by the arc abj the difference
between N. N; Q. and N« G. i £• is 2i points =: the angle AC B, mea-
sured by the arc 6 g, and the difference between N. B. i E. and £. by N.
(the opposite point to W. by S. the ship's course,) is 2f points = the angle
ABC, measured by the arc g d :-— t^nee, the side A C may be found by
tl)e abqve-pi^ntioned Problem ; as thus :
•
As the ang^e A C B =s .3} poinU Log. coHBeeiiita 10. 868008
Is to the side AB =: 15 miles, Log. = . • .1. 176081
Soiatheang^eABC =: SfpoiaULog.sines . 9.7110M
Tpdue»HeACf3l8*93mi}ef/. Log, c ., , }.^{f6149
Now, in the. triangle ADC there are pven, the side A D a 15 utiles }
the side A C =s18p 03 miles, and the included angle D AC, 6 points s the
difference between N. N. £• and N. W. mea/Blir^ by thelare eb,Ui find the
angles ADC and A C D, and the side D C.-r^Hence, by trigonometry^ Pror
blem III., page 178,
As the sum of A C and A B =: 33. 03 miles, I^g» air. co)npt.=s8. 481081
IstodiffcrenceofACandAB = S.03. Log, = . • 0.481443
Soi8 4sumofangs,ADCandACD=56n5f Ori4)g.tang.= 10,l75107
To I difference of those angles :? : . 7?49.' 2rLog.taDg.=: 9, 137fi41
Angle ADC s= . . 64? 4f 2f
Angle ACD= . . 49?3S'.SBr
To find the Side DC r—
As the angle A C D = 48?25f 58f Log. co-secant = . i 10. 125995
Is to the side AD =15 miles. Log. = ..... 1. 176091
Soi8theangl9DAC-= 6 points Log. sine 7 . , . . 9.965615
To the side DC = 18.52 milesj Log. = . . . . , 1.267701
Hence, the distance betweeb the two head-lands is 15^ miles.
s 2
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To find the relative Bearings of the two given Head-lands z-—
To the angle A C D = 48?25 ' 58r add the course or bearing from A to
C = 2 points, or 22?30' and the sum = 70?55^58r is the bearing of D
from C = S. 70956^ W., or W. by S. J S. nearly— And, to the angle
A D C = 64?4:2r add the bearing from A to D - 4 points, or 45 de-
grees, and the sum = 109^4^2^ being taken from 180? gives 70?55C58f
s the bearing of C from D s N. 70?561 E. or B. by N. i N. nearly.
Esample 4.
Being desirous of ascertaining the exact position of a head-land, with
respect to latitude and Idngitude, it was carefully set, by an azimuth com-
pass, and found to bear N. Jb. E., and after sailing N.W. b. W« 12 miles,
it was again set, and observed to bear E. b. N. i N., due allowance being
made for the variation of the compass. Now, the correct latitude of the
ship at the last place of observation was 21?50C.2K N., and the longitude
85?9'6T W. ; required the latitude and longitude of the said head-land ?
Solution.— In the oblique angled
triangle ABC, where the side AC
represents the first bearing of the
head-land, the side B C the second
bearing, and the side A B the dis-
tance sailed; given the three angles
and the side AB = 12 miles, to find
the side B C = the ship's distance
from the headland at the second sta-
tion. Thus, the difference between
N. b. E., and N.W. b. W., is 5 points
= the angle.C AB, measured by the
arc ad; the difference between. N.
W. b. W., and E. b. S. i S., the opposite point to E. b. N. J N., is 4|
points = the angle ABC, measured by the arc de, and the difference
between E. b. N. i N., and N. b. E., is 5 J pointe = the angle AC B,
measured by the arc ab. Hence, by. oblique angled trigonometry. Pro-
blem I., page 107, to find the side BC = the ship's distance from the
head-land at the second station.
As the angle A CB = SJ points. Log. co-secant = 10. 054570
Is to the side AB =: 12 miles, Log. = . . . 1.079181
Sois tlieangleCABr: 6 points, Log. sine = . 9.965615
To the side BC =* 12. 57 miles, Log. = . . . 1.099366
OBLIQUJB SAILING, 261
Hence, the distance of the ship from the head-land at the second station
is 1 2^ miles, nearly.
To find the Difference of Latitude and Difference of Longitude between
the Ship's Place at B, and the Head-Land C*:—
In the right angled triangle BCD, given the angle C; 6^ points = the
bearing of B from C, and the distance BC = 12..57 miles, to find the
difference' of latitude C D, and the difference of longitude BD 3 therefore,
by Mercator's Sailing, Problem IL, page 239,
As radius = 90<?0C Log. secant =: . v 10.000000
Is to distance B C = 1 2. 57 miles^ Log. == 1 . 099366
Sq is the course C=6§ points, Log. co-8iner=9!. 462824
To the diff. of lat CDszS. 65 miles, Log.=: &. 562190
As radius =: 90?0' Log.co-secant =: . ; 10.000000
Is to mer. diff. of lat, = 3.9 miles. Log. c: . 0.591065
So is the course C =: 6} points, LfOg. tangent= 10. 518061
To the diff. of long. = 12: 85 miles, Log. = 1. 109126
Lat.of8hip=:21?50^2KN. M.pts=1343.3 Lon.ofship=85? 9^ 6rW.
Diff.lat.3.65,or3^39rN. Diff.lon.l2.S5,orl2.51 B.
Lat.ofhd.W.21?54^ OrN. M.pt8=1347.2 Lon.ofhd.ld.84?66:i6rW.
Meridional difference of latitude =3.9 miles.
Hence, the latitude of the head-Und is 2if54;0? N., and its longitude
84?56M5fW.
Note. — ^The foregoing examples contain all the oases in oblique sailing
that are of any immediate import to the mariner. Other examples, indeed,
might be given ; but since they would rather tend to the exercise of the
mind on trigonometrical subjects, than to any useful nautical purpose, they
have therefore been intentionally omitted.
The two last examples will be found particularly useful in maritime sur-*
veying, when the operations are conducted on board of a ship or vessel.
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262 NAVlGAttOH.
. ,SOLtJTION OP CASBd IN WiNDWARD SAILINCJ.
Windward Sailing ia the method of reachihg the pott or pla^e bbund to
by the shortest roiit^^ when the wind is in a direction Contrary to*the direct
course between the ship and the pUc6 to whieh ()he is boundi
When the wind is opposed to th6 course which a ship should steer from
any one port to another^ she is obliged to sail upon different tacks^ close-
hauled to the witid^ in order to reach the port bound to. The object,
therefore^ of this method of sailing, is to find the prbper course to be
conned on feach tadk, sp that the ship may.ikrrite at the place to which
she is bound, lathe shortest time possible.
EtBomfle 1%
A ship tliat can lie within 6 points of the wind is bound to a port 50
miles directly to windward, which it id intended she shall reach on two
tacks ; the first being oa the starboard tack) and the wind steady a,t N. b. £• j
required the course and distance to be run upon each tack ?
jSblutton.-— Since the ship cto •
lie within six points of the wind,
vMnh iA lit N. b. E., the course oh
the starboard tapk will be N.Wi
b. W., and that on th# larboard
tack E. b. N. Now, in the annexed
diaigtratn, let the side A C represent
the course and distance between
the ship and her intended port;
A B the course and distance to be
ttiade good on the starboard tack;
end B C the course and distancis to
be made good on the larboard tack. Then^ in the triangle ABC, Uni
three angles ane given' to find the side AB or BC> which sides ait muto-
ally equal to each other, because the triangle is isosceles, and its Verted at
B =r this .ahgte comprehended bet^eeh those sides^ Thus^ the diflfierence
between N. b. B., And N.W. b; W.^ is 6 points =i the angle B AC, mea-
sured by the arc ai; the difference between E. b. N., and S.E. b. E. (the
opposite point to N.W. b. W.), is 4 points, measured by the arc d e ; and
the difference between N« b. E., and E. b. N., is 6 poiQts, measured by the
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WIKDWAIID iAILING.
263
are Ad ; and since Che distance AC is ^ven =: 50 miles^ the side A B^ or
its equal B C^* may be readily determined by oblique angled trigonometry.
Problem I.^^ page 177 1 as thus :—
As the angle B '== 4 points. Log. co-secant =s 10. 150515
Is to the distance AGs 60 miles. Log. s 1.698970 •
So is the angle C s= 6 points. Log. sine , = # 9. 965615
To the distance A B s 65. 33 miles, Log. = 1. 815 100 .
Hence^ it is evident that the ship must run 65. 33 miles on the starboard
tack^ and 65.33 miles on the larboard tack, before she can reach her
intended port.
Example 2.
A ship that can lie within 6 points of the wind is bound to a port bearing
NJB. b. N., distance 90 miles, which it is intended she shall reach on three
lacks, with the wind steady at north ; required the course and distance to
be run upon each tack, the first course being on the larboard tack ?
<SoIti(um.<i— Since the wind is
at north, and that the ship can
lie within 6 points thereof, the
course on the larboard tack will be
EJI.E., and that on the starboard
tackW.N.W.
In the annexed diagram, let the
NJE. b.N. line AB =± 90 miles,
represent the bearing and distance
between the ship and her intended
port^ let the E.NJBi. line AD re-
present the first board on the lar-
board tack| and, parallel thereto,
the line B C = the second board on that ti^^k. And, since the ship is to
make her port in three tacks, it is evident that the board on the starboard
tack, represented by the W.N.W. line CD (parallel to dg), must bisect
the line A B in the point F; and that, therefore, AF and FB.are equal to
one another, each being equal to 45 miles == half the line, or distance- A B.
N0W5 since the straight line A B falls upon the two parallel straight lines
CB and AD, it makes the alternate angles equal to ofie another; there-
* Since Uie aog^les A and C are equal to one another^ the sides which subtendy or are
oppotite to those socles (vii., B C and A B)^ are also equid to oae anotber^<-£uclid, Book
I., Prop. 6.
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264 NAVIGATION.
fore the angle A B C is equal to the angle BAD.— -Euclid^ Book I., Prop.29.
And because the straight line C D falls Upon the two parallel straight lines
CB and AD, it makes the angle AD B equal to the angle BCD, by the
aforesaid proposition. And because the two triacigles A D P and B C F
have, thus, two angles of the one equal to two angles of the other, viz., the
angle F A D to the angle F B C, and the angle A D F to the angle B C F ;
and the side A F of the one equal to the sid^ B F of the other t therefore the
remaining sides A D and D F of the one are equal to the remaining sides
B C and C F of the other, each to each ; and the third angle A FD of the
one equal to the third angle B F C of the other. — Euclid, Book I., Prop. 26.
Now, since the two triangles A F D and B P C are, thus, evidently equal to
one another, we have only to compute the unknown sides of one, viz., of
the triangle AFD, where the three angles are given, and the side AF, to
find the sides At) and FD3 thus, the diJQference between N.E.b. N. and
E.NJB., is 3 points, zr the angle FAD, measured by the nxcab: the differ-
ence between E.N.E. and E.S£. (the opposite point to W.N.W.), is 4
points = the angle A DP, measured by the arc bg ; and the difference
between WJI.W. and N.E« b.N., is 9 points == the angle AF D^ measured
by the arc ad: hence, by oblique angled tr^onometry. Problem I.^
page 177,
To find the Side A D .—
As the angle D = 4 points. Log. co-secant =10. 150515
Is to the side AP = 45 miles, Log. =5 . 1. 653213
. So is the angle P = 9 points, Log. sine = 9. 991574
To the side AD = 62.42 miles, Log. s 1.795302
To find the Side FD:—
As the angle D ;= 4 points, Log. co-secant = 10. 150515
Is to the side AP = 45 miles, Log. = • . . 1.653213
So is the angle A = 3 points^ Log. sine = • 9. 744739
To.the side P D = 35 . 35 miles, Log. =? • • 1 . 548467
Side DC ss. . . 70. 70 miles..
Hence it is evident that the ship must first run 62. 42 miles on the lar-*
board tack ; then 70. 70 mihes on the starboard tack ; and, again^ 62. 42
miles on the larboard tack, before she can reach her intended port.
Example 3.
A ship that can lie within 6 points of the wind is bound to a port bear-
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WINDWAED SAILING.
2M
ing N,N.W., distance 120 miles, which it b intended she shall make o
four tacks, with the wind at N.b. W. The coast, which is to the east
wai-d, trends in a direction nearly parallel to the bearing of the port, so*
that the ship must go about as sopn as she reaches the straight line joining
the two ports ;• required the course and distance to be run upon each tack,
on the supposition th^t ti^e ship's progress is not affected by either leeway^
or currents ?
Solution. — Since the wind is
N. b« W., and the land trends in a
N.N.W. direction, the first board,
therefore, must be on the starboard
tack ; and, as the ship can lie .
within 6 points of the wind, the
course on the starboard tacks will
be W. b. N., and that on the lar-
1>oard tacks N.E. b. E.
In the annexed diagram, let the
N.N.W. line AB, 120 miles, re-
present the bearing and distance
between the ship and the port to
which she is bound; let the W.'b.N. S
line A D represent the first board
on the starboard tack, and FC, parallel to AD, the second board on that
tack ; let the N.E. b. E. line D F represent the first board on the larboard
tack, and, parallel thereto, the line CB = the second board on this tack*
And, since the ship is to make her port in four tacks, without going to the
eastward of the line AB, therefore, at the end of the second tack, she
must reach the point F, which bisects or divides the distance A B int6 two
equal parts, of 60 miles each ; thus making A F = to A B.
Now, because the strught line A B falls upon the two parallel straight
lines AD and FC, it makes the angle BFC equal to the interior and
opposite angle FAD : and, because the straight lin^ AB falls upon the
two parallel straight lines FD and CB,' it makes the angle A FD equal
to the interior and opposite angle C B F,< — ^Euclid, Book I., Prop. 29. And,
since the two triangles AFD add FB C have, .thus, two angles of the one
equal to two angles of the other, viz., the angle A F D to the angle F B C,
and the angle F A D to the angle BFC, and the side A t^ of the one equal
to the side F B of the oth^r,— ^therefore the remaining sides *A D and D F
of the one, are equal to the remaining sides FC and C B of the other,, each
to each ; and the third angle A D F of the one equal to the third angle
p C B of the other.— Euclid, Book 1., Prop. 26. .
The two triangles A D F and F C B, being, th^s, clearly equal to one
S66 NAVlGATtOM.
•
another in «very reipect, we have only to compute ttie unknown sides of
one^ vIb., of thd triangle A PD, where the three angles are given, and the
side A F bi 60 miles^ to find the sides AD and D F^ thus the diflPerehee
between.N.N.W. and W.b.N., is 6 points =3 the angle FAD, measured
by the arc 06} the difforenoe between W. b. N. and S.W, b.W. (the
opposite point toN.E;b.E.)> is 4 points =a the angle AD P, measured by
the arc a e ; and the difference between N.N.W. and N.E. b. E.^ is 7 points
= the angle A F D, measured by the arc b d.
Hence, by oblique angled trigonometry, Problem L, page 177^
To find the Side AD s= P C :—
As the angle D = 4 points, Log. co«secant ac 10. 150515
Is to the side A F == 60 miles, Log. « • . . 1.778151
So is the angle F = 7 points, Lo!g. sine s= • 9. 991574
To the side AD » 83. 22 miles. Log. = . . L 920240
To find the Side D P = C B :—
As the an^e D = 4 points, Log. co-secant = .10. 150515
IstothesideAF=60iniles,Log. = ... 1.778151
So is the angle A =s 5 points. Log. sine = • 9. 919846
To the side D F & 70. 50 miles^ Log. >= . 1 .8485 12
From this it is manifest, that the ship must first run 83. 22 miles upon
the starboard tack j then 70. 5$ miles upon the larboard tack ; then 83% 22
miles again upon the starboard tack ; and 70. 55 miles upon the larboard
tacki before she can reach the port to which she is bound.
SOLUTION OF CASES IN CURRENT SAILING.
Current Sailing is the Aiethod of determining the true course and distance
made good by a ship, when her own motion is affected or combined with
that of the current in which she stuls.
A current Is a progressive motion of the water, causing all floating
bodies thereon to move in the direction to which its stream is impelled.
The setting of a current is that point of the compass towards which the
water runs ; and the driftot a current is the rate at which it runs per hour.
When a ship saik in the direction of a current, her Velocity will be equal
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CtfftRSNt SAtLlNG. i67
to the sum of her own proper motion and the current's drift ; but when
she sails directly against a current, her velocity will be expressed by the
difference between her own proper motion and the drift of the ourrent : in
this case^ the absolute motion of a ship will be a-headj if her proper velo-
city exceeds the drift of the current i but if it be leisi, she will make stern-
way. Whea a ship's course is oblique to the direction of a current, her
true eoun^ and distance will be compoutjded of thfl course tind distance,
given by the log, and of the observcid setting and drift of the current.
When a ship's course and distance by the log, and the setting and drift
of the cuRcnt in which she sails are given, the true course atid distance
made good may be found by a trigonometrical solution of the triangles
forming the figure } but the easiest and most eKpeditibus method of finding
the course and distaifce made gobd^ particularly when a ship sails upon
different courses, is by resolving a traverse, in which the dettmg and drift
of the cutrent are to be esteemed as an additional course and distance to
thos^ exhibited by the log.
Example L
If a ship sails S.W. b, W.> at the ratfe of 4 knots an hour, in a current
setting S.SJi.1 E., at the rate of H miles ah hour; required the course
Md tMtaAdi madegdod in 24hmir8 \
iSbZuHoit.-^r X 24t = 96 miles, the distance aailtd^ by log» in 84
hoars}
And U? X 24 = 42 miles, the observed drift of th< current in 24
hours.
In ^e ume^ed diagrani^ tet Che
aide AB of the tiriangle ABC-
represent the course and distance
sailed by the log, and the ride 6 C
parallel to d& the setting atid drift
of the current; then^ the side AC
Will represent the course And dis-
tance made good in the given time.
Now, in the triangle A B 6, ^ven
the side A B ^ 96 miles, the side
B C = 42 miles, aiid the included
angle B «= 8i points, being the
(fifference bet^veen S.S Jl. i E. and N.E. b. E. (the opposite point to
S.W.b.W.), Pleasured by the arc a 5, to find the.aiigles A and C, and the
true distance A C. Hence^ by oblique angled trigonometry, Problem III.,
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To find the Angles A and C :—
As AB + BC = 138 miles. Log, ar, comp. = 7. 860121
IstoAB-BC = 54 miles, Log. = . /' . L732S94
So is i sum of the angIes=43?35<37^T Log. tangent=9. 978673
To i diff. of the angles - 20. 25 . 59 ' Log. tangent=9. 57 1 188
Angle C=.
Angle A =
64° 1^36ir
239 9r38if
To find the true Distance = AC : —
As the angle C = 64?l'.36|r Log. co-secant = 10.046241
' Is to the side A Br: 96 miles. Log. = . . • . 1.982271 .
So is the angle B = 8i points, Log. sine = ... 9. 999477
To the true distance = AC r: 106. 7 miles. Log. = 2. 027989
To find the Course made good :— -
From the angle SAB = S.W. b. W., or 56? 15^, subtract the an^^le
C AB = 23?9^38ir,and the remainder, 33^5^21*^ = the angle SAC,
is the course made good.
Hence the course made good is S. 33? 5' W., or S,W. b. S. nearly, and
the .distance 106} miles nearly.
To find the Course and Distance made good by the Traverse Table :—
Travbrsb Table. 1
Corrected
Courses.
2
Difference of Latitude.
Departure. 1
N.
S.
E.
W.
S.W.b.W.
Current
S.S£fE.
96
42
:
53. S
36.0
21.6
79.8
Diff.lat.=
•
89.3
21.6
Dejpart =
79.8
21.6
58.2
CURRBNT SAILING.
269
Now, by Problem II., page 108,
The difference of latitude 89. 3, and the departure 58. 2, are found to
agree nearest abreast of 33?, under or over distance 107.
Hence the course made good is S. 33? W., or S.W. b. S,, and the dis*
tance 107 iniles ; which nearly agrees with the above result.
Mxample 2.
Suppose a ship sails N.W. 65 miles, W.N.W. 70 miles^ and N. b. E. 71
miles, in a current th^t sets S.E. b. S. 36 miles in th^ same time ; required
the true cQurse. and distance made good ? ,
Teavjihsb Tablb
«
Corrected
Courses.
O
S."
•
Difference of Latitude.
Departure.
N.
S.
E.
W.
N.W.
65
46.0
— ■
46.0
W.N.W.
70
26.8
—
■— ,
64.7
N.b.E.
71
69.6
—
13.9
•«
Carfent
SJl.b.S.
36
•^
29.9
20.0
—
142.4
29.9
33.9
110,7
29.9
33.9
112.5.
76. 8
So&ifioii.— With the difference of latitude and departure, thus found, the
course and distance made good may be djetermined by Problem II., page
108 ; as thus :
The difference of latitude 112. 5, and the departure 76. 8, are found to
agree nearest abreast of 34? under or over 136.
Hence the direct course made good is N. 34? W., or N.W, b. N. nearly,
and the distance 136 miles.
To find the Course and Distance made good by Calculation : —
This may be done by means of the 5 th analogy, page 237 > as thus :
To find the true Course :— .
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AS ine ain« oi iat« =
U to radiuf ^ • »
So is the depairture *«;
1 iz. D jLiOg. af« compx
90?Q: Log. sine, s ,
76,8 liog. w . .
la 000000
h 885361
Tp the true course = 34? l9'A2iLog. tangent = 9. 834209
To find the true Distance :—
As radius = . . . 90?0; Log. co-s^cant =;: 10.000000
Istodiff.oflat. = .112.5 Log,= . . . 2.051152
So is the true course=34? 19' 12rLog. secant = , 10. 083072
To the distance 3= 136. 2 miles. Log. = .
2.134224
Hence the course made good is N. .34? 19' W., or N.W. b. N. neariy^
and the distance 136 miles.
Example 3.
There is a harbour 2 miles broad, in which the tide is running N.W* b. N.
at the rate of 3 miles an hour. Now, a waterman who can pull his boat
at the rate of 5 miles an hour, wishes to cross the harbour to a point 09
the oppoiSte side bearing E.N.E.; required the direction in whidi he should
pull^ so as to meet with the Iqast possible resistance from the force of the
tide in gaining the intended poinl^ Md the time that it wiU take him to
reach that point }
SpluUott^'^ince the principles
of this Problem are but little un-
derstood by the generality of young
navigators, a brief account of the
geometrical construction will be
given, with the view of elucidating
and rendering familiar the nature
of the corresponding calculations.
Thus,
With the chord of 60? describe the arch N E S W j draw the porth and
south line NS, and, at right angles thereto, the east and west line WP;
make the arc N a = 3 points, and draw the N.W.b. N. line A a D, which
make equal to 3 ladles (t^en from a.ny scale pf jsqu^l pants), to xepresent
the direction of the harbour ; perpendicular thereto draw the N.E. b. E. line
AbC, whiA wake e^ tp 2 miles, to reprcseoC die bxttadth of Ae har-
bour ; and, from the point C, draw the line C G parallel to AD, which lines
will represent the ctastern ^d w^fim ^tmf^ ^ tiw harbour respectively.
CTTRUtfT f AILINO. 371
Make N c equal to 6 points^ and draw the E.N.E. line Ac¥, cutting C G
in B; then will 6 represent the point to which the waterman intends to
cross. Take 5 miles in the compasses ; place one foot on the point D ; and
where the other falls upon the E.NJB. line A F, there make a point, as at F,
and draw the line DF; parallel to which, draw the line A G, and it will
represent the distance and direction in which the waterman must pull to
gain the point B : for in the time that he would reach the point G, by pull-
ing at the rate of 5 mUes an hour, the tide, running at the rate of 3 miles
an hour, would carry him to the poirft B ; because B G bears the .same pro^i
portion to 3 miles an hour that A O does to 5. Now, AG, being applied
to the same scale of equal piurts firom which the other sides were ti^en^
will measure 2, 95 miles, and the angle G AE^ or eAB, being applied to
the line of chords^ will measure 13?33t ; benee the direction in which he
should puU, is £• 18?33 C S^ or E. b. S. ^ S« nearly*
Now, in the triangle A D F, given the side A D s: 3 milesy the side D F
=t 5 miles, and the angle D A F :=: 9 points (being the difference between .
E.N.E« and N.W. b. N.,. measured by the arc a c), to find the angle A F D.
Hence^ by oblique aisled trigonometry, Problem L, page 1 779
As the side DF = 5 miles. Log. ar. comp. =: 9.301030 '
Is to the angle A :: 9 points. Log. sine =: . 9. 991574
So is the side A D = 3 miles, XiOg. =: ... 0. 47712 1
To the angle A FD = 36?2:55^ Log. = . 9769727
Now^ because the straight line A F falls upon lik two parallel straight
lines D F and AG, it makes the alternate angles equal to one another ;
therefore the angle DFA is equal to the angle FAG,---Euclid, Book L,
Prop. 29 j but the angle DFA is known to be 36?2'55'/; therefore the
angle FAG, meosuredby thearece, is also equal to 36? 2 '55.^; ancUif to
the ang^ FAG weaM the angle BAG =: U?i5' (being the difference
betwe^i NJS. b. E# and S.N.E., measured by the arc j& c)^ the sum ^
47?17-55^is the angle C AG, measured by thearc be. Then^
In the right angled triangle A CG, given the angle C A G == 47? 17'55r
and the side A C z= 2 miles, the breadth of the harbour, to find the ride
AG equal to the distance which the waterman must pull befiare he ean
reach the pomt B. Hence, by rigbt angled trigonometry, Probleip U*,
page 172, making A C radius^
As radius =: . , 90?0;: Log. co-secant = 10.000000
IstothesideAC= 2miles,Log. = * .; . 0.301030
So is the angle CAG=:47?17'55^Log.secant=10. 168657
V»thedHtaice s AO s 2«949 Mies. Leg* 9 0.4aM»7
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To find the Tinje requisite ts reach the Point B :*-
M distance 5 miles. Log. an comp. = 9.301030
Is to 1 hour^ or 60 minutes. Log. == . 1. 778151
So is AG = 2.949 miles, Log. = . . 0.469687
Tothetime=35r23;. 34=35". 389 Log.= 1.548868
To find the Dii^ction in. which he should pull or steer :— -
. From the angle 6Ae = 47?17'55T, take away the angle 6AE =
33?45f, and the remaining angle £ A e = 13?32f55r is the direct course
which he should steer .j viz., E. IS?3S'. S., or.E. b. S. i S. nearly.
Hence it is evident, that if the waterman pulls in the direction of E.
IS'tSS' S. or E. b. S. i S. nearly, he will reach the intended point in the
space of about 35 minutes and 23 seconds.
SOLUTION OP PROBLEMS RB^iATlVE TO THE ERRORS OF
THE LOG- LINE AND THE HALF -MINUTE GLASS, BY
LOGARITHMS.
The instruments generally employed at sea, for finding the distance run
by a ship hi a given time, are the log-line and the hdf-minute glass.' Now,
since a ship's reckoning is kept in nautical miles, of which 60 make a
degree, the distaiice between any two adjacent knots on the log-line should
bear Ihe same proportion to a nautical mile that half a minute does to an
hour ; viz., the one hundred and twentieth part. Arid, since a nautical
mile contains 6080 feet, the true length of a knot is equal to 6080 divided
by 120; that is, 50 feet and 8 inches : but, because it is advisable at all
times to have the reckoning a-head of the ship, so that the mariner may
be looking out fbr the land in, sufficient time, instead of his making it
unexpectedly, or in an unpre|3ared moment, 48 feet, therefore, is the cus-
tomary measure allowed to a knot. And, to make up for any time that
may be unavoidably lost, in turning the half-nrinute glass, its absolute
measure should not exceed twenty^-nine seconds and a half.
The method of finding the hourly rate of .sailing, or distance run in a
given time, by the log-line and the half- minute glass, is subject to
many errors : thus, a new log-line, though divided with the utmost care
and attention, is generally found to contract after being fii^t used;
■ 0..: ■ /V- '-. >-■ ^ ^■^'-''
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LOG-LINB AND HALF-MINUTB GLASS. 273
and^ after some. wear, it stretched so very considerably as to be out of due
proportion to the measure of the half-minute glass. Nor is the 'half-
minute glass itself free from error : for this instrument is. so very liable to
b^ affected by various chuiges of weather, from moist to dry, and con-
verseLy, that notwithstanding its being perfectly correct when first taken on
board, yet it alters so sensibly at sea, that at One time it will run out in
the short space of 26 or 27 seconds, and at another not till it has passed
the half-minute by several seconds. • Hence it becomes indispensably
necessary to examine those instruments frequently; and, if found erso^
neous, to correct the ship's run accordingly. . This may be done by means *
of the following rules, whioli are adapted to a log-line of 48 feet to a
knot, and to a glass measuring 30 seconds.
Problem I.
Given the Distance sailed by the Log, and the Number of Seconds run
by the Glass; to find the true Distance, the lAne being truly divided.
Rule.
To the arithmetical complement of the logarithm of the number of
seconds run by the glass, add the logarithm of the distance given by the log,
and the constant logarithm 1. 477121* ; the sum of these three logarithms,
abating 10 jn the index, will be the logarithm of the true distance sailed.
Eaiample 1.
Let the hourly rate of sailing be 1 1 knots, and the time measured by
the glass 33 seconds ; required the true rate of sailing I
Seconds run by the glass = 33, Log. ar. conip. = 8. 481486
Rate of sailing, by log = 11 knots, Log. =: . . L 041393
Constantlog. = . .'. '''. 1.477121
True rate of sailing = 1 0 knots. Log. = . . * . 1 . 000000
Example 2.
If a ship sails .198 miles by the log, and the glass is found, on exam-
ination, to runout in 26 seconds, required the true distance sailed ? '
Seconds run by the glass z= 26, Log. ar. comp. zz 8. 585027
Distan(5e sailed by log = 198 miles, Log. =: • 2.296665
Constant log. = ........... 1.477121
True distance sailed = 228. 46 miles. Log. = . 2. 35881*3
* This is the logariihm of 30 secoDds^ the tru^ measure of the half-minute glass.
T
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274 NAVIGATION.
. . PaOBL£M II.
Given the Distance sailed by the Log, and'the measured Length of a Kffot;
tofindthetnie Distance, the Glass beimg correct.
Rule.
To the logarithm of the distance given by the log, add the logarithm of
the measui-ed length of a knot, and the constant logarithm 8.318759*;
the sum of these three logarithms^ rejecting 10 in the index, will be the
logarithm of £he true distance sailed.
Example 1.
Let the hourly rate pf sailing be 9 Hnot^, by a log-line which measures
53 feet to a knot ; required the true rate of sailing ?
Hourly rate of sailing IT 9 knots. Log, = . 0.954243
Measured length of a knot = 53 feet, Log. =: . U 724276
Constant log. = ........-• 8.318759
True rat^ of sailing = 9. 937 knots, Log. = . . 0. 997278
Example 2.
Let the distance sailed be 240 miles, by a log-line which tneasures 43
feet to a knot ; required the true distance sailed ?
Distance sailed by Jog e 240 miles. Log. »2. 38021 1
MeaiBured length of a knot ss 43 feet, Log. = 1. 633469
Constant log. = .......... 8.318759
True distance sailed s= 215 miles, Log; s , 2.338439
PaoBLBM IIL
Given the measured Length of a Knot^ the Numher of Seconds run hy the
Glass f and the Distance sailed by tlie Log ; to find the true Distance
sailed^
Rule. '
To the arithmetical complement of the logarithm of the number of
seconds run by the glass, add the logarithm of the measured length of a.
knot, the logarithm of the' distance sailed by the log, and the constant
* Ttils is the arithmetical complement of the logarithm of 48, the gcneraHy»approyed
IcD^hof aknot. • •
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LOG-LINB AND HAJLF-MIN'UTB GLASS. 275
logarithm 9. 795880*; the sum of these four logarithms, rejecting 20 from
the index, will be the logiarithm of the trae distance sailed
Example 1.
Let the hourly rate of sailing be 12 knots, the measured length of a
knot 44 feet, and the time noted by the glass 25 seconds ; required the
true rat^ of sailing ?
Seconds run by the glass = 25, Log. ar. comp.=8. 602060
Measured length of a knot=44 feet. Log. = 1 . 643453
Rate of sailing by log :fs 12 knots. Log. = !• 079181
Constant log. = g. 795880
True rate of sailing = 13. 2 knoU, Log. = . 1 . 120574
Example 2.
Let the distance sailed by the log be 354 miles, the measured length of
a knot 52 feet, and the interval run by the glass 34 seconds ; required
the true distance sailed ?
Seconds run by the glass = 34, Log. ar. comp.= 8.468521
Measured length of a knot ^ 52 feet. Log. = • 1. 716003
Distance sailed by log = 354 miles, Log. ss 2. 549003
Constant log. a • • • 9. 795880
True distance = 338. 38 miles, Log. s . . . 2. 529407
Pboblem IV.
Gioen the Number, of Seconds run by any Glass whatever, to find the
correiponding Length of a Knot, which shall be truty proportional to
the Measure of thai Glass.
RULB.
To the logarithm of 10 times the number of seconds run by the glass,
add the constant logarithm 9. 204120, and the sum, abating. 10 in the
index, will be the logarithm of the proportional length of b, kndt, in feet,
correqM>nding to the given glass.
* This is thesQin of the two precedioi: constant lo^thmt ; thus 1.477121 .-f 8.318759
« 9. 795880.
T 2
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276
NAVIGATION.
Example 1. * '
Required the length of a knot corresponding to a glass that runs 27
seconds ? ■
Number <}f seconds 27 x 10 = 270 Log. - 2.431364
Constant log. = ...... . N • • 9-2^120
True length of a knot, in feet, = 43. 2 Log. = 1 . 635484
Example 2.
Required the length of a knot corresponding to a glass that runs 34
seconds?
Number of seconds 34 x 10 = 340 Log. = ?. 531479
..Constant log. = 9.204120
IVue lengtli of a knot, in feet, = 54. 4 Log. = 1 . 735599
SOLUTION OF A PROBLEM IN GREAT CIRCLE SAILING,
Very usefid to Ships going to Van Diemen's Land, or to New South fValeSf
by the way of the Cape of Good Hope.
Great Circle Sailing is the method of finding the successive latitudes
and longitudes which a ship nuist make ^ with the courses that she must
steer, and the distances to be run upon such courses, so* that her track may
be nearly in the arc of a great circle, passing through the place sailed firom
and that to which she is bound.
The angle of position is an angl6 which a great circle, passing fhrough
two places on the sphere, makes with the meridian of one of them ; and
shows the true position of each place, in relation to the intercepted arc of
the great circle and the respective meridians of those places.
The polar angle is an arc of the equator intercepted between the meri-
dians, or circles of longitude, of two given places on the sphere.
On the sphere, the shortest distance between two pl^es is expressed by
the arc of a great circle intercepted between those places: consequently
the spiral, or rhumb line, passing through two places on the sphere, can
never represent the shortest distance between those places, unless such
rhumb line coincides with the arc of a great circle ; and this can never
hi^pen but when the places are situate under the equator, or under a
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GRBAT CIRCLB SAILING. 277
meridian. Hence^ although Mercator's Sailing resolves correctly all the
cases incident to a ship's course along the rhumb line passing through two
places^— yet, since there is no case in which the course, along the direct
rhumli line indicates the shortest distance between those places^ except
when they both lie under the same meridian, or under the equator, the
distance, therefore, obtained by that method of sailing, must always exceed
tlie truth (the above-mentioned positions excepted) ; and the nearer the
places are to a parallel of latitude, and the farther they are removed from
the equator, the greater will be the error in distance.
Now, since it is f)-equently an object of the greatest importance, to the
commander of a ship, to reach the port to whi^h he is bound by the short-
est route, and in the least tiitie possible,-^particularly to the commander of
a ship bound from the Cape of Good Hope to Van Diemen's Land, or to
His Majesty's Colony at N<ew South Wales, where the length of the voyage
generally occasions a great scarcity of fresh water, — the following Problem
is, therefore, given^ by which all the particulars connected with the shortest
possible route between those places will be fully and clearly Exhibited,
Were a ship to sail exactly in the arc of ^ a great circle (not under the
equator or upon a meridian), the navigator would be obliged to keep con-
tinually altering her course ; but, as this would be attended with more
trouble and inconvenience than could be reasonably admitted into the
general practice of navigation, it has been deemed sufficiently exact to
determine, a certain number of latitudes and longitudes through which a
ship should pass, with the relative courses and distances between them ; so
that the track, thus indicated, though not exactly in the arc of a great
circle may, notwithstanding, approximate so very near thereto,, as not to
produce any sensible difference between it and the true spherical track.
Problem.
Given the Latiiudea and Lmigitudes of two Places on the Globe, to deter-
mine the true spherical Distance betwfien iliem ; together with the angu-
lar Position of those Places with respect to edch other, and the successive
Positions at which a Ship should arrive when saiUng on or near to the
Arc of a great Circle, agreeably to any proposed Change of Longitude.
Rule. .
1 • Find the true spherical distance between the two given places, by
oblique angled spherical trigonometry. Problem IIL, page 202.
2. Fmd the highest latitude which the great circle touches that passes
through tl|e two given places 5 th^t is, find the perpendicular froii> th^ pole
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878 NAVIGATION.
tQ that circle by right angled spherical trigonometry, Problem 11., pag^ 185 ;
and (ind, also^ the several polar angles (made by the proposed alterations
Qf longitude^) contained between the perpendicular, thus^ found, and the
several meridians corresponding to the successive changes of longitude.
' . 3.' . With the co-latitude or perpendicular, sa found, and the several
polar angles, compute as many corresponding co-latitudes by right angled
spherical trigonometry, Problem IV,, page 188.
4. With {he several latitudes and longitudes through which the ship is
to pass, compute die corresponding courses and distances by Mercator's
Sailing, Problem I., page 238 ;^ and they will indicate the path along which
a ship must sail| so as to keep nearly in the arc of a great circle.
Note, — ^The smaller the alterations are in the longitude, the nearer will
the track, thus determined, approximate to the ti^th; because, in very small
arcs, the difference between the are and its corresponding chord, sine, or
tangent, is so very trifling, that the one may be substituted for the other,
in most nautical calculations, without producing any sensible di£ference in
the result. •
. Example I,
A captain of a ship bound from the Cape of Good Hope (in latitude
34?24: S., and longitude 18?32^ E.) to New South Wales, being desirous
of making the north point of King's Island, at the western entrance to Bass'
Strait (in latitude 39?37' S., and longitude 143?54; E.), by the shortest
possible route, proposes, therefore, to sail as near to the arc of a great
circle as he can, by altering the ship^s course at every 5 degrees of longitude ;
required the latitude at each time of altering the ^course, and, alsoj the
respective courses and distances between those several latitudes and longi-
tudes made by the proposed changes ?
Cape of Good Hope, Latitude = 34?24CS. Longitude = 18?32^E.
King's I^&d^ N. point, Latitude = 39. 37 S. Longitude = 143. 54 E.
Difference of longitude = 1 25 ? 22 !
Stereographic Projection*
With the chord of 60 degrees, describe the primitive circle SENQ on
the plane qf the meridian, or circle of longitude passing through the Cape
of Good Hope i dra^ the line E Q to represent the equaitor, and, at right
angles thereto, the line S N for the earth's axis ; then S represents the
south, or elevated pole, and N the north, or depressed pole. Take the
latitude of the Cwpe of Good Hope in the oompassei from the line of
ehorda = S4?1t4C, and. lay it off from Q to A ; draw the diameter A C ^
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GREAT CiaCLS I{AILIN6.
279
and, at right angles thereto, the diameter ^c v. Take the latitude of King's
Island = 39?37' in .
the compasses from ^
the line of chords,
and- lay it off from
Q to r, and also from
£ to r ; and, with the
tangent of its com-
plement = 50?23':
draw the parallel cir-:
cle rr. Take the
difference of longi-
tude 125? 22^ from
the scale of semi-
tangents, and lay it *
off on the equator
from Q to m: thus
90? will reach from
Q to C; then the
excess above 90?, viz., 35?22C, will r^ach from C to m. With the secant
of the complement of the excess of the difference of longitude above 90?
= 54?38' (being the supplement of the difference of .longitude to 180?),
describe the great circle S m N ; the intersection of which with the' parallel
circle rr at B shows the position of King's* Island* Then, the great circle
SBmN reprcjsents the meridian of King's Island. Through the three
points AB / describe a great circle, and then will the arc A B represent the
true spherical distance between the Cape of Good Hope and King's Island ;
in which A represents the place of the former, and B that of the latter.
Through P, the pol^ of the great circle A B f, draw the great circle 8 F P N;
then the arc S F will be perpendicular to the arc A B. Hence, S P repre-
sents the least co-latitude at which the ship should arrive in her spherical
passage from the Cape of Good Hope to King's Island ^ which, being
reduced to the primitive circle, and measured on the scale of chords, gives
about 31 J degrees. The arc AB, reduced to the primitive circle, arid
measured on the line of chords, shows the true spherical distance to be
about 90| degrees. The angle S A B is the aingle of position which the
meridian of the Cape of Good Hope makes with King's Island; and the
angle SB A is the angle of position which the meridian of King's Island
makes with the Cape of Good Hope. These angles, being reduced to the
primftive circle, and measured on the line of chords, give about 39? for the
former, and 42^9 for the latter.
Note.-^Tbe remaining parts of the projection will be explained here-
after.
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Calculntion.
In the oblique angled spberical triangle A S B, there are given two sides
and the included angle, to find the remaining angl^ and the third side;
viz.) the side AS = 55?36C, the co-latitude of the Cape of Good Hope ;
the side B S = 50?23<, the co -latitude of King's Island; and the angle
A S B = 125^22'., the difierence of longitude between those places, to find
the true spherical distance A B, and the respective angles of position SAB
and SB A. The distance may be readily found by Remark 1 or 2, to
Problem III., page 203 or 204 ; as thus :
DifF.oflong,ASB =
125922: H- 2 = 62?4llTwicelog.sine 19. 897300
Co-lat. of Cape of
Good Hope=AS 55. 36 Log. sine =z 9. 9165 14
Co-Iat. of King's
l8land=BS . 50. 23 Log. sine = 9.886676
Sum = 39. 70049a
Diflf.ofco-lats. = 5n3: Half= 19.850245 . . 19.850245
Half diff. of do. ^ 2?36:30'^Log.sine= 8. 658090
Arch=i, . . . 86?19^27':'Log,T.= 11.192155 Log.sine9. 999106
HalfthesideAB = 45. 13. 8| Log. sine =: 9.851139
Side A Be . . 90^26^7^; which is the true spherical distance
between the two given places.
To find the Angle of Position at Cape of Good Hope s^ Angle S AB :-"
This is found by Problem I., page 198 ; as thus :
As Uie distance AB . 90?26n7^ Log. co-secant=10. 000013
Is to' diff. of long. A SB 125.22. 0 Log. sine = , 9.911405
So.is the co-lat. - BS . 50.23. 0 Log. sine =: • 9.886676
To the ang. of posit. SAB 38955 C 1 r Log. sine = . 9. 798094
To find the Angle of Position at King's Island == Angle SB A : —
This 18 found by Problem I,, page 198 ; as thus :
As the distance A B . 909 26^ 1 7^ Log. co-secant = 10. 000013
Istodiff. oflong. ASB 125.22. 0 Log. sine = . 9.911405 .
So is the co-lat. AS « 55.36. 0 Log. sine = . 9.916514
To the ang. of pos. SB A 429l7:20irLog,rine = . 9.827932
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GREAT CIRCLE SAILING. 281
To find the Perpendicular FS = the Complement of the highest southern
Latitude at which the Ship should arrive in the proposed Route : —
Here we have a choice of two right angled spherical triangles, viz., ASlF
fUid B S F ; in ' each of which the hypothenuse and the angle at the base
are given, to find the perpendicular. Thusj in the triangle ASF, given
the hypothenuse AS, 55°36' = the co-latitude of the Cape of Good
Hope, and the angle at the base, S AF 38?55 ' K = the -angle of position
at that place^ to find the perpendicular F S = the complement of the
highest latitude at which the ship should arrive. Hence, by right angled
spherical trigonometry. Problem II., page 185,
As radius = 90? 0^ OT Log. co-sec.= 10. 000000
Is to co-lat. C. Good Hope A S = 55. 36. 0 Log. sine = 9. 9165 14
Soistheang.of position .SAF= 38, 55. 1 Log. sine = 9.798094
To the perpendicular PS = . 31. 13. 13|- Log. sine = 9.714608
Highest lat. at which the ship
should arrive = . « . , » 58?46'.46|^ south.
* Hence the true spherical distance between the Cape of Good Hope and
the north point of King's Isltuid, is 90?26.' 17^, or 5426.3 miles 3 the
angle of position at the Cape of Good Hope, is 38?55'1''; and that at
King's Island, 42? 17-20^ ; and the highest southern latitude at which the
ship should arrive, 58 ?46C 46'/. Now, by Mercator's Sailing, the course
from the Cape of Good Hope to King's Island is S. 87?r. E., or E. i S.
nearly^ and the distance 6011. 2 miles; whence it is evident, that if a ship
sails on the direct rhumb line indicated byMercator's Sailing, she will have to
run a distance of no less than 585 miles more than if her course had been
shaped along the arc of a great circle passing through the two given places.
Now, since it is extremely difficult for persons unacquiunted with the
doctrine of spherics to reconcile a route to their senses, as the shortest
distance between two places, which carries them nearly 22 degrees to the
southward of the middle latitude between the two given places ; and since,
in sailing on the arc of a great circle;, the course ought to be changing
constantly, with the view of keeping the side of the polygon on which the
ship sails as near to the arc of its circumscribing circle as possible, or that
the difference between the arc and its chord may be so small that the one
may be substituted for the other without sensibly affecting the result in
nautical operations,— -I shall, . therefore, show the successive latitudes at
which the ship should arrive at every 5 degrees of longitude, as proposed
(which is sufficiently near to preserve the desired ratio between the arc and
its chord) i together with the respective courses and distances, by Merca-^
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282. NAVIGATION.
tor's Sailing, between- those several successive latitudes and longitudes :
then, if tbe jsum of the several distances coincide, or nearly so, with the
true spherical distance found as ab^ve, the senses must become reconciled
to the propriety of adopting that high southern route at which they
originally seemed to recoil.
In order to determine the several successive latitudes at which the ship
muist arrive, we must previously compute the vertical or polar angles ASF
and B S : then, if the sum of these angles make3 up the whole difference
of longitude, or polar angle between the two ^ven places, it will be a
convincing and satisfactory proof that, for so far, the operations will have
been properly conducted. Now, in the right angled sphtrical triangle
ASF, given the hypothenuse A S, 55? 36' = the co-latitude of the Cape
of Good Hope, and the perpendicular F S, 31 ° 13' 13^^ == th^ complement
of the highest latitude at which the ship should arrive, to find the vertical
or polar angle F S A. And, in the right angled spherical triangle B S F^
given the hypothenuse B S, 50?23' =? the co-latitude of King's Island, and
the perpendicular F S, 3 1 ? 1 3^ 13^r, to find the vertical or poliir angle B S F«
Hence, by right angled spherical trigonometry. Problem I., page 184,
To find the Polar Angle A S F .-^
As radius ±5 90? 0' 0? Log. co-secant = 10.000000
Is to the CO' latitude A S s S5. 36. 0 Log. co-tangent a 9. 83550d
So is the co-latitude FS =: 31.13. 13^ Log. tangent « 9.782550
To the polar angle ASF = 65-? 28 C48r Log. co-sine = 9.618059
• . To find the Polar Angle BS F:—
As radius = . . • . . 9#? 0^ 07 Log. co- secant = 10.000000
Is to the co-latitude B S = 50. 23. 0 Log. co-tangent = 9. 91^906
So isthe co-latitude PS =a 31. 13. 13^ Log. tangent = . 9.782550
To the polar angle B S F = 59?53 H 27 Log. co-sine = . 9. 700456
And, since the sum of the polar angles, thus obtained, viz., ASF
65?28U8r + BSF59?63U2'r = 125^22^07, makes up the whole dif-
ference of longitude between the two given places expressed by the whole
angle A S B, it shows that thus far the work is right.
Now, on the equator, from Q to m, lay off the proposed changes of Ion*
gitude, viz., 5?, 10% 15?, 20?, 25% &c. These are to be taken respectively,
in the compasses, from the scale of semi-tangents, reckoning backwardB
from 90? t&wards 0?, till the proposed changes, of longitude reach the
centre C ; and then forv^ards on that scale, or from 0? tawards 90?, till
those changes of longitude meet the point m; thns, the extent from 90?
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GREAT CIRCLB SAILING. 283
to 85? will reach from Q to 5? ; the extent from 90? to 80?, will reach
from Q to 10?, and so on to the centre C ; then, the extent from 0? to- 5?,
will reach from C to 95? ; the extent from 0? to 10?, will reach from C to
100?, and so on to the point m. Through the points S and N, and the
several points made by the proposed changes of longitude on the equator,
draw arcs of great circles, viz., S 1, 5? ; S 2, 10? ; S3, 15? ; S 4, 20?, &c.
&c. ; and then the arcs S 1, S 2, S 3, &c. &c., will represent the respective
complements of the several latitudes at which the ship should arrive at the
given changes of longitude ; the true values of which may be found in the
foUoiving manner, viz..
From the polar angle ASF, subtract the proposed changes of longitude
continually ; and the several polar angles made by those changes, and con-
tained between* the perpendicular FS and the- co-latitude of the Cape of
Good Hope = S A, will be obtained. Thus, from the polar angle A S F =±
65?28U8r, let 5? be continually *uA/racfed, and the results will be FS 1
= 60?28!48r;FS.2= 55?28M8r-FS3 =; 50?28:48r, &c. &c. And,
since the last . subtraction in this trif^ngle leaves the remainder, or polar
angle, FS 12= 5?28M8r, which is 28'.481 greater than the proposed
alteration of longitude, therefore, in the triangle HSF, where the polar
angle S is 59?53n2? (and where the several polar angles contained
between the perpendicular F S and the co-latitude of King's Island are to
be determined by a contrary process to that which was observed in the pre-
ceding triangle), the first polar angle is expressed by 5? — 28'48f =
4?31M21^ = the angle FS a; to which let the proposed alterations of
longitude be continually added, and the sums will be PSA = 9?31M2f ;
FSc = 14?81'12T, &c. &c. Those various results are to be arranged
agreeably to the form exhibited in the first column of the following Table ;
and, since they respectively express the true measures of the several polar
angles contained between the meridians of the given places and those of
the several co^latitudes to which they correspond, it is, therefore, manifest
that those results reduce the two right angled spherical triangles (ASF
and BS F)' into a series of right angled spherical triangles; to each of
which the perpendicular FS is common. Then, in each of these triangles,
we have the perpendiciriar aiid the angle adjacent, to find the hypothenuse
or co-latitude. Thus, in the right angled spherical triangle F S 1 , right
angled at F, given the perpendicular FS = 31? 13 '131^, and the polar
angle PS 1 = 60?2SM6r, to find the hypothenuse or co-latitude S 1 ; in
the right angled spherical triangle FS2, given the perpendicular FS =
31?l3nSjr, and the polar angle FS 2 = 55?28C48r, to find the hypo-
thenuse or co-latitude S 2, &c. &c. Hence, by right angled spherical
trigonometry. Problem IV., page 188,
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284 NAVIGATION.
• To find the Hypothenuse^ or Co-Latitude = S 1 :—
As the perpendictirar FS =» 31913^ 13^^ Log. co- tangent = 10. 217450*
Is. to the radius = . • 90. 0. 0 Log. sine = . 10. 000000
So is the angle F S 1 = . 60. 28. 48 Log. co-sine = 9. 692607
To the co-latitude S 1 = 50. 53. 28 Log, co-tangent = 9. 910057
' First latitude == • ., .. 39? 6'32/rS.^ at which the ship should arrive.
To find the Hypothenuse^ or Co^Latitude = S 2 : —
As the perpendicular FS = 31?13a3jr Log. co-tahgent = 10. 217450*
Is to the radius = . . 90. 0. 0 . Log. sine = . . 10^ 000000
So.is theangleFS2" = 55.28.48 Log. co-sine == . 9.753349
To the co-latitude S2 = 46.55.29 Log. co-tangent = 9.970799
Second latitude = . . 43? 4'3KS.^ at which the ship should arriye.
Hencc^ the first latitude at which the ship should arrive^ is 39?6^32^S.;
and the second latitude 43?4'3KS. : and', since it is the latitude itself^
and not its complement, that is required, if the log. tangent of the sum of
the three logarithms be taken, it will give the latitude direct ; and, by
rejecting the radius, the work will be considerably facilitated. Proceeding
in this manner, the several successive latitudes corresponding to the pro-
posed alterations of longitude will be found, as in the third column of the
following Table.
Npw, let the several successive lohgi.tudes be arranged (agreeably to the
proposed change, and to the measure of the corresponding polar angles,) as
given in the second column of the following Table ; cmd find the difference
between every two adjacent longitudes, as shown in the fourth colunan of
that Table. Find Ihe difference between every two successive latitudes^
and place them in the .fifth column of the Table. Take out from Table
XLIII. the meridional parts corresponding to the several successive lati-
tudes, as given in^ column 6, and find the difference between every two
adjacent numbers, as given in the seventh column. Then fuid, by Merca*
tor's Sailing, Problem I., page 238, the respective courses and distances
between the several successive latitudes and longitudes; and let those
courses and distances, so found, be arranged as in the two last columns of
the following Table : viz..
* The log. co-tangent is used, so as to avoid the trouble of findin|^ the arithmetical com-
plement of the log. tangent.
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Now, the sum of the several successive differences of longitude = 7522
miles, rAakes up the whole difference .of longitude between the two given
places; the sum of the successive differences of latitude == 2612* 53 miles,
is equal to the whole difi^rence of latitude comprehended under the high-
est latitude at which the ship should arrive, and the latitudes of the two
given places, viz. 34?24!0r S., 58?46^46ir S., and 39?37'0r S.— And,
'the sum of the several meridional differences of latitude = 3973. 85 miles,
coincides exactly with the whole meridional difference of latitude corres-
ponding to the highest latitude, and the latitudes of the two given places ;
which several agreements, form an incontestable proof that the work has
been carefully conducted.
The sum of the several distances measured on the consecutive rhumb
lines intercepted between the successive latitudes and longitudes, as exhi-
bited in the last column of the Table, is 5426.46 miles;— but the true
spherical distance on the arc of a great circle is 5426. 30 miles ; the diff-
erence, therefore, is only 0' . 16 ; or, about ^ of a mile ; which is very tri-
fling, considering the extent of the arc< — ^The distance byMercator's sailing
is 6011 . 2 miles ; which is 583 miles more than by great circle sailing.
Hence, it is evident that the shortest and most direct route from the Cape
of Good Hope to King's Island is bythelatitude of 58?46M6f^ S.; and that
the ship must make^ successively, jthe several longitudes and latitudes con-
tained in the 2nd and 3rd columns of the Table, in the same manner, pre-
cisely, as if they were so many headlands, or places of rendezvous, at which
she was required to touch.— The first course, therefore, from the Cape of
Good Hope is S. 40?22^ £. disUnce 371 miles, which will bring the ship
to longitude 23?32; £. and latitude 39i?6'32r S.;--the second course is S.
43?31' £. distance 328 miles, which brings the ship to longitude 28?32'.
E. and latitude 43?4^3ir S.; the third course is S. 46?56: E. distance
292 miles, which brings the ship to longitude 33?32^ E. and latitude
46 ? 23' 39 'r S ;— and so on of the rest.— Whence, it is evident that if the
ship sails upon the several courses, and runs the corresponding distances
respectively set forth in the two last columns of the Table, she will, most
assuredly, arrive at the several successive longitudes and latitude^ pointed
out in the 2nd and 3rd columns of that Table ; and thus will she reach
King's Island, the place which it is intended she shall make, by a track
585 miles shorter than if such track had been determined agreeably to the
principles of Mercator's sailing.
And, in a long voyage, like the present, in which ships generally expe*
rience a great scarcity of fresh water, particularly those bound to His Ma-
jesty's colony at New South Wales with troops, or convicts, the saving of
585 miles run at sea becomes a consideration of no inconsiderable import-
ance.
Nor is there ^any more difficulty in sailing on the arc of a great circle^
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GBBAT cmCLV SAILING. 287
thus detennined, than there is in saijifig on a parallel' of latitude ; for, if
the ahip's compate be but tolerably good, the variation thereof carefully
attended to, and proper attention paid to the steerage, the courses and dis«
tances expressed in the two last columns of the Table will, undoubtedly,
cany the ship direct from the Cape of Good Hope to the north point of
King's Island, without ever referring to celestial observation for either lati-
tude or longitude ; provided^ indeed, that the ship's way is not affected by
current? : — butj since the courses contained in the 8th column of the
Table, express the true bearings between the several successive latitudes
and longitudes through which the ship must pass ; these must, therefore,
be reduced to the magnetic, or compass course, by allowing the observed
variation to the right handthereof if it be westerly, but to the left hand if
easterly ; this being the converse process of reducing the magnetic, or
course steered by compass, to the true course. — And, if the spherical
track, so determined, be delinelEited on a Mercator's chart, it will, perhaps,
not only simplify the navigation, but also point out to the mariner
any known land that may be adjacent thereto * ; and thus enable him to
alter his course as occasion may require. — ^The spherical track may be rea-
dily ddineated on a chart by means of the angles of meeting made by the
several latitudes and longitudes, which show the places or points where the
ship is to alter her course :— -Now, those points being joined by right lines
will indicate the triie courses and distances, or the absolute route on which
the ship must sail from the Cape of Good Hope to King's Island ; then, if
each day's run be carefully measured on the track, so delineated, (he navi-
gator can always know his distance from the place to which he is bound,
without resorting to the trouble of calculation.
I have dwelt at considerable length upon this Problem for the express
purpose of simplifying a sulyect which is but very little understood by the
generality, of maritime people: — and, with the view of rendering it still more
familiar^ another example will be given by which the approximate spherical
route, as performed' by I^is Majesty's ship Dauntless, under the command
of .George Cornish Gambier, esq. on her voyage from Port Jackson to Val-
paraiso, in the year 1822, will be clearly illustrated.
Example 2.
His Majesty^s ship Dauntless being bound from Po^t Jackson, in latitude
33?52^ S. and longitude 151? 16^ E. to Valparaiso, in latitude 33^^ S.
and longitude 71^52' W., the captain, G. C. Gambier, Esq., proposed to
navigate bet as near to the arc of a great circle as he could, by altering
her course at every 5 degrees of longitude ; required the latitude at each
* It Is pres amed that ther^ is not any laxxl to intercept a ship's progress in this track.
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288
NAVIGATION.
time of altering th^ course^ together with the respective courses and dis-
tances between those severaMatitudes and longitudes, occasioned by tlie
proposed changes ?
Port Jackson,
Valparaiso,
Latitude 33?52^S. Longitude = 151?16' E.
Latitude 33. 1 S. Longitude = 71.52 W.
Sum =
223? 8'
Difference of longitude between the two given places = 136?S2'
Calculation.
Since the elements
of this Example are
analogous to those of
the last; it is not,
therefore, deemed ne-
cessary to repeat the
mode of projection ;
the only difference in .
the. construction being *l'
that, in the preceding
diagram, because the
ship is bound to a place
to the eastward of that
from which she is to
sail ; the latter is, there-
fore, for the sake of uni-
formity, placed on the
primitive circle in the
western hemisphere.:-— and, in the present diagram, because the ship is
bound to a port to the westward of that from which she is to sail, the lat-
ter (for the sake of uniformity also) is placed on the primitive circle in the
eastern hemisphere : — the letter Q representing the western hemisphere in
the forjner case, and the eastern hemisphere in the latter.
Now, the figure being thus constructed on the plane of the meridian
passing through Port' Jackson-; let the point A represent that place; the
point B, the place of Valparaiso, and the arc AB, the true spherical dis-
tance between those places ; — then, A S represents the co-latitude of Port
Jackson ; B S, that of Valparaiso ; S AB, the angle of position at the for-
mer place, and S B A, the angle of position at the latter place. — ^The arc
F S, which is drawn perpendicular to A B, represents tbe complement of the
highest latitude at which the ship should arrive ; and the several arcs S j ;
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GRBAT CIRCLE SAILING. 289
S2j S3; S4; &c. &c. &c., represent the complements of the successive
latitudes through whith the ship must pass. -^HencCj in the oblique angled
spherical triangle A B S, two sides uid the included angle are given to find
the third side and the remaining angles; viz., the side AS =s 56? 8^ the
co-latitude of Port Jackson; the side BS= 56?59' the co-latitude of
Valparaiso, and the angle ASH = 136^^52^ the difference of longitude
between those places ; to find the spherical distance AB, and the respect-
ive angles of position = SAB and SB A : — the distance may be readily
found by Remark 1, or 2, to Problem IIL, page 203 or 204 ; as thus :
Diff.long.ASB136?62C ^-2 =
68?26: Twice log. S.=:19. 936958
Co^atitude of Port
Jackson » A S 56? 8'. Log^ sine = 9. 919254
Co-Iat. of Valpa-
raiso :^ B S 56?59' Log. sine = 9. 923509
Sum ST. .39.779721
Diff. of co-lat. s 0?5H Half =: . .19. 889860^ 19. 889860}
Halfdiff.of ditto= 0?25:30r Log. S. = 7. 870262
Arch=± . • . . 89?25U7!rLog.tang.=12.019698iLbgS.=:9. 999979
HalfthearcAB=50?53:56rLog.sine= 9.889881}
Side A B rs . • 101 ?47 '52^= the true spherical distance between the two
given places.
To find the Angle of Position at Port Jackson ss Angle S A B :'—
This is found by Problem L, page i98 ; as thus :
As the distance A B = 101?47^52r Log. co-secant = 10. 009273
Istodiff. long. ASB= 136.52. 0 Log. sine = . . 9.834865
So is the co-latitude B S= 56. 59. 0 Log. sine ;^ . : 9. 923509
Toangleofpos.=S^AB=35?50C59rLog.sine=5 . . 9.767647
To find the Angle of Position at Valparaiso = Angle S B A :—
This is found by Problem L^ page 198 ; as thus i
V
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290 NAVIOATIOH.
As the distance AB« 101?47^52r Log. co-secant s 10.009273
Istodiff.long.ASBa 136. &2. 0 log. sine a . . 9.834865
Soi8theco-lat.ASs: 56. 8. 0 Log. sines • • 9.919254
Toanglebfpos. SBA = 35?26^50r Log. sine « . .9.763392
To find the Perpendicular F S k the Complement of the highest Southern
Latitude at which the Ship should arrive :«— ^
Here we have a choice of tWo right angled spherical triangles^ viz. ASF
and B S F ; in each of which the hypothenuse and the angle at the base
are given to find the perpendicular 5 — thus, in the triangle A S F^ given the
hypothenuse AS = 56?8^ the co-latitude of Port Jackson; and the angle
at the base S A B = 35?50^59r the angle of position at that place^ to
find the perpendicular F S » the complement of the highest southern la-
titude afwhich the ship should arrive :—
* •
Hence, by right angled spherical trigonometry, Problem XL, page 185>
As radius = .... . 90^ 0! 0? Log. co-secants 10.000000
Is tocp-lat. Port Jackson s AS 56. 8. 0 Log. sine = . .9.919254
So is ang. of position =:^ S A F 35. 50. 59 Log. sine =£ . . 9. 767647
To the perpendicular P S = 29? S^Sir Log. sine = . .9. 686901
Highest lat« at which the
ship should arrive = 60?54^ 91 south.
From the above calculations it appears evident that the true spherical
distance between Port Jackson and Valparaiso is 101 ?47 '*52r, or 6107* 87
miles; the angle of position at Port Jackson = 35?50^59C^| and that at
Valparaiso r= 35?26^50f, and the highest southern latitude at which the
ship should arrive = 60?54'9^. — Now, by Mercator's sailing, the course
from Port Jackson tp Valparaiso is N. 89^34 C E. and the distance 6853. 16
miles ;-^whence it is manifest, that if a ship sails on the direct rhumb
line between, those places, as indicated by that mode of sailing, she will
have to run 745| miles more than by shaping her course along the arc of a
great circle. •
To compute the vertical, or Polar Angles ASP, and B S P :—
In the right angled spherical triangle ASF, given the hypothenuse A S
56 ?S' = the co-latitude of Port Jackson, and the perpendicular FS
29?5'51T c: the complement of the highest latitude at which the ship
should arrive ; to- find the vertical, or polar angle A S P.— And, in the
right angled triangle B S P, given the hypothenuse B S, 56?59' == the Co-
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GRBAT CIACUI SAILING. 291
latitude of Valparaiso^ and the perpendicular FS, 29?5'51f ; to find the
polar angle B S F, — Hence^ by right angled spherical trigonometry^ Prob. I.,
page 184,
To find the Polar Angle A S F :-«
As radius = 90?0' Or Log. co-secant = 10- 000000
Is to the co-latitude AS =z. . 56. 8. 0 Log. co-tangent = 9. 826805
So is the co-latitude F S == . 29, 5. 5 1 Log. tangent = • 9. 745493
To the polar angle A S F r: Q8W.B1 Log. co-sine s: . .9. 572298
■ I
To find the Polar Angle B S F :^
As radius == ...... 90? 0^ 0? Log. co-«ecant s: 10.000000
Is to the co-latitude B S = 56. 59. 0 Log. co-tangent = 9. 812794
So is the co-latitude FS = 29. . 5. 5 1 Log. tangent =s 9. 745493
To the polar angle B S F = 68?47'55r Log. co-sine = . 9. 558287
Now, since the sum of the polar angles^ thus obtained, viz. J0Ff
68?4^5r + BSF, 68?47C55r = 136?52^, makes up the whole differ-
ence of longitude between the two given places, expressed by the whole
apgle A S B, it shows that, thus far, the work is right.
To find the several successive Polar Angles made by the proposed
changes of Longitude.
From the polar angle ASF, subtract the proposed alteration of longi^
tude continually, as far as subtraction can be made ; ^nd the several polar
angles occasioned by those alterations, and contained between the perpen-
dicular F S, and the co-latitude of Port Jackson = A S, will be obtained.
— Th^s, from the polar angle A S F = 68?4'5C', let 5? be continually sub-
tracted, and the results will be FS 1 = 63?4C5r FS 2 = 68?4^6r ;
FS 3 = 53?4'5'r, &c. ic, the last remainder being 3?4'5r r: the polar
angle F S 13.-^Now, the polar angles contained between the perpendicular
F S, and the co«latitude of Valparaiso . == B S, are to be determined by a
contrary process ; and, since the last subtraction in the triangle F S A, left
the remainder, or polar angle F S 13 ^ 3?4^5r, which is l?55'55r, less
than the proposed alteration of longitude ; therefore, the first polar angle
in the triangle FBS^ must be l'?55:55'r = the polar angle F S a ; to
which, let 5? be continually added, as far as the measure of the angle FSB
will allow, lindweshallhaveFSA = 6?55^55r3 FSc= ll?55^55r;
F S d =; 16?55 ^55T, and so on ; as expressed in the first column of the
folk>wing Table.
v2
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292 NAVIGATION.
To compete the successive Latitudes at which the Ship should arrive : —
Since the several successive polar angles^ obtained as above^ evidently
reduce the two right angled spherical triangles AFS and BFS, intoa
series of right angled spherical triangles, to each of which the perpendicu-
lar F S is jcommon ; therefore, in each triangle of this series we have the
perpendicular and the angle adjacent, to find the hypothenuse, or co-la-
titude.— ^Thu8,in the right angled spherical triangle F S 1, right angled at
F, given the perpendicular FS = 29?5'5K, and the polar angle FS 1 =
.63?4^5^ ; to find the hypothenuse, or co-latitude S 1 ; — In the right an-
gled spherical triangle FS 2, given the perpendicular FS = 29?5'5K,
and the polar angle F S 2 = 58?4'5? ; to find the hypothenuse, xyf co-
latitude S 2, &c. &c. &c.
• Hence, by right angled spherical trigonom.etry^ Problem IV., page 188,
To find the Hypothenuse, or Co-latitude S 1 :—
As the perpendicular F S = 29? 5 ^ 5 K Log. co-tang* = 10. 254507*
Is to the radius = . . . . 90. 0. 0 Log. sine = . .10. OOOQOO
So Hthe i[>olar angle FS 1 s 63. 4. 5 Log. co-sine . 9.656033
To the co-latitude S 1 =: 50. 5 1 . 36 Log. co-tangent = 9. 9 10540
First latitude = • • 39? 8^24^ S. at which ship should arrive.
To find the Hypothenuse, or Co-la(itude S 2 :—
As the perpendicular F S := 29? 5 ^ 5 K Log. co-tang. = 10. 254507*
Is to the radius = . ... 90. 0, 0 Log. sine = . .10.000000
So is the jpolar angle F S 2 = 58. 4. 5 Log. co-sine = . 9. 723383
To the CQ-latitude S 2 = 46. 27. 28 Log. co- taifg. = 9. 977890
Second latitude = 43?32^32? S. at which ship should arrive.
Hence, the first latitude at which the ship should arrive is 39?8' 24r S. ;
a^d the second latitude 43?32'32^:S.— And since it is the latitude, and
not its complement that is required ; therefore, if the log. tangent of the
sum of the three logs, be taken, it will give the latitude direct ; and, by re-
jecting the radius from the calculation, the work will be considerably faci-
litated.-=-Proceeding in this manner, the several successive latitudes cor-
• The loff. co-taDg^ent is used, so as to save the trouble of finding the arithmetical (
plemeut of the log. taageut.'
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GRBAT CIRCU 8AfLIN6. 293
responding to the proposed alterations of longitude will be found as shown
in the 3d. column of the following Table.
Now, let the several successive longitudes be arranged (agreeably to the
proposed change, and to the measure of the corresponding polar angles,)
as exhibited in the 2d column of the following Table ; and find the
difference between every two adjacent longitudes^ as shown in the 4th
column of that Table.— 'Find the difference between every two adjacent
latitudes, and place those differences in the 5th column.— Find the
meridional parts corresponding to the several successive latitudes^ which
place in the 6th Column ; and find the difference between every two adja-
cent meridional altitudes^ as shown in the 7th column.— Then, find^ by
Mercator's sailing, Problem I., page 238, the respective courses and dis-
tances between the several successive latitudes and longitudes ; and,
let. the courses and distances, so found, be arranged in regular succession,
as exhibited in the two last columns of the Table.^Then, will thiff Table
be duly prepared for navigating a ship on the arc of a great circle, a^ee-
ably to the proposed alterations of longitude. — ^And, should the sum of the
several successive differences ef loingitude, contained in the Table, coin-
cide with the whole difference of longitude between the two given' places ;
—the sum of .th^ several successive differences of latitude be found to
agree with the whole difference of latitude comprehended under the mean,
or highest latitude, and its corresponding extremes ; — ^tfae sum of the se-
veral meridional differences of latitude to be equal to the whole meri-
dional difference of latitude corresponding to the mean, or highest lati-
tude, and its respective extremes, — and the sum of the several successive
distances to make up the whole spherical distance (or nearly so,) be-
tween the two given places; then, those several concurring equalities
will be so many satisfactory proofs that the work is right.
JVbite.*— In the spherical track laid down in the following Table, it is pre-
sumed that there is not any land to intercept a ship's progress ; but since
this track will take the navigator into high southern latitudes, it will be in-
dispensibty necessary to keep a sharp look-out at all times^ particularly
during the night, so as to guard against any of the ice-bergs that may be
floating to the northward of the Antarctic cirele ;— though if the track be
made in .the months of November, December, January, or February, there
will be no real night or darkness to experience ; for during these months
there wiH'be a strong twilight between the latitudes of 53, and 61 degrees
south ; and thus the navigating at night will be attended with very little
n^ore danger than that by day.
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GRBAT ClRCtV SAILING. 295
Nov^, the sum of the Sereral successive differences of longitude, viz. 8212
miles^ coincides exactly with the whole difference of longitude between
the two given places ; the sum of the successive differences of latitude =
3295« 30 miles,, agrees with the whole difference of latitude comprehended
under the highest latitudeat which the ship should arrive, and the latitudes
of the two given places; viz. 33?52^07 S; 60? 54' 9'/ S, and 33? T. Or S:
—and, the sum of the sever^ meridional differences of latitude » 501 1, 80
miles, makes up the whole difference of latitude corresponding to the higft-
est latitude and the latitudes of its respective extremes :— these several
concurrences or agreements, form, therefore, the most satisfact<^ and in-
disputable proofs that the work has been properly conducted.
The sum of the several distances, measured' on the respective rhumb-lines
intercepted between the successive longitudes and latitudes, as given in the
last column of the Table, is 6108. 73 miles ; — but the true spherical dis-
tance dn the arc of a great circle is 6107- 87 miles ; the difference, there-
fore, is only 0. 86, or a little more than three fourths of a mile ; which is a
very close approximation in the measure of so great an are.
The distance by Mercator's sailing is 6853. 16 miles ; which is 745. 29,
or about 745^ miles more than by great circle sailing.— HencCj it is evi-
dent that the shortest and most direct route from Po^f Jackson to Valpa-
raiso is by the latitude of 60? 54 '9^ S ; and that the ship must make, suc-
cessively, the several longitudes and latitudes contained in the 2nd and Srd
columns of the Table, .in the same manner precisely, as' if. they were so
many ports or places of rendezvous, at which she was directed to touch.
The first course, therefore, from Port Jackson to Valparaiso, is S* 37^ 1 8(
E. distance 398 miles; which will bring the ship to longitude 156?16C0r
K and latitude 39? 8 ^24? S ;— the second course is S. 40?26^ E. distance
347 miles ; which brings the ship to longitude 161? 16 ^OfEk and latitude
43?32'32r S ;— the third course is S. 43?53^ E. distance 304 miles, which
brings the ship to 166? 16^0^ E. and latitude 47?U 'S6f 8.
Whence it is evident that Captain Gambier saved a distance of 745^
miles in that judicious and well-planned route : And this saving of distance
should be an object of the highest consideration to every captain who wishea
to recruit the strength and spirits of his ship's company by a generous sup-
ply of fresh provisions ufter a fatiguing and tedious voyage ; the measure
of which falls very little short of being equal to one fourth of the earth's
circumference as taken under the equator, or to the one third of that cir-
cumfenence if taken under the given parallel of latitude.
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!296
NAUTICAL ASTRONOMY.
SOLUTION OF PROBLEMS IN NAUTICAL ASTRONOMY.
Nautical Asteonomy is the method of findings by celestial t>bservation^
the latittide and loiigitude of a ship at sea ; the variation of the conipass ;
the apparent time at ship ; the fdtijtudes of the heavenly bodies, &c. &c. &c.
—Or, it is that branch of mathematical astronomy which shows how to
solve all the important Problems in navigation by means of spherical oper-
ations,, when the altitudes^ or distances of the celestial objects are under
consideration.
Introductory Problems to the Science of Nautical Astronomy.
Problbm I. '
To convert Longitude or Parts of the Equator into Time.
Rule. . .
Multiply the given degrees by 4, and the product will be the correspond-
ing time :— -observing that seconds multiplied by 4 produce thirds ; mi-
nutes multiplied by 4 produce seconds, and degrees multiplied by 4 pro-
duce minutes, which, divided by 60, give hours, &c.
Exampfe !•
Required the time correspond-
ing to 12?40U5r ?
Given degrees = 12?40M5r
Multiply by • • • . 4
Corresp.time = 0*50:43!0f
Example 2.
Required the time corresponds
ingto76?20^30r?
Given degrees = 76?20C30r
Multiply by . . . . 4
Corresp. time t= 5*5?22;0f
Pro]$L£m II.
To conoeri Time into Longitude, or Parts of the Equator.
Rule.
Reduce the hours to minutes, to which add the odd minutes, if any;
then, the minutes divided by 4 give <}egrees ; the seconds divided by 4 give
minutes, and the thirds divided by 4 gjve seconds.
Example I.
Required the degreea corres-
ponding to 0*47"36!?
Given time = . 0*47"36!
Divide by .
Corresp. deg. =
4) 47^36!
Ilf54: or
Example 2.
Required the degrees corres-
ponding to 9^25^37! ?
Given times? . . 9*25^37!
Divide by . 4) 565r37!
Corresp. deg, = 141?24: ISr
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INTRODUCTORY PROBLBlfS.
297
Note,f~^pie two preceding Problems are readily solved by means of
Table I^ — see explanation^ pages 1 and 2.
PaoliLEM IIT.
Qioen the Time under any known Meridiany to find ike corresponding
T%7ne at Greinwichn
RVLB.
Let the given time be reckoned from the preceding noon^ to which apply
the longitude of the place in rfme (reduced by Problem I., as above,} by
addition if it be west, or subtraction if east; and the sum, or difference will
be the corresponding time at Greenwich.
Example 1, .
Required the tiine at (}reenwich,
when it is 4M0ri3! at a ship in
Iongitudi5 80?53:i5r W.?
Time at ship = . . 4*40rre:
Long.80?53a5rW. ,
in times • . \r + 5.23.33
Corresp. time at Green-
wich = ... 10* 3?46!
Example 2.
Required the time at Greenwich,
when it is 20* 1 1T41 ! at a ship in
longitude 98? 1 4 USrE?
Time at ships: . . 2Q*llr4i!
Long. 98n4U5rE.
in time =s . . - 6.32.59
Corresp. time at Green-
wich = . . . . 13?38r42!
Problem IV. .
Given the Time at Greenvneh, to find the corresponding Time under a
known Meridian,
Rule.
Let the given time be reckoned from the preceding noon^ to which apply
the longitude of the place in time (reduced by Problem I. as above,} by ad-
dition if it be east, or subtraction if west; and the sum, or difference will
be the corresponding time under the given meridian.
'Example 1.
When may the emersion of the first satellite of Jupiter be obser\'ed at
Trincomalee, in longitude 81 ?22' E., wifich, by the Nautical Almanac, hap-
pens at Greenwich, March 4th, 1825, at 9*9T28! ?
Apparent time of emersion at Greenwich = . • . 9t 9f28*
Longitude of Trincomalee 8 1 ? 22 ^ E., in time =: . 5 . 25 . 28
Apparent time of emersion at Trincomalee = • • 14*34T56!
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2d8 ' NAUTICAL ASTRONOMY*
Example 2.
When may the immersion of the first satellite of Jupiter be observed at
Port Royal, Jam^ca, in longitude 76?52'30r W., which, by the Nautical
Almanac, happens at Greenwich Nov. 1st. 1825, at 18^ 17 "'45 ' ?
Apparent time of immersion at Greenwich = • . 18^ 17*45!
Longitude of Port Royal 76?52:30r W., in time == 5. 7. SO
Apparent time of immersion at Port Royal r= • • 13* 10? 15 '
^ ■ I I ■■ ■' « ■ ■ ■ ■■■■ ■ ■■■■ !.■ ■' » ■ ■ ■ ■ ■ ■ ■ ■«. III! i»— „. m. ■ ,■
Pboblem V.
To reduce the Sun^s Longitude^ Bight Asc^iswUf and DecUnation; and^
alsoy the Equation qf Time, €is given in the Nautical Almanac, to any
other Meridian,' and to any thne under that Meridian;
RUJLB.
Let the given apparent time at ship, or pl^ce, be always reckoned from
the preceding noon; to which apply the longitude in time (reduced by Pro-
blem I., page 296,) by addition if it be West, or subtraction if ea^t, and
the sum or difference will be the corresponding time at Greenwich.
Take, from page U. of the month in the Nautical Almanac, the sun's
longitude, right ascension and declination, or the equation of time, as the
case may be, for the noons immediately preceding and following the
Greenwich time, and find their difference; then,
To the proportional log. of this difference, add the proportional log. of
the Greenwich time (reckoning the hours as minutes, and the minutes as
seconds), and the constant log. 9. 1249; * the sum of these three logs,
rejecting 10 from the index, will be the proportional log. of a correction
which is always to be added to the sun's longitude, or right ascension, at
the noon preceding the Greenwich time ; but to be applied by addition or
subtraction to the sun's declination, or the equation of time at that noon^
according as these elements may be increasing or decreasing.
Bemark. — Since the daily difference of the equation of time is eiq)ressed^
in the Nautical Almanac, in seconds and. tenths of a second ; if,, there-
fore, these tenths be multiplied by 6 they will be reduced to thirds : hence,
the daily difference will be obtained in seconds and thirds.— Now, if those
seconds and thirds be esteemed as minutes and seconds, ti^ operation of
reducing the equation of time wijU become as. simple as that of the sun^s
declination ; — observing, however, that the minutes and seconds. Corres-
ponding to the sum. of the three logs., are to be considered as seconds and
thirds.
* This is the arithmetical complement of the proportional log. of 24 hours esteemed as
winuteif
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304 . NAUTICAL ASTRONOMY.
To^find the Moon's Declination :-—
Diflference in 12 hours = . . • . 2?47'57^ . Prop. log. = .0301
Greenwich time = 7*20r53' Prop. log.= 1.3891
Constant log. = ....'..•......•,.. 8.8239
Correction of moon's declination = + 1 ?42C5 1 r Prop. Iog.= 0. 2431
Moon's declin. at noon March 6th .= 7> 58. 6 south.
Moon's declination^ as required ss • • 9?40^57^ soutli.
To find the Moon's Semi-diameter : —
DiflF.in 12 hours = ... 4r ...... . Prop. log.. =: 3.4314
Greenwich time = .. . 7'20r53! .... Prop. log. = 1.3891
Constant log. = . . . . ..... . . . ..-.,. 8.8239
Correction of the moon's semi-diameter =s —21 Prop. log. =: 3. 6444
Moon's Semi-diam. at noon March 6th =s 16^38^
Moon's senii-diameter^ as required = . 16C36T
To find the Moon's Horizontal Parallax : —
Difference in 12 hours = 16^ Prop. log. s 2. 8293
Greenwich time = 7*20r53! Prop. log. = 1.3891
Constant log. = . 8.8239
Corr. of the moon's horizontal parallax = 10^ Prop, log, ^ 3. 0423
Moon's horiz. par. at noon, March 6th 61^ 2T
Moon's horizontal parallax, as required = 60^52^
Reniark.'^When much accuracy is required, the proportional part o(
the moon's motion in 12 hours, found as above, must be corrected by the
equation of second difference contained in Table ^VII., as explained in
pages 33, 34, and 35. And, in all cases, the moon's semi-diameter, so
found, must be. increased by the augmentation given in Table IV., as ex-
plained between pages 8 and 11.
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INTRODUCTORY I'ROBisMS, 305
Example 2.
Required the moon's longitude,- latitude, right ascensioh, declination^
semi-diameter, and horizontal parallax, March 26th, 1825, at l*30?47!,
in longitude 94?15:30r east; the apparent altitude of that object being
.24??
Apparent time at ship or place = ......... 1*30"47'
Longitude of the ship or place = 94? 15 C30r E., in time =s . 6. 17.' 2
Greenwich time past midnight, March 25tb, 1825 = .. . • 7t 13?45 !
To find the Moon's Longitude :—
Diff. in 12 hours = 6?22:57^ t- 3 = 2?7^39r Prop, log: = . 1493
Greenwich time = 7* 13T45! • Prop. log. = L 3962
Constant log. = • . • 8. 8239
One-third of the proportional part s 1? 16C53r Prop, log, a 0.3694
Multiply by 3
Prop, part of J 's motiofi in long. = 3?50'39?
Equation from Table XVII. = . . - 33
Proportional part corrected = . . 3?50' 6T
J 's long, at midnt., March 25th = 2! 15?31 C56r
Moon's true longHude s . . . 2!19?22: 2?
To find the Moon's Latitude :—
Difference in 12 hours = *.. . . . 33137^ Prop. log. = .7287
Greenwichtime=:7M3?45! . ...... Prop. log. = 1.3962
Constant log. = • * . 8.8239
Proportional part of D 's latitude = - 4i0' ISr Prop. log. « 0. 9488
Equation from Table XVII. =s . . • — 6
Proportional part corrected sa . . — 20' 9^
J 's lat. at midnight, March 25th = 0?*46^36r north.
Moon's true latitude = * • . . . 0?26'.27r north.
X
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806 MAOTICAL A8TB0MOMT.
To find the Moon's Right AAcension :—
Diff.ml2houn«6?59n:. + 3= 2?19'40J? Prop. log. « .U02
Greenwich time- 7* 13T45! Prop.log.= 1.3962
Conttunt log. = . , . ' ^'^239
One-third of the proportional part = 1?24< 81 Prop. log. = 0. 3303
Multiply by 8
Prop, part of J 's motion in right asc^*? 12'24?
Equation from Table XVH. = . . - 36 .
Proportional part corrected = . • 4? 11 '48^
])'8rightasc.atmidnt./March25thas74. 1U56
Moon's true right ascension = . 78^ 23 ' 44^
To find the Moon's Declination :—
Differencein 13 hours « I'.Or Prop. log. » 2.2553
Greenwich time = 7M3r45! Prop. log. = .1.3962
Constant log. = . 8.8239
Proportional part of J 'a declination = - 0' 36? Prop, log, = 2. 4754
Equation from Table XVII. =s . . . - 10
Proportional part corrected tt • . . — 0'26?
)) 's dec. at midnt., March 25tii = 23^26^53? north.
' I
Moon's true declination •= . . 23^26:27? north.
To ifind the Moon*s Semi-diameter :—
Difference in 13 hours = . ... . , . .6? Prop, log, fai 3. 2553
Greenwich time -> 7M3W5! .... ... Plop. log. a 1.3962
Constant log. =. . . : 8.8239
Pfopoftlowil part of ) 's semUdfemeter + 4r Prop. log. » 3.4754
D 's semi-diam. at midnt., March 25^th. = 151 19^
Moon's apparent semi-diameter =B .. ..15128?
Augment, from Tab* IV., for alt. 24 ? p. 6
Moon's true semi-idiBnietec =:: « .. .1 15129^
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IKTHODUCTORT PROBLXMS. ^307
To find the Moon's Horisontol Ptoalliix:~
Diflference ih 12 hours = 22r Prop. log. = 2.6910
Greenwich time c* 7M3?45; . . . , . . . . Prop. log. ^ ^3962
Constant log. = ......•.•.'.•,,.,, 8.8239
Proportional part of 'j 's hor. parallax = + 13f Prop. lo^. s= 2. 91 U
> 's hor. par. at midnight, March 25th ps 56*. ISf '
Moon's true horizontal parallax = • • 56' 26^
Note. — ^The examples to the foregoing Problem may \}e very correctly
solved by means of Table XVL^JSee explanation, page 8O4
Problbm VIL
To reduce the lUghi AscenAm, and Declination of a Planet, as given ifi
the Nautical Jlmandc, to any given ^me under a known Meridian.
Ruu«
Let the apparent time at the ship or place be reckoned from the preced*
ing Jloon, to which apply the longitude In time, (reduced by Problem I.,
page 296,) by addition if it be west, or subtraction if east j and the $\m
or difference will be the corresponding time at Greenwich.
From page IV. of the month in the. Nautical Almanac, take out the
planet's right ascension and declination for the nearest days preoeding and
following the Greenwich time, and find the difference; find, also, the
diflference between the Greenwich time and the nearest preceding day;
then,
To the proportional logarithm of this diflference, esteemed as minutes
and. seconds, add the proportional logarithm of the diflference of right
ascension, or declination^ and the constant logarithm 9.9031*; the sum
of these three logarithms, rejecting 10 from the index, will be the propor*
tional logarithm of a correction ; which being applied, by addition or subr
traction, to the right ascension or declination (on the nearest day preceding
the Greenwich time), according, as it may be increaring or decreasing, the
sum or diflference will be the correct right ascension or declination.
Example I. -
Required the right ascension and declination of the planet Mars, Marfth
16th, 1825, at 4*40r apparent time, in longitude 68?12f west of the
meridian of Greenwich ? "
--—■---■»- I - I , I . - . ■■ ■
« Tbit to the arithmetical complement of the proportkmal logarithm of 144 hours » 6
daytff fttesmed M »tfiirlr« ; and, hie Ace> taken M S hovrt sad 24
X 2
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308 NAUTICAL ASTRONOMY.
Apparent time at ship or place =s .■ . . March, 16 days, 4!40T 0!
Longitude of -ship or place = 68^ 12 C W,, in time = . . . 4. 32. 48
Greenwich time = . . •.. . . . ... . 16days, 9M2T48!
To find the Right Ascension : —
R. A. of Mars, March.l3th=0*4ir 13f 0* Or 0!
Ditto I9th=0.38Gr.time=16. 9.12.48
Differences: .... 0*17^ Diff. = 3f 9M2M8!= 81M2r48!
Piff. oftime = 8lM2r48r, or l*21?12M8f Prop. log. = .3456
Difference of right ascension 5= 0M7* . . . Prop. log. = 1.0248
Constant log. = . 9.9031
Correction of right ascension = . . + 9T35! Prop. log. = 1.2735
Planet's right ascension, March I3th = OM 1 ? 0 !
Planet's right ascension, as required « 0*50T35 !
To find the Declination : —
Dec.ofMars, March 13th=3?53^N. 13f 0* Or 0!
Ditto 19th=5.43N.Gr.time=16. 9.12.48
Dlfferehc€= \ . . 1950: 3f 9M2r48'.— 81M2r48!
Differenceof.timfe==81*12r48',or H2iri2!48f Prop. log. = .3456
Difference of declination = 1?50' Prop. log. = .2139
Constant log. = /........ 9.9031
Correction of declinations . . + 1^ 2' 2r Prop. log. = 0.4626
Planet's declinatbn, March 13th = 3.53. 0 north.
Planet's declination, as required ^ • 4?55' 2? north.
Example 2*
Required the right ascension and declination of the planet Mars, Sept.
23d, 1825, at l*23ri9!, apparent time, in longitude I00?40:30r east of
the meridian of Greenwich ?
Apparent time at ship or place = ... Sept. 23 days, 1 *23rl9!
Lonptude of ship or place = 100?40^30r E., in time = . 6. 42. 42
Greenwich time (paist noon of the 22d Sept.) = 22 days, 18U0?37'
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INTRODUCTORY PROBLEMS. 309
To find the Right Ascension : —
R.A.ofMar8,Sept.l9th=9*37r 19f 0* Or Of
Ditto 25th=9. 52 Gr. time 22, 18. 40. 37
Difference = . . . 0*15r Difr.=:3f I8t40?37! = 90*40?37!
Difference of time = 90*40?37!, or It30r40!37f Prop.log. = .2977
Difference of right ascension = 0*15? . . . Prop.log. = 1.0792
Constant log. = ^ 9.9031
Correction of right ascension s . + 9?27!. Prop. log. = 1.2800
Planet's right ascension, Sept. 19th i= 9?37*r 0:
Planet's right ascension, as required5=9t46r.27!
To find the Declination :—
Dec.ofMarsVSept.l9th=15'?30'N. 19f 0! 0? 0!
Ditto • 25th=tl4. 18 N. Gr. time 22. 18. 40'. 37
Differences . . . l9l2^ Diff. = 3fi8M0?37'=90!40r37!
Difference of time = 90M0?37 ', or 1 *30?40!37 f Prop. log. = . 2977
Difference of declination = 1?12' * Prop. I05. =3 .3979
Constant log. = • . • • 1 • . . • 9.9031
Correction of declination = . — 45'2K Prop.log. s 0.5987
Planet's declination, Sept. 19th = 15. 30. 0 north.
Planet's declination, as required.=14?44'39T north.
Problem VIII.
To compute tlie Apparent Time of the Moon's Transit over the Meridian
of Greenwich.
Since the moon's transit over the meridian of Greenwich is only given
to the nearest minute in the Nautical Almanac ; and, since it is absolutely
necessary, on many astronomical occasions, to have it more strictly deter-
mined : the following rule is, therefore, given, by which the apparent time
of the moon's transit over the meridian, of Greenwich may be obtained true
to the decimal part of a second.
Rule.
From the moon's right ascension at noon of the given day (converted
into tipie, and increased by 24 hours if necessary,) subt^ct the sun's xi^t
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310 KAi;tIOAL AttRONOMY.
ascension at that noon, and the remainder will be *the approximate time of
the moon's transit over the meridian of Greenwich.
Find the excess of the moon's motion in right ascension over the son'a
in 12 hours; then say, as 12 hours, dimitiished by tliis excess, 'is to 12
hourSj so is the apptoxiitiate time of transit to the apparent time of transit.
Notfi.'^U the three tetms be reduced to seconds, the operation may be
teadily performed by logarithms.
Example 1.
Required the apparent time of the moon s transit over the meridian of
Greenwich, March 26th, 1825 ?
Moon's R.A. at noon of given day = 81?10^57^ in time = 5*24r43\S
Sun's right ascension at that noon = .^ 0. 20. 24 . (F
ApproK..time of the moon's tr. over the merid*. of Greenw. =s 5 1 4T19' • 8
* Son's right ascension at noon^ March 26th, s 0^20^24!
Son's ditto 27th s 0.24. 9
Sun's motion in 24 hours =s Ot drSS!
Sun's motion in 12 houi^ = . . . • r . 0* 1T49!
Moon's R. A. at noon, March 26th, = 81?10:57r, in time = 5?24r43'. 8
Moon's ditto, at midnt, March 26th,=88?14^ 15f, in time= 5. 52. 5? .0
Moon's motion in 12 hours == • 0*28T1'3'.2
Siiti'd nidtion in 1^ hoUr3 3 • . • • . ; . • « • 0. 1,49 .0
Excess of the moon's motion over t)ie sun's in 12 hours s 26?24\2
As 12? -* 26r24' i 2 .« 1 1 *33?35 • . 8,
in seconds s « « , « , . 41615.8 Log. ar.c6.mp.s'5. 380742
Is toa2 hours, in seconds s. . . 43200. Log. s * .. 4.635484
So is the approximate time of transit
= 5*4?19'.8, in seconds = . 18259.8 Log. = . . 4.261496
To the ^parent time of the moon's
t»uunt?s5M5:54\99iiiaecs« a; 18954»B Log^ = , » 4.277722
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l3fTft02>UCrORy PROBUMS. 8U
Example 2.
Required the apparent time of the moon's transit over the meridian «f
Greenwieh, April 10th, 1825 ?
Moon's R, A. at noon of given day=:i94?59'lK^intime=l9!39r56\7
Sun's right ascension at that noon = 1.15,1.6
Approx* time of the moon's tr. over the merid. of Oreenw* s= 1 8. 24 • 55 4 1
Sun's right ascension at noon, April 10th, 9 1M5? l'«6 .
Sun's ditto 11th, = L 18.41 .6
Sun's motion in 24 houra s Ot 3r40',0
Sun's moti(Hi in 12 hours = • . • • • . Ot i?50'.0
Moon's R. A. at midnt, April 10th a son 15 C6'r, in time a 20 1 57 0\4
Moon's ditto at noon, April 1 1th = 307. 20. 12, in time = 20. 29. 20 . 8
Moon's motion m 12 hours Si .••.«..•. 0^24?20^4
Sun's motion in 12 hours = 0« 1.60 .0
Excess of the moon's motion over die son's in 12 houn == 0. 22. 30 • 4
As 12? -22r30\ 4=1 1 *37r29\ 6,
in seconds ss 41849.6 liog. ar.comp.sS. 378909
Is to 12 hours, in seconds rss . . 4^200. Log. s « . 4.B35484
So is the approximate time of transit
ss 18^24755 M, in seconds tt 6G295.1Log.« • • 4.821481
To the apparent time of the moon's
transits^ 1 9*0734 •. 8, in sees: =s 68434.3 Log. ss . . 4.835274
•
Note.^ln strictness the apparent time of transit, thus found, should be
corrected by the equation of second difference answering thereto, and the
mean Second difference of the moon's place in right ascension ; but, at sea,
this correction may safely be dispensed with.
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312 NAUTICAL ASTRONOMT.
Problem IX.
Given the Jpparent Time of the Mm! 9 Traimi over the Meridian of
Greenwichf to find the Jpparent Time of Dransit over any olher
Meridian.
RULB.
Takej from page VI. of the month in the Nautical Almanac, the moon's
transit over the meridian of Greenwich on the given day, and also on the
day following if the longitude be west^ but on the day preceding it if it be
east, and find the difference ; which difference will be the daily retardation
of transit : then say.
As the sum of 24 hours and the daily retardation of the moon's transit,
thus found, is to the daily retardation of transit ; so is the longitude of the
given meridian, in time, to a correction, which, being applied by addition
to the apparent time of transit oVer the meridian of Greenwich on the
given day, if the longitude bq west, but by subtraction if east 5 the Stum, or
difference, Will be the apparent time of transit oVer the given meridian.
JVbte.— This proportion may be readily performed by proportional loga-
rithms, esteeming the hours and minutes in the J&*«f and third terms as
minutes and seconds.
Example 1.
Required the apparent time of the moon's transit over a ' meridian
94?30^30r west of Greenwich, March 26th, 1825, the computed apparent
time of transit at Greenwich being 5 ! 15 ?54 '. 9 ?
Moon's transit over the mend, of Greenwich on the given day =: 5t 16T
Moon's ditto on the day ybOotnn^^: 6. 11
Daily retardation of moon's transit == . • 0^557
As 24 hours+0*55r (daily retard.)=24*5.5? Prop. log. ar. comp.±9. 1412
Is to the daily retard, of transit == 0.55 Prop. log. = . . 0.5149
So is the Ion. 94?30:30^W;, in tiiAe^e* 18r2! Prop. log. = 1. 4559
■ * f ' '
To the correction of retardation = . + 13T54' . 5 Prop. log.=: 1. 1 120
Computed apparent time of moon^s
transit over the mend, of Greenwich=:5 * 1 5 ?54 ' . 9
App. time of ]) *s tn ovet the given mer.=;5 * 29T49 ' . 4
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IMHODUCTOEY PROBLBMS, 3 IS
Example 2.
Required the apparent time of the moon's transit over a meridian
105?10M5r east of Greenwich, April 10th, 1825, the computed apparent
time of transit at Greenwich being 19t 0T34 ' • 3 ?
Moon's transit over the merid. of Greenwich on the g^ven day =: 19 1 1?
Moon's ditto on the day preceding:^, 1 8. 13
Daily retardation of moon's transit =: • • • 0!48?
As 24 hours+0*48r (daily retard.) =24? 48? Prop. log. ar.comp.=9. 1392
Is to the daily variation of transit =: 0. 48 Prop, log. =: • • 0. 5740
SoisthelongJ05? 10(45 ^E., in timez=7* 0r43! Prop. log. = 1.4094
To the correction of r^ardation =z 13'?34'.4 Prop, log. = lil226
Computed apparent time of moon's tr.
over the merid. of Greenwich = 19 1 0?34 ' . 3
App. time of ]) 's tr. over given men = 18! 46T59 * . 9
Note. — ^Thc above problem may be readily solved by means of Table'
XXXVIII.— See explanation, page 100,
Problem X.
To compute the Apparent Time of a Ptanet'e JVanrit over the Meridian
of Greenvokh.
Rule.
Find the planet's right asoension at noon of the given day, by Problem
VII., page 307 ; from which (increased by 24 hours if necessary), subtract
the sun's, right ascension at that noon, and the remainder will be the
approximate time of the planet's transit over the meridian of Greenwich.
Take the difference of the sun's and the planet's daily variations, or
motions, in right ascension, if the planet^s motion be progressive, but the
sum if it be retrograde* : then say.
As 24 hoursj diminished or augmented by this difference or sum (accord-
ing as the planet's diurnal motion in right ascension is greater or less than
* When the daily variation of the planet's rt^bt ascension Is greater than that of tl^e
sun*8y its motion i$ jnvgreitwe ; bat whenless; its mot}oo is retrograde^
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the sun's), is to 24 hours, so is the approximate time of transit to the
apparent time of the planet's transit over the meridian of Greenwich.
M>le.— -If the terms be reduced to seconds, the operation may be easily
performed by logarithnis. .
Example 1.
Required the apparent time that the planet Mars will'pass the meridian
of Greenwich, March l6th, 1825 ?
Planet's right ascension at noon of the given day = • • 0M9T30' « 0
Sun's right ascension at that noon = ••••.• •23.44. 0.3
Approx. time of the planet'^ transit over the mer. of Greenw.=s 1 1 5*29 ' . 7
Planet's right ascension at noon> March .16th, ss 0M9T30!
Planet's ditto 1 7th, =^ 0. 52. 20
Planet's motion in 24 hours = . . • . . 0? 2?50! 0* 2r50!
Sun's right ascension at noon, March I6th, sr 28M4" O'.S
Sun's ditto 17th, = 23.47.39 .3
Sun's motion in 24 hours = . . . . . .0? Sr39*.*0 0* 3rS9!
Sum of the motions s 0* 6*29!
^To^e.^^The sum is taken because the planet's motion is retrograde^
As 24ho.+6r29?=:24',i3t29?, In secs.s:86789 LQg.ar.comp.^5. 061535
Is to 24 hours, in seconds = ... 80400 Log. = . . 4, 936514
So is the approximate time of transit
= I*5?29' . 7, in seconds = . . 3929. 7 Log. = S. 594359
To the apparent time of the planet's .
transit = 1!5712\ I, m seconds s 3912. 1 Log. a 3. 592408
Example 2.
Recpiired the apparent time that the planet Venus will pass the meriditfi
of Greenwich, Sept. 23d, 1825 ?
Planet's right ascension at noon of the given day s . » 9^34^40' • 0
Sun's right ascension at that noon = ••••••.. 12. 0. 29 • 7
Approx. time of the planet's tr. over the mer. of Gieenw. =3 2l!34?10\3
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HrrROPOCTOlY FR0BL8US. 815
Planet's right ascension at noon^ Sept. 23d, = 9^34^40!
Planet's ditto 24th,= 9.39.20
Planet's motion in 24 hours = .... 0* 4?40! Ot 4?40!
Sun's right ascension at noon, Sept. !23d, as 12t 0?29' • 7
Sun's ditto 24th,=5 12. 4. 5 .5
Sun's motion in 24 hours K ... . . Ot 8T35\8 0. 3.35 .8
Difference of motion » 0^ ir 4'.2
iVate.--<-The difference, ia taken because tbe planet's motion is pro-
gressive.
As 24*-.ir4*. 2e23*38r55\ 8,in8ec8.s: 86335. 8 Log.ar.co.=5. 063809
Is to 24 hours, in seconds :st . . . S6400. Log. ^ . 4.936514
So is the approx* time of '
traQsit=21*34riO\3Jin8ecs.= . 77650. 3 Log. = . 4.g90143
To the apparent time of the planet's
trai8it=21*35?8'. 1, in sees. =' • 77708. 1 Log. = . 4. 890466
Problem XL'
Given the Apparent Time of a Planet'tl Ti'anHt over the Meridian of
Greemdch, to find the Jpparenit Time qf Tramit over anjf other
Meridian^
Rule.
Take, from page IV. of the month in the Nautical Almanac, the appa-
rent times of the planet's transits over the meridian of Greenwich oh the
days nearest pf eceding and following the given day, and find the interval
between those times ; find, also,' the difference of transit in that interval :
then say.
As the interval between the times of transit is to the difference of transit,
so is the longitude, in time, to a correction ; which, being added to the
computed apparent time of transit^ if the longitude be west and the plilnet's
transit increasing, or subtracted if decreasing, the sum or difference will be
the apparent time of transit over the meridian of the given place; but, if
the longitude be east, a contrary process is to be observed 3 that is, the
.correctioa is ta be subtracted from the approximate time of transit, if the
transit be increasing, but to be added thereto if decreasing.
Note. — ^If the first and third terms of the proportion be esteemed as
mmtes and seconds, the operation may be performed by proportional
logaritbnis.
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316 Nautical astronomy.
Example 1.
Required the apparent time that the planet Mars will pass the meridian
of a place I45?30^ west of Greenwich, March 16th, 1825, the computed
apparent time of transit at Greenwich being lt5?12M ?
Time of preceding transit = . . • • 13f 1* 8?
Time of following transit s • » • • 19. 1. 3
Interval between the times of transit ss 5 f 23^55?
Difference of transit in that interval s= • * 5 minutes.
As the interval=5f23*55r=143*55'r=2*23r55! P.l6g.ar.co.=9.9028
Is to the difference of transit = * « 5? Prop. log. = . 1. 5563
So is the long. 145?30'. W. in time =«: 9*42r Q! Prop. log. = 1. 2685
To the correction of transit = , . . - 20! Prop. log. = 2. 7276
Computed time of planet's transit over
the riieridian of Greenwich = • 1 * 5 ? 1 2 ' . 1
Apparent time of planet's transit over
the given meridian = .... It 4T52M
Example 2.
Required the apparent time that the planet Venus will pass the meridian
of a place 175^40' east of Greenwich, Sept. 23d, 1825, the computed
apparent time of transit at Greenwich being 21t35T8'.l?
Time of preceding transit ss . . . 19^21*317
Time of foHowing transit =s . .. • . 25.21.37
interval between the times of transit =s 6f Ot 6*
Difference of transit in that interval s • 6 minutes.
As the interval=6f0*6'r=t=144*6r=2*24r 6! P. log. ar, comp.=9. 9034
Is to the difference of transit = 6T Prop. log. = . 1. 4771
Soisthelong.l75?40^E.,intiroe=5ll*42?40! Prop. log. = . 1:1867
To the •correction of transit s= ... . /^ 29! Prop. )og.s:2. 5672
Computed time of planet's transit over the
meridian of Greenwich s . . . . 21^35? 8 '.1
Apparent time of planet's transit over the
given meridian s «•«,,. 21t34T39M
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INTRODUCTORY PROBLEMS. 317
Probijsm Xlf.
To find the Apparent Time of a Starts Transit, or Passage over the
Meridian of any known Place*
Since the plane of the meridian of any given place may be conceived to
be extended to the sphere of the fixed stars,— therefore, when the diurnal
motion of the earth. round its axis brings the plane of that meridian to any
particular star, such star is then said to transit, or pass over the meridian
of that place. This observation is applicable to all other celestial objects.
The apparent time of transit of a known fixed star is to be computed by
the following
Rule.
Reduce the right ascension of the star, as given in Table XL1V., to the
given day ; from which (increased by 24 hours if necessary,) subtract the
sun's right ascension at noon of that day, as given in ttie Nautical Almanac^
and the remainder will be the approximate time of transit.
Turn the longitude of the given meridian or place into time, by Problem
I., page 296, and add it to the approximate time of transit if the longitude
be west, but subtract it if east ; and the sum, or difference, will be the
corresponding time at Grreenwich ; and let it be noted whether that time
precedes or follows the noon of the given day.
Pind^ in the Nautical Almanac, the variation of the sun*s right ascension
between the noons preceding and following the Greenwich time ; then.
To the proportional logarithm of this variation, add the proportional
logarithm of the difference between the Greenwich time and the noon of
the given day (esteeming the hours as minutes, and the minutes as seconds),
and the constant logarithm 9» 1249* *, the sum of these three logarithms,
abating 10 in the index, will be the proportionaHogarithm of a correction,
which^ being added to the approximate time of transit if the Greenwich
time precedes the noon of the given day,, or subtracted therefrom if it
follows that noon, the sum or difference will be the apparent time of the
star's transit over the given meridian.
Example 1.
At what time on the 2d of January, 1825, will the star Rigel transit, of
come to the meridian of a place 165?30C east of Greenwich ?
* This U the ariihoMCical complement of the proporHontl logarithm of 24 hours, esteemed
as wtwtUfit
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Right ascension of Rigel^ reduced to the given day, = . . 5* 6T 8!
Sun's right ascension at noon of the given day =s • . • . 18. 5 1 • 44
Approximate time of transit » ••«••••«•• 10M47S4!
Longitude 165 ?30C east^ in time a , . , » 11. 2. 0
Greenwich time past noon of January 1st so 23M2724!
which it 47*36! brfpre noon of the given day*
Sun's right ascension at nooUj January 1st =; • • • • • 18? 47*19 f
Sun's ditto 2d s . ^ . . . 18.51.44
Variation of right ascension in 24 hours ss Oi 4T2S'.
Variation of right ascension s 4T25 ! Prop. log. = 1. 6102
Diff. of Gr . time from noon = 47. 36 Prop. log. = 2. 355 8
Constantlog. « •,•,..•«...• 9.1249
Correction of star's transit s + 0? 9! Prop. log. = 3. 0909
Approximate time pf transits lOM 4?24 ?
Apparent time of transit = 10? 14?33!^ as required.*
Example 2.
At what time on the 2d of January, 1825^ will the star Markab transit,
or come to the meridian of a place 140? 40' west of Greenwich ?
Right ascension of Markab, reduced to the given day, as * 22!56? 3!
Sun's right ascension at noon of the given day =& . . • • 18.51«44
Approximate time of transit ac • • • . • 4 . • • • 4* .4*19'
Longitude 140? 40'. west,. in time a . ....... 9.22.40
— J. ■ ■
Greenwchtime =5 , , . 13i26T59!
which, of course, is post the nooa of the given day«
Sun's right ascension at noon, January 2d, ss • . • • 18M1T44!
Sun's ditto 3d, s . . • . 18.56. 8
Variation of right Mceorion in 24 houi«s «...•. Ot 4?24!
* If 12 hourst dimioished by half tlie variation of t&e sun's right ascension, ht added to
te apparent time of tnarit. tte« foaod* ike stna, abalUic 24 hoafs if aacasiao* wttl give
the apparent time of transit below the pole.
IVTROBUCTOEY PROBUTMS. 319
Variation of right ascensions 4?24; IVop.log. s 1.6118
Diff.ofGr.timcfjromttoonal3*26?59! Prop.log. « 1.1266
Constant log. m . ^ . ^ 9. 1249
Correction of star's transit B= .-2T28! Prop. log. = 1.8633
Approximate time of transit a= 4U?19!
Apparent time of titmsit a . 4M rS 1 !| as required.*
Noter-^Th^ correction of a star's, approximate time of transit may be
readily found by means of Table XV., in the same manner^ precisely, as jf
it were the proportional part of the sun's right ascension that was under
consideration. — See explanation, page 25, and examples, pages 26 and
28.
PnoBLXM XUL
Tojmd what Siar$ mil be on^ar neate$i to^ th^ Meridian ai any gteen
lime.
RULB.
To the sun's right ascension, at noon of the given day, add the apparent
time at ship, and the sum will be the right ascension of the meridian or
mid-heaveh; with which enter Table XLW., and find what stars' right
ascensions correspond with, or come nearest thereto, and they will be the
stars required.
If much accusacy be required, the sun's right ascension at noon of the
given day must be previously reduced to the given time anci place, by
Problem V., page 298 ; at sea, however, this reduction may be dispensed
with.
Example 1.
What star 1^ be nearest to tiie meridian, April 6th, 1825, at 9U0r20!
apparent time }
Sun's right ascensbn at.ndon of .the given day = 1 ^ 0?24 !
Given apparent time at ship or place = • • 9. 40. 20
Right ascension of the meridian or mid-heaven= 1 0M0T44 !
Now^ this being looked for among the right ascensions of the stars^ in
* See Note, page 318.
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320 NAUTICAL ASTRONOMY.
Table XLIV., it will be found, that the star's right ascension corresponding
neatest thereto, is that of ij Argds Navis; which, therefore, is the star
required, or the one nearest to the meridian at the giren time.
'Example 2.
Wliat star will be neifTest to the meridian, December 31st, 1825, at
10^ 12r41 : apparent time ? *
Sun's right ascension at noon of the given dayss 18?41!'4«V
. Given apparent time at ship or place = • • 10. 12. 41
Right ascension of the meridian or mid-heaven= 4*54'r30!
Now, this being looked for among the right ascensions of the stars, in
Table XLIV,, it will be found that the staf's right ascension corresponding
nearest thereto, is that of j3 Eridani ; which, therefore, is the star required,
or the one nearest to the meridian at the given time.
2Vbf€«--*When the sum of the sun's right ascension and the apparent
time exceeds 24 hours, let 24 hours be subtracted therefrom 5 and the
remainder will be the right ascension of the meridian, as in tlie last
example.
Probijbm XIV.
Given the observed Altitude of the hwer or upper Limb of the Sun^ to
find the true Altitude of its Centre. »
• Rule.
For the Fore Observation.
To the observed altitude of -the sun's lower limb (corrected for index
error, if any,) add the difference between its semi-diameter * and the dip
of the horizon f 3 and the sum will be the apparent altitude of the sun's
centre: or, from, the corrected observed altitude of the sun's upper limb
subtract the sum of the semi-diameter* and the dip of the horizon fj and
the remainder will be the apparent central altitude.
For the Back Observation.
From the observed altitude of the sun's lower limb subtract the difi^r-
* Page III. of the month in the Nautical Almanac. t Table II.
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INT&ODUCTORT PR0BUM8. 321
ence between its semidiameter and the dip of the horizon : or^ to the
observed altitude of its upper limb add the sum of the semi- diameter and
the dip of the horizon^ and the sun's apparent central altitude will be
obtained.
Now^ from the apparent altitude of the sun's centre, thus found, sub-
tract the difference between the refraction* corresponding thereto, and the
parallax in altitudef, and the remainder will be the true altitude of the
sun's centre.
Example \.
Let the observed altitude of the sun's lower limb, by eifore observation,
be 16?29^, the height of the eye above the level of the sea 24 feet, and
the sun's semi-diameter 16H8?^ required the sun's true central altitude?
Observed altitude of the sun's lower limb = 16? 29^ 01
Sun's semidiameter =s , . ^^'^^^In'^r
Dip of the horiz. for 24 ket^ 4. 42 J^^^' - + 1 1 . 36
Apparent altitude of the Sim's centre = • • 16?40'36T
Refraction = 3^ 8r n .^ ^ ^
Parallax =:: 0,8 / DiflFerence = ... -^. 3. 0
True altitude of the sun's centre = • • . 16?37'36r
Kxample 2.
Let the observed altitude of the sun's upper limb, by hfore observation,
be 18?37 •) the height of the eye above the surface of the water 30 feet,
and the sun's semi-diameter 15M6'' ^ required the true central altitude ?
Observed altitude of the sun's upper limb s 18^37* 0?
Sun's semi-diameter s . . 15' 46^1 _9i i *
Dipof the horizon for 30 feet = 5. 15 / ^^^ ■' *
Apparent altitude of the sun's centre = • • 18? 15 '.591
Refraction = 2^5K
Parallax = 0. 8
j Differences. . . — 2.43
True altitude of the sun's centre :? . . • 18? 13^6?
Example 3,
Let the observed altitude of the sun's lower limb, by a bach observation,
be 20? 10' , the height of the eye above the level of the sea 25 feet, and the
sun's semi-diamete/ 15^55^ ; required the true central altitude ?
• Table Vni. t Table Vn.
Y
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322 MAUTIOAL AlTROlfOirr.
Observed altitude of the aun's lower limb = 20M0i 0^
Sun's semi-cUameter ss . . \^''^^'XryM— \\ g
Dip of the horizon for 25 feet = 4.47 J *
Apparent altitude of the aun's centre = . . 19'?58'.52*
Refractional 2?35^
Farbllax = . 0. 8
Refractions 2?35 ^ | ^^^^^^ ^ ^ ^ ^ ^ 2.27
True altitude of the sun's centre =s . • • 1 9? 56 ' 25 ?
Example 4.
Let the observed altitude of the sun's upper limb^ by a back obsenaiianj
be 25?31', the height of the eye above the surface of the water 27
feet^ and the sun's semi-diameter 15 M9?; required the true central
altitude ?
Observed altitude of the sun's upper limb s 25 "^Sl i Ot
Sun's semi-diameter = . 15C49T1 ^ oa 47
Dipofthehori2onfor27feet=4.58 J^^- +^U-47
Apparent altitude of the sun's centre =s . . 25?51 M7^
Refractions 1'57^1-..^ , ^^
Parallax = 0. 8 /I>iflference« , • - 1.49
True altitude of the sun's centres . • • 25?49'^58f
Semark.'^ think it my duty, in this place, to caution the mariner
against the mistaken rule for the back chservaHony given in some treatises
on Navigation ; — because, if that rule be adopted^ the ship's place will,
most assuredly, be affected by an error in latitude equal to the full measure
of the sun's diameter, or about 32 miles : and this, to a ship approaching
or drawing in with the land, becomes an object of the most serious con-
sideration, since it so very materially affects the lives and mlpreits of those
concerned. To set the mariner right in this matter^ I will here work an
Examfie.
December 25th, 1825, in longitude 35"? W., the meridiw altitude of tke
sun's lower limb, by a back observation^ was 16?28C south, the height of the
eye being 20 feet 3 required the latitude ?
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IKT^ODUCrofiY PROBLEMS. d23
Observed altitude of the sun's lower limb = 16?28! 0?
Sun's semi-diameter = . 1 6 U 8^ i
Dip of the horizon for 20 feet=4. 17 /^^^•== ^ *2. 1
Apparent altitude of the sun's centre = . . 16^15^59^
Refraction = 3'13?1
Parallax = .0. 8 /Differences , . . - 3. 5
True altitude of the sun's centre a: . . i 16?12'64^
Sun's meridional zenith distance =: • i * 73M71 6^ north.
Sun's corrected declination s . • « ; . « 2di 24^ 46 south.
Required latitude c= ........ 60?22ia0r north.
By the old rule, the latitude is only 49?50' norths wbieb is evidently
erroneous^ it being 32 miles and 20 seconds less than the truth.
Problbm XV.
Given the oSserved Altitude of the upper or tower Limb qfihe Moon, to
find ihe trw central AUitude.
RULB.
Turn the longitude into time, and add it to the apparent time of
observation if it be west, or subtract it therefrom if east, and it will give
the corresponding time at Greenwich.
To this time let the moon's semi-diameter and horizontal parallajc be
reduced, by Problem VL, page 302, (or by Table XVI., as ekplaincsd iri
pages 30 and 33,) aiid let the reduced semi-dimneter be increased by the
correction contained in Table IV., answering to it and the observed alti-
tude; then.
To the observed altitude of the moon's lower limb (corrected for index
error, if any), add the difference between the true semi-diameter and the
dip of the horizon ; dr, from the observed altitude of the upper limb
subtract the sum of the semi-diameter and dip, and the apparent central
altitude of the moon will be obtained ; to which let the correction (Table
XVIII.) answering to the moon's reduced horizontal parallax and apparent
central altitude be added, and the sutn will be the altitude of the moon's
cetitre.
Example 1.
In a certain latitude, March 10th, 1825^ at3M0T2O! apparent time,
y 2
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324 NAUTICAL ASTRONOMY*
the observed altitude of the moon's lower limb was 20?10U0T9 and the
height of the eye above the level of the sea 24 feet ; required the true
altitude of the moon's centre^ the longitude of the place of observation
being 35^40'. west? ^
Apparent time of observation = •
Longitude 35'?40' W., in time =
Greenwich time =
Moon's reduced semi-diameter =s ,
Augmentation, Table IV. s • •
Moon's true semi-diameter s . •
Moon's reduced horizontal parallax s=
3*40T20*
2.22.40
6* 3? 0!
15^401:
0. 6
15<46r
57 '32^
Observed altitude of moon's lower limb = 20?10'.40r
Moon's true semi-diam. = 15 ^46T '
Dip of the horiz. for 24 feet=4. 42
}Diff..= +11. 4
Apparent altitude of the moon's centre = 20? 2 T. 44 ^
Correction to altitude 20?21M4r, and
horiz. parallax 57 ' 32'r , Table XVIII. = + 5 1 . 24
True altitude of the moon's centre = • 21 ? 13' 8r
Example 2,
In a certain latitude, March 26th, 1825, at 1 ^30^47 • apparent time,
the observed altitude of the moon's upper limb was 30? 17 '30?,- and the
height of the eye above the level of the sea 30 feet ; required the true
altitude of the moon's centre, the longitude of the place of observation
being 94^15 '.30r east?
Apparent time of observation = . . . Ii30"47*
Longitude 94n5:30^ E., in time = . 6.17. 2
Greenwich time past midnight, March 25th=7 ^ 13T45 !
Moon's reduced semi-diameter = ... 15'.23T
Augmentation, Table IV. = . • • • 0. 9
Moon's true semi-diameter s=. • . • . 15 '3 K
Moon's reduced horizontal parallax == • 56'.2(>T
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INTRODOCTORY PROBLBMS* 325
Observed altitude of moon's upper limb = 30? 17C30^
Moon's true semt-diaro. = 15 '3K 1
Dipofthehoriz.for3(>feet=5. 15 /^^"">= -20.46
Apparent altitude of moon's centre = . 29?56'44r
Correction to altitude 29?56M4?, and
horiz. parallax 56 \ 26r, Table XVIII. = + 47. 16
True altitude of the moon's centre = . 30? 44^ Or
iVb/6*— ^In the above examples, the altitudes are supposed to be taken by
the fore obseroatum; and since this mode of observing is not only the
most naturiU, but, also, the most simple, it will, therefore, be constantly
made use of throughout the subsequent parts of this work. Hence the
necessity of making constant reference to the particular mode of observa-
tion may, in future, be dispensed with.
Pbobum XVI.
Gicen the observed AUUude qf a Planet's Centre, to find its true
JUitude. ^
Rule.
From the planet's observed central altitude (corrected for index error, if
any,} subtract the dip of tlie horizon, and the remainder will be the apparent
central altitude.
Find the difference between the planet's parallax in altitude (Table VI.}
and its refraction in altitude (Table VIll.} ; now, this difference being
applied by addition to the apparent central altitude when the parallax is
greater than the refraction, but by subtraction when it is less, the sum or
remainder will be the true central altitude of the planet.
JBspample I.
Let the observed central altitude of Venus be 16?40^, the index error
2^30' tvbtractive, and the height of the eye above the level ot the sea 28
feet; required the true altitude of that planet, allowing her horizontal
parallax to be 31 seconds ?
326 NAUTICAL ASTRONOMY.
Observed central altitude of Venus = . 16?40' Or
Index error = — 2. 80
Dip of the horizon for 28 feet = . . . — 5.5
Apparent central altitude of Venus = . 16'?32'25T
Refraction, Table VIII.,=3 '. 10^ \j^.^_ n . i
Parallax, Table VI., = 0.29 J ^^^^^ " ^^^*
^^/c^. Apparent central altitude of Venus = . 16?29^24r
Example 2.
Let the observed central altitude of Mars be 17'?29'40r, the index error
aUSr additive, and the height of the eye above the surface of the water
26 feet \ required the true central altitude of that planet, allowing hi8
horizontal parallax to be 17 seconds ?
Observed central altitude of Mars » . 1 7?39 ' 40^
Index error = ....... • +3.45
Dip of the horizon for 26 feet = . . — 4. 52
Apparent central altitude of Mars = . 17°28'33'!f
Refraction,TableVIII.,=2?59lf1_.„ _ ^ .^
Parallax, Table VI., = 0.16
/yH^4^ Apparent central altitu4^ qf Mars = • 17^25^50^
'Reffwvrk^ — ^In taking the altitiick pf a planet, its centre should be
brought down to the horizon. Neither the semi-diameters nor the hori-
zontal parallaxes of the planets are given in the Nautical Almanac, but h is
to be hoped that they soon will be. If the parallaxes of the plaueta be de-
termined by means of a comparison of their respective distances (from the
earth's centre) with the earth's semi-diameter, they will be found to be as
follows^ very nearly ; viz.,
Venus' greatest horizontal parallax, about S3 seconds \ and her least
parallax about 5 seconds.
Mars' greatest horizontal parallax, about 17 seconds \ and bis least paral*
lax, about 3 seconds.
Jupiter's mean horizontal pai^aiUKx^ i^btfiit 2 seconds \ and that of Saturn
about 1 second.
Thj» pairallaxed of the two last planets are subjeet to i»ry iiMie alleiatioQ,
because the dislances at which those otjeels ate placed froHL tks esvlL'b
oentm are so exQeedi^g^y great as to raidor any variatiMiB id tkisu panal-
laxes almost insensible.
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INTRODUCrOftY PHOBLBMS. 327
Problem XVII.
Gioen the observed Altitude of a fixed Star, to find the true Altihide.
RULB.
To the observed altitude of the star apply the index error^ if any; from
which subtract the dip of the horizon, and the remainder will be the
star's apparent altitude.
From the apparent altitude, thus found, let the refraction corresponding
thereto be subtracted, and the reminder will be the true altitude of the
star*
Example 1.
Let the observed altitude of Spica Vir^inis be 18?30^, the index error
3! 20^ subtractive, and the height of the eye above the level of the water
18 feet 'j required the true altitude of that star ?
Observed altitude of Spica Virginis = 18?30^ 01
Index error « — 3. 20
Dip of the horizon for 18 feet = • — 4. 4
Apparent altitude of Spica ^rginis s 18? 22 '36?
Refraction = — 2. 50
True altitude of Spica Virynk ar . 18? 19M6r
Example 2.
Let the observed altitude of Regulus be 20?43', the index error 1'47?
sdcKtive, and the height of the eye above the level of Ihe sea 20 feet ;
reqiured the true altitude of that star ?
Observed aUtode of Reguhiss 20?43C Oi
Index errors ..... +1.47
I>ipofthehori»»lbr20ieet8i - 4.17
Apparent altitude of Reguhis s 20?40C30r
Refraction s -. 2. 29
Tnie«ititwleofiUg«lBa«: . a0t38^ K
JVbte.— The fixed stars do not exhibit any apparent semi-diameter, nor
any sensible parallax ; because the immense and inconceivable distance at
which they are placed from the earth's surface causes them to appear, at
all times, as so many mere Ittwomit indivisible points in the heavens.
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328 NAUTICAL ASTRONOMY.
SOLUTION OF PROBLEMS RELATIVE TO THE LATITUDE.
The Latitude of any place on the earth is expressed by the distance of
such place from the equator, either north or south, and is measured by an
arc of the meridian intercepted between the said place and the equator.-—
Or,
The Latitude of any place on the earth is equal to the elevation of the
pole of the equator above the horizon of such place ; or (which amounts to
the same), it is equal to the distance of the zenith of the place from the
equinoctial in the heavens. The complement of the latitude is the distance
of the zenith of any place from the pole of the equator, and is expressed
by what the latitude wants of 90 degrees. The latitude is named north or
south, according as the place is situate with respect to the equator.
Problem I.
Given the Sun's Meridian AUiiude, to find the Latitude oftlie Place qf
Observation.
Rule.
Find the true altitude of the sun's centre, by Problem XTV., page 320,
and call it north, or south, according as that object may be situate with
respect to the observer at the time of observation ; which, subtracted from
90?, will give the sun's meridional zenith distance of a contrary denomi-
nation to that of its altitude. - • •
Reduce the sun's declination to the meridian of the place of observation,
by Problem V., page 298, or, more readily, by Table XV. Then, if the
meridional zenith distance and the declination are both* north or both south,
their sum will be the latitude of the place of observation; but if one be
north and the other south, their difference will be the latitude, and always
of the same name with the greater term.*
* The principles upon which this role is founded may he seen hy referring^ to ** The
Youni^ Navigator's Guide to the Sidereal and Planetary Parts of Nautical AstroimiQj/*
page 98 1 readiog;, howerer, the word «m instead of star.
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OF FINDING THE I.ATITUDB BY A MBRIDIAN ALTITUDB. 329
Example 1.
April 10th, 1825, in longitude 75? W., the meridian altitude of the
sun's lower limb was 57?4U'.30^ S., and the height of the eye above the
level of the sea 22 feet ; required the latitude^
Observed altitude of the sun's lower limb = S7?40C30r S.
Sun's semidiameter =s . ^^'^S'' I ^.^ _^
Dipofthehoriz.for22feet=4.30 /^»"- - + "-29
Apparent altitude of the sun's centre = • 57?51 C59T S.
Refraction = 0^35ri^.^ • ,, ^^
Parallax = 0. 5 / Difference = . - 0.30
True altitude of the sun's centre s « . . 57^5K29rS.
Sun's meridional zenith distance = . . . 32? 8C3irN. 32?8:3KN.
Swi's declination at noon, April 10th = 7^56 M2? N,
Correction for longitude 75? W. = . + 4. 36
Sun's reduced declination = ... 8? I'lSTN. 8?ia8rN.
Latitude, as required = • ; • • • 40?9C49?N.
Note. — ^The meridional zenith distance and the declination are added
tc^ther, because they are both of the same name : hence, the latitude is
40?9M9r N.
Example 2.
October 24th, 1825^ in longitude 90? east, the meridian altitude of the
ran's lower limb was 27?31^20^ S., and the height of the eye above the
surface of the sea 23 feet ; required the latitude ?
Observed altitude of the sun's lower limb = 27?31 ^ 20r S.
Sun's semi-diameter = . 16' 8^1 ^
Dip of the horiz. for 23 feet = 4. 36 J *^'^- •" + ^ * • ^^2
Apparent altitude of the sun's centre = • . 27?42*52T S.
?*'*^°°=i''*f\ Differences ... - 1.40
Parallax = 0. 8 J
True altitude of the sun's centre = . . • 27 ?4 1 U 2r S.
Sun's meridional zenith distance ss , . « 62?18'48rN,
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330 NAOTICAt ASTRONOMY.
Sun*8 declination at noon, Oct 24th = 1 1 ?45 ^ 42r S.
Correction for longitude 90? east = • — 5. 15
Sun's reduced declination = • • • 1 1 ?40^ 27^ S.
Sun's meridional zenith distance ss • 62.18.48 N.
Latitude, as required ^ • • • * 50?38'2KN.
Note.— The difference between the meridional zenith distance and the
declination is taken^ because they are of contrary names : hence^ the lati*
tudei8 50?38:2irN.
Problem IL
Given the Mom's MenSmial AUitude, ioJM the Lufiltnde of the Place
of Observatioju
Bulb.
Reduce the moon'^ passage over the meridian of Greenwich, on tb«
given day, to the meridian of the place of observation, by applying thereto
the correction in TaUe XXXVIIL, by addition or subtraction, accoidiiig
as the longitude is west or east; as explained in examples 1 and 2, pages
tOl and 103.
To the time of the moon's passage over the meridian of the place of
observation, thus found, let the longitude of that meridian, m time, be
added if it be west, or subtracted if east; and the sum, ox differeiice, will be
the corresponding time at Greenwich : to which let the moon's decUiuitkM^
horizontal parallax, and semi-diameter, be reduced by Problem VI., page
302, (or by means of Table XVI., as explained in page 30,) and let the
moon's reduced semi-diameter be corrected by the augmentation contained
in Table IV.
Find the true altitude of the moon's centre, by Problem XV., page 323^
and call it north or south, according as it may be situate with respect to
the observer at the time of obeenration | which, sobtiacted from 90?, will
give the moon's meridional zenith distance of a ccmtrary detwmBHiation to
that of its altitude.
Then, if the meridional zenith distance and the declination are of the
same name, their sum will be the latitude of die place of observation ; but
if they are of contrary names, their difference will be the fattitud^ of the
same name with the greater term.
Note^-^ln strictness, the moo»'s dccliBatioo siM>uld be oovreeted by the
equation of second difference contained in Table XVII, as explained
between pages 33 and 37*
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OF FINDING THB MTITTOB BT A' MBBIDIAN ALTITUDB. 331
Exampie I.
Jnnviuy 27th9 1825, in longitude 55"? W., the meridian altitude of the
moQp'a lower limb was 58?40t S., and the height of the eye above the
level of the sea 26 feet ) required the latitude )
Time of ]) 'a passage over the meridian of Greenwich s= , • 5^54? 0!
Correction, Table XXXVIII., for longitude 55? W. = . . +7.23
lime of )> *8 pass, over the merid. of the place of observation = 6^ 1^23^
Iiongitude 55? W., in time = + 3.40. 0
Greenwich time = 9?4lr23!
Moon's horizontal parallax at noon, Jan. 27th ss 55 '20^
Correction of parallax for 9*41 r23t » • • +0.17
Moon's reduced horizontal parallax s • « • . 55 '37^
Moon's semi-diameter at noon, Jan. 27th = , . 15^ 5T
Correction of semi-diameter for 9*41 ?23! = . +4
Augmentation of semi-diameter. Table IV. 9 , +12
Moon's true semi-diameter = 15^31^
Mom's declination at nooo, Jan. 27tb a 18? 19! 181: N.
Cbntctioaofdeclinatioafbr9MlT2S!s: +1. 16. 7
Moon's reduced declination ::;: « . . 19?35 1 25 r N.
Obeenred altitude of the mooo^s lover Kmb s . • 58?40? OfS.
Motn'a true semi-diameter = 15!2P l ^
I>ipofthehoriz.for26feet = . 4.52/*^*" + IU.2»
Apparent altitude of the moon's centre =
CorrectioQ of altitude. Table XVHL =
Thie altitude of the moon's centre s=
\
Moon's meridional zenith distance =
Moon's reduced declination = • •
Latitude of the place ot obeeivation ^
.. 58t50^2»*S.
. + 28. 12
. 5»?I8UirS.
. 30?4in9rN.
. 19.35.25 N.
• 50?16:44rN.
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332 KAtrricAL astronomy.
Example 2.
February 3d, 1825, in longitude 65? E., the meridian altitude of the
moon's upper limb was 62?45^ north, and the height of the eye above
the level of the sea 29 feet j required the latitude ?
Time of ]) 's passage over the meridian of Greenwich = « 12t25? 0!
Correction, Table XXXVIIL, for longitude 65? east = , . - 9. 44
Time of }) 's pass, over the merid. of the place of observations 12 1 15T16'
Longitude65?E., in time = .....•....— 4.20. 0
Greenwich time = 7*55?16!
Moon's semi-diameter at noon, February 3d = 16^341^
Correction of semi-diameter for7*55?16! s -f 1
Augmentation of semi-diameter, Table IV. = +16
Moon's true semi-diameter = ]6^5K
Moon's horizontal parallax at noon, February 3d = 60U9T
Correction of parallax for 7*55716! = . . . + 5
Moon's reduced horizontal parallax = . . • • 60' 54^
Moon's declination at noon, February 3d s 1 2^ 52 ! 50^ N.
Correction of declination for 7*55?16! =5 - 1. 47. 29
Moon's reduced declination s 11?5!21TN.
Observed altitude of the moon's upper limb = 62?45C OTN.
Moon's true semi-diameter =5 16'51^1
Dip of the horiz. for 29 feet = 5. 10 J ^""^ = - 22. 1
Apparent altitude of the moon's centre = . . . 62?22'59r N.
Correction of altitude. Table XVIII. == . . . +29.33
True altitude of the moon's centre s • • • 62?52^32rN.
Moon's meridional zenith distance ^ . • 27? 7-28? S.
Moon's reduced declination s= • , • . 11; 5.21 N.
Ladtude of the place of observation = • • 16? 2C 7' S.
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OF FINDING THE JLATITUDB BY A MERIDIAN ALTITUDE.
Remark. — Although this method of finding the latitude at sea is strictly
correct when the longitude of the place of observation is well determined;
yet, in some cases, it is subject to such peculiarities as to render it inconve-
nient to the practical navigator : this happens in high latitudes, and when
the variation in the moon's declination is very considerable ; because, under
such circumstances, the moon's altitude sometimes continues to increase
after she has actually passed the meridian* To provide against this, the
observer should be furnished with a chronometer, or other well-regulated
watch, to show the instant of the moon's coming to the meridian of the
ship or place 3 at which time her altitude should be taken, without waiting
fcr its ceasing to rise or beginning to dip, as it is generally termed at sea :
then this altitude is to be considered as the observed meridional altitude of
that object^ and to be acted upon accordingly.
PUOBLSM IIL
Given the Meridional Altitude of a Planei, to find the Latittide qfthe
Place of Observation.
Rule.
To the apparent time of observation (always reckoning from the preceding
noon,) apply the longitude, in time, by addition or subtraction, according
as it is west or east ; and the sum, or difference, will be the corresponding
time at Greenwich, to which let the planet's declination be reduced, by
Problem VIL, page 307.
Find the true altitude of the planet's centre, by Problem XVI., page
325 ; and hence its meridional zenith distance, noting whether it be north
or south : then, if tlie meridional zenith distance and the declination are of
the same name, their sum will be the latitude of the place of observation ;
but if they are of contrary names, their difference will be the latitude, of
the same nam'e with the greater term.
Example I.
February 3dj 1825, in longitude 80? W,, at II*28?30! apparent time,
the meridional central altitude of the planet Jupiter was 58?22' S., the
height of the eye above the level of the sea 24 feet, and the planet's hori-
zontal parallax 2 seconds ; required the latitude?
Apparent time of observation, February = 3 f 1 1 1 28T30'
Longitude 80? W., in time = .... + 5.20. 0
Greenwich time =5 ....... 3fl6*48r30!
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884 KAimcAL AiTRONoiiy.
Jupiter's declination^ February Ist » , 19? S' OI'N.
Correction of ditto for 2fl6M8730! « + 5.51
Jupiter's reduced declination tt , « . 19? 8'5KNi
Jupiter's observed central altitude = 58?22C Or S.
Dip of the horizon for 24 feet = — 4.42
Jupiter's apparent central altitude = 58? l/'lSf 8.
RefracUohjTab.VII
Parallax, Table VI.
RefracHoh,Tab.VIII.=0:S4r \
1.= 0. 1 /^*^'= -"-^
Jupiter's trae eentral altitude » • 68? 16'. 45? S.
Jupiter's meridional zenith distance =3 1 ?43 M 5 ' N.
Jupiter's reduced declination ss ; 19. 8.51 N^
Latitude of the place of observation sa 50?52' 6? N.
Bsampk 2.
March 16th, 1825, tn lon^tude 75? E., at iU9? apparent time, the
meridional central altitude of the planet Venus was 31? IOC N., the height
of the eye above the level of the horizon 18 feet, and the planet's horizon-
tal parallax 23 seconds ; required the latitude ?
Apparent time of observation, Mareb ta 16f 2M9?
toingltude,75?B.^in time =s « « « ^ 5^ 0
(}reenwich time = «••«••• 15 f 21^49?
Venus' declination, March 13th = 17?I5( OrN,
Correction of ditto for 2f 21*49*?= + 1. 5. 57
Venus* reduced declination =3 • • 18?20'57?N.
Venus' observed central altitude = . 3 1 ? 10 ! 0? N.
Dip of the horizon for 18 feet a • . — 4. 4
Venus' i^parent central altitude = 4 3 1 ? 5 C 56? N.
Refraction, Table VIII. = 1 • 83 ^ 1 p..^ _
Parallax, Table VI. = 0*20 i^*"' ^'^^
Venus' true central altitude = • . • 3 1 ? 4 ' 4 1 ?N.
Venus' meridional zenith distance = s 58?55' 19? S.
Venus' reduced declination = . • . 18.20.57 N.
Latitude of the place of observation = 40?34 ' 22? S.
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OF FINDING THB LATITUDB BY A IIBRIDIAN ALTITUDB. 835
Note.^-^e principles of finding the latitude by the meridional altitude
of a celestial object may be seen by referring to ^* the Young Narigator's
Guide to the Sidereal and Planetary Parts of Nautical Astronomy/' between
pages 98 and 105.
PnoBiJUlff IV*
Given the Meridional Jltiiude qf a fixed Star, to find the Latitude qfthe
Place of Obeervation,
RtTLB.
Und the true altitude of the star^ by Problem XVIL, page 327; and
hence its meridional zenith distance^ noting whether it be north or south.
Take the declination of the star from Table XLIV., and reduce it to the
time of observation. Now^ if the star's meridional zenith distance and its
declination be of the same name, their sum will be the latitude of the place
of obtervadoD ; but if they are of contrary names^ their diffeience will bo
the latitude, of the tame name with the greater term.
Example 1.
January Ist, 1825, fn longitude 85?3f W., at 12f 39r26! appartot time,
the meridian altitude of Procyon was 44?49^ S., and the height of the eye
atbove the level of the horizon 16 feet ; required the true latitude ?
Observed altitude of Procyon ss •
Dip of the horizon for 16 feet = •
Procyon's apparent altitude = • •
Refractions:^ .......
Proeyon's true altitude
• • •
44?49i OrS.
- 3.50
44?45n0rS^
— 0.57
44^44: 13rS.
Pipocyon'a meridional zenith distance =a 45? 15' 47 ^N.
Procyon's reduced declination =3 . • 5.40.16 N.
Latitude of the place of observation 8 50?56C 37N*
Example 2.
January 2d, 1825, in longitude 165 ?30' R, at I0M4r3d! i4>parent
time^ the meridian altitude of Rigel was 30^39^ S., and the height of the
eye abo^e tiM level of the sea 21 twt} jctfired Ibe true lMiCitde7
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336 NAUTICAt ASTRONOMY.
Observed altitude of Rigel= . . . 30?39t O^S.
Dip of the hwizon for 21 feet «= . . — 4. 24
Rigel's apparent altitude = . . . 30?34:36rS.
Refraction = — 1.37
Rigel's true altitudes 30?32:59fS. ^
Rigel's meridional zenith distance s 59? 27' KN.
Rigel's reduced declination s= . . • 8.24.35 S.
Latitude of the place of observation = 5 1 ? 2 ( 26rN.
Note. — The principles upon which the above rule is founded, are given
in ^^ the Young Navigator's Guide to the Sidereal and Planetary Parts of
Nautical Astronomy," between pages 98 and 105.
Problem V.
Given tlie Meridimal Altiiude of a Celestial Object observed below iJie
Pole, to find the Laiiiude of the Place of Observation.
Rule.
Find the true altitude of the object, as before ; to which let the polar
distance of that object, or the complement of its corrected declination, be
added, and the sum will be the latitude of the place of observation, of the
same name with the declination.
Example I.
June 20th, 1825, in longitude 65? W., the meridian altitude of the
sun's lower limb, observed below the pole, was 9?12l, and the height of
the eye 20 feet; required the latitude ?
Observed altitude of the sun's lower limb = . 9?12' OT
Sun's semi-diameter = . ^5'46^1 ^ ^
Dip ofthe horizon for 20 feet=4. 17 J^^^*-^ +11.29
Apparent altitude of the sun's centre = • . 9?23'29^
Refractions 5^34r
Parallax s 0. 9
I Difference =3 ... — 5.25
True meridian altitude below the pole s • • 9?18^ 4r
Sun's corrected polar distance, or co-declination =66. 32. 17 N.
Latitude ofthe place of obeervation =s • . . 75?50C21fN«
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OF FINDING THE ULTITUBB BT THB NORTH POLAR STAR, 337
Example 2.
June 1st, 1825^ in longitude 90? £.^ at 11^26r40t apparent time^ the
observed altitude of Capella^ when on the meridian below the pole^ was
11?48^ and the height of the eye above the level of the sea 25 feet;
required the latitude ?
Observed altitude of Capella as 11?48< Of
Dip of the horizon for 25 feet = — 4. 47
Capella's apparent altitude ==• •• . •• • lI?43n3T
Refraction = — 4. 29
Capella's true meridian altitude below the pole ss ll?d8M4f
Capella's corrected polar distance^ or co-declination=44. 1 L 28 N.
Latitude of the place of observation s .... 55?50U2rN.
Eemarks.^^1^ When the polar distance or co-declination of a celestial
object is less than the latitude of the place of observation (both being of
the same name), such celestial object will not set, or go below the horizon
of that place : in this case, the celestial object is said to be circumpolar^
because ii revolves round the pole of the equator^ or equinoctial, without
disappearing in the horizon.
2. If 12 hours, diminished by half the daily variation of the sun's right
ascension, be added to the apparent time of the superior transit of k fixed
star, it wU give the apparent time of its inferior transit over the opposite
meridian; that is, the apparent time of its coming to the meridian below
the pole.
3. The least altitude of a circumpolar celestial object indicates its being
on the meridian below the pole.
Probubm VI.
Given the AltUude of the North Polar Star, taken at any Hour qfthe
Night, to find the Latitude of the Place of Observation.
Although the proposed method of finding the latitude at sea is only
applicable to places situate to the northward of the equator, yet, since it
can be resorted to at any time of the night, it deserves the particular atten*
tion of the mariner.
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388 NAUTICAL ASTRONOMY,
Of all the heavenly bodies^ the polar star seems best calculated for find-
ing the latitude in the northern hemisphere by nocturnal observation;
because a single altitude^ taken at any hour of the night by a careful
obaerveri will give the latitude to a sufficient degree of accuracy^ provided
the apparent tinpe of observation be but known within ^^few minutes of the
truth : however, an error in the apparent time, even as considerable as 20
minutes, will not affect the latitude to the value of half a minute, when
the polar star is on the meridian, either above or below the pole ; nor will
it ever affect the latitude more than about 81 minutes, even at the star's
greatest distance from the meridian.. But, as it is highly improbable, in
the present improved state of watches, that the apparent time at the ship
can ever be so far out as five minutes, the latitude resulting from this
method will^ in general^ be as near to the truth as the common purposes
of navigation require.
RULB.
To the Mn's right ascension, as given in the Nautical Almanac, or in Table
XII. (reduced to the meridian of the place of observation, by Problem V.,
page 298,) add the apparent time of observation ; and the sum (rejecting
24 hours, if necessary,) will be the right ascension of the meridian, or
mid-heaven ) with which eater Table X., and take out the corresponding
correction. Find the true altitude of the star, by Problem XVIL, page
327 ; to which let the correction, so found, be applied by addition or
subtraction, according to the directions contained in the Table, and the
sum or difference will be the approximate latitude. •
Enter Table XL, with the approximate latitude, thus found, at top of
the page, and the right ascension of the meridian in one of the side
columns ; in the angle of meeting will be found a correction, which, being
applied by addiiion to the approximate latitude, will give the true latitude
of the place of observation*
iZemorft.— Since the corrections of the polar star's altitude, in Table X.,
have been computed for the beginning of the year 1824, a reduction there-
fore becomes necessary, in order to adc^t them to subsequent years and
parts of a year. The method of finding this reduction is illustrated in
exampkfis 1 and 2, pages 17 and 18.
Example h
January 2d» 1825, in longitude 60? west, at 8M0r40! apparent time,
the observed altitude of the polar star was 52?15^20f, and the hdght of
the eye above the level of the sea 16 feet ; required the latitude ?
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OF FINDING THE LATITUDB BT THB NORTH POLAR STAR. 839
Sun's reduced right ascension ss 18^53?58t
Apparent time of observation = 8.10.40
Right ascension of the meridians: •...«.«. 3* 4?38!
Correction of altitude, Table X.^ answering to 3? s». « • 1?94' 16f
Proportional part to 4T38! of right ascension s • . f — 1. I
Annual var. of correction = 13^. 67 ; which x by 1 year, gives — 0. 14
Correcticm of altitude, reduced to time of observation = • 1 ? 23 ' 1 T ;
which is iubtracHve, because the right ascension of the meridian falls in
one. of the left-hand columns.
Observed altitude of the polat star xs .
Dip of the horizon for 16 feet s . «
Apparent altitude of the polar star ss .
Refraction s : ' •
True altitude of the polar star, ss • •
Correction from Table X., ans. to 3!4?38
Approximate latitude s , . . « •
Correction of ditto from Table XI. s .
Latttilde of the place of obsenration s
Example 2.
*i
52? 15 ^20?
- 8.50
52?II{3df
- 0.44
S2?10M6f
- ].23. 1
60?47'45rN,
+ 0^28?
5Q?48U8TN.
January 1st, 1830, in longitude 75? W., at 9^3? i^parent time, let the
olMerred altitude of the north polar star be 19? 15', and the height ef the
eye above the level of the horizon 28 feet; reqtured the latitude?
Sun's R.A,, Table XII., reduced to ^ven times 18)49?
Apparent time of observation =: , . . . \ 9. S
Rig^t ascension of the meridian = . . . . 8*SS?
Correction ofaltitudp, Table X., answering to 8*50? s= . . U\V.3K *
Proportional part to 2 T of right ascension = ..•••— 0. 34
Annual var. of correction^ 1 0*' . 06, which X by 6 years, gives —1.1
Correction of altitude, reduced to time of observation, = . 1? 9' 57^ I
which is subtractive, because the right ascension of the meridian falls in
one of the left-hand columns.
z2
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340
NAUTICAL ASTRONOMY.
Observed altitude of the polar star = •
Dip of the horizon for 23 feet = . •
Apparent altitude of the polar star
Refraction ss
True dititude *of the polar star =s . •
Correction of altitude from Table X. =
-Approximate latitude = • ..• • • •
Correction of ditto. Table XI. s • •
Latitude of the place of observation ss
. i9?i5ror
. . — 4.36
. 19?10124r
. - 2.43
. 199 7'4ir
-1. 9.57
• 17?57U4rN.
. + 0. 15
. 17?57^59rN.
Example 3.
Let the true altitude of the north polar star, January Ist, 1854, be
50?5Uir^ and the right ascension of the meridian 17 • 13? i required th^
latitude ?
True altitude of the north polar stars 50"^ 5'4K
Cor. from T«ai.X., answ.to 17M0T=0?44^24n
Proportional part to as . . 3= — 1. 8 /Addit.= +0.41.36
Annual var.=s -3''. 34x30 years = - 1.40 J
Approximate latitude s 50?47-17^N.
Correction of ditto from Table XL, answ. to 17 * 13? = . + 1 • 17
True latitude, as required s • • . 50?48'34?N.
Noie.^-The true latitude, computed with the most rigorous degree of
accuracy, by spherical trigonometry, is 50?48'13^ N.3 the difference,
therefore, between the true spherical latitude, thus deduced, and that
resulting from Tables X. and XI., as above, is only 21 seconds in the long
period of 30 years : hence it is evident, .that the latitude may 1)e always
•determined by mea^s of those Tables, to every degree of exactness desir-
lible in most nautical operations.
The elementajy principles of computing the latitude by an altitude of
the north polar star, are given in " the Young Navigator's Guide to die
Sidereal aitd Planetary Parts of Nautical Astronomy,'" between pages 144
and 156, where a diagram may be seen, illustrative of the star's apparent
fnoHon round its orbit.
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OP FIKDIM6 TttB LATTTDDB BT BOVBLB ALTITTTBBS. 341
Paoblbm VIL
Given the Latitude by Account^ the Sun's DecUnation, an^ two observed
Altitudes of its lower or upper Limby*the elapsed Time^ and the Course
and Distance nm between the Observations; Jo find the Latitude of
the Ship at the Ime qf ObservaUm qf the greatest Altitude.
Ruus.
' To reduce the least Altitude to what it would be, if taken at the PlAce
where the greatest Altitude was observed :— - '
' Find the angle contained between the ship's coiurse, (corrected for lee-
way, if any,) and the sun's bearinjp at the time of taking the least altitude ;
with which, if less than 8, or with what it wants of 16 points if it be more
than 8, ester the general Traverse Table, and find the difference of latitude
corresponding thereto and\he distance made good between the observa-
tions, which call the reduction of altitude*
Now, if the kast altitude be observed in the forenoon, the reduction of
altitude is to be applied thereto by addition when the above angle is less
than 8 points, but by subtraction when it is tnore than 8 points 3 the sum,
or difference, will show what the less altitude would be if observed at the
same place with the greater altitude. Again, if the less altitude be
observed in the afternoon, a contrary process is to be observed ; viz.,' the
reduction of altitude is to be subtracted therefrom, when the above angle
is less than 8 points, but to be added thereto when it is greater.
To compute the Latitude:—
Reduce the sun's declination to the time and place where the greatest
dtitude was observed; then, to the log. secant of the latitude by account,
add the log. secant of the corrected declination ; the sum, rejecting 20
from the index, will be the logarithmic ratio.
To the log. ratio, thus found, add the logarithm of the difference of the
natural co-versed sines of the two corrected altitudes, and the logarithm of
the half-elapsed time (Table XXX.) ; the sum of these three logarithms
will be the logarithmic middle time. Find the time corresponding to this
in Table XXXI. ; the difiS^rence between which and the half-elapsed time
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342 NAUTICAL A8TROVOMY.
will be the time from noon when the grieatest altitude was observed.*
From the log. rising (Table XXXII.), answering to this time, subtract the
log. ratio ; and the remainder wiH be the logarithm of a natural number,
which, being subtracted from the natural co-versed sine of the. greatest
altitude, will leave the natural versed sine of the sun's meridional zenith
distance ; to which let the corrected declination be applied by addition or
subtraction, according as it is of the same or of a contrary name : and the
sum, or difference, will be the latitude of the ship at the time that the
greatest altitude was taken ; which may be reduced to noon, by means of
the log, if necessary.
If the latitude, thus found, differ considerably from that by account, the
operation must he repeated^ using the computed latitude in place of that by
account, until the latitude last found agrees nearly with the latitude used
in the computation*
Remarks. — 1. Since this method is only an approximation to the truth,
it requires to be used under certain restrictions ; vis., the observations must
bt tak^n between nine o'clock in the fortnoon^ and three in the afterno<Mi.
If both observations be in the forenoon, or both in the afternoon, the
elapsed time must not be less than the distance of the observation of the
greatest altitude from noon. If one observation be in the forenoon, and the .
other in the afternoon, the elapsed time must not exceed four lioura and a
half; and, in all cases^ the nearer the greater altitude is lo noon, the better.
2. If the sun's meridional zenith distance be less than the latitude^ the
limitations are still more contracted. If the latitude be double the meridian
zenith distance^ the observations must be taken between half-past nine in
the forenoon and half-past two in the afternoon ; and the elapsed time
must not exceed three hours and a half. The observations must be taken
still nearer to noon, if the latitude exceeds the meridian zenith distance in
a greater proportion.
Esample 1.
At sea, January 9th, 1825, in latitude 50"? 121 N., by account, and lon-
gitude 30? IOC W., at 21 ^30T0! apparent time> the observed altitude of the
sun's lower limb was 10? 27 -30?, and the bearing of its centre, by azimuth
compass, S.E. f S.3 and at 23^ lOTlO! the observed altitude was 17?6C40r«
and the height of the eye above the level of the sea 20 feet ; the ship's
course during the elapsed time was S.S.E.^ at the rate of 10 knots an hour;
required the latitude of the ship at the time of observing the greater alti-
tude?
* When the middle time is §preater than the half-elapsed time, hoth olMsrvaUons will be
on the same side of the meridian \ otherwise, on different sided.
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OF FINDING THB LATITUDB BY DOUBLE ALTITUDBS. ii$
Sun's bearing at let observation sb S.EL | S.^ or 3| points.
Ship's course = S.S.E.9 or 2 points.
Contained angle = 1| point.
Time elapsed between the observations = 1 *40T10!.— And,
Asl^ : lOr :: lUOriOt : 16U2r, or 17 miles nearly = the distance
run between the two observations.
Now, to course 1| point, and distance 17 miles, the difFerence of latitude
is 16. 5 miles, the reduction of altitude; which is additioe to the least alti-
tude/because the contained angle is less than 6 points, and the observation
made in the forenoon.
Time of observing the greatest altitude = . 23 MOT 1 0 !
Longitude SOUO^W., in time = ... -f 2. 0.40
Greenwich time past noon of the 10th Jan. =s 1M0?50!
Sun's decHnatipn at noon, January 10th & . • 2 1 ? 57 ' 50^ S.
Correction of declination. Table XV., for 1 \ 10?50! ^ -<« 0* 28
Son'a corrected declination ae • 21?£7«22?S.
Fint observed altitude of the sun's lower lunb = 10? 27 ' SOT
Sun's semi-diamcter = . 16' ISr I ^.^ , ^ ,
,4 jy jDiftaa + 12. 1
Dip of the horiz. for 20 feet :
II ■ m
Apparent altitude of the sun's centre s: • • 10?S9'SI T
Refractions: 4'56TI-..^ . .-
Parallax = 0. 9 }l>^«fe'«^<^' • • • ' *^*^
Reduction of altitudes .•«#«.. 'f 16*80
Reduced altitude s , 10?51M4r
Second observed altitude of the sun's lower limb as 17? 6140?
Sun's semi-diameter s= . • 16' 18? I .^_^ x 12 I
Dip of the horiz. for 20 feet = 4.17 P*^*-* + *^*
Apparent altitade of the nm's centre 3s • • • 17!18«41?
Refraction = 3ar } ^.^ « ko
RiraU«c» . 0.8 5DiflFerence= .... - 2S3
True altitude of the sun's centre =»..... 17U5'.48? S.
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344 KAtrriCAL astronomy.
Latitude by account = 50? 12' OrN.Log.secslO. 193746
Hme of obfl. Altitude. Nat co-v. sine. Red. dec
21?30r 0! 10^51U4r 811695. 2 l?57'22fS. Log. 8ec.= 10.032700
23.10-10 17.15.48 703190. Log. ratio = . . 0.226446
lUOr 10'. elap8.tiine.Diff.=z 108505. Log. = . . . . 5.035450
Ot50? 5! half-elapfedtime. Log. half-elapsed Uine = 0.663950
1. 39. 43 middle time. Log. middle time = . . 5. 925846 .
0M9T38! time from noon when the #
greatest altitude was taken. Log. rising = • . • • 4. 368450
Nat. co-versed sine of the greatest alt. = 703190. Log.ratio=0. 226446
Natural number = .4 13868. Log. = 4.142004
Sun's mer.z^.di8.=7 1 ? 54 i OrN. Nat.V.S.=689322
Do. reduced dec.= 21. 57. 22 S.
Latitude of shipss 49 ?56C38f north. And, since this latitude differs so
much from that by account, it will be necessary to repeat the cperatian.
Computed latitude = . . . 49?56^38r Log. secant a 10.191427
Reduced declination ^ . . 21 . 57. 22 Log. secant s 10. 032700
Log. ratio = 0.224127
Diff. of nat. co-versed sinesss 108505. Log. =s 5.035450
Half-elapsed time = . • Ot50r5! Log. half-elaps. time = 0. 663950
Middle times . . . . 1.39.9 Log. middle time = 5.923527
Time from noon when great-
est altitude was taken s 0M9?4 ! Log. rising a ... 4. 358520
Nat co-versed sine of the greatest alt. s 703190. Log.ratio=0. 224127
Natural number s 13627. Log. s . 4.134393
Sun's mer. z.dis.=71 ?54C52rN. Nat.V.S.=:689563.
Sun's red. dec. = 21. 57. 22 S.
Lat. of the ship s49?57'.30r north. And, since this latitude differs
only 52 seconds from the last, it may, therefore, be esteemed as the true
latitude.
iVbte.-»The correct latitude^ by spherical trigonometry, is 49?56^0r
north.
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OF FINDING THB LATITtlDB BT J>OUfiLB ALTITUDBS. S45
Example 2.
At sea^ April 14th, 1825, in latitude 43?47' S., by account, and longi-
tude 60? 25^ E., at 23^20^40! apparent time, the observed altitude of the
sun's lower limb was 35?54^, and at 2* lOTlO' apparent time, April 15th,
the observed altitude of that limb was 28?42'. 15?, and the bearing of the
sun's centre, by azimuth compass, N.W. } N. ; the height of the eye above
the level of the horizon was 24 feet, and the ship's course during the elapsed
time S.W., at the rate of 9 knots an hour ; required the latitude of the ship
at the time of observation of the greater altitude ?
Sun's bearing at 2d observation = N.W. } N., or ±: 3i points.
Ship's course = S.W.. or s= 4 points.
Contained angle = 8| points.
Time elapsed between the observations = 2*49?30! And,
Asl* : 9V :: 2*49?30! : 25^26?= the distance made good between
the observations.
Now, to course 7i points, and distance 25 miles, the difference of latitude
is 3. 7 miles, the reduction ofaUUitde; which is additive to the least altitude,
because the contained angle is less than 8 points, and the observation was
made in the afternoon.
Time of observing the greatest altitude s= . . , 23*20^40!
Longitude 60? 25 ^ E., in time = 4. 1.40
Greenwich time past noon, April 14th, = . . • 19M9? 0!
Sun's declination at noon, April 14th, s . . D?24n5?N.
Correction from Table XV., for \9t 19T0! = . +17. 19
Sun's reduced declination = 9?4 1'34?N.
First observed altitude of the sun's lower limb = 35?54' OrN.
Sun's semi-diameter =
Dip of the horiz. for 24 feet -
. . 15'58r). ^
eet= 4.42 i^'^'^ + "-^^
Apparent altitude of the sun's centre s • , . 36? 5n6VN.
Refraction = l'18r I __ , ,,
„ „ ^ - {Difference =5s « • • • — 1.11
Parallax, s 0. 7 ^
True altitttdeoTthe sun's centres . . . . 36? 4' srN.
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346 NAUTICAL ASTftONOMY.
Second observed altitude of sun's lower limb = 28M2C 15?
Sun's semi-diameter = . 15^56?
Dip of the horiz. for 24 feet= 4. 42
JDiff. = + 11.16
Apparent altitude of the sun*s centre = • • 28?5d^3lT
Refraction = l'43r
Parallax = 0. 8
Reduction of altitude =2 +3.42
j Difference = • • • — 1 . 35
Reduced altitude » « 28?55'.38r
Latitude by account » 43?47' O^Log.sec. =::10« 141486
Time of obs. Altitude. Nat. co-v. slue. Red. dec.
23*20r40* 36? V 51 411255 9941 ^34rLog. sec, =10, 006244
2.10.10 28.55.38 516302 Log. ratio = • . . 0.147730
2. 49. 30 elaps. time. Diff.=105047 Log. = .... 5. 021384
1 1 24T45 ! half-elapsed time . . Log. half-elaps. time =: 0. 441990
0.47. 8 middle time .... Log. middle time s 5.611104
0*37 "37* time from noon when the
greatest altitude was taken. Log* rising ss • • 4. 128390
Natural co-vened sine of the greatest alt. ss 41 1255Log.rakio80. 147730
Natural number s 9564Log. si 3. 980660
Sun's mer.z. dist.=53?15' 4?S.Nat.ver.S.=401691
Sun's red. dec. = 9.41.34 N.
Lat of the ship s 43^33^30? S. But^ since this latitude differs so much
from that by accounti it becomes necessary to repeat the operation.
Computed latitudes . . 43?33^30rS. Log. secant sz 10.139858
Reduced declination = . 9.41.34 N. Log. secant = 10.006244
Log. ratios . 0.146102
Diff. of nat. co-versed sines = 105047 Log. s • • 5.021384
Half-elapsed time = . . 1*24T45! Log.i-el^^s.timesO. 441990
Middle time » » . . , 0,46,57 Log. middle timesS* 609476
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OF FINDING THB LATITUPB BY TBB ALTITUDES OF TWO STARS. 347
Time from noon when the
greatest alt. was taken s 0^37*48 f Log. rising = . 4* 132610
Nat. co-vers. sine of the greatest altitude = 411255Log.ratio3:0. 146102
Natural number = 9694Log. = . 3. 986508
Sun's meM.dis.tt33rl4^30rS.sNat.V. S.si401561
Sun's red. dec. a:» 9.41.34 N.
Latitude = • 43?32^56'' south. And, since this latitude only differs
34 seconds from the last, it may be considered as being the latitude of the
ship at the time of observation of the greater altitude. The correct lati-
tude, however, by spherical trigonometry, is 43?29'30^ south : hence the
method by double altitudes, even after repeating the operation^ differs from
the truth by 3 minutes and 26 seconds.
Noie^^The method of Sliding the latitude by double altitudes, being a
very tedious and indirect operation^ and generally a very inaccurate one,
uftless the llmiutidni pointed out in the remarks (page 342) are strictly
attended to, no notice, therefore, would have been taken of it in this work,
w«r« it tiot fcr the purpose of giving the most ample illustration of the
general use of the IVbles. And, notwithstanding what has .been said in
favour of double altitudes by ^Aeoretteol torifer^, this method of finding the
latitude at sea is evidently far from being one of the most advantageous in
pf actteal navigation : for the operation, besides being rather circuitous,
requires a considerable portfoaof time to go through with it correcdy;
a^d, after all) it fVequentiy happens, that although every seeming precau^
tion has been taken, the mariner's hopes are disappointed in the result.
We will now proceed to a more direct and universal method of finding
the latitude^ either at sea or on shore.
PaoBLBM VIII.
GwenAke Altitudes of two known fixed Stars observed at the same instant^
af otiy Time ^ike Night, to find the Latitude qfthe Place of Obeerv-
tftimi, tfidi}ieitcfett^ qfthe LatUnde by Jccounif the Idmgitside, or the
Af^fOfent Time*
In tlte preceding problems for finding the latitude (the two last excepted),
tlie meridional aHitudes of the celestial objects were the principal elements
under consideration : however, since it fj^equently happens that, in conse-
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348 NAUTICAL AfiTRONOMT,
quence of the interposition of clouds, or other causes, the altitudes of the
heavenly bodies cannot always l)e taken at their respective times of transit,
the-present problem is, therefore, proposed, which possesses the pecuKar
advantage of enabling the mariner to determine the position of his ship,
with respect to latitude, by the altitudes of two known fixed stars, observed
at the same instant and at any hour of the night, either before or after their
passing the meridian, and independent of the latitude by account, the lon-
gitude, or the apparent time of observation. Nor will the mariner^ in this
method, be subjected to the necessity of repeating the operation^ or of puz-
zling himself with a variety of cases and corrections, in finding an approx-
imate latitude.
Rule.
Let the altitudes of two stars be observed, at the same moment, whose
computed spherical distance asunder is given in Table XLIV. ; and let those
observed altitudes be reduced to the true by Problem XVIL^ page 327.
Take the right ascensions and declinations of the two stars^ and also their
computed spherical distance, from Table XLIV., and let these be reduced,
respectively, to the night of observation. Let the star which is adjacent
or nearest to the elevated pole, be distinguished by the letter A, and that
which is remote^ or farthest, by the letter R.— Now,
To the log. sine of the tabular distance between the two stars, add the
log* secant of the decl^iation of the star A, and the log. half-elapsed time
of the difference of right ascension ; the sum, rejecting 20 frcmi the ind^x,
will be the log. half-elapsed time of arcA thejirst.
From the natural co-versed sine of the altitude of the star A^ subtract the
natural co-versed sine of the sum of the tabular distance between the sUrs
and the altitude of the star R, and find the logarithm of the remainder; to
which add the log. co-secant of the tabular distance, and the log. secant of
the altitude of the star R ; — the sum of these three logarithms, abating 20
in the index, will be the log. rising of arch the second; the difference
between which and arch the first, will be arch the third.
To the log. rising of arch tlie third, add the log. co-sines of the declina-
tion and altitude of the star R, and the sum, abating 20 in the index, will
be the logarithm of a natural number ; which, being added to the natural
versed sine of the difference between the altitude and declination of the star
R, when its polar distance is less than 90?, or to that of their sum \riien
it is more than 90% the sum will be the natural co-versed sine of the
latitude.
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OP PIKDIN6 THE LATITUDB BY THB ALTITUDES OF TWO 8TAB8. 349
Example L
January Ist, 1825^ in north latitude^ the true altitude of the star
Alphard was 16?0H2T^ and, at the same instant, that of Regulus waa
27?14^8T J required the latitude of the place of observation ?
A,orRegulu8'red.RA=9^59r S!«and reduceddec.= 12?49' lOr « N.
R, or Alphard's ditto = 9. 18. 59 and reduced dec. =: 7. 54. 13 S.
Tabular distance between the two stars = 22^59 '22r*
DiflF. of right asc, = 0*40? 4! Log. half-elapsed time =: 0.759620
Di8t.beUhe two 8tars=22?59'22r Log. sine = .... 9.591690
Dec. of star A = . 12. 49. 10 Log. secant = .... 10. 010962
Arch the fir$t::z . It42r57' Log. half-elapsed time = 0.362272
Di8t.beUhetwo8tars=22?59^22r Log. co-secant = . . 10.408310
Altitude ofthe star R = 16. 0.12 Log. secant == ... 10.017165
Sum= .... 38?59C34rNat.co-V.S.=370778
Altitude ofthe star A= 27. 14. 8 Nat.co-V.S.=54235l
Diff. = 171573 Log.=5. 234449
Arch the second =: 3MSr27! Log. rising = .... 5.659924
Arch the first =: • 1.42.57
Arch the third = . 2* 5r30r Log. rising = . . . 5.165010
Dec. of the star R =: 7^54^ 13rS. Log. co-sine = . . . 9. 995855
Altitude of ditto = 16. 0. 12 Log. co-sine =: . . . 9. 982835
Sum = 23?54'. 25rNat. vers. S.= 085795
Natural number = .... 139220 Log.=:5. 143700
True latitude = • 50948! 13rN.Nat.co-V.S.=225015
Example 2.
January 1st 1825^ in north latitude, the true altitude of a Arietis was
27?12'9T, and,, at the same instant, that of Aldcbaran was 51M5!28r 5
required the latitude ofthe place of observation ?
* The method of re duciDip the right sseensioni^ dedioationsy and computed spherical dis-
ofthe itan, to a giTea period, U shown in the explanation to Table XLIV., pagell4.
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350 NAUTICAL ASTRONOMT.
True spherical distance between the two stars, reduced to night of
observation = 35?32:7r»
A,or«Arietis'red.R-A.= l»57ri9!* reduced dec,=:22?37'50^N.*
R, orAldebaran's ditto=4 . 25 • 53 reduced deer: 1 6. 8. 57 N.
Diff. of right asc. =: 2^28^34! Log. half-elapsed timers 0.219110
Dist.bet.thetwo8tar8=35?32'. 7". Log. sine = .... 9.764329
Dec. of star A = .. 22.37.50 Log. secant =: • . . 10,034796
Arch the first = . 4^S4T 3! Log. half-elapsed time =: 0.018235
Di8t.bet.thetwostars=35?32^ 7? Log. co-secant = • . 10.235671
AltitudeofthestarR;=51.45.28 Log. secant ^ . » . 10.208S18
Sum=: • . . . 87?17'.35^Nat.co.V.S.=001116
AltitudeofthestarA=27. 12, 9 Natco.V.S.=:542863
Difference = 541747 Log.=5. 733796
Arch the second = 8* 1T33! Log. rising = • . . . 6.177785
Arch the first = . 4. 54. 3
Arch the third = . 3 1 7"30! Log. rising = . . . . 5.500250
Dec. of the star R= 16? 8^57^ Log. co-sine = . . . 9.982516
Altitude of ditto = 5 1 . 45 . 28 Log. co-sine = . . , 9. 79 1 682
Difference = • . 35 ?36 ^ 3 1 TNat. vers. S. = 1 86987
Natural number = 188126 Log.=:5. 274448
True latitude = 88?40C26^N.Nat.co.-V.S.=375113
Example 3,
March Ist, 1825, in north latitude, the true altitude of Rigel
27?9'7':', and, at the same instant, that of Sirius 28?55^391^ j required the
latitude of the place of observation ?
* See Note, pa^ 349.
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OF FINDING THE IJITITUDB BY THE ALTITODES OF TWO STARS, 851
True spherical distance between the two given stars, reduced to night
of obser>'ation = 23?40M3^ *
A,orRiger8red.R.A.=:5* 6T 8! reduced dec.= 8?24^S5rS.
R,orSiriu8'red.R,A. = 6. 37. 26 reduced dcc.= 16. 28. 58
Diflf. of right asc. = l*3iri8! Log. half-elapsed time = 0.411262
Dist-bet.thetwostar8=23.40.40 Log. sine = . • • . 9.603786
Dec. of the star A = 8?24:35r Log. secant = .... 10. 004695
Arch the first = . 4*5 1 r25 ! Log. half-elapsed time s; 0. 019743
Dist.bet.thctwo8tars=23?40U3r Log. co-secant = . . . 10.396200
Alt. of the star R=: 28.55.39 Log. secant =. . . ; 10.057877
Sum = . . . 52?36:22rNatfco-V.S.=205520
Alt. of the star A = 27. 9. 7 Nat.co-V.S.=543648
Difference = 3^8128 Log. = 5. 529081
Arch the second = 5t51'ri7' Log. riising = .... 5.988158
Arch the first = . 4. 5 1 . 25
Arch the third = . 0*59^52! Log. rising = .... 4.530500
Dec. of the star R = 16928^58^ Log. co-sine = . . . . 9. 981775
Altitude of ditto = 28. 55. 39 Log. co-sine = . . . . 9. 942123
Sum= .... 45?24^37^Nat.ver8.8inc=:297975
Natural number ^ 28471 Log.= 4. 454398
True latitude = 42?20^3KN. Nat.co.V.S.326446
* The distance between Rigel and Sirius, as given for the year 1822, at the end of the
Namical Almanac for 1825, Table II., is 23''40'35'% aod Uie change in 10 years + (K5".
This is, evidently, a mistake ; for the distance between those two stars, at the beginning of
1832, was 23«40'42'': and, since th« annual variation of distance is — 0'^56, the change,
therefore, in 10 yean, is — 0' 5'^6 ; being mbtr active instead of addUwe, A similar remark
is applicable to the stars Fonlalhaut and Achemar ; for, by the above-mentioned Table, it
appears that the distance between those stars, at the beginniDg of 1822, was 39^20", and
the change in 10 years — 0^ 1" : whereas the true distance, at that period, was 39^7'13" ; and,
since the annual variation of distance is ^ 0^^ 17, the change, therefore^ in 10 years, is
— tf Y\ 7, being very nearly two seconds of a degree. The distances aod annual variations
of the remaining stars in the said Table wiU be found equally incorrect, as may be seen
by comparing them willi those contained in this work. Tabic XUV.
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352 NAUTICAL ASTRONOMY.
Example 4.
September let, 1825, in south latitude, the true altitude of Fomalhaut
was 63?6^ 18r, and, at the same instant, that of Achernar 37?441 ISr ;
required the latitude of the place of observation ?
True spherical distance between the two given stars, reduced to the night
of observation, = 89?7'13r
A,orAchernar'sred.RA.= l*3iri3! reduced dec.==58e 7^27^8.
R,or Fomalhaut's ditto=22. 48. 0 reduced dec.=:30. 32. 38
Diff. of right ascensions 2U3T13! Log. half-elapsed time = 0.184770
Dist..bet.thetwostars=39°10'.37r Log.sine= . . . . 9.800523
Dec. of the star A = 58. 7.27 Log. secant = . • . 10.277300
Arch the first =x . . 2*12r27! Log. half-elapsed time = 0.262593
Dist bet. the two stars=39? 7 ' 13^ Log. co-secant = . . 10. 200004
Alt. of the star R = 63. 6.18 Log. secant = . . . 10.344518
Sum= 102?13^3KNat.co.V.S.=022677
Alt. of the star A = 37.44. 18 Nat.co.V.S.=387944
Difference = 365267 Log.=5. 562610
Arch the second = . 7^ 4r59! Log. rising = . . . 6.107132
Arch the first = . . 2. 12. 27
Arch the third . . . 4*52r32! Log. rising = . . . 5.851160
Dec. of the star R = 30?32'38r Log. co-sine = . . . 9.935124
Altitude of ditto s . 63. 6.18 Log. co-sine = . . . 9.655481
Difference =s . . . 32?33U0'/Natv.sine= 157182
Natural number s 276544 Log.s 5. 441765
True latitude = 34^29^27^ S. Nat.co-V. S.=433726
Note. — ^The principles from which the above method is deduced, will be
found in ^^ The Young Navigator's Guide to the Sidereal and Planetary
Parts of Nautical Astronomy,'' between pages 136 land 144.
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OF FINDING THE LATITUBE BY THE ALTITUDES OF TWO STARS. 353
Thus^ then, is the mariner proyided with a direct and most accurate
m^^Aod of finding the latitude at sea; and^ since it prevents the uncertainty
and confusion arising from an error in the assumed latitude, or that by
account, and, besides, being free from all ambiguity, restriction, and
variety of cases whatever, — it may, therefore, be employed with a certainty
of success, at any hour of the night, whenever two known fixed stars are
visible. Indeed, if the altitudes of the objects be determined with but
common attention, the latitude resulting therefrom will be always true to
the nearest second of a degree, without the necessity of rqfeating the
operation, or of applying any correction whatever to the result.
Remarks. — ^Although it is at all times advisable for two observers to
take the altitudes of the stars at the same moment of time, yet, should one
person be desirous of going through the whole operation himself, he is to
proceed as follows ; viz., — Let the altitude of one star be taken, and the
time of observation noted by a watch that shows seconds ; then let the
altitude of the other star be observed, and the time noted also ; and let
the altitude of the first observed star be again taken, and the time of
observation noted.
Now, find the difference between the first and last times of observation,
and the altitudes of the first observed star ; and find, also, the difference
between the first time of observation of the first star, and the time of ob-
serving the second star. Then say, as the interval or difference of time
between the two observations of the first star, is to the difference of altitude
in that interval ; so is the interval, or difference of time between the observ-
ations of the first and second star, to a correction ; which, being applied by
addition or subtraction, to the first observed altitude of the first star,
according as it may be increasing or decreasing, the sum or difference will
be the altitude of that star reduced to the time that the altitude of the
second star was taken. This part of the operation may be readily per-
formed by proportional logarithms ; — see example, page 75. The interval
between the observations ought, however, to be as much contracted as
possible, on account of guarding against any irregularities in the change of
altitude.
Caution. — In order to guard against falling into any error, by working in
an imposrible triangle, it will be advisable to make choice of two stars
whose computed spherical distance, in Table XLIV., is not less than 20?,
and difference of right ascension not less than a quarter of an hour ; and,
since the Table contains an extensive variety of distances and differences
of right ascension greater than those values, the mariner can never be
at a loss in finding out two eligible stars for observation. The distances
in that Table are all computed to the greatest degree of accuracy 3 and,
2 A
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854 NAUTICAL ASTRONOMY,
notwithstanding that some of those whieh are but of small meoiuse
ought not to be employed in the determination of the latitude by the
above method^ yet they will be found extremely useful on many oceasions ;
particularly in assisting to distinguish the stars to which they are annexed^
when the latitude is to be inferred from their meridional ^dtitudes^ agree*
aUy to Problem IV., page 335.
Problbm IX.
Owen the Latitude by Accmni, the Jltitude of the Sun*8 loiper or upper
Limb observed near the Meridian, the apparent Time of Observation,
,^nd the Longitiide ; to find the true Latitude.
Since it frequently happens at sea, particularly during the winter months
of the year, that the sun's meridional altitude cannot be taken, in conse-
quence of the interposition of clouds, fogs, rains, or other causes ; and since
the true determination of the latitude becomes an object of the greatest
importance to the mariner when his ship is suling in any narrow sea
trending in an easterly or a westerly direction, such as the British Channel;
the present problem is, therefore, given, by means of which the latitude
may be very readily and correctly inferred from the sun's altitude taken at
a given interval from noon^ within the following limits i viz., — The num"
ier of minutes and parts of a minute^ contained in the interval between
the time of observation and noon, must not exceed the fmmber of degrees
and parts of a degree contained in the object's meridional zenith distance
at the place of observation. And^ since the meridional zenith distance of
a celestial object is expressed by the difiference between its declination and
the latitude of the place of observation, when they are of the same name,
or by their sum, when of contrary names, the extent of the interval from
noon, within which the altitude should be observed^ may, therefore, be
readily ascertdned, by means of the difiference between the latitude and
the declination, when they are both north or both south, or by their sum
when one is north and the other south : thus, if the latitude be 60 degrees,
and the declination 23 degrees, both of the same name, the interval
between the time of observation and noon ought not to exceed 37 minutes;
but if one be north and the other south, the interval may be extended, if
necessary, to 83 minutes before or^ after noon. The altitude^ however^
may be taken as near to noon as the mariner may think proper ; the <mly
restriction being, that the observation must not be made without the above*
mentioned limits.
The interval between the ^parent time of observation and noon must be
^curately determined : this may be always done^ by means of a chrouch-
meter or any well-regulated watch showing seconds; proper allowanm
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LATITUDB BY AN ilLTITTTDE TAUN MBAR THB MBRIDIAN. 955
being made for the difference of time answering to the ehange of longi-
tude^ if any^ since the last observation for determining the error of such
watch or chronometer.
Now, if the sun's altitude be observed at any time imihin the abcve-men"
Honed limits^ the latitude of the place of observation may then be deter-
mined, to every degree of accuracy desirable in nautical operations, by the
following rule ; which, being performed by proportional logarithms, ren-
ders the operation nearly as simple as that of finding the latitude by &i€
meridional altitude of a celestial object.
See cflcplanation to Tables LI. and LIL, between pages 188 and 149.
RULB.
Reduce the son's declination to the time and place of observation^ by
Problem V., pag6 298 ; and let the observed altitude of the sun's lower or
upper limb be reduced to the true central altitude, by Problem XIV., page
320. Then, with the sun's reduced declination, and the latitude by
aecount, enter Table LI. or LIL, (according as the latitude and the decUh-
ation are of the same or of a contrary denomination,) and take out the
corresponding correction in seconds and thirds, which are to be esteemed
as minutes and seconds ^ agreeably to the rule in page 139. Now,
To the proportional logarithm' of this correction, add tvsice the propor-
tional logarithm of the interval between the time of observation and noon^
and the constant logarithm 7 • 2730; the sum of these three logarithms,
abating 10 in the index, will be the proportional logarithm of a correctioni
which being added to the true altitude of the sun's centre, the sum will be
the meridional altitude .of that object : hence the sun's meridional zenith
distance will be known ; to which lei its declination be applied by addition
or subtraction, according as it is of the same or of a contrary name, and
the sum or difference will be the latitude of the place of observation.
Example 1«
At sea, January 1st, 1825, at 22U5r24? apparent time, in latitude
51?36f N., by account. Mid longitude 10?45^30^ W., the observed alti-
tude of the sun's lower limb was I8?33fd4f *, and the height of the eye
above the level of the horizon 25 feet; required the latitude of the place of
observation?
Apparent'time of observation = . • • 22?45T24t
Longitude 10^45^30^ in time = . . + 0. 43. 2
Greenwich time = ....... 23*28?26f
• This is the mean of seversl aMtudes df the sun's lower limb.
2a2
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556* NAUTICAL ASTRONOMY.
Sun's decliQation at noon, January 1st ss 23? 0'59^ S.
Correction of ditto for 28*28r2e! = . - 5. 6
Sun's reduced declination = * , . . 22? 55^ 53^ S.
The observed altitude of the sun's lower limb, reduced to the true central
altitude, is 13?41'24rS.
Cor. in Table LIL, answering to lat. 50?N., and dec. 22?S. s KIS*'. 8
Difference to 2? of lat.=r -3-^. 9; now, 3-^. 9 x 96'. h- 120C = -3.1
Diff. to 1? of dec.= -(r. 9; now, O-'. 9 x 55'53^ ^ 60C = -0.8
Cor. to lat. 51936C N. and declination 22?53:53r S. = . . K 9^ 9
Computed corrections 1?9'?'.9, Proportional log.ss . . • 2.1889
Time of obs. from noon 1 M 4 T36 ! , twice the prop. log. s • . 0. 7650
Constant log. = 7.2780
Correction of the sun's altitude =
True altitude of the sun's centre =
Sun's meridional altitude = . .
Sun's meridional zenith distance =
Sun's corrected declination =
l?46M5r Prop.log,=0.2269
13. 41. 24 S.
15?28' 9rS.
74?3K5irN.
22.55:53 S.
Latitude of the place of observation == 5 1 ?35 '. 58TN. ; which differs but
2". from the truth.
Example 2. -
At sea, March 21st, 1825, at 0^50^25! apparent time, in latitude 51 ?5C
N., by account, and longitude 35?45' W., the observed altitude of the sun's
lower limb was 37?55^27^*, and the height of the eye above the level of
the sea 21 feet; required the true latitude of jthe place of observation ?
Apparent time of observation = . » . 0^50^25!
Longitude 35?45^ W., in time :s • . . 2. 23. 0
Greenwich time = 3?13'25r
Sun's declination at noon, March 21st = 0? 14(30^ N.
Correction of ditto for 3 ? 13T25 * = . . +3.11
Sun's reduced declination = . . . • 0? 1 7 ' 4 1 r N.
* See Note, pa^e 355^
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J^TITUDB fiY AN ALTITUDE TAKBN KBAR THE MERIDIAN. 357
The observed altitude of the sun's lower limb, reduced to the true central
altitude, is 38?6'2rS.
Cor. in Tabh LI., answering to lat. 50"? N. and declin. 0? s . 1 rSS"^* 8
Difference to 2? of lat; = - 6^8; now, 6*^.8x65^-4-120'=: -3,6
Diff. to l«. of declin, = + ^.5; now I'^.Sx 17MKh.60; = + 0 ,4
Correction to lat. 5 1 ?5 ^ N. and declination 0° 1 7 ' 4 1 r N. = . 1 r35-^. 6
Computed correction =7 . 1^35*'. 6, Prop. log. = • . • 2.0530
Time of observ. from noon = 0*50? 25!, twice the prop. log. = 1. 1054
Constant log. =3 , , 7*2730
Correction of the sun's altitude » . . 1? 6 '40? Prop. log.sO, 4314
True altitude of the sun's centre =s « « 38. 6. 2 S«
Sun's meridional altitude = . . . .39?12U2rS.
Sun's meridional zenith distance = • . 50?47' 18?N.
Sun's reduced declination =s • • • . 0. 17*41 N.
Latitude of the place of observation =.51? 4'59?N. ; which differs but
1 ? from the truth.
Hence it is evident, that the latitude may be determined by this method
to all the accuracy desirable in nautical purposes. It possesses a decided
advantage over that by double aUitudes ; and, since the operation is so
extremely simple, the mariner will do well to avail himself thereof on
every occasion ; because the latitude, thus deduced, will be equally as cor-
rect as that resulting from the observed meridional altitude, provided the
observation be made within the prescribed limits. When, however, the
latitude and the declination are of different names, it will not produce any
sensible error in the result, if the altitude be observed a few seconds without
those limits, as may be seen in Example 1, above.
But it is to be remembered, that the apparent time of observation must
be well determined.
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358 VAUTICAl* ASTRONOlfV.
Problem X.
Bwm the La^iude by Account ^ the AUiUide of the Mom's lower or upper
Limb observed near the MeridiOMy the apparent Time of Ob$ervatkm,
and the Longitude ; to find the true Latitude*
To the apparent time of observation apply the loiigitudei in time, by
addition or subtraction^ according as it is west or east, and the corre-
sponding time at Greenwich will be obtained ; to which let the sun's right
ascension be reduced, by Problem V., page 298 ; and let the moon's right
ascension, declipation, semi-diameter, and horizontal parallax^ be also
reduced to that tim^, by Problem Vf., page 302. Let the observed altitude
of the moon's limb be reduced to the true central altitude, by Problem XV.,
page 323.
To the apparent time of observation add the sun's reduced right ascen-
sion, and the sum (abating 24 hours, if necessary,) will be the right ascen-
sion of the meridian ; the difference between which and the moon's reduced
right ascension, converted into time, will be the moon*s distance from the
meridian at the time' of observation. Now, with the mooji's reduced
declination, and the latitude by account,, enter Table LI. or LIf., according
9s they are of the same or of a contrary denomination, and taHe out the
corresponding correction, agreeably to the rule in page 139 ; with which,
and the moon's distance from the meridian, compute the correction of
altitude ; and, hen^e, the latitude of the p(ace of observation, by Problem
IX., page 354.
Note. — ^The limits within which the altitude of the moon should be
observed, are to be determined in the same manner, precisely, as if it were
the sun that was under consideration ; observing, however, to estimate the
interval from the .time of transit over the meridian of the place of obsenra*
tion, instead of from noon.
See the explanation to Tables LI. and LII., between pages 138 ancl 143.
Example 1.
January 23d, 1825, at 3*55ri7? apparent time, in latitude 51?l5i N.,
by account, and longitude 45? W., the observed altitude of the moon's
lower limb was 39? 27 "30^*, and the height of the eye above the level of
the horizon 24 feet; required the true latitude of the place of observation ?
* This is the mean of several altitudes.
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LATITUDE BV AN ALTtTtoB TAXKN NBAR THB MERIDIAN. 859
Apparent time of obserration =: . .... 3*55T17!
Longitude45? W., intima =: 3. 0. 0
Greenwich time = 6*55T17!
Smi's right ascension at noon, January 23d, =: 20! 22*23!
Correction of ditto for 6!55ri7! =: ... +1.12
Sun's reduced right ascension = . . . , 20! 23735!
Apparent time of observation = .... 3.55.17
Right ascension of the meridian = , , . 0!18T52!
Moon's R. A. at noon, January 23d,=:349?47^55r
Correction of ditto for 6!55ri7! = 4- 3. 6. 56
Moon's reduced right ascension =: 352?54!51!'r: 23*Slr39f
Right ascension of the meridian = 0. 18. 52
Moon's distance from the meridian = . . . . 0!47ri3!
Moon's declination at noon, January 23d, =: 1?10'39!^N.
Correction of ditto for 6!55r 17! = . . + 1 . 22. 36
Moon's reduced declinatioii =••««» 2?d3 • 15f N.
Observed altitude of moon's lower limb = . 89? 27 '30' S.
Scmi-diaoieter 14^52^ — dip4C42f =: . • + 10.10
Apparent altitude of the moon's centre = . 39?37:40f S.
Cor.^ Table XVIIL, ans. to hot. parallax 54 ' 4f r= 40. 36
True altitude of the moon's centre = • . . 40? 18! 16T S.
Cor. in Table LI., answerhie to lat 50?N. and declin. 2?N. s 1^41'^. 8
Diff. to 2? of lat. =s -7*.25 now,7*'.2x75:-i.l20: = . - 4 is
Diff. to l?of decUn.r: + 1^6; now,!''. 6x33! 15^-^60! sr + 0 .9
Cor. answering to lat. 5 1 ? 15 ! N. and declin. 2?33 n5rN. =: . 1 r38^ 2
Computed correction = 1!'38*'.2, Prop. log. = , . . . 2.0413
Moon's mer.distance =: 0M7*13!, twice the prop. log. = , 1. 1624
Constant log. =: ••-.*••••••-•,,. 7.2730
Carrectkm of the moon's altitude =: 1? 0 3r Prop»log, := 0.4767
True altitude of the moon's centre = 40. 18. 16 S.
Moon's meridional altitude ss , . 4i;i8!19!rS.
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360 NAUTICAL ASTRONOMT.
Moon's meridional altitude x= • • 41?18M9?S.
Moon's meridional zenith distance = 48?4 1 ' 4 1 '/N.
Moon's reduced declination = • 2. 33. 15 N.
Latitude of the place of observations 5 1 ? 14 ' SGI'N. ; which. differs but 4f
from the truth.
Example 2*
January 30th, 11325, at 9M5rl2! apparent time, in latitude 57"? 40^ S.,
by account, and longitude 60? east, the observed altitude of the moon's
lower limb was 5?37' 12^*, and the height of the eye above the level of
the horizon 26 feet] required the true latitude of the place of observation ?
Apparent time of observation = • • • • 9t45T12!
Longitude 60? east, in time s= . . . 4 . 4. 0* 0
Greenwich time = . ..•..-.. 5M5ri2!
Sun's right ascension at noon, January 30th, = 20*5I?25!
Correctioaofdittofor 5M5ri2! = . . . . + 0.59
Sun's reduced right ascension = . . • . 20?52?24!
Apparent time of observation = • • . « 9.45.12
Right ascension of the meridian s • . . 6t37*36!
Moon's R. A. at noon, January 30tli, = 76?2H55r
Correction of ditto for 5 M5r 12: = + 2.22.52
Moon's reduced right ascension = . 78?44'.47T = 5n4?59!
Right ascension of the meridian s ^ 6. 37. 36
Moon's distance from the meridian =.•••• 1?22T37!
Moon's declination at noon, January SOth, a= . . 23?57M6fN,
Correctionof dittofor5M5?12! = .... — 4. 3
Moon's reduced declination ss ..•,•, 23?53^4S?N.
« See Note, page 358.
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LATITUDE BT AN JkLTTTVVM TAKBN NEAR THB MERIDIAN. 361
Observed altitude of the moon's lower limb = • . 5?37' 12^N.
Semi-diameter 15'.52r- dip 4^52'/ = . . . . +11.0
Apparent altitude of the moon's centre = • • . 5?48' 12rN.
Correc.,TableXVIII.,au8.tohor.parallax,58^5r=s + 49. 6
True altitude ofthe moon's centre =5 • * . . 6?37'18C'N.
Cor. in Tab. LIL, answering to lat 56? S. and declin. 23? N. =: 1 r l*'. 8
Difference to2?oflat. = - 3^6; now, 3*\ 6 x 100^-4-120^ = - 3 .0
Diff. to 1? of declin. ss - 0*'.7j now,0^.7x53C43r-«-60^= -0.6
Cor. to lat. 57?40'. S. and declination 23?53'.43r N. * . 0^58^2
Computed correction = 0^58*^. 2 ftop. log. = 2. 2685
Moon's merid. distance = 1 *22r37 ' Twice the prop. log. = . . 0.6764
Constant log. = ...•.?! 2730
Correction of the moon's altitude = . l?48'59f Prop. log.sO. 2179
True altitude of the moon's centre = . 6. 37. 18 N.
Moon's meridional altitude = ... 8?26^ 17^N.
Moon's meridional zenith distance = • 8 1 ?33 ^ 43 C" S.
Moon's reduced declination = . • • 23.53. 43 N.
Latitude of the place of observation = . 57?40' 01 S.*; which is exactly
right.
Hence it is evident, that, by this method, the latitude may be inferred
from the true altitude of the moon's centre, to every degree of accuracy
desirable in nautical operations, provided the altitude be observed within
the proper limits ; which, for the sake of assisting the memory, will be here
repeated, — ^viz.. The number of minutes and seconds, in the moon's dis-
tance from the meridian at the time of observation, must not exceed the
number of degrees and minutes contained in the meridional zenith distance
of that object at the place of observation. Thus, in the above example,
where the moon's meridional zenith distance is 81?34^ nearly, the interval
between the time of observation and the time of the moon's transit, or
passage over the meridian of the place of observation, must not exceed'
81T34! ; though the moon's altitude may be taken at any time within that
interval, or as near to the time of transit as the observer may think proper.
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$69 NAtrncAL astronomy.
Problbm XL
Given the Latitude by Account ^ the observed central Altitude qfa Planet
near the Meridian, the apparent Tiine of Observation, and the Longir
tude : to find the true Latitude.
RUIJB.
To the apparent time of observation apply the lonptttde, in tamei by
addition or Bubtraction^ according aa it is west or east; and th« tiini) or
difference, will be the corresponding time at Greenwich* To this time let
the planet's right ascension and declination be reduced, by Problem VII.,
page 307 ', and let the sun s right ascension at noon of the given day be
also reduced to that time hy. Problem V., page 298.
Let the observed central altitude of the planet be reduced to its true
central altitude, by Problem XVT., page 325. Then, to the apparent time
of observation add the sun*s reduced right ascension, and the sum (abating
24 hours, if necessary*) will be the right ascension of the meridian } the
difference between which and the planet's reduced right ascension, will be
that object's distance from the meridian at the time of observation. Now,
with the latitude by account, and the planet's reduced declination, enter
Table LI. or LII., according as they are of the same or of contrary deno-*
minations, and take out the corresponding correction, agreeably to the mle
in page 139; virith which, and the planet's distance from the meridian^
compute the correction of altitude, and, hence, the latitude of the place
of observation, by Problem IX., page 354.
Note, — The measure of the interval between the time of observation and
the time of transit, — that is, the number of minutes and seconds contained
in the planet's distance from the meridian, must not exceed the number of
degrees and minutes contained in that object's meridian zenith distance at
the place of observation.
See explanation to Tables Lt. and Lit., page I3S, and thence to 143.
Example K
. January 4th, 1825, at 12t3lT30! apparent time, in 65?28CS., by
account, and longitude 60? east, the observed central altitude of the planet
Jupiter was 5? 14 ^35^*, and the height of the eye above the level of the
horizon 25 feet ; required the true latitude of the place of observation ?
* This is the mean of several altitudes.
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LATITUDB B7 AN AI.TITUDB TAXBH NVAR THE MBRIDIAN. 363
Apparent time of observation ^
Longitude 60? £., in time 2= .
Greenwich time
12*31T30!
- 4. 0- 0
8f31?30!
Sun's right ascension at nooti^ Jan. 4tb| = 19^ 0?32!
Correction of ditto for 8?3ir30t =s . » + 1^34r
Sun's reduced right ascension = • . 19? 2? 6t
Apparent time of observation =; , . , 12.31.30
Right ascension of the meridian = . . 7*33?36!
Jupiter's right ascension at noon, Jan. Istss 8t58? Of
Correction of ditto for 3f8*31?30r = . - 1^ 6r
Planet's reduced right ascension ■« • . 8t56?54!
Right ascension of the meridian = . . 7* 33. 36
Planet's distance from the meridian = . 1 ? 23? 1 8 !
Jupiter's declination at noon^ January 1st an 17^56' OTN.
Correction of ditto for 3f8*3ir30! = . . +6.43
Planet's reduced declination ss
18? 2MS?N.
Correction ip T«ble UL, answ. to lat. 64?$. and dec. I8?N. m Or49''. 6
Difference to 2? of lat.=: - 3^8} qqm^ 3'*.8 x 88^ ^ 120? as - 2 .8
Difference to 1? of de«.= - 0^. 4; now, 0^. 4 x 2M3r-4- 60^ = 0.1
Correction answ. to lat. (55?28? S. and dec. 18?2U3r N. = . 0r46*'. 7
Computed correction = . . . 0?46*'. 7 Prop. log. = .2. 3642
Jupiter's dist. fr.mer.attimeofobs.rs 1 *23?18! Tw. the prop. log.=0. 6692
Constant log. =: 7*2730*
Correction of Jupiter's altitude == . . 1 ? 28 ? 54? Prop. log.=;0. 3064
Jupiter's obs« alt^ red. to tnie centr. alt, is = 5 • 0. 1 0
jupiler't meridional altitude 8 .... 6?29? 4?N.
Jupiter's meridional zenith distance =: • 83?30'56? S.
Jupiter's reduced decimation =s '. . . 18. 2.43 N.
Latitude of the place of observation b . 65?28( 13? S.; which differs 13?
from the truth*
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364 NAUrrcAL astronomy.
Example 2.
February 4th> 1825^ at d?36T20' apparent time, the observed central
altitude of the planet Venus was 36?24^25r*, in latitude 52? 12'. N., by
account, and longitude 45? 40^ W., and the height of the eye above the
level of the horizon was 26 feet ; allowing the horizontal parallax of the
planet, at that time, to be 17^9 the true latitude of the place of observation
is required ?
Apparent time of observation == . • . 3^36?20'
Longitude 45?40^ W., in time =: . + 3. 2. 40
Greenwich time =••••••« 6. 39. 0
Sun's right ascension at noon, Feb. 4th = 21M1T45!
Correction of ditto for 6*39r0! = . . + T. 7?
Sun's reduced right ascension = • , • 21M2T52:
Apparent time of observation =: • • • 3^3fr?20!
Sun's reduced right ascension =: • . • 21. 12. 52
Right ascension of the meridian = . . 0M9T12!
Venus' right ascension at noon, Feb. lst,= 23M5T 0'.
Correction 9f ditto for 3f 6 ?39T0! = • + 13. 6
Venus' reduced right ascension = • . 23*58? 6!
Right ascension of the meridian =: . . 0. 49. 12 '
Planet's distance from the meridian =: • 0*5 IT 6!
Venus' declination at noon, Feb. 1st, = 2? 7' OTS.
Correction of ditto for 3f 6*39?0! = • - 1. 42. 41
Venus' reduced declination = . . . 0?24n9fS.
Observed central altitude of Venus = 36?24'.25r S.
Dip of the horizon for 26 feet =: ^4.52
Apparent central altitude of Venus =: 36?19^33rS,
Refrac. (Tab.Vin.)in7r-Parall.(Tab.VI.)Oa4r= - 1. 3
True central altitude of Venus = 36?18:30rS.
* This is the mean of several altitudes. The altitude of Venus may be taken very cor*
rectly when the son is above the horizon, provided the atmosphere be fine and clear.
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LATITUDE BY AN ALTITUDE TAKKN NEAR THE MERIDIAN^ 365
Cor. in Tab. LII., ans. to lat 52?N. and dec. 0^ = ... 1^32*'. 0
Diff.to2?onat.= -6''.45now,6^4xl2^-Hl20'=: . . . - 0 .6
Diff.toI?ofdec.=:-l-'.2;now,1^2x24n9r^60C= . . . - 0 ,5
Correction answering to lat. 52?12'. N. and dec. 0?24n9rS.= KS(r.9
Computed correction = H30*'.9 Prop. log. = 2.0749
Venus' merid. distance = 0^ 5 1 Te ! Twice the prop. log. r= . 1 . 0938
Constant log. = 7.2730
Correction of Venus' altitude = . . 1 ? 5 ^ 6r Prop. log.= 0. 44 1 7
True central altitude of Venus =: . . 36.18.30 S.
Venus* meridional altitude = . . . 37^23'.36rS.
Venus' meridional zenith distance = . 52?36 '. 24rN.
Venus' reduced declination = • . . 0. 24. 19S.
Latitude of the place of observation = 52? 12' 5rN.; which differs but
5T from the truth.
Problem XII.
Cfioen tlie Latitude by Account, the Altitude of a fixed Star observed near
the Meridtauy the apparent Time of Observation; and the Longitude, to
find the true Latitude.
Rule.
Turn the longitude into time, and apply it to the apparent time of
observation, by addition or subtraction, according as it is west or east ;
and the sum, or difference^ will be the corresponding time at Greenwich.
To this time let the sun's right ascension at n6on of the given day be
reduced by Problem V., page 298.
Let the star's right ascension and declination (Table XLIV.) be reduced
to the night of observation, by the method shown in page 115; and let the
star's observed altitude be reduced to the true altitude, by Problem XVIL^
page 327.
To the apparent time of observation add the sun's reduced right ascen-*
aion, and the sum (abating 24 hours, if necessary,) will be the right ascen-
sion of the meridian ; the difference between which and the star's reduced
right ascension will be that object's distance from the meridian at the time
of observation.
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366 NAUTICAL ASTRONOMY.
Now^ with the latitude by account, and the star's reduced deeUoation,
enter Table LI. or LII., according as they are of the same or of a contrary
denomination ; and take out the corresponding correction, agreeably to the
rule in page 139; with which, and the star's distance from the meridian,
compute the correction of altitude ) and, hence, the latitude^ by Problem
IX., page 354.
Note. — ^The interval between the time of observation and the time of
transit must not exceed the limits pointed out in the three preceding
Problems ; viz., the number of minutes and parts of a minute contmed in
the star's distance from the meridian, is not to exceed the number of
degrees and parts of a degree contained in that object's meridional zenith
distance at the place of observation.
See explanation to Tables LI. and Lll., from page 138 to 143.
Example 1.
January Ist, 1825, at S?52T17! apparent time, in latitude 52?46' N.,
by account, and longitude 56? 15'. W., the observed altitude of the star
Menkar was 39?42'.40'f, and the height of the eye above the level of the
sea 26 feet ; required the true latitude of the place of observation ?
Apparent time of observation = ....•«... 8^52"17'
Longitude56?15' W., intime = ........ + 3.45. 0
Greenwich time = 12*37*17!
Sun's right ascension at noon, January 1st = 18*47*19'
Correction of ditto for 12*37"17' = ... +2.19
Sun's reduced right ascension = .... 18*49?38!
Apparent time of observation = . . . . 8. 52. 17
Right ascension of the meridian = . • . • 3*41T55'
Menkar's right ascension, January 1st, 1824 =: 2*53T 5'
Correction of ditto for 1 year =: . . . . + 0' 3^
Menkar's reduced right ascension =: • . . 2*53? 8!
Right aseensioii of the meridian = . . • 3. 41. 55
Star's distance from the meridian = « • « 0*48?47 •
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JLATITUDE BY AN ALTITUDH TAKfiN N^^R THB MERIDIAN. 8W
Menkar'a declination, January l«t, 1824 = . S?23'4lrN.
Correction of ditto for 1 year =..... 0. 15
Menkar*8 reduced declination =: . . . . , 3?23'58rN.
Star's observ. alt., reduced to ite true alt, is = 39?36'39rS.
Correction in Table LI. answering to lat. 52?N. and dec, 3?N. = KSG*'. 0
Difference to 2? of lat.= -7^ 0; now, 7^ 0 x 46^ h- 120^ = -2.7
Differenceto l?of dec.rr +1^2; now,l-'.2x2S^56r-«-e0r= +0.5
Correction to lat, 52*4e^N. and declination 3?23:56rN. =: lr33^8
Computed correction =: K33*'. 8 Prop. log. = . . , . 2.0612
Star's mend, distance = G*48?47* Twice the prop. log. = 1. 1340
Constant log. =: 7.2730
Correction of Menkar's altitude == . 1? IHSf Prop. log. = 0.4682
True altitude of Menkar = ... 39. 36. 39 S.
Menkar*s meridional altitude =: . . 40^37 '.54rS.
Menkar's meridional zenith distances: 49^22^ 6rN.
Menkar's reduced declination =: . . 3. 23. 56 N.
Latitude ofthe place of observation =: 52^46' 2rN.; which Offers but
2f from the truth.
Example 2.
September Ist, 1825, at 13^28^42! apparent time, in latitude 49?30^S.
by account, and kmgitude 22?I0'30'!r E., the observed altitude of the star
fi Pegasi, or Scheat, was 1 1 ?37 ' 59?, and the height of the eye above th«
level of the sea 19 feet ; required the true latitude of the place of observa-
tion?
Apparent time of observation = .... 13'28"42!
Longitude 22? 10^30? E., fai time = . . . - 1 . 28. 42
Greenwich time =....,.... 12* 0? 0!
Sun's right ascension at noon, Sept. 1st, = . 10M1T16!
Correction of ditto for 12* OTO! = . . . . + l'49r
Son's vedueed right ascension ±a • . • • 10*43? 5!
Apparent time of observation sa . • « . 13. 28. 42
Right ascension of the meridian = • • . 0MIT47'
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368 NAUTICAL ASTRONOMY.
Scheat's right ascension, January Ist, 1824, s= 22^55? 5 '.
Correction of ditto for 1 year and 8 months &= +0^5
' i:'^
Scheat's reduced right ascension = • • . 22^55? 10!
Right ascension of the meridian = • • . 0, 1 1. 47
Star's distance from the meridian = . • • 1M6?37 •
Scheat's declination, January 1st, 1 824, s . 27? 7 -SS^N,
Correction of ditto for 1 year and 8 months =: + 0.32
Scheat's reduced declination ss . • • • 27- 8C 7^N.
Scheat's observed altitude = ll?37'59rN,
Dip of the horizon for 1 9 feet s • • • . ^4,11
Scheat's apparent altitude = . . . . . 1 1 ?33 U8rN.
Refraction = • • • • • *- 4. 33
Scheat's true altitude = • . . , • . . ll?29'15rN.
Correction in Table LII., answ. to lat. 49?S. and dec. 27?N. = K 1 1-». 0
Difference to 1 9 of latitude = - 1*'. 8 j now, ^ . 8 x 30^ h-60^ = -0.9
Difference to 1? of dec.= - 0^. 8 j now, 0^. 8 x 8^7r-*-60' = -0.1
Correction to latitude 49?30CS. and declination 27?8!7^N. = "KIO^'.O
Computed correction = KIO'^.O Prop. log. = .... 2.1883
Star's mend, distance = 1 * 16r37 ' Twice the prop. log. = . 0. 74 1 8
Constant log. = 7. 2730
Correction of the star's altitude =s .* 1?52U(!? Prop. log. =£ 0.2031
Scheat's true altitude =: . . . . 1 1 . 29. 15 N.
Scheat's meridional altitude =: . . 1 3 ? 22 ^ KN.
Scheat's meridional zenith distance = .76?37'59rS.
Scheat's reduced declination = . . 27. 8. 7 N.
LAtitudeof the place of observation s 49?29^52rS. ; which differs ST
from the truth.
Remark. — ^The latitude may be also very correctly inferred from the
altitude of a celestial object observed near the meridian below the pole.
In this case, the meridian distance of the object is to be reckoned from the
apparent time of its transit below the pole ; the correction answering to the
latitude and the declination is always to be taken out of TcJ>le LILj in the
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LATITUDE BT AN ALTITUBB TAKEN NEAR THE MERIDIAN* 369
same manner as if those elements were of different denominations ; and the
correction of altitude is to be applied by subtraction to the true altitude of
the object, deduced from observation, in order to find its meridional altitude
below the pole. Then, with the meridional altitude below the pole, thus
found, and the declination, the latitude is to be determined, by Problem V.^
page 336.
The interval, or limits within which the altitude should be observed, is
to be determined in the same manner as if the celestial object were near the
meridian above the pole.
Example 1,
June 20th, 1825, at 11M8?30! apparent time, in latitude 71?50'N.,
by Account, and longitude 65? W., the observed altitude of the sun's lower
limb was 5?30^50^, and the height of the eye above the level of the sea
20 feet 3 required the true latitude of the place of observati^?
Interval between the time of observation and midnight =s • OMlTSO!
Sun's observed reduced to its true central altitude = 5?S3C371'
Sun's corrected declination S3 23?27'36?N.
Sun's north polar distance s 66?32'24r
Cor. in Table LII., answering to lat- 70? and declin. 23? = . 0^37*^. 1
Diff. to2? oflat.= -3^5; now,3'".5xllO^H-120C = . - 3 .2
Diff. to 1? of dec.= - O*'. 2; now, O*', 2 x 27'.36r-H60 = . -0.1
Ck)rrectiontolat.71?50' anddec. 23?27'36? = .... 0r33''.8
Computed correction =5 0^33''. 8, Proportional log. = . . . 2. 5045
Sun's dist. from midnight = 0*4lr30!, twice the prop. log. = 1. 2744
Constant log. = 7. 2730
Correction of altitude =: . . . . - 0?15:58r Prop.log,= 1.0519
True central altitude of the sun = . 5. 33. 37
Sun's meridian altitude below the pole = 5? 17 ' 391^
Sun's north polar distance = ... 66. 32. 24
Latitude of the place of observation = 71?50' 31 N.j which differs but
3? from the truth.
2b
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370 NAUTICAL AJTROMOMT.
In case of a Fixed Star :—
Find the apparent time of the star's superior transit above the pole, at
the given meridian, by Problem XIL, page 317 > to this time let 12 hours,
diminished by half the variation of the sun's right ascension on the given
day, be added, and the sum will be the apparent time of the star's inferior
transit below the pole. Then, the rest of the operation is to be performed
exactly the same as that for the sun in example I, as above.
Example 2.
January 1st, 182S, at ll^SOrOt apparent time, in latitude 71^90^ N.,
by account, and longitude 84?9'S0^ W., the observed altitude of the star
Albireo was 9?3S!, and the height of the eye above the level of die
horizon 19 feet ; required the true latitude of the place of obaenration ?
Apparent time of star's transit above the pole » • 0tS5732'
To which add 12t - 2rl2! (halfvar.of S.R. A,) = 11.57.48
Apparent time of the star's transit below the pole = 12!33T20'
Apparent time of observation n •••••• 11.50. 0
Star's distance from the meridian a: . • • • ; 0M3T20!
Observed altitude of the star Albireo s • 9?SS{ Of
Dip of the horizon for 19 feet s • • • — 4. 11
Star's apparent altitude = 9?28U9r
Refractions . . « -i- 5,83
Star's true altitude cs 9?23'17r
Cor. in Table LIL, answering to lat. 70? and declin. 27^ = . 0r36*', 2
Diff. to 2° of lat.= - 3*^.4 J now, 3''.4x90^h-120C = .-2.5
Dittto l?ofdec,» --0^.3} now,0^.3x35:54r-^60 a • - 0 •!
Correction to latitude 71?30'. and declin. 27?35C54f ss . , Or33*', 6
Computed correction = . 0^33*. 6, Prop. log. = . . 2. 5071
Star's merid. distance = , 0*43T20 !, twice the prop, log. = 1 . 2870
Constant log. = , 7.2730
Correction of the star's altitude s= . .- On7n8? Brop.k)g.aB L0I71
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LATITUDE BT AN ALTXTUBS TAKSN NEAR THB MBRIBIAN. 871
Correction <^ the atar's altitude a . - 0? 17 U87
True altitude of the star b . , . . . 9. 23. 17
Star's meridiau altitude below the pole = 9? 5 '59"?
Star's north polar distance =s • • . .62.24. 6
Latitude of the place of observation = . 71?30^ 5rN. j which differs but
f f from the truths
^ole.— From the abov^ examples, the method of finding the latitude by
an altitude of the moon, or of a planet, observed near the meridian below
the pole, will appear obvious.
Remark.
The following ingenious problem for determining the latitude^ either at
sea or on shore, has been communicated to the author by that scientific
and enterprising officer. Captain William Fitzwilliam Owen, of His
Majesty's ship Eden, who is so highly renowned for his extensive know«
ledge in every department of science connected with nautical subjects.
Pboblbk.
Given the Latitude by Jccount, the true AUiiude qfthe Sun's Centre, and
the apparent Time; to find the true Latitude qf the Place qf Qi-
eervathn.
RuLb.
Find the mean between the estimated meridian altitude, and th^ altitude
deduced from observation, which call the middle altitude; then,
To the log. rising of the apparent time from noon, add the log. eo-sine
of the latitude, the log. co-sine of the corrected declination, the log. secant
less radius of the middle altitude, and the constant logarithm 7» 536274 j*'
the sum of these five logarithms, abating 30 in the index, will be the loga-
rithm of a natural number, which is to be esteemed as minutes, and which,
being added to the sun's true central altitude, will give his correct meri-
dional altitude ; and, hence, the true latitude of the place of observation ?
. Eceample 1*
December 22d, 1825, in latitude 890' south, by account, at 23*41?15:
q)parent time, the true altitude of the sun's centre was 74? 16' j required
the true latitude ?
* Thif is the log. sscaat of one minute, with a modified index.
2b2
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872 NAUTICAL ASTRONOMY.
Apparent time from noon = • 0M8r45! Log. riring = 3-524365
Latitude by account = ... 8? 0' O^S. Log. co-sinc = 9. 995753
Sun's corrected declination = . 23. 27. 0 S. Log. co-sine = 9. 962562
Estimated meridian altitude = 74?33^ Or Constant log.= 7.536274
True central altitude = . . 74.16.0 74?16: Or
Middle altitude == .... 74^24 '.30r Log, secant == 0.576604
Correction of altitude = +'39^ 0rLog.= 1.589558
Sun's correct meridional altitude s • . • 74?55C Or
Sun's correct declination = 23. 27. 0 south.
True latitude of the place of observations . 8?22^ Orsouth; which ex-
actiy agrees with the result by spherical trigonometry.
Jfote. — ^By this method of computation, an error of one degree in the
latitude by account^ in places within the tropics^ will produce litde or no
effect on the latitude resulting from calculation : thus, if the latitude by
account be assumed at 7?0^, or at 9?0^, the resulting latitude, or that
deduced from computation, will not differ more than one minute from the
truth ; and the same result would be obtuned, if the altitude were observed
at the distance of an hour from noon : provided, always, that the measure
of the interval from noon be very correcdy known.
Esdmpie 2.
December 23d, 1825, in latitude 50?0^ N., by account, at lM4ri5!
apparent time, the altitude of the sun's centre was 13?58' ; required the
true latitude ?
Time from noon = . lM4rl5! Log. rising = . . . 4.716200
Latitude by account =: 50? O^N. Log. co-sine =: . . . 9.808068
Sun's corrected dec. = 23. 27 S. Log. co-sine =: . . . 9. 962562
Estimated merid. alt.= 16?3d( Constant log. =: . • 7.536274
True central altitude = 13.58 . . 13?58!
Middle altitude = . 15?15^30r . . .'. Log. secant =: 0.015586
Correction of altitude = ... + l?49' = 109C=Log.=2. 038690
Sun's correct meridional altitude =: . I5?47^
Sun's correct declination = ... 23. 27 soutii.
Co-latitude of the place of observation=:39? 14 C north ; hence the true lati-
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OF FINDINO TH£ JLATITUDB BT TH£ SUN's DIAMBTBIU 373
tude is 50?46' north, which is 2' less than the result by spherical trigono-
metry : the correct latitude being 50"? 48^ north.
If die latitude by account be assumed at 51^48 ^, the latitude by com-
putation will be 50^50^ ; being, in this instance, only two minutes more
than the truth.
Note.— The above method of finding' the latitude is, as far as I am
aware, perfectly original ; it is exceedingly well arranged, and it affords a
direct and general solution to the problem given, for the same purpose, in
page 354 : the apparent time, or the measure of the interval from noon,
must, however, be very correctly known ; although, in places distant from
the equator, or where the sun does not come very near to the zenith of the
place of observation, an error of a few minutes in the time' will not very
materially affect the latitude : thus, in the last example, an error 6f two
minutes in the interval from noon would only produce an error of six
minutes in the latitude; and in the first example, where the sun passes nearer
to the zenith, it would produce an error of eight minutes in the latitude.
As this method does not labour under any restraint, or since it does not
require that the interval from noon should be governed by the object's
meridional zenith distance, the observation may therefore be taken at any
hour before or after the sun's transit ; and this is a peculiarity that gives
it a most decided advantage over the method contained in the above-
mentioned page.
PftOBLBM XIII.
Given the Lmgiiude of a PUxcCy the Sun*s Declination and Semi-diameter,
and the Intertal of Time between the Instants of hig Limbs-being in th6
Horizons tofnd the Latitude of that Place.
Rule.
Reduce the apparent time, per watch, of the rising or setting of the
sun's centre to the corresponding time at Greenwich, by Problem III.,
page 297 ; to which time let the sun's declination be reduced, by Problem
v., page 298.
To the logarithm of the sun's semi-diameter, reduced to seconds, add
the arithmetical complement of the logarithm of the interval of time, ex-
pressed in seconds, between the instants of the sun's limbs being in the
horizon, and the constant logarithm 9. 124939; the sum of these three
logarithms, rejecting 10 in the index^ will be the logarithmic co-sine of an
arch. Now,
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374 MAtrricAL asthonomt.
To the logarithmic sine of the sum of this arch and the sun'd reduced
declination, add the logarithmic sine of their difference ; half the sum will
he the logarithmic sine of the latitude of the place of observation*
Example 1.
July 13th, 1824^ in north latitude, and longitude 120? west, the mm's
lower limb, at the time of its setting, was observed to touch the horizon at
7t59?581 apparent time, and the upper limb at 8M?4! ; required tha
latitude of the place of observation }
Apparenttimeofsun's setting = 7*59768! + 8MT4: ^2s= 8t 2T i:
longitude 120? west^ in time s 8. 0. 0
Greenwich time of sun's setting = • . . • ^ • • • 16. 2* 1
Sun's declination at noon^ July ISth, 1824, sa 21?49C5ir N.
Correctionof ditto for 16^ 2?i: c . . • . — 6. 0
Sun's reduced declination = 21?43^5irN.
Sun's semi-diameter = 15 U5^. 8, in seconds = 945*, 8 Log.s2. 975799
Interval of time between the setting of the sun's
lower and upper limbs s 4T6?, or 246! Log. ar. comp. =» 7. 609065
Constant log. (the ar. comp. of the prop.
log. of 24 hours esteemed as minutes) =s 9. 124939
Arch= 59? 9^39r Log. co-sine = 9.709803
Sun's reduced declination s 21. 43. 51 N.
Sumn ...... • 80?5S^30r Log. sine »= « , 9.994489
DifferencQ« ,,«., 37.25.48 Lpg«sine::« , « 9.783755
Sums . . . 19.778244
Latitude of the place of obs.s 50? 46 ! 34 TN. Log. sine = . • 9. 889122
Bsmmple 3.
October 1st, 1824, in north latitude, and longitude 105? eastp the sun'a
upper limb, at the time of its rising, was observed to emerge from the
horizon at 6^3T43', and the lower limb at 6^6T32'. ; required the latitude
of the place of observation ?.
Apparent Ume of sun's rising » 6^3r43: + 6^6732? -^^ 2 a 6t 5? 7|!
Longitude 105? east, in time 9 • , 7, 0, 0
Greenwich time pa^ noon, September 30th =; « • • • 11' 57 7|!
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OP FIimiKO THB APPABBMT TIME. 375
Sun's declination at noon, September 30th, 1824, s 2?52U6? 8.
Correction of ditto for 11 *5?7i! =« +10.48
Sun's reduced declination = 3? 3:341^8.
Sun's semi-diameter s 16^ 1'. 2, in seconds = 961"". 2 Ix>g.=2. 982814
Interval of time between the rising of the
sun's upper and lower limbs=:2?49!, or 169! Log. ar. comp.s7. 7721 18
Constant log. A 9.124939
Arch= .••..••. 40^40'54^ Log. co-sine = 9.879866
Sun's reduced declination = • 3. 3.34S.
Sum= . 43?44^28r Log. sine = . 9.839730
Differences 37.37.20 Log. sine = . 9.785652
Sum = 19.625382
Latitude of the place of observation=40?31 ' N. Log. sine = . 9. 8I269I
A^marfc.— -In this method of finding the latitude, it is indispensably
necessary that the interval of time (per Mratch) between the instants of the
sun's lower and upper limbs touching the horizon be determined to the
nearest eecond ; otherwise the latitude resulting therefrom may be subject
to a considerable error, particularly in places where the limbs of that object
rise or set in a vertical position ; which is frequently the case in parts
within the tropics.
SOLUTION OP PROBLEMS RELATIVE TO APPARENT TIME.
Tim^, as inferred directly from observations of the sun, is denominated
either apparefii or mean iolar time. Apparent time is that which is
deduced from altitudes of the sun, moon, stars, or planets. Mean tme
arises from a twpposed ufAform motion of the sun : hence, a mean solar
day is always of the same determinate length ; but the measure of an
apparent day is ever variable,— being longer at one time of the year, and
shorter at another, than a mean day ; the instant of ^parent noon will,
therefore, sometimes precede, and at other times follow, that of mean
noon. The difference of those instants is called the equation of time;
which equation is expressed by the difference between the sun's true right
ascension and his mean longitude, corrected by the equation of the Equi-
noxes in right ascension, and converted into time at the rate of 1 minute
to every 15 minutes of motion, &c. &c. The equation of time is always
equal to the difference between the times shown by an nniform or equable
going clock, and a true sun-dial.
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376 NAUTICAL ASTRONOMY.
The sun's motion in the Ecliptic is constantly varying, and so is his
motion in right ascension ; but since the latter is rendered further unequal,
on account of the obliquity of the Ecliptic to the Equator, it hence follows
that the intervals of the sun's return to the same meridian become unequal,
and that he will gradually come to the meridian of the same place too late,
or too early, every day, for an uniform motion, such as that shown by an
equable going watch or clock.
It is this retardation, or acceleration of the sun's coming to the meridian
of the same place, that is called the equation of time ; which implies a cor*
rection additive to, or subtractive from, the apparent time, in order to
reduce it to equable or mean time.
The equation of time vanishes at four periods in the year,— which hap-
pen, at present, about the 15th of April, the 15th of June, the 31st of
August, and the 24th of December ; because, at these periods, there is no
difference between the sun's true right ascension and his mean longitude :
hence the apparent noon, at those times, is equal to the mean noon. When
the sun's true right ascension differs most from his mean longitude, the
equation of time is greatest: this happens, at present, about the 11th of
February, the 15th of May, the 27th of July, and the 3d of November.
But, since at those times the diurnal motion of the sun in right ascension
is equal to his mean motion in longitude, or 59' 8 T, the length of the appa-
rent day, at these four periods, is, therefore, equal to that of a mean day ;
at all other times of the year, the lengths of the apparent and mean days
differ ; and it is the accumulation of those differences that produces the
absolute equation of time.
The equation of time is additioe from about the 25th of December to
the 15th of April, and, again, from the 16th of June to the 31st of August;
because, during the interval between those periods, the sun comes to the
meridian later than the times indicated by a well-regulated clock : but it
is subtractive from about the 16th of April to the 15th of June, and, again,
from the 1st of September to the 24th of December; because, during the
interval between these periods, the sun comes to the meridian earlier thaa
the times indicated by an equable going clock*
The equation of time is contained in page IL of the month in the
Nautical Almanac ; but, since it is calculated for the meridian of the Royal
Observatory at Greenwich, and for noon, a correction, therefore, becomes
necessary, in order to reduce it to any other meridian, and to any given
time under that meridian. This correction is to be found by Problem V.^
page 298 ; or by means of Table XV.^ as explained in page 25.
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OF nKPIKO TBB AFPAB8KT TIMB. 377
Problem I.
To find the Error of a Watch or Chronometer ^ by equal Altitudes of
the Sun.
Rnus.
In the morning, wh^n the sun is nearly in the prime vertical, or at least
when he is not less than two hours distant from the meridian, let several
altitudes of his upper or lower limb be taken, and the corresponding times
(per watch) increased by 12 hours, noted down in regular succession. In
the afternoon, observe the instants when the same limb of the sun, taken
in the morning, comes to the same altitudes, and write down each, aug-
mented by 24 hours, opposite to its respective altitude. Take the means
of the morning and of the afternoon times of observation ; add them toge-
ther, and half their sum will be the time of noon, per watch, incorrect.
The difference between the means of the morning and afternoon times will
be the interval between the observations : with this interval, and the
latitude, enter Table XIIL, and with the interval and the declination,
corrected for longitude, enter Table XIV.; take out the corresponding
equations, noting whether they be affirmative or negative, agreeably to the
rule in page 23 : then, with the sum or difference of those two equations,
according as they are of the same or of contrary signs, and the variation
of the sun's declination for the given day, compute the equation of equal
altitudes, by th6 said rule in page 23. . Now, to the time of noon, per
watch^ incorrect, apply the equation of equal altitudes, by addition or sub-
traction, according as its sign is affirmative or negative, and the sum or
difference will be the time, per watch, of apparent noon, or the instant
when the sun's centre was on the meridian of the place of observation ;
the difference between which and noon, or 24 hours, will be the error of
the watch for apparent time.
If the watch be regulated to mean solar time^ such as a chronometer, let
the equation of time (as g^ven in the Nautical Almanac, and reduced to
the meridian of the place of observation by Problem V., page 298,) be
applied to noon, or 24 hours, by addition or subtraction, according to its
title, and the mean time of noon will be obtained ; the difference between
which and the time, per watch^ of apparent noon, will be the error of tiie
watch for mean solar time.
Esample 1.
March 1st, 1 825, {civil time) in latitude 50948 ^ N., and longitude SO?W.,
the following equal altitudes of die sun were observed 3 required the error
of the watch?
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378'
ItAVTttAt AmtOMOMT.
n'g Lower Limb.
Forenoon Ttmeti p.
Watel
k°.
11?56^ .
• •
19^59?47J
28* 0?58!
12. 1 .
► •
20.
0.23
»
28. 0.22
12. 6 .
» •
20.
0.59
27.59.46
12.11 .
1 / •
20.
1.35
27.59.10
12.16 .
1 •
•
20.
2.11
•
» •
27.58.34
Mean = . .
20*
0'r59!
Mean ae
27i59T46'.
Afternoon mc
^ans^
•
27.59.46 1
forenoon meam
Sum =
b20. 0.59
Interval = <
7^58r47'.
48i Or45: .
Time of noon, per watch, uncorrected s= • . 24^ 0^22^!
Equation, Table Xlli., ans. to lat.
50?48'. and interval 7'58T47' = -*-16r59r; negative, because the sun is
advancing towards the
Equation, Table XIV., ana. to dec. elevated pole.
7?82'25^S.andint.7*58r47*= — 0.55; negative, because the sun's
' dec. is decreasing.
Sum of the equations s » • • — 17^^54^ Prop. log. a . 1.0024
Variation of sun's declination a 22 '. 46}^ * Prop. log. » . 0. 8979
Equation of equal altitudes ss , -^22r39r Prop. log. » . 0. 9003
Timeof noon, p. watch, uncor.cs24t 07221?
Time, per watch, of app.noon=24t Of 0! • 24* 0? 0!
Apparent noon = ... 24. 0. O+Eq. of time 12738!
ss mean noon s 24. 12. 38
Watch true for apparent time=sO* 07 0* Watch slow for
mean time s I27S8!
Example 2.
August 2d, 1825, {dvU or nmUcal iiine) in latitude 50?48^ N,, and
longitude 30? W., the following equal altitudes of the sun were observed |
required the error of the watch ?
* Since the morning observationft belong, astronomically, to Febmaiy 28th, therefore^
half the Bum of the variation of the sun^t declination^ for the days preceding and following
the given one, is to be taken for the true variation of declination.
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OP FIMDIN6 THU API»ARSMT TIME.
379
Alt of Suu's Lower Limb.
Forenoon Timesy p. Watch.
32?18: .
. . 20* 3T52!
32.23 .
. . 20. 4.25
32.28 .
. . 20. 4.57
32.33 .
. . 20. 5.30
32.38 .
. . 20. 6. 2
Afternoon Times^ p.Watch.
. . 28* 3743!
• . 28. 3.11
. • 28. 2.39
. . 28. 2. 6
. . 28. 1.34
Mean« • • . 20^4757^21 Means 28* 2r38!36.
Afternoon mean s 28. 2.38.36Foren.niean=:20. 4.57.12
Interval =
7i57'r4r.24;
Sum= 48 1 7"35!48f
Time of noon, per watch, uncorrected = • . 24* 3^47 ' 54 f
(Equation, Table XIIL, ans. to lat.
50?48C and int. 7 '57741! 24!= +16'r58r; affirmative, because thetun
is receding from the ele-
Equa.»Tab,XIV.,an8.todec.l7^47 C9'r vated pole.
and interval 7*5774r.24f = — 2. 14 ; negative, because the sun's
dec. is decrea.sing.
Pifference of the equations s=: , + IVAiV Prop. log. s . 1.0870
Variation of sun's declination =: . 15i23i« Prop. log. » » 1.0685
Equation of equal altitudes ss • +12^35^ Prop. log. = • 1.1555
Timeofnoon, p.watch,uncor.ss24 1 3747 1 54 !
Time, p. watch, of app. noon = 24* 4? 0!29! .... 24* 47 0!29!
Apparent noon = ... 24. 0. 0. 0+Eq. of times
5754!=mean
noon = . 24. 5.54. 0
Watch fast for apparent times 47 0\ 29 ! Watch slow for
mean times l753!31f
Now, since the equal altitudes in the two preceding examples have been
observed at the same place, and the times of observation specified by the
same watch, the daily rate of that machine may therefore be readily esta-'
blished, upon the assumption of an uniform motion j as follows, viz.,
March 1st, 1825, watch slow for mean time at noon s 12738!
August 2d, 1825, watch slow for mean time at noon s 1. 53|
Interval s 154 days.
Difference s 10744|!
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380 KAUTICAL ASTRONOMY.
Now, 10r444!, divided by 154 days, gives 4*. 185 ; which, therefore, is
the daily rate gaining.
Remarks.
In finding the rate of a watch or chronometer, if it be too fast at the
time of the first observation, and the error increasing, the machine will
evidently be gaining on mean time; but if decreasing, it will be losing for
mean time. Again, if the watch or chronometer be too slow at the first
observation, and the error increasing, the machine will be losing for mean
time 3 but if decreasing, it will be gaining on mean time, as in the case or
example above.
Since the method of finding the apparent or mean time, by equal alti-
tudes of the sun, does -not indispensably require that the latitude of the
place of observation and the value of th^ sun's declination be strictly deter-
mined, as these elements are only employed in taking out the equations
from Tables XIII. and XIV. ; and since any trifling error therein will not
sensibly affect the resulting equation,-— this method, therefore, is the best
adapted for practice on shore, where the altitudes may be taken with a
sextant, by means of an artificial horizon, and the corresponding times
determined with the greatest exactness. Nor is it absolutely necessary
that the instrument be very rigidly adjusted, provided only, that it shows
the same altitude at botli observations.
In taking equal altitudes, it will be advisable for the observer to fix the
index of his sextant or quadrant to some particular division on the arch,
and then wait till the contact of the images takes place.
Problbm JI.
To find the Error of a Watch or Chronometer ^ by eqml Altitudes of
a fixed Star,
Rule.
Let several altitudes of a known fixed star be observed when in the east-
ern hemisphere, and the corresponding times, per watch, noted down in
regular succession. When the star is in the western horizon, observe the
instants when it comes to each of the former altitudes, and write down
each opposite to its respective altitude. Take the means of the eastern
and of the western times of observation ; add them together, and half their
sum will be the time, per watch, of the star's transit over the meridian of
the place of observation.
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OP FINDTKG THB APPARBMT TIMB.
381
Compute the apparent time of the star's transit over the given meridian,
by Problem XII., page 317 ; the difference between which and the observed
time of transit will be the error of the watch, which will be fast or slow
according as the observed time of transit is greater or less than the com-
puted time of transit.
Example I.
April 24th, 1825, in latitude 50?15^ N., and longitude 60?45^ W., the
foUowing equal altitudes of Arcturus were observed ; required the error of
the watch for apparent time ?
titode* of Aretoni
26? 4'.
26.19
26.34
20.49
27. 4
1*.
Intern Timei, per Watcli.
. 7'- 6r41! . . .
. 7. 8.14 . .
. 7. 9.47 . . .
. 7.11.21 . . .
. 7.12.55 . .
1 1
1 4
Wcftero Timet, perWaU
. . 16M9r49!
. . 16.48.16
, . 16.46.43
. 16.45. 9
, . 16.43.35
Mean = 7* 9r47^36f
Mean of eastern times = . .
Mean
• «
= 16M6r42r24f
7. 9.47.36
23!56r30! Of
Time of star's transit over the given mer., per watch,sIlt58Tl5! Of
Star's R. A., reduced to night of obs. =s 14? 7?4 1' • 3
Sun's R. A. at noon of the given day = 2. 6. 55 • 3
Approximate time of the star's transit == 1 2 1 0*46 ' . 0
Longitude 60?45 : W., in time = • -f 4, 3. 0
12t 0r46f
Girresponding time at Greenwich =: 16* 8T46!
Correction of transit answering to Greenwich timd
1 6 13?46 ! , and variation of sun's right ascension 3r45 ' . 5 s - 2TS 1 ?
App, time of star's transit over the merid, of the place of obs.s= 1 1 158?15 !
Apparent time of transit, per watch^ = 11. 58. 15
Wiatch true for apparent time =5 . 0? OT 0:
Example 2.
January 1st, 1825, in latitude 30?45 * S.^ and longitude 75?30C E., at
7 M7*23! apparent time^ per watch, the observed altitude of Sirius, in the
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882 NAUTICAL A8TEOKOMY.
eastern hemisphere, was 33?43U0^, aiid at \Si47Tl9', when the atar
was in the western hemisphere, it was observed again to have the same
altitude ; required the error of the watch for apparent time ?
Apparent time, per watch, of the obs. equal alt. in east. hemi8.=37M7*23;
Apparent' time, per watch, of the obs. equal alt. in west. hemis.= 15. 47* 19
Sum =s 23?34?42!
Apparent time,, per watch, of star's transit over the given mer.:^ 1 1 147?21 !
Star's R, A., reduced to night of observ.=6t37"25'. 6
Sun's right ascension atnoon, Jan. lst,=sI8.47« 19 . 1
Approximate time of star's transit = 1 1 ?50r 6*. 5 . ll?50r6*.S
Longitude 75?30^ E., in time = . . — 5. 2. 0
Corresponding time at Greenwich = . 6M8T 6'. 5
Correction of transit answering to Greenwich time
6 M8r6i? and variation of sun's right ascension 4724*. 8s— lrl5!
Apparent time of star's transit over merid. of place of obs.s: 1 1 148*5 1 ' • 5
Apparent time of transit, per watch, = 11. 47. 21
Watch tfZoto for apparent time = • • » 1T31',5
Remarks — In ascertaining the error of a watch by equal altitudes of a
fixed star, it will be advisable to select one whose declination is of the same
nafie with the latitude, and which exceeds it in value. In high latitudes,
the altitude most advantageous for observation may be computed Ky the
second part of the rule in pages 120 and 121,^ as exemplified in the second
example of those pages.
In this case, if the latitude of the place of observation be considerably
distant from the Equator, the interval between the times of taking the
equal altitudes will be sensibly contracted; and, therefore, any probable
irregularity in the going of the watch, during that interval, will be propor-
tionably diminished.
Example,
May 1st, 1825, in latitude 70?30^ N., and longitude 35?45< W., at
9^36718! apparent time, per watch, the observed altitude of the star
Kochab, in the eastern hemisphere, was 77°33C20r, and, at 14?577ll!
that star, in the western hemisphere, was again observed to have the same
altitude $ required (be error of the watch for apparent time i
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OF FINDING THB ^PPARBNT TIME. 383
Apparent time, per watch, of obs. equal alt. in the east. hemis.=9?36?18!
Apparent time, per watch, of obs« equal alt. in the west, hemis^^ 14. 57* 1 1
Sum= 24*33:29!
Apparent time, per watch, of star's transit over given merid.s: 12 ? 16T44J !
Star's R, A., reduced to night of observ. = 14 15 1 ri 8 * . 6
Sun's right ascension at noon, May lst= 2, 33. 23 ,9
Approximate time of transit = • . 12* 17:54*. 7 . 12fl7:64'.7
Longitude 35^45 '. W., in time = + 2. 23. 0 . 0
Corresponding time at Greenwich s= 14*40?54'.7
Correction of transit answering to Greenwich time .
14M0?54\ 7, and var.of sun's right ascension 3r49! s - 2:20\ 1
App. time of star's transit over the merid. of place of obs. s 12* 15T34' . 6
Apparent time of transit, per watch, = 1.2.16.44 .5
Watchyiwt for apparent time = •..."...• 1? 9*.9
M>te.— -In this example, since the interval between the observations is
only 5t20?5S! (the star being in the prime vertical; that is, bearing due
east and due west at the equal altitude,) it is, therefore, evident that any
probable irregularity in the going of the watch, during that interval, is less
liable to affect the resulting error for apparent time, in any sensible man-
ner, than if such error had been determined from observations compre-
hending an interv^ of 9 ^36:55!, as in the case of Example 1, page 381.
Pkoblbm III.
Oioen Hie LcAitude qfa Place, and the Altitude and DecUnation of the
Sunf to find the apparent Time of Observation, and, thence, the Error
of a Watch or Chronometer.
Method I.
Rl7U,
Reduce the sun's declination to the time and place of observation, by
Problem V., page 298 ; which being applied to 90?, by addition or sub-
traction, according as it is of a different or of the same denomination with
the latitoiky tbt sum or remainder will be the sun's polar distance.
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384 NAUTICAL A9TRONO»fT«
Reduce the observed altitude of the sun's limb to the true central alti-*
tude, by Problenr XIV., page 320.
Now, add together the sun's true altitude, its polar distance, and the
latitude of the place of observation ; take half the sum, and call the differ-
ence between it and the son's true altitude the remcAider. '
Then, to the log. co-secant of the polar distance, add the log. secant of
the latitude, the log. co-sine of the half sum, the log. sine of the remain-
der, and the constant logarithm 6.301030: the sum of these five loga-
rithms, abating 20 in the index, will be the log. rising answering to the
sun's distance from the meridian ; which will be the apparent time at ship
or place, if the observation be made in the afternoon ; but if in the fore-
noon, its complement to 24 hours will be the apparent time ; the diffv-
ence between which and the time of observation, per watch, will be the
error of the watch, and which will be fast or slow according as the time
shown thereby is later or earlier than the apparent time.
JRemcark. — In practice, it •becomes absolutely necessary to take several
altitudes of the sun's limb, and to note the corresponding times per watch;
then, the sum of the altitudes, divided by their number, «gives the mean
altitude,-— and the sum of the times, so divided, gives the mean time.
Example I.
January 1st, 1825, in latitude 40? 27' N., and longitude 54 940 C W., the
following altitudes of the sun's lower limb were observed^ the height of the
eye above the level of the sea being 20 feet ; required the apparent time
of observation and the error of tiie watch ?
Mean time of observation, per watch, := ... 3? 2? 0!
Longitude 54?40^ W., in time = .... +3.38.40
Greenwich time = 6MOT40!
Sun's declination at noon, January Ist, = • • • 23? 0'39?S.
Correction of ditto for 6 M0T40t = -1.27
Sun's reduced declination = .....;• 22?59:32rS.
Sun's north polar distance ss * I12?59.32r
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OF FINDING THB APPARBNT TIMB. 885
Time, pe^ Watch. Altitude of San'g Lower Limb,
3* 0r30! ' 13?49M0r
3. 1.15 . 13.44. 0
3. 2. 0 13.38.10
3. 2.45 18.32.30
3. 3.30 13.26.40
10? 0! 191 : 0?
Mean= 3? 2T Of Mean= 13?38n2r
Sun'sseini.diam.l6a8r-dip4tl7r= 4-12. 1
Sun's apparent Altitude = • . . 13?50n3i:
Refraction 3C48r-Parallax 0^9^- — 3. 39
Sun's true central altitude s . . 13?46 C34r Constant log. =6. 301030
Sun's north polar distance = . . 1 12. 59. 32 Log. co-sec«*sO. 035949
Lat. of the place of obsenration = 40. 27. 0 Log. secant* =0. 1 18631
Sum = . . • 167? 13: 6C
Half s{im = . . 83?36133^ Log. co-sines 9. 046534
Remainder =s . 69.49.59 Log. sine = 9.972523
Apparent time of observation = . 3* 1*45 ! Log. rising = 5. 47466. 7
Time of observation^ per watch, = 8. 2. 0
Watcb^l fpr apparent time = • 15 seconds.
Example 2.
.June 9th, 1825, in latitude 50?40^ N., and longitude 47^56^ 15? E.^ the
following altitudes of the sun's lower limb were observed, the height of the
eye above the level of the. horizon being 23 feet ; required the apparent
time of observation, and the error of the watch ?
Time of observation, per watch, = .... 19*22725!
Longitude 47'?56: 15? E., in time = . . -3.11.45
Greenwich time » ^ . , 16M0?40!
* The lOf are rejected from the indices of the lo^rithmic secant and co-secant ; and,
with the view of facilitating' the future operationa in this work, the same plan will be pur-
sued in all the coDputations.
2c
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886 NAUTICAL AlTRONOMT«
Sun's declination at noon, June 9tbji = , , 22?56^37?N.
Correction of ditto for 16* 10?40! =.. • . +3.16
Sun's reduced declin«tiQa » 22?59C53rN.
Sun'a north polar dUtance =s . . . • , 67? Of 7'
Hme, per Watch. . Altitude of aon'i Lower Uub.
19i20T45! 29?33aor
19.21.35 • 29.25.10
19.22.25 29.17.30
19.23.15 29. 9.40
19.24. 5 29, 1.50
112r 5! 87^207
Mean aq l'9?22r25! Mean = 29?17;28r : hence the
traa cenbral Altitude is 29?27f 7?
Sun's true central altitude s . 89^^7-7'
Sun's north polar distance ss . 67< 0. 7 Log- oo-secahtssO; 035967
Latitude of the place of observ.B 50. 40. 0 Log. Meant = 0.198026
Sum = . . . 147? 7n4r
Half sums. . 78?33C37? Lpg. «o^ k 9.45)797
Remainders . 44. 6.30 Log. sine = . 9.842620
Constant log. k 6. S01080
Sun's distance from the meridian = 4t44?l 1 ' Log. rinng s 5. 82944. 0
Apparent time of obaervatiou s 19M5*49!
Tlo^e of observation^ per watch^ a 19. 22. 25
Watehjtut for apparent time ■■ 6?9e!
Note, — SiocQ the log. rising^ in Tibbie XXXIL, is only computed to five
places of decimals^ therefore^ in taKing out (he meridian distance of a
celestial oBject from that TaUe^ answering to a given log. rising,* the sixth
or right-hand figure of such given log. rising, is tn be rej«<;ted; observing,
however, to increase the fifth or preceding figure by unity or 1,. when the
figure so rejected amounts to 5 or upwards : thus, in the preceding example,
where the log. rising is 5, 474667^ th^ meridiftQ distan<:e ia tatoi oiH for
5. 47467 i and so on of others*
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OF FINDING THB APPARBNT TIMS, 387
For the principles on which the meHdian distance of a celestial object
is computed, and hence the apparent time, the reader is referred to '^The
Young Navigatbr's Guide to the Sidereal and Planetary Fterts of Nautical
Astronomy,'' page 156.
Eemarks.
Altitudes for ascertaining the ^rror of a watch ought to be taken by
means of an artificial horizon : one produced by pure quicksilver should be
preferred, because it shows, at all times, when placed in a proper position^
a truly horizontal planer and, therefore, the angles of altitude taken therein
are always as correct as the divisions on the sextant with which those
angles are observed ; whereaa, altitudes taken by means of the sea horizon
are generally subject to some degree of uncertainty, owing to its being
frequently broken or ill-defined, by atmospherical haze, at the time of
observation ; though such dtitudes are, nevertheless, sufficiently correct for
finding the longitude at sea«
In taking altitudes by means of an artificial horizon, it is to be observed,
that the angle shown by the sextant will be donble the altitude of the ob-
served limb of the object ; which is to be corrected for index error, if any :
then, half the corrected angle will be the observed altitude of the object's
limb above the true horizontal plane; to which, if its semi-diameter,
refraction, and parallax be applied, the true central altitude of the observed
object will be obtained. There is no correction necessary for dg>, because
the quicksilver shows a truly horizontal plane, as has been before remarked.*
The position of a celestial object most favourable for determining the
apparent time with the greatest accuracy, b, when it is in the prime
vertical ; that is^ when it bears either due east or due west at the place of
observation, or, if it be circumpolar, when it is in that part of its diurnal
path which is in contact with an azimuth circle ; viz., when the log. sine
of its altitude = log. sine of the latitude + radius — log. sine of its declina-
tion ; because, then, the change of altitude is quickest, and the extreme
accuracy of the latitude not very essentially requisite^ The nearer a celes-
tial object is to either of these positions, the nearer will the apparent time,
deduced from its altitude, be to the truth $ as, then, the unavoidable small
erron which generally creep into the observations, or a few miles differ-
ence in the latitude, will have little or no eiFect on the apparent time so
deduced^
Table XLVII. contains the time or distance of a celestial object from
the meridian at which its. altitude should be observed, in order to determine
the apparent time widi the greatest accuracy ; and Table XLVIII. contains
* Tfae difsct mlis for applying the necessary corrections to altitudes taken on shore by
means of an aititfcial baHcon, wiU be found at the end of tlM Gstopcndiiai of Practical
^laTigatioDy towards the latter part of this Tolame.
2c2
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388 . NAimCAJL ASTRONOMY,
the corresponding altitude most advantageous for observation. But, since
those Tables are adapted to the declination of a celestial object when it b
of the same name with the latitude of the place of observation, th^y will
not, therefore, indicate either the proper time or the altitude when those
elements are of contrary denominations : in this, case, since the sun or other
celestial object comes to the prime vertical before it rises, atid therefore
does i\pt bear due east or west while above the horizon, the observation
for determining the apparent time from its altitude 'must be made while
the object is neax to the horizon ; taking care, however, not to take an
altitude for that purpose under 3 or 4 degrees, on account of the uncertain
manner in which the atmospheric refraction acts upon very small angles of
altitude observed adjacent to the horizon.-— See explanation' to the above-
mentioned Tables, pages 119 and 120..
Mbthod II.
Of computing the horary Distance of a celestial Object from the
Meridian.
RULBi
If the latitude of the place of observation and the declination of the
celestial object be of diflFerent names, let their sum be taken,— otherwise,
their difference, — and the meridional zenith distance of the object will be
obtained; to which apply its observed zetiith distance, by addition and
subtractioii, and let half the sum and half the difference be taken ; then,
To the log. secant of the latitude add the log. secant of the declination,
the log. sine of the half sum, the log. sine of the half difference, and the
constant logarithm 6.301630; the sum of these five logarithms, abifting
20 in the index, will be the log. rising of the object's horary distance from
the meridian ; and if this object be the sun, the apparent time will be
known, as in the last method; and, hence, the error of the watch, if
necessary.
Example 1.
January lOth, 1825, in latitude 40^30^: N. and longitude 59?2'30r W.,
the mean of several observed altitudes of the sun's lower limb was
14?31 '47'', that of the correspondmg times, per watch, 3* IT45!, and the
height of the eye above the level of the horizon .18 feet; required the
apparent time of observation, and the error of the watch ?
Time of observation, per watch, = . . ' 3? IT45!
Longitude 59?2^30TW., in time = . + 3. 56. 10
Greenwich time =s . . • .. • . . 6*57*55!
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OF FINDING THB APPARENT TIMB, 389
Sun's declination at noon, January lOth^ =s 21?57 -50^ S.
Correction of ditto for 6t5i7"55! =5 . • — 2,40
Sun's reduced declination s= 21?55'10^S.
Obs. alt of sun's U limb=14?31M7^3 hence, iU true cent, alt isl4?41CS6!
Sun's true zenith distance at time of observation s . ; ; 75 ? 1 8 ' 24 T
Latitudes. • 40?30' OfN Log. secantsO. 118954
Declinations* 21.55. 10 S. . • . • . Log. secant=0. 032588
Sun's mer.z.dist= 62?25 C lOT
Obs. zenith dist. = 75. 18. 24 Const log.s 6. 301030
Sum= . . . 137M3'.34rHalf=68?51M7^ Log. sine = 9.969752
Difference = . I2?53a4r Half= 6.26.37 Log. sine s 9.050091
Sun's dist from the mer.=theapp. time=3M?15!Log.risings5.47241.5
Time of observation, per watch, = • • 3. 1. 45
Watch ya«< for apparent time = • • 0?30!
Example 2.
January 20th, 1825, in latitude 37^20^8. and longitude 49?45' £.,
the mean of several altitudes of the sun's lower limb was 26?39C 157, that
of the corresponding times, per watch, 19M 1 ?45 !, and the height of the
eye above the level of the horizon 16 feet 3 required the apparent time of
observation, and the error of the watch ?
Time of observation, per watch, = • • 19*1 1?45 !•
Longitude 49?45^ £., in time s . • - 3. 19. 0
Greenwich time = ' 15*52T45!
Sun's declination at noon, January 20th,= 20? 7 ' 1 K S.
Correction of ditto for 15*52?45! = . . - 8.45
Sun's reduced declination == • . • • 19?58'26rS.
Obs. alt of sun's 1. limb s 26?39^ 1575hence,its truecentalt is26?49i58^
Sun's true zenith distance at the time of observation s . • 63? 10' 1"
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990 NAUTICAL ASTKONOMY.
Latitude =». . 87 -20^0^8 Logr. MCiuit^O. 099567
Declinations:. 19.58.26 S Log. secaatsO. 026942
Sun'smer.s. clist.» 1 7? 21 1 34?
Obs. zenith di8t. = 63. 10. 2 Const. 1(^.= 6.301030
Sum* . . . .80°81^36r Halfa40?15M8r Log. sine «= 9.810435
Differences .45.48.28 Half=22. 54. 14 Log. sine = 9. 590158
Sun's horary distance from the merid.=4M3?42!Log.rising=5.82813.2
Apparent time of observation = . . 19^ 16T 187
Time of observation, per watch, = . 19.11.45
Watch *Ioto for apparent time = . 4?8S!
Method UI.
C>f compuHng (Ae Aorary Ditttance of a celestial Object jrom th^
Meridian.
Rule.
•
If the latitude of the place of observation and the declination of the
celestial object are of different names, let their sum be taken, — othenrise,
their differ ence^ — and the meridional zenith distance of the object will be
obtained; the natural versed sine of which, being subtracted from the natural
co-versed sine of the object's true altitude, will leave a remainder* Now,
to the logarithm of this remainder add the log. secants of the latitude and
the declination, and the sum will b^ the log. rising of the object's horary
distance from the meridian ; and if this object be the sun, the apparent
time willbe known, and, hence, the error of the watch, if required, as
shown in the first method, page 384.
Eaanq^le !•
May 1st, 1825, in latitude 40?35' S., md longitode 63? 15 f EL, the
mean of several altitudes of the sim's lower limb was 19?43^581'; that of
the.correspondipg times, per watch, 20^57*45?, and the height of the eye
ahove the level of the sea 14 feet; required the apparent time of observa-
tion, and the error of the watch ?
Time of observation, per watch, = . . 20* 57 "45 !
Longitude 63? 151 £;, in time s^ • . — 4, 13.. 0
Greenwich time s= • . # . , ♦ 16*44?45*
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OF FINDINO TBB APPARBNT TIME. 891
Sun's declination at noon, May Ist, ^ 15? 4C19?N.
Correctionof ditto for 16*44r45! =+ 12.84
Sun's reduced declination ss . • . • 15?16'53^N.
Obs. alt. of sun's I Umbial9?43^58r| henoe^ the true cent, alt isl9?53U7^
Latitude =. . 40?85{ OTS. ..... Log. secantvO. 119495
Reduced dec. =: 15. 16.58 N. ..... Log. secantsO. 015634
Sun's mer. z.dis.= 55?5 1 ;53rNat.V.S. = 43885 1
Sun's true alt. » 1 9. 53. 47 Nat.co- V.S.=659680
Remainder =» 220829 Log. s 5.344056
Sun's horary distance from the merid.s: 3^ 2T45'Log.risings=5. 47918.5
Apparent time of observation = • . 20*. 57* 15 !
Time of observation, per watch, = • 20. 57- 45
Watch y<M/ for apparent time =s . . 0*30!
Example 2.
November 10th, 1825, in latitude 49? 13^ S., and longitude S6t50' W.^
the mean of several altitudes of the sun's lower limb was 22?28'30?, the
mean of the corresponding times, per watch, 5M?25!, and the height of
the eye above the level of the horizon 20 feet ; required the apparent
time of observation, and the error of the watch ?
Time of observation, per watch, = • . 5 ? 4?25 !
Longitude 36?50' W., in time ±= . + 2. 27. 20
Greentnch time ss 7*3ir45f
Sun's deelination at noon, Nov. 10th, « • 17? 9^50r S.
Correction for 7*31?45! = + J. 1*
Sun's reduoed decimation » . . / . . 17^15' 5r S.
Ob6« alt. of rail's 1. fimbss22^?28:30r; heoce^ its tiroi cent. alt.is22?38^ 17"^
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392 NAUTICAL ASraOKOMY.
Latitudes. . 49^3' OrS Log. secantsO. 184954
Reduced dec. = 17. 15. 5 S Log. 8ecant=0. 019991
Suii*8mcr.z.di8t.=s31?57^55rNat.V.S. = 151631
Sun's true alt. = 22. 38. 17 Nat.co-V.S.=615091
Reminder = 463460 Logi- 5.666012
Sun's dist. from the mer.ssthe appar. time=5* 0T25!Log.ri8.=:5. 87095.7
Time of observation, per watch, = • • 5. 4. 25
Watch^o^ for apparent time = ... 4? 0!
Mbthod IV.
Cf computing the horary Diiicmce qf a cekitial Object from the
Meridian.
Rule.
If the latitude of the place of observation and the declination of the
celestial object be of different names, let their mm be taken, — otherwise,
their dtj^erenee,— and the meridional zenith distance of the object will be
obtained ^ from the natural co-sine of which, subtract the natural sine of
the object's true altitude, and to the logarithm of the remainder add the
log. secants of the latitude and the declination;. and the sum will be the
log. rising of the object's horary distance from the meridian. . Now, if
this object be the sun, the apparent time is known, and, hence^ the error
of the watch, if required, as shown in the first method, page 384.
Example I.
July 4th, 1 825, in latitude 39?47 ' S., and longitude 60?50' E.^ the mean
of several altitudes of the sun's lower limb was ld?2'30T, that of the
corresponding times, per watch, 3M0T45!, and the height oT the eye
above the level of the horizon 22 feet ; required the apparent time, and
the error of the watch ?
Time of observatioiij per watch, = • . 3M0T45!
Longitude 6Q?50^ E., in tune = • • - 4. 3.20
Greenwich time past noon of July 3d s 23 1 7*25'
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OF FINDING THB APPARBNt TIME. 393
Sun's declination at noon, July 3d, = 22?59C30?N,
Correction of ditto for 23 ^ 7*25 ! = . — 4. 48
Sun's reduced declination s . . . 22^54^ 42rN.
Obs. alt. of the sun's 1. limB= 13?2^30^; hence, its true cent. alt. is 13?9'53r
Latitudes. . 39?47' OrS. Log. sccant=0. 114373
Reduced dec. as 22. 54. 42 N • Log. secaiits=0. 03^90
San's mer. z. dist.=:62?41 '42?Nat. co-8ine=458727
Sun's true alt. = 13. 9.53 Nat sine = 2*27751
Remainder = 230976 Log. = 5.363567
Sun's dist. from the mer.sithe appar. time=3* 10?35 ! Log.ris.s5. 5 1363. 0
Time of observation, per watch, =s . . 3. 10. 45
Watch /(wKor apparent tirtie= . • . OTIO!
Example 2.
July 19th, 1825, in latitude 40? 10^50'/ N., and longitude 53920^ W.,
the mean of several altitudes- of the sun's lower limb was 33^23^5?, that
of the corresponding times, per watch, 19*47*30', and the height of the
eye above the level of the horizon 15 feet; required the apparent time,
and the error of the watch ? . ^
Time of observation, per watch, = • . 19*47"30!
Longitude 53? 20 'W., in time = . . + 3.33.20
Greenwich time = 23*20?50!
Sun's declination at noon, July 19th, = 20?53nKN.
Correction of dittoVor23»20?50! = - 10.44
Sun's reduced declination :^ . . . 20?42^271^N.
Obs. alt. of sun's I. limb=33? 23 '15^; hence, the true cent. alt. is 33 ?34 '. 1 Z
Latitude = . . 40?10^50?N Log. secant=0. 1.16898
Reduced dec. = 20.42.27 N Log. secant=0. 029004
Sun's mer. z. dist.s 19?28^ 23r Nat. co.«ne= 942798
liun's true alt. s= 33i.34. 1 Nat. sine s 552011
Remainder = 389887 Lbg.= 5. 590939
Sun's horary distance from the merid.s=4! 1 lr53! Log. rising=:5. 73684. 1
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394 NAUTICAL ABTRONOICT.
Sun's horary distance frdm the merid.=4 't 1 1 ?53 !
Apparent time of Qbseiyation = • 19*48? 7'
Time of observation^ per watch^ s= 19. 47* 30
Watch tUno for apparent time a: . 0737 '
Problek IV..
Given the Latitude of a Place, tlw JUitude, Right jtscenrion, and
Declination of a known fixed Star, and the Sun'i Right Ascension;
to find the apparent Time, atid, hence, the Error qfthe iVatch,
RULB.
Find the true altitude of the star, by Problem XVIL, page 327 ; and let
its right ascension and declination, as given in Table XLIV., be reduced to
the night of observation ; then.
With the latitude of the place, the star's true altitude, and its reduced
declination, compute its horary distance from the meridian, by any of the
methods given in the last problem. . .
Now, if the star be observed in the western hemisphere, let its meri-
dian distance, thus found, be added to its reduced right ascension, but, if
in the eastern hemisphere, subtracted from it, and the sum or remainder
will be the rijght ascension of the meridian ; from which, (increased by 24
hours, if necessary,) subtract the sun's right ascension at noon of the given
day, and the remainder will be the approximate time of observation,
fleduce this to Greenwich time, by Problem 111., page 297> and find the
proportional part of the variation of the sun's right ascension, for the
given day, answering thereto and 24 hours, •by Problem XII., page 317 ; .
which, being subtracted from the approximate, will give the apparent time
of observation : hence the error of the watch may be known.
Note. — For the principles of this nile^ see ^^ The Young Navigator's
Guide," page 156.
Example U
January 1st, 1825, in latitude 40?29' N., and longitude 59?45f W., the
mean of several altitudes of a Arietis, west of the mcHdian, waa 80?29 '481|
that of the corresponding times, per watch, 11*9T29!, and the height of
the eye above the level of the sea 19 feet; required the apparent time, and
the error of the watch }
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OF FINBIKO THB APPARBNT TIMB. 395
Reduced R. A. of a Arietis=l ^57^19^ and iu reduced dec,s22'?37'50rN.
Observed alt. of a Arieti8=r36?29C48r; hence, its true alt. is 36?24f 20'r
True zenith distance of • Arietis s •.•••• 53 ?35U07
LatiUides . . 40?29' OrN. Log. secant :=:0. 118847
Star's red. dec. =a 22. 37. 50 N. .... . Log. secant^O. 034796
Star's mer. 2. dis.= 17?51 ClOi:
Star's obs z.dist.=:53.35.40 ....... Const. log.= 6.^1030
Sum = ... . 71?26^50r Half=35?43:25r Log. sine = 9,766321
Differences • 35.. 44. 30 Half=:17.52. 15 Log. sine = 9:486958
Star's horary dist^-west of the merid.^ 4^ 2T45 1 L(«.Tindg»5.70795. 2
Star's reduced right ascension » .. • L57«19
Right ascension of the meridian s= . 6? OT 4!
Sun'sR. A. atnoon^January Ist =s . 18.47.19
Approximate time = ..... ll*12?45! .... llM2r45!
Longitude59?45^W., in time = +3.59. 0
OreennHcli time s ...... 15Ml?457
Correction of approximate time, ans. to Qreenwich time
15nir45!,aiidva£iatbn6fsun'sR.A.4?24'-.8, tft • • ^ 2?48!
Apparent time of observation = '; . • . 11V9"57-
Time of observation, per watch, = 11. 9. 29
Watch Woio for apparent time = 0T28'
Example 2.
January 1st, 1 825, in latitude 89?20r30rS., and Iqngitnde 75?40' B.,
the odeanof aereral altitudes of Procyon,east of the meridian, was 27? 15 ^47^>
th&t of the corresponding times, per watch, 9^30T23!, and the height of
the eye above the level of the sea 19 feet ; required the apparent time of
observation, and the error of the watch ?
Procyon's reduced R. A.=7'30r8!, and its reduced dec.=5?39C58fN.
Observed alt. of Procyon=27?i5U7^, hence, its true alt. is=27?9:46r
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396 NAUTICAL ASTRONOMY.
Latitudes. . 39?20i3<KS. ..... Log. 8ecant=0. 1 1 1607
Star's red. dec. = 5,39.58 N Log. secant^O. 002128
Procyon'8m.z.dis.=45° 0^28'Nat.vers.S.=292989
Procyon*8trueaIt.=27. 9.46Nat.c.o-V.S.=548480
Remainder s 250491 Logi = 5.398792
——^1^ I ' I III*
Procyon's horary dbU, cast of the iner.=3i 10T20! Log,ri8ing=55. 5 1252, 7
Procyon's right ascension = ... 7* 30. 8
Right ascension of the meridian == • 4 * 1 9T48 !
Sun'sjrightasclttnoon^ Jan. Ist^ = 18.47.19
Approximate time = 9*32r29t . . • . 9i32r29:
Longitude 75940^ K, in time = . - 5. 2.40
Greenwich time =...... 4?29T49!
Correction of approximate time» answering to Greenwich time
4t 29r49!, and variation of sun's right ascension 4r24' . 8 = — OTSO!
Apparent time of observation = • • • 9^31?39!
lime of observation, per watch, s . r 9. 30. 23
Watch floto for apparent time = • • 1*16!
fiote^ — When the star's horaiy distance ecist of the meridian exceeds the
right ascension, the latter is to be increased by 24 hours, in order ^ find the
right ascension of the meridian.
In finding the error of a watch by sidereal observation, two dr more stars
should be observed, and the error of the watch deduced from each star
separately. And, if an equal number of stars be observed on different sides
of the meridian, and nearly equidistant therefirom, it will conduce to still
greater accuracy ; because, then, the errors of the instrument and the
unavoidable errors of observation will have a mutual tendency to correct
each other. The mean of the errors, thus deduced, should be taken for the
absolute error of the watch. •
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OP FINDING THE APPARENT TIME. 397
Problem V.
Oiven the JjsHiude and Longitude qf a Place^ and the Altitude of a
Planet, to find the Apparent Time of Observation.
Rule.
Reduce the estimated time of observation to the meridian of Greenwich,
by Problem III., page 297 ; to which time let the planet's right ascension
and declination be reduced, by Problem- VI L, page 307 ; and let the sun's
right ascension, at noon of the given day, be also reduced to that time, by
Problem V., page 298. Reduce the observed central altitude of the planet
to its true central altitude, by Problem XVI., page 325.
l*hen, with the latitude of the place, the planet's reduced .declination,
and its true central .altitude, compute its horary distance from the meridian,
by any of the methods given in Problem III., pages 384 to 392. Let
die planet's horary distance from the meridian, thus found, be applied
to its reduced right ascension, by addition or subtraction, according as it
may be observed in the western or in the eastei?! hemisphere, and the right
ascension of the meridian will be obtained; from which (increased by 24
hours, if necessary,) subtract the sun's reduced right ascension, and the
remainder will be the apparent time of observation*
Note, — ^When the planet's horary distance east of the meridian exceeds
its right ascensiopy the latter is to be increased by 24 hours, in order to find
the right ascension of the meridian.
Example 1. '
January 34, 1825, in.latitude 50?30C N., and longitude 48?45^W., the
mean of several altitudes of Jupicer^s centre, east of the meridian, was
23?41^55?, that of the corresponding times, per watch, 9Mr, and the
height of the eye abov| the level of the sea 16 feet; required the apparent
timfc of observation ?
. . Time of observation, per watch, = • 9^ 1" 0!
Longitude 48 ?45^W.,. in time = +3.15. 0
Greenwich time = 12**16? 0!
Sun's right ascension at noon, January 3d, = 18?56T 8!
Correctionof ditto for 12 M6? = . ... +1.39
Sun*3 reduced right ascension* =5 • • . • 18*57?47*
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398 NAUnCAL ASTRONOMY,
Obs. cent«alt.of Jupiter=:23?41 ^55r; hence, its true cent..alU is 23?35 C56^
Jupiter's right ascension^ Jan. Ist, =s 8t58T 0'
Correction of ditto for 2f 12*16T = -- 0.50
Jupiter's reduced right ascension \s= 8*57*10!
Jupiter's declination, January Isty =: 17^561 OTN.
Correction of ditto for 2f 12M6? =: + 5. 1
Jupiter's reduced declination = • • '18? IC KN.
Ditto north polar distance = . . 71?58'59f
Jupiter's true central altitude s 23^35 .56?
Jupiter's north polar distance rs 7 1 • 58. 59 Liog. co-secantssO. 021836
Lat. of the- place of observationsSO. 30. 0 Log. sec^^nt aa 0. 196489
SumLd: 146? 4:551: Constant log. =6.301030
Hilf8um= ...... 73? 2;25^r Log. co-sine = 9.464933
Remainders ..•..". 49?26:29|r Log. sines . 9.880667
Jupiter's horary dist, east of the mer.= 4?58T 0!Logjrisings5. 86495,5
Jupiter's reduced right ascension = .8. 57^ 10
Right ascension of the meridian =: • 3*59T10!
Sun's reduced right ascension = • . 18. 57- 47
Apparent time of observation = . . 9* 1?23!
Example 2.
January 16th, 1825, in latitude 34?45^ S., and longitude 80?30^ E.,
the mfean of several altitudes of Venus* centre, west of the meridian^ was
22?53C25r, that of the corresponding times, per watch, 7*20^45 !, and the
height of the eye above the level of the sea 18 feet f required the apparent
time of observation ?
Time of obseryation, per watch, = . 7*20r45!
Longitude 80?30!E., in time = . -^5.22. 0
Greenwich time = I*58r45!
Sun'sright ascension at noon, January {6th, == 19^52*41 f
Correction of ditto for l*58r45f = . . . +0.21
Sun^ reduced right ascension =: • • • « » I9t53? 2!
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OF FINDINO THB APPAUMT T1MB« ' 899
Venus' right ascension^ January 13th, = 22^23? 0!
Correction of ditto for 8f I *58?45! = . + 13. 52
Flanet'ji reduced right ascension ^ • « 221367521
Venus' declination, January 13th, = . 1K39! 01 S:
Correction of ditto for 3 f 1 ?58T45 ! = - 1 . 29, 24
Planet's reduced declination as . • 10? 9CS6r.S.^
Obsenwd central altitude of Venus sob 22?53.25?; hence, her true eeih*
tral altitude' it 22?47'24f, on the assumption that her horiiontal parallax,
at the time o^ observation, was 1 8 seconds of a degree.
Latitudes. . 34?45C OfS. ..... Log. secant=0. 085315
Planet's red. dec. =10. 9.36 S , . Log. 8ecwt=:0^b06864
Planet's m.a.di8t.=24?35:24f Nat. co-sine=3909309
Planet's true alt.=22. 47. 24 Nat. sine = 387355
Remainders 521954 Log. = 5.717632
Venus* horary dist.^west of the meiid.ar4t36?S5? L(^. rislngs5, 809^1 . 1
Venus' reduced rig^t ascension = . 22. 36. 52
Right ascension of the meridian a SMSr47?
Sun'a reduced right ascension a . 19. 53. 2
Apparent time of observation a • 7*20T45!
Bemark. — Should the horisontal parallaxes of the planets be ever given
in the Nautical Almanac, the mariner may then deduce the apparent time
from their altitudes, by the abpve Problem, to a very great degree of
accuracy, provided' the longitude of the place of observation be known
within a few minutes of the truth, or that there be a chronometer on board
to indicate the time at Orfetawieh. However, even admitting that those
parallaxee are still to remain unnoticed, the apparent time, computed as
abow, wiH always be sufficiently near the truth ffH* the purpose of deter*
mining the longitude at sea.
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400 NAUTICAL ASTRONOMY.
Probusm VI. .
Given the Latitude dnd Longitude of a Place, the estimated Time at that
Place, and the Altitude of the Moon's Lmb ; to find the apparent Jhne
of Observation.
RULB.
. Reduce the estimated time of observation to the meridian of Greenwich,
by Problem III., page 297; to which let the sun^s right ascension be
reduced, by Problem V., page 298 ; and let the moon's tight ascension,
declination, semi-diameter, and horizontal parallax be reduced to the same
time, by Problem VI., page 302. Reduce the observed altitude of the
moon's limb to the true central altitude, by Problem XV., page 323 ; then.
With the latitude of the place of observation, the moon's reduced
declination, and her true . central altitude, compute her horary distance
from the meridian, by any of the methods given in Problem III., pages
384 to 392. Now, let the moon's horary distance from the meridian,
thus found, be applied to her reduced right ascension, by addition or
subtraction, according as she may have been observed in the western or
eastern hemisphere, and the right ascension of the meridian will be ob-
tained ; from which (increased by 24 hours, if necessary,) subtract the sun's
reduced right ascension, and the remainder will be the apparent time of
observation.
Note. — ^When the moon's horary distance, east of the meridian, ei^ceeda
her right ascension, the latter is to be increased by 24 hours, in order to
find the right ascension of the meridian.
And it is to' be borne in mind,, that the moon's right ascension and
declination inust be corrected by the equation of second difference. Table
XVXI., as explained between pages 33 and 38.*
^Example 1.
January 4th, 1825^ in latitude 50?I0^ N., and longitude 60? W., the
mean of several observed altitudes of the moon's lower limb, east of the
meridian, was 29?25'23^, that of the corresponding times, per watch,
7 •28? 181, and the height of the eye above the surface of the water 17
feet ; required the apparent time ?
* For the eflfecU resulting from the eqaation of the mean secood difference of the moon's
place In right ascension and declination, see *' The Young Navigator's Guide to Ac Side-
real and Plsnetary PMrt^ of Nautical Astronomy/' page 171,
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OF FINDING THB APPARENT TIME. 401
Time of observation, per watch, = . 7*28ri8!
Longitude 60?W., in time = . -f 4. 0. 0
Greenwich time = U»28ri8!
Sun's right ascension at noon, January 4th, = I9t 0T32'
Correction of ditto for 11*28?1 8! = . . . + 2. 6
Sun's reduced right ascension = , • . . 19t 2T38!
Moon's right ascenuon at noon, January 4t'h,= 98? 6'5df
Corrected prop, part of ditto for 1 1 J28ri8! = +7. 17. 28
Moon's corrected right ascension =: . . . 105?24'.21?
Moon's semi-diameter at noon, January 4th, =: 16 C 9?
Correction of ditto for 1 1 ^28T18! = . . . +5
Augmentation, Table IV., =: + 8
Moon's true semi-diameter =: 16^22?
Moon's declination at noon, January 4th, =: 22?35'39?N.
Corrected prop, part of ditto for lli28?18!= -1. 10. 18
Moon's corrected declination = . . . . 2 1 ? 25 ' 2UN.
Moon's horizontal parallax at noon, January 4th,= 59^17^^
Correction of ditto for 11 *28?18! = . . . . + 16
Moon's true horizontal parallax =: . • . • • 59^33^
Observed altitude of the inoon'sIowerlimb=:29?25'23^j hence, her true
central altitude is 30? 27 ' 55 r.
Latitude = . . 50?10^ OrN. . . .. • Log. secant=0. 193442
Moon's corr. dec. = 21.25.21 N. .... Log. 8ecant=:0. 031091
Moon's mer.z,dist.= 28?44 ', 39? Nat. vers. sine= 1 23225
Moon's tfue alt. = 30. 27. 55 Nat. co-V. S. = 492988
Remainder = 369763 Log.s 5. 567923
MoonV horary dist, east of the merid.:::4 ^30?41 !Log«risingr:5. 79245. 6
2 D
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402 NAUTICAL ASTRONOMY.
Moon's horary dUt. cast of the merid.=:4t 30*41 !
Moon's cor. R. A. 105 ?24 ^ 2 H, in tim€=7. 1 . 87
Right ascension of the meridian = . 2?80'?56!
Sun's reduced right ascension = . .19. 2. 38
Apparent time of observation = • « 7^28?18!
Example 2. ^
January SOth, 1825, in latitude 10?20^ S., and longitude J00?50^E.,
the mean of several altitudes of the moon's lower limb^ west of the meri-
dian, was 7 ^ 23 ' 30'% that of the corresponding times, per watch, 1 3 ^ 33 T20 ' ,
and the height of the eye above the surface of the water 20 feet; required
the apparent time ?
Time of observation, per watch; = 13*33?i20!
Longitude 100^50! E«, in time =1-6. 43. 20
Greenwich time = 6^50^0!
Sun's right ascension at noon, Jan. 30th, = 20t51T25!
Correction of ditto for 6^50? « . « • . ^1.10
Sun's reduced right ascension = • . « • 20?52?35!
Moon's right ascension at noon, Jan. 30th, = 76?21'55T
Corrected prop, part of ditto for 6t50r = + 4.13.38
Moon's corrected right ascension i= . ... 80^35 !33?
Moon's semi-diameter at noon^ JanuatydOth/ ts 15(46?
Correction of ditto for 6*50? zr ..... +5
Augmentation^ Table IV., zz +2
Moon's true semi-diameter = 15(53T
Moon's declination at noon, January 30th, =t 23?57-46rM.
Corrected prop, part of ditto for 6? 60? = — 8.44
Moon's corrected declination tz . . * 23*54! 2TN.
Moon's horizontal parallax at noon, Jan. 30th, = 57'51?
Correctioiiof dittofor6?S0r £2: ..... +17
Moon's true horistontal jporallax =: » < • * 58r 8f
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OP FINDING THB ALTITCTDSS OF THB BBAVENLT BODIES. 408
Observed altitude of the moon's lower limb = 7?23'30T; hence^ her true
central altitude is 8?25^54r.
Latitude s= . . 10?20C OrS. .... Log. secantnO. 007102
Moon's corr. deer: 23.54. 2.N Log. secantzzO. 038935
Moon's m. z. dist. == 34 ? 14 ^ 2r Nat. co^ine= 826748
Moon's true alt. = 8.25.54 Nat. sine =' 146630
Remainder = 680118 Log.=5. 832584
Moon'shorarydistwestof themerid.= 5^ 3r33! Log. rising^: 5. 878621
Moon's cor. R. A. 80?35 ^3Sr, in time=5. 22. 22
Right ascension of the meridian =: 10^25755 '
Sun's reduced right ascension = • 20. 52. 35
Apparent time of observation = • 13?33?20!
RemarJe.'^lf there be a chronometer on board to indicate die time at
Greenwich^ the apparent time of observation, at any given place, may be
very correctly ascertained by the above problem. But, since the chrono-
meter shows the equable or mean time at Greenirich, this time must be
reduced to apparent time, by applying the equation of time thereto with a
contrary sign to that expressed in the Nautical Almanac. Thus, in the
above example, if the chronometer give the mean time at Greenwich zz
7?3?44!, then the reduced equation of time, vis., 13T44!, being subtracted
therefrom, shows the apparent time at that meridian to be 6t50?0t.
Hence, when the equation of lime in the Nautical Almanac is marked
additive, it is to be applied by subtraction; but when marked subtracHve,
it is to be applied by addition to the mean time (per chronometer) at
Greenwich, in order to reduce it to apparent time.
SOLUTION OP PROBLEMS RELATIVE TO FINDING THE
ALTITUDES OP THE HEAVENLY BODIES.
It sometimes happens at sea, particularly in taking a lunar ohseroaiion,
that the horizon is so ill-defined as to ftnder it impossible to observe the
altitudes of the objects to a sufficient degree of exactness ; or, perhaps,
that one or bodi of die objects are directly over the land, at the time of
measuring ^ Imutf distance^ and the ship so contiguous thereto as to
2i>2
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404 NAUTICAL A8TRONOMV.
render the absolute value of the horizontal dip uncertain : in such cases,
therefore, the altitudes of the objects must be obtained by computation, as
in the following problems ; the principles of which will be found amply
illustrated in " The Young Navigator's Guide to the Sidereal and Planetary
Parts of Nautical Astronomy," page 237.
Paoblbm I.
Gwen the Latitude and Longitude of a Place, and the AppareiU Thne at
that Place; to find the true and the apparent Altitude of the Sun's
Centre.
Rule.
Reduce the given apparent time to the meridian of Greenwich, by
Problem III., page 297 ; to which let the sun's declination be reduced, by
Problem V., page 298.
If the latitude of the place and ^he sun's declination are of different
names, let their sum be taken ; otherwise, their differefice: and the meri-
dional zenith distance of that object will be obtained. Then,
To the logarithmic rising answering to the sun's distance from the
meridian, (that is, the interval between the given apparent time and noon,)
add the logarithmic co-sines of the latitude of the place and of the sun's
reduced declination : the sum, rejecting 20 from the index, will be the
logarithm of a natural number ; which, being added to the natural versed
sine of the sun's meridian zenith distance, found as above, will g^ve the
natural co-versed sine of its true altitude.
To the sun's true altitude, thus found, let the correction corresponding
thereto in' Table XIX., be added.; and the sum will be the apparent altitude
of the sun's centre.
Example L
Bequired the true and apparent altitude of the sun's centre, January 10th,
1825, at3Mr45! apparent time, in latitude 40?30^ N., and longitude
S9?2C30rW.?
Apparent time at ship or place = • • • 3* 1T45!
Longitude 59*? 2 ^ 30'' W., in time = • + 3, 56. 10
Greenwich time 22 i 6 1 57*55!
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OF FINDING THS^ ALTITCDBS OF THB HEAVENLY BODIES. 405
Sun's decimation at noon, January lOth^s 21^57 -50^ S.
Correction of ditto for 6?57T55! = \ - 2. 40
Sun's reduced declination = ... 21?55M0rS.
Sun's hor. men di8t.=:3^ lr45! .... Log. risings . 5.474670
Sun's reduced dec.ss21. 55. 10 S. . . . . Log. co-sine = 9. 967412
Lat. of the place = 40. 30. 0 N. . . . . Log. co-sine = 9. 881046
Sun's mer. z. dist. = 62?25 '. lO^Nat. vers. 8ine= 537005
Nat. number = 210440 Log.s 5.323128
Truealt.ofsun'scen.= 14?37 '48TNat.co.V.S.= 747445
Corrcc.,TableXIX.= + 3.26
App.alt.ofsun'scen.=14?4i: 9'r
Example 2.
Required the true and apparent altitude of the sun's centre, January
20th, 1825, at 19* 16ri8! apparent time, in latitude 37^20^ S., and Ion-
gitude49?45:R?
Apparent time at ship or place == . . 19M6T18*
Longitude 49? 45 ^ £., in timers . . —3. 19. 0
Greenwich time = Ij5*57"18!
Sun's declination, January 20th, = • 20? 7 ' 1 1 r S.
Correction of ditto for 15*57'ri8! = - 8.48
Sun's reduced declination = . . . 19?58f23rS.
Sun's hor. dist. fr.mer. = 4*43r42!* . . . Log. rising = 5.828140
Sun's reduced dec. = 19.58.23 S. . . . Log. co-sine= 9. 973060
Latitude of the place = 37. 20. OS.... Log. co-sine= 9. 900433
Sun's mer. zen, dist. = 1 7 ? 2 1 C 37 ''Nat. v. sine=045552
Nat, num. = 503075 Log.=5. 701633
True alt. of sun's cent.= 26?49'.55^Nat.co.V.S=548627
Reduc. of do.,Tab. XIX.,= + 1 . 44
App.alt.of8un'scentre=: 26?51 ^89r
• %i hoixrs - 19*16*18* ^ 4*43*42' , the sun's Uorwy distance from the meridian,
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406 NAUTICAL ASTROHOMT.
Probijem IL
Given the apparent Time at a known Place, tojind the true and apparent
Jmtude of a, fi^ed Star.
Rule.
Reduce the given apparerit time to the meridian of Greenwich^ by
Problem III.^ page 297 1 to which let the sun's right ascension, at noon
of the given day, be reduced, by Problem V., page-298. •
Let the star's right ascension and declination (Table XLIV.) be reduced
to the given period, hy the method shown in page 1 15. To the sun's
reduced right ascension let the given apparent time be added, and the sum
will be the right ascension of the meridian ; the difference between which
and the starts reduced right asQcnsion will b^ the horary distance of the
latter from the meridian. Now, with the star's horary distance from the
meridian, thus found, its reduced declination, and the latitude pf the place,
compute the true altitude of that object, by the last problem. Then, to the
star's true aldtude, thus found, let the correction corresponding thereto, in
Table XIX.^ be added ; and the sum will be the star's apparent altitude.
Example 1.
Required the true and apparent altitude of a Arietis, January Ist, 1825,
at llt9T29! apparent time, in latitude 40? 29^ N., and longitude
59?45^W..?
Apparent time at ship or place = . « , 11? 9T29t
Longitude59?45^ W., in time= . + 3.59. 0
Greenwich time = 15t 8?29!
Sun's right ascension at noon, January Ist, = 18M7*19*
Correction of ditto for 15 ?8T29! =: ... + 2.47 •
Sun's reduced right ascension s • , , • 18?50? 6!
Given apparent time = 1 1 . 9. 29
Right ascension of the meridian s . • • « 5?59?35!
Star's reduced declination - « « • . 22?37'50?N.
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OF FINDING THE ALTITUM8 OF THS BEAVBNLY BODIES. 407
Star's reduced R. A.= l ?57r 19!
R. A. of the merid. = 3* 59. 35
Star'shor.di8fr.mer.=?4t 2ri6! .... Log. rising = 5.706360
Star's redii4:^d«c.»22';37*50rN. , • . Log. co-sine 3: 9. 965204
Lat. of tbe place 9 40. 29, ON..., Log. co-sine :« 9. 8S1 153
Star's mer.zen.dist.= 17?5 1 '. 10rNat.vers.sine= 048153
Nat. number = 357040 Log.= 5. 552717
True alt. of the 8tar=36?29:56rNat. co-V. S.= 40S19S
Reduc.of do.TabJClX= + 1.17
App.'a]t.of giv. 8tar«36?31 HSr
Example 2.
Required th« true and apparent altitude of Procyon, January Ist, 1825,
at 9^SirS9t apparent tiine> in latitude 39?20:30^S., and longitude
75?40CB.J
Apparent time at ship or place =s . . • • 9*31TS9!
Longitude73?40CB., intime 38 .... 5^ 2.40
Gr^nwich time » 4t28T59:
Sun's right ascension at noon^ January Ist^ = 18M7*19!
Correction of ditto for 4*28r59! = . , . + 0, 49
Sun's reduced^right ascension = . . , • 18M8? 8!
Given apparent time = • • • • . . . 9.31.39
, Right aseensioh of the meridian a • • • • 4M9T47'
Procyon's reduced declination s .... 5?39'587N.
Procyon's red. R. A = 7*30? 8 !
R. A. of the meridian==4. 19. 47
Star'8hor.dis.fr.mer.=3M0?21! .... Log. rising = 5.512600
Star's reduced dec. = 5?39^58'rN. . , . Log. co-sine = 9. 997872
Latitude oftheplace=39, 20. 30 S« . . . Log. co-sine = 9. 888393
Star's mer. z. dist. = 45? 0^28rNat. vers.sine=292990
Nat. number =250533 Log.= 5.398865
True alt. of giv. star =27? 9^36'rNat. co-V. S.=543523
Reducof do.Tab JCIX.= + 1 . 50
App. alt. of giv. star=27? 1 K 26r
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408 NAUTICAL ASTRONOMY.
Problem III.
Given the Latitude atid Longitude of a Places and the apparent Time at
that Place; to find the true and apparent Altitude qfa Planets
Rule-
Reduce the given apparent time to the meridian of Greenwich, by Pro-
blem III., page 297 ; to which time let the sun's right ascension be
reduced, by Problem V., page 298 ; and let the planet's right ascension
and declination be reduced to the same time, by Problem VIL, page 307*
To the sun's reduced right ascension let the given apparent time be
added, and the sum will be the right ascension of the meridian ; the differ-
ence between which and the planet's reduced right ascension will be the
horary distance of the latter from the meridian. Now, with the planet's
horary distance from the meridian, thus found, its reduced declination, and
the latitude of the place, compute the true altitude of that object, by Pro-
blem I., page 404. Then, with the planet's true altitude, thus found, by
computation, enter Table XIX., and take out the quantity corresponding^
to the redaction of a star's true altitude ; the difference between which
and the planet's parallax in altitude. Table VI., will leave a corrccdon^
which, being added to the trucj will give the oppare?!^ altitude of the
planet.
Example 1.
Required the true and apparent altitude of the planet Jupiter, January
3d, 1825, at9MT23! apparent time, in latitude 50?30' N., and longitude
48^45 :W.?
Given apparent time at ship or place = • • 9* l?23!
Longitude 48?45^ W., in time = . . . + 3. 15. 0
Greenwich time = . 12*l6r23!
Sun's right ascension at noon, January 3d, = 18^56? 8!
Correction of ditto for 12* 16T23 ! = , . , + 1 . 39
Sun's reduced right ascension = .... 18*57"47*
Given apparent time = • . • . • ... 9. 1 . 23
Right ascension of the meridian ss , • • • 3*59T10!
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OF PlxNDlNG THB ALTITUDES OF THE HEAVENLY BODIES. 40&
Jupiter*8 declination at noon, January 1st, 3= 17? 56' 0?N.
Correction of ditto for 2fl2M6r23'. = . . + 5, 1
Jupiter's. reduced declination s • . . • 18? IC KN..
Jup/8lLA.atnoon,Jan,l=8*58r 0*.
Cor.ofdo.for2fl2;i6T23!= -0.50
Jupiter's reduced R. A. = 8*57TlOt
R. A. of the meridian = 3. 59. 10
Planet's hor. dist. fr. mer. = 4*58? 0'. . . . Log. rising = 5. 864960
Planet's reduced dec. = Ip. 1. 1 N. . . Log. co-8ine= 9.978164
Latitude of the place ^ 50. 30. ON... Log. co-sine= 9. 80351 1
Planet's mer. zen. dist. = 3^?28^59^Nat.V.S.= 156449
Nat. num.=443236Log.=5. 646635
Jupiter's true central alt.=: 23?35 : 52^N.co-V.S.=599683
Red.Tab.XIX.2nK) jj.^_ g, g^
Par.Tab.vi.= 0. 2 J
Jupiter's app. central alt.= 23 ?38 ! 1 r
Note. — Jupiter's horizontal parallax is assumed, in the preseiU instance^
at 2 seconds of a degree.
Example 2.
Required the true and apparent altitude of the planet Venus, January
16th, 1825, at 7^20^45! apparent time, in latitude 34?45^ S., and longi-
tude 80?30C £., admitting her horizontal parallax, at that time, to be 18
seconds ?
Apparent time at ship or place = , , . . 7 *20"45 *
Longitude 80?30' E., in time = . . . • 5.22. 0
Greenwich time = 1?58?45:
" Sun's right ascension at noon, January 16th, = 19*52T41 !
Correction of ditto for 1 *58T45! =: . . . - 0. 21
Sun's reduced right ascension = • . • • 19t53? 2!
Given apparent time = * /. 20. 45
Right ascension of the meridian =: , « ^ 3M3T47*
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410 NAUTICAL ASTEOVOMT.
VcQua' declination^ January 13th, » . . . 1K39'. OrS.
Correction of ditto for 3f 1 ?S8r45 '. . . ^ U 29. 24
Veni»' reduced declination 55 10? 9' 36* S,
Venus' R- A., Jan. 13th,=22*23T 0!
Cor. of do.for3f 1 ?58r45 ! = + 13. 52
Venus' reduced R. A. = 22*36T52!
R. A. of the meridian = 3. 13. 47
Planet's hor. dist. fr. iner. == 4 ?36T55 ' ... Log. rising as 5 . 8098 10
Planet's reduced dec. =^ 10? 9^36^ S. . . Log. co-8ine= 9. 993136
Latitude of the place = 34.45. OS.,* Log. CQ-sine= 9. 914685
Planet's mer. zen. dist. = 24?35^24Wat.V. S.= 090691 -r
Nat.num.= 521953Log.=5. 717631
Venus' true central alt. = 22?47'24^'N,co.V.S=612644
Red.Tab.xix.2n5r)jj.jf^^^j 59
Par.Tab.vi.=0. 16 )
Venus' app. central alt. = 22?49'23r
Remark. — In these problems, a cipher is annexed to the logarithmic
rising taken from Table XXXII. : this is done with the view of reducing it
to six places of decimals ; so that there may be no mistake in property
applying thereto the logarithmic co-sines of the latitude and of the
declination.
PnofiLSM rv.
Given the Latitude of a Place, and the apparent Time at that Place,
with the Longitude; to find the true and apparent Altitude qf the
Moofii^e Centre.
Rule.
Reduce the given apparent time to the meridian of Oraenwieh, by Pro*
blem III., page 297 ; to which let the sun's right ascension be reduced, by
Problem V., page 298; and let the moon's right ascension, declination, and
horizontal parallax be reduced to the same time, by Problem VI., page 302.
To the sun's reduced right ascension let the given apparent time be
added, and the sum will be the right ascension of the meridian ; the differ-
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OF FINDING THB AtTITUDBS Q9 THV HBAVBNLY BODIBS. 4U
ence between which and the moon's reduced right ascension^ will be the
horary distance of the latter from the meridian*
Now, with the moon's horary distance from the meridian, her corrected
declination, and the latitude of the place, compute her true central altitude,
by Problem I., page 404. Then,
From the moon's true central altitude, thus found, subtract the correction
corresponding thereto and her reduced horizontal parallax, in Table XIX.,
and the remainder will be the apparent central altitude.
Note. — ^The moon's right ascension and declination must be corrected by
the equation of second diflPerence contained in Table XVII., as explained
between pages 33 and 37*
Example 1.
Required the true and apparent altitude of the moon's centre, January
4th, 1825, at 7 ^28r 18 '.apparent time, in latitude 50? lO'.N., and longitude
60?W.?
Apparent time at ship or place = . • . . . 7 -28" 18!
Longitude 60? W., in time = 4. 0. 0
Greenwich time = Il*28ri8!.
Sun's right ascension at noon, January 4th, t= 19 1 0?32!
Correction of ditto for 11!28?1 8! = . • . -f 2, 6
Sun's reduced right ascension = • . • • 19* 2?38'
Given apparent time = 7* 28. 18
Right ascension of the meridian =s . . . . 2^30T56!
Moon's horizontal parallax at noon, January 4th, = 59M7^
Correction of ditto for ll?28r 18: =....+ 16
Moon's true horizontal parallax = 59!33T
Moon's declination at noon, January 4th, =? 22?35:391^N.
Cortrectcd prop, part of ditto for Ili28ri8:= - 1. 10/ 8
I* 111 M^— —
Moon's corrected declination = • • • • 2 1 ? 25 '. 3 1 ^N.
Moon's right ascension at noon, January 4th, = 98? 6^.53^
Corrected prop, part of ditto for 1 1 ?28r 18! = + 7. 17. 28
Moon'8CorrectedR,A.=7* 1"37! = . . • . . 105?24:2H .
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412 NAUTICAL ASTRONOMY.
Moon's corrected R. A.= 7 * 1 "37 '
R.A.of the meridian = 2.30.56
J '8 horary dis. fr. mer, = 4 ?30T4 1 ! ... Log. rising = 5. 792450
Moon's corrected dec.=21?25'3ir N; . . Log. co-sine = 9. 968909
Latitddeof theplace = 50. 10. ON... Log. co-sine == 9. 806558
Moon'8mer.zen.di8t.=28?44^39r NatV.S. = 123225
Nat. num. = 36975 7Log.=:5. 567917
True alt. of ]) '8centre=30?27'.55^NatiCo-V.S.=492982
Reduc,ofdo.Tab.XIX.= -50. 8
App. alt. of ]) 's cent.= 29?37U7^
Example 2.
Required the true and apparent altitude of the moon'a centre, January
30th, 1825, at 13t33r20! apparent time, in latitude 10?20^ S., and lon-
gitude 100?50: E.?
Apparent time at ship or place r= • . • • 13^33^20!
Longitude 100?50' E., in time = . . . - 6. 43. 20
Greenwich time = . . 6?50r 0!
Sun's right ascension at noon, January 30th, = 20*5lT25 !
Correction of ditto for 6?50r =z .... + 1. 10
Sun's reduced right ascension =1 .... 20t52?35!
Given apparent time = 13. 33. 20
Right ascension of the meridian = • • • . 10t25?55!
Moon's horizontal parallax at noon, January 30th,=:57^51'!
Correction of ditto for 6* 50T =: -f 17
Moon's true horizontal parallax =: 58^ 8T
Moon's'declination at noon, January 30th, =: 23 ? 57 ' 46 ^N.
Corrected prop, part of ditto for 6 ?50? = . . —3.44
Moon's corrected declination = • . . . . 23?54' 2^N.
Moon's right ascension at noon, January 30th,=76?21 '55^
Corrected prop, part of ditto for 6?50'r = + 4. 13. 38
Moon's corrected R.A.=5*22:22! :;: , . . • 80?35^33f
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OF FINDING THB LONGITUDE. 413
Moon's corrected R. A.=:5*22r22!
R. An of the meridian = 10. 25. 55
}) 's hor. dist. fr. mend. = 5 1 3T33 ! . . . Log. rising =: 5 . 878620
Moon'* corrected dec. =23^54' 21 N. . . Log. co-sine = 9. 961065
Latitude of the places 10.20. OS.,. Log. co-sine = 9. 992898
D sm«r.2en.di8tence=:34?14^ 2r Nat. V. S. = 173252
Nat. num. = 6801 16Log.=5. 832583
True alt. of }) 's centrer: 8?25 '. 54^Nat. co.V.S.= 853368
Reduc.ofdo.Tab.XIX.= -50.47
App. alt. of J) 's centre =: 7?35 '. T".
Remarh^^The natural sines may be used in the solution of the four
preceding problems, instead of the versed sines : in this case, if the natural
number be subtracted from the natural co-sine of the object's meridional
zenith distance, the natural sine of its true altitude will be obtained. Thus,
in the last example, the moon's meridian zenith distance is 34? 14 '2?. Now,
the natural co-sine of this is 826748 ; from which let the natural number
680116 be subtracted, and the remainder =: 146632 is the natural sine
of that object's tru6 altitude ; the arch corresponding to which is 8?25 '54T.
These problems are, evidently, the converse of those for finding the
apparent time, as given in pages 383, 394, 397, and 400.
SOLUTION OF PROBLEMS RELATIVE TO THE LONGITUDE.
The Longitude of a given place on the earth, is that arc or portion of
the equator which is intercepted between the first or principal meridian
and the meridian of the given place ; and is denominated east, or west,
according as it may be situate with respect to the first meridian.
The ^gt or principal meridian is an imaginary great circle passing
through any remarkable place and the poles of the world : hence it is
entirely jarbitrary ; and, therefore, the British reckon their first meridian to
be that which passes througli the Royal Observatory at Greenwich ; the
French esteem their first meridian to be that which passes through the
Royal Observatory at Paris *, the Spaniards, that which passes through
Cadiz, &c^ &c. &c. Every part of the terrestrial sphere may be conceived
to have a meridian line passing through it,' cutting the equator at right
angles : hence there may be as many dififerent meridians as there are points
in the equator.
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414 NAUTICAL ASTBONOMT.
Every meridian line, with respect to the place through which it pASses,
may be said to divide the surface of the earth into two equal parts^ called
the eastern and western hemispheres. Thus, when the face of an observer
is turned. towards the north pole of the world, the hemisphere which lies
on his right hand is called east, and that on his left hand west; and, vice
versay when the face is directed towards tlie south pole of the world, the
hemisphere which lies on the left hand is called east^ and that on the right
hand west.
The longitude is reckoned both ways from the first meridian, east and
west, till it meets with the same meridian on the opposite part of the
equator : hence the Ibngitude of any place pa the earth can never exceed
180 degrees. The difference of longitude between two places on the earth
is an arc of the equator contained between the meridians of those places,
showing how far one of them is to the eastward or westward of the other,
and can never exceed 180 degrees, or half the earth's circumference.
All places that are situated under the same meridian have the same
longitude ; but places which lie under different meridians have different
longitudes : hence^ in sailing due north or due south, since a ship does not
change her meridian, she keeps in the same parallel of longitude ; but, in
sailing due east or due west, she constantly changes her meridian^ and there-
fore passes through a variety of longitudes.
When the meridian of any place is brought, by the diurnal revolution of
the earth round its axis, to point directly to the sun, it is then noon or mid-
day at that place.
The motion of the earth on its axis is, at all times, equable and uniform;
and, since it turns round its axis eastward once in every 24 hours, all parts
of the equator, or great circle of 360 degrees, will pass by the sun, or
star, in equal portions of time : therefore the twetity-fourth part of the
equator, viz., 15. degrees, will pass by the sun in one hour' of time : for,
24 1 X 15? or 1 hour, := 360 d^ees; and^ conversely^ 360 d^ees -i-
24 hours =: 15 degrees or 1 hour.
fivery place on the earth, whose meridian is 15 degrees east of the Royal
Observatory at Greenwich, will have noon and every other hour <me hour
sooner than at the meridian of that observatoryj if the meridian be 30
degtces east of Greenwich, it Mrill have noon and every other hour two
hours sooner than at the meridian of that place, and so on ; the time always
diflering at the rate of 1 hour for every 15 degrees of longitude, I minute
of time for every 15 minutes of longitude, and 1 second of time for every
15 seconds of longitude. . Again, every place whose meridian is 15 degrees
west of the Royal Observatory at Greenwich will have noon and every other
hoar one hour later than at the meridian of that observatory; if the meri«
dian be 30 degrees to the westward of Greenwich, it will have noon and
every other hour tu)o hours later than at the meridian of th«t place, and
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OF PIKDtNG THB MVGtTUDB* 415
SO on. Hence it is evident, that if the time at the meridian of a ship or
place be greater than the time, at the same instant, at the meridian of
Greenwich, such ship or place will be to the eastward of Greenwich; but if
the time at a ship or place be less than the time, at the same instant, at
Greenwich, such ship or place will be to the westward of Greenwich.
Since tlie longitude of any place on the earth it expressed by the differ-
ence of time between that place and the Royal Observatory at Greenwich ;
therefore^ to determine the longitude of a given place, we have only to find
the time of the day at that place, and also at Greenwich, at the same
mstant; then, the diflerenee of these times being converted !nto motion,
by allowing 15 degrees for every hour, &c., or, more readily, by Table I. in
this work, the longitude of such given place will be obtained.
The readiest, and, indeed, the most simple method of findii^ the longi-
tude at sea, in theory , is by a chronometer, or other machine, that will
measure time so exaetly true as to go uniformly correct in all .places, sea-
sons, and dimates : for, such a machine being once regulated to the
meridian of the Royal Observatory at Greenwich, would always show the
true time under that meridian, though temoved in a ship to the most
distant parts of the globe,-^even to the utmost extent of longitude.
Although such a perfect piece of mechanism can scarcely be hoped for
or expected to result from the ablest and best applied course of human
industry, — yet, on the supposition that the chronometers used at sea are
sufficiently correct for the measurement of time in short VQjfoges, we will
now proceed to show how the longitude is to be found by means of those
instruments.
PaoBLBM L
To convert «tppare»it Time in^ mean Time.
Rule.
Reduce the equation of time^ as given in page II. of the month in the
Nautical Alraanai^ to the time and place of observation^ by Problen^ V.,
page 298 ; then, let this reduced equatioit be applied to the given apparent,
time of observation, by addition or subtraction, according to the sign
expressed against it in tfie Bphemeris^ and the stam or difference will be
the corresponding mean time.
Ejcample 1.
' Jffliuary 24tb, 1S25> in longitude 75? W.^ the apparent time of observa-
tion was3U0?10' ; required the mean tame i
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416 NAITTICAL ASTRONOMY.
Equation of Ume at noon^ January 24th> == + 12?29' . 5
Correction of ditto for 8U0r 10! = . . . . +5.1
Reduced equation of time = + 12"34'.6
Apparent time of observation sr ' . • • • 3 MO? 10! . 0
Mean time, as required, = 3t52?44'.6
Example 2.
October 6th, 1825, in longitude 80? E., the apparent time of observation
was 20^ 10T40! ; required the mean time ?
Equation of time at noon, October 6th, == — 11T49|.5*
Correction of ditto for 14* 50?40! • . • . +10.5
Reduced equation of time = . . • . • — 12? 0\0
Apparent time of observation 20MO?40'.0
Mean time, as requi^d, =s •••••• 19*58?40'«0
Paoblsm IL
To amoert mean Tltne, at Oreenwichy into apparent Time.
RULB.
Reduce the equation of time, page II. of the month in the Nautical
Almanac, to the given mean time at Greenwich, by Problem V., page 298 ;
then, let this reduced equation be applied to the mean time, with a contrary
sign to that which is expressed against it in the Ephemeris; that is, by
addition when the sign is negative, but by subtraction when affirmative }
and the corresponding apparent time will be obtained,
Example 1.
January 1st, 1825, the mean titne at Greenwich, per chronometer, was
lot I3r45 ! ; required the appai^nt time ?
Equation of time at noon, January lst,*= + 3T56' • 7
Correction of ditto for 1 0* 13?45 ! = . , + 12 .0.
Reduced equation of timer:: . . . • — 4T 8'.7
Mean time at Greenwich = . . . • 10* 13T45 * . 0
Apparent time at Greeniwchs • . . 10? 9?36'.3
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OF FINDING THB 'LON6iTUD£ BY A CHRONOMETBE. 417
Example 2.
September 19th^ 1825^ the mean time at Qreenwicb^ per chronometer^
was 18M5*30! ; required the apparent time ?
Equation of time at noon, Sept. 19th, = — 6T 14 \ 3
Correction ofditto for 18*45 T30! = • • + 16 .4
Reduced equation of time =3 . • . . + 6T30'.7
Mean time at Greenwich = . . . • 18*45T30\0
Apparent time at Greenwich SB • • • 18^52? 0\7
PUOBLBM IIL
Given the LatHude qf a Place, the observed Altiiude of the Sun's Umb,
and its DecUnation; to find the Longitude of thai Place by a Chrono^
meter or Jlme-Keeper.
RULB.
Let several altitudes of the sun*8 limb be observed, at a proper distance
from the meridian,* and the corresponding times, per chronometer, noted
down ; of these take the means respectively.
Let the mean attitude of the sun's limb be reduced to the true central
altitude, by Problem XIV., page 320.
To the mean of the times of observation apply the original error of the
chronometer, by addition or subtraction, according as it was slow or fast
for mean time at the meridian of Greenwich, when its rate was established;
to which let its accumulated rate be applied affirmatively or negatively,
according as the machine may be losing or gaining, and the result will he
the mean time of observation at Greenwich, which is to be converted into
apparent time, by Problem II., page 416.
To the apparent time at Greenwich, thud^found, let the sun's declination
be reduced, by Problem V,, page 298. Then, with the sun's true central
altitude, reduced declination, and the latitude of the place, compute the
apparent time of observiation, by any of the methods given in Problem III.,
page 383 ; the difference between which and the apparent time at Green-
wich will be the longitude of the place of observation in time;— east, if the
former time be greater than the latter ; otherwise, west.
* See remarks on the most favourable times for observatioDy page 387.
2 s
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418 NAUTICAL AtrROIfOlfT.
Uote. — If the meridian of the place where the error of the chronometer
was determined be different from that of Greenwich, let its longitude in
time be applied to the mean time of observation, per chronometer, by
addition or subtraction, according as it is west or east, and the mean time
of observation at Greenwich will be obtained.
Example L
April Tthy 1825, in latitude 4S?43^ N., the mean of several altitudes of
the sun's lower limb was 9? 1 H42^, and that of the corresponding times
9^37*55!, by a chronometer, the error and rate of which had. been
established at noon, January 1st, when it was found 4?37- fast for mean
time at (SreMiwicb, and gaining 1'.75' daily; the height of the eye above
the level of the horizon was 20 feet } required the longitude of the place
of observation ?
I
Mten time of obsermtioii at Greenwich ss 9^37755!
Original error of the dironometer a • • — 4. 37
Accumulated rates 1 ' . 75 x 96 days ss . ^ 2. 48
Mean time at Greenwich s • . • . 9? 30T30!
Reduced ecpiation of time as • • • • — 2. 5
Af^Mrent time of observation at Greenw.s 9t28?26!
Sun's declination at noon, April 7th =s . 6?49'38rN.
Correction of ditto for 9* 28^25 ! = . . + 8. 53
Sun's reduced deelinatioa ss . • « . 6?58^3KN.
Obs.altof the sun'slr.limb=9?llM2r; hence, its true cent. alt. is9?i8C0f
Lat. of the place = 48?43' OfN. . , , liog, secant = 0.180599
Sun's re4|iQeddec.= 6.58.31. N, . . . Log. secant = 0, 003226
Sun's mer.z.dist. s= 41?44f29r Nat. vers. 8»atf253843
TrucaIt.ofsuii*8cen.=:9. 18. 0 Natco-V.S.s 838396
Remainder s 584553 Log.»5. 766824
Apparent time at the place of observations 5?35?20!Logjis,=5.95064.9
Apparent time of observation at Greenw.= 9. 28. 25
Longitude of die place of obs<» in time s 3!537 5!«s 58?16t5r west.
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OF FINDING THJI LONOITCTDV BT A CHRONOMBTBR. 419
Example 2.
May Ut^ 1825, in latitude 30? 15 ^ S., the mean ofseveral altitudes of Olf
sun's lower limb was 11"^ 17 -14?, and that of the corresponding times
13 ViSTlO!^ by a chronometer^ the error and rate of which were established
at noon, February Ist, when it was found 3T25? slow for mean time at
Greenwich, and losing 0*.97 daily; the error of the sextant was 2 '30^
subtractive, and the height of the eye 23 feet ; required the longitude ?
Mean time of observation at Greedmeh n » 13^23710!
Original error of the chronometer :«s • • » «f 3,25
Accumulated rate = 0'.97 x 89^ days = • i< 1.27
II I ■■■■■>■ I
Mean time at Greenwich « 13^28? 2!
Reduced equation of time =,••,.» +3.8
Apparent time of observation at Greenwich =s 13t31?10!
Sun*s declination at noon, May Ist, ss • • IS? 4< 19fN.
Correction of ditto for 13 1 3 1?10! = . . . + 10. 9
Sun's reduced declination s . . . , . 16?14'287N,
Obs. altof the sun's l.limb= 1 1 ? 17' 14^; hence, its true cent alt.\sl 1?24 ^5?
Lat, of the place s dO?15^ OrS. . . . Log. secant » 0.0685iS9
Sun's reduced dec.s 15. 14. 28 N. . . . Log. secant s 0. 015550
Sun's mer. z. dist. = 45?29^28r Nat. co-sine=701020
Truealt.of8un'8cen.=:lU24. 5 Nat. sine =5 197681
Remainder = 503339 Log.=s 5.701880
Smi's horary dist. from the merid. or noon=4*26r40!Log.ris.=5. 78097.9
Appai^nttime atthe place of observation=19^33t20!
Apparent time of observation at Greenw«=13. 31. 10
Long, of the place of observ., in time = 6* 2rl0!=90?32l30? eiwt.
2b2
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420 NAXmCAL ABTROMOMT.
Problem IV.
Given the Latitude of a Place, and the observed AUUude of a knowfi fixed
Star; to find the LongUude of the Place of Observation^ by a Chrono-
meter or Thne-Keeper.
Rule.
Let several altitudes of the star be observed^ at a proper distance from
the meridian,* and the corresponding times, per chronometer, noted down;
of these, take the means respectively. •
Let the mean altitude of die star be reduced to the true altitude, by Pro-
blem XVIL, page 327.
To the mean of the times of observation apply the original error of the
chronometer, by addition or subtraction, according as it was slow or fast
for mean time at the meridian of Greenwich when its rate was established ;
to which let its accumulatiBd rate be applied affirmatively or negatively,
according as the machine may be losing or gaining, and the restdt will be
the mean time of observation at Greenwich ; which is to be converted into
apparent time, by Problem IL, page 416.
To the apparent time at Greenwich let the sun's right ascension be
reduced, by Problem V., page 298 ; and let the star's right ascension and
declination, as given in Table XLIV., be reduced to the period of observa-
tion. Then, with the star's true altitude, its declination, and the latitude
of the place^ compute its horary distance from the meridian, by any of the
methods given in Problem III., page 383.
Now, if the star be observed in the western hemisphere, its horary
distance from the meridian, thus found, is to be added to its reduced right
ascension ; but if in the eastern hemisphere, subtracted from it : the sum,
or remainder, will be the right ascension of the meridian j from which,
(increased by 24 hours, if necessary,) subtract the sun's reduced right
ascension, and the remainder will be the apparent time at the place of
observation ; the difference between which and the apparent time at Green-
wich will be the longitude of the place of observation in time : — east, if
the computed apparent time be the greatest ; if otherwise, west.
Example 1 •
January 29th, 1825, in btitude 40?30' N.. the mean of several altitudes
of the star Aldebaran, west of the meridian, was 24?&7 ' OT, and that of the
• Sec Note, pa[fe 417.
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OF FINDING THB LOfGfTUBB BT A CHRONOMBTBR. 421
corresponding times 16?56?3t, by. a chronometer, the error and rate of
which were determined at noon, January 1st, when it was found 7*29! fast
for mean time at Greenwich, and losing 1*.53 daily; the error of the
sextant was 3' 10^ additive, and the height of the eye above the level of
the horizon 22 feet; required the longitude ?
Mean time of observation at Greenwich = , 16t56? 3!
Original error of the chronometer = • •
Accumulated rate = 1 ' • 53 x 28i days =
Mean time at Greenwich =
Reduced equation of time == • . • •
— 7.29
+ 0.44
16?49ri8:
- 13,38
Apparent time of observation at Greenwich = 16t35T40!
Sun's right ascension at noon, January 29th^ = 20 M7" 1 9 !
Correction of ditto for 16?35?40! = . . , +2.60
Sun's reduced right ascension s . , . , 20^50? 9!
Aldebaran's reduced right ascension = • • 4t25?54!
Aldebaran's reduced declination ==.... 16? 8f57*N.
Aldebaran's north polar distance = • • • • 73^51, 37
Observed altitude of Aldebaran =
24?57'0r; true altitudes- • 24°53^387
Aldebaran's north polar distances: 73. 51. 3 Log. co-secantsO. 017484
Lat of the place of observation=40. 30. 0 Log. secant ss 0. 118954
Sum= • 139? 14 Mir Constant 105. = 6.301030
Half sum =• ...... 69?37^20§'Log. cfo-sine = 9.541836
Remainders 44. 43. 42i Log. sine = . 9.847417
Star's horary distance^ west of the mer.=4?43?10!Log.rising=5. 82672.1
Star's reduced right ascension = , . 4. 25. 54
Right ascension of the meridian • . 9* 9T 4!*
Sun's reduced right ascension s . . 20. 50. 9
Apparent time at the place of observ.s 12M8T55!
Apparent time of observ. at Greenwich= 16. 35. 40
Longitude of the place of obs., in time= 4* 16T45 ! =64?1 1 : 157 w^t
? The right ucension of the meridian is to be considered w beiD|^ increaaed by 24 booing
because if is lest than the sun's reduced right ascension.
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422
NAUTICAL ASTRONOMY*
Example 2.
January 29th, 1825, in latitude 39"? 15 ^ S., the mean of several altitudes
of the star Regulus, east of the meridian, was 10?28'48^, and that of the
corresponding times 3^36*46', by a chronometer, the error and rate of
which had been established at noon, December 1st, 1824, when it was
found 4T37- slow for mean time at Greenwich, and gaining 1 ' . 17 daily;
the error of the sextant was 1 '34T subtractive, and the height of the eye
above the level of the sea 21 feet; required the longitude of the place of
observation ?
Mean time of observation at Greenwich =
Original error of the chronometer s « •
Accumulated rate=l', 17 x 59 days a
Mean time at Greenwich =
Reduced equation of time =
3*36r46*
+ 4.37
- 1. 9
8i40?U-
- 13.33
Apparent time of observation at Greenwich « 3*.26T41 !
Sun's right ascension at noon, January 29th,
Correction of ditto for3*26r41! =
Sun's reduced right ascension ts
Star's rediictd right ascension «
Star's reduced declination =:
Star's south polar distance s:
20»47?19;
+ 0.35
20t47r54t
9J59? 3!
i2?49norN
I02?49n0^
Observed altitude of Regulus =
10?28U8r; true altitude = . 10?17M6r
Regulus' south polar distance = 102. 49. 10 Log. co-secant=:0. 010962
Latitude of the place of observ. == 39. 15. 0 Log. secant = 0. II 1039
I I r ,
Sums: 152"? 2 H56r Constant log. = 6.301030
Half sum = • 76n0^58r Log. co-sifte = 9.378080
Remainders . . . . /. 65.53.12 Log. sine = . 9.960347
Star's horary distance, east of the merid.a4 f 20? 0!Log.risiiigoA« 76145.8
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OP PINDIN6 THB LOVQITUDB BT A CHRONOMBTBR. 4S3
Star's horary distance, east of the inerid.=4 120? 0!
Star's reduced right ascension =s , • 9. 59. 3
Right ascension of the meridian as , 5t39" 3!
Sun's reduced right ascension = • • 20. 47. 54
Apparent time at the place of observ. = 8?51? 9!
Apparent time of observ. at Greenwich^ 3. 26. 41
Long.of the place of observ., in time » 5{24r28! » 81?7-0? east.
PaoBLBM V.
Given the,]jatUude of a Place, and the observed Altitude of a Planet ;
to find the lAmgUude of the Place qf Observation^ by a, Chronometer or
Jlme-Keeper.
RULB.
Let several altitudes of the planet be observed, at a proper distance from
the meridian,* and the corresponding tfanes, per chronometer^ noted down;
of these take the means respectively.
Let the moMi altitude of the planet be reduced to its true centiial altitude,
by Probiaa XVL, page 325.
To the mean of the times of observation apply the original error and
the accumulated rate of the chronometer, as directed in the last Problem :
hence the mean time of observation at Greenwich will be obtained ; which
is to be converted into apparent time, by Problem II., page 416.
To the apparent time of observation at Greenwich let the sun's right
ascension be reduced, by Problem V., page 298 1 and let the planet's right
ascension and declination be reduced to the same time, by Problem VIL,
page 307* Then, with the latitude of the place, the planet's reduced
declinatioiii and its true central altitude, compute its horary distance from
the meridian, and, hence, the apparent time at the place of observation,
by ProUem v., page 397.
Now, the difference between the computed apparent time of observation
and the apparent time at Greenwich will be the longitude of the place of
observation in time;— east, if the former exceed tfie latter; otherwise,
west.
. •SstNeiSim;e4m
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424 NAUTICAI. ASTRONOMY.
Example 1.
February 4th, 1825, in latitude 39?5^ N., the meflln of several altitudes
of Jupiter's centre, east of the meridian, was 31?25^291', and that of the
corresponding times 12t6?47*9 by a chronometer, the error and rate of
which were determined at noon, January 1st, when it was found 3T7 * fast
for mean time at Greenwich, and gaining 0'.71 daily; the error of the
sextant was 1'30^ subtractive, and the height of the eye above the level
of the horizon 19 feet ; required the longitude of the place of observation ?
Mean time of observation at Greenwich = • 12 1 6^47 1
Original error of the chronometer =z • . • — 3. 7
Accumulated rate = 0\71 x 34 days =: • — 0.24
Mean time at Greenwich =: ' • 12 1 3?16?
Reduced equation of time r: ....•• —14.20
Apparent time of observation at Greenwich = 11 t48T56! .
Sun's right ascension at noon, February 4th, ri 21 M 1 T45 1
Corrcctionof ditto for 11 ?48?56! =: • ... + 1.59 '
Sun's reduced right ascension c • • • • 21 ? 13T44!
Jupiter's right ascension at noon^ February l8t,=:8M3? 0'
Correction of ditto for 3 f 1 1 *48?56 ! = • . . - 1 . 45
Jupiter's reduced right ascension = • . ^ 8 14 1 ? 15 !
Jupiter's declination at noon, February 1st, =: 19? 3' 07N.
Correction of ditto for 3n 1 ?48?56! = . . +7.34
Jupiter's reduced declination = • • . . 19?10C34rN.
Observed central altitude of Jupiter =3 1 ? 25 ' 29 f; hence, the true central
altitude of that planet is • • . 31?18'.16r
Zenith distance at time of observation zz . , 58?4 1 U4 ?
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OP FINDING THfi'LOKGirODB BY A CHRONOMETER. 425
Lat. of the place = . 39? 5'. O^N. ... Log. secantrrO- 110010
Planet's red. dec. =: . 19. 10. 34 N. . . . Log.secant=:0. 024792
Flanet'Biner.z. di8t.= ]9?54^26r .... Conat. log.=:6. 301030
Zenith dist. by obs. == 58. 41 . 44
Sum= .... 78?36:i0^Half=39?18' 5^Log.sine=9. 801678
Difference = . . . 38.47. 18 Half=: 19. 23.39 Log.8ine=:9. 521223
Jupiter's horary dist., east of the merid.=4^ 19T 5!LfOg,rising=:5. 75873.3
Jupiter's reduced right ascension = • 8. 41. 15
Right ascension of the meridian = • 4 * 22? 1 0 '.
Sup's reduced right ascension = . • 21. 13.44
Apparent time at the place of observ. = 7^ 8T26!
Apparent time of obs. at GFreenwich =11. 48. 56
Longitude at the place of obs., in time= 4*40^30! z: 70?7*30r west.
Example 2.
October Ist, 1825, in latitude 26?40' S., the m^n of several altitudes
of Saturn's centre, east of the meridian, was 10?25<40'', and that of the
corresponding times 6^36724?, by a chronometer, the error and rate of
which had been established at noon, August 1st, when it was found 3?51 !
slow for mean time at Greenwich, and losing 0' . 49 daily ; the error of the
sextant was 2^20^ subtractive, and the height of the eye above the surface
of the sea 18 feet; required the longitude ?
Mean time of observation at Greenwich =: 6*36*24!
Original error of the chronometer = . • + 3, 5 1
Accumulated rate = 0'. 49 X 61^ days . +0.30
Mean time at Greenwich =: .... 6*40"45!
Reduced equation of time = . . • • + 10. 24
Apparent time of observation at Greenwich=:6*51T 9!
Sun's right ascension at noon, Oct. 1st, = 12^29^*21 !
Correction of ditto for 6*51 r9! = . . + 1. 2
Sun's reduced right ascension == . • . 12*30T23!
Saturn's right ascension at Greenwich timers * 25? 0!
Saturn's declination at Greenwich time =: 21?4H OrN.
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426 NAUTICAL AaraovoMy*
Observed altitude of Saturn's centre = 10^25 '40^; true alt. zz 10? 14'. 12?
Zenith distance = 79?45M8r
Lat of the place = • 26?40^ O^S. . . . Log. secant=0. 048841
Saturn's declination = 21.41. ON. .'. • Log. 8ecant=0. 031872
Saturn's men z, di8t.= 48?2n 0'/
Zenith dist, by ob8.=: 79.45.48 . • . . Const. log.=6. 301030
Sum= 128? 6C48?Half:t64? 3:24rLog.sine=9.953869
Differences . . ^ 31.24.48 Half=: 15. 42.24 Log.fiine=:9. 432508
Saturn's horary dist., east of the nierid.=:4^227l5!Log.rifting=:5. 76812.0
Saturn's right ascension 13 ... 5.25 ,*»0
Right ascension of the meridian r: .It 2T4S*
Sun's reduced right ascension = .. • 12.30.23
Apparent time at the place of observ.rz IS t32T22!
Apparent time of obs. at Greenwich ;= 6. 5 1 . 9
Longitude of the place of d)6.^ in timez: 5 Ml?13! = 85?18'15?eaat.
Peoblxm VI.
Gwen the Latitude of a Place, and the observed JWtude of the^Mom^e
. Limb / to find the Longitude ofthePlace of Observation, by a Cfcno-
norneteror Time-Keeper.
RULB.
Let several altitudes of the moon's limb be.observed^ at a proper distance
frojn the meridian^* and the corresponding times^ per chronometer^ noted
down ; of these take the means respectively.
To the mean of the times of observation apply the original error and
the accumulated rate of the chronometerj as directed in Problem III.^ page
417 : the result will be the mean time of observation at Greenwich, which
is to be converted int(> apparent time, by Problem IL, page 416.
To the apparent time of observation at Greenwich let the sun's right
ascension be reduced, by Problem V., page 298 ; and let the moon's right
ascension, declination, semi-diameter, and horizontal parallax be, also,
reduced to that time, by Problem VI., page 302. To the moon's reduced
s^mi-diameter apply t&e augmentation. Table IV., and the true semi-
diameter will be obtained.
• See Note, paje 417.
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OF FINDING THB LONGITUDB BY A CHRONOMBTBR. 427
Let the mean altitude of the moon's limb be reduced to the true central
altitude, by Problem XV., page 323.
Then, with the latitude of the place, the moon's corrected declination,
and her true central altitude, compute her horary distance from the faieri-
dian, and, hence, the apparent time at the place of observation, by Problem
VI., page 400. The difference between the computed apparent time of
observation and that at Greenwich, will be the longitude of the place of
observation in time ; — and which will be east, if the computed time be the
greatest 3 if otherwise, west.
Example !•
April 21st, 1825, in latitude 50? 48' N., the mean of several altitudes of
the moon's lower limb, west of the meridian, was 29?30^26T, and that of
the corresponding times 12ti6?58?, by a chronometer, the error and rate of
which had been established at noon, February 1st, when it was found 7*^46!
fast for mean time at Greenwich, and losing 6' . 79 daily ; the error of the
sextant was 2 '25? additive, and the height of the eye above the level of
the horizon 17 feet; required the longitude of the place of observation ?
Mean tfme of observation at Greenwich = 12* 6*58!
Original error of the chronometer s . . — 7^ 46
Accumulated rate S5 0\ 79 x 79| days = +1.3
I
Mean time at Greenwich = • • • • 12? 0?I5!
Reduced equation of time s , . . , + 1 . 27
Apparent time of observ. at Greenwich ss 12 1 l?42t
Sun's right ascension at noon, April 2l8t,= 1?55'!'41 '. 5
Correction of ditto for 1 2 * 1 T42 ! = . • 4-1,52.3
Sun's reduced right ascension = • . • 1*57 "33 ' . 8
Moon's semi-diameter at midnt., April 2l8t^ 15 f 14T*
Augmentation of ditto. Table IV., » * . . 4- 7
Moon's true semi-diameter s • • • • • 15'2K
* The apparent time at Greenwich belo^ so rery close to mldnifht, and the Tariation In
the nooa'e Jeriiaatioa/ — H-dwaeter, and horiaontal parallax but triHio;, no comctkui
lor these elements becomes necessagr in the present instance.
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428 NAUTICAL A8TRONOMT,
Moon's right ascension at midnight, April 2l8t s 70?57'59^
Corrected proportional part of ditto for 0MT42! s= -f 0. 56
Moon's corrected right ascension = . . • . 70*? 58' 55 T
Moon's declination at midnight, April 2l8t, = • 23? 9:28rN*
Moon's horizontal parallax at midnight, April 2l8t 7= 55'54T*
Observed altitude of the moon's lower limb=2d?30'. 26^'; hence, her true
central altitude is '30^31 '. 191.
Lat. of the place • . 50?48^ OfN. • . , Log. 6ecant=0. 199263
Moon's corrected dec,=23. 9.28 N. . . . Log. secant=0. 036483
Moon's men zen.dist.=27?38'32rNat.yers.S.sI 14138
Moon's true cent. alt. =30.31. l9Nat.co-V.S.=492131
Remainders 377993 Log.=5. 577484
Moon's horary dist, west of the mer.= 4?38?10! Log. rising=5. 81323.0
Moon's corrected right ascension =
70?58;55r,inUme= .... 4.43.56
Right ascension of the tneridian = • 9? 22? 6 !
Sun's reduced right ascensioa ^ • • 1 . 57. 34
Apparent time at the place of observ.=7*24T32!
Apparent time of obs. at Greenwich^: 12. 1. 42
Longitude of the place of obs., intime=4J37"10! = 69'?l7'30r west/
Example 2.
September 2d, 1825, in latitude 40? 10' S., the mean of several altitudes
of the moon's lower limb, east of the meridian, was 9?8'36?, and that of
the corresponding times 6t39T0!, by a chrpnometer, the error and rate of
which were determined at noon. May 1st, when it was found 4?10! slow
for mean time at Greenwich, and gaining 1'.37 daily; the error of the
sextant was 1 ', 20'' subtractive, and the height of the eye above the surface
of the sea 14 feet ; required the longitude ?
* See Note; pa^ 427.
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OF FINDING THB LONGITUDE BY A CHRONOMSTER. 429
Mean time of observation at Greenwich =: • • 6t39? 0'
Original error of the chronometer = . • ' • + 4. 10
Accumulated rate = 1'.37 x 124iday8s • — 2.50
Mean time at Greenwich = 6*40r20!
Reduced equation of time ss + 0. 33
Apparent time of observation at Greenwich »= 6M0?53!
Sun's right ascension at noon, September 2d, ^ 10?44T53*. 5
Correctionof ditto for 6*40?53! = . . • + 1. 0 .5
Sun's reduced right ascension = . . • . 10*45 T54 ' . 0
Observed altitude of the moon's lower limb = 9?8^36^ ; hence, the true
central altitude of that oliject is 10?6^23r.
Moon's right ascension at noon, September 2d,=:S0?5i3' 8^
Ck)rrected prop, part of ditto for 6?40r53! = + 3. 19. 9
• Moon's corrected right ascension*=d . • • 34? 17* 17^
Moon's declination at noon, September 2d, = 1 6 ? 1 2 ' 13 f N.
Corrected prop, part of ditto for e*40r58! = + 54. 3
« ■ 111 !■■
Moon's corrected declination = . . . • .17? 6'16rN.
Moon's semi-diameter at noon, September 2d2 = 14^46f
Correctionof dittofor 6*40T53! = . . • . 4-1.
Augmentation, Table IV., = ....... +2
Moon's true semi-diameter = • . 14^491^
Moon's horizontal parallax at noon, Sept. 2d, = 54'. 11 •
Correction of ditto for 6 M0To3' = .... +4
Moon's true horizontal parallax = . . . . . 64' 15^
Lat. of the place . . 409 10' OlS. . . . Log. secant=0. 1 16809
Moon'8correcteddec.= 17. 6. 16 N. . . * Log. secant=0. 019647
Moon's mer. z. dist. = 57?16'. 16rNat,vers. S.=459.336
Moon's true cent, alt.= 10. 6.23 Nat. co.V.S.=824524
* Remainder = 365 188 Log.sS. 5625 16
Moon's hotary dist., east of the merid.=4 * OT 01 Log. rising=5. 69897. 2
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430 MAirncAL astronoiit*
Moon'8 horary dist, east of the inerid.rs4 1 0? 0?
Moon's reduced right ascension
349l7'l7?,intime« • • . . 2.17. »
Right ascension of the meridian as • 22 M 7* 9'
Sun's reduced right ascension s= • 10. 45. 54
Apparent time at the place of ohsery.=l 1 1 31 715 !
Apparent time of obs. at Greenwich = 6. 40. 53
Longitude of the place of obs., in time=4!50r22! = 72?35 ^SOr east-
Remark 1. — ^The longitude, thus deduced from the true central altitude
of the moon, will be equally as correct as that inferred fipom the mm's
central altitude, provided the moon's place in right ascension and declina-
tion be carefully corrected by the equation of second difiference, as explained
between pages 33 aad 38. Whatever little extra trouble may be attendant on
this particular operation, will be infinitely more than counter-balanced by the
pleasing reflection that it affords the mariner an additional method of find-
ing the longitude of his ship, either by night or by day, with all the accu-
racy that can possibly result from the established rate or going of hia
chronometer.
Remark 2.— It frequently happens at sea, that, owing to clouds, rains,
or other causes, ships are whole days without profiting by the presence of
the sun,, or obtaining an altitude of that object for the purpose of ascer-
taining either latitude or longitude ; but it must be remembered, that there
are few tiights, if any, in which some fixed star, a planet, or the moon,
does not present itself for observation, as if intended by Providence to
relieve die mariner from the great anxiety which the doubtAil position of
his ship must naturally excite in him, particularly when returning from a
long voyage, and about to enter any narrow sea, such as the English Chan-
nel. -Under such circumstances, the three preceding problems will be
found exceedingly useful; because they exhibit safe and certain means of
finding the true place of a ship, so far as the going of the chronometer
used in the observation can be depended upon. In this case, since a know-
ledge of the heavenly bodies becomes indispensably necessary, the reader
is Kfeired to '^ The Young Navigator's Guide to the Sidereal and Planetary
Parts of Nautical Astronomy," where a familiar code of practical directions
is given for finding out and knowing al! the principal fixed stars and planets
in the firmament.
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OF FINDING THX LOMGITODB BT tOKAR OBSERVATIONS. 431
PaoblbK VIL
To Jind th0 JLongiiude qfa SUp or Place by celestial ObMrvaltmiy com^
mimhf coiled a hmar ObeervQtiom.
The direct progressive motion of a ship at sea is so liable to be disturbed
by various unavoidable and often imperceptible causes,— such as a frequent
aberration from the true course, by the ship's continually varying a little,
in contrary directions, round her centre of gravity ; high seas with heavy
swells, sometimes with and at other times against, or in directions oblique
to the true course ; storms, sudden shifts of wind, unknown currents, local
magnetic attraction, unequal attention in the helm^-men, with many other
casualties which cannot possibly be properly provided for, — that the place
indicated by the dead reckoning is frequently so erroneous as to be whole
degrees to the eastward or westward of the actual position of the ship. Of
this every person must be fully aware, who has navigated the short run
between England uid the nearest of the West Indian Islands.
As the best account by dead reckoning is evidently but a very imperfect
kind of guess-work, it should be employed only as an amdliary to the
elementary parts of navigation, and never confided in but with the utmost
<»ution. Hence it is that celestial observation should be constantly
resorted to, because it is the only certain \^ay of detecting the errors of
dead reckoning, and of ascertaining, with any degree of precision, the
actual position of the ship.
If a chronometer or time-keener could be so constructed as to go
uniformly correct in all seasons, places, and climates, it would immediately
obviate all the difficulties attendant on a ship's reckoning, atid thus render
the longitude as simple a problem as the latitude ; for, such a machine
being once regulated to the meridian of Greenwich, would always show
the absolute time at that meridian ; and, hence, the longitude of the place
of observation, as has been illustrated in the four preceding problems; but
those pieces of mechanism are so exceedingly complicated, and sd extremely
delicate, tiiat they are liable to be affected by the common vicissitudes of
seasons and climates, and also by any sudden exposure to a higher or lower
d^ree of atmospheric temperliture than that to which they have been
ftccvstomed : the celestial bodies ought, therefore, to be consulted, at
all times, in preference to machines so subject tonnitabitity, and should
ever be confided in by the mariner, as the only immutable and unerring
time- keepers.
Of all the apparent motions of the heavenly bodies, in the zodiac, with
which we are acqucdnted, that of the moon is by far the most rapid ; it
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432 NAUTICAL ASmONOMY.
being, at a mean rate, about 13? 10^ in 24 hours, or nearly half a minute
of a degree in one minute of time. Hence, the quickness of the moon's
motion seems to adapt her peculiarly to the measurement of small portions
of corresponding time; and, therefore, careful observations of the angular
distance of that object from the sun, a plan.et, or a fixed star lying in or
near the zodiac, afford the most eligible and practicable means of deter-
mining the longitude of a ship at sea : for the true distance deduced from
observation, being compared with the computed distances in the Nautical
Almanac, will show the corresponding time at Greenwich ; the difference
between which and the apparent time at the place of observation will be
the longitude of that place. in time; and which will be east if the time at
the place of observation be greater than the Greenwich time, but west if it
be less.
The method of finding the longitude at sea, by lunar observations, is
very familiarly explained, by geometrical construction and by spherical
calculation, in '^The Young Navigator's Guide to the Sidereal and
Hanetary Parts of Nautical Astronomy," between pages 172 and 212^
where it will be seen that in a lunar obser\'ation there are two oblique
angled spherical triangles to work in, for the purpose of finding the true
central distance; in the 'first of which the three sides are given^ viz.^ the
apparent zenith distances of the two objects, and their apparent central
distance, to find the angle at the zenith,— that is, the angle comprehended
lietween the zenith distances of those objects ; and, in the other, two sides
and the included angle are gi^n, to find the third side, viz., the true zenith
distances of the objects ; and their contained angle, to find the side oppo-
site to that angle, or the true central distance between those objects. The
solution of the first triangle, falls under Problem V., page 207> and that of
the second under Problem III., page 202. This is the direct spherical
method of reducing the apparent central distance between the moon and
sun, a planet, or a 'fixed star, to the true central distance ; or, in other
words, diat of clearing the apparent central distance between those, objects
of the effects of parallax and refraction : but, this being considered by some
mariners as rather a. tedious operation, the following methods are ^ven,
which, being deduced direcdy from the above spherical principles, will be
always found universally correct ; and, since they are not subject to any
restrictions whatever, they are genei^al in every case wliere a lunar observa-
tion can be taken. Besides this, they will be found remarkably simple and
concise, particularly when the operations are performed by the Tables
contained in this work.
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OF FINDING THB IX>NGITUDB BT LUNAR OBSBRVATIONS.
43?
JMbthod I.
* . . .... t
Cf reducing the apparent to the true central Distance.
Rulb. •
Take th^ auxiliary angle from Table XX.,. and let it be corrected for the
sun's, bUht% or planet's apparent altitude, as directed in pages 41 and 45.
Find the difference of the .apparent altitudes of the objects, and, also, the
difference of their true altitudes.
Then, to the natural versed sines supplement of the sum and the differ^
en^e of the auxiliary angle and the .difference of the apparent altitudes,
add the natural versed sines of the sum and the difference of the auxiliary
angle and the apparent distance, and the natural versed sine of the differ-
ence of the true altitudes :. the sum of these five numbers, abating 4 in the
radii or left-hand place^ will be the ntttural versed sine of the true central
distance.
•
Example 1.
Let the apparent central distance between the moon and sun be
6fi?48'S41f, the sun's apparent altitude 60? IS'.SSr, the lAoon's ^parent
altitude 17? 15* 151^, and her horizontal parallax 59M3^ $ required the
true central distance ?
Sun'^ apparent alt. = 60?15'.35r-Correc. 0C29^=true alt.=:60?15C 6r
Moon's apparent art.=: 17* 15. 15+Correc. 54. 0 =true alt.=:18. 9. 15
Diff. of the app: alts.sr43? 0^20"!^
Auxiliary angle = . 60. 9.* 27
Apparentcentraldist.r=66. 48, 34
Diff. of the true alts. = 42? .5'5K
Suip of auxiliary' angle
anddiff.ofap.alts.=:103? 9^47^
difference of ditto .== 17. 9. 7
Sum aux.ang.&ap.dist.126. 58. I
Difference of ditto = 6.39. 7
Diff. of the true alts.= 42. 5.51
Tme central di8tance=66? 2'20r Nat. versed sine = * 0,593882
2 F
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Nat. versed sine
sup. r=
0.772277
Nat. Versed sine
sup. =s
1.955526
Nat. versed sine
= .
1.601354
Nat. versed sine
•-™ • •
0.006730
Nat. versed 'sine
"~ •
0.257995
434
NAUTICAL ASTRONOMY.
General Remarks.
1.' The correction of the moon's apparent altUude is contained in Table
XVIIl., and is to be taken out therefrom agreeably to the directions given
in page -39.
2. The correction of the sun's apparent altitude is the difiference between
the refraction and the parallax Corresponding to that altitude in Tables
Vlll.andVIl.
. 3. The correction of a planet's apparent altitude is the difference between
the refraction and the parallax answering to that altitude in. Tables VQI.
aUdVI. And,
4. The correciuni (ff a starts apparent aUitude is the refraction corre-
sponding thereto in Table VIIL' The fixed stars have not any sensibk
parallax.
Matample 2. .
Let the apparent, central distance between the moon and a fixed star be
37^12'40'r, the star's apparent aUitude H?27'50?, thempon's apparent
altitude 40?55'. 15?, and her horizontal parallax 54' 10"?^ required the
true central distance ? • '
•Star'« apparent alt.= ll?27<50r.^Correc. 4f35r=truealt.= ll?23M5r
M'oon'sapparentalt,=s40.55. 15 +Correc.39.51 =true alt.=41.3S. 6
f^ II - -•
Difr.oftheapp.alt8.=39?27'25? Difference ofthetruealts.=30?l 1^51 *:
Auxiliary angle= . 60* 19, 30
Apparent cent. dist.=37. 12^ 40
Sum of auxiliary angle
and diflF.of ap.alUi-=89?46^55?
Difference of ditto =30. 52. 5
Sumaux.^ilg.&ap.dist.97.32. 10
Difference of ditto =23. 6.50
Nat. versed sine sup.
Nat. versed ^inc sup. .
Nat. versed sine =
Nat. versed .sine =
Diff.of the true alts.=?30. 11.51 Nat. versed sine ss
1.003806
1.85835^2
1.131151
0.080274
0. 135703
True central dist. =37°44^52? Nat. versed sine = . . 0.209286
Remark 1. — InsteaH of the natural versed aiiies supplement of the fipt
two terms in the calculation, the natural versed sines of the supplements of
those terms to 180? may be taken : for it is evident that the natural
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OF FINDING TUB LONGITUDE BT LUNAR OBSBRVATIONS. 485
versed sine of the tupplefitent of an arch is the natural versed sine supple-
ment of that arch. Thus^ in the above example^ the supplement of
89?46^55r is 90^3^5^, the natural versed sine of which is 1.003806;
and the supplement of 30?52^5? is 149^7 -55?^ the natural versed sine of
which is 1. 85 835 2, the same as above. By this transformation of the first
two terms, all the tabular numbers that enter the calculation will become
affirmative.
Remafk 2.-«rWhen the sum of the auxiliary angle and the apparent
centra] distance exceeds a semi-circle, or 180 degrees, the natural versed
sine supplement of its excess labove that quantity is to be taken, or, which
is the same thing, the natural versed sine of its supplement to 360 degrees.
Remark 3>-Instead of using the natural versed sines supplement of the
first two terms in the calculation, as above, or the natural versed sines of
their supplements to ISO?, as mentioned in Remark 1, the natural versed
sines of those terms may be employed directly 3 as thus : — Let the sum of
the natural versed sines of the first two terms be subtracted from the warn
of the natural versed sines of the last three terms, and the remainder will be
the natural versed sine of the true distance. "^
Example.
Let the apparent central distance between the moon and sun be
119?53^58?, the sun's apparent altitude 22?10C35T, the moon's apparent
altitude 15^51:22^, and her horizontal parallax 58U0r ; required thf true
central distance ?
Sun'sapparentalt. «22?10'.3.5r--Corr^. 2'! K« true alt =22? 8C24r
Moon'sapparentalt.= 15.51.22 +Corree;53, 7 Fstruealt.sl6.44.29
Diff.ofapparentalts.=i 6?19n3r Diff. of true altitudes s 5?23^6Sr
Auxiliary angle = . 60. 8.25
Apparent cent. diat.=s 1 19. 53. 58
Bum of the aux. angle
anddiff.ofap.alts.=66?27'38r Nat.V.S.^0.600620> « _, 01029«
Difference pf ditto ^ 53. 49. 12 Nat.V.S.=0. 409676) ^°*^ *' ^^
Sum of auxiliary angle
andapp.dist. = 180. 2.23 Nat.V.S.=2.000000l
Difference of ditto= 59. 45. 33 Nat.V. S.==0. 496365 >Sum=2. 500800
Diff. of true alts. = 5.23.55 Nat.V.S. =0.004435)
True cential dist. = 1 ig*? 22 i 25 r Nat. versed sine = . . . 1 . 490504
2f2
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436
NAUTICAL ASTRONOMY.
Mbthob II.
Of redttdi^ tA6 opporcnt to tAe !rtt6 centra! IKfiance.
RUJLB.
Take the auxiliary angte from Table XX., and let it be corrected for the
ftun's, star's^ or placet's apparent altitude^ as directed in pages 44 and 43.
Find the sum of the apparent altitudes of the objects, and, ako, the sum
of their true altitudes ; then,
To the natural versed sines of the «um and the difference of the auxiliary
angle and the sum of the apparent altitudes, add the natural versed sines of
the sum and the difference of the auxiliary angle and the apparent distance,
and the natural versed sine supplement of the sum of the true altitudes ; the
sum of these five terms, abating 4 in the radii or left-hand place, will be
the natural verised sine of the true central distance.
Example 1.
Let the apparent* central distance between the moon and Venus be
53?49j:54r3the apparent altitude of Venus 19?10<40r, and her horizontal
parallax 23T ; the moon's apparent altitude 37?40^20^, and her horizontal
parallax 59^47? ;'reqiiired the true central distance 7
Venus' apparent alt.=:19?10:40^-Correc. 2^2K=truealt.= 19? 8^9^
Moon's apparentalt.=37. 40. 20 +Ck>rrec. 46. 5 ^true alt.=38. 26. 25
Sum of the app. alts.s 36? 5 1 ' Or Sum of the true alts, s
Auxiliary angle = 60.20.14
Apparent cent. dist.= 53. 49. 54
57^34 U4r
Sum of auxiliary angle
&sumofap.alts.=:117?lIM4r
Difference of ditto = 3. 29. 14
SumauxAng.&ap.dis.ll4. 10. 8
Difference of ditto =: 6. 30. 20
Sum of the true alts.s=:57. 34. 44
Nat. versed sine =
Nat. versed sine =
Nat. versed sine =
Nat. versed sine =
Nat. versed sine sup<
1.456899
0.001851
1.409423
0.006439
1.536138
0.410755
True central dist. = 53?53M8r Nat. versed sine s .
Note.'^'For the correctioas of the apparent altitudes of the objects, see
remarlbs, pa^ 434.
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OF FINDING THB L0N6ITUDB BY LtTNAR OBSERVATIONS.
487
Example 2.
Let the apparent central distance between the moon an4 sun be
119?57'56^, the sun's apparent altitude 18?10'50'', the moon's apparent
altitude 10?30nOf, and her horizontal parallax 60^37^; required the
true central distance ?
Sun's apparent alt.= 18?10^50r-Correc. 2M4'r=truealt.= 18? 8'. 6t
Moon's app^ent alt.s 10. 30. 10 +Correc. 54. 35 =strue alt= 1 1 . 24. 45
Sumoftheapp.aIts.=2.8?41^ Oir Sum of the true alts. = 29?32'51^
Auxiliary angle a • 60. . 5. 34.
Appar. central dist.= 119.57-56
Sum of auxiliary angle
& sum of app. alts.=:88?46^34r
Difference of ditto = 31. 24. 34
Sum aux.ang.&ap.dis. 1 80. 3. 30
Difference of ditto = 59. 52. 22
Sum of the true alts.=29.32.51
Nat. versed sine =3 • .
Nat. versed sin^ = • .
Nat. versed sine ^ • «
Nat. versed sine = • <
Nat. versed sine sup. ==
True central dist. = 1 19?33C 4r Nat versed sine
0.978640
0.146535
1.999999
0.498076
1.869948
1.493200
Remark 1. — Instead of using the natural* versed sine supplement of the
sum of the true altitudes, the natural versed sine of that term may be
employed : in this case, if from the sum of the natural versed sines of the
first four terms in the calculation, the natural versed sine of the last term
be taken, the remainder, abating 2 in the radii or left-hand place^ wiU be
the natural versed sine of the true central distance.
Hemark 2. When the sum of the auxiliary angle and the apparent cen-*
tral distance exceeds a semi-circle, or 180% the natural versed sine supple-
ment of its excess above that quantity is to be tak^n, or, which amounts to
the same, the natural versed sine of its supplement to 360?, as in the above
example. The same is to be observed in the event of the aggregate of the
auxiliary anglQ and the sum of the apparent altitudes exceeding 180
degrees : this, however, will but very rarely happen, .
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498 IIAUTICAL ASTRQNOinr*
mbthod hi.
Of reducing the apparent to the true central Distance.
Rule.
Take the logarithmic difference from Table XXIV., and let it be cor-
rected for the tun's, star's, or planet's. apparent altitude, aa directed in
pages 49, 51, and 52.
Find the difference of the apparent altitudes of the object^, and, also,
the difference of their true altitudes.
.Then, from the natural versed* sine of the apparent distance, subtract
the natural versed sine of the difference of the apparent attitudes j tp the
logarithm of the remainder let the logarithmic difference be added, and the
sum (abating 10 in the index^ wil) be the logarithm of h natural number;
which, being added to the natural versed sine of the difference of the true,
altitudes, will give the Natural versed sine of the true central distance.
£4faf^ple 1.
Let th9 apparent central distance between the moon and Mara be
83'?10'.23r, the apparent altitude of Mars 17?10^20r, and his horizontal
pi^all^x 15? ; the moon's apparent altitude 31 ?20'.30T, and her horizontal
parallax 58' 531f | required the true central distance ?
Mara' apparent alts l7?10'20?-^Ck>rrec. aUfir=true alusl?? 7'3ir
Moon's appt. alt. = 31. 20. 30 +Correc. 48. 44 stru« alt,:=:82, 9. 14
Diff.ofappar.alts.= 14?10M0r^ Diff. of true altitudes =: 15? lM3r
Apparent disuneet: 83?10^28rNat.V.S.=:881129
Diff« of appar. alto.:3 14. 10. 10 NatV. 8.2^030436 tog. diff«=39. 996299
Remainder =: 850693 Log. =s 5. 929773
Natural number = 843476 Log. =: 5.926072
Diff. of the true alts.=15? lUSrNat.V.S.=034204
True central dist = 82?58^26rNat.V.S.=877680
Nofe.— For the corrections of the apparent altitudes of the objects, see
remarks^ page 434.
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OF FINDING THg tONGITyBB BY LUNAR OBSBRVATIONS. 439
Example 2.
Let the apparent central distance between the moon and sun be
I18?56'40^, the sun's apparent altitude 16?40flOf^ the moon's apparent
altitude 9?39'50f^ and her horizontal parallax 59' 19r i required the true
centra] distance ?
Sun's apparent alt* a 16? 40n or -Correc. S^ Ors:truealt.sl6?37nOr
Moon's apparent altss 9. 39. 50 +Correc. S9. 3 «:trae alt,ss 10. 32. 53
Diff. of the app. alts.= 7? 0' 20r Diff. of the true alts. » . 6? 4^ 17r
Apparent distance =U8?56^40r N.V.S.= 1.483961
Diff. of appar. alts. = 7. 0,20 N.y.S.= .007466Log.diff.=9. 998919
Remainder = 1. 476495 Log. = 6. 169232
Natural number = ^ . 1.472825 Log. a» 6. 168151
Diff. of the true alt8.:ce* 4M7?Nat.V.S.= .005609
True central dist. =118°35C lrNat.V.S.= 1.478434
Mbthod IV.
Of reducing the apparent to the true central Distance.
RULB. '
Take the logarithmic difference from Table XXIV., and let it bo cor-i
racted for the sun's, star's, or planet's apparent altitude, as directed in
pages 49| 51, and 52.
Find the sum of the apparent altitudes of the objects, and, also, the sum
of their true altitudes 3 then,
From the natural versed vine supplement of the sum of the apparent
altitudes, subtract the natural versed sine of the apparent distance; to the
logarithm of the remainder let the logarithmic difference be added, and
the sum (abating 10 in the index,) will be the logarithm of a natural num-i
ber; which, being subtracted from the natural versed sine supplement of
the sum of the true altitudes, will leave the natural versed sine of the true
central distance.
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440 NAUTICAL ASTRONOMY.
Example 1.
Let the apparent central distance between the moon and sun be
1 10?53^34'r, the sun's apparent altitude 38?1 1 '59?^ the moon's apparent
altitude 15?51'22'r^ and her horizontal parallax 58 MO''; required the
true central distance 7
Sun's apparent alt.=38? 1 1^59^ - Cofrec. 1 ' IS^^-true alt=38?10tS4r
Moon's appar. alt.=: 15. 5 1 . 22 -^ Correc. 53. 7 =tnie alt.== 16. 44. 29
Sumoftheap.alts.=:54? 3^2K Sum of the true altitudes = 54?5S: 23 r
Sumofap.aIt8.=54° 3 :2KNat.V.S.suj).= 1.586997
Appandist. = 110.53.34 Nat. vers. S.= 1. 356620 Log.diff.=^9. 998150
Remainders: . 230377 Log. = 5.362439
Natural number = . 229398 Log. = 5.360589
Sum of true alts. 54?55^23rNat.V.Su8up.=:l. 574676
Truecent.di8,= 110?ir.56rNat. vers. S.= 1.345278
^o^e.— See remarks^ page 434, relative to the corrections of the
apparent altitudes of the objects.
Example 2.
Let the apparent central distance between the moon and a fixed star be
4^11^7^, the star's apparent altitude 43?10'.20r, the moon's apparent
.altitude 56?48'16^, and her horizontal parallax 59'.25T; required the
true central distance ?
Star's apparent alt.=43? 10^ 20r-. Correc. 1'. ir=true alt=439 9U9^
Moon's appar. alt.s:;56. 48. 16 + Correc. 31, 56 strue alt.=57. 20. 12
Sumoftheapp.alt8.=99958'36^' •Sum of the true altitudes=!00?29^3ir
Sumoftheapp.alt8.=99?58'36r N.V.S.8up.= 826753
App.centraldist.^ 41. 11. 7 N. vers. S.=2474 16 Log,diff.=9. 998895
Remainder = 579337 Log. = 5.762931
Natural number = . 571250 Log. = 5.756826
Sumoftruealt8.=100^29^3KNat.V.S.fiup.=^817902
True cent. dist. = 41° 7- 8^Nat, vers. S.= 246652
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OP FINDING TM£ L0N6ITUDB BV LUNAR OBSBRVATIONS* 441
Method V.
Ofredudng tlie apparent to the true central Distance.
Rule.
To die logarithmic sines of the sum and the difference of half the
apparent distance and half the difference d( the apparent altitudes^ add the
logarithmic differencie, Table XXI V.^ and the constant logarithm 6. 301030:
the sum of these four logarithms (rejecting 30 in the index,) will be the
logarithm of a natural number; which, being added to the natural versed
sine of the difference of the true altitudes^ wiH give the natural versed sine
of the true central .distance;
Example 1.
Let the apparent -distance between the moon and a fixed star be
37?56'43^, the star's apparent altitude 19?32'^ th^ moon's apparent alti-
tude 56?33', and her horizontal parallax 61 ' 161} required the true cen-
tral distance ?
Star's apparent alts 19?32: Or-Corrcc. 2U0?=true alt=19?29'20r
Moon'sappar.alt.^ 56.33. 0 -t-Correc. 33. 9 ^tmetiu^SJ. 6. 9
Diff.oftheapp.alt8.=?37? V. Or Diff. of the true altitudes^: 37^36^49?
Half diff. of ap. alts.=: 18?30:30r
Half the appar« dist.= 18.58.21^
Log. diff. = .... 9.993713
Sum= • . . . 3r?28'51jr Log.^sine = .... 9.784259
Differences , . 0.27.51^ Log.^sine = . , • . 7.908677
Constant log. ±: . . • 6.301030
Natural number s= «•.......« 9720 Log,s3. 987679
Diff.of thetruealts.=^37?36:49r Nat. vers. S.=207855
True central dist. =s 38?31'. ir Nat. vers. S.=2 17575
^Urample 2.
Let the apparent central distance between the moon and sun be
L06?22M8r^ the sun'e d^iparent altitude 39'?25^^ the moon's apparent
altitude 19?56^^ and her horizontal parallax 58^0^; required the true
central distance ?
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442 VAxmcih Awnmour.
Sun's apparent alt. = 39?25^ 0^— Correc. 1^ 3r=trucalt,=39?23:57^
MooD'sapparentaIt.= 19.56. 0 +Conrec. 51.56 =:truealt.=20.47.56
Diff. of the app. alu.=s: 19?29' Or Diff, of the true altitudes^^lS^SG^ i '.
Half diff. of app; alt8.= 9944C30r
Half the app. di8t« = 53.11.24
Liog,diflF. =; . • , • 9,997672
Sums . . . . 62?55^541f I^.aines^ ^ . . , 9.949616
Differences: . . 43.26,54 Log. sine =5 . , • , 9/837399
Constant log. =: . . , 6,30103U
Natural number =s • 1. 218196 Log.=6^ 085717
Diff. of the true alts.= 18?36' ' 1 ^Nat. vers. S.= 052234
True central dist.=105?41^24rNat.ver8.S.= l. 270430
Mbthod VI.
Of reducing the apparent to the true central Distance.
Rule.
To the logarithmio cocaines of tho sum and the diffarenca of half the
apparent distance and half the sum of the apparent altitudes^ add the
logarithmie diffnence^ Table XXIV., and the constant logarithm 6. SOI 090:
the sum of these fouir logarithms (rejecting 30 in the index,) will be the
logarithm of a natural number; which, being subtracted from the natural
versed sioe supplement of the sum of the true altitudes, will leave the'
natural versed sine of the true distance.
. Example 1.
Let the apparent central distance between the moon and a fixed star be
69?21'.25r, the star's apparent altitude 27tS2{S7?, the moon's apparent
altitude 22?28'56r, and her horizontal parallax 56 M 7^; required the
true central distance ? •
Star's apl)arent alt.=;27?32:37?-Corwc. 1 U9r=true alt.=27?30:48T
Moon's appar. alt.= 22. 28. 56 +Correc. 49.43 =true alt.=:23. 18.39
Sttmoftheap.alts.= 50? 1^33? Sum of the tfue4ltitudes^50?49C27r
Half8umofap.alt8,=25? 0M6ir
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OF FINDING THB liONQITUDB BT LUNAR OBSERVATIONS. 443
U«lf9iimofap<alt8.c:a5? 0'A^1
Halfiip.coiit.di8t,:;: 34.40.42^ .
Log. diflF. =: • • . , 9. 997468
Sum=. . , , 59M1C291: Log. co-sine = , , . , 9.702997
PiflFerence = ^ . 9.39.56 Log.co-6ine= , . ,., 9.993791
Constant log. = . , . 6. 301030
Natural number z=: . , . 989204 Log.=:5^995286
Sumof truealte.=50?49',27* Nat.V,S.iup,=:l. 631703
True cent. di8t.=69? 3' 1 1 jrNat. vers. S. = . 642499
Example 2.
Let the apparent central distance between the moon and Jupiter be
116'?40'.28^, Jupiter's apparent altitude. 10^40' 20r, and his horisontal
parallax 2*!, the moon's apparent altitude 15?10'30'!f, and her horizontal
parallax 59^ 13? ; required the true central distance 7
Jupitey's appar. aU.= 10?40'.20':-Correc. 4:54'/fctnie alt,= 10.?35^26r
Moon's appar. alt. = 15. 10. 30 +Correc. 53. 41 =true alt.^^16. 4. 1 1
Sum c^theap. alts, s 2^5?50'.50r Sum of the true altitudes sc 26?S9f37r
Halfsum of ap. alts.=12?55:25?
Half app. ccfnt. dist.s 58. 20. 14
— i».*~^ Logidiff. « 9.998220
Sum = . . .. 71^15^39? Log. cQ-sine =i . . • • 9.506857
Difference =' . . 45. 24. 49 Log. co-sine ?=.... 9. 846327
Constant log. » .... 6.301030
Natural number s ^ ...... . « 449194 Log.=5. 652434
Sum of true alts. =26? 39: 37^ Nat.V. Si sup.ss 1 . 893683
True cent. dist=; U6?23 ' 26? Nat. vers. S. s= 1 . 444489
Method VIL
Cf reducing the apparent io the true central Distancei
JlULB.
To the apparent central distance add. the apparent altitudes of the
cribjects^ and take half the ^um; the difference between which and the
apparent distance> call the remainder ^ then.
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444 KAtJTicAL ASTRoyomr.
To the logarithmic difference^ Table XXIV., add the logarithmic co-aines
of the above half sum and remainder : the amn of these three logarithms
(rejecting 20 in the index,) will be the logarithm of a natural number.
Now, twice this natural number being subtracted from the natural versed
sine supplement of the sum of the true altitudes, will leave the natural
versed sine of the true central distance.
Smtmrks, — If the remaining index of the three logarithms (after 20 is
rejected) be 9, the natural number is to be taken out to. six places of
figures; if 8, to five places of figures ; if 7^ to four places of figures ; if 6,
to three places of figure8,^and so on.
The logarithmic difference is - to be corrected for the sun's, star's, or
planet's apparent altitude, as directed in pages 49, 51, and 52 ;*-this, it is
presumed, need not l>e again repeated.
Example 1.
«
Let the apparent distance between the moon and a fixed, star be
48?20'2K, the star's apparent altitude 11 ?33' 29^, the moon's apparent
altitude U?10'35?, and her horizontal parallax 55 '32^; required the
true central distance ?
Star's apparent alt = ll?33^29r-Correc. 4^33^=:truealt.=;ll?28^56r
Moon's appar. alt. ss 11.10.35 -f Correc.49.46 =truea]t.=3l2. 0.21
Appar. central dist.= 48. 20. 21 Sum pf the true altitude»=23?29^ 17^
Sum= . . . . 71? 4^25r
Log. diff. = .... 9. 998827
Half sum = . . . 35?32n2ir Log. co-sine = . . , . 9.910487
Remainder es . . 12*48. 8| Log. co-sine = . . . . 9.989067
Natural number = 791372 Log.=9. 898381
Twice the natural number s; • • . . . 1. 582744
Sum of true alts.=23?29M7^ Nat.V. S. sup.= 1 . 917143
True cent, dist.= 48? 16^ 17^ Nat vers. S. = . 334399
Example 2.
Let the apparent distance between the moon and sun b6 108?42' Sf,
the sun's apparent altitude 6? 28% the moon's appairent altitude 54?12r,
and her horizontal parallax 56 H9?; required the true central distance?
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OF FINDING THB LONGITUDE BY LUNAR OBSSRVATIONS. 445
Sun's apparent alt. = 6?28' Or-Correc. 7'.45r=:truealt= 6^20^5^
Moon's apparcntalt.=54, 12. 0 +Correc,3l.40 sstrue alt.=54. 43. 40
Appar. central dists 108. 42. 3 Sum of the true altitude8=6 1 ? 3'55T
Sum= . . . . 169?22'. 3f
Log. difr.= 9.994507
Half sum = . , . 84?4n Ijr Log. co-sine = . ... 8.966858
Remainders . . 24. 1. 1^ Log. co-sine = • • • . 9.960673
Natural i)umber = . .......... 83568 Log.=8. 92203d
u
Twice the natural number =: ...... 167136
Sumoftruealts.= 61? 3^55? Nat.V.S.sup.= l;483813
True cent dist.=sl08?27M3r Nat. vers. S.= 1.316677
Method VIIL
Of reducing the apparent to the true central Distance. .
Rule.
To the logarithmic sines of the sum and the difference of half the
apparent distance and half the difference of the apparent altitudes^ add the
logarithmic difference : half the sum of these three logarithms (10 being
previously rejected from the index,) will be the logarithmic sine of an arch.
Now, half the sum of the logarithmic co^sines of the sum and the differ-
ence of this arch and half the difference of the true altitudes, will be the
logarithmic co-sine of half the true central distance.
Example 1.
Let the apparent central distance between the moon and a fixed star be
41?24'22^, the star's apparent altitude 12? 4 '27^, the moon's apparent
altitude 7^47^47^, and her horizontal parallax 57'24f ; required the true
central distance ? ^ '
Star's apparent alt.= .12? 4'27?-Correc. 4122^s=true ait.=12? O'.SI
Moon's appar. alt. = 7. 47. 47 +Correc. 50. 13 strue alt.= 8. 38. 0
Diff. of the app. alt8.= 4 ? 1 6 f 40^ Diff. of the true altitudes = 3? 22 ^ 5 ^
Half diff. of app. alt8.= 2? 8 C20r Half diff. of the trtte alts.= 1 ?4 1 C 2jr
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446 HAUTICAL A8TR01C0MT.
Halfdiff.ofap.altt.« 2? 8;20r
Half theBp.cent.di8.s20. 42. 1 1
Log. diff. = ... . 9.999201
Sams , . . • 22. 50. SI Log. sine s ... . . 9.589045
Differences . . 18.33.51 Log. sine s «... 9.502927
Sum =19.091173
Arch= .... 20?33!'44|f Log. sines .... 9.545586}
Half diff. of true alts.s 1.41. 2|
Bums .... 22?14U7? Log. co-sines . . . 9.966406
Differences . . 18.52.42 Log. co-sine s . . . 9.975987
Sums 19. 942393
Half the true dist. s 20?38^ l5r Log. co-sine s . . . 9. 9711961
True central dist. = 4m6'30r
Example 2.
Let the apparent central distance between the moon and Satmn be
110?l4'34r, Saturn's apparent altitude 9?40f48r^ and his horizontal
parallax 1% the moon's apparent altitude 15^40^6?^ and her horizontal
parallax 58'.43T.; required the true central distance ?
Saturn's apparent alt.as9?40:48r-Correc. 5'24rstrue aIt.s9?35C24r
Moon's apparentalt.s 15. 40. 6 +Correc.53. Ilstruealtsl6.33. 17
Diff. of the app. alts.s 5 ^59 ' 1 8r Diff. of the true altitude8s6?57 ' 53r
Half di£of app. alu.s^?59^39r Half diff. of the true alto.=:8e28C56|r
Half app. cent dist.s55, 7« 17
Log. diff. s. .... 9.998176
Sums .... 58? 6(S61 Log. sine s 9.928966
Differences . . 52. 7.38 Log. sine = 9.897284
19.824426
Aiehs . . . 54?47' 21 Log. sine s ..... 9.91221S
Half diff.of true alts.«3. 28. 56i
Sums . . . 58? 15 :58ir Log. co-sine s .... 9.720963
Differences . . 51. 18. 5^ Log. co-sine s .... 9.796035
§um= 19.516998
Half the truedist s, 55? 0'30jr Log. co-sines .... 9.758499
True centra! 'dist.= nor IM*
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OF FINDING THB LONGItUDB BY LUNAR OBSSRVATIONS. 447
Method IX.
Cff reducing the apparent to the true central Distance,
RULK.
To the logarithmic co-sines of the sum and the difference of half the
apparenl distance and half the sum of the apparent altitudes^ add the
logwithmie difference : half the sum of these three logarithms (10 Seiof
previously rejected from the ihdex^) will be the logarithmic ico-sine of a&
acch. No% half the sum of the logarithmic sines of the sum and differ-
ence of this arch and half the sum of the true altitudes, will be the loga-
rithmic sine of half the true central distance.
Example I. -
' Let the apparent central distance between the moon and a fixed star be
41?!!9'58r, the star's apparent altitude ll?3l'2r, the moon's apparent
altitude 8?44<35T, and her horizontal parallax 57^247 ; required the true
central distance ?
Star's apparent alt.=rll?31( 27-Correc. 4^34r=truealt.= ll?2jSf28:
Moon's appar. alt. a= 8.44.35 ;-f€k>rrec.50.46 struealt.as: 9.35.21
Sumoftheapp.a]ts.=:20? 15 ^37^ Sum of the true altitudes^21 T 1 '49r
Halfsumofap.alts.»10? 7U8^r Half sum of the^«ealts.=:10?30^54i?
Half ap. cent dist.^ 20. 44. 59
Log. diflF. = 9. 999083
Sum = ... 30?52M7ir Log. co-sine = ..... 9.933612
Difference ss • • 10. 37* 10} Log! co-sine = .... '9.d92497
19.925192
Arch a . . . 23?26:23^ Log. co-sine » t • . . 9.962596
Halfsumoftr.alU.=s 10.30.541 . .
Sum.r:: . . . . 33?57n7|^Log. sitic = ..... 9.747053
Differences: .' . 12. 55. 28} Log. sine = 9.349604
. ^ Sums 19.096657
Halfthetraedist.a8 20?4lC54^r Log. sine £= . . . . . 9.548328|
Truecentraldist.= 41?23'49r
Example 2.
JjBt the apparent central distance between the moon and sun be
101?54:51?» the #un's apparent altitude 39?34^35r, the moon's apparent
altitude 29?23C2% and her horiaontal parallax 58^53^^ ; requured die true
central distance ?
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448 NAUTICAJ. ASTRONOMY.
Sun's apparent alt.a 39?34'35r-Correc. 1^ 3r=traealt.s39?33^32r
Moon's appar. alt. = 29.23. 2 -f Correc.49.38 =truealt.=30. 12.40
Somof the app.alts.s68?57'37" Sum of the true aldtudes= 69?46'. 12^
Halfsumofap.alts.=:34?28'.48ir Halfsumof thetnie alt8.=: 34?53^ 6?
Halfapp.cent.dist.= 50.57.25i .
.-. ■ • — Log.diff. = . .... 9.996517
Snin= . . . 85?26'14r Log. co-sine = . . . . 8.900647
Differences . . 16.28.37 Log. co-sine = . . . . 9-981789
18.878953
Areh=. . . . 74? 1'57' Log. co^sine = . . . 9.439476^
Halfsumoftruealt8.=:34.53. 6 .
Sum= . . . . 108?55^ 3r Log. sine s \. . . . 9.975885
Differences . . 39. 8.51 Log. sine = '. . . . 9.800249
Sum= 19.776134
Half the true di8t.= 50?36(22r Log. sine = .... 9.888067
True central dist.=: 101? 12M4r
. MEtHOD X.
* Of reducing the apparent to the true central Distance.
' Rule.
To the logarithmic sines of the sum and the difference of half the
apparent distance^ and half the difference of the apparent altitudes^ add
the lit^arithmic difference^ its index being increased by 10 : from half the
sum of these three logarithms subtract the logarithmic sine of half the
difference of the true akitudes, and the remainder will be the logarithmic
tangent of an arch ; the logarithmic sine of which, being subtracted from
the half sum of the three logarithms, will leave the logarithmic sine of
half the true central distance.
Example 1.
Let the apparent central distance between the moon and b fixed star be
55^4'.531f the star's apparent altitude 10?8'.6f, the moon's apparent
altitude 8? 1^25^, and her horizoatal parallax 58^?; required the mte
central distance ?
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or riNDINO THB LON6ITUDB B7 LTTNAR OBSERVATIONS. 449
Star's Bpparenfalt= 10? 8' 6r-Correc. 5ni7=trueitlt.s=10? 2.55?
Moon's apparoit alt.s=8. 1.25 +CoiTec. 50. 58 =:truealt.= 8.52.23
Diff.oftheapp.aIts.=2? 6'4ir Diff. of the true altitudes = 1°10'327
Halfdiff.ofapp.alts.sl? 3t20^t Half diff. of true aldtudes = 0?35n6^
Halftheap.cent.dis.=27. 32. 26i
•Log.diff.sa 19. 999162
Sums. ... 28?35'.47?LQg.8iiie=> 9.680006
Difference a . . 26.29. 6 Log.8ine=*9. 649299
Sum= 39:328467
Half sum 3 . . 19.664233^ . 19.664233|
Half diff. of true alts.=0?35n6rLog.Bme= 8. 011083
ArchB . .... 88?43^367Log.tan.sll.653150iLog.d.9. 999893
Half the true distances • . .. . 27?29M4r Log. 8ine=9.664340i
True central distance s . . . . 54?59'28;
ExanqtU 2.
Let the apparent central distance between the moon - and sun be
91?26'8', the sun's apparent altitude WHS'AMy the moon's apparent
altitude 53?41 H 7, and her horizontal parallax 58C 297 ', required the true
central distance ?
Sun's ^parent alt.= U?45M17-Correc. 3:267strue alt.= 14?42'157
Moon's appar.alt.= 53.41. 1 -{-Correc. 33. 56 =true alt=54. 14)57
Diff.of the ap. alts.=38?55^2p7 Diff. of the true alts. =s . 39?32'.427
Half diff.of ap.aite.= 19?27 M07 Half diff. of thefpie alts.= 19?46'. 217
Halfap. Cent. (Ust.= 45.43. 4
Log.diff.= 19. 994220
Sums .... 65:?lO<447Log.sines 9. 957905
Differences . . 26^15.24 Logcsine = 9. 645809
Sum s 39.597934
Half8nm=s ......... 19.798967 . . 19.798967
Half diff.oftrue «lto.l9?46^217Log.sine= 9.529285
Archs . . , 61?44^48?Log.tanU= 10. 269682 Log. sine 9. 944902
Halflhe true distances •, . . . 45?36:38i7 Log, 8ine=9. 854065
True central distance a .'v . . 91?13''17'
2 o
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450 HAimCAL ASTRONOMT*
Method XI.
Of reducing the apparent to the true cmUral Distance.
Rule.
To the logarithmic difference (its index heing {ncreaaed by 10,) add the
logarithmic co-sines of the stmi and the difference of half the apparent
distance and half the sum of the apparent altitudes ; from half the sum of
these three logarithms subtract the logarithmic co^sine of half the sum of
the true altitudes, and the remainder will' be the logarithmic sine of an
arch ; the logarithmic tangent of which^ being subtracted from the half
sum of tlie diree logarithms^ will .leave the logarithmic sine of half the
tnie central distance^
Example I.
Let the apparent central distance between the moon and a fixed star
be 68?5i2M0r, the starts apparent altitude 10?52U7^ the moon's
apparent altitude 6*39^28^, and her horizontal parallax 58<3ir} required
&e true central distance,?
Star's apparent a]t.=10?52< 17^-Correc. 4^507= true ait=10?47^27^
Moon's appar. alt.« 6. 89. 26 +Correc. 50. 26 a true alt.s 7. 29. 54
Sum of the ap. alts.= 1 7?d 1 ' 45 r Sum of the true dtitndes = 1 8? 1 7 ' 2 1 r
HaIfsumofap.alts.=: 8?45^52ir Half sum of the true alts. =: 9?8M0ir
Half ap. cent. dist. = in. 26. 20
Log.diff; 19. 999826
Sam= . . . . 48?12a2irLog.co-si.9. 862684
Difference = • • 25. 40. 27^' Log.co-si.9. 954856
■III I 1^
Sum= . 39.816866
Half mm 3= 19.908433 . . 19.908433
Half8umoftruealts.s9? 8'40irLog.co-n.9.994445
Ai«h« .... S5T 7'. 4'. ho^jiae»9,9l»998hag.T.=:l0.\i6679
Half the tniedutances .... 34?22^34r Lo^.tkio :s9,7S1758
TVue central distance ts ,■ ', . . 68?45( 87
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OF FINDING THB LOM6ITUDB BY LUNAR OBSBRVATIONS. 451
Example 2.
Let the apparent central distance between the moon and sun be
120?I0'4(K, the sun's apparent altitude 13?30^0r, the moon's apparent
altitude 6?10;0r, and her horizontal parallax 61(12^5 required the true
central distance ?
Sun's apparent alt,c:ql3?30' Or-Cofrec. 8M5?== true alt,«13?26I15f
Moon's appar. alt,?? 6.10. 0 +Correc. 52.36 = true alt.= 7. 2.36
Sumoftheap,alts.=:19?40C 0? Sum of the true altitudes « 20?28'64r
Halfsum of ap,alts.=9?50^ 0? Half spm pf the true ftlt8.= 109l4'25ir
Half ap, cent. dist.= 60. 5.20
-. Log, diff.a 19. 999345
Sums. . , . 69?55 i20r Log. co-sine 9.535668
Difference = . . 50, 15. 20 Log.co-sine 9. 805749
Sum^i. . /. .39.340762
H^lfsums . ......... 19.670381 . 19,67088!
Half8umoftrtieaIu.l0?l4^25ji:Log,cor6iiie 9.993026 •
Arch== . . . 28M4'.23r Log. sine = 9.677355 Log.T. 9. 738070
Half the true distance = . • . . 59? 56 r 59'r Log. sine=9: 9378 1 1
True centnd distance an . . . U9?53C58r
AterHoo Xljr •
Of redwing the apparent to the true central Distance.
RuLB.
From the natural versed sine supplement of the sum of the apparent
altitudes^ subtract the natural versed sine of their difference^ and call the
remainder arch first. Proceed in a -similar manner with the true altitudeb^
and call the remainder arch second; and from the natural versed sine
supplement of the sum of the apparent altitudes, subtract the natural versed
sine of the apparent distance, and call the remainder aroA tlArd,
Now, td the arithmetical complement of the logarithm of arch first add
the logarithms of arches second and third, and the sum (rejecting 10 from
the index,) will be the logarithm of a natural number ; which, being sub-
tracted from the natural versed sine supplement of the sum of the true
altitudes, will leave the natural versed sine of the true central distance.
2a2
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452 ' NAUTICAL ASTRONOMY.
Example 1.
Let the apparent central distance between the moon and a fixed star be
83? 15 '19", the star's apparent altitude 7? 39' 4 r, the moon's apparent
altitude 10?57'36% and her jiorizontal parallax 58(53'; required the
true central distance ?
*'sap.alt.= 7?39' 4?-Cor. 6M5?=Truealt. 7^32119^
])'sap.«lt=10.57.36 +Cor.53. 3 <=iTruealt.ll.50.39
Sum = . ISeseUO'^iJ^Jl. 947707 Sum = I9?22(58r••;:i^}1..943322
Dlff. « . 8.18.32N.V.S..001668Diff. = 4. 18.20N.V^..002822
Arch first = 1 . 946039 Areh second = 1 . 9405QO
Sumofaplalt8.=sl8?36M0r N.V,S. sup. = 1. 947707
Ap. cent. di8t.s=83? 15 a9r Nat. V. S. = . 882554
Arch third = 1. 065 153 Log. s; 6.027432
Arch seconds 1. 940500 Lqg. = 6.287914
Arch firsts ......... 1.946Q39Log.ar.co.3. 710848
Natural number s . . 1. 0621^9 Log. s 6i 026194
Sum of true alts. 19?22(58: N.V.S. sup. = L 943322
TVue cent. dist. 83?10(28? Nat. vers. sine- .881153 .
Example 2.
Let the apparent distance between the moon and sun be II 1?27' If,
the sun's apparent altitude 24^40' 16?, -the moon's apparent aldtude
16?52.3K, and her horizontal parallax 54'56r j required the tone central
distance ?
©'sap.alt.=24?40n6r-Cor. l'56r=Truealt.24?38(20r
])'sap.alt.=16.62.31 +Cor.49.28 =TrueaIt.l7.41.59
Sums , 41?32U7rY,f}l. 748419 Sum = 42?20<19r'UV}l. 739177
DiflF. = . 7.47.45N,V.S..009242 Diff.= 6. 56. 21N.V.S. . 007325
Aroh first = 1. 739177 Arch second = V. 731852
Sumofap.alts.=4 1 ?32 U7?N.V.S.8up.= 1 . 7484 19
App. central dist. U 1 ° 27^ ? Nat.V.S.= 1 . 365694
Arch thirds ....... .382725 Log. s . 5.582887
Archseconds ..-.■.... 1.731852 Log. = . €.238511
Arch firsts ........ 1.739177 Log. ar.co.=3. 759638
Natural number = 381097 Log. s . 5.581036
Sumoftruealts.42?20( 19f N.V.Sjup.sl. 739177
True central dbt. U0?58(56rN.V.S.5=l. 356080
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OF FINDING TUB LONGITUDE BT LUNAR OBSERVATIONS. 453
Method XIII.
To the apparent distance add the apparent altitudes of the objects ; take
half the sum, and call the difference between it and the apparent distance,
the remainder. Then,
To the logarithmic difference (its index being augmented by 10,) add t^e
logarithmic co^sines of the half sum and the remainder ; from half the sum
of these jthree logarithms subtract the logarithmic co-sine of half the sum
of thq true altitudes^ and the remainder .will be the logarithmic sine of an
arch.' Now, the logarithmic co-sine of this arch, being added to the
logarithmic co-sineof half -the sum of the true altitudes (rejecting 10 from
the index), will give the logarithmic sine of half the true4:entral distance*
Example !•
Let the apparent central* distance between the moon and Spica Virginia
be 37^1 2 UOr, the stat's apparent altitude ll?27'50r, the moon's
apparent altitude 40^55 '. 15 T, and her horizontal parallax 54C'10? ; required
the true central distance ?
Star's apparent alt.= 11? 27 '50!r^Correc. 4^35r=-:true alt.=ll?23n5r
Moon's appar.alt.=40. 55. 15 -f-Correc, 39.51 =strue alt,=41.35. 6
Appar. cent, dist, = 37. 12. 40
Sum = . . . 89^35 USr
— rLog.aiff.=r 19. 995703
Half sum =: . , 44?47 ' 52i:Log.co-sin, 9. 85 1012
Remainders • 7.35. 12* Log.co-sin.9.996\81
Sum= . . 39.842896
Half sum = 19.921448
Halfsumoftruealts.26?29'.10irLog.co-sin.9.951844 • . 9.951844
Arch s . . . 68?48'.45r Log. sine=:9.969604Log.co-8i.9. 558014
Half the true disunce :=^ . • . • 18?5a|26ir Log. 8ine=9. 509858
True central distance =3 . . . • 37^44(53^1:
Example 2.
' Let the apparent central distance between the moop and sun be
U7?42'28^, the sun's apparent altitude 10? 19^ 19^, the moon's apparent
altitude 42?55'.1% and her horizontal parallax 60C2Cj required the true
antral distance?
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454 WAOTIGAL AtTAONOMY.
Sun'Bapparentalt=10?19a9r-Correc. 4^56r= true alt.=10?14^23^
Moon's app. alt. = 42.55. 1 +Currec. 42. 57 = true alt=43.37.S8
Appar.cent.dist.=:117.42. 28 .
Stttns • . • 170?56^48r
^ — Log.diff.= 19. 995005
Half turn » . . 85^28124rLog.co-Mii. 8. 897204
Remainder z= • 32* 14* 4 Logxo-sin*9.927305
Suin'=: 38.819514
Half sum 19.409757
Half8umoftruealt8.26?S6nOi*Log.co.8.=9. 950127 . • . 9.950127
Arch= . • . 16?44^52^ Log. 8me=9.459630Log.co.8i.8.981l77
Half the true distance s .... 58?36'58r Log. 8ine=9. 931304
True central Stance — 117?13'.56.r
Note. — There are some curious properties peculiar to the limar obsenra^^
tions, with which the mariner ought to be acquainted^ but which the
general tenor of this work, will not allow of being touched upon here :—
these properties or peculiarities may, however, be readily seen, by making
reference to the General Remarks contained between pages 208 and 212
of '^ The Young. Navigator's Guide to the Sidereal and Planetary Parts of
Nautical Astronomy."
Probum VIIL
Given the apparent Thne and the true central Distance between the Moon
and Sun, a fixed Siar, or a Planet; to determine the Longitude qf the
Place of Observation.
Rule. .
If the true central dbtance can be found in the Nautical Almanac, the
corresponding apparent timflr at Greenwich will be seeii standing drer it
at the top of the page ; bat if the true central distance cannot be exactly
found, which in general will bt the case, take o^t the tiiro distaaoee firom
the Nautical Almanac, one of which is next greater and the other next less
than the true central distance, And find their difference ; find, also, the
difference between the true central distance and* the' ^irf of .tiie'two
distances so taken from the Nautical Almanac ; then, froip the proportional
logarithm of this difference, subtract the proportional logarithm of the.
former difference, and the remainder will be the proportional lcs;aritlun of
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OF FINDING THB LONGITUDB BY LUNAR OBS£RVATION(S. 455
B portion of time, which, being added to the time corresponding to the
first of the two distances taken from the Nautical Almanac, will give the
apparent time of observation at Greenwich* Now, the differeniee between
the apparent time at Greenwich, thus, found, and the apparent time at the
place of observation^ being turned into degrees, will be the longitud^of
the latter place ;— and which will be east, if the time at the ship be greater
th^p that at Greenwich ; if otherwise, west.
Example I.
At sea, January 9th, 1825, in longitude (by account) 54?48^. east, at
23t40?47' apparent time, the true central distance between the moon
and sun was 107^ 19 '56^; required the corresponding apparent time at
Greenwich, and the longitude of the place of observation ?
Traeccntdtst.atship=:107?l9^56^{ Djff^io 7/32*; Prop. log.=:4258
Distance at 18 hours = 108. 27, 28 >
Distance at 21 hours = 1Q6. 48. 12 j Diff.=: 1. 39. 16 Prop. log.=:2585
Portion of time = ....... 2*2^27*. Prop, log.zi 1673
lime corresponding to first distance =: » 18. 0. 0
Apparent time of observ. at Greenwich = 20 1 2T27'
Apparent time oJP observation at ship = 23. 40. 47
Longitude of the ship, in time = . . . 3*38r20* =54^85 ( east.
Example 2.
At sea, MaiehSd, 1825, in longitude (by account) 47?55^ west^ at
10M2T43! apparent time, the true central distance between the moon and
Spica Virginis, was 50?3^2S^; required the corresponding apparent time
at Greenwich, and the longitude of the place of observation ?
True central disUince=50? 3 ' 23? I l)iff.=0?53 ' 20r Prop. log. =^5288
Distance at 12 bourse 50. 56. 43 < .
Di8tanceatl5houn=49. 2.48 jDiff.= 1.54. 0 Prop. log. = 1984
Portion of time = .• l*24rlS! Prop. log. = 3299
Time corresponding to first distance = 12. 0. 0
Apparent time of observ. at Greenwich := lS^24ri3'
Apparent time of obaervation at ship = 10.12.43 * .
Longilttde of the ship, in time = . . ^ 3! 1 lT30;;=47?52130r west.
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^56 NAUTICAL A8TR0K0MY.
Problbm IX.
Given the LatUude of a Place and Us Longitude by accdtmt^ the observed
Distance between the Moon and Sun, a fixed Star, or a Planet, and
i^ observed JUiiudes of those Objects ^ to find the true Longitude qf
the Place of Observation.
Rule.
Reduce the apparent time of observation to the meridian of Greenwich,
by Problem III., page 297 ; to this time let the moon's horizontal parallax
and semi-diameter be reduced, by Problem VI., page 302, and let the
moon's reduced semi-diameter be increased by the augmentation (Table
IV.) answering to her observed altitude. *
Find the apparent and the true altitude of each object's centre, by the
respective problems, for that purpose, contained between pages320and 327*
To the observed distance between the nearest limbs of the moon and
sun, corrected for index error, if any, add their respective semi-diameters^
and the sum will be the apparent central distance. But, if the distance
be observed between the moon and a fixed star or planet, then the moon's
true semi-diameter is to be applied to that distance by addition when it is
measured from the nearest limb, but by subtraction, when it is measured
from the remote limb : in either case, the result will be the apparent
central distance. With the apparent and the true altitudes of the objects,
and their apparent central distance, compute the true centra] distance^ by
any of the methods given in Problem VII., between pages 433 and 454 ;
and find the apparent time at Greenwich corresponding to this distance,
by Problem VIII., page 456.
Now, the difference between the apparent times of observation at the
ship and at Greenwich, being converted into degrees, will be the longitude
of the place of observation ; which will be east or west> according as the
time at the ship is greater or less than the Greenwich time*
Remarks*
It the watch be not well regulated to the time of observation, the
apparent time may be deduced from the true altitude of the sun, moon,
star, or planet, used in the computation, provided the object made choice
of for this purpose be sufficiently far from the meridian at the time of
measuring the lunar distances; if not, the error of the watch must be
inferred from the true altitude of one of those objects^ when in a more
favourable position with respect to the meridian : then the error .of the
watch, thus found, being applied to the mean time of measuring the lunar
distances, by addition or subtraction, according as it is slow or fast, the
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OF FINDING THR LONGmiDB BY LCNA& OBSBRVATIONS. 4&7
sum or difference will be the apparent time of taking the lunar observa«
tion, agreeably to the meridian under which the error of the watch was
obtained. The error of the watch is to be found by Problems III., IV.^ V.,
or VL, between pages 383 and 400, according as the object may be the
sun, a fixed star, a planet, or the moon.
In taking a lunar observation, it is necessary that several distances be
meaauredj — that the corresponding times, per watch, be carefully noted
down, — and that the altitudes of the objects be observed at the same
instants with the distances : then, the respective fiujns of the times (per
watch) of the observed distances and of the altitudes, being divided by their
common number^ will give the mean time of observation, the mean
observed distance, and the mean observed altitude of each object*
Example L .
January 9th> 1825, in latitude 19?30'. N;, and longitude 5?45^ E.^ by
account, the following observations were taken ; the index error of the
sextant by which the distances were measured Was 2^S0lf subtractive, and
the height of the eye above the level of the sea 20 feet ; required the
longitude of the place of observation ?
Apparent time of
observfttion.
Obsenred distance between
nearest limbs
of. Mooh and Sun.
Altitude of Sun's
lower limb.
Altitude of Moon's
lower limb.
19* 10-45 •
• 11. 50
• 12. 55
• 14. 0
.15. 5
1070 48' 30''
. 47. 50
. 47. 15
. 46. 40
. 46. 0
70 9' 0*
7.22. 30
7.36. 0
7.49.30
8. 3. 0
530 13' 30"
53. 2. 0
52. 50. 30
52. 39. 0
52. 27. 20
Mean..». 19M2-55'
Longitude r 23. ^
in time f ^* ^
Red. time 18M9-55«
Mean 10r» 47' 15"
Index error.. — 2. 30-
D 'a semi-diam. +16. 25 ■
Q'i semi-diam. +16. 18
Metti..7^36' 0"lMean 52°50' 28''
Moon's semi-tdiameter +16. 25
Dip of the horizon • • —4.17
Appar.di8t,..108<' 17' 28"
Moon's appar. altitude 53° 2' 36"
Correction of ditto. ... +35. 1
Moon's true altiiude, • 53® 37' 37"
Observed altitude of sun's lower limb = 7^361 01
Sun's semi-diameter = . . • • • » •+16.18
Dip of the horizon = ..,.•• — 4. 17
Sun's apparent altitude = ..•••• 7?48! K
Gorrection of ditto «= • • • • . » — 6.29
Sun's true altitude ==
7Mr.32:
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458
KAtrnCAL ASTRONOMY.
Moon's rednoed horizontal parallax »
Diff. of the app. alts.=45 ? 1 4 ^ 35 'r
Auxiliary angle = . 60. 26. 25
App. central dist. =. 108. 17. 28
59C25?
Sum of aux. ang. and
difl:oNpp.alts.^l05?4K 0^
Difference of dittos 15.11.50
Samaax.ang.&ap.di8J68. 4S. 53
Differenoeofdittoa47.5.1. 3
Diff. of tnie alts« s 45. 56. 5
Nat. versed sine sup* a=
Nat. versed sine sup. ^
Natural versed sine s .
Natural versed sine = .
Natural versed sine s= ..
Natural versed sine =
True central di8t=107^59^32r )
JDi
.729680
1.965029
1.980722
.328937
.3045122
1. 308890
Dist jai 1 8 hours = 108. 27. 28 ] Diff.=0? 27 i 56r Prop. log. = 8091
Dist, at21 hours =106.48. 12 } Diff.= L39. 16 PropJc^. = 2585
Portion of time = • . Ot50?39! Prop. log. = 5506
Hme corresponding to first distance = 18. 0. 0
Apparent time of observ. at Greenwich = 1 8 tSOTSS '
Apparent time at the place of observ. =;; 19. 12. 55
Longitude at the place of obs., in time a Ot22ri6! = 5?34' east.
Example 2.
February 1st, 1825, in latitude 45?40'.N., and longitude 59?10^W., by
account, the following observations were taken ; the height of the eye above
the level of the horizon was 22 feet, and the index error of the sexftmt by
which the. distances were measured K30? additive j required the true
longitude?
Apparent time of
obserration.
Obgerved dUtance of
Moon'g remote limb.
Altitucle of
Reguktt.
Altitude of Moon's
lower limb.
8*50«10»
.51. 25
. 52. 40
. 53. 55
. 55. 10
350
7' 40"
6. 50
6. 10
5. 20
4. 40
28^50' 50"
29. 4. 30
29. 18. 20
29. 32. Q
29. 45. 50
Mei^n.... S*52«40«
Mean 35o 6^ 8'
Index error. . +. 1. 30
]) 'b semi-diam. —16.26
Mean 29o IS' 18'
10«44' 40^
11.40. 0
12. 35. 20
13. 30. 50
14. 26. 10
Mean 12o35' 24"
Red. time 12H^»20»
Apparent dist. 34'? 51' 12^'
Moou'bsemi^ameter*. 4-16. 26
Dip of the horizon .... — 4. 30
Moon's appar. altitude.. 12^ 4r 20*
Correction of ditto... .. •h54. 27
Moon's true altitude.,.. 13« 41' 47"
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OF FINDING THB J^MGITUDB BY LUNAR OBSERVATIONS.
4S9
Obsenred altitude of B^guluii =
Dip of the horisQH for 22 feet ^
• Apparent altityde of Regulut ==
Correction of ditto = . • •
True altitude of Regulus =:
29? 18' 18?
- 4.30
29? ISMS'?
- 1.41
29?12'. 7"
Moon's reduced horizontal parallax =
60^3:
Sumoftheapp.ftlt8:=42? V. 61
Auxiliary angle == fiO. 6.49 '
App. central dist. = 34.51. 12
Sum of aux, ang. and
sum of app. dts.= 1029 7^57? Natural versed sine rs
Difference of do. = 18. 5.41 Natural versed sine =
S4UBaiii|4uig:&ap.dist.94.58. 1 Natural versed sine =
Difference of do. =r 25. 15. 37 Natural versed sine =
1.210173
.049456
1.086581
.095622
Sum of the true alti«s:42. 53. 54 Natural versed sine sup. = 1 . 732563
Natural versed sine = . . 174395
True central dist. = 34?2r. 0? ) : —
Dist.atl2hours = -34.51. 16 j Diff. = 6?30n6r Prop. log. = 7743
Di»t..at 15 hours = 33. 2.35 } Diff.= 1.48.41 Prop. log. = 2191
Portion of time r= ••...• .
Hme corresponding to first distance :r
0*50r 8!Prop.iog.=5552
12. 0. 0
Apparent time of observation at Greenwichz: 12^50? 8!
Apparent time ait the place of observation = 8. 52. 40
Longitude at the place of obsery., in time ;= 3^57^28! ={9?22'01' west.
EteampU 3. • •
March 9th, 1825^ in latitude 43n7' S., and longitude 57?55: E., by
account, at 20M4T per watch, not regulated^ the mean of several observed
distances between the moon and sun was 107? 28' 17'' ; at the same time
the mean of several altitudes of the sun's lower limb was 26?39'40^, that
of the moon's upper limb 39?30'45^, and the height of the eye above the
level of the sea 19 feet ; required .the longitude of the place of observation ?
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460
NAiniCAL ASTHONOMY.
Time, per watch, = 20*14r 0'.
Long.57?55CE.= -3.51.40
Reduced time = . 16t22r20'
Alt sun's lower limb= 26?39C40r
Sun's semi-diameter=; +16.. 8
Dip of the horizon = — 4. 1 1
Sun's apparent alt. = 26'?51f37r
CorrecUonz: . . — 1.44
Sun's true alt. = . 26'>.49'.53'.
Dist of nearest limbs
of moon and sun=107?28'. 17?
Sun's semi-diameter = +16. 8
Moon's semi-diam. = +16. 0
Apparent distance = 108*! 0'.25r
Alt. J 'slower limb = 39?30'45r
Moon's 8emi-diameter= +16. 0
i)ip of the horizon == — 4. 1 1
Moon's apparent alt.:=39?42^34?
Correction = . . +43.33
Moon's true altitudes 40?26t 7?
Moon's reduced horizontal parallax = 58'47
App. cent. dist.= l08? 0<25rN.V. S.=:l. 309132
Diflf. of ap. alts. = 12. 50. 57 N.V. S.= . 025041 Log. diff. = 9. 995483
Remainders 1.284091. Log.s . 6.108596
Natural number = 1.270809 Log.= . 6.104081
DifiF.oftruealts.= 13°36n4?N.V.S.= .028054
Truecent.di8.= 107?23;2ir »N.V.S.= 1. 298863
Dist. at 15 hrs= 108. 6.55 jDiff. = 0?43'.34r Prop. log. = 6161
Dist.atl8hr8=106.32.42 jDiff. = 1.34.13 Prop. log. = 2811
Portion of time = ^_ 1^23T14! Prop. log. = 335a
Time corresponding to first distance — 15. 0. 0
Apparent time of obs. at Greenwich = 16?23?14;
To find the apparent Timfe at the Place of Observation t—
Lat.oftheplaceofob8.=43»17' OrS. . . . Log. 8ecant=0. 137885
Sun's reduced declinations 4. 13. 10 S. . . . Log.8ecant=0.001179
Sun's toerid.zen. dist. =39? 3^50rNat.V.S.=22.^556
Sun's true central alt. = 26'. 49. 53 N.co-V.S.=548634
Remainders 325078Log.s5.5U98S
Sun's horary distance from noon s 3 MSToS : Log. rising = 5. 65 105.2
Apparent time at the place of ob6.s20* 14T 5 !
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OV FINDING THE LONGITU0B BY LUNAR OBSERVATIONS.
461
Apparent thne at the place of obs.=20M4? 5 !
Appar. time of obs. at Greenwich =161 23. 14
Lwig, of the place of obs., in tiine=:3*50T51 ! = 57?42'45r east.
Example 4.
April 1st, 1825, in latitude 49?30^ S., and longitude 61?30^ E., by
account, at 1 1 t29T per watch, not regulated, the mean of several distances
between the moon's remote limb and the star Antares was 76948' 27'', and,
at the same time, the mean of an equal number of altitudes of the moon's
lower limb was 39"? 10' 12?, and that of the star, east of the meridian,
37?56^3? ; the height of the eye above the level of the horizon was 23
feet; required the longitude of the place of observation ?
Time, per watch, = ll?29r 0'.
Long.61?30^B.= - 4. 6. 0
Reduced time :r: . 7^23? 0!
Altitude of Antares = 37?56^43r
Dip of the horizon =: — 4. 36
Apparent altitude = 37?52^ 7?
Correction =: . . — 1; 14
Star's true altitude = 37?S0^53r
Dist.]) 's remote limb=:76?48^ 27?
Moon's semi-diam. ^ — 16. 50
Apparent distance = 76^31 137^
Alt. ]> 's lower limb = 39? lOU 2?
Moon's semi-diam. = + 16. 50
Dip of the horizon r: — 4. 36
Apparent altitude = 39^22'. 26!
Correction = . '. H-46. 9
Moon's true altitude= 40? 8^35?
Moon's reduced horizontal parallax = 61< 11?
Sum of app. alts.=:77? 14^33?N.V. S. sup.rr 1 . 220825
App. cent dist = 76. 31. 37 Nat.V. sine = . 767012Log.diff.=:9. 995271
^ Remainder == .453813 Log.=5. 656877
Natural number = 448898 Log.=5. 652148
Sum of true alt8.=:77?59^28?N.V. S. 8up.= 1. 208063
Truecentraldi8t.=76? 3^55? ? Nat.V. 8.= .759165
Dist. at 6 hours = 76. 57. 9 f^'^- = 0?53M4? Prop. log. = 5291
Dist at 9 hours = 75. 3. 29 J D^^- = ^^ 53. 40 Prop. log. = 1996
Portion of time = ....... It24ri7! Prop. log. = 3295
Time corresponding to first distance =: 6. 0. 0
Apparent time of obs. at Greenwich =: 7 J24r 1 7 !
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468
NAimCAL ASTRONOMY.
To find the apparent Tune at the Place of Observation : —
Lat. of the place of obs. = 49?S0' OrS. .
Star's reduced declin. = 26. 2. 2 S. .
Log. secantrrO. 187456
Log. 8ecant=:0. 046465
Star's merid. zenith dist.=23?27-581'Nat.co-sin. 917296
Star's true altitude = . 37. 50. 53 Nat. sine =613570
Remainder ^ 303726 Log.=::5. 482482
Star's horary distance, east of the mer.=4? 5?23!Lbg. rising=5. 71640.3
Star's reduced right ascension ^ » 16. 18. 42
Right ascension of th6 meridian =
Sun's reduced right ascension = .
12M3ri9!
0.43.19
Apparent time at the place of obsarv.ss 1 1 130* 0!
App. time of observ. at Greenwich = * 7. 24. 17
Longitude of tfie place of obs., in time= 4t 5?48; ^ 61?25U5f eatt.
Example 5.
April 22d, 1825, in latitude 40?10< N.,. and longitude 55? 17' W., by
account, at Ot23? per watch, not regulated^ the mean of several observed
distances between the nearest limbs of the sun and moon was 48?47-46fy
and, at the same time, the mean of an equal number of altitudes of the
sun's lower limb was 61?26'44^, and that of the moon's upper limb
48<?46'32r I the height of the eye above the level of the borismi was 21
feet; required the longitude of the place of observation ?
Time, per watch, = 0*23r 0!
Long. 55?17'.W.=: +3.41. 8
Reduced time =
4i 4? 8!
Alt. sun's lower limb= 61^26^4:
Sun's semi-diameter = -h 15. 56
Dip of the horizon = — 4.24
Sun's app. altitude = 61?38'16r
Correction =s . . — 0. 27
Sun's true altitude = 6l?37'49r
Dist. nearest limbs of
mooti and sun s 48?47 • 46f
Sun's 8emi*diameter s -f 15. 56
Moon's semi-diameter=: + 15. 32
— »
Apparent distance s 49?19a4f
AlL of J 's upp. limb=:48?46^32r
Moon's semi-diameter= — 15. 32
Dip of the horizon = — 4. 24
Moon's apparent alt.=48?26^36f
Correction = . . +36. 29 '
Moon's true altitude ;;s49? 3' 5'
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OF FINDING THB LOKOITVDB »» LJWAR OBSERVATIONS. 408
Moon's reduced horisoDtal parallax = 56' 16?
Half the apparent central di>tance=s24?39'37r
Half difference of the app. alts. = 6. 35 . SO
_ Log.diff. = 9.994848
°™n= 31?15i27? Log. sine =5 9.715071
Differences 18. 3.47 Log. sine = 9.491451
Const, log. = 6,301030
Natural number = ... . . . 317980 Log. « . . 5.502400
Diff. of true alts.= 12?34< 44 rN.V. sine 024008
True cent di8t.=48?61i 4r ) N.V.siiie341983
Di8t,at3h«iirB«48. 19.26 JDiff. ~ 0?3H38r Prop. log. s» 7551
I>ist.«t6h(mr9=49.47.57 jDi£^ :* 1.28.31 Prop. log. s 3082
Portion of time =s ...... li 4?19! Prop, log, = 4469
Time corrosponding to first distance « 3, 0. 0
Apparent time of obs. at Greenwich ^ 4 1 4? 1 9 !
To find the apparent Time at the Place of Obaervation i^
. Aote.— Since the apparent time cannot be inferred with sufficient
aocaracy from the sun's altitude^ on aiseonat of the proximity of that object
to the meridiap^ \t must be deduced from the moon's true central altitude ;
88 thus I-*
Lat. o/ place of obs.=40n0< O^N. . . , . . Log.8ecant=. 116809
Moon's correct dec.= 23. 13.46 N Log. secants:, 036716
J'smer.zen.dist. = 16?56n4r Nat. V. 8»043376
Moon's true cent. alt,«49. 3. 5 Nat. co-V.S= 244702
Remainder s 201326 L<^.s5. 308900
Moon's horary dist., east of the mer. = 2?57"59? Log.ri8ing=5. 45742.5
Moon's eorrected right ascension ss . 2.51. 1
Right ascension of the meridian = . 2t23T 2!
Sun's reduced right ascension =r • . 2. 0. 4
Apparent time at the place of observ.= 0*22r58!
Apparent time of obsenr. at Greenwich=4, 4. 19
Longitude of the place of ohs., in time=3?4i?21 ! = 55?20( 15f west.
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464
NAUTICAL ASTRONOMY.
Example 6.
May 6th, 1825, in latitude 34?45^ S., and longitude 33?30^ E-, by
account, at 21 ?30T per watch, not regulated^ the mean of several observed
distances between the nearest limbs of the sun and moon was 11 9? 50' 38 T;
and, at the same time, the mean of an equal number of altitudes of the
sun's lower limb (imperfeetly observed, owing to an obstructed horizon,)
was 27? 13^27^, and that of the moon's upper limb (also imperfectly
observed,) 19?24M2^ ; the index error of the sextant used in measuring
the distances was 2^25? subtractive, and the height of the eye above the
level of the horizon 19 feet ;- required the true longitude of the place of
observation ?
Time, per watch, = 21*30r 0!
Long,33?30'. E., = -2.14. 0
Reduced time = . 19M6r 0!
Alt of sun's low. limb=27? 13^ 27^
Sun's semi-diam. = +15.52
Dip of the horizon = —4.11
Sun's apparent alt. = 27^25^ 8*
Correction = . . — 1.41
Dist. nearest limbs
of moon and sun= 119?50^3S1!
Index error = • . — 2. 25
Sun's semi-diameter = +15.52
Moon's semi-diameter =s + 15. 30
Apparent distance = 120M9^35r
Alt. of ]) 's upp. limb= 19?24C 12f
Moon's semi-diameter=:--15.30
Dip of the horizon = —4.11
Moon's apparent alt=19? 4131^
Correction =8 . . +50. 44
Sun's true altitude = 27^23^27'^ ' Moon s true altitude^ 19?55; 15f
Moon's reduced liorizontal parallax == 56^34T
Half the app. central distance = 60e 9'.47i^
Half sum of the appar. altitudes— 23. 14. 49^
Log.difF. = 9.997841
Sum = 83?24^37^ Log. co-sine=9. 059787
Difference == 36. 54. 58 Log. co-sine=:9. 902827
Constant log.=6. 301030
Natural number = 182593 Log.sS. 261485
Slim of truealte.=47n8M2r N.V.S. sup. = 1.678010
True cent dist= 1 19?41 ^ 50r ) NatV. S.=: 1 . 495417
Distatl8hours=120. 19.47 jDiff. = 0?37^57r Prop. log. = 6761
Distat21hours=U8.50.22 JDiff. = 1.29.25 Prop. log. = 3039
Portion of time = 1 * 16r24^ Prop. log. =t S722
Time corresponding to first distance = 18. 0. 0
Apparent time of observ. at Greenwich = 19^ 16T24!
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OF FIKBINO THS IjONGITUBE BT LUNAR OfiSBRVATIONf. 465
Since the obstruction of the horizon prevented the altitudes of the
objects from being taken to that degree of accuracy which is so essentially
necessary to be observed when the apparent time is to be inferred from
their altitudes (though sufficiently exact to be employed in the reduction of
the apparent to the true central altitude), — ^therefore^ in the afternoon^
that is, on May 7th, at 1^53* per same watch, the sun's altitude was
again observed, and, when reduced to the true, was found to be31°44^28^;
at that time the latitude of the ship was 34?50' S. Hence the apparent
time, the error of the watch, and the longitude of the place of observation,
are obtained as follows : —
Time of observing the sun's altitude, per watch, = 1^53T 0!
Hme of observing the lunar distance, per watch, = 21. 30. 0
Interval = 4»23r 0!
Apparent time of lunar observation at Greenwich = 19. 16. 24
App. time at Greenwich of observing the sun's alt.=: 23?39r24!
Sun's declination at noon. May 6th, ss . • . . 16?3lC5KN.
Correction of ditto for 23*39?24! = • . • • +16.27
Sun's reduced declination =s 16948a8rN.
Lat. of place of obs.:=34?50' 0?S Log. secant^O. 085754
Sun's reduced ^dec. = 16.48. 18 N Log. seeant=0. 018955
Sun's mer. zen. dist. = 5 1 ?38^ 18rNat.V. S.=379376
Sun'strue altitude = 31.44.28N.co.V.S.=473918
Remainder = 94542 Log. = 4. 975625
App. time of observing the sun's altitude = 1 ?53?35 7Log.rising=5. 08033.4
App. time at tjreenw. of obs. sun's alt.= 23. 39. 24
Longitude, in Ume = 2*14Tll! = 33?32M5reast.
JVbf^.— Hiis is the longitude of the meridian where the sun's altitude
was observed for the purpose of finding the apparent time.
Remark, — In place of finding the interval between the time of observing
the lunar distance and that of taking the sun's altitude, as above, this part
of the operation may be performed as follows ; which, perhaps, may be
more intelligible to those who are not very conversant with this subject.
2h
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466
NAUTICAL ASTILONOlfT.
Apparent time of observing the sun's altitudes: 1 ^ 5S?35 *
Time of observing dittOj per watch, s . • 1 • 53. 0
Watch slow for apparent time c= . . « « 0T3$'.
Timei per watch, of gbs. the lunar distances 21. 30. 0
Apparent time of observing the lunar dist. = 21 ?30T35 !
Apparent time of ditto at Greenwich = • 1 9. 1 6. 24
2*14?li:=33?32M5rE.
Longitude of the place where the error of
the watch was found, in time = • . •
JExample 7*
June 22d, 1825, in latitude SOMO^N., and longitude 45?7^W.,by
account, at 3^0T5! apparent time, the mean of several observed distances
between the moon's remote limb and the planet Venus was 118M1'48T,
and, at the same time, the mean of an equal number of altitudes of the
moon's upper limb was 30? 18^ 25 T, and that of the planet's centre
15? 1 1 M7^; the index error of the sextant used in measuring the distances
was 2' 10? additive, and the height of the eye above the level of the sea 18
feet; required the true longitude of the place of observation ?
Apparent time ss . 3^ 0" 5!
Long. 45?7' W. = 3. 0.28
Reduced time
6? 0r33!
Alt of Venus' centre=5l5?l 1 U7t
Dip of the horizon ss -* 4. 4
Venus' apparent alt.= 15? 7'43r
G)rrection ^s . « ^ 3. 9
Venus' true altitude= 15? 4C34r
Dist. of Moon's remote
limb from Venus= 118?41 :481
Index error ss . . + 2. 10
Moon's semi-diamet^s -*16. 16
Apparent distance s 11 8? 27 '42?
Alt. of 3) 'sup. limb = 30? 18'. 25?
Moon's 8emi-diameter= ^16. 16
Dip of the horixon s& — 4. 4
Moon's apparent alt.=29?58C 5?
Correction = . • +49, 40
Moon's true altitudes 30? 47^45?
Moon's reduced horizontal parallax s= 59n3?
Apparent distance = . ^ . 118?27'42?
Venus' apparent altitude = • 15. 7*43
Moon's apparent altitude = • 29. 58. 5
Sum
163?SS:30?
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OP FINDING THB LONOITUOB BY hVVATL OBSBRVATIOKS. 46/
Sam = 163^33 ^aO'r
Log, diff. =2 9. 996430
Half8um= 81?46U5r Log. co-sine = 9. 155302
Remainder = ^ 36. 40. 57 Log. co-sine = 9. 904152
Natural number = ..••;.... 113/32 Log.=:9. 055884
Twice the natural number s= . . • . • . 227464
Sum of true alt8.=45?52^ l9i:Nat,V. S, 8up.= 1. 696264
True cent di8t=117?57'23T ) Nat. V. sine=l,468800
Di8t.at6hour8=117-36.42 *DiflF.»: 0? O'All Prop. log. = 2. 4206
Dist. at 9 bourse 119. 39. 7 i^^'"^ ^•*2-25 Prop, log. ^ .244»
Portion of time « Ot 1?12! Prop. log. ^k 2, 1757
Time corresponding to first distance » 6. 0. 0
Apparent tiioe of obs. at Greenwich sb 6 ? 1 ? 1 2 ?
Apparent time at the place of observ, =s 3. 0, 5
Longitude of the place of obs.^ in time = 3i 1? 7- == 45?I6'45T west.
i2^mar&.— In taking a lunar observation, it is customary to have three
assistants, two of whom are to observe the altitudes of the objects at the
moment that the principal observer measures the distance ; the third is to
be provided with a watch^ showing seconds^ and to note down carefully the
respective times of observation, with the corresponding distances and alti-
tudes. But, since it sometimes happens, particularly in small ships, that
the necessary assistant observers cannot be in readiness, or at liberty to
attend, the following example is given, by which it will be seen how one
person may take the whole of the observations himself, without any other
assistant than merely a person to note down the times of observation^ par
watch, with their respective distances and altitudes.
Example 8.
July 6th, 1825, in latitude 49M3: N., and longitude 42?22C W., by
aoeo^Bt, the following observations were made, in order to determine the
true longitude ; the index error of the sextant used in measuring the dis-*
tances was 1 '40? subtractive, and the height of the eye above the level of
the sea 17 feet.
Appar. Time. Mean Time. Meao Altitude.
21 ^ 8-32? Alt.of sun's low. limb=46?5S^ 01-]
21. 9.32 Ditto 47. 7. 0 S21f 9T32? 47? 7^ 0
21.10.33 Ditto 47.16. 0 J
2h2
i^
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468 NAUTICAL ASTRONOMY.
Appar. Time. Mmui Time. Mean Altitude.
21 M ir37' Alt. of J 's upp. limb=23";51130r-|
21.12.37 Ditto 23.42.20 ^21M2?37^ 23?42:20r
21.13.37 Ditto 23.33.10 J
21.14.50 Observed distance s 98.58.50 '^
21.16. 0 Ditto 98.58.20 I
21.17.10 Ditto 98.57.50 >21. 17. 10 98.57.50
21.18.20 Ditto • 98.57.20
21.19.30 Ditto 98.56.50 ^
21.20.43 Alt.of )>'8upp.limb=20. 9.40 -j
21.21.43 Ditto 20. 0. 0 J>21. 21.43 20. 0. 3
91.22.43 Ditto 19.50.30 J
21.23.48 Alt.of8un'8low.limb=49. 14. 0 ^
21.24.48 Ditto 49.22.40 J>21.24.48 49.22.40
21.25.48 Ditto 49.31.20 J
To find the Sun's Altitude at the Time of talcing the mean Distance :—
l8ttime21» 9r32! l8talt.47? T- 07 1sttimc21* 9r32' l8talU7? 7' Or
2dtime21.24.48 2d alt.49. 22. 40 .^r4it}21. 17- 10
As 0»15?16^ isto 2?15U0r so is 0^ 7 "38 f to + 1. 7.50
Reduced observed altitude of the sun's lower limb = . . 48?14'50r
Sun's semi- diameter = +15.46
Dip of the horizon = — 3.57
Sun's apparent central altitude =s 48?26'397
Correction of the sun's apparent altitude = — 0. 44
IVue altitude of the sun's centre = 48?25'55'
To find the Moon's Altitude at the Time of taking the mean Distance : —
I8ttime21*12r37^ lstalt.23?42'20? l8ttime2lM2r37t l8talt.23?42C20r
2dUme21.21.43 2dalt.20. 0. 3 J2:Vl}21. 17. 10
As 0? 9T 6: isto 3?42n7^ so is 0* 4r33: to - 1.51. 8
Reduced observed altitude of the moon's upper limb = . . 21?51'12f
Moon's true semi-diameter = , —14.52
Dip of the horizon = — 3. 57
Moon's apparent central altitude s 21?32'23T
Correction of the moon's apparent altitude = +48. 4
True altitude of the moon's centre s= 22? 20' 27'
Moon's reduced horizontal parallax ss . . . . 541 147
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OF FINDING THB LONGITUBB BY LUNAR OB8BRVATIOH8. 469
Obs.di8t.betw. J & ©=s98?57'50r
Index error of 8extant= — 1.40
Sun's semi-diameter = + 15. 46
Moon's 8emi-diameter= +14.52
Appar. central dist. = 99?26'43!
App. time of ob8erv.=21*17"10!
Longitude42?22'W.,
in time ^ . . + 2. 49. 28
Reduced time past noon,
July 7th, = . . 0? 6r38!
Half the app. central di8tance= 49^43'. 24"^: •
Half the diflF. of the app. alts. =r 13.27. 8
Log. diflF. =? . 9. 997660
Sum = 63?10'32'r Log. sine = . 9.950556
DiflFerence = 36. 16. 16 Log. sine = . 9. 772033
Sum =: 19. 720249
Archss 46?26^2ir Log. sine = . 9.860124i
Half, the diflF. of the true alu.=: 13. 2. 44
Sum = • .
DiflFerence =
59?29^ 51 Log.co.8ine= 9.705665
33. 23. 37 Log. co-sine = 9. 921639
Sum= 19.627304
Half the true distance — .. 49?22130? Log. tso-sine s 9. 813652
True central distance == . . 98?45! Or?
Distance at 0 hour, or noon, = 98. 48. 3 iDiff-O^ 3^ 3rP.log.= 1.7710
Distance at 3 houra = . . 97.26. 27 j^*ff-^- 21-36 PJog.= .3436
Portion of time = 0? 6r44! Prop.log.=5l.4274
Hme corresponding to first distance =s . 0. 0. 0
Apparent time of observ. at Green^ch = 0* 6T44*
Apparent time at the place of observations 21. 17* 10
Longitude of the place of observ., in time = 2*49^34! =42? 23^ 30r west
Note, — Proportional logarithms will be found very convenient in the
reduction of the altitudes of the objects to tlie time of taking the mean
lunar distance : thus, to the arithmetical complement of the proportional
logarithm of the first term, add the proportional logarithms of the second
and third terms ; and the sum, abating 10 in the index, will be the propor-
tional logarithm of the reduction of altitude.--See Example, page 75 or 76.
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4^0 HAUTICAT. ASTRONOMY.
Remarlci.-^ln taking the means of the several observations^ those which
are evidently doubtful or erroneous ought to be rejected. A doubtfiil
altitude or distance may be readily discovered^ by observing if the successive
differences of altitude or distance be proportional to those of the times of
observation. If, however, the time (which is supposed to be accurately
noted,) and two of the observations be correct, the erroneous observation
may be easily rectified by the rule of proportion.
In order to attain to the greatest accuracy in deducing the mean from a
series of observations, these ought to be taken at equal intervals of time, as
nearly as possible ; such as, one minute, one minute and a hatf, or two
minutes.
Problsm X.
Owen the apparetit Time, the observed Distance between the Moon and
Sun, a fixed Star, or a Planet, the Latitude, and the Longitude iy
accctmt; to find the true Longvtudx.
RULB.
Compute the true and the apparent altitude of each object's centre, by
Problem I., IL, III., or IV., between pages 404 and 410, according as the
mood is compared with the sun, a fixed star, or a planet.
Reduce the observed to the apparent central distancei by the rule to
Problem IX., page 456 ; with which, and the computed altitudes of tlie
objects, let th^ true central distance be determined, by any of the methods
given in Problem VII., betweeki pages 433 and 453 ; and find the apparent
time at Greenwich answering to the true central distance, thus computed,
by Problem VIIL, page 454. Then, the difference between the apparent
times of observation at the ship and at Greenwich will be the longitude of
the ship or place, in time ; which is to be called east or west, according as
the apparent time at the place of observation is greater or less than that at
Greenwich.
Example 1.
August 4th, 1825, in latihide 40^25^ N., and longitude 56?S6; W.^ by
account, at 19M0T35!, apparent time, the mean of several observed
distances between the nearest limbs of the sun and moot) was 107^3' 47^;
requited the true lon^^tude of the place of observation }
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OF FINDING THB LOKOITUDB BT LUNAR OBSBRVATION8. 471
Appar* time of obs.s: 19t 10735 !
Long.56?36^W.,in
times . • . 3.46.24
lUduced time =
22*56759:
Obs. dist. between
moon and sun = 107° S'ATI
Sun's 8emi-diameter= +15.48
Moon's semi-diam. == +14.57
Appar. central diBt= 107^34^321!
To find the Sun's true and apparent Altitude :^
Sun^s dist. frommerid.=4*49725! • • . • Log. rising s 5.843150
Lat.ofplaceofob8. s=;40?25: OIN. . . . Log. co-sine = 9.881584
Stto's reduced dec. ^ 17« 1.43 N. . . . Log. co-sine = 9. 980530
Sun*s mer. aen. dist.= 23?23^ 17^ Nat. V. S.=082163
Natural number=507299 Log.=5. 705264
Sun's true central alt.=:24?14'. 197N. oo-V. S.s589462
Correction of altitudes + 1.59
Sun's apparent alt. s> 24?16U8^
To find the Moon's true and apparent Altitude &—
App. time of observ.s:19^ 10*35!
Sun's reduced R. A. = 9. 0.28
R. A. of the merid. =
Moon's red. R. A. =
Moon's dist. from mer.
Lat. of place of obs. =r
Moon's reduced decs
4Mir 3!
1.28.35
=2*42T28!
40? 25! O-fN.
13.44. 4 N.
Moon's red.horiz. par. s. 54! 167
Moon's red. semi-diam.s 14! 47 7
Augmentation of ditto = +10
Moon's true 8emi-diam.=14!577
. . Log. rising « 5.381870
. . Log. co«sine3s9. 881584
• . Log. co-sine=s9. 987402
Moon's mer. «en«dist.s26?40!567 NatV. S.a 106489
Natural number %: 178174 Log.s5. 250856
Moon's true cent, alt.:
Correction of altitude^
=45?40! 157 N.co-V. S.C284663
s ^37. 0
Moon's apparent alt.^ 45? 3! 157
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472 NAUTICAL ASTRONOMY.
To find the true central Distance, and, hence, the Longitude of the
Place of Observation :—
Half the app. cent. dist.=53?47' I6r
Half sum of the ap. alts.=34. 39. 46|
Log.diff. = . . . fl. 995327
Sum= 88?27' 2ir Log, co-sine = . . 8.431961
DiflFerence = ... 19. 7. 29f Log. co-sine = • . 9. 975343
Sum = 18.402631
Aiy;h= 80?5in0f Log. co-sine = . . 9.2013151
Half sum of true alts.si: 34. 57. 17
Sum= 115?48:27^ Log. sine :?= . . . 9.954369
Difiference = . « r 45.53.53 Log. sine ^ . . . 9.856186
Sums: 19.810555
Halfthe true distances 53 ?3K 3f Log. sine == . . . 9.905277|
True central distaiice= 107? 2'. 6r }
Dist. at 21 hours = 107.55. 2 iDiff.=0?i2^56rProp.log.= .5315
Dist.at24hours,ornoonl06.33.30 }DiflF.= 1.21.32 Prop. log.= . 3439
Portion of time = 1^56r52!Prop.log.= . 1876
Time corresponding to first distance = • 21. 0. 0
Apparent time of observ. at Greenwich = 22^56?52!
Apparent time at the place of observations 19. 10. 35
Longitude of the place of observ., in time = 3*46?17* = 56?34i 15r W.
Example 2.
September 27th, 1825, in latitude 36? 151 S., and longitude 47?301 K,
by account, at 14?58?10! apparent time, the mean of several observed
distances between the moon's remote limb and the star Aldebaran was
55?17^36f ; required the true longitude of the place of observation ?
Apparent time of observation z= . . • • 14t58?10!
liongitude 47?30' E., in time s . . — 3. 10. 0
Reduced times 11M8?10!
Observed distance of moon's remote limb s 55?17'36?
Moon's true semi-diameter ss • • • • —14.49
Apparent central distance = • • • • « 55? 2C47?
/Google
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OF FINDING THE LONGITUDB BY LUNAR OBSERVATIONS. 473
Sun's right ascension at noon, Sept. 27th, = 12^ 14T54!
Correction of ditto for 11 *48r 10! = ... +1.47
Sun's reduced right ascension s . . • • 1 2 M 6T4 1 '
To find the true and apparent Altitude of the Star Aldebaran :—
Apparent time of observation = . • «
Sun's reduced right ascension = • • •
Right ascension of the meridian = . • .
Aldebaran's right ascension =:' ; . . .
Aldebaran's distance from the meridian =
14*58-10!
12.16.41
» >
3*14?51!
4.'25;55
mir 4!
Aldebaran's dist. from merid.s 1 M 1 r 4 ! . . Log. rising = 4. 67S460
Lat. of the place of observ.=36? 15 '. 0':rS. . Log. Qp-sine=9. 906575
Aldebaran's reduced dec. = 16. 9. 1 N. . Log. co-sine=:9. 982513
Aldebaran's mer. zen. dist.=52?24! KN.V.S.=389859
Natural number = 36944 Log.as4. 567548
Star's tniealtitude=:34?58'24r Nat.co.V.S.= 426803
Correction of alt. = + 1.21
Star's appar. alt = 34?59'.45r
To find the true and the apparent Altitude of the Moon :«
Moon's reduced semi-diameter rs . • •' 14^42?
Augmentation of ditto = +7
Moon's true semi-diameter =: 14!49T
Moon's reduced horizontal parallax = 53'55T
R. A. of the meridian = . 3* 1475 1 !
Moon's reduced right ascen.s 0. 42. 28
Moon's dist. fi-om the merid.=2^32r23! . . Log. rising = 5. 328420
Lat. of the place of ob8erv.=36? 15 '. OrS. . Log. co.sine= 9, 906575
Moon's reduced declination^ 9. 19. 30 N. . Log. co-sine= 9. 994223
Moon's merid. zenith di8t.= 45?34!30rN.V.S.=300025
Natural number = 1 695 19Log.= 5 .229218
Moon's true altitude = 32? 2^10!' N.co.V.S.=469544
Correction of altitude = — 44. 32
Moon's apparent alt. = 3 1 ? 17 ' 38r
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474 NAOTICAL AtTRONOMT.
To find the true central Distance, and, hence, the Longitude of the
Place of Observation : —
Half the ap.cent.di8t.=27?31 ' 23^?
Halfdiff.ofapp»iUts. 3* 1.51. 3f
Log.diff.l9. 996651
Sum= 29?22'27rLog.si&e 9.690648
Differences . . . 25.40.20 Log. sine 9.636711
Sum= 39.324010
Half sum a 19.662005 . . 19.662005
Half diff. of true alts. = l?28f 7?Log.sine 8.408737
Arehor e6?48'20rLog.taa.ll.253868Log.sin.9.9993S5
Halfthe true distances 27?22^54ir .... Log. sinesd. 662680
TVtt6 ctotral distance *» 54 M5 'A91 \
Di8tenceat9hou«.=s56. 8. 7 iDiflF.=:l?22:i8r Prop. log,= • 3399
Di8t.atl2hr8,orm{dnt.54.40. 8 J I>lff.«1.87.59 Prop. log.* . S109
Portion of time = 2M8r 32! Prop, log.s. 0290
Time corresponding to first distance = 9. 0. 0
Apparent time of obsehr. at Greenwich se 1 1 M8?22! ^
Apparent time at the place of observ. = 14. 53. 10
Longitude of the plabe of observ.^ in time— 3^ 9*48! 3± 47? 27' east.
Example 3.;
December 25th^ 1825, in latitude 39? 13^ N.. and longitude 42?55^W.,
by account, at 14M9T27' apparent time, the mean of several obsenred
distances between the moon's nearest limb and the planet Mars was
94?40^28^ ; required tl^e longitude of the place of observation ?
Apparent time of observation = .... 14M9T27I
Longitude 42? 55 ^ W., in time = • . • + 2. 5 1. 40
Reduoad times 17-4lr 7!
/
Observed distance of moon's nearest limb =: 94?40^ 22?
Moon's true semi-diameter =: -fi5«41
Apparent central distance = » • < • . 94?56! 3?
/Google
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OF FINDING THB LONGlTODt BY LtTtlAR OBSBRVATIONS. 475
Sun's right ascension at noon, December 25th,= 18? 15T13!
Correction of ditto for 17 Mir? ' « • . • 4 3. 16
Sun's reduced right ascension = • . . • 18M8r29!
Apparent time of observation = • • * • 14.49.27
Right ascension of the meridian ::£.•• 9? 7*56!
To find the true and the apparent Altitude of the Planet Man i«
Horizontal parallax of Mars ^ 5 seconds.
R. A. of the meridian a 9t 7^56?
Reduced R. A. of Marss 13.12. 23
Mars' dist. from merid.= 4? 4r32! . . . Log. rising = 5.7136S0
Lat. of place of obs. == 39? 13'. OrN. . • Log. co-sines 9. 889168
Reduced dec. of Mars=: 8. 52. 53 S. . . Log, co-sine= 9. 994761
Max«'mer.zen.dist. -48? 5^53r Nat.V.S.=i:332142
Natural number = 395921 Log.= 5« 597609
IVii6 altitude of Mars£3l5?46U6rN.co.V. S.=728063
Correction of altitudes* + 3. 14
App. alt. of Mars = 15^50^ 0*
To find the true and the apparent Altitude of the Moon's Centre :—
Moon's reduced horizontal parallax s= 56^48C
Moon's reduced < semi-diameter 15C28^ + augmentation 13T = 15 '4 It
R.A.ofthemerid.=9i 7^56!
Moon's fed. R. A.S 7. 1.30
])'8di8t.frommerid.2* 6r26! Log. rising = 5. 171280
Lat.ofplaceofobs.39?13^ 0?N Log. co-sine = 9.839168
Moon's red. dec.= 1 9. 49. 23 N- .... Log. co-sine = 9. 9734/2
> *8 mer. zen. dist. 19?23'S7^Nat. vers. sine=056740
Natural number = 108123 Log.=5. 033920
J's true cent. alt.=56937'48r Nat.co.V.S.= 164863
Correction of alt. == — 3 1 . 7
Moon's ap. alU« 56? 6:41?
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476 NAUTICAL A8TAONOMT.
To find the true centrid DiBtance, and, hence, the Longitude of the
Place of Observation :—
Half ap. cent. di«t.=47?28: Hi
Half sumof ap.alt8.=:35. 58. 2o|
Log.diff. 19.994221
Sum= .... 83?26122rLog.co-8in. 9. 057868
Differences . . 11.29.41 Log.co-Bin. 9. 991201
Sum s 39. 043290
H^lfsums: 19.521645 . . 19.521645
Half sumof true alt8.=36? 12' 17^Liog.co.sin. 9. 906826
Ai^ch= .... 24?19:33rLog. 8ine=9.614819Log.tang.9.655197
Half the true dist = 47? 19U8ir Log. sine = ... 9. 866448
True central dist. = 94?39'.37^ )
Dist. at 15 hours = 96. 4. 2 /Diff.= l?24<25r Prop.log.= .3288
Dist. at 18 hours = 94.30. 7 (l^^ff-^ 1-83.55 Prop. log. = .2825
Portion of time = 2Mir47* Prop. log. = .0463
Time corresponding to first distance == 15. 0. 0
Apparent time of observ. at Crreenwich =17-41T47'
Apparent time at the place of observ. =: 14. 49. 27
Long, of the place of observ.^ in time = 2 t52r20! « 43?5 ^ west.
Example 4.
December 30th, 1825, in latitude 46?30^S., and longitude 84?15CE.,
by account, at 21 MO? 15' apparent time, the mean of several observed
distances between the nearest limbs of the sun and moon was 107?2'7^9
and, at the same time, the mean of an equal number of altitudes of the
moon's upper limb was 15?40'24^ ; but, for want of the necessary assist*
ants, the sun's altitude could not be taken ; the height of the eye above the
level of the horizon was IS feet; required the true longitude of the place
of observation?
Apparent time of observation = 21M0T15!
Longitude 84? 15' E., in time = 5.3/. 0
Reduced time = ; 15t33T15!
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OF FmDfNO THB IX>NGIT0I>B BY LUNAR OBSERVATIONS. 477
Observed distance between the moon and 8un = . 107? 2' 7^
Sun's semi-diameter = • +16.18
Moon's semi-diameter = +16.5
Apparent central distance s: 108?34'30^
Observed altitude of the moon's upper limb == . 15?40^24!
Moon*s true semi-diameter = . —16. 5
Dip of the horizon =: —4.4
Moon's apparent altitude = 15?20C15r
Correction = +53. 16
True altitude of the moon's centre = • ... 15? IS^Sir
Moon's reduced horizontal parallax = • • • • • 58' 47^
To find the true and the apparent Altitude of the Sun's Centre : —
Sun's horary dist. from mer.=:2*49r45 ! . . Log. rising = 5. 418280
Lat.ofthe place of observ.=46?30! OTS. • Log. co-sinen 9.837812
Sun's reduced declination =23. 8. 14 S. • Log.co-sine= 9.963583
t ___
Sun's mer. zenith di8tance=:23?21U6rN.V.S.=081988
Natural number = 165835 Log.=:5. 219675
Sun's true cent. alt.=:48?46'46r Nat. co-V. S.=247823
Correction of ditto= + 0.43
Sun's apparent alt.=: 48?47'29^
To find the true central Distance, and, hence, the Longitude of the
Place of Observation :—
O's ap.alt.48?47^29r ©'s true central alt.48?46'46r
})'sap.alt.l5.20. 15 D 'strue central alt. 16. 13.31
Sum = 64? 7'.44r ^tif"} 1.436349 Sum64?59^59r '^:*}. 1.422622
DiflF. = 33.27.14 N.V.S. .165671 Diff.32.33.15 N.V.S. . 157117
Arch first = . . 1 . 270678 Arch secondzz 1 . 265505
/Google
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478 VAVncAM. asteonoiiy.
Sum of app. alto.=:64? 7 • 44 rNatV. S.= 1 . 436349
App. cent. di8t.= 108?34'30rNAt.V.S.= 1.318546
» ■ ■
Arch third = 117803 Log. = 5.071156
Arch seconds 1.265505 Log. = 6.102264
Arch first = 1. 270678 Log.ar.co=: 3. 895965
Natural number == . 117323 Log. X 5.069385
Sum of true aIt8.=64?59'59rN.V.S.8up.l. 422622
True cent. di8t.= 107M6'.34r J N.V.sinel. 305299
Di8t.atl5hour8=108. 4.31 jDiflF. = 0^17^57^ Prop. log.= 1.001 2
Dist. at 1 8 hour8= 106. 27. 28 } Diff- = 1 - 37. 3 Prop. log.= . 2683
Portion of time =: 0?S3?17' Prop. log.=: .7329
Time corresponding to first distance = 15. 0. 0
Apparent time of observ. at Greenwich^ 15 133?17 •
Apparent time at the place of observ. zz 21. 10. 15
Longitude of the place of obs.^ in time = 5?36r58! = 84?14^30r east.
Remark.-^ln Problem XXIX, page 320, of « The Young Navigator's
Guide to the Sidereal and Planetary Parts of Nautical Astronomy," there
is an interesting method given for reducing the apparent central distance
between the moon and sun, or a fixed star, to the true central distance, by
an instrumental operation ; it being a correct mechanical mode of working
the lunar observaUons by Gunter's scale and a pair of compasses.
PaoBLBM XI.
To^ftd the LmgUude of a Place by the Eclipses of JupUer's Satellites.
First,
To know if an Eclipse wiU be visiffle at a gieen Place.
Rule.
f
Convert the mean time of the eclipse at Greenwich (as given in page ITI.
of the month in the Nautical Almanac,) into apparent time, by Problem
II., page 416; and let this time be reduced to the meridian of the place
of observation, by Problem IV., page 297.
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OF FINDING THB JLONOITUDl BY BCUFSES. 479
Now^ if at this reduced time Jupiter be not less than 6 degrees above
the horizon of the given place, and the sun be as many below it^ or stars
of the third $j(Uignitude be visible to the naked eye, the eclipse may be
observed at that placet this, it is presumed, does not require to be
illustrated by an example.
SmoNP,
To find the Longitude of the Place of Observation of an Eclipse.
RULB.
Reduce the mean time of the eclipse at Greenwich into apparent time,
by Problem IL, page 416. Then, to the observed time of the eclipse, at
the given place^ ^PP^y ^he error of the watch for apparent time, deduced
from obaervationa of the sun's altitude, or from those of a fixed star, a
planet, or the moon : hence the apparent time at the place of observation
will be known. Now, the difference between this time and the apparent
time at Greenwich will be the longitude of the place of observation in
time ; which will be east or west, according as the former is greater or less
thun the latter*
Example I.
January 8th, 1825, in latitude 39?5( N., and longitude 28?3f W., by
account, an immersion of the first satellite of Jupiter was observed at
8M2t59!, by a watch which was 1T46! fast for apparent time; required
the true longitude of the place of observation ?
Mean time of the eclipse at Greenwich = 10M0T29!
Equation of time ss _ 7, 16
Apparent time of the eclipse at Greenwichss 10? 3713!
Time of observation, per watch, a: ... 8* 1 2759 !
Watch fast = p • . . - 1.46
Apparent time at the place of observation ^ 8M 1 7 1 3 1
Apparent time at Greenwich = , . «, • 10. 3. 13
Longitude of the place of observ., in time = I ?527 0! =5; 28?0'0r west.
Hote. — If Jupiter be far enough from the meridian at the time of
observing an immersion or an emersion of one of his satellites, the
apparent time of observation may be inferred directly from his altitude; and.
If the altitude be taken at the same instant of observing the immersion or
emersion of the satellite, the use of a watch will then become unnecessary.
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480 NAtrriCAL astronomy.
Example 2.
January 2d, 1825, in latitude 39?5i;i0r N., and longitude 4?15: E.,
by account, an emersion of the first satellite of Jupiter was observed; and,
at the same instant, the altitude of that planet's centre, east of the meri-
dian, was found to be 28?49'30T ; the height of the eye above the level of
the sea was 20 feet ; required the true longitude of the place of observation ?
Mean time of the emersion at Greenwich s=
Equation of time =
Apparent time at Greenwich = • • .
Observed altitude of Jupiter's centre =
Dip of the horizon for 20 feet s • .
Jupiter's apparent altitude s • . •
Refiraction= l'43r)
Parallax = -0. 2 JDiffe'^»<^e = •
True altitude of Jupiter's centre =
7V3?50:
-14.31
6f49ri9!
28»49^80'r
- 4.17
28?45n3r
- 1.41
28?43'.32r
Lat.ofplaceofobs.=39?5in0rN. .... Log. 8ccant=0. 114812
Jupiter's red. dec. =3 19. 5.48 N Log. secant^ 0. 024583
Jupiter's mer.z. dist. 20?45'22r Nat,V. S. = 064902
Jupiter's true cent. alt28. 43. 32 N.co-V.S.= 519385
Remainder = 454483 Log.s 5. 6575 1 8
Jupiter's horary dist., east of the merid.=:4'32rl6!Log.rising=5. 79691.3
Jupiter's reduced right ascension = . 8. 43. 39
Right ascension of the meridian == . 4* 1 1 T23?
Sun's reduced right ascension = • » 21. 4. 52
Apparent time at the place of ob^erv^ =7- 6T31 1
Apparent time at Greenwich == •• 6.49. 19
Longitude of the place of obs., in umes:0^17"12! = 4?18'.0f east.
ttemarlcs.
An immer$ion of a sateUUe is, the instant of its entrance into the shadow
of Jupiter; and an emersion is that of its re-appearance out of the shadow.
The instant of an immefsion is known by the last appearance qf the
satellite; that of an emersion, by its^r^f appearance.
Tlie eclipses of Jupiter's satellites afford the readiest, and, for general
practice, the best method of determining the true Icmgitudes of places on
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OF FINDING THB LONGITUDE BT LUNAR BCLIPSBS. 481
shore : but, since those eclipses cannot be distinctly observed except by
means of telescopes of a high magnifying power, — and since these cannot
possibly be used at sea, on account of the incessant motion of the vessel,
which continually throws the object out of the field of view, — ^this method,
therefore, though the very best at land, will be but of little, if any
advantage to the mariner. It is to be observed, however, that this method
of finding the longitude is not always available ; because Jupiter passes so
apparently close to the sun at certain intervals, that, for about six weeks
in every year, both himself and his satellites are entirely lost in the superior
splendour of the solar rays.
Problem XIL
To find the Jjmgitude of a Place by an Eclipse qfthe Moon.
Rulb.
Observe the times, per watch, (regulated to apparent time,) of the
beginning and the end of the eclipse : the mean of these times will be
the apparent time of the middle of the eclipse 5 the difference between
which and that given in the Nautical Almanac, will be the longitude of the
place of observation in time ; which will be east or west, according as it
is greater or less than the time at Greenwich.
Note. — If only the beginning or the end of the eclipse be observed, the
apparent time of observation must be compared with the time answering to
the corresponding phase in the Nautical Almanac ; but, it must be remem-
bered, that it will always be conducive to greater accuracy to observe the
instants of both phases.
Example 1.
May Slst, 1825, in latitude 38?24' N., and longitude 26^0^ E., by
account, the beginning of the lunar eclipse was observed at 13^35?32!
per watch^ and the end at 14t4T47' j the error of the watch was 2T13!
slow for apparent time; required the true longitude of the place of
observatioit ?
Beginning of the eclipse, per watch, = . . 13!35r32!
End of ditto ditto ^ . . 14. 4. 7
Middle of the eclipse, per watch, = . . . 13 *50r 9^ !
Error of the watch s +2.13
Apparent time of the middle of the eclipse = 13?52?22i!
Apparent time of ditto at Greenwich s . . 12. 8. 30
Longitude of the place of observ., in time = lM3T52i!s25?58^7irE.
2 I
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482 NAUTICAL AITMNOlfy.
EsMtnple 2.
November 25th, 1825, ia latitude J6?40' N»»^nd IqPgiMe 54°40^ E.,
by account, the beginning pf the lunar ellipse was observed at 7*6"40!,
an4 the end at 9J0T55; per wat«b, not regulate,d^ la order to ftud the
error of the w^tch, the altitude of Aldebiirau, ea^t of th? meridlao, was
taken at8?7T30:, and found to be-28'?42:30r5 the height of the eye
above the level of the horizon waa 22 feet; required the true longitude oi
the place of observation ?
Time, per watch, of observing the star's altitude = 8* 7^0*
Longitude 54?40^E., in time MBS 3.38.40
Reduced times . . . 4^28750!
Sun^s right ascension at noon = 16^ 3T44!
Correction of ditto for 4^28750' «■ . . . . +0.45
■irw
Sun's reduced right ascenaipu v^ • , , ^ » , 16 1 4729 1
Observed altitude of Aldebaran =?..... 28? 42 'SOT
Dip of the horizon for 22 feet = , -^^ 4. 30
Star's apparent altitude ?= • ^ • . • • » , 28?86^ 07
Refraction SF »««•*»•«•#•• -» L44
Star's true altitude = • . • , 28?36'167
Lat.ofthe place of obs.=16?40^ O'N. , . . Log. secant— 0. OI80S9
Aldebaran's red. dec. = 16. 9. 5 N. . • . Log. secant=sO. 017489
AldebMan'si»er,z.dist.= 0?30iWr N»t,V. 8,=3000040
Aldebarw'struealt* P9 28.36. }6N.cp-V.8»»i$3l240
Remainder = 521200Log.;=5. 717004
Star's horary distance, east of the mpr.=54 M 7713 ' hof, ris)in;^5, 7531341
Star's reduced ri^ht ascension = • 4. 25. 57
Right ascension of the inusridian = . 0 * 8744 J
Sun's reduced rigbt ascension =. . 16, 4. 2^
Apparent time of obs. the star's alt, m 8! 4716!
Time of observation, per watch, :s , 6. 7«80
Watch fast tor apparent time as , 8? IS !
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OF FINDING THB VAEIAtlOH Q? rHB COMPASS.
Begmuing Qf th^ eclipse, per watph, 5= 7t 6-4Q5
Bncl of ditto ditto =9 9- 0.56
Middle of the eplipse, pef watch, = 8* 3?47i-
Errorof thn watch = . . . , . - 3,15
App. time of the middle of the eclipse = 8? (iT32i'.
Apparent time of ditto j|t Greenwich = 4. 22. 0
Longitude of the place of obs., in timer=8;38782f ! = 54?3817i? east
Hemarks.
From the two preceding examples^ it is evident that the beginning )vnd
the end of the eclipse are the principal phases frqm which the longjtpde is
to be found. If the observer be provjded with a sextant, those phases may
be observed to a tolerable degree of accuracy^ by inean^ of the largest
telescope belonging to that instrument; or they may be observed with a
good night telescope.
This method of finding the longitude at sea is evidently the most simple
of any of the astronomical methods that have been proposed for that
purpose ; however, since the lunar eclipses happen so very seldom^ there
are but few opportunities of carrying it into practice : nevertheless, when-
ever such eclipses take place, the prudent mariner will do well to avail
lumself thereof, and to determine his longitude by them accordingly.
SOLUTION OP PROBLEMS RBLATIVB TO THE VARIATION
OP THE COMPASS.
The variation of the eomposs is the deviation of the points of the
mariner's compass from the. corresponding points of the horizon, and is
denominated east or west variation accordingly.
East variatum is, when the north point of the compass is to the
eastward of the true north point of the horizon ; west variation is, when
the north point of the compass is to the westward of the true north point
of the horizon. «
The variation of tfie compass may be found by various methods, such as
amplimdes, azimuths, transits, equal altitudes/ rising and setting of the
celestial objects, &c.
2i2
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484 NAtrricAL astronomy.
The true ampUluie of any celestial object is, an arch of the horizon
intercepted between the trae east or west point thereof, and the object a
centre at -the time of its rising or setting.
The magnetic amplitude of an object is, the arch of the horizon that
is intercepted between its centre, and the east or west point of the com-
pass, at the time of its rising or setting; or, it is the compass bearing of
the object when in the horizon of the eastern or western hemisphere.
The true amplitude of a celestial object is found by calculation; and the
magnetic amplitude is found by an azimuth compass.
The true azimuth of a celestial object is, the angle Contained between
the true meridian and the vertical circle passing through the object's
centre.
The magnetic azimuth is, the angle contained between the magnetic
meridian and the azimuth, or vertical circle passing through the centre of
the object; or, in other words, it is the compass bearing of the object, at
any given elevation above the horizon.
The true azimuth of a celestial object is found by calculation; and the
magnetic azimuth by an azimuth compass.
Problem I.
Given the Latitude of a Places and the Sun'§ magnetic AmpUtude; to find
the Variation of the Compose.
Rule.
Reduce the apparent time of the sun's rising or setting to the meridian
of Greenwich, by Problem III., page 297 ; to which time let the sun's
declination at noon of the given da%be reduced, by Problem V., page 298.
Then, to the logarithmic secant of the latitude, add the logarithmic sine
of the sun's reduced declination; and the sum, .abating 10 in the index,
will be the logarithmic sine of the true amplitude, — to be reckoned north
or south of the true east or west point of the horizon, according to the
name of the declination. Now, if the true amplitude, thus found, and
the ma^etic amplitude, observed per azimuth compass, be both north or
both south, their difference is the variation; but if one be north and the
other south, their mm is the variation : — and to know whether it be east
^or west, let the observer look directly towards that point of the compass
representing the /m« amplitude; then, if the magnetic amplitude be to
the left hand of this, the variation is easterly ; but if to the righi hand, it
is'westeriy.
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OF FINDING THB VARIATION OF THB COMPASS, 485
Example 1. ..
May 20th, 1825, in latitude 48?50^ N,, and longitude 6?30: W,, ajk
about 7*40T, the sun was observed to set W. 56?42^ N,; required the
variation of the compass ? .
, Estimated time of observation ^ 7^40?
Longitude 6?30' W., in time = .••... +26
Redweddmess • • 8^ 6?
Sun's declination at noon, May 20th, . • . 19?58M3rN.
Correction of ditto for 8*6? = +4.11
Sun's reduced declination s ••••.• 20? 2^54fN.
Latitude of the place of observ.= 48?50'N. Log. secantslO. 181608
Sun's reduced declination = . 20? 2'54rN. Log. sine = 9.535057
True amplitudes . • . W. 31?23< 8rN. Log.sine = 9.716665
Magnetic amplitude = . W. 56. 42. 0 N.
Variation =?..«.: • 25? 18^.52? ; which is weet, because the
magnetic amplitude is to the right hand of the true amplitude.
Example 2«
July 10th, 1829; in latitude 18?40'. N., and longitude 73?45f W., at
about 17^29T, the sun was observed to rise E. 30?12^ N.; required the
variation of thcf compass ?
Estimated time of observation == • . • • • 17^29?
Longitude73?45' W., intime = . . • . +4.55
Reduced time = 22*24?
Sun's declination at noon, July 10th, . . 22?16n6?N. *
Corrcetionofditto for 22^24?= . . . . - 7. 13
Sun's reduced declination =s 22? 9 C 3fN,
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486 NAttxiCAt AistltONbttYi
Latitude of the place of observ.ac 18°40'. OrN. Log. seCant= 10. 023468
Sun's reduced declination = . 22. 9; 3 N. Log. sine = 9. 576395
true amplitude = . . . E. 23927' 7^1*: Log. Sine =s 9.599863
Magnetic amt)litu(ie = . fi, 30. ll b N.
Variations ...... 6?44C53r; which is e(M<, because the
magnetic kmplitu^e is to the l^ Kdfid ttf the trtie Aniplitudci
Example 3.
October 17th, 1825, in latitude 42? lO: N., and longitude 14?30^ W.,
at about 5!277, the sun was observed to set W. 7'?33'. N.j required the
variation of the compass r
EsUniated time of observation = . . • 5*27" 0*.
Longitude 14 ?30'. W., in time = . . +58.0
kediiced tiine = 8^25? 0!
Sun's declination at noon> October 17th, = 9? 15^9: S.
Correction of ditto for 6! 26 T » • • • + 5.5^
Sun'srednceddeclmatiohs . . . . '9^2inKS.
Latitude oif the place olf observ. = 42? 10 • O?^. Log. secahkrs lO. 1S0067
Sun's reduced declination = . 9. 2 1 . 1 1 S. Log. sine = 9.2 10901
True ampKtude = . . . W. 12?39:57'S. Log. sine = 9.340968
MHghetic wni)llW(fc * . W. ?.3l 0 N. •
Variations 20?12'.57?; #hM* isioert, beonne the
magnetic aqaplitude is to the right hand of the troe amplitude.
In finding the variation of the compass by this method, the sun's
amplitude shduld hk tHketi, With an azimuth compa^, Wheh HM altitude-
of his lower limb is equal to the sum of his semi-diameter and the dip of
the horiibn. Thus, if the stin's GiehiUdiamet^ be 16'5fj and *e dip of
the hor&in 4<17^ (for 20 feet), the khtik st 20^22^ is Ae hqgbt which
the lower hmb of that object should be above the horizon, at the time
of obsefrVing its 'Amplitude.
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OF FINDING TBS VAtUATlOK OF TB£ COMPASS. 467
If the index of the quadrant be set to the altitude^ thus determined^ the
sun's magnetic amplitude may be taken when his lower limb attains that
altitude, either at rising or setting ; for, although the sun is apparently so
el^vatied) f^ty m nctOMtit lof tile atoid^beritd reftticUoii, his centre is
MMdlythi^niiithehorfauiii^fiiieplftMoroltaemrtioiu *
Note. — For the principles of finding the variation .of the compass by
the amplitude of a celestial object, see '^ The Young Navigator's Guide to
the Sidereal and Planetary Parts of Nautical Astronomy,'' page 261.
Probj^m II.
Given the Laiktideofa Placey the Sim's Altitude^ and his magnetic
JzimtUhj to find tke Variation of the Compass.
RuLfi,
Reduce the appattnt time of observation to the meridian of Greenwich,
by Probleni III., page 297 ; to which time let the sun's declination, at noon
of the given day, be reduced, by PhiUem V.^ page 298, '
Find the true central altitude of the sun, by Probkm XtV*, pi^SSO)
now.
To the sun's polar distance, add its true ceitfral altitude and the latitude
of the jplace of observation ; take half their sun^ and call the differenc#
Wtween it and the jpolar distance the remainder.
llien, to the logarithmic secanls, less radius, of the tcift central altitude
and the latitude, add the logarilimiic co^^sines of the half-sum and the
remainder : half the sum of theS6 four logarithifis will be the logarithmic
€o-^Qe of an arch ; which, being doubled, will be the true azimuth^ to be
redumed from the northin north latitude, but from the south in south
latitude; towards the east in the forenoon, and towards the west in the
t(teni«Diii
Now, if the true azimuth, th^is found, and the magnetic azimuth,
bb^retv^d per iKzMnfth eompttss, k^ oh t)ie "same side of the mimiian, their
difference is the variation; but if on different sides, their sum is the
variation : — and to know whether it be east or west, let the observer look
directly towards that point of the compass which represents the true
^aimiiUhj tfiei^ if the ttii^etic azittiutli be to the l^ hand ef thts^ the
variation, is ewteidly j but if t# irtie right Aond, it is wescerlf •
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488
KATTTICAL AtTBOMOICT.
Example 1.
April 15th, 1825, in latitude 39^40^ N., and longitude 14?0^ W., at
4 M 0? per watch, the observed altitude of the sun's lower limb was 27 ? 1 1 - »
and the bearing of his centre, by azimuth compass, N,-80?37-30T W. ;
the height of the eye above the level of the sea was 24 feet ; required the
variation of the compass ?
Time of observation, per watch, :
Longitude 14?0' W^ in time as
Reduced time =
4*10r 0!
+56. 0
5* 6? 01
Observed altitude of sun's lower limb ss 27?11 • 0!
Sun's semi-diameter = • J . •
Dip of the horizon for 24 feet ^ «
Sun's apparent altitude =s . . • .
Refraction l(50r - Parallax 81 =
Sun's true central altitude = • ,
+ 15.57
- 4.42
27^22'. 15r
- 1.42
27?20^33r
Sun's dec. at noon, April 15th,=9?45'46rN.
Correction of ditto for 5 ?6r = + 4. 32
Sun'a reduced declination = 9?50a8rN.
Sun's north polar distonce = 80? 9M2r
Sun's true central altitude = 27.20.33 Log. secant = 0.0514S1
Latitudeof the place of obs.= 39.40. 0 Log. secant b Q. 113638
Sum = \i7nO'.l51
Half sum =
Remainder =5
Arch = • . • ••
True azimuth = •
Magnetic azimuth =:
Variation = • .
. 73?35C 7V Log- co-sine = 9.451150
• 6.34.^4^ Log. co-sine s 9.997133
Sums 19.613872
. 50? 9^ 9r Log. co-sine = 9.806686
N.100?18;i8rW.
N. 80.37.30 W.
19?40U8r; which is wett, because the
magnetic azimuth is to the right band of the true azimuth.
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OF FINDINO THB VARIATION OF THB COMPASS. 489
Example 2.
March lOUi, 1825, in latitude 42?4i: S.» and longitude 14895^ E., at
19^25? per watch, the observed altitude of the sun's lower limb was
18?3' , and the bearing of his centre, by azhnuth compass, S.I08?37 -SOrE. ;
the height of the eye above the level of the sea was 19 feet ; required the
variatioa?
Tim%of observation, per watch, =
Longitude 148?5 i K, in time =: •
Reduced time =: • • • • •
Observed altitude ofthe sun's lower limb=: 18? 3' Or
Semi-diameter = '
Dip of the horizoh := *••••.
Apparent altitude = ••.•••
Refraction 2^5ir - Parallax 8 r = . .
Sun's true central altitude = . « • .
Sun's dec. at noon, March 10th, = 4? 5'43r S.
Correction of ditto for 9i32T40!= - 9.21
19?25? o:
9.52.20
9^32T40!
:18» S'. or
+ 16. 7
- 4.11
18?l4<56r
- 2.43
I8?i2n3r
Sun's reduced declination = . . 3?56'.22rS.
Sun's south polar distance = . 86? 3^38r
S.un's true central altitude = . 18. 12. 13 Log. secantr: 0.022298
Latitude of the place of observ. = 42.41. 0 Log. secant= 0.133647
Sum=: 146?56fTir
Half sum ;= . 73?28^25ir Log. co-siner: 9.454014
Remainders . . • • . • 12.35. 12^ Log. co-sine= 9. 989435
Sum= 19.599394
Archr: 50?54'42r Log. co^inei: 9. 799697
TVue azimuth = . • • . S. 101 ?49^ 24? E.
Magnetic azimuth = . . . S. 108.37.30 £.
Variations 6M8'7 6! ; which is ecue, because the
magnetic azimuth is to the left hand of the true azimuth.
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490 IIAVTICAIi AflmOIiOMT.
Note. — After this manner may the variation be deduced from the true
altitude and magnetic bearing of a fixed etar, a planet, or the moon, as
will be seen by referring to ^^ The Voung Navigator's Guide to the Sidereal
attd Pl&il^tA^ Pdrtt tdt Ndtotlcal AdtfrOH«my/' jMige 263 ; whtM^ the prb-
eiples of this methed Are fimiiliaHy explained hf % ttemogniphia
proj^tiob»
^ •-'• --
J new MMod oft&mputing ike CfM JMmyHh «(^(i todoBtUl Otjecty andy
ikm^y finding the Fi»Uaim ^Ifo Ctoni|MM«
RctLB.
From the natural versed «ine supplement •f the stun of the latitude and
the true altitudei subtract the natural versed sine of the olgect's polar
distance : to the logarithm of the remainder add the logarithmic secants
of the latitude^ and the true altitude : the sum of these three logarithms,
rejecting 20 ftt)M til( index, will be the logariihai «lf' the MsUiral versed
sine supplement of the true azimuth; to ^e ret^k^ti^d (torn thfe north in
north latitude, but. ttom the south in south latitude ; the difference between
which and the magnetic azimuth will be the variation ot the compass^ as
before*
Example 1.
October 17th, 1825, in IfttitXnfe 42?10; N., imd hmgitwl* U^dO'. W;,
at 3^2? per watch, the menn of several observed altitudes of' the sun's
lower limb was 23 939 '34^, And the miean of an equal ttttmb^r of his
central bearings, by *tiiftiuth compass^ N. lO$?28!56? W.; the height of
die eyh above th« level x^t the hroHzmi \^as 17 feet} leqtltted thfe ^iffatioti
of the compass ?
^Tlttie Of obseiVation, pet tjlitch, = ; . 3? 2? 0^
Longitude 14^30 J W,, iti, time == . • +58. t)
Redtrced time = 4t 0? 0!
Sun's declin&tibn at noon, October 1 7th, = 9? 15 '19? S.
Correction of dilftofi* 4 *0*«! ds . . -f S.«9
Sun's reduced declination zz . . . . 9? 18'58r S«
Sun's north poUrdistMice a . ^ > % Mn8<i8? .
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OP FINDING TUB V^UItAtlOlC W TBB COMPASS. 491
Ob«erV«d ftltittide of Sun's lotin^r limb = 2S?89C34?
Semi-diatneter = ; • . i • ^ • +16. 5
Dip of tht horizon r: i 4 v * . • — 3^57
Appfrretil altitude =: ....*. 23^51 : 49^
Refraetion 2C9? - Parallax %: . . , - 2. 1
Sun's t^ue central altitude = . . •• . 23?49Ui?
Lat.of placeofobs.=:42?10' Or ... . Log. secant = 10. 130067
Sun's true cent. alt.=:23. 49. 4 1 . . . . Log. secant =: 10. 038692
Sum=. . . . 65?59Ulrrl.V.S,8up.= i. 406821
Sun's N. polar dist.=99. 18. 58 NatV. sine=l. 161881
Remainder = . 244940 Log.=:5. 389060
Thiea^muth=:N.12»?41-63rW. N.V.SvStip.2:3612o9 Li«.=:a. 557819
Magnetic do^z N>109-. 28. 56 W.
Variations: . 20?12'57rj which is west, because t\\e magnetic
azimuth is to the right hand of the true or computed azimuth.
*£!»r(itnjMv fi.
Pecember 9th| 18255 in latitude i9?40^ N., Ihe true attitucle of ^he
star Capella was 20*? 10'^ and his bearings by azimutli compass,
N. 41 ?0' E. ; required the variation ?
Latiifplaceof obs.=:l9?40^ 0* . . . . Log. af^cant =: It). 0261(18
dapelfe's true ilt. = 20. 10. X) . . . .' Log. ^ckdt = 10.0fi747»
Sum= . . . . 39*50^ OrN.V.S.sup.zz 1.767911
Capella'sN.pol.dis.=:44. 11.24 N.Vwsin'd = .282968
ReteiuAder := I-. 4949ii!iiogA9. mHO
Trueazimuthr:N.47? 9U5rE. N.V.S.sup'.= 1.67«»28L^.3«»225a8ft
Magnetic do.= .N. 41. 0. 0 E.
Variation =t . 6? 9 '43 1''; which is east, because the magnetic
aan^mith ts 1» the le/t hakd tf iHe true or counted wHBmth.
Remarks — Instead of finding the natural versed sine supplement of the
aiitt tif the tlire« k^garithnn^ that mem may lie considered as a l<^garithmie
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492 NAUTICAL ASTRONOMY.
rising. In this case^ if the supplement of the time corresponding thereto
be taken from Table XXXII.^ and converted into degrees, by Table L or
otherwise, the result will be the true azimuth. Thus, in the last example,
the sum of the three logarithms is 6. 225289 ; the time corresponding to
this, in the Table of Logarithmic Rising, is 8t5lT21', which, taken from
12 hours, leaves 3*8T39V; and this, being converted into time, gives
47^9 '45^ for the true azimuth, which is precisely the same as above.
Problem IIL
To find the Variation of the Compois by Observations of a circunqH>lar
Star.
Rule.
From the log. co-sine of the star's declination, (the index being increased
by 10,) subtract the logarithmic co-sine of the latitude : the remainder will
be the logarithmic sine of the star's greatest eastern or western azimuth
(according as it may be situated with respect to the meridian) ; to be
reckoned from the north in north latitude,, but from the south in south
latitude. Then,
From the logarithmic sine of the latitude, (the index bring increased by
10,) subtract the logarithmic sine of the star's declination, and the
remainder will be the logarithmic sine of the star's true altitude when at
its greatest eastern or western azimuth. Set the index of the quadrant -Co
this altitude, and, when the stir has attained it, let its bearing be taken by
the azimuth compass ; the difference betweefn which and the computed
azimuth, when they are of the same name, or their sum when of contrary
names^ will be the variation ; which will be east^ if the observed or magnetic
azimuth be to the left of the computed azimuth ; otherwise, west.
Example 1.
January 1st, 1825, in latitude 41^53'. S., the greatest eastern azimuth
of the star Canopus, by azimuth compass, was S. 72?50^ E. ; Required the
variation of the compass ? * -
To find the Star's Altitude when at its greatest Azimuth :—
Latitude of the place of observ. = 41 ?53 ' Or S. Log. sine=:9. 824527
Reduced declination of Canopus =: 52. 36. 10 S. Log. sinez=9. 900063
Star's alt. at greatest azimuth = 57?10UOr Log. Btne=9t 924464
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OF FINDING THB VARIATION OF tHB COMPASS. 493
To find the Star's greatest eastern Azimuth : —
Reduced declination of Canopu8=52?36^ lOr S. Log. co.8ine=9. 783430
Lat. of the place of observation = 41. 53. 0 S. Log. co-8ine=:9. 871868
Greatest eastern azimuth = S. 54?39'45rE. Log. sine = 9.911562
Magnetic azimuth = • . S. 72.50. 0 E.
Variation == ...... 18? 11/. 45 ^ ; which is ea«^ because the
magnetic azimuth is to the left hand of the computed azimuth.
Example 2-
December 31st, 1825, in latitude 43?45< N., the greatest western
azimuth of the star Dubhe, by azhnuth compass, was N. 16?56^ W. ;
required the variation of the compass ?
To find the Star's Altitude when at its greatest Azimuth:—
Latitude of the place of observ. =r 43?45'. OrN. Log. sine= 9. 839800
Reduced declination ot Dubhe = 62. 4 1 . 19 N. Log. sine=9. 948670
Star's altitude at greatest azimuth=:51? 6^ 8r Log. 8ine=9.891130
To find the Star's greatest western Azimuth :—
Reduced declination of Dubhe = 62^1^ 19?N. Log. co-8iue=9. 661648
Lat. of the place of observation = 43. 45. 0 N. Log. co-8ine=9. 858756
Greatest western azimuth = N. 39?25 ^58rW. Log. sine = 9. 802892
Magnetic azimuth = . . N. 16.56. 0 W.
Variation = '• 22?29'58^ ; which is west, because the
magnetic azimuth is to the right hand of the true or computed azimuth.
- Itemarks*
In the above method of finding the variation of the compass, the star's
declination must be greater than the latitude of the place of observation,
and of the same name.
A star, or other celestial object is said to be circumpolar when its
^distance from the elevated pole is less than the latitude of the given place
(the declination and latitude being of the same name) ; because, under
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494 WAUTICAL ASffftONOMr.
such circumstances, the object comes within the circle of perpetoal
apparition, and r-evolves round the celestial pole without erer settidg^ or
going belpw the horizoi^ of t^at place.
The variation of the coinpass may be found by equal altitudes of the
fixed stars J as thus :—
Let the star'^ altitude be observed i|i the eastern hemispher^j wb^V) it is
at least two hours distant from t\\e meridian } and, at the s^m^ ivistant, l^t
its bearing be taken with an azimuth compass : then, when the star comes
to the same altitude in the western hemisphere, }et its azimi|th be -again
taken. Npw» half the di^ience between the eastern a^d western
azimuths will be the variation ; which, when the observations are reckoned
from the south point of the compos, will be east or west according as
the eastern or western azimuth is the greatest ; but if they be reckoned
fropi the nortl^ point of the copip^ss, a poptrary prqpf ss is to be ol^^rv^ :
that is, the variation js tp be called^^o^^ if the western aaiimth be th^
greatest ; but west if the eastern azinmfh be th^ ^eatest. The v^at^qii
also may be found, by observing the points of the compass upon which a
fixed star rises and set^ ; then, half the difference between those points
will be the variation of the compass, as before.
Nof^.— The above met^iod of finding the variation of the compass by
observations of a circumpolar star, is clearly illustrated in *' The Young
Navigator's Guide tp the Sidereal and Planetary P^ts of Nautical ^strpr
nomy,'' between pages 267 and 27 1 •
PRO^LEIf IV,
Tojind tJie Variation of the Compass by the magnetic Bearing ofajixei
SlaTf or Planet, taken at tfie Tim^ of its Transit, pr Passage over any
known Meridian.
Rnd the apparent time of the star's transit or passage over the meridian
of the given place, by Problem XII,, p^ge 317 ; but if the object selected
for observation be a planet, its apparent time of transit, as given in the
Nautical Almanac, is to be reduced to the meridian of the place of
ob&ervation, by Problem XI., page S15. Let the watch be well regulated
to apparent time under the meridian of the given place, and it will show
the instant of the star's or planet's transit over that meridian ; at which
instanjt its bearing, by asimuth compass^ ia to be carefully taken : the diftr-
f )ice betw^n wb^ck md tbe.noidi or SQUtb point of the fiompats (aefiordiog
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OF FINDING TRX VAftUTIOV 69 VHB COMPASS. 495
to the hemisphere In whioh the star may be posited), wilLshow the deviation
of the needle from the true corresponding point of the hoiizon $ then, if
the observed or magnetio aaimuth be to the left hand qf the meridian, the
^aviation is easterly y but if to the figki hmd,- it is westerly.*
Example \.
January 2d, 1885, In latitude 20?)0f N., and longitude IMtaOf S., at
IP 28? 15! apparent time, the star Canopus was on the meridian, and
bore, by azimuth compass, S. 9?30' E. ; required the variation ?
Sokdion.— The observed or magnetic bearing of the star 9?30' is the
variation ; and is easterly, because it is to the left hand of the meridian.
JExqmpk 2.
January 1st, 1825, in latitude 34?25' S., and longitude 18?52' E., at
I4?8TI5 ! apparent time, the planet Jupiter was on the meridian, and bore,
by aoimuth compass, N. S5f Sd' B. ; required the variation )
Sobitum^^The observed bearing pf t|^f planetj^ ?5?36^ is the variation;
which is west, because the magnetic bearing or azimuth is to the right
hand of the meridian.
The lesr the altitude of die star or planet^ and the greater its dedinatioSi
the more accurately will the variation be obtained. When ^e north polaf
star is in (be sam^ vertical circle wi(h the star Aliofh or the 9^ Cor Caroli,
it will be on the meridian, or nearly so ; and if its azimuth be observed
at that time, the variation will be pbt^ned as before. If two stars be
observed to be vertical, whose right ascensions are either equal or differ
180 degrees, they will be on the meridian: the azimuth of either may
then be taken, but that which is nearest to the elevated pole should be
preferred ; whence the variation may 4)e inferred, in the same manner as
if i^ apparent tin^e of transit ha4 been computed.
* In like manner may the yarlatioa h$ d$$^rfalntA at nopfi ; vis., I^y Qbferving tb^
mapietic bearing of the sun at the time of its being on the meridian : and, if the place
of ffbfepr^tiaD If^ canai4erably distant fvofn th^ equ^r^ ^ very rig^ degree of accuracy is
no^ necessary in the moment of observing the sun's bearing ; since, in such a place, an
error of 5 minutes in the time, before or after noon, wHl only produce ao error of about
kai/a fuarier of apMii in the variation $ which comes suftcicntly peer the triftb for mosf
9Atttic^ ptti|^%»f f If finpe, tfiis piieth9d va^y often prove useful ip ques where the fnariner
is prevented, by clouds or other unavoidable causf 8| from ascertaining, fn thf forenoon^ the
tnie value of the magnetic variation.
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496 NAUTICAL ASTRONOMY.
The variation may also be deduced from the magnetic azimuth of a
fixed star at the apparent time of its transit below the pole : this time may
be always known, by adding 12 hours, diminished by halfthe.variaiian qf
the mm'g right asceruiion an the given day, to the computed apparent time
of the star's superior transit above the pole.
The number of brilliant stars which pass over the meridian of a ship at
night, and the readiness with which their respective times of transit may
be found, render the above method of finding the variation of the compass
at sea both desirable and convenient to the practical navigator.
Problem V.
Given the true Course between two Places^ and the Fariaiionqfthe
Compass; to find the Magnetic or Compass Course.
Rule.
' When the variation is westerly, let it be allowed to the right hi^id of the
true course ; but when easterly, to the left hand : in either case, the
magnetic or compass course will be obtained.
Example 1.
Required the course, per compass, from Scilly to Cape Gear, the true
course being N. 52?55^ W., or N.W. } W« nearly, and the variation 2i
points westerly ?
Solution.— The variation 2| points, being allowed to the right hand of
the true course, because it is westerly, shows the magnetic course to be
N.N.W. i W.; which, therefore, is the course which a ship must steer by
compass from Scilly to Cape Clear, provided the variation be a3 above.
Example 2.
Required the course, per compass, from Port Royal, Jamaica, to Santa
Martha, Columbia; the true course being S. 21?42^ E., or S«S£. nearly,
and the variation about half a point easterly ?
Solution, — ^The variation i a point, being allowed to the left hand of the
true course, because it is easterly, shows the magnetic course to be
S.S.E. i E. ; which, therefore, is the course which a ship must steer by
compass from Port Royal to Santa Martha, provided the variation be as
above,— and independent of currents.
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\
OF FINDING THB VARIATION OF THE COMPASS. 497
Problem VI.
Given the magnetic Course, or that steered hy Compass, and the
Variation; to find the true Course.
Rule.
If the variation be westerly, it is to be allowed to the left hand of the
course steered by compass; but if easterly, to the right hand: in either
case, the true course will be obtained.
Example 1.
Let the magnetic, or course steered by compass, be E. by N. | N., and
the variation 1^ point westerly ; required the true course ?
Solution, — ^The variation 1| point, being allowed to the left hand of the
compass course, because it is west, shows the true course to be N.E.
byRJE.
Example 2.
Let the course steered by compass be N.W. f W., and the variation one
point and three-quarters easterly ; required the true course ?
Sobition, — ^The variation 1|^ point, being allcwed to the right hand of
the magnetic or compass course, because it is easterly, shows the true
course to be N.W. by N.
AZIMUTH COMPASS;
The card being graduated on an improved principle, so as to be more
particularly adapted to the taking of amplitudes and azimuths, the measur-
ing of horizontal angles, &c. &c. ; being thus rendered far more applicable
to nautical purposes in general than that which is now in common use at sea.
The azimuth compass, as well as the mariner's compass, is an artificial
representation of the horizon of any place on the terrestrial globe : it
consists of a circular card, divided into 32 equal parts, called points or
rhumbs; and, since the circle contains 360?, each point is equal to
2 K
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NAUTICAt A6TR0N0MT«
11915' : for 360? h- 32 = ll?15' .♦ The four principal points of the
compass, viz., N,, E., S,, and W., are called cardinal points; the others
are compounded of these, and are named according to the quarter in which
they ore situated.
To the under side of the card, and in the direction of its north and
south line, a bar of hardened steel is attached, called the needle, which,
being touched by a load-stone, acquires the peculiar property of pointing
north and south, and thus directs the different points on the card to the
correspondent points of the horizon. In the centre of the needle there is
a small socket, by means of which it is placed, with its attached card, on
an upright pin called the pivot or supporter, which is fixed in the bottom
of a circular or conical brass box : on this pin the needle turns freely, and,
by its magnetic property, the several points of the compass card keep
always in the same <Urection, very nearly; though these do not always
indicate the true correspondent points of the horizon, beeause of the
aberration which the needle suffers, owing to that secret and unknown
agency which causes its north and south poles to deviate more or less from
the respective correspondent poles of the world.
However, since the compass is an instrument with which mariners are
well acquainted, it is not deemed necessary, in this place, to enter any
farther into its description. Hence, I shall merely point out some of the
many advantages which a compass card, graduated on the above principle,
possesses over those now in general use at sea. In this card, the circular
ring of silvered brass is to be sufficiently broad to admit of four concentric
Bpaees* The outer edge of the ring is to be graduated, nvnthematicaUy
correcHy^ to every 20th minute of a degree (though, for vrdnt of room, the
present card is only graduated to every 30th minute of a decree), to which
a vernier is to be adapted, containing 20 divisions on each side of its nonius
for the purpose of subdividing the divisions on the card into minutes of a
degree.
The interior surface of the vernier should be ground concave to the
segment of a circle, whose radius is equal to that of the card. The remote
edge of the inner concentric space, on the silvered brass flat ring, may be
graduated similarly to that of the outer edge, so as to render it more
convenient in reading off amplitudes according as they may be reckoned
from the prime vertical, or from the meridian.
The first space on the broad ring of silvered brass, viz., that next the
points of the compass, is particularly adapted to taking amplitudes when
the observations are reckoned from the east or the west points of the
* Tabki XXXIII. contains the different ang^les which ereiy point and quniter^peiDt'of
the compass makes with the meridian; and Table XXXiV. contains tfaeiogaithaucaiBflB>
taofent^ and secants of ev€i;y point and quartor-poiAt of the compass*
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OF FINDING THB YABIATIOM OP THE COMPASS. 499
horizon; and^ therefore, it b numbered both ways, from those points,
towards the meridian : that is, from 0? to 90?. The second space being
adapted to Jiorizontal azimuihSf viz., to amplitudes reckoned from the
meridian, is therefore numbered both ways, from the north and south
points of the horizon towards the east and west points thereof: that
is, from 0? to 90?, in a contrary order to the last. The third space is
intended for the accommodation of an azimuth when the obserratioQ is
reckoned from the south in north latitude, or from the south la south
latitude : hence, it is numbered both ways from the south to the north
point of the compass, or from 0? to 180?. The fourth, or outer space, is
designed for azimuths reckoned from the north in north latitude, or from
the north in south latitude, according to the will of the observer} an^
therefore, it is numbered both ways from the north to the south^ or from
0? to 180?, &c.— See the Frontispiece to this volume.
Besides the evident uses of a compass card, graduated after this maimer^
in observing amplitudes and azimuths, it will also be found of the greatest
utility in taking correct surveys of coasts and harbours, and in settling the
true positions of places on shore from a knoum position at sea. It may,
moreover, be applied successfully to many astronomical purposes ; nay, it
may even be applied to the determination of the longitude by Imar
observations, as thus :— Let two observers, with two good compasses of
the above description, take the azimuths of the moon and sun, or a fixed
star, &c., at the same instant ; then, if those two azimuths be reckoned
from the same point of the horizon, their sum, subtracted from 360?, will
be the angle at the zenith comprehended between the zenith distances ci
the objects ; with which, and the true zenith distances of the objects, the
true central distance may be found by oblique angled spherical trigoao*
metry, PirobTem III., Remark I or 2, page 203 or 204 ; and, hence, the
longitude of the place of observation, by Problem VIIL, page 454.
An azimuth compass of this description would be of real advantage to
the practical navigator ; whereas, the one now in common use at sea is so
very ill adapted to the important purposes for which it is designed, that
it is very seldom resorted to for those purposes; and, therefore, it is
scarcely ever seen upon deck, except for the simple purpose of comparing
its parallelism with that of the binnacle, or steering compass.
2k2
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500 NAUTICAL ASTROMOMT.
SOLUTION OP PROBLEMS RELATIVE TO PINDING THE
APPARENT TIMES OF THE RISING AND SETTING
OF THE CELESTIAL BODIES.
Problem L
Given the Day of the Mmth, the Latitude of a Place, and the Height of
the Eye above the Level of the Horizon : to find the apparent Times qf
the &m*s Rising and Setting.
RULB.
Let the sun's declination^ at noon of the given day^ be reduced to the
meridian of the given place, by Problem V., page 298 ; then, to the
logarithmic tangent of this reduced declination, add the logarithmic
tangent of the latitude ; and the sum (abating 10 in the index,) will be
the logarithmic co-sine of an arch ; which, being converted into time, will
be the approximate time of the sun's rising, and its supplement to 12
hours will be that of the sun's setting, the latitude and the declination
being of the same name ; but if these elements be of contrary names, the
above arch, reduced into time, will be the approximate time of the sun's
setting, and its complement to 12 hours that of the sun's rising.
Reduce the approximate times of rising . and setting, thus found, to the
correspondent times at Greenwich, by Problem III., page 297 , to which
times, respectively, let th6 sun*s declination be reduced, by Problem V«,
page 298; then,
To the aggregate of 90 degrees,* the horizontal refraction,t and the
dip of the horizon, diminished by the sun's horizontal parallax,^ add the
sun's polar distance, and the co-latitude of the place of observation : take
half the sum ;^ the difference between which and the first term, call the
remainder.
Now, to the logarithmic co-secants, less radius, of the polar distance,
and the co-latitude, add the logarithmic sines of the half sum, and of the
remainder : half the sum of these four logarithms will be the logarithmic
sine of an arch ; which, being doubled, and converted into time, will be
the apparent time of the sun's rising. In the same manner the apparent
time of the sun's setting is to be computed ; but, in this case, the half sum
of the four logarithms is to be considered as a logarithmic co-sine.
Example 1.
Required the apparent times of the sun's rising and setting, July 13th,
1824, in latitude 50?48' N., and longitude 120? W., the height of the eye
above the level of the sea being 30 feet ?
« The sun's distance from the zenith when hU centre is in the horizon,
t The horizontal refraction of a celestial object is 33 minutes of a de|^e.
t The sun's horiaontal parallax is about 9 seconds.
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APPARENT TIMB OF RISING OR SETTING OF A CBLBSTIAL OBJECT. 501
Sun's dec. at noon, July 13th, = 2IM9^5HN.
Con of do. for long. 120? west = -- 2. 59
Sun's reduced declination = 21 ?46'.52rN. Log. tangent= 9.601613
Lat. of the given place = . 50.48. 0 N. Log. tangents 10. 088533
Arch= 60^39 U7r Log. co-sine = 9.690146
Approx. time of sun's rising =s 4 1 2T39 ', Appr. time ©'s set 7 ? 57 ?2 1 !
To find the apparent Time of the Sun's Rising :— c. '^
Approximate time of the sun's rising = 4* 2T39f
Longitude 120? W., in time =s . . + 8. 0. 0
Greenwich time past noon of given day = Ot 2?39!
Sun's declination at noon, July 13th, = 21?49'5irN.
Correction of ditto for 0?2?89! = . . — 0. 1
Sun's dec., reduced to Greenwich time =s 21?49'50fN.
Sun's north polar distance = . • • . 68? 10. 10?
90? + 33^ + 5'.15r - 9r = 90?38: 6?
Sun's polar distance = . . 68. 10. 10 Log, co-secant= 0, 032317
. 39.12. 0 Log. co-secantss 0. 199263
Co-latitude =
Sums
• • •
Half sum = •
Remainder ^
198? on6r
99? 0< 8r Log. sine =. 9.994617
8.22. 2 Log. sines. 9.162914
Sum = 19.389111
Archs 29?39^57i? Log. sine == . 9.694555^
Arch doubled = . . . . 59?19^55r = 3J57?20t j which, therefore,
is the apparent time of the sun's rising.
To find the apparent Time of the Sun's Setting :~
Approximate time of the sun's setting = 7^57*21!
Longitude 120? W., in time ==: . . . + 8. 0. 0
Greenwich time past noop of given day =5 15 f 57T21 J
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- K'
508 NAtmCAL ABTROKOMY.
Sun's declination at noon, July 13th^ s 2 1 ?49 1 5 KN.
Correction of ditto for 15!57"2n =s . — 5. 58
Sun's dec.^ reduced to Greenwich time =s 21?43^53fN.
Sun's north polar distance =s .... 68? 16' Ti
90? + 33^ + 5n5r — 9r = 90?38' 6?
Sun's polar distance s . . 68. 16. 7 Log. co-secant^ 0. 032017
• 39.12. 0 Log. co-secant= 0. 199263
Co-latitude =
Sum = • •
Half sum =s
Remainder b
198? 6n3r
-'
99? S: 6jr Log. sine =3 . 9.994558
8,25. Oi Log. sines: . 9.165461
Sums 19.391299
Arch= 60?15^ 6r Log. co-sine = 9.695649$
Arch doubled =3 . . . 120?30^ 12r = 8*2? 1 ! j which, therefore, is
the apparent Ume of the sun's setting.
Example 2.
Required the apparent times of the sun's rising and setting, October 1st,
1824, in latitude 40?30^ N., and longitude 105? E. ; the height of the eye
aboye the level of the sea being 29 feet ?
Son's declination at noon, Oct. lst,=:3? 16^ 6rS.
Correc. of ditto for long. 105? E. = - 6. 48
Sun's reduced declination = • 3? 9M8rS. Log.tang.s: 8.741316
Latitndeof the given place == • 40.30. 0 N. Log.tang.= 9.931499
Ai«h m • 87?18f 6r Log. co^sine 8.672815
Appr^Xf time of the sun's setting = 5 149712! Appr.time O'sris.6M0T48!
To find the apparent Time of the Sun's Setting :—
Approximate time of sun's setting » . . . 5 * 49? 1 2 !
Longitude 105? B., in time »»....- 7. 0. 0
Greenwich time past noon, September 30th, s 22t49?12!
Sun's declination at noon, September 30th^ =r 2?52U6? S.
Correction of ditto for 22U9?12! s . v . +22. 11
Sun's declination, reduced to Greenwich time :& 3? 14C57? S.
Sun's north polar distance as • . • • • 98? 14^57?
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APPARBNT TIMB OF RISING OR SITTING OF A CELESTIAL OBJECT. S03
90? + 381 + snor - 91ss 90988^ K
Sun's polar distance =
Co-latituda c3 • • •
Sum =
• • • •
Half sum = .
Remainder =
98. 14. 57 Log. co-secantsO. 000699
49.30. 0 Log. co-8ecaat«0« U8954
233^22158r
116Ml^29r Log. sine = . 9.951065
26, 3,28 Log. sine = . 9.642739
Sum= 19.713457
Archs= 44? IMl^? Log. co-sine = 9.856728|
Arch doubled = .... 88? 3^23? = 5*52?13i! ; which, there-
fore^ is the apparent time of the sun's setting.
To find the apparent Time of the Sun'sfiising :—
Approximate time of sun's rising = . , • 6M0T48?
Longitude 105? E., in time = .... — 7* 0. 0
Greenwich time past noon^ September SOth^s: 1 1 ! 10?48!
Sun's declination at noon, September 30th, a 2?52'46f S.
Correction of ditto for UU0T48! ss . . « +10.52
Sun's decIinaUon, reduced to Greenwich time, == 3? 3'38? S.
Sun's north polar distance = 93? 3^38r
90? + 331 + 5'lOr - 9? = 90?38' 1?
Sun's polar distance s . • 93. 3.38 Log. eo-secantssO. 000620
Co-latitudes 49.30. 0 Log. co-secant sO. 118954
Sum» ..<... « 23S?1U39?
Half sum = 116?35U9i? Log. sine » , 9.951423
Remainders 25.57.48^ Log. sine = . 9.641274
Sum= 19.712271
Archas 45?53<28r Log. sine a « 0« 8561351
Arch doubled = .... 91?46^56f = 6*5T8! ; which, therefore,
is the apparent time of the sun's rising.
See Examples 1 and 2, page 125j and, also, the Example, pagesl26,1274
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504 NAUTICAL A8TROKOMT*
Remark, — ^If the equated or mean times of the sun's rising and setting
be required, then, to the apparent times, found as above, apply the reduced
equation of time, as directed in Problem I., page 415 ; and the result will
be the mean times of that object's rising and setting.
Probijbm IL
Given the Latitude of a Place, and the Height of the Eye above the Level
of the Horixon ; to find the apparent Times of the Rising and the Setting
of a fixed Star.
RtJLB.
Compute the apparsnt time of the star's transit, or passage over the
meridian of the given place, by Problem XII., page 317 ; then.
To the aggregate of 90 degrees,* the horizontal refraction,t and the
dip of the horizon, add the star's polar distance and the co-latitude of the
place of observation : take half the sum ; the difference between which and
the first term, call the remainder.
Now, to the logarithmic co-secants, less radius, of the polar distance,
and the co-latitude, add the logarithmic sines of the half sum and of the
reminder : half the sum of these four logarithms will be the logarithmic
co-sine of an arch ; which, being doubled and converted into time, will be
the star's semi-diurnal arc, or half the time of its continuance above the
horizon ; which is to be reduced to apparent solar time, by subtracting
therefrom the proportional part corresponding to it and the variation of
the sun's right ascension for the given day : this is done by Problem V.,
page 298. Now, the apparent semi-diurnal arc, thus found, being applied
by subtraction and addition to the apparent time of the star's transit over
the given meridian, will give the respective apparent times of its rising and
setting at that meridian : these may be reduced to the mean times of rising
and settings by Problem L, page 415.
Example 1.
' Required the apparent times of the rising and setting of the star « Arietis,
January 1st, 1824, in latitude 50^48' N., and longitude 30?0! E. ; the
height of the eye above the level of the sea being 16 feet ?
* This is the star's distance from the zenith when its centre is in the horizon,
t The horizontal refraction of a celestial,object is 33 minutes of a degree« The fixed start
have no sensible paraUaxt
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APPARBNT TIMB OP RISING OR SETTING OP A CELESTIAL OBJECl*. 505
*'8dec,,red.togivenday, 22?S7C33^N.,&it8R^.= l JSr^lG!
Sun's right ascension at noon of the given day s= 18. 43. 58
Approx. time of star's transit over the meridian = 7*13?18'.. 7M3T18!
Longitude of the given place 30?0^ E.^ in times — 2. 0« 0
Corresponding time at Greenwich = • • • • 5M3?18!
Reduc.oftrans.aiis. to 5M3T18!&4T24! the var. of sun's R.A.s - 0.57
Apparent time of star's transit over merid. of the given place ss 7^ 12721 !
90 degrees + 33^ + 3^50r = 90?38^50r
Star's north polar distance = 67. 22. 27 Log. co-secant=0. 034781
Co-latitude of the given place=39. 12. 0 Log. co-secant=0. 199263
Sum = 197?13n7C
Half sum = 98?36:38jr Log. sine = . 9.995077
Remainders 7«57.48i Log. sine = • 9.141580
Sum =19.370701
Arch= 61? 0C58r Log. co-sine = 9.685350^
Star's semi-diurnal arc = . 122? lC56r, in time = . . 8? 8? 8!
Prop, part of variation of sun's R. A. 4T24 ? ans. to 8^ 8?8 ! = — 1 . 30
Star's apparent semi-diurnal arc = 8i 6T38!
Apparent time of the star's transit over the given meridian = 7« 12. 21
Apparent time of the star's rising past noon, Dec. 3Isty 1823^=: 23 * 5?43 '
Apparent time of the star's setting past noon of the given day=15M8T59!
Example 2.
Required the apparent times of the rising and setting of the star Sinus,
January 1st, 1824, in latitude 40?30: N., and longitude 120?01 W.; the
height of the eye above the level of the horizon being 46 feet ?
*'8dcc.,red.togivenday,16?28^53^S.,&it8R.A.=6*37?23!
Sun's right ascension at noon of the given day = 1 8. 43. 58
Approximate time ofstar's transit over the merid.= 1H53T25! Ilt53t25!
Long, of the place of observ. 120?W., in time=: 4- 8. 0. 0
Corresponding time at Greenwich = . . • . 19* 53? 25!
Reduc.oftrans.ans.tol9?53r25!&4T24!thevar.ofsun'sR.A.= - 3r39!
Apparent time of the star's transit over the given meridian = 1 1 *49T46 !
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506 NAUTICAL ASTRONOMY.
90 degrees + 33J + 6C30r a 90?39f30?
Star's north polar distance » 106. 26. 53 Log. co^secants 0. 018220
Co-latitude of the place = • 49.30. 0 Log. co-8ecant=: 0.118954
Sum= 246?38^23r
Half sum = 123? 19Ul|r Log. sine » . 9.922008
Remainders 32.39.41§ Log. sine a . 9.732132
Sum ;r= 19.791314
Arch = 38? 8^501'/ Log. co-sine =s 9.895657
Star's semi-diurnal arc 2= . 76 ? 1 7 ' 4 11^, in time = . . 5 * 5 T 1 1 !
Prop, part of var. of sun's R. A. 4 ?24 ! answering to 5 ?5 T II ! = — 0. 56
Star's apparent semi- diurnal arc = 5? 4?15!
Apparent time of the star's transit over the given meridian = 1 1. 49. 46
Apparent time of the star's rising past noon of the given day = 6^45T31 !
Apparent time of the star's setting past noon of the given day= 16^54? 1 !
See Example 2^ page 129 ; and^ also^ the Example^ page 130.
Problem IlL
Ghen the LatUude of a Place, and the Height of the Eye above the
Level of the Horixon ; to find the apparent Times qfa Planet's Bisuig
and Setting.
RULB.
Compute the apparent time of the planet's transit over the meridian of
the given place, by Problems X. and XI., pages 313 and 315 ; reduce Ais
time to the meridian of Greenwich, by Problem IIL, page 297 ; to which
let the planet's declination be reduced, by Problem VII., page 307 J then.
To the logarithmic tangent of the latitude add the logarithmic tangent
of the planet's reduced declination, and the sum (abating 10 in the index,)
will be the logarithmic sine of an arch ; which, being converted into time,
and added to 6 hours when the latitude and the declination are of the same
name, but subtracted from 6 hours when of contrary names, the sum or
difference will be the planet's approximate semi-diurnal arc^ or half the
time of its continuance above the horizon.
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APPARENT TIME OF RtSISG OR SETTING OF A CELESTIAL OBJECT. 507
Liet this time be applied^ by subtraction and addition, to the apparent
time of transit ; and the approximate times of the planet's rising and setting
will be obtained.
Reduce the approximate times of rising and setting, thus found, to the
correspondent times at Greenwich, by Problem IIL, page 297 j to which
times, respectively, let the planet's declination be reduced, by Problem.
VII., page 307; then.
To the aggregate of 90 degrees,* the horizontal refraction,t and tiie dip
of the horizon, diminished by the planet's horizontal parallax,^ add the
planet's polar distance at the approximate time of rising or setting, and
the co-latitude of the ^ven place : take half the sum; the difference
between which and the first term, call the remainder.
Now, to the logarithmic co-secants, less radius, of the polar distance
and the co-latitude, add the logarithmic sines of the half sum and of the
remainder : half the sum of these four logarithms will be the logarithmic
co-sine of an arch ; which, being doubled and converted into time, will be
half the time of the planet's continuance above the horizon, or its semi-
diurnal arc.
Find the proportional part of the variation of the planet's transit over
the meridian, answering to half its continuance above the horizon, by
Problem XI., page 315, in the same manner as if it were the reduction of
transit to a different meridian that was under consideration. Now, this
proportional part being added to half the time of the planet's continuance
above the horizon when the planet's transit is increasing, but subtracted
therefrom when decreasing, the sum or difference will be the planet's
apparent semi-diurnal arc ; which being applied by subtraction to the
apparent time of transit, the remainder will be the apparent time of the
planet's rising. In the same manner let the apparent semi-diurnal arc for
the time of setting be computed ; which, being added to the apparent time
of transit, will give the apparent time of the planet's setting. The apparent
times of rising and setting, thus found, may be reduced to the mean
times of rising and setting, if necessary, by Problem I.^^ page 415.
Example 1.
Required the apparent times of Jupiter's rising and setting, January
4th, 18^4, in latitude 36*? N., and longitude 135? W.j tiie height of the
eye above the level of the horizon being 23 feet ?
The apparent time of Jupiter's transit over the meridian of the given'
* This is the zenith distance of a planet when its centre is in the horizon,
t The horizontal refraction of a celestial object is 33 minutes.
X For the parallaxes of the planets, see pa^e 326.
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508 NAimCAL ASTRONOBCY*
place 18 1I!20T30! ; his declination^ being reduced to this time and the
given longitude^ is 23? 18^55? north.
Latitude of the given place = 36? 0^ O^N. Log. tangent = 9. 861261
Jupiter's reduced declination=23. 18. 55 N. Log. tangent = 9.634461
Arch= 18?14152i: Log. sine = . • 9.495722
Arch, converted into time^ = 1 ^ 12T59 ! ; this^ being added to
6 hours^ gives the approximate semi-diurnal arc = •■ • 7- 12T59!
Apparent time of Jupiter's transit over the given meridians • 11. 20. 30
Approximate time of Jupiter's rising =.••-..•• 4^ 7*31'.
Approximate time of Jupiter's setting == 18?33?29!
The longitude^ in time, being applied to those times by addition, because
it is west, shows the approximate time of the planet's rising at Greenwich
to be 13*7*31 ' past noon of the given day, and that of its setting 3^33729!
past noon, January 5th« The declination reduced to these times respectively,
is 23?18:46rN. at the time of rising, and 23?I9UrN. at the time of
setting.
To find the apparent Time of Rising :—
90? + 33r-»- 4^36r « 2r = 90?37'34r
Jupiter's polar distance =s • 66.41.14 Log. co-secant == 0.036988
Co-latitude of the place 3= . 54. 0. 0 Log. co-secant = 0.092042
Sum=: 211?18'48r
Half sum = ...... 105?39:24r Log. sine = . . 9.983580
Remainder = • . • • • 15. 1.50 Log. sine = .- . 9.413860
Sum = 19.526470
Archs 54 ?34 ^ 3r Log. co-sine a . 9.763235
Semi-diurnal arc = . . . 109? 8^ 6r, in time = . . 7M6T32!
Variation of transits=30? decreasing; the proportional part of
which, answering to 7* 16r32^ is = -1.31*
Apparent semi-diurnal arc = 7M5? 1!
Apparent time of Jupiter*s transit = 1 1 . 20. 30
Apparent time of Jupiter's rising = ' 4t 5T29!
* If the transit had beeo pro^^ressivei or increaslDi^y the proportional part would be
addUivtM
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APPARBKT TIME OF RISING OR S£TTINQ OP A CBLBSTUL OBJECT. 509
To find the apparent Time of Setting :—
90? + 33< + 4^36r - 2r = 90?37^24r
Jupitcr'i polar distance = . 66. 40. 56 Log. co-secant =: 0. OS7004
Co-latitude of the place =s . 54. 0. 0 Log. co-secant = 0.092042
Sums 211?18'.20r
Half sum s 105?39n0r Log. sine = . . 9.983588
Remainders 15. 1.46 Log. sine = . . 9.413829
Sums 19.526463
Arch s 54?34^ 5r Log. co-sine s 9.763231}
Semi-diurnal arc = . . . 109? 8:iOr, in time = . . 7*16?33!
Variation of transit = 30?, decreasing; the proportional part
ofwhich, answering to 7 M6T33!, is = — l?3l!»
Apparent semi'-diumal arc s 7M5T2!
Apparent time of Jupiter's transit = • • • 1 1 . 20. 30
Apparent time of Jupiter's setting == 18?35?32!
Example 2.
Required the apparent times of the rising and setting of the planet Mars,
January 16th,. 1824, in latitude 40? N., and longitude 140? E.; the height
of the eye above the level of the sea being 26 feet ?
The apparent time of Mars' transit over the meridian of the given place
is 16M3T37^ ; now, his declination, being reduced to this time and the
given longitude, is 0?55M5? south.
Latitude of the given place=40? 0^ OrN. Log. tangent = 9.923814
Mars' reduced declination = 0.55.45 S« Log. tangent = 8.210009
Arch= 0?46M71' Log. sine = . 8.133823
Arch, converted into time = 0^ 3? 7 • ; this, being subtracted
from 6 hours, leaves the approximate semi-diurnal arc = 5756T53!
Apparent time of Mars' transit over the given meridian = • 16. 43. 37
Approximate time of Mars' rising = 10M6T44!
Approximate time of Mars' setting s 22*40T30!
• See Note, pa^ 508.
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51.0 NAUTiCAX. AST&ONOMT.
The longitude, in time, being applied to those times by subtraction,
because it is east, shows the approximate time of the planet's rising at
Greenwich to be 1*26T44! past noon of the given day, and that of its
setting 13*20T30! past the same noon. The planet's declination reduced
to these times, respectively, is 0?54^20?S. at the time of risix^, and
0?57'9^ S. at the time of setting.
To find the Planet's apparent Time of Rising :—
90? + 38: + 4^52* - lOr =r 90?37M2r
Planet's polar distance = • 90. 54. 20 Log. co-secant=:0. 000054
Co- latitude of the place = . 50. 0. 0 Log. co-secant=:0. 115746
Sum =: 231?32^ 21
Half sum =: I15?46: It Log. sine = • 9.954517
Remainder =: 25? 8'19r Log. sine = . 9.628195
Sum= 19.698512
Aiths: ..,,,,. 45? VA9t Log. oo-sim = 9. 849256
Semi-diiimal arc =: . . . 90? 3C3dr, in times • . 6^0715!
Variation of transit =: 20T, decreasing ; the proportional part
of which, answering to 6*0? 15!, is = —0.50*
Apparent semi-diurnal arc = 5t59r^!
Apparent time of transit =: • • • • * 16. 43. 37
• ■ ■ ■ ■■
Apparent time of Mars' rising = 10!44rl2!
To find the Planet's apparent Time of Setting v^
90? + 33'. + 4:52r - lOr = 90?37'42r
Planet's polar distance =: . 90. 57. 9 Log. co-seeant=0. 000060
Co-latitude of the place == . 50. 0. 0 Log. co-secant=:0. 1 15746
Sum= 231?34^5K
Half sum =. . ; , • . 115?47'.25ir Log. sine =: . 9.954432
Remainder = 25? 9 M3i? Log. sine =5 . 9.628573
Sums 19.698811
Arch=: ....... 45? 0^38r Log. co-sine =: 9.8494051
• SeeNo«e>pa{:e508.
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APPABBNT TIMB OF ItlSlNO OA SETTING 09 A CBJLBSTIAL OBJECT. Sll
Arch= 45? 0'.38i:
Semi-diurnal arc = ... 90? i: 16'/, in time = . • 6? 0? 5!
Variation of transit =: 20?, decreasing; the proportional part
of which, answering to 6 tOT5!, is =z — 0.50 *
Apparent semi-diurnal arc =: 5^59^15!
Apparent time of transit = • 16.43.37
Apparent time of Mars' setting = 22*42?52!
^ Problem IV.
Given the Latitude of a Place, and the Height of the Eye above the
, Level of the Horizon ; to find the apparent Times of the Moon's
Bising and Setting.
Rule.
Compute the apparent time of the moon's transit over the meridian of
the given place, by Problems Vill. and IX., pages 309 and 312; and
reduce it to the meridian of Greenwich, by Problem IIL, page 297 ) to
which let the moon's declination and horizontal parallax be reduced, by
Problem VI., page 302 ; then.
To 90 degrees, diminished by the difference between the moon's hori-
zontal parallax and the sum of the horizontal refraction and the dip of the
horizon, add the moon's polar distance and the co-latitude of the giveB
place. Find the difference between half the sum and the first term, which
call the remainder.
Now, to the logarithmic co-«secants, less radius, of the polar distance,
and the co4atitude, add the logarithmic sines of the half sum and of the
remainder: half the sum of these four logarithms will be the logarithmic
co-sine of an arch; Which, being doubled, and converted into time, will
be the moon's approximate semi-diamal arc : this being subtracted from
and added to the apparent time of the moon's transit, the respective
approximate times of her rising and setting will be obtained.
Reduce the approximate times of the moon's rising and setting, thm
found, to the correspondent times at Greenwich, by Problem III., page
297 ; to which times, respectively, let the moon's declination and hori-
zontal parallax be reduced, by Problem VI., page 302; and let the moon's
• See Nolt, imge 909.
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512 NAUTICAL ASTRONOMY.
declination^ at each tioie^ be corrected by the equation of second difference;
then,
M^th 90 degrees^ ditnimshed as before, the moon's respective polar
distances, and the co-latitude, compute the approximate semi-diurnal arcs
corresponding to the times of rising and setting.
Find the proportional part of the daily variation of the moon's transit
answering to each semi^diurtidl arCy and 24 hours augmented by the
variation of transit, by Problem IX., page 312, in the same manner as if it
were the reduction of transit to a different meridian that was under consi-
deration. Now, these proportional parts, being cMed to their corre-
sponding semi-diurnal arcs, will give the apparent semi-diurnal arcs at
the times .of the moon's rising and setting: the former being subtracted
from the apparent time of transit, and the latter added thereto, the
respective apparent times of the moon's rising and setting will be obtained.
These may be reduced to the mean times of rising and setting, by Problem
!•> page 415, if necessary.
Example 1.
Required the apparent times of the moon's rising and setting, January
17th, 1824, in latitude 51?28M0?N., and longitude 75?W*; the height
of the eye above the level of the horizon being 30 feet ?
The computed apparent time of the moon's transit over the meridian of
the given place is 13*43r46! ; now, her declination, being reduced to
this time, and the given longitude, is 10? 29' 27 TN., and her horizontal
parallax 60^ 59r.
90?-60:59r+33' + 5n5r=89?37:i6f
Moon's north polar distance = 79.30.33 Log. co-secant=:0. 007321
Co-latitude of the given place == 38. 3 1 . 20 Log. co-secant=:0. 205639
Sum = 207^39: 9".
Half sum =: 103^49^34^^ Log. sine = . 9.987230
Remainder =: . . • , . 14. 12. 18i Log. sine = . 9.389864
Sum=: 19.590054
Arch= 51?24^28r Log. cosine = 9.795027
J's approx. semi-diurnal arcn 102?48^56r, in time =: . . 6t51?16!
Moon's apparent time of transit over the given meridian =: • 13. 43. 46
Approximate time of the moon's rising = 6J52T30!
Approximate time of the moon's setting =: • . • • . •20^357 2*.
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APPARBNT TIMB OF RISING Oft SETTIK6 OF ▲ OBLBSTIAL OBJECT, 513
The longitude, in time, being added to those times, because it is west^
shows the approximate time of the moon's rising at Greenwich to be
ll?52r30! past noon of the given day, and that of her setting 1 t35T2!
past noon of the 1 8th. Now, the moon's declination and horizontal
parallax, reduced to these times respectively, (the former being corrected
by the equation of second difference,) gives the declination at the time of
rising 12?12'0lfN., and the horizontal parallax 61 ^7^; and the declina-^
tion at the time of setting 8?46^ 8f N., and the horizontal parallax 60'49r«
To find the apparent Time of Rising : —
90?-.61'7^+33: +5a5r=89?37' 8r
Moon's polar distance == • 77*48. 0 Log. co-secants 0.009921
Co-latitude = 38.31.20 Log. co-secants 0.205639
Sums 205?56:28r
Half sums 102?58'14r Log. sine s . 9.988775
Remainders 13.21. 6 Log. sine s . 9.363475
Sum= 19.567810
Archs 52?33a7i^ Log. co-sine s 9.783905
Semi-diurnal arc s ... 105? 6:35r, in time s . . 7? 0r26!
Variation of transit s 53?, the proportional part of which,
answering to 7 *0T26!, is +14.55
Apparent semi-diurnal arc s 7*15 ?2l!
Apparent time of moon's transit s . . 13. 43. 46
Apparent time of the moon's rising s ««.««. k • 6 . 28T25 !
To find the apparent Time of Setting :—
90?-60M9r+33^+5U5'/s89^37^26r
Moon's polar distance . . 81 . 13. 52 Log. co.»ecant;=0. 005 106
Co-latitudes 38.31.20 Log. co-secantsO. 205639
Sum = 209?22:38r
Half sums 104?41U9'r Log. sines . 9.985570
Remainders 15. 3.53 Log. sines . 9.414824
Sums 19.611139
Archs 50? 16:30§i: Log, co-sines 9, 805569*
2 L
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514 NAOTICAL ASnONOMT.
Aroh« 60?16J30i^
Semi-diurnal uc » « . . 100?33M% in time s • • 6!42?12!
Variation of transit a 53T, the proportional part of which,
aniweringto6*42Tl2:, ias ; . . . +14.17
Apparent semi-diurnal arc s « . » « • b • • • • 6?56729!
Appairent time of moon's transit a •••..••• 13. 48» 46
Apparent time of the moon's setting = ^ 20?40?15!
Example 2.
Required the apparent times of the moon's rising and setting, January
20th, 1824, in latitude 40?30^N., and longitude 80?B.5 the height of the
eye above the level of the horixon being 30 feet ?
The computed apparent time of the moon's transit over the meridian of
tlie given place is 15*55?5! ; now, her declination, being reduced to this
time, and to the given longitude, is 5?55'.35rS., and her horizontal parallax
58^53f.
>*^
90?-58J53r + 33C+5a5*'=89?39122*
Moon's north polar distance a 95.55.35 Log. co-s6cant=± 0.002329
Co-latitude of the given place=:49. 30. 0 Log. co-secant=> 0.118954
Sum = 235^ 4^57^
Half sums: 117?32C28ir Log. sine s . 9.947767
Remainders 27.53. 6$ Log. sine = • 9.669968
Sum= 19.739018
• J I -T -^
Airchssa «««•.., 42?13!42? Log. co-sine as 9.869509
}) 's approx. semi-diurnal arc = 84?27' 24r, in time = . • S!37"50!
Apparent time of the moon's transit over the given meridian sl5. 55. 5
Approximate time of the moon's rising = 10' 17*15*
Approximate time of tiie moon's setting = 21 i32?55!
The longitude, in time, being subtracted from those times, because it b
east, shows the approximate time of tiie moon's risbg at Greenwich to be
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APPARBNT TIMB OF KIIINO OR SBTTINQ Of A CfilJBSTIAL OBJBCT. 515
4*57"15! past noon of the given day, and that of her setting 16M2?65!
past the same noon. Now^ the moon's declination and horizontal parallax,
reduced to these times respectively^ (the former being corrected by the
equation of second difference,) gives the declination at the time of rising
4?32^3rS., and the horizontal pafallftx 59'6? ; and the declination at the
time of setting 7? 18' 12rS., and the horizontal parallax 58'40r.
To find the apparent Time of Rising :—
90?- 59^6^4- ssc + snsrssg^sg^ 97
Moon's polar distance cs « 94.32. 3 Log. co-secantaO. 001361
Co-latitudes 49.30. 0 Log, co-secant»0, 118954
Sums ...... • 233?4ia2?
Half sums 116?50^36? Log. sine = . 9.950484
Remainders 27.11.27 Log. sine = . 9.659874
Sum=5 19.730673
Archs 42?49'.42? Log. co-slnc = 9. 8658861
Semi-diurnal arc =: . . . 85?39:24^ in time s . . 5M2r38:.
Variation of transit s 49T, the proportional part of which^
answering to 5*42r38!, Is « +11.16
Apparent semi'*diunial arc a • • « 5^53?54!
Apparent time of the moon's transit as .•••••• 15.55. 5
Apparent time of the moon's rising ^ • • • lOt ITli;
To find the apparent Hme of Setting i^
90?-58U0r+33^ + 5n5r=89?39^35r
Moon's polar distance = • . 97.18.12 Log. co-secant* 0.003538
Co-latitude = . . •
Sum =3 . . • . •
Half sum = •
Remainder ^
Arch:
^9.30. 0 Log. co-secant= 0,118954
236°27M7r
1 1 8? 13 : 53if Log. sine =s . 9. 944998
28.34. 18J Log. sines: . 9.679664
Sum = 19.747154
41*S7'5ir Log.co-sine = 9.873577
2l2
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516 NAtrriCAL astronomV.
Arch=r ; . 41^37'.5ir
Semi-diurnal arc = . . . 83^5 '42^ in time = . . 5*33r 3!
Variation of transit = 49?, the proportional part of which^
answering to 5i33T3!, is SB +10.57
Apparent semi-diurnal arc = 5*44T 0!
Apparent time of the moon's transit = 15.55. 5
Apparent time of the moon's setting = 21*39? 5!
See Examples 1 and 2, pages 134 and 136 ; and^ also, the example or
work, pages 137 vid 138.
Problem V.
Gken the Latitude and Longitude of a Place, and the Day of the Month;
to find the Time qfthe Beginning and of the End of TwiUgkt, and the
Length of its Duration.
Rule.
Reduce the sun's declination, at the midnights preceding and following
the noon of the given day, to the meridian of the given place, by Problem
v., page 298; then.
Add together the constant quantity 108 degrees,*^ the sun's polar distance,
and the co-latitude of the given place : take half the sum ; the difference
between which and the constant quantity call the remainder: Now,
To the logarithmic co-secants, less radius, of the polar distance, and the
co-latitude, add the logarithmic sines of the half sum and of the remainder:
half the sum of these four logarithms will be the logarithmic sine or
logarithmic co-sine of an arch ; which, being doubled, and converted into
time, will be the apparent time of the beginning or of the end of twilight
accordingly.
Compute the apparent times of the sun's rising and setting, by Problem
L, page 500 ; then, the interval between the time of the cdiiimencement of
twilight and that of sun rising, will be the duration of the morning twilight;
and the interval between the time of sun setting and the end of twilight,
will be the duration of the evening twilight.
Note. — If much accuracy be required, the sun's declination must be
reduced to the meridian of the given place, at the respective times of the
• 90<> + 18° » 108^ See Remarks, page 518,
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OF FINDING THB BEGINNING OR END OF TWILIGHT.
517
commencement and of the end of twilight^ found as above | then^ the
operations being repeated, the correct apparent times of the beginning
and of the end of twilight will be obtained. This degree of accuracy may,
however, be dispensed with, — unless in cases of mere speculative inquiry^
or where some philosophical object is under consideration.
Example^
Hequired the apparent times of the beginning and of the end of twilight,
and its duration, October 1st, 1824, in latitude 40?30C north^ and longitude
105? east?
To find the Beginning of Twilight :—
Sun's declination at midnt. September 30th,=:3? 4C26r S.
Reduction of ditto for longitude 105? E. ^ — 6.48
Sun's reduced declination =
2?57'38rS.
Constant quantity = 108? 0' Or
Sun's polar distance = 92. 57. 38
Co-latitudes . . . 49.30. 0
Sum = . •
Half sum =
Remainder =
. 250?27^38f
. 125?13U9r
. 17.13.49
Log* co-secant
Log. co-secant
Log. une s
Log. sine =
= 0.000580
= 0.118954
• 9.912137
• 9.471604
19.503275
Arch
34?2r.54ir Log. sine = . . 9.751637J
Beginning of twilight = 68?43M9r, in time =
Apparent time of sun-rbing on the given day =:
4*34?55!
6. 5. 8
Duration of morning twilight = 1^80?13t
To find the End of Twilight :—
Sun's declination at midnight, October Ist^ s 3?27'45r S.
Reduction of ditto for longitude 105? E. = — 6. 48
Sun's reduced declination =
• • • •
3?20:57rS.
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518
NAUTICAL ASTEONOMYf
Constant quantity s 108? Ot Or
Sun's polar distance = 93. 20. 57 Log* co-aecaat
Co-latitude ^s ... 49. 30. 0 Log, co-secant
Sum = 250?50:57r
Half sum = .
Remainder =
0.000743
0.U8954
. 125?25^28|r Log. sine = .^. . 9.911093
. 17.25,284 Log. sine = . . . 9.476324
Archss 55?27U0r Log. co-sine = •
End of twilight = . . 110?55:20r, in time = . . .
Apparent time of sun-setting on the given day a . . •
Duration of evening twilight
• • • • •
19.507114
9.753557
7*23T41!
5.52.13
1*31T28!
Remarks.
Twilight, technically called the crepusculumj is that faint light which
we perceive before the sun rises and after he sets. It is produced by the
rays of light being refracted in their passage through the earth's atmo-
spherci and reflected from the different particles thereof.
The morning twilight commences when the sun wants 18 degrees of
appearing in the horizon of the eastern hemisphere, and the evening
twilight ends when he is depressed 18 degrees below the horizon of the
vi^tem hemisphere.
When the. sun's deolinatifMi exoeeds the difference between the
co-latitude of any given place and 18 degrees, there will be no real
doribteit or night at that place, but continual day and twilight; as is the
case at London, from the 22d of May to the 21st of July.
When the sun is on the same side of the equinoctial with the derated
polcj the duration of twilight will constantly increase as he approaches
that pole, till he enters the tropic ; at which time the duration of twilight
will be the longest. It will then decrease until some time after the sun
passes the ec^uinox, but vrill increase again before he arrives at the opposite
tropic : hence, there must be a point within the tropics where the duration
of twilight b the shortest. This point may be found by the following
problem.
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OF FINDINO TOB TIMB OF THB 9IIOETB8T TWILIGHT. (18
PftOdLBM VI.
Cfiveti the LoUtude qfa Place ; tojind the Time of the ehorte$t Tmligla,
and ite Duration.
Rule.
To the logarithmic tangent of the half of 18 degrees^ add the loga«
rithmic sine of the latitude ; and the sum (abating 10 in the index,) will
be the logarithmic sine of the sun's declination at the time of the shortest
twilight, of a contrary name to the latitude : the day corresponding to this
declination will be that required.
Again, to the logarithmic sine of the half of 18 degrees, add the loga-
rithmic secant of the latitude ; and the sum (abating 10 in the index,) will
be the logarithmic sine of an arch, which, being doubled and converted
into time, will be the duration of the shortest twilight;
Example.
Required the time of the shortest twilight, and its duration, in the year
1824, in latitude 50?48^N.?
Half of 18 degrees = 9? OC Of Log. tangent=:9. 199713
Latitude of the placezzSO. 48, 0 Log. sine = 9. 88927 1
Sun's declination =: 7? 3^ K Log. sine s 9. 088984 j
which is south, of a contrary name to the latitude.
Half of 18 degrees =z 9? 0^ 01 Log. sine = 9. 194332
Latitude of the place= 50. 48. 0 Log. secants 10. 199263
Aichzi • . . . 14?19C497 Log« sine = 9.393595
Duration of twilights 28^39^38?^ in time s 1^54?39;.
The days, in the Nautical Almanac, corresponding to the sun's decUna^-.
tion 7?3a? S«, are March 2d and October 1 1th, which, therefore, are the
days of the shorteat twiHght in the year 1824, in latitude 50?48C north |
and the duration of the twilight^ on those days, is l!54T39!*
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520 nautical astronomy*
Problem VI^.
Given the Latitude of a Place between 48'!32: and 66 ?32^ (the lAmiis
of regular TbnUght) ; to find when real Night or Darkness ceases,
and when it commences. •
RULB^
The complement of the latitude, diminished by 18 degrees, will be the
declination of the sun, of the same name as the latitude, at the time when
it ceases to be real night, and also when real night commences.
Example.
Required the interval of time, in the year 1824, during which there will
be no real darkness or night, in latitude 50?48^ north ?
Solution.— The complement of the latitude 39^21 N.-18?=21?12'N,
= the sun's declination. Now, the days answering to 21? 12' of north
declination are. May 26th and July 17th. Upon the first of these days,
therefore, real night ceases, and it commences upon the last. During this
interval there is no real darkness, because the sun is less than 18 d^rees
below the horizon ; and so on for any other latitude within the limits.
pROBf^M VJIL
Gioen the Sun*s DecUnation and Semi-diametet ; to find the Interval
between the Instants of his lower and upper lAmbs being in the Horizon
of a known Place.
Rule.
Rnd the appronmate time of the sun*s rising or setting, by Problem I.,
page 124; to which time let the sun's declination be reduped, by Ph>bleia
v., page 298.
To the logarithm of the sun's semi-diameter, expressed in seconds, add
the constant logarithm 9. 124939, and call the sum a reserved logarithm;
then.
To the logarithmic co-sine of the sum of the latitude and declination,
add the logarithmic co-sine of their difference : half the sum of these two
logarithms, being subtracted from the reserved hgariihm, will leave the
logarithm of the interval of time, in seconds, between the instants of the
sun's lower and upper limbs being in the horizon of the given place.
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OF THB TIME THAT THE SUN's DIAMETER TAKES TO RISE OR SET. 521
Example 1.
Required the interval between the instants of the sun's lower and upper
limbs being in the horizon, at the time of its setting, July ISth^ 1824, in
latitude 50?48C N., and longitude 120? W. ?
Apparent time of setting in Table L,, to latitude 50?48' N.,
and declination 2 1?49^ 51 rN.= . . . * 7*57"12'.
Longitude 120? west, in time s= 8. 0. 0
Greenwich time of sun's setting = 15*57T12!
Sun's declination at noon, July 13th, 1824, = 21 ?49^ 5 1 ?N.
Correction of ditto for 15*57T12!= ... —5.58
Sun's reduced declinatioa = 21?43'53?N.
Sun's semi-diameter 15M5\8=945\.8 Log.=2.975799
Constant logarithm = 9.124939
Reserved logarithm s 12. 100738 . . 12. 100738
Sun's red. dec. = 21?43^53rN.
Lat.ofthcplace=50.48. ON.
Sum= . . . 72?3K53r Log. co-sine = 9.477387
Differences . 29. 4. 7 Log. co- sine = 9.941531
Sum a 19.418918
Half sums « 9.709459 .. .9.709459
Interval, in seconds, s= 246. 195 = Log. s 2.391279
Hence, the interval between the instants of the sun's limbs touching the
horizon^ is 4 minutes and 6 seconds.
Sxample 2.
Required the interval between the instants of the sun's upper and lower
limbs touching the horizon, at the time of rising, October Ist^ 1824, in
latitude 40?30C N., and longitude 105? E. ?
Apparent time of rising, in Table L., to latitude 40?30'N.,
and declination 3? 16C6r S. = 6M0?48!
Longitude 105? east, in time = 7. 0. 0
Greenwich time past noo% September SOth^ 11M0T48!
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523 FEAGTICAL ASTRONOIfT.
Sun's declination at noon^ Sept. SOth, 1824^= 2?52U6? S.
Correction of ditto for 1 1 ? 10T48 ? = . . + 10. 53
Sun's reduced declination = 3? 3'39rS.
■ •
Sun's 8en)i-diameter«=: 16'. 1 ''. 2=^961 '. 2 Log.=:2. 982814
Constant logaritbm » « • . 9.124939
Reserved logarithm = 12. 107753 •. 12. 107753
Sun's red. dec. = 3? 3^39rS.
Lat.oftbeplace=:40.30. 0 N.
Sum as • . 43?3SC39r Log. co-sine a 9.860124
Differences . 37.26.21 Log. co-sine s 9.899820
Sum as 19. 759944
Half sum = . ; 9.879972 .. .9.879972
Interval, in seconds, = 168. 958 = Log. = 2. 227781
Hence, the interval between the instants of the sun's limbs touching the
horizon, is 2 minutes and 49 seconds.
Note* — ^The constant logarithm made use of in thk problem is the arith*
metical complement of the proportional logarithm of 24 hours esteemed
as minutes. If the sun's diameter be used, instead of its semi-diameter, it
must be expressed in minutes and decimal parts of a minute : in this caie^
the same result will be obtained by employing the constant logarithm
8. 823909 1 vi£., the arithmetical complement of the oommon logarithm
of 15 degrees, or the motion corresponding to one hour of time.
SOLUTION OF PROBLEMS IN GNOMONICS OR DIALLING.
Dialling, or GnomonicSf is a branch of mixed mathematics, which
depends partly on tiie principles of geometry and partiy on those of
astronomy; and it may be defined as being the method of projecting on
the surface of any given body, whether plane or otherwise, a figure called
a sun-dialj — the different lines of which indicate, by the shadow of a s^le
or gnomon, when the sun shines thereon, the apparent time of the day.
The upper edge of the style or gnomon, which projects the son's shadow
on the plane of the dial, must be parallel to the earth's axis : hence, it k
sometimes called the axis of the dial.
The plane of the gnomon must be perpendicular to that of the dialf
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GNOMONICS^ OR PIAiXtNG. 523
The plane on which it is erected is called the sub-style i in horizontal dials
it may be called the meridian, or 12 o'clock line.
The angle comprehended between the style and the sulhstyle, is called
the elevation of the style: this angle, in horizontal dials, is always equal to
the elevation of the pole, or the latitude of the place for which it is com-
puted; but, in erect direct north or south dials^ it is equal to the comple-
ment of the latitude of such place.
Those dials whose planes are parallel to the plane of the horizon, are
called horizontal diab i but such as have their planes perpendicular to the
plane of the horizon, are called vertical or erect dials.
Those vertical dials whose planes are either parallel or perpendicular to
the plane df the meridian, are called direct erect dials. One of these must
always face one of the cardinal points of the horizon, according as it may
be a north, south, east, or west, erect dial.
All other erect dials are called declining dials* Those dials whose planes
are neither parallel nor perpendicular to the plane of the horizon, are called
reclining dials.
In this place, however, we shall only show the method of constructing a
horizontal dial, and, also, that of a north or south erect direct dial ; these
being by far the most useful, and, indeed, the most common of all the
varieties in dialling.
PaoBLEM I.
Given the Latitude of a Place; to find the Angles which the Hour Lines
make uAth the Sub- Style or Meridian Line of a Horizontal Sun-Dial.
Gbneral Proposition.
In every right angled spherical triangle, radius is to the sine of one of
the legs containing the right anglcy as the tangent qf the angle adjacetit
to that side is to the tangent of the other containing side ^ tlie triangle.
This is merely a variation of the equation for finding the leg BC, in
Problem IV., page 189 : bence the following
RULK.
To the logarithmic sine of the latitude, add the logarithmic tangent of
the sun's horary angle iWtm noon ; and the sum (abating 10 in the index,)
will be the logarithmic tangent of the angle comprehended between the
correspopding hour line and the sub^style, at the centre of the dial.
Note, — Sbce the sun's apparent motion in the ecliptic is at the rate
of 15 degrees to an hour, therefore at one hour from noon the sun's horary
angle is 15? ; at Iwo hours from noon it is 30:^ J and so on.
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524
PRACTICAL ASTRONOMY.
Example.
Required the angles which the hour lines make with the sub-style, or
meridian line of a horizontal dialj in a place situated in 50?48' 15 f north
latitude ?
To find the Angle at one Hour from Noon : —
Latitude of the place = . . 50?48'. 15r Log. sine = 9. 889296
Sun's horary ang. at 1 * from noon= 15. 0. 0 Log. tangent=:9. 428053
Hourlineofl, or 11 o'clock = . ll?43C52r Log. tangent=9.317349
To find the Angle at two Hours from Noon : —
Latitude of the place = • . . 50?48: 15^ Log. sine = 9. 889296
Sun's horary angle at 2? from noon =15. 0. 0 Log. tangent^ 9. 761439
Hour line of 2, or 10 o'clock = 24? 6 : 20^ Log. tangent=9. 650735
Proceeding in this manner, the several angles which the respective hour
lines make with the meridian will be found to be as follows ; viz.^
Hour lines of L and XL =: ll?43^52r
Ditto IJ. and X. = 24. 6.^0
Ditto in. and IX. = 37.46.31
Ditto IV. and VIIL = 53.18.53
Ditto V. and VU. = 70.55.39
Ditto VI. and VI. = 90. 0. 0
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GNOMONICS^ OR DIALLING. 525
The hour lines of VII. in the evening and V. in the morning, make the
same angles with the meridian, on the opposite Hde of the VL o'clock hour
line^ as the hour lines of VII. in the morning and V. in the evening. In
the same manner the hour lines of VIII. in the evening and IV. in the
morning make the same angles with the meridian as the hour lines of VIII.
in the forenoon and IV. in the afternoon; and so on.
The angles for the halves, quarters, or other subdivisions of the hours,
are to be determined in the above manner.
The angles which the diiferent hour lines, &c. make with the meridian,
being thus determined, the dial may then be very readily constructed, by
means of a pair of compasses, and the line of chords on a common Gunter's
scale, or of that on a Sector : the latter, however, should be preferred,
because the degrees thereon are generally divided into halves, and some-
times quarters, which gives it a decided advantage, in point of accuracy,
over that on Gunter's scale.
Construction.
On the proposed plane draw the meridian, or XII. o'clock hour line, a b ;
parallel to which, at a distance equal to the intended thickness of the
gnomon or style, draw the line c d : perpendicularly to these draw the VI.
o'clock hour line e/. Open the Sector to any convenient extent, and
ti^e the transverse distance 60? to 60? (on the line of chords) as a radius
in the compasses, and, from a as a centre, describe the arc g h : with the
same radius, and from c as a centre, describe the arc i ft ; and, since the
hour lines are less distant from each other about noon than in any other
part of the day, it is advisable to have the centres of those quadrants or
iarcs at a little distance from the centre of the plane of the dial, on the side
opposite to XII., so as to allow of the hour distances being enlarged near
the meridian under the same angles in the plane of the dial : thus, the
centre of the plane is at A ; but the centres of the quadrants or arcs are
taken a little below it, at the points a and c.
Take the transverse distance 1 1 ?43 ^ 52? to 1 1 ?43 ^ 52^, in the compasses,
from the line of chords, and set it off from ^ to^l, and, also, from i to 6 :
take the transverse distance 24?6^20?, in the compasses, and set it off
from ^ to 2, and from i to 7 ; and proceed in the same manner with the
remaining horary angles.
Now, from the centre a draw the forenoon hour lines a 1 XI., a 2 X.,
a 3 IX., a 4 VIII., a 5 VII. ; and, from c as a centre, draw the afternoon
hour lines c6I., c7II., cSIII., c9IV., cOV. : produce .a 5 VII. and
a 4 VIII. for the hour lines of VII. and VIII. o'clock in the evening; and
produce c9IV. and cOV. for the hour lines of IV, and V. in the
morning. In the same manner may the quarter and half-hour lines be
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526 lHACriCAL ASTltOKOMY.
drawn (and minutes if necessary), by setting off the compnted corre-
sponding angles from the meridian : these, however, have been omitted
in the above diagram, with the view of preventing embarrassment.
Take the latitude 50?48n5T in the compasses, vis., the transverse
distance 50^48^ ISr to 50^48' 151f, and set it off from g to L, and draw
the hypothenuse line a L P for the axis of the style or gnomon.
The style may have any shape the artist pleases, provided its edge a L P
be a perfectly straight line.' It should be a metallic substance, and must
be of an equal thickness with the breadth of the space comprehended
between t^e two parallel straight lines ab and cd; in which space it must
be erected truly perpendicular to the plane of the dial : then, since the
angle B a P is equal to the latitude, the straight edge of the style = a L P
will be directed to the elevated pole of the world, and, hence, parallel to
the earth's axis when the dial is truly set ; the shadow of which, when the
sun shines, will indicate the hour of the day.
Note. — Since the hour of the day indicated by a sun-dial is expressed in
apparent solar time, it must be reduced to mean time, by Problem I.,
page 415, so as to make it correspond with that shown by a well-regulated
watch or clock.
Probubm II.
To Jind the Jngles on the Plane qfan erect direct sciUh Dial for aiiy
proposed north Latitude, or on that qf an erect direct north JXal for
any proposed south Latitude.
RVLE.
To the logarithmic co-sine of die latitude, add the logarithmic tangent
of the sun's horary angle from noon ; and the sum (abating 10 in the index,)
will be the logarithmic tangent of the angle comprehended between the
corresponding hour line and the sub-style, at the centre of the dial«
Example.
Required the angles which the hour lines on an erect direct south
dial make with the sub-style or 12 o'clock line, in latitude 50?48:15?
north ?
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QNOMOKlCSi OR DlALLtKQ.
5i7
To find the Angle at one Hour from Noon :—
Latitude of the placer: . . . 50^48^5' Log. co-sinezz 9. 800699
Sun's horary ang. at 1^ from noon=: 15. 0. 0 Log. tangent=: 9. 428053
Hour angle of 1^ or 1 1 o'clock =: 9?S6M0r Log. tangentr:9. 228752
To find the Angle at two Hours from Noon :*-*
Latitude <^ the place = . . . 50?48n5r Log. co-sine=: 9. 800699
Sun'8horaryangleat2?fromnoon=30. 0. 0 Log* tangent=::9. 761439
Hour angle of 2^ or 10 o'clock =: 20? 2M4r Log. tangent=9. 562138
Proceeding in this manner^ the several angles which the respective hour
lines make with the meridian will be found to be as follows { viz..
Hour lines of
I.
and
XI.
;^
9?36M0r
Ditto
II.
and
X.
^
20. 2.44
Ditto
III.
and
IX.
^Z
32.17.30
Ditto
IV.
and
VIII.
*"*
47.35.10
Ditto
V.
and
VII.
ZZ
67. 1.25
Ditto
VI.
and
VI.
zz
90. 0. 0
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528 MSNSURATION OF HBIGHTS AND DISTANCES.
Construction,
On ^ the proposed plane draw the XII. o'clock hour line ab; parallel to
which, at a distance equal to the intended thickness of the style, draw the
line c d : at right angles to the sub-style, or XIL o'clock line, draw the VI.
o'clock hour line «/• Open the sector to any convenient extent, and take
the transverse distance 60^ to 60? (on the line of chords) as a radius in
the compasses, and, from a as a centre, describe the arc gh; with the
same radius, and from c as a centre, describe the arc i k. Take the trans-
verse distance 9?36'40^ to 9?36'40r in the compasses, and set it off from
g to 1, and, also, from i to 6. Take the transverse distance 20?2U4^ to
20? 2' 44 f in the compasses, and set it off from g to 2, and from i to 7 ;
and proceed in the same manner with the remaining horary angles. Theiv
from the centre a, draw the forenoon hour lines a 1 XI., a 2 X., &c. &c. ;
and, from c as a centre, draw the afternoon hour lines c 6 I., c 7 II*^ &c. &c.
Take the complement of the latitude in the compasses, viz., the trans-
verse distance 39? 1 1 M5 T to 39? 1 T. 45 f ; set it off from ^ to L, and draw
the hypothenuse line aLP for the axis of the style or gnomon.
Now, when the dial is placed vertically, with its plane duly facing the
south, the VI. o'clock hour line e/will be parallel to the plane of the
horizon ; and the style B a L P, directed downwards, making an angle with
the sub- style or XIL o'clock hour line equal to the complement of the
latitude, will be truly parallel to the earth's axis.
Since the sun cannot shine any longer on a dial of this description than
from VI. in the morning until VI. in the evening, it is not necessary to
describe hour lines upon it before or after those periods of time.
Note. — ^An erect direct north dial for a place in north latitude, is con-
structed exactly in the same manner as an erect direct south dial ; but the
position of the dial must be reversed : that is, the VI. o'clock hour line
must be at the bottom instead of the top of the dial ; and the style or
gnomon must be directed upwards instead of downwards.
SOLUTION OF PROBLEMS RELATIVE TO THE MENSURA-
TION OP HEIGHTS AND DISTANCES,
Since it is frequently of the greatest importanpe to the mariner, but at
all times to the engineer or other military officer^ to be able to ascertain
the heights and distances of remote objects with precision, the following
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MBN8URATIOM OF HEIGHTS AND DISTANCES^
529
problems are given for their general guidance in such cases. In solving
these problems, it is the logarithmical mode of calculation that will be
attended to, with the view of showing the direct application of the prin-
ciples of plane trigonometry to such cases. . To the imagination of the
ingenious, however, many other modes of obtuning an approximate value
for the heights and distances of remote objects will soon present themselves :
such as, by means of shadows, mirrors, unequal vertical staves, &c. &c. ;
but, since these methods entirely depend upon the principles of similar
triangles (as demonstrated in Euclid, Book VI., Prop. 4), they admit of
direct sdutions without the assistance of trigonometrical tables : hence^ no
notice can be taken of them in this work.
Problem L
To find the Hetght qfan accessible Object.
Rule.
Let A B, in the annexed diagram, be the
object : from B measure any convenient dist-
ance to C ; take, at C, with a quadrant or other
instrument, the angle ADE; then, in the
triangle ADE, given the side D E s= B C, and
the angle at D ; to find the side A E : to which
let the height of the observer's eye above the
horizontal plane == CD or B E be added, and
the sum will be the true height of the object
AB.
Example.
Let the horizontal distance & C be 250 feet, the angle of elevation ADE
s 41945C, and the height of the eye CD =5 feet; required the height
of the object AB?
This comes under Problem II. of right angled plane trigonometry, page
172 ; and, by making D E radius, it will be
As radius =: 90? Log. co-secant = 10. 000000
Is to the distance DE:sCB= 250 feet Log. =s • . « 2.397940
So is the angle of elevation A D E=:4 1 ?45 '. Log. tangent = . 9. 950625
To the part A E = . .
Height of the eye C D =3
Height of the object A B :
223. 13
5.
Log.ss
2.348565
228. 13 feety as required.
2 M
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5S0 MBN8URATION OF HBIOSVS AND B18TAVOB9.
JZemorfc.— By removing either tQwavds or from the objeet, uatil the
quadrant shows the angle of altitude to be 45 degrees, the measure of the
distance between the toot of the observer and that of the objeel, aug^
mented by the height of the eye, will become the altitude or height of that
object.
Probj^m II.
Owen the Angle of Elevation, and the JSeighi ofm Okfee$; t^fi^ Ifce
Observer's horizontal Distance from thai Object*
At any convenient distance, as at C, let the angle of
elevation A D E be taken ; then, in the triangle A D E,
given AE s: the height of the object AB, dipciinished
by the height of the eye CD, or iU equal BE, and the
angle at D j to find the horizonUl ({istaiice DE = Cfi.
Example.
Let the height of the object AB be 175 feet, the angle of elewHioa
A D E 37?20' , and the height of the observer's eye C D » 5 feet> leqiured
the horizontal distance B C ?
This f^ls under Problem IL, Qf rigl^t ^ngM pl«Be trigoaoiii?t(j, jM^e
172 > and by ipokuig AB radius, tbe proportiop wiU be
As radius =. ...... 90? Log. co-secant = 10.000000
Is to height of the object AB 175 ft.^BE 5 ft.= 17Q Log, = 2.230449
So is the angle of elevation A D E=37?20^ Log. co.tangent= 10. 1 17637
T9thehQrizontaldist.l)£==CB=^22.89 I^g[. ^ . , . 8.348086
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msm8uration ov hbi6hts and distancat* 581
Paoblbm IIL
To find the Height of an inaccessible Object^ asAB^
Rule.
At any convenient points, as C and D (these
being in the same vertical plane with A B),
observe the angles of elevation AFE and
AGE; and measure the distance C D : then^
because the exterior angle AFE is equal to
the two interior and opposite angles A G F
and G A F (Euclid, Book I., Prop. 32), if from
the angle AFE the angle AGE be subtracted, the remainder will be the«
angle GAP. Now, in the oblique angled triangle AGP, given the side
G F = D C, and the angles A and G ; to find the side A F : and, in the
right angled triangle AFE, given tlie hypothenuse AF, found as above,
and the angle AFE; to find the perpendicular AE: to which let the
height of the observer's eye above the horizontal plane be added^ and the
sum will be the height of the ob)ect AB.
Example.
In the above diagram let the angle of elevation at C = A FE be 49?2&^ ,
and, after receding 200 feet in the same vertical plane, to the point D, let
the angle of elevation AGEbe31?20^; now, admitting the height of the
observer's eye above the horizontal plane = DGor BEtobeS feet^ it is
required to determine the height of the object A B ?
The angle AFE 49?28C - the angle AGF 31?20: a the aif^
OAF 18?8<.
Now, in the oblique angled triangle AGF, since the angles and one
side are given, the side A F is found by oblique angled plane trigonometry.
Problem I., page 177; and, in the right angled triangle AEF, since the
hypothenuse A F is now kno\vn, and the angle at F given, the perpen-
dicular AE is found by right angled plane trigonometry, Problem L, page
171. Hence,
To find the Side A P :—
Asthe angle GAP SB . « . . 18? &C Log. co^secants 10. 506919
IstothesideGF = DC= . . 200 Log. = . . . 2.301030
So is the angle AGF = • . . 31?20: Log. sine = . 9,716017
TothestdeAFs 834,17 Log-= . . . 2.52396&
2m2
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583 IIBMSURATION OP HBI6BT8 AND DI8TAMCB8.
To find the Perpendicular AB :—
As radius ss • • • • .
Is to the hypothenuse AF =
So is the angle A FE = •
To the perpendicular AE zs
Height of the eye B E = .
Height of the object A B =
90^ Log. co-8ecant= 10. OOOOOO
334.17 Log.= • . . 2.523966
49?28^ Log. sine = . 9. 880830
253.98 Log. = ' . . 2.404796
5.
258. 98 feet| as required.
Remark^-lf it be required to know the horizontal distance B C or B D,
it may be readily determined by means of the last problem.
PROBLBM IV*
To Jind the Distance of an inaccemble Object tohkh the Observer can
ndther advance towards nor recede from in Us vertical Line of Direction.
Rule.
Let the point A be any inaccessible object,
and B and C two stations from which the
<fistance of that object is to be determined :
measure the distance B €» and, with a sextant
or other instrument, observe the horizontal
anglesABC and AC B; then, in the triangle (*4
ABC, the angles and the side BC are given; to find the other two sides^
viz., ABand AC»
JExatnpIe*
Let the horizontal angle ABC, measured with a sextant, be 59? 15 C,
the angle ACB 42?45^, and the measured baseline BC 350 yards;
required the respective distances A B and AC?
The angle ABC 59915^ + the angle ACB 42945^ = 102?; and
180? - 102? =; 78?, the ang)le CAB.
Now, the angles and one side being thus known, the remaining sides
are to be determined by oblique angled trigonometry, Problem L, page
177* Hence, the following proportions :-?- .
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MSNStJRATION OF BB1QHT8 AND DI8TANCB8. SS3
To find the Distance A C :'—
As the angle C A B = • . . « 78? Log. co-secant s^ 10. 009596
Is tothesideBC^s «... 350 Log. » • « « 2.544068
So isthe angle ABC s • ,. . 59? 15^ Log. sine =» • * 9.934199
To the distance A C = • . • 307.51 Log. =s « • • 2.487863
To find the Distance A B : —
As the angle CAB = .... 78? Log. co-secant= 10. 009596
Is to the side BC = • • • « 350 Log. =s . . . 2.544068
So is the angle A CB=: . . . 42?45^ Log. sines . 9.831742
To the distance A B = . • . . 242.89 Log. = • « • 2.385406
Remark. — This problem will be found of very essential service to His
Majesty's ships and vessel^ of war, on many hostile occasions : for, when
it is intended that a squadron of those ships should cannonade a fort to
eifect, or batter a breach in the sea-defences of a town, the distance at
which the ships should be placed, abreast of such fort or town, with the
view of opening their fire to the greatest advantage, may be readily deter-
mined in the above manner. Thus, let two competent persons, provided with
sextants, in two ships, observe the angles subtended between the fort and
each ship respectively; and let the distance between the two ships be care-
fully ascertained, which is readily done by Problem II., page 530, provided
the height of the masts be known ; or it may be found by means of a boat
sent from one ship to the other, with instructions to pull at an uniform rate :
then, if the interval, per watch, be noted between the time of the boat's
pulling off from one ship and that of her arrival at the other, and her
velocity or hourly rate of sailing be duly determined by the log, the distance
between those ships may be easily obtained by the rule of proportion.
Now, with the distance between the two ships as a base line, thus found,
and the angles subtended between the fort and each ship, the respective
distances of those ships from the fort may be very readily comput^d^
agreeably to the principles of the present problem.
Note. — The most convenient distance for commencing a cannonade, is
about 300 yards ; that is, about a cable and a quarter's length from the
object at which the guns are directed. On such occasions, however, the
captains of His Majesty's ships of war always make choice of a much
closer position, provided there be a sufficient depth o| water.^
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S34 1ISN8UBATI0N OF HBIGHTS AND DI6TAMCBS.
This problem is also extremely useful in military movements: because,
when a general is determined on the reduction of a to^n or garrison, his
engineer is thus enabled to apprize him of his absolute distance from any
point of the enemy's defences against which he may be desirous of com-
mencing eperetiens, and of the most advantageous position for throwing
up batteries which may produce the greatest possible eflfect on the fortified
works pf the besieged. »
In military' operations, the battering guns are generally placed at about
375 paces (312J yards) from the works intended to be breached. —
A military pace is reckoned at 30 inches.
Problem ¥•
To^nd the Distance between two inaccessible Objects.
Rule.
Let A and B be any two inaccessible
olgects, the distance between which is
required. Measure any base line, as C D ;
at the point G observe the angles A C B,
BCD; and, at the point D, observe the
angles B D A, A D C. Now, in the triangle
A C D, in which the angles and the side C D t"^^— ^D
are given, compute the side AD, by oblique angled trigonometry, Problem
I., page 177. In like mumer, in the triangle BCD, where the angles and
the side C D are given, compute the side B D by the above-mentioned
problem. Now, in the triangle ABD, the sides AD and BD, and the
included angle ADB, areghren; with which the distance A B is to be
computed, l^ oblique angled trigonometry. Problem III., page 179.
Example.
Wanting to know the distance between the two inaccessible objects A
and B, in the above diagram, I measured a base line C D of 360 yards : at
C, the horizontal angle A C B was observed with a sextant, and found to
be 53^30'., and the angle BCD 38?45' ; at D, the horizontal angle
BD A was 67?20^ and the angle ADC 44?30^ $ required the distance
between A and B>
Angle ACB 53?30'. + angle BCD 38?45^ « angle ACD 92?15!;
aiid^e ACD 92?15'. + angle ADC 44?S0^ » 136?45:. Nov,
180? - 136?45; ^ the angle CAD 43n&f.
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MUfSllRATlOK or BJBieHTS AND DI8TANCBS. 6SS
Again: Atigle BDA 67?80: + angle ADC 44?30: = ahgleBDC
lll^SOCj and angle BDC 111°50C + BCD 38?45f = 160?85';.
Now, 180? - 150?35: =^ the angle CBD 29?25:.
In the Triangle A CD, to find the Side AD:—
As the angle CAD = . . . 43915^ Log. co-secant = 10. 164193
Is to the side CD = • ... 360 Log. = • . . 2.556303
So is the angle A C D = . • . d2? IS ^ Log. sine =: « . 9. ^9^665
To the side AD =: 5iS.b Log. =f • • . 1720161
In the Triangle BCD, to find the Side BD:—
As the angle C BD = . • . 29?25? Log. co-secant =5: 10. 306^79
Is to the »ide CD == ... 360 Log. =: . . . 2.5563^3
Soistheaftgi^BCDzr . • . 38^45 C Log. sine =i . . 9.796521
To the side BD =1 . • . . 45^,78 Log. = . . . ld6l60S
In the Triangle ABD, to find the Angle DAB or DBA, and
the Side AB:—
180^ - the angle BD A 67?20^ = 112?40' -♦- 2 = 569205 = half
the sum of the angles DBA and DAB.
Ai» the sum of the sides A D and D Br:988. 78 Log. an eoini^«=a7^ 007 102
Is to their difFerenee = .... 66.28 Log. =i: . . 1.821388
So is ^ sum of tfBglesDBAand DAB=s56?20: Log. tangents: 10. 176476
To half di£ferenee of ditto =: • . 5 ?46 ^ 83? Log. tang.:: 9< 00496 1
Angle DBA = 62? 6^33r
Angle DAB = S0?33:27?
To find the Distance ABs—
As the angle DAB = « . . 50933C37? Log co.secant=lO. 112218
Is to the side BD r: . . . 458.78 Log. r: . . 2.661603
Soistheangle ADBn . . 67^20^ 0? Log. sine = . 9.965090
Totheside ABist . . . 54^.10 Log.fi^ ^ , 2*788911
wl^b^ tbefefote, is lli« f eqiilred dfittaacii^
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536 MBN8URATI0N OF HXI6HT8 AMD DI8TANCB8.
^ote.— This problem is very useful in taking surveys of coasts^ harbours^
bays, islands^ &c.
Problem VJ,
Given the Distances between three Objects, and the angular Distances
between these Objects taken at any Point in the same horizontal plane j
to find the Distance between that Point and each of the ObjectSm
RULB»
Let Ay B, and C be any three objects whose
distances from each other are given, and E the
place of the observer : at E, observe the angles
C E A and C E B ; connect the points A, B, and C,
by right lines; make the angle AB D equal to the
observed angle C E A, and make the angle DAB
equal to the angle C E B : hence the point D is
found; then, through the three points A, D, and
B, describe the circle AD BE; join CD, and
produce this line till it meet the circle at the point
E, tlie place of the observer ; join E A and E B.
Now, in the triangle ABC, of which the three sides are given, find the
angle B A C. In the triangle A B D, in which the angles and the side A B
are known, find the side A D. In the triangle A C D, of which two sides,
AC and AD, and the included angle CAD are known, find the angle
A C D. In the triangle A E C, of which the angles and the side A C are
given, find the sides E A and EC. And in the triangle ABE, the sides
AB, AE, and the angles AEB and EAB are given; to find the sideEB.
Exanq^le.
Let the points A, B, and C, in the above diagram^ be three known
objects : the distance between A and B, 290 yards ; between B and C, 195
yards ; and between A and C, 240 yards : let E be the place of an observer,
where the angle C E A was measured with a sextant and found to be 30?5',
and the angle CEB'25?45^ ; required the distances E A, EC, and ED ?
In the triangle ABC, the three sides are given ; to find the angle B A C.
Hence^ by oblique angled plane trigonometry^ Problem lY.j page 180,
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MBN8URATI0N OP RBIGHTft AND DISTANCES.
5S7
Side BC (opposite the required angle) =195
Side A C (containing the required angle) = 240
Side A B (containing the required angle) = 290
Log. ar.co.=:7. 619789
Log. ar.co.= 7. 537602
Sum=:
725
Half sum =: .
Remainder =:
Arch =
362.5 Log. =:
167.5 Log.=
2.559308
2. 224015
Sum= 19.940714
20?55<46r Log. co-sine =9.970357
Angle BAC= 41?51^32f
Angle BAD = the angle CEB= 25.45. 0
Angle DAC =
16? 6^32r
In the triangle A B D, the angles and the side A B are given ; to find
the side AD : thus, the angle ABD (= the angle CEA) = 30?5'. +
the angle DAB (= the angle CEB) = 25?45^ = the angle AEB
55?50C ; and 180? - 55?50C = the angle ADB =*124?10: : for, the
angle A DB is evidently the supplement of -the angle AEB; because the
opposite angles of every quadrilateral figure described in a circle are equal
to two right angles. — Euclid, Book III,, Prop. 22. Hence, by trigo-
nometry.
As the angle A D B = 124? 10' Log. co-seeant = 10. 082281
Is to the side A B = 290 Log. = . . . 2.462398
So is the angle A B D= 30? 5 C Log. sine = . 9. 700062
To the side AD = .175.69 Log. = • . • 2.244741
In the triangle ADC, the two sides AC, AD^ and the included angle
DAC, are given; to find the angle ACD: hence, by oblique angled
trigonometry. Problem III., page 179,
Log.ar.co.7.381230
Log. = 1,808279
DAC 16?6^32r J 81?56:44TLog.tang.lO, 849213
As side AC 240 + side AD 175. 69 = 415. 69
Is to side AC 240- side AD 175. 69 = 64. 31
So is 180? - angle DAC 16?6^32f
= 163?53:28f
Tohalfdiflf.ofanglesADCandACDzz 47.33. 3 Log.tang.lO. 038722
Angle ACD = 34?23Mi:
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S88 ftCBNSUEATIOIf Of BVtOHTS AlTD StlTAKCtt.
In the triangle AEQ the unglet and the tide AC lu« gWdn j to And
the sides EAand ECt thus^ the angle CEA dO?6^ + the angle ACfi
d4?23MI? 2 64?28UK ; and 180? - 64^28^1'' a the atlgle E AC
115?31 ' 19?. Hence^ by oblique angled trigonometry. Problem I.^ page
177,
To find the Side E A >—
As the angle AEG = 30? 5 C Or Log, co-secant = 10.299938
Is to the side AC = 240 Log. = . . . 2.380211
SoistheangleACEs: 34?2dMir Log. sin^ = . . 9.751905
TothesideEA= . 270.47 Log. =: . . . I4d21l4
To find the Side EC:--
As the angle AEC = 30? 5 C Or Log. = . . . 10.299938
I* to the side AC = 240 Log. :± . . • 2.3S02tl
Si is the angle EAC =:115?3lU9r Lbg.Sine =: . . $.955407
To the side fiC r: 432.07 Log. r: . . . 2.685356
In the triangle ABE, the sides AB, AE, and the angles A EB, EAB,
are given ; to find the side E B : thus, from the angle E A C li5?3U 19r,
take the angle BAC 41?5H32r, and the remainder is the angle EAB
=: 73?39'47r. Hence, by trigonometry.
As the angle A EB <= 55?50^ Or Log. oo-seeant =: 10.082281
Is to the side A B = 290 Log. = . . . 2.462398
So is the angle E AB = 73?39U7^ Log. sine = . . 9. 982101
To the ride EB» 336.34 Log. » « • , 2^526780
Hence the distance of the object A from the observer lit E!^ is 270*47
yards ; that of C, 432. 07 yards ; and that of B, 336. 34 yards.
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MBNSURATION OF HBIGMTS AND DISTAKCBS.
580
Problem VII.
Given the Distances between three Objects , and the angular Distances
between these Objects^ taken at any Point within the Triangle formed
by the right Lines connecting the Objects ; to find the Distance
between that Point and each of the Objects.
Rule.
Let A, B, and C be any three objects whose
distances from each other are given, and E the
place of the observer: complete the triangle ABC;
at E, observe the angles AEC, AEB, and BEC;
make the angle BAD equal to the supplement of
the angle B E C ; in like manner, make the angle
A BD equal to the supplement of the angle A E C :
hence the point D is found. Through the three
points A, B, and D, describe a circle; join D C, and
it will cut the circle in E, the place of the observer ;
connect the points A E, BE, and the construction will be completed; the
calculations in which will be nearly similar to those in the preceding
problem.
Example.
Let A, B, and C, in the above diagram, be any three known objects
whose distances from each other are as follow : viz.^ A B, 620 yards ; A C^
570 yards ; and B C, 460 yards. At a point E, within the triangle formed
by those objects, the angle A E C was measured with a circle, and found to
be 125?151 ; the angle AEB, I24?15'. ; and the angle BEC, I10?80'. ;
required the distances E A, EC, and EB ?
In the triangle A B D, the angles and the side A B are given ; to find
the side AD: thus, the angle BAD 69^30 C sf the supplement of the
angle BEC; the angle ABD 54°45C = the supplement of the angle
AEC; and the angle AD B 55M5'. =: the supplement of the angle AEB.
Hence^ by trigonometry.
As the angle ADB = 55?45C
Is to the side AB = 620
So is the angle ABD = 54?45C
Log. co-secant=10. 082710
Log. = . . . 2. 792392
Log. sine = . 9.912032
To the side AD
612.54 Log. = . . . 2.787134
In the triangle A B C, all the sides are given ; to find the angle B A C :
which, being added to the angle BAD, will give the obtuse angle CAD.
Henca^ by trigonometry^ Problem IV.^ pi^ 180.
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540 MENSURATION OF HBIOIITS AND DI8TANCB8.
Side BC = 460
Side AC = 570 Log. ar. comp.= 7. 244125
Side AB = 620 Log. ar. comp.= 7. 207608
Sum s= . 1650
Half sums 825 Log. = . . 2.916454
Remainder = 365 Log. = . . 2. 562293
Sums 19.930460
'' Arch= . 22 ?37 ^ 9'Log.co-8ine=9. 965240
Angle CAB=45?14n8? + angle BAD= 69?30' s angle
CAD114?44a8r.
In the triangle A C D, the sides A C, A D, and.the included angle CAD
are given ; to find the angle A C D : hence, by oblique angled trigtmometiy,
Problemlll., page 179,
As the side AD 612. 54+the side AC 570=1182. 54 Log.ar.co.6. 927IS4
Is to the side AD 6 12. 54 -the side AC 570= 42. 54 Log. = 1.628798
Sois 180?- angle CAD H4?44n8r J 32,37,5,, Log. tang. 9. 806374
To half diff. of angles A C D and A D C= 1.19.10 Log. tang. 8. 362356
AngleACD= •••...• 33?57^ K
In the triangle ADC, all the angles and the side A C are gitren ; to
find the sides AE and EC : thus, the angle A EC 125?15'. + angle ACE
33?57n'/ = 159?12'K; and 180? - I59?12!ir = the angle CAE
20? 47 •59?. Hence, by oblique angled trigonometry. Problem I.^ page
177,
To find the Side A E :—
Astheangle AEC= 125?15^ 0* Log. co-secant =10. 087968
Is to the side A C = 570 Log. = • . • 2. 755875
So is the angle A C E= 33? 57 ' K Log. sine = • 9. 747002
Tothe8ideAE= 389.80 Log. = . . . 2.590845
To find the Side EC:—
As the angle A E C = 1 25 ? 1 5 ' 0^ Log. co-secant = 1 0. 087968
Is to the side AC = 570 Log. = • . . 2. 755875
So is the angle C A E= 20? 47 ' 59f Log. sine = • 9. 550359
To the side EC = 247.86 Log* = , . « 2.394202
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In the triangle BEC^ given the sides BC, CE, and the angle EEC;
to $nd the angle BCE, and, thence, the side BE : the angle BC E is
found by oblique angled trigonometry. Problem IL, page 178; and the
side BE by Problem L, page 177« Hence,
To find the angle B C E :—
As the side B C =r 460 Log. ar. comp. = 7. 337242
Is to the angle B E C s= 1 10?30: Log. sine =r • 9. 971588
So is the side EC = 247. 86 Log. «... 2. 394202
To the angle C BE = 30?18f4KLog. sine = • 9.703032
Angle B£C= . . 110.30. 0
Sum= 140?48Uirj and 180? ^ 140?48'41f =
the angle B C E z 39m n9f
To find the Side BE :—
As the angle 3 ECs 110?30^ Or Log. co-secants 10. 028412
Is to the side B C = 460 Log. = . . . 2. 662752
So is the angle BCEr:39?lin9r Log. sines: . 9.800631
TothesideBE= 310.31 Log. =s . . . 2.491795
Hence the required distances are, EA, 389.80 yards; EB, 310.31
yards; and EC, 247. 86 yards.
Problem VIIL
Given the Distances between three Objects situated in a straight Line^
and the angular Distances of these Objects taken at any Point in the
same horizo7ital Plane; to find the Distance between that Point and
each qfthe Objects »
RuiJs.
Let the points A, B, and C be any three
objects situated in a straight line : make the
angle ACD equal to the observed angle
AEB, and make the angle DAC equal to
the observed angle B EC : hence the point D
is found. Through the three points A, D,
and C describe a circle ; join D B, and produce
it till it cuts the circle in E ; then E will be
the place of the observer, and EA, EB, and
£ C the required distances.
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542 IISNSUEATION OF HXIOHTS AND DItTAKCXS.
Example.
Let A, B, and C, in the above diagram, be any three knofvm chjeets
situated in a straight line, whose distances from each other are as follow :
viz., A B, 490 yards ; B C, 300 yards ; and A C, 790 yards : at a point E,
the angle BBC was observed^ and found to be 43"^, and the angle BE A
33?45'. I required the distances £A^ EB, and SC?
In the triangle A D C, all the angles and the side A C are given ; to find
the side AD: thus, the angle D AC 43? = the observed angle BEC;
the angle ACD 33?45^ = the observed angle BE A; and, consequently^
the angle AD C =: 103? IS ' : hence the Me A D may be found.
As the angle AD C = lOSnSC Log. co-secant=10.011718
Is to the side A C = 790 Log. = . . , 2. 897627
SoistheangleACD » d3?45: Log. sine s . 9.744739
TothesideADs 450.94 Log. » . « . 2.654084
In the triangle A B D, given the sides A B, A D, and the included angle
DAB; to find the angle ABD: hence, by trigonometry. Problem III.,
page 179,
As the side AB 490+the side A D 450. 94=940. 94 Log.ar.co.7. 026438
Is to the side A B 490- side A D 450. 94=: 39. 06 Log. = 1 . 591732
Soisl80e-aBgleDAB43? = 137?^2=68?30^ Or Log. tang.lO. 404602
Tohalfdiir.ofanglesADBandABD= 6? 0'57rLog.tang. 9.022772
Angle ABD = 62?291 3rj and, since the two
straight lines AC and DE intersect each other in the point B, the opposite
angles are equal to one another (EucKd, Book I., Prop. 15) : therefore the
angle E B C is 62?29^3?, equal to the angle ABD. In like manner, the
angles D BC and ABE are equal to one another; and because DBC is
the supplement of the angle D B A, it is equal to U 7 ?S0^ 57' : hence the
angle ABE is also equal to 117?30C57r.
In the triangle ABE, all the angles and the side AB are given ; to find
the sides EA and EB : thus, die an^e BE A 33?45( + the angle ABB
117?30:57r = 151?l5^57f ; aod 180? - l51?lSC57r = the aagle
B A E 28?44 ^ 3r. Hence, by trigonometry^ Problon i, page 177>
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To find the Side EA:—
AstheangleBEA = 33?45f 01 Log. co-secant= 10. 255261
Is to the aide A B =; 490 Log. =» « , , 2, 690196
Soi8theangleABE=117?30:57^ Log. sine = . 9.947867
To the side £A =3 782.21 Log. =: . . . 2.893324
To find the Side EB^^
As the cuigW B£A s 3SHi'. 01 Log. eo-ieeontss 10. 255261
Jltptb^«i4^AB s? 490 Log.>R. . • 2,690196
So b the angle BAE=: 28^44' 3? Log, aioe pi . 9.681917
To the side EBs 424.01 Log. = . . . 2.627374
^ In th^ triangle SBC, given the sides £ B, B C^ and all the angles ; to
i find the side EC.
As the angle B EC » 43? Oi OZ Log. oo-sfioant^ia 166^17
Is to the side BC = 300 ' Log. = . . • 2.477121
So is the angle E B C =62^^29' 31 Log. sine = . 9. 947866
TathesideEC^. 390.19 Log. » . . . 2.591204
Hence the required diatanees are, EA^ 782.21 yards; EB, 424.01
yardft; wi EC, 390. 13 yards.
JZemorilr.— The above problem, together with that given in page 536,
wiU be found exceedingly useful to a general or other officer employed in
conducting the military operations of a iiege ; because, if he cati only
preeure a correct map of the town or garrison which he may have occasion
to invest, so. aa to ascertaia the relative dietances between any three
desirable positions, the above problems will enable him to find his absolute
distance from those positions without the trouble of measuring a base line :
nor is it necessary to» resort ta trigonometrical calculation for this partieular
purpose, since the distances may be readily determined by geometrical
projection, to every degree of accuracy desirable in such operations.
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544 MENSURATION OP HBI6HTS AND DI8TANGB8.
Problem IX.
Given the Height qfthe Eye; to find the Distance of the visiile Horizon.
. RULB.
Let the earth's diameter, in feet, be augmented by the height of the
eye ; then, to the logarithm thereof, add the logarithm of the height of
the eye ; from half the sum of these two logarithms subtract the constant
logarithm 3.783904,* and the remainder will be the logarithm of the
distance in nautical miles ; which is to be increased by a twelfth part of
itself, on account of terrestrial refraction.
Esample.
Chimbora$o, the highest part of the Andes, is said to be 20633 feet
above the level of the sea: now, admitting that an observer be placed upon
its summit, at what distance can he see the visible horizon^ allowing a
twelfth part of that distance for the effects of refraction ?
Diameter of the earth, in feet, = . • • 4 1 804400
Elevation of Chimbora9o 20633+5 feet,? 20638 Log.= 4.314668
the height of the observer's eye =3
Sum= 41825038 Log.= 7.621436
Sum s= 11.936104
Half sum = 6.968052
Constant log. =5 • « • . 3.783904
Distance uncorrected by refraction ^ • • « 152.81 Log.=2« 184148
Allowance for terrestrial refraction = • . . 12. 40
Dist. at which the visible horizon may be seenss 165. 21 miles#
• This U the logarithm of 6080, the number of feet in a nantiGal mile.
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MBNSURATION OF HBI6HTS AMD DISTANCES.
545
Problem X.
Given the measured Length qf a base Line ; to find the Allowance for the
Curvature or spherical Figure of the Earth.
Rule,
Let £ B F represent the arc of a great
circle on the earth ; C^ the earth's cen-
tre; C B its semi-diameter ; and A B the
measure of a base line^ on an apparent
level or horizontal plane on the earth's
surface : join C A, and it will cut the arc
of the great circle in D ; then A D will be the excess of the apparent level
of the horizon above its true leveK
Now, in the right angled plane triangle ABC, given the perpendicular
B C and the base A B ; to find the hypothenuse A C : which is readily
determined by Euclid, Book I., Prop. 47. Then, the difference between
A C, thus found, and C D = C B, will be equal to D A, or the absolute
value of the true level behm the apparent level : and, if this value be
expressed in miles and decimal parts of a mile, it may be reduced to inches,
if necessary, by being multiplied by 63360 = the number of inches in an
English mile.
Example.
Let the base line A B, in the above diagram, be 1 English mile, and
the earth's semi-diameter BC = 3958.75 miles; required the allowance
for the earth's curvature answering to that base line, or. the difference
between the true and apparent levels on the earth's surface expressed by
the measure of the line AD?
BC3958. 75 xB 03958.75:^15671701.5625
AB= 1 X AB= 1= . 1.
Sum of the squares = . . 15671702. 5625 ; the square
root of which = C A, IS • • • 3958.7501263
Subtract C D = C B, the earth's semi-diameter, = . . 3958. 7500000
Remainder == the line A D, the allowance for curvature,= 0000. 0001263
Multiply by the number of inches in an English mile, = 63360.
Number of inches which the true level is below the
apparent level in one mile =s 8. 0023680
2n
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546 MBNSURATION OF HBIOHTS AND DI8TAMCB8.
^Now, since the curvature answering to A B is known, that corresponding
to any other base line on the earth's surface may be readily determined by
the following proportion: —
As the square of A B, is to AD; soisthesquareofBG, to GH: whence
it is manifest, that the curvature answering to any given distance^ as B G^
is in the duplicate ratio of that distance to A B :
And, since AB is expressed by unity or 1, and that AD is a constant
quantity, the proportion may be reduced to a logarithmical expression ; as
thus : —
To twice the logarithm of the given base line, expressed in miles and
decimal parts of a mile, add the constant logarithm 0. 903219 (the log. of
8. 002368 inches) ; and the sum will be the logarithm of the number of
inches and decimal parts of an inch which the true horizontal level at sea
is below its apparent level.
Example L
Required the curvature of the earth answering to a distance of 2 miles
on its surface ?
Distance = 2 miles ; twice the log. = « . • 0. 602060
Constant log* s . ; • 0.903219
Curvature, in inches, = 32.009 Log. = 1.505279
Hence, the curvature answering to a distance of 2 miles on the surface
of the earth, is 32. 009 inches ; or 2| feet, nearly.
Example 2»
Required the curvature of the earth answering to a distance of 15 miles ?
Distance = 15 miles ; twice the log. = . . 2. 352182
Constant logarithm = ........ 0.903219
Curvature, in inches, = 1800.533 Log. = 3.255401
Hence, the curvature answering to a distance of 15 miles on the earth's
surface, is 1800§ inches ; or 150 feet and half an inch.
Remark. — If to twice the logarithm of the ^ven base line, in mile% tbe
constant logarithm 9. 824037 be added, the sum (abating 10 in the index,}
will be the logarithm of tha excess of the apparent above the true lefe^ m
feet*
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M&NSt7itATIDN DP HtilCHTS AND DiSTANCSd. 54^
Example.
Required the curvature of the earth, or the excess of the apparent above
the true level, answering to a base line of 15 English miles in length ?
Given base line =15 miles; twice the logarithm =r • • . !2. 352182
Constant log. =z log. of 8. 002368 inches, diminished by the
log.of 12 inches, = 9.824037
Bxcess of the app« above the true level^ in ft.,s: 150. 044 hog.tsi. 176S19
Note, — ^This problem will be found useful to land-surveyors, engineers,
and others employed in the art of levelling, cutting canals, and conducting
water (by means of pipes, &c.) from one place to another.
Problbm XI.
QiDen the meoiured Length of a Base Line on any deoaJted Level $ to
find its true Measure^ when referred to the Level of the Sea.
Rule.
In the annexed diagram, let the arc AB represent the A^
measured length of a base line, at any given elevation
above the level of the sea expressed by the arc D E ; let
C D be the radius of the earth, or the distance from its
centre to the surface of the sea ; and let C A be the earth's
radius referred to the level of the measured base line A B<
Now, because the arcs A B and D E are concentric and similar, and that
similar arcs of spheres are to each other as their radii, we have the following
analogy j viz,,
As the radius C A, is to the radius C D ; so is the arc A B, to the arc D E:
that is, as the earth's semi -diameter, augmented by the height of the base
line above the level of the sea, is to the earth's true semi-diameter ; so is
the measured length of the given base line, to the true measure of that
line at the surface of the sea.
Example*
Given a base line of 36960 feet in length, measured on a horizontal plane
which is elevated 1 20 feet above the level of the sea j required the measure
of that base line at the surface of the sea ?
2n2
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548 MENSURATION OF HEIGHTS AND DISTANCES. .
As CD = 20902200 + DA = 120 = C A 20902320=AB 36960 ::
CD 20902200 : DE = 36959.787813. Hence the given base line,
reduced to the level of the sea, is 36958. 787813 feet; which is about 2^
inches less than the measure on the elevated horizontal plane.
But, since the probable elevation of ^ny horizontal plane on the earth,
above the level of the sea, can bear but a very insignificant proportion to
the earth's semi-diameter,^ — if, therefore, the product of the measured base
line by its height above the level of the sea be divided by the earth's radius,
the quotient will be the excess of the measured base above the correspond-
ing arc at the surface of the sea. This may be reduced to a logarithmical
expression, in the following manner 5 viz., to the constant logarithm
2. 679808, add the logarithms of the base line and of its elevation above
the level of the sea, both expressed in feet : the sum will be the logarithm
of a natural number, which, being taken from the measured base line, will
leave the measure of that line at the surface of the sea, sufficiently near the
truth for all practical purposes. Thus, to work the last example.
Constant log.=ar. co. of the log. of the earth's semi-diam. in fit.=2. 679808
Elevation of given base line above level of sea= 120 feet. Log.=2. 079181
Measured length of the given base line = 36960 feet. Log.=4. 567732
Excess of the given base line above the
arc at the surface of the sea = ... — 0. 212188 Log. 9. 326721
Given base line, reduced to level of sea, = 36959. 7878 1 2 ; which approx-
imates so very closely to the true result by the direct method of computa-
tion, as scarcely to admit of any sensible difference.
Remark. — In consequence of the spherical figure of the earth, no two
points on its surface can be situated exactly on the same horizontal plane;
for it is the chord of the arc, and not the arc itself, that measures the
horizontal distance between two points. Hence, when philosophical
inquiries are under consideration, it becomes necessary to apply a small
correction to the measured base line on a horizontal plane, so as to reduce
it to the corresponding terrestrial arc ; though, in general, this correction
is so very inconsiderable, that, even in the most extensive trigonometrical
surveys, it may be safely disregarded. If, however, it be deemed necessary
to find its value, or (which amounts to the same thing) if the excess of
the arc over its chord be required, it may be very readily determined by
the following rule, to every desirable degree of accuracy ; viz..
From thrice the measured 'length of the base line, in feet, subtract the
constant logarithm 16. 020595 : the remainder will be the logarithm of the
excess of the arc over its corresponding chord, expressed by the given base
line.
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Let it be required to find the excess of the terrestrial arc over its chords
answering to a measured base line of 36960 feet in length, or seven English
miles?
Given base line = 36960 feet ; thrice its logarithm = . ♦ 13. 703196
Constant log. = log. of 24 times the square of the earth's
radius, in feet, = 16. 020595
Excess of the arc over its chord, in feet,=0. 004815 Log.= - 7. 682601
Hence it is evident, that the extent by which a terrestrial arc of 36960
feet exceeds the chord of the same arc, is only the small decimal fraction
. 004815 of a foot, — an excess so very trivial, as to be scarcely worth taking
into account, even where the greatest accuracy is required : for, in the
present instance, though the base line is 7 English miles in length, it
amounts to no more than the two hundred and sixteenth part of an inch.
Note. — ^The index of the logarithm of the excess comes out a negative
quantity, because the index of the constant logarithm, is greater than that
of the term from which it is subtrac*ted.
Problem XII.
To find the Height and Distance qfa IBll or Mountain.
Rule.
Let the point A, in the annexed
diagram, be the summit of a hill,
the height of which, A B, is to be
determined; and let the point C
be the place from which its dist-
ance is to be found : at C, observe
the vertical angle ADE; then
measure any convenient distance
for a base line, as C F ; at the point
C, observe the inclined angle ACF,
and, at F, observe the angle AFC. Now, in the inclined triangle ACF,
given the angles and the side F C ; to find the side A C, which may be
considered as being essentially equal to the side A D : and, in the vertical
or right angled triangle A ED, the angle ADE and the hypothenuse or
side A D are given ; to find the perpendicular AE: to which, the height
of the eye B E =: C D being added, gives the required height A B.
S$Q MB^rSURATIOH^ OF II9I6HT3 At^B 0I3TAKCB6.
Wanting to know the height of the hill ^ P, and the distance of its
summit A from the poipt C^ the vertical angle A D E was ohsefved, afid
found to be 14?30^ ; at the points C and F, 500 feet asunder, the inclined
angles ACF and AFC were measured, and found to be 80?5', and
73?30' respectively; Required the distance of the point A from the
pbserver at C, and its height above the level of the horizontal plane C B ?
Ii^ the inclined trij^ngle ACF, the angles are given, and the side FC =s
500 feet ; to find the side or distance A C : thus, the angle A C E 80?5 ^ +
the angle AFC 73^80^ = 158?35^ ; and 180? - 153^85'* = 26?25^
the measure of the angle C A F. Hence, by trigonometry,
To find the Distance AC s the Side AD :—
A^ the a^gje C AF =; 26?25; Log. co-seeaut= 10. 351742
Is to the side C F = . 500 Log. = . . 2. 698970
So is the angle CF A = 73?30^' Log. sine = . 9. 981737
To the distance A C = 1077. 5^5 hog. ^ . . 3. 032449 ;
and, since A C and A D are essentially equal, tha aide A D is also IQ77«5a
feet.
To find the Height A B :—
As radius = ... 90? Log. co-secant=l 0.000000
Is to the side ADnAC 1077.5$ Log. = . . 3.032449
So is the angle A D E = 14?30: Log., fine s: .9. 39860Q
To the perpendicular AE=269. 84 Log. = . . , 2.431049
Height of the eye BE=C D 5.
Height of A 3= . . 274.84 feet
PROBLEM XIIL
To find the Height of a Mountain^ hy means of two Barometers and
Thermometers.
RULB.
Let two observers (pravided with barometers and thermpqiet^rs of equal
constructio^i,) carefully note dpwn^ at the sain^ in^tanty the r?^>ective
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MBNAtnUTlON OF HSIGHTS AKD DI8TAKCS8. 551
heights of the barometers at the top and bottom of the mountain, or other
eminence intended to be measured^ with the temperature of thewquicksilver
in each instrument by means of attached thermometers, and also the tem-
perature of the air, in the shade, hj means of detached tliermoitietei% ;
then.
Find the diflference of the logarithms of the observed heights of the
barometers, the first four figures of which, besides the index, are to be
considered as whole numbers. To this difference apply the product of
0. 454, by the difference of the altitudes of the two attached thermometers.^
by subtraction if the temperature of the quicksilver at the bottom station
exceed that at top ; otherwise, by addition : and the sum or difference will
be the approximate height, in fiatboms, English measure.
Now, to the logarithm of the approximate height, thus found, add the
logarithm of the difference between the mean of the two temperatures of
the detached thermometers and 32?, and the constant logarithm 7. 387390:
the sum of these three logarithms will be the logarithm of a correction,
which being added to the approximate height when the mean temperature
exceeds 32"?, but subtracted if it be less> the sum or difference will be the
true elevation of the mountain, expressed in fathoms; which may be
redueed to feet, if necessary, by being multiplied by 6.
JVbto. — ^This m)e is deduced firom that given by Dr. HutCon, in the
second volume of his ^^ Course of Mathematics/' page 255.
Example 1.
Let the observations at the top and bottom of a mountain be as follow ;
required its height ?
Attschftd Detadked
thermometer* thermom. Barometer.
Obs- at bottom=57 57 29- 68 Log.r: 1 . 472464
Ditto at top - 43 42 25. 28 Log.=: 1 . 402777
Differtnee =r 14 S«Bn = 99 Difference = 0.069&87
Multiply by .454 Mean=: 49| Product =i - 6.36
Product = -6.356 32 Approx. alt> n 690. 51 Log.2« 839170
Diff.sr \7ior\7.5 hog.zz . . . 1.243038
CfHMKtant log. =r 7.387390
Correction of the approximate altitude =: • • • + 29.48 Log.l. 469598
True altitvte of tbe momtaiD^ in fotbooae, =: . 719. 99,or 4319. 94 h.
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552 MENSURATION OF HSI6HT8 AND DI8TANCB8.
Example 2.
Let the observations at the top and bottom of a mountain be as follow ;
required its height ?
Attached Detached
thermometer, thermom. Barometer.
Obs. at bottom=38 31 29. 45 Log.= 1 . 469085
Ditto at top = 41 35 26. 28 Log.= 1 . 428459
Diflference = 3 Sum = 66 Difference =0.0406.26
Multiply by .454 Mean= 33 Product = +1-36
Product = + 1 . 362 32 Approx. alt. = 407. 62 Liog.2. 610255
Diff. = 1 Log.= 0.000000
Constant logarithm = 7.387390
Correction of the approximate altitude = , • + 0. 99 Log.9. 997645
True altitude of the mountain^ in fathoms, =: • 408. 6 1, or 245 1 . 66 ft.
Problem XIV.
To find the Distance of an Object, by observing the Interval of Time
between seeing the Flash and hearing the Report of a Gun or of a
Thunder-Cloud.
RULB.
To the logarithm of the number of seconds elapsed between seeing the
flash and hearing the report, add the constant logarithm 9. 273762* ; and
the sum (abating 10 in the index,) will be the logarithm of the distance in
nautical miles ; or, if the constant logarithm 9. 335032t be made use of,
it will give the distance in English statute miles.
* This is the sum of the arithmetical complement of the logarithm of 6080, the nnmher
of feet in a nautical mUe, and the log^thm of 1 142, the number of feet which sound travels
in one second of time.
t This is the sum of the arithmetical complement of the los;arithm of 5280, the mimber
of feet in an English mile, and the logarithm of 1142 feet, the established velocity of aound.
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OF THE COURSE 8TEBRBD BY A SHIP 8BBN AT A DISTANCE. 553
Example 1.
A ship at eea was observed to fire a gun, and 43 seconds afterwards the
report was heard ; required the distance of the ship in nautical miles ?
Interval betw. seeing the flash and hearing the gun=43? Log.=:l. 633469
Constant logarithm = 9. 273762
Distance in nautical miles
8.076 Log.=0. 907231
Example 2.
A flash of lightning was seen, and after a lapse of 18 seconds the repdrt
reached the ear of the observer ; required the distance of the thunder-
cloud in English miles ?
Inter, betw. seeing the flash and hearing the thunder= 1 8 ! Log.= 1 . 255273
Constant logarithm = 9. 335Q32
Distance of the thunder-cloud in English miles=3.894 Log.ssO. 590505
PaOfiLBM XV.
Given three Bearings of a Ship sailing upon a direct Course, and the
Intervals of Time between those Bearings ; to find the Course steered
by that Ship, and the Time of Iter nearest Distance from the Observer.
Describe the circle N E S W,
and from the centre C, th6
place of the observer, draw the
lines CA, CB, and CD, to
represent the three bearings
of the ship : through the centre
C draw the line FG, at right
angles to the second bearing
CB; make CF equal to the
first interval, and C G equal to
the second, each being ex-
pressed in minutes, and taken
firom any convenient scale of
equal parts; from the points F
Rule.
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554 IfEKSUBATlON OP HBI6HT8 AND DI8TANCJK8.
and G draw the lines FA, GD, parallel to the second bearing CB^ and
meeting CA and CD in the points A and D; join AD, and it will
represent the ship's track ; through C draw C K, parallel to A D, and the
arch S K will be the measure of the ship's course. From C let fall the
perpendicular C H upon the line A D, produced if necessary ; and from H
let fall the perpendicular H I upon the line F G^ produced also^ if necessary |
then the measure of CI will give the interval between the dme of the
second bearing and that when the ship was nearest to the observer.
Make A L equal to the difference betweeh the perpendiculars A F and
D G ; then, in the right angled triangle A L G, given the perpendicular
AL and the base LD; to find the angle LAD, which is evidently equal
to the angle a C K ; to this let the inclination of C B to a parallel be
applied, and the result will be the apparent course of the ship.
Example.
At 1 ^20T past noon a ship, sailing upon a direct course, was observed to
bear N.W. b. N. ; at 2M0r, she bore N. ^ W. ; and at 3*25?, the bearing
w«3 N.E. b. EL; required ttie apparent course steered by that ahip> and the
time when she was nearest to the observer ?
Sobition. — ^The circle being described and quartered, and the three given
bearings laid down as above directed^ through C draw F G perpendicular
to the second bearing C B ; make F C equal to 50 minutes, the interval
between the first and second bearings, and C O equal to 75 mkiQleey Ae
interval between the second and third bearings : tfiese may be taken from
any scale of equal parts. Then proceed with the other parts of the con-
struction, agreeably to the rule ; now, the ship's apparent course, repre-
sented by the angle S C K, being applied to the line of chords, will be
found to measure 72j degrees : hence the course is & 73?SO^ Bb^ or
E. b. S. I S. nearly. The perpendicular GD, being sfqplied to the scale
of equal parts from which the intervals were taken, will be fomd to
measure 40, and the perpendicular F A 93^ ; tbe diffetence between wliich
= 53}, is the measure of A L. Then C I, measured upon the same scale,
gives 26 minutes ; which is evidently, by the construction, past tbe tnne of
the second bearing : hence the time of the ship's nearest approach to tbe
observer at C, is 2M0T + 26r = 2^36? past noon. Now, the figare
being thus completed, the required parts may be obtained by trigono-
metrical calculation, in the following manner : —
In the right angled triangle AFC, given the an^es and the base FC 50
minutes ; to find the perpendicular FA. Thus, since the sln^ht line AC
falls upon the two parallel straight lines C B and PA, it midkea the ahemate
angles equal to one another (Euclid, Book I., IVop^ 29) : therefbie the
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OP THB COUBSB STKBRBB BY A SHIP SBBN AT A DISTANCB. 558
angle FAC ia equal to the angle ACB; but the angle ACB is given,
being equal to 2^ points, viz., the difference between N.W. b. N. and
N. I W. : henee the angle F A C is also equal to 2^ points.
In the same manner it may be shown (in the right angled triangle DOC,
where the angles and the base C G 75 minutes are given ; to find the
perpendicular G D,) that the angle G D C is equal to the angle BCD;
and ^ since BCD is given, being equal to 5^ points, viz., the sum of
N.E. b« E., and N. | W., therefore the angle G D C is also equal to 5^
points. Hence,
To find the Perpendicular GD :—
As radius ^ • • . • 90? Log. co-secant = 10.000000
Is to the base CG» . 75? Log. =i . . . 1.875061
So is the angle G D C^H poinU, Log. eo-tangent ^ 9. 727957
Tothep6i7endicularGD=40.09Log.= . . . 1.603018
To find the Perpendicular F A : —
As radius = .... 90? Log. co-secant 10.000000
Isto the base FC . . 50r Log. = . . . 1.698970
So is the angle F A C =x 2^ points, Log. co-tang. = 10. 272043
I ....
1^ the perpendicular FA=s93. 54 Log. » . . . 1.971013
Perpendicular QD a 40.09
Difference =;«.,, 53* 45,^ which is equal tQ the part A L*
In the ri^ht angled triangle ALP, given the base LD = FG 125
minutes, and the perpendicular AL 53.45; to find the angle LAD:
therefore,
M the perpendicular A L =^ 53. 45 winytes Log. ar^ co(np.=:8. 272052
IstoMiwB== . . « . 90.0 Log. sine *=i 10.000000
Soiatbe baseLD = . . 125 minutes, Log. = . . 2.096910
To the angle LAD = . 66^50^54'/ Log. tangent 55^10,868962
Now^ ttpce CK is paraUel to AD, and Ca to AL, the angle aCK is
equal to the angle L AD ; but the angle L ADis found, by computation,
to b^ $Q?50'54'' ; wherefore the angle a C K is al^o equal to 66?50'.54^ :
to thij^ let the angle a C S ==^ the angle N C B Oi point, or 5?37 '30r, be
added; and the sum 72^28" 24'/ = the angle S C K is the apparent course
of the ship between the south and the east, vLz.^ S. 72'^28'24^'E, or
E. b. S. i S. nearly.
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556 Mj^SURATION OF HBfGHTS AND DISTANCES.
We have now to determine the measure of the base C I^ in the right
angled triangle C I H ; to do which, we must first find the value of the
hypothenuse A C in the right angled triangle AFC, and that of the base
CH in the right angled triangle A H C. Thus,
To find the Hypothenuse AC: —
As radius = .... 90? Log. co-secant =10. 000000
Is to the base FC= . . SOT Log. = . . . 1.698970
So is the angle FAC = 2? points. Log. co-secant= 10. 326613
To the hypothenuse A C = 106. 07 Log- =s • • • 2. 025583
Tofind the Base CH:—
As radius ^ .... 90? Log. co-secant= 10. 000000
Is to the hypothenuse AC- 106. 07 Log.= • • . 2.025583
So isLAD-FAC=CAH=38?43^24rLog.sine = 9.796269
To the base CH= . . 66.35 Log. = . . . 1.821852
Now, in the right angled triangle C I H, given the hypothenuse C H =
66. 35 minutes, and the angle CHI; to find the base C I. The measure
of the angle C H I is thus determined. In all quadrilateral or four-sided
figures, the sum of the four angles is equal to four right aisles, or 360
degrees. Now, in the quadrilateral figure A H I F, since three of the
angles are given, the remaining or obtuse angle A H I is known by sub-
tracting the sum of the given angles from 360 degrees : thus, the angle
HIF90? + I FA 90? + FAH 66?50^54r = 246?50:54r5 and 360?
- 246?50^54r = 113?9^6r, is the measure of the angle A HI; from
which take away the right angle AHC 90?, and the remainder =
23?9^6T is the absolute measure of the angle CHI. Hence CI may be
readily found ; as thus :-*
As radius = 90? Log. co-8ecant=l 0.000000
Is to the hypothenuse C H s= 66. 35 minutes, Log. = . . . 1 . 821852
So is the angle C H I = . 23?9'.6r Log. sine = . 9.594572
To the interval or base C 1= 26. 09 minutes. Log. rs . . . 1. 416424
Time of second bearing = 2M0T.0
Sum= 2*36r.09; which is the time of the ship's
nearest approach to the observer.
Note. — ^This interesting problem is thus worked at length, trigono-
metrically, with the view of adapting it to the use of mariners in
general j though, indeed, in such cases, calculation need not be resorted
to, since the solution deduced from geometrical construction will always
be sufiiciently near the truth.
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PRACTICAI, GUMNBRT. 557,
SOLUTION OP PROBLEMS IN PRACTICAL GUNNERY.
Gonnery is the art of projecting balls and shells from great guns and
mortars ; of finding the ranges and times of flight of shot and shells ; and
of determining the different degrees of elevation at which those bodies
should be projected, so as to produce the greatest possible effect.
Problbm L
Qiven the Diameter of an iron Ball; to find its Weight.
RULS.
The diameter of an iron ball of 9 lbs. weight is 4 inches, very nearly;
and, since the weights of spherical bodies, composed of the same
materials, are as the cubes of their diameters, (Euclid, Book XII.,
Prop. 18,) it wil^ be, — as the cube of 4, is to 9 lbs. ; so is the cube of the
diameter of any other iron ball, to its weight. Hence the following rule : —
To thrice the logarithm of the diameter of the given ball, add the con-
stant logarithm 9. 148063 ; and the sum (abating 10 in the index,) will be
the logarithm of the required weight in lbs.
Example 1.
Required the weight of an iron ball, the diameter of which is 6. 7 inches ?
Given diameter = 6. 7 5 thrice its log. = 2. 478225
Constant log. = 9.148063
Weight in pounds = 42. 295 Log. = . 1. 626288
X Example 2.
Required the weight of an iron ball, the diameter of which is 5. 54
inches ?
Given diameter =: 5. 54 ; thrice its log. ==: 2. 230530
Constant log. = 9.148063
Weight in pounds = 28.91 Log. = . 1.378593
Note. — ^The constant logarithm used in this problem is expressed by the
arithmetical complement of the logarithm of the cube of 4, added to the
logarithm of 9.
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558 PllACtlCAL fitNNERY,
Problbm II.
Givefi the Weight of an iron Ball; to find its Diameter.
This problem being the converse of the last^ we obtain the following
logarithmieal expression : —
To the logarithm of the weight of the given ball^ add the constant
logarithm 0. 851937 ; divide the sum by 3^ and the quotient 1^11 be the
logarithm of the required diameter.
Note. — The constant logarithm given in this rule is expressed by the
arithmetical complement of the logarithm of 9, added to the logarithm of
the cube of 4.
Example 1.
Required the diameter of a 42 lb* iron baU ?
Given weight sis 42 lb. Log. = . • • 1 . 623249
Constant log. = 0.85 1 937
Divide by 3) 2.475186
Diameter in inches ^ 6. 685 Log. ^ 0. 825062
Example 2.
Required the diameter of a 24 lb. iron ball ?
Given weight = 24 lb. Log. = •.,!. 3802 1 1
Constant log. = 0.851937
Divide by 3) 2. 232148
Diameter in inches = 5.547 Log. =: 0.744049^
Problbm III.
Given the Diameter of a leaden Ball; to find Ue Wdghi.
RULB.
A leaden ball of 1 inch in diameter^ weighs -r\ of a lb.; wfaiclt^ reduced
to a decimal fraction, is . 2143, very nearly : and, as the weights of spherical
bodies are as the cubes of their diameters, it will be,— as the cube of 1, is
to . 2143 ; so is the cube of the diameter of any other leaden ball, io its
weight in lbs. Whence the following logarithmieal rule :*-
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BALLS AND SHSLLS. 559
To thrice the logarithm of the diameter of the given leaden ball, add
the constant logarithm 9. 331022 1 and the sum (abating 10 in the index^}
will be the logarithm of the required weight.
Example I.
Required the weight of a leaden ball> the diameter of which is 6, 68
inches ?
Given diameter := 6. 68 ; thrice its log. = 2. 474331
Constant log. = 9.331022
Weight in pounds = 63.88 Log. = 1.805353
Example 2.
Required the weight of a leaden ball^ the diameter of which is 5, 32
inches ?
Given diameter = 5.32; thrice its log. = 2. 177736
Constant log. = 9.331022
Weight in pounds b 82.26 Log« a 1.508758
JNb/e.— The constant logarithm used in this problem is the logarithm
of the decimal fraction • 2143.
Problbm IV.
Qioen the Weight of a leaden BaU; to find its Diameter.
RtLB.
Since this problem is merely the converse of the last, we obtain the
following logarithmical expression; viz., to the logarithm of the weight of
the given leaden ball, add the constant logarithm 0. 668978 ; divide the
sum by 3, and the quotient will be the logarithm of the required diameter.
Example 1.
Required the diameter of a 64 lb. leaden ball ?
Given weight = 64 lb. ... Log. j= 1. 806180
Constant log. = 0.668978
Divide by 3) 2. 475 158
Diameter in inches a 6*68 Log* ^ 0.82505^1
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500 PRACTICAL GUMNBRT.
Example 2.
Required the diameter of a 32 lb. leaden ball ?
Given weights 32 lb. . . . Log. = 1.505150
Constant log. = 0.668978
Divide by 3) 2. 174128
Diameter in inches = 5.305 Log. = 0.724709^
I
Note. — ^The constant logarithm made use of in this rule is the arith-
metical complement of that used in the preceding rule.
Probi^m V.
Given the intemal and estemal Diameters of an iron Shell; to find
Us Weight.
RULB.
Find the difference of the cubes of the intemal and external diameters
of the shell, to the logarithm of which add the constant logarithm
9. 148063 ) and the sum (abating 10 in the index,) will be the logarithm
of the required weight in pounds.
Note. — ^The constant logarithm used in this rule is the same as that
given in Problem I., page 557.
Example 1.
Let the external diameter of an iron shell be 12, 8 inches, and its
internal diameter 9. 1 inches ; required its weight ?
12. 8 X 12. 8 X 12. 8=2097. 152, cube of the external diameter.
9. 1 X 9. 1 X 9. 1= 753. 571, cube of the internal diameter.
Differences . , 1343.581 Log.= 3.128264
Constant log. = , 9. 148063
Weight in pounds ^ 188.94 Log.s 2.276327
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BALLS AND 8HBLL8. 561
Example 2.
Let the external diameter of an iron shell be 9. 8 inches^ and its internal
diameter 7 inches ; required its weight ?
9. 8 X 9. 8 X 9. 9=?94L 192^ cube of the external diameter.
7 X 7 X 7 =343. cube of the inteitial diameter.
Difference =.. 698.192 Log. = 2.776841
Constant log. = 9. 148068
Weight in poundss84. 12 Log. = 1.924904
Problem VL
To find how tmich Powder will fill a Shells
RULB.
To thrice the logarithm of the internal diameter of the shelly in inches,
add the constant logarithm 8.241845; and the sum (abating 10 in the
indeX;) will be the logarithm of the pounds of powder.
ExcmvpU 1.
How much powder will fill a shell, the internal diameter of which is 9. 1
inches 7
Internal diameter, 9. 1 inches ; thrice its log. = . • • .2. 677123
Constant log. = 8.241845
Powder, in pounds, s 13, 15 Log. = ••••'•• 1. 118968
Example 2.
How much powder will fill a shell, the internal diameter of which is 7
inches ?
Internal diameter, 7 inches ; thrice its log. = 2.535294
Constant log. a= 8.241845,
Powder, in pounds, =s 5.986 Log. = ..... .0.777139
Note.— The constant logarithm made use of in this problem is the arith-
metical complement of the logarithm of 57.3, the established divisor for
filling shells.
2 o
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5fl2 PRACTICAL OIJMIBRY.
PaOBtBM VII.
To find iht Size of a SheU to contain a given Weight of Powder.
RULB.
Thi8 problem being the convene of the last, we obtain the following
logarithmical eaipression : —
To the logarithm of the given weight of powder, in pounds, add the
constant logarithm 1.758155; divide the' sum by 3, and the quotient wiU
be the logarithm of the internal diameter of the shell, in inches.
Example 1.
Required the internal diameter of a shell that will hold 13. 15 lbs. of
powder ?
Given weight = 13. 15 lbs. Log. =s . . 1 . U8926
Constant log. = 1.758155
Divide by 3) 2* 877081
Internal diameter, in inches, = 9. 1 Log. = 0. 959027
Example 2.
Required the internal diameter of a shell that will hold 5. 986 lbs. of
p6wder?
Given weight = 5. 986 lbs. Log. = . . 0.777137
Constant log. =a 1.758155
Divide by 3) 2. 535292
Internal diameter, in inches, = 7. 0 Log. =0. 845097i
Problbm VIII.
To find how much Powder will fill a rectangular Bos.
Ruus*
To the logarithms of the length, breadth, and depth of the box, in
inches, add the constant logarithm 8.52287.93 and the sum (abatiqg 10
in the index,) will be the logarithm of the pounds of powder.
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POWDBR AND BOXUf. 56S
Example 1.
How much powder will fill a box, the length of which is 15, the breadth
1 2, and the depth 10 inches ?
Length s= 15 inches. Log. es ... 1. 176091
Breadth ^ 12 do. Log. b= . . , 1.079181
Depth = 10 do. Log. = . . • 1.000000
Constant log. s= 8.522879
Powder, in pounds, = 60. 0 Log. =:•:!. 77815 1
Example 2.
Required the quantity of powder that will fill a cubical box, the side of
which is 12 inches?
Side of the cubical box =12 inches. Log. ss 1. 079181
• Multiply by 3
Logarithmicalcubeof 12= • . , . . . 3.237543
Constant log. = . .• . 8. 522879
Powder, in pounds, « 57. 6 Log. as « . 1, 760422
No^e.— The constant logarithm made use of in this problem is the arith-
metical complement of the logarithm of 30, the established divisor for
filling, rectangular powder boxes. *
Problem IX.
To find the Size of^ ciMcal Box to contain a gio^ Weight of Powder.
RULB.
To the logarithm of the given weight of powder, in pounds, add the
constant logarithm 1.477121 ; divide the sum by 3, and the quotient will
be the logarithm of the side of the box, in inches.
Example 1.
Required the side pf a cubical box that will hold 60 lbs. of gunpowder ?
Given weight s 60 lbs. Log. s^ • • » • 1.778151
Connant log. = 1.477121
Divide by 3) 3. 255272
Side of the box, in inches, 12. 16 Log. = 1. 085090f
2o2
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564 PRACtlCAI. GUNNBRY.
Example 2.
Required the side of a cubical box that will hold 120 lbs. of gun-
powder ?
Givenweight=5 1201b8, Log, = . • . 2.079181
Constant log. =5 . 1.477121
Divide by 3) 3. 556302
Side of the box^ in inches, 15.32 Log. = . 1.1 85434
Note, — Since this problem is the converse of the last, the constant loga-
rithm made use of is the logarithm of 30, the established divisor and
multiplier for filling rectangular boxes.
Problem X.
. Tojind how much Powder wiUJlU a Q/linder»
RuL^.
To twice the logarithm of the diameter of the cylinder, add the logarithm
of its length and the constant logarithm 8.417937; the sum (abating 10
in the index,) will be the logarithm of the pounds of powder.
Example 1.'
How much powder will a cylinder hold, the diameter of which is 13
inches, and the length 26 inches ?
Diameter of the cylinders 13 inches $ twice its log. = 2. 227886
Length of ditto = . . 26 ditto. Log. = . 1.414973
Constant log. = 8.417937
Powder, in pounds, =s • 115.02 Log. = . *2. 060796
Example 2.
How much powder will a cylinder hold, the diameter of which is 4
inches, and the length 12 inches ?
Diameter of the cylinder = 4 inches j twice its log. er 1. 204120
Length of ditto = . . 12 ditto. Log. .= . 1.079181
Constant log. = . 8.417937
Powder, in pounds, = 5. 026 Log. = . . . . 0.701288
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POIjVDBR AND CYLINDBES. 565
Note. — The constant logarithm made use of in this problem is the
arithmetical complement of the logarithm of 38. 2y the established divisor
for filling cylinders with gunpowder.
Problbm XI.
Tojind what Length of a CyUnder vM be filled with a given Weight
of Gunpowder.
Ruj-B.
To the arithmetical complement of twice the logarithm of the diameter
of the cylinder^ or caliber of the gun, add the logarithm of the ^vei) weight
of powder iii pounds, and tW constant logarithm 1.582063: the sum
(abating 10 in the index^) will be the logarithm of the length of the
cylinder, in inches.
Example 1.
What length of a 24-pounder gun, of 5. 66 inches caliber, will be filled
with 8 lbs. of gunpowder ? . .
Caliber of the gun ±s 5. 66 Ar. comp, of twice its log. = 8. 494368
Given weight of powder=8 lbs. Log. =s • 0.903090
Constant log. s •••... 1.582063
Length, in inches, = 9.539 Log. s , .... . . 0.979521
Example 2.
What length of a 42-pounder gun, of S. 23 inches caliber, will be filled
with 1 Of lbs. of gunpowder ?
Caliber of the gun = 6. 23 * Ar. comp. of twice its log. ss 8. 411024
Given weight of powder= 10. 666 Log. = ..... 1. 028002
Constant Jog. =: ......•».• 1.582063
Length, in inches, = 10.497 Log. = . • . ^ . 1.021089
Note, — ^This problem being the converse of the last, the constant loga-
rithm is the logarithm of 38. 2, the established divisor and multipiur for
filling cylinders with gunpowder.
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586 PRACTICAL OtTVNBRY*
Problbm XII.
Tojmd the Numbisr of Balk in a triangular Pile,
RUJLB«
To the logarithm of the number of balls in the bottom row, add the
logarithm of that number increased by 1, and also its logarithm increased
by 2y and the constant logarithn> 9. 221849 : the sum (rejecting 10 in the
index,) will be the logarithm of the required number of balls.
Example !•
Required the numWr of balls in a triangular pile, each side of its base
containing 30 balls ?
Balls in one side of the base = 30 Log. = 1.477121
Ditto, increased by 1, = • . 31 Log. = 1.491362
Ditto, increased by 2, = . . 32 Log. ^ 1.505150
Constant log. 33 9.221849
Number of balls =t . . . 4960 Log. = 3. 695482
Example 2.
Required the number of balls in a triangular pile, each side of its base
c6ftUuiiin|(20baU8} '
Balls in one side of the base = 20 Log. s L 301030
Ditto, increased by ), = • . 21 Log. = 1.322219
Dittoi inertesed by 2, e= . . 22 Log. ^ 1.34S42S
G>nstant log. = .......»»• 9.221849
Niimfaecofb«IIa» . . . 1540- Logi » S. 187521
^^•^^The constant logarithm employed in this problem is the aritli-
metical complement of the logarithm of 6, the established divisor for
triangular, square, and rectangular piles of shot
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To find the Number of Balk in a square Pikr
Rule.
To the logarithm of the number of balls in one side of the bottom row,
add the logarithm of that number increased by I, the logarithm of twice
the same number increased by 1, and the constant logarithm 9. 221849 :
the sum (abating 10 in the index,) will be the logarithm of the required
number of balls.
Example 1.
Required the num*ber of balls in a square pile, each side of its base
containing 30 balls ? .
Balls in one side of the base :z 30 Log. =: 1. 47f 121
Ditto, increased by 1, =^ 31 Log, z:: L 491362
Twice ditto, increased by 1, z: 61 Log, =. 1.785330
Constant log. = . • 9.221849
Number of balls = . . 9455 Log, ;=: 3. 975662
Example 2.
Required the number of balls in a square pile, each side of its base
containing 20 balls ?
Balls in one dde of the base = 20 Log. == 1.301030
Ditto, increased by 1, = . . 21 Log. z= 1.322219
Twice ditto, mcreased by 1, ss 41 Log. =: 1. 612784
Constant log. = . . . ... . . . . 9.221849
Mimberofbftll*-: • , • 2870 Log. =; i3.457S82
Probwm XIV,
To find the Number of Balls in a rectangular Pile.
RULB.
From three times the nnmber of balls contained in the loigtfa of the
bottom row, subtract the number of balk, less by. 1, contained in the
breadth of that row ; then, to the logarithm of the remainder, add the
logarithm of the number of balls eontained in the breadth of the bottom
row, the lognrtthm of l^at number increased by 1, and the constant lo|;a-
rithm 9. 221849*: the sum (rejecting 10 in the index,) will be th^e logarithiQ
of the required number of balls.
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568
PRACTICAL GUNMBRy.
Example 1.
Required the number of balls in a rectangular pile, which contains 46
balls in the base row of its longest side, and 15 balls in that .ofi cs shortest
side?
Balls in length 46 x 3= 138
Balls in breadth 15—1= 14
Remainder = . • . 124
Balls in breadth row =t 15
Ditto, increased by 1, = 16
Constant log. !=..«•.
Number of balls = . 4960
Log.
Log.
Log.
2. 093422
1. 176091
1.204120
.9.221849
Log. == 3.695482
Example 2.
Required the number of balls in a rectangular pile, which contains 59
balls in the base row of its longest side, and 20 balls in that of its shortest
side ?
Balls in length 59*x3=:177
Balls in breadth 20-1= 19
Remainder = . . . 158 Log. = 2. 198657
Balls in breadth row = 20 Log. = 1.301030
Ditto, increased by 1, = 21 Log. = 1.322219
Constant iog. = 9.221849 .
Number of balls = 11060 Log. = 4.048755
Problem XV.
To find the Number qf Balls in an incomplete triangular PUe.
Rule.
Find the number of balls in the whole pile, considered as complete, by
Problem XII., page 566 ; and find also, by the same problem, the number
of balls answering to the triangular pile, the side of whose base is repre-
sented by the number of shot in the side of the top course of the incom-
plete pile diminished by 1 ; then, the difference of the two reiults will be
the number of shot remaining in the pile.
Example.
. Required the number of shot in an incomplete triangular pile ; each side
of its bottom course containing 40 balls, and each side of its top course
conUuning 20 balls ?
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PJtlNG PF BALLS, 569
To find the Number of Balls in the complete Pile : —
Balls in one side of bottom course = 40 Log. = 1 . 602060
Ditto^ increased by 1^ ^ .*' . ., 41 Log. = 1.612784
Ditto, increased by 2, = .... 42 Log. = 1. 623249
Constant log. =± . . 9«^221849
. Number of balls for the whole pile ^ 11480 Log. = 4.059942
To find the Number of Balls deficient :-^
Balls in each side of top course=20- 1 = 19 Log.= 1 . ^78754
Diminished course, or 19, increased byl,=20 Log.zs ] . 301030
Ditto, increased by 2, = .... 22 Log.= 1.322219
Constant log. = . • 9.221849
Number of shot wanting = • . . 1330 Log.=3. 123852
Now, 11480— 1330=10150 is the number of shot in the incomplete pile.
Problem XVL
To find the Number of Balls ui an incomplete square Pile.
Rule.
Find the number of balls in the whole pile, considered as complete, by
Problem XIII., page 567 ; and find alsoj by the same problem, the number
of balls answering to the square pile, each side of whose base is represented
by the number of shot in each side of the top course of the incomplete pile
diminished by 1 } then, the difference of the two results will be the number
of shot remaining in the pile.
Example.
Required the number of shot in an incomplete square pile ; each side of
its bottom course containing 24 balls, and each side of its top course 8
balls?
To find the Number of Balls in the complete Pile : —
Balls in one side of the base = .24 Log. = 1. 38021 1
Ditto, increased by 1, = ... 25 Log. = 1.397940
Twice ditto, increased by 1, = . 49 Log. = 1.690196
Constant log. = «... 9.221849
Number of balls for the whole pile = 4900 Log. = 3. 690196
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570 PRACrtCAL OlfKKBRY.
To find the Number of Balk deficient :-*-
Balls in each side of top course = 8—1 =7 Log.=0. 845098
Diminished course, or 7^ increased by 1^=8 Log.=0. 903090
Twice ditto, increased by 1, = . . 15 Log.=l. 176091"
Constant log. = ............ 9.221849
Number of balls wanting = . .140 Log. = 2. 1 46 1 28
Now, 4900— 140 = 4760 is the number of shot in the incomplete pile.
PBOBtBM XVII.
To find the Nuvuber o/BqIIs in an tncomplete rectangular Pile.
RULB.
Find the number of balls in the whole pile, considered as complete^ by
Problem XIV., page 567 ; and find also, by the same problem, the number
of balls answering to the rectangular pile, whose ?ides are represented by
the respective sides of the top course of the incomplete pile, the number of
shot in each side being diminished by 1 ; then, the difference of the two
results will be the number of shot remaining in the pile.
Example,
Required the number of shot in an incomplete rectangular pile; the
length of its bottom course being 40 balls, its breadth 20, and the length
of its top course 29 balls, and its breadth 9 >
To find the Number of Balls in the complete Pile :~
Bottom course, 40 X 3 » 120
Breadth, 20 -- 1 = 19
Remainder s .... 101 Log.^ 2, 004321
Balls in breadth row = . 20 Log.= 1.301030
Ditto, increased by 1, :^ . 21 Log-=: 1.322219
Constant log. =: \ . . . 9.221849
Number of balk fiw whole piles7070 Log.s3» 8494 19
To find the Number of Balls deficient :—
Top row, 29-1 = 28 X 3 ='84
Breadth, 9-1 = 8 - 1 sb 7
Remainder = 77 Log.s: 1.886491
Balls in breadth row = . • 6 Log.= 0. 903090
Ditto, increased by 1, = . . 9 Log.= 0. 954243
Constant log. s= ».••«,,» 9.221849
. Number of balls wantmg s 924 . Log.ss 2. 965673
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VBLOC1TIB8 OF. SHOT AND SHELLS. 571
Now, 7070 — 924 = 6146 is the number of shot in the incomplete
pile. .
Note.^n triangular and square piles, the number of horizontal rows or
courses is always equal to the liumber of balls in one side of the bottom
row ; and, in rectangular piles, the number of horizontal rows is equal to
the number of balls in the breadth of the bottom row. In these piles, the
number of balls in the top row or edge is always one more than the differ-
ence between the number of balls contuned in the length and the breadth
of the bottom row.
PfiowsM XVIIL
To find the Felocity of any Shot or Shell,
RULB.
. From the logarithm of twice the weight of the charge of powder, in
pounds, subtract the logarithm of the weight of the shot : to half the
remainder add the constant logarithm 3. 204120, and the sum (rejecting
5 in the index,) will be the logarithm of the velocity in feet, or the
number of feet which the shot or shell passes over in a second.
Example 1.
With what velocity will a 24-poandB ball be projected by 8 lbs. of
powder?
Twice the charge s 16 lbs. Log. s 1. 2D4120 •
Weight of the shot = 24 lbs. ' Log. s 1.380211
Remainder = 9. 823909
.Half the remainder ^ . . ... ... 4.9119544
Constant log. = . . . . . . . . 3.204120
Velocity of shot, in feet,i: 1306 Log. f= 3. 1 16074|
Example 2.
With what velocity will a 13-inch shell, weighing 196 lbs.> be discharged
by 9 lbs. of powder ? '
Twice the char^ s 18 lbs. Log. s 1.255273 •
*Weightofthe shell =19(5 lbs. Log. = 2.292256 '
Remainders 8.963017
Half the remainder == ...... 4.481508^
Constant log. = 3.204120
Velocity of sheH, in feet, 485 Log. a 2.685626^
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572 PRACTICAL GUNNBAY.
Noie. — ^The constant logarithm made use use of in this problem is the
logarithm of 1600 feet, which is the velocity ifcquired by a 1 lb. ball, vrhen
fired with 8 ounces of powder.
Probj^m XDC
To find the terminal Velocity of a Shot or Shell; that », the greaiegt
Velocity it can acqukre in descending through the Air by its cum
Weight.
RULB.
For Balls.-^-To half the logarithm of the diameter of the ball, in inches,
add the constant logarithm 2. 244277 ; and the sum will be the logarithm
of the terminal velocity of tUe ball.
And, for Sliells. — ^To half the logarithm of the external diameter of the
shell, in inches,' add the constant logarithm 2. 168203 3 and the sum will
be the logarithm of the terminal velocity of the shell.
Example 1.
Required the terminal velocity of a 24 lbs. ball, its diameter being 5. 6
inches? ' . .
Diameter of the ball = 5.6 Log. s 0.748188.
Half the log. = 0.874094
Constant log. =^ «.......» 2. 244277
Terminal velocity r± . 415 Log. == 2.618871
Example 2.
Required the terminal velocity of a shell weighing 196 lbs., its external
diameter being 1 2^ 8 inches ?
Diameter of the shell =^ 12. 8 Log. c 1. 1Q7210
Half the log. =r ....:... 0. 553605
Constant log. = .2. 168203
Terminal velocity = 527 . . Log. = 2.721808
Note. — The constant logarithms made use of in this problem are the
respective logarithms of 175.5 and 147.3, the established multipliers for
shot and Shells. It is by this problem that the terminal velocities contained
in Tables A and B, following^ have been computed.
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VBLOCITIBS OF SHOT AND SDBLLS. 573
PilOBLBM XX.
To find the. Height from which a Body mustfaUf in vacuo^ in
. order to acquire a given Velocity.
. R(7LB.
Since the spaces descended by falling bodies are as the squares of the
velocities, and as a fall of 16tV f®«^ produces a velocity of 32} feet, —
therefore, as the square of 32 J feet, is to 16^^ feet; so is the square of
any other given velocity, to the altitude from which it must fall, to acquire
such velocity. Hence the following logarithmical expression : —
To twice the logarithm of the given velocity, in feet, add the constant
logarithm 8; 191564 ; and the sum, (abating 10 in the index^) will be the
logarithm of the required altitude, or height
Example 1.
From what height must a body fall, in order to acquire a velocity of
1340 feet per second?
Given velocity = 1340^ Iwice its log. = 6.254210
Constant log. = 8.191564
Altitude, or height, = ' 2791 1 Log.. = 4. 445774
Example 2.
From what height must a body fall, in order to acquire a velocity of 1670
feet per second?
Given velocity s: 1670; twice- its log. = 6.445434
Constant log. r= ........ 8. 191564
Altitude, or height, = 43352 Log. = 4. 636998
Note — It is by this problem that the altitudes in Tables A and B,
following, have been computed; but, since the fractional parts beyond J 6
and 32 were omitted, and the constant logarithm, in consequence thereof,
assumed at 8. 193820, the respective altitudes, in these Tables, are some-
thing beyond the truth.
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574
PRACTICAL 60NNBRT.
CONCISE TABLES
FOR DETERMINING THE GREATEST HORIZONTAL RANGE OF A SHOT OR SHELL, 1IFHE>'
PROJECTED IN THE AIR WITH A GIVEN YELOCITY;
TOG&TBER WITH THE ELEVATION OF THE PIECE TO PRODUCE THAT RANGE.
Table A. — For Great Guns.
Weight of
Shot.
Diameter, in
inches.
Terminal
Velocity.
Log^tbm.
Altitude.
Logarithm.
1.
1.94
244
7.612610.
930
2.^69483
2
2.45
275
7. 560667
. 1182
3.072618
3.
2.80
294
7.531653
1360
3. 133539
4
3.08
308
7.511449
1482
3. 170848
6
3.53
330
7.481486
1701
3.230704
9
4.04
353 .
7. 452225
1958 .
3.291613
• 12
4.45
370
7.431798-
2139
3.330211
18
5.09 <
396
7. 402305
2450
3.389166
24
5.60
415
7.381952
2691
3.429914
32
6.17.
436
7. .360514
2970
3,472756
36
6.41
444
7.352617
3080
3.48KMi
42
6.75
456
7, 3*41035
3249 •
3.511750
Table B. — For Mortars.
SizeofSlien.
in inches.
Weight of
Shells filled.
Diameter, in
inches.'
Terminal
Velocity.
Logarithm.
Altitude.
Logarithm.
f
10
13
9
18
47
•91§
201
4.53
5.72
7.90
9.84
12. 80
314
352
414
i62.
%2r
7.503070
7.453457
7.383000
7. 335358
7.278189
1541
19S6
2678
3335
4340
3. 187803 ,
3.286905
3.427811
3. 523096
3.637490
Table C^For Great Guns and Mortars.
Initial Velocity,
divided by
TerminalVclocity
Lpgarithm.
Elevation.
Range divided by
Altitude.
Legarithm.
0.6910
9. 839478
44<> 0'
0.3914
9.592621
0.9i45
. 9.975202
43.15
0.&850
0.767166
1. 1980
0. 078457
42.30
0. 7787
9. 891370
L4515
0.161817
41.45
0.9724
'9.987845
1.7050
0.231724
. 41. 0
1. 1661
0.066736
1. 9585
0. 291924
40.15
1.3590
0. 133475
2.2120
0.344785
39.30
1.5535
0.191311
"2.4655
0. 391905
38.45
. 1.7472
0.24234a
. 2.7190
0. 434409
38. 0
1.^09
0.288003
2.9725
0.473122
37.15
2. 1346
0,329317
3.2260
0.508664
36.30
2. 3283
0.367039
3.4795
0.541517
35.45
. 2.5220 '
0.401745
. 3.733a
0.572058
35. 0
2. 7157
0.433882
3. 9865
0. 600592
34.15
2.9094
0.463S03
4.2400
0. 627:^66
33.30
3.1031
0. 491796
4.4935
0.652585
32.45 .
^.2968
0. 518093
4. 7470
. 0.676419
32. 0
3.4905
0.542888
5.0000
0. 698970
31.15.
3.6842
0.566343
JVb^e.— These Tables are deduced from those given id the third voltime
of Dr. Huttoh's " Course of Mathematics.'*
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RANGES OF BAU.8 AND 8HBtI^. 575
Problem XXI.
Tojind the greatest Range of a Ball or Shell, and the Eleoafion of the
. Piece to produce that Range.
Rule.
Enter Table A or B^ and take out the logarithm of the tenninal velocity
answering to the given ball or shell, as the case may be^ and also the
logarithm of th6 corresponding altitude ; then.
To the logarithm of the velocity with which the ball or shell is projected,
add the logarithm of its terminal velocity ; and the sum (abating 10 in the
index,) will be the logarithm of the quotient of *the -initial velocity of the
ball or shell, divided by its terminal velocity. With this logarithm, enter
the second column' of Table C, and iji the adjoining or middle qolumn will
be .found the corresponding degree of elevation to produce the greatest
range ; abreast of which, in the last column of the same table, will be found
the logarithm of the range divided by the altitude. Now, to this logarithm
add th^ logarithm of the altitude taken from Table A or B,, as above'
directed; and the sum will be the logarithm of the greatest range.
Note^-^K great accuracy be required, proportional parts must be taken
for the excess of the given above ^e next less tabular numbers in Table C.
Example 1.
Let it be required to find the greatest range of a 24 lb. ball, when
discharged with a velocity of 1640* feet, and the elevation of the piece to
produce that range ? :
Log« of terminal velocity of a 24 IK ball/ Table A, ss 7. 381952
Given velocity of the balls: 1640 . . / Log.s:3. 214844 •
Ans>«reririg to which, in Table C, is 34« \5\ . Log.=0. 596796
«
Log. of corra^onding altitude. Table A, = • « .- 3. 429914
Abreast of 34? 15 C, in last column of Table C, stands 0. 463803
Greatest range, in feet^ = 7829 . . • . . Log.B3. 8937 17
Hence the |;reatest range of a 24 lb. ball, when, projected with a velocity
of 1640 feet, is 7829 feet^ which is nearly an English mile. and a half] and
the elevation to produce that range, is 34 ? 1 5 ^
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576 PRACTICAL GUNNSRT.
Example 2.
Let it be required to find the greatest range of a 13-inch shell, when
projected with a velocity of 2000 feet per second, and the eleTation to
produce that range 3 the diameter of the'sl^ell being 12.^0 inches ?
Log. of terminal veloitity of a 13^ inch shell. Table B,=s7. 278189
Given velocity of the shell = 200(r . . . Log.=3. 301030
Answering to wliich, in Table C, is 34^49: Log.=0. 579219
Log.. of corresponding altitude, Table B, = • . . 3. 637490
Corresponding to 34?49', in Table C, is . . • . .0. 441196
Greatest range, in feet, = 1 1986 .... Log.=4.t)78686
Hence the greatest range of a 13*inch shell, when projected with a
velocity of 2000 feet, is 1 1986 feet, which is 2i miles and 106 feet; and
the elevation to produce that range, is 34? 49'.
- Not^. — In this example, proportion is made for the excess of the given
above the next less numbers in Table C.
Problbm XXII.
Gioefi the Range at one Eletaiion; to find the Range at another
Elevation.
RaLB.' .
As the logarithmic sine of twice the first elevation, is to the logarithm
of its corresponding range $ so is the logarithmic sine of twice the other
elevation, to the logarithm of its correspondii^ range.
Example I., -
If a 13-i;ich shell be found to range 1 1986 feet, when discharged at an
elevation of 34?49% how far will it orange when the elevation is 45 degrees;
the charge of powder being the same at both elevations ?
As twice 34?49^ = 69?38r Log. co-secantn 10. 028036
Is to its ranges 11986 feet, Log. = . • . 4.078674
. Soi8twice45?0' = 90? 0' Log. sine = . . 10.000000
To the required range, in feet, = 12785 Lojj. = 4. 106710
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. RANGES OF SHBLtS. 577
Example 2.
If a shell be found to range .4760 feet; when discharged at an elevation
of 45 d^rees^ how far will it range when the ^elevation is 30^5' | the
charge of powder being the same at both elevations I
As twice 45?= 90? O:- Log. co-secant=s 10. 000000
Is to its range = .... . 4760 feet, Log. = . . 3.677607
So is twice.30?45; = .... 61?30^ Log. sine = . 9.943899
To the required range, in feet, = 4183 Log. s , * • 3. 621506
Problem XXIIL
Given the. Elevation for one Range; to find the Eletmtionfor another
Range. •
Rule. •
As the logarithm of the first range, is to the logarithmic sine of twice.its
corresponding elevation ; so is the logarithm. of any other given range, to
the logarithmic sine of an arch. Now, the half of this arch will be the
elevation required. "
Example 1.
If a shell be found to range 11986 feet. when projected at an elevation
of 34?49'., at wh^t elevation must it be discharged to strike an object at
the distance of 1378.5 feet, with the same .charge of powder ?
As the first range =: 11986 Log. ar.comp.=5. 921326
Is tA twice 34?49^ = 69?38C Log. sine = . 9.971964
So is the other range=12785 .Log. = . . 4,106710
Toanchrr • • . 90? Oi Log. sine a . 10. 000000
Half the arch =: • 45? 0^, the elevation required.
Example 2. '
If a shell be found to ranjge 4760 feet when discharged at an elevation
of 45?, at what elevation must it be projected to strike an object at the
distance of 4 1 83 feet, with the same charge of powder ?
2 p
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578 PRACnCAi; GUNNSR7*
As the first range = 4760 Log. ar. t:oinp.=6. 322393
Is to twice 45? =: 90? Log. sine s 10. 000000
So is the other Taiige=s4183 Log. sine s « . 3. 621488
To arch = • . 61?29:45r Log. tifte » • 9.943881
Half the arch = . 30?44^52^^9 the elevation required,
. Problem XXIV.
Qioen the Charge for one Range; to find the Charge for awAher Range.
RULK.
Since th^ ranges at the same elevation are nearly proportional to the
charges^ therefore — ^As the logarithm of the first range, is to the logarithm
of its corresponding charge ; so is. the logarithm of the other range^ to the
logarithm of the charge corresponding thereto.
Exaifnpte 1*
If, with a charge of 12 lbs. of powder, a shell range 5334 feet» what
charge will be sufficient to throw it 2667 feet ; the elevation being 45? in
both cases ? - .
As the iirst tange ss « « 5334 Log. ar. eoinp.sc6. 272947
Istoitscha^eto , « .* 12 lU Log. » . . 1.0791S1
So is the odier range ss . 2667 Log. a . . 3. 426023
To the required charge^ in lb8.=:6« 0 Log. =? « « 0. 77815 1
*
If, wtthacharge.of dibs, of powder, a shell range 4000 feet, what
charge will be sufficient to throw it 8000 feet ; the elevation being 45? in
both cases ? • . •
As the first range = • . 4000 Log. ar. comp.=6. 397940
Is to its charge. = • . .9 lbs. Log. = . . 0.954243
So is the other range s ,3000 Log.* , ,3*477121
To the required charge, in Ib8.ss6. 75 Log* » « t 0. 829304
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KAM6B8 OF SHBIXk: 579
Problbm XXV.
Gwen the Range for one charge; tojwd the Range far another Charge.
Ruis.
As the logarithm of the first charge^ is to the logarithm of its corre-
sponding ranges so is the logarithm of the other charge^ to the logarithm
of its coitesponding r^g^ ; the elevation being the same in both ease6«
Example K
If a shell be projected 5334 feet by a charge of 12 lbs, of powder^ at
what distance will it strike an object when . discharged with 6 lbs. of
powder;; the elevatiaH being the same in both cases ?
As the first charge =: ' • 12 lbs. Log. ar. comp.==8. 920819
Is to its range == • . . 5334 Log. = . . 3.727053
So is the other charge ?s 6 lbs. Log. as « • 0. .778 15 1
T^the required range, in ft.,= 2667 Log. = . • 3* 426.023.
Example 2,
If a shell be projected 4000 feet by a charge of 9 lbs. of powder^ at what
distance will it strike an object when discharged with 6i lbs. of powder;
the elevation being the same in both cases?
As the first charge = . • 9 lbs. Log.*at.corop.=9. 045757
Is to its range =•.••■ 4000 Log. s , . 8. 602060
So is the other charge s . 6. 75 Log. = . . 0. 829304
To the required range, in ft,, = 3000 Log.= • • 3.477121
—
• PaoBiBM XXVL
Given the Range and tlie Elevation; to find the Impetus. .
Rule.
As the logarithmic sine of twice the angle of elevation, is to the loganthm
of half its corresponding range ; so is radius, or the logarithmic sine of
90?, to the impetus^
2p2
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S8Q PKAtrriCAL flUNNBRT.
■
Example L
With Vfh9t impetus must a shell be discharged at an elevatidn of 34?49^^
to strike an object at the distance of 2986 feet ?
As twice 34^49^ = . 69?38^ Log. co-secants 10. 028036
Is to half the range =s 1493 Log.s:. . .. 3.1740^
So is radius, or ... 90? Log.sine= . 10.000000
To the impetos, in feet, ss 1592 Log. s • . . 3. 202096
Example 2.
With what impetus must a shell be discharged at an elevation of 25?, to
strike an object at the distance of 2760 feet ?
As twice 25? S3 .... 50? Log.co-secants: 10. 115746
Is to half the ranges . . 1380 Log. s . . 3.139879
Sow radius, or .... 90? Lojj.sines .10.000000
To the impetus, in feet> s 1804 Log.s. . . 3.255625
Problem XXVII.
Given the Elevation and the Bange; to find the Time of the tVght.
.Rule.
As radius, is to the logarithmic tangent of the elevation ; so is the loga«
rithm of the rahge, in feet, to' a logarithmic number j which, bein^ divided
by 2, will give the logarithm of 4 times the number of seconds taken up in
the flight.
• • * '
Es^mple I.
In what time will a shell range 1 1986 feet, at an elevatbn of 34?49'. ?
Asrwtiuss . . . . 90? OC L6g. co^secants 10* 000000
Is to the elevation = . 34?49f Log. Ungent = 9. 842266
So is the ranges . . 11986 Log.=: . . . 4.078674
Divide by 2) 3. 920940
■■■■I '■■ 1 1 '■
Four times the flight s= 91.30 Log. s . . ; 1.960470
Number of seconds a • 22. 825,athe time of flight.
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JLxample 2.
In what time will a afaell range 3250 feet, at an eievation of 32 degrees ?
Asradiwas . . • , . 90? Log. co-secantslO.OOOOOO •
Is to the elevation =# . . 32? Log. tangent = 9. 795789
So 18 the range = . . .3250 Log. » « , . 3.511883
Divide by 2 J 3. 307672
Four times the flight s , 45.06 Log. » . « • 1.653836
Number of seconds = « II. 26^=s the time of flight
Note. — ^From this it b manifest that when the elevation of the piece u
45?, half the logarithm of the range will be the logarithm of 4 times the
number of seconds taken up in the flight. .
Probusm XXVIII.
Given the Range and the Elevation : tojlnd the greateet Altitude qfthe
, SheU.
RVLBs
As radius, is to the logarithmic tangent of the elevation ; so is the loga-
rithm of one-fourth of the range, to the logarithm of the required altitude.
Example 1. .
If a shell range 11986 feet, when projected at an elevation of 34?49^ ;
required the greatest altitude which it acquires during its flight ?
As radius^ .« . . .90? 0' Log. co-sec.= I 0.000000
Is* to the elevation = 34 ?49 ^ Log. tangent = 9. 842266
So ia i ot the range = 2996. 5 Log. = 4.3, 476614
To the altitude, in feet,=20S4 Log. =:
3.318680
Example 2.
If a shell range 4760 feet, when projected at an elevation of 45?;,
required the. greatest attitude which it aqquires during its flight? .
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582 mucrtCAL otrnvBRT.
As radius = . . 90? 0'. Log. co-sccantnl 0.000000
Is to the elevatioVi = 45? 0^ Log. tangent = 10. 000000
So is i of the ranges 1190 Log.= . . . 3.075547
Tothe altitude, in fcet,= 1190 Log. :^ , . .• 3.075547
. Note^r^TOOi this it is manifest, that when the elevation of the mortar
is 45 degreesj one-fourth of the range will .be equal to the greatest altitude
at which the ahell can amve.
Probum XXIX.
Gwenfhe Inclbuiiion qf the Plofie, the Elevaiian of the Piece, and
the Impetus f to find the Range*
Rule.
. To twice the logarithmic secant of the inclination' of the plane, add the
logarithmic sine of the elevation of the piece above the plane, the loga-
rithmic co-sine of the elevation of the piece above the horizon, and the
logarithm of 4 times the impetus : the sum of these four logarithms
(rejecting 40 iii the index,) will be the logarithm of the required range.
tiow far will It shell range on a jplane which ascends 10? 15', and also
en another plane which. descends 10? 15^ ; the impetus being SOOO feet in
both cases, and the elevation *of the mortar 3 1 ?45 ' ?
. fSb&ilion. 31 ?45 C - 10? 15 '. =21 ?30^ the elevation of the piece above
the aflcending plane ;
and, 31?45^ +10? 15^ =42? 0', the elevation of the piece above
the descending planew
To find the Range on the ascending Plane : —
Inclination of tba plane = 10^15': Twice the log. secant=SO. 013974
Elevation above the plane f= 21. 30 Log. sine = . . * . 9. 564075
Elevation above the hofizon=z3 1.45 Log. co-sine = . . 9.929599
tour times the impetus = 8000 Log. = , ... . , 3«. 903090
Range, in feet, =s • • • 8575 Ug. sb ^ « , ^ , 3.410788
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RANGB8 ^VD iimrros or srblls. 583
To find the Range on the descending Plane :~
Inclioailion of the plane := 10?I5C Twice the log. flecant=20. 018974
Bleyation above the planes 42. 0 Log.aines . • • 9.82S511
Elevalion above the horizoa=S 1. 45 Log. co-sine =: . • 9. 929509
Four times the impetus ^ 8000 Log. =: . • . • • 3.90S090
Range, in feet, =: • . % 4701 Log« = . ^ .• .,> 8.672174
PROBLSM XXX,
Gwm the belinqfion qf the Piane^ the Elevation qfthe Pieci,.and the
. Range J to Jind the Impetus.
RULB.
To twice the logarithmic co-sine of the inclination of the plane, add the
logarithmic co^secant of the elevation of the piece above the plane, the
logarithlnic secant of the elevation of the piece above the horizon, and the
logarithn^ of the one-fourth of the range : the sum of these four logarithms
(abating 40 in the index,) will be the logarithm, of the impetus*-
EaAmple.
With what impetus mttsi a shell be dischatgsd to strike an object at the
distance .of 2575 feet, on an inclined plan^ which ascends 10^15^, and,
also, another olgeet at the distance of 4701 feet, on an inclined plane
which descends 10? 15^ ; the elevation of the piece being. 31?45' in both
cases ?
Solution. 31?45!-.10?15;=21?30i; is the elevation of the piece,
abbver the ascending plane;
and, Sl?45' + 10?15^r:42? 0', is the elevation of the piece
above the' descending ]dane»
To find the Impetus on the ascending Plane: —
Inclination of the plane = 10? 15! Twice thelog.co-sine= 19. 986026
Elevation above the plane =r 21 . 80^ Log; cor-secant :i , 10. 435925
Elevation above the horizohzrd 1 . 45 Log. secant =: . . 10. 070401
One-fourth of the range = 043: 75 Log. = •,.,. 2. 808717
Impetus, in feet, c:* i» • % 8000 Log, c * / ; . . 3.301069
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584 PRACTICAL GUKMBAy.
To find the Impetus on the descending Plane :—
Inclination of the plane = 10?15< Twice the log. co-siner: 1 9.966026
Elevation above the.plane = 42. 0 Log. co-secant =i , 10. 174489
Elevation above the horizon=3 1 . 45 Log. secant = • . 10. 070401
One-fourth of the range = 1 1 75 . 25 Log. = .... 3. 070 130
Impetus, in feet^ =. . • 2000 Log. = . . , . . . 3.301046
Problem XXXI.
Gix^m the Weight of a Ballj the Charge of Powder twlfc which it itjtfed,
and thekfioum Velocity of that Ball; to find the FelocUy ofaSheli,
when projected' with a gioen Charge qf Powder.
RlJLB.
To the arithmetical complement of half the logarithm of the weight of
the ^hell) add half the logarithm of twice the weight of the charge, in
pounds, and the constant logarithm 3. 204120 : the sum (abating 10 in the
ind^x,} will be the velocity of the shell answering to the given charge.
JExample,
If a ball of 1 lb. \^eight acquire a velocity of 1 600 feet per second, when
fired with 8 ounces of powder, it is required to find with what velocities
the several kinds of shells will be projected by the respective '*chaige8 of
powder expressed against them in the following table ?
For the Ifl-inch Shell :—
Weight of the shell = . . .196 Ar. comp. of i its log.=8. 853872
Twice the weight of the charge =: 18 Half its log. = ... 0. 627636^
Constant log. = • .' . . . . . . . . .' . . . 3.204120
Velocity, in feet, =; . .. . . 485 Log. =: • • • • 2.685628|
For the 10-inch Shell :—
Weight of the shell . . . .90 Ar. comp. of | its log.s: 9. 022879
Twice the weight of the charge =: 8 Half its lo^. = • . . 0. 45 1545
Constant log. = •.....,. 3,204120
Velocity, in feet, = • • • • 477 Log,c? ...•...• . 2,678544
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IMPBTUS^ VELOGITY^ AND* CHARGSB OF POWDER.
585
For the 8-inch. Shell 2—
Weig;ht of the shell = • . . . 48 Ar. confj). of 4 its log.sQ. 159380
Twice Ifce weight of the charge = 4 Half its log. = . .0. 301030
Constant 1(^. = 3.204120
Velocity, in feetj = . ... 462 Log. s . . . . 2.664530
Nofe.— The. same results will be obtained* by computing agreeably to
th^ rule in Problem XVIIL, page 571.
Table D^Showhig the Felocities^ofthe diffjerenl sized SheUs, when
projected with given Charges qf Powder i
SizeofSbeU,
in inchM.
WeiphtofShclI,
in pounds.
Charge of
Powder,
in lbs.
Loga/itbpa.
V€iIocity, .
in feet.
1
Logarithm.
13
10
8
196
90
48
16
8
- 9
4
2
1
0.477124*
0.301030
0. 150515'
0. 000000
0.^9485
485
477
462
566
566
7.314258t
7.321482
7.335368 .
7.247184
7.247184
Problem XXXII.
Given tlie JSI^a<ian and the Range ; to find the Impetus^ Velodty, and
Charge of Pjowder. .
Rule.
Find the impetus^ by Problem XXVI., page 579 ; to the logarithm of
which add the constant logarithm 1. 206376 ;[ : take half the sum, and it
will be the logarithm of a natural number ; which, being doubled, will be
the required velocity. Now, to the logarithm of the velocity, thus found,
add the logarithms from Table D answering to the charge and the velocity
of the givcfn shell: the sum of these three logarithqis (abating 10 in the
index,) being doubled, will give the logarithm of the required charge of
powder, in pounds. •
* The numbers in tim column are the logarithms of .the square voots of the respective
charges. . •
t The numbers in this column are the aiithmetical complemeots of the logarithms of the
lespectiye velocities. .
t This is the logarithm of l^ feet, the descent of a. falling body in the first second of
time. . • . • -
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696 MAcncAL mjvtmBY^
Example*
With what impetiiSy felocity, and charge of pofwder^ miMt a IS^itich
shell be fired ataneleratioii of 34?49^^to strike an object at the distance
of 11986 feet?
. . To find the Impetiu and the Velocity t~ '
Twiee the elevatbn = . . , , 69?38: Log. co-secants 10. 028036
Half the range = ...... 5993 Log. =s . . •3.777644
Impetus, in feet; = ..... 6392 Log. = . . . 3.805680
Constant i0g. » ................ 1.206376
V ^ Divide by 4)5.012056
Nataral.namber s ...'•. 321 Log.= . • . 2.506028
Velocity, in feet, = :.... 642
To. find the Charge of Powder : — *
Velocity, in feet, = 642 Log. == . . . 2. 807535
Log. of charge for a 13-ineh shell, from Table D, = . . . 0.477121
Log. of velocity for a IS-inch shell, from Table D,^^ ... 7. 314258
Sum IS 0.598914
Charge, in pounds, <s , . . 19.77 Log. :s . , . K 197828
Hence the impetus is 6392 feet, the velocity 642 feet, and the charge of
powder, 15. 77 Ib'B.,' or 15 lbs. 12§ oz. nearly.
PROBLKM XXXIIL
Gioen the /ncKnatiofi of the Plane^ the Eleoaiwn of the Piece, and the
Range; tojbad the Charge t^ Powder.
RlTLB.
Find the impetus, by Problem XXX., page 583 ; with which proceed as
directed in the last problem.
Example 1.
How much powder will throw a 10-inch shell 6760 feet, on an inclined
plane which ascends 7?30C ; the elevation of the mortar being 33? 14! }
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above the ascending plane.
inclination of the plane = . 7?30' 'IVice the log. co-8irie= 19. 992538
Elevation above the pkne = 25. 44 Log. co-secant =» . 10. 362327
Elevatioti ab6ve the horizon = 33. 14 Log. secant = . . 10. 077562
One-fourth of the range = 1690 Log. = . . . . 3.227887
ImpetUA^ in fe«t^ » 4 . / 4574 Log. = . . . . 3.660314
Constant log. = ..•..•.... L?06376
Divide hy 2)4.86(5690
Naturd number. =
271
Log. s=
2.433345
Velocity = ... • . . 542 feet.
To find the Charge of Powder :—
Velocity = .542 feet, Log. - .... 2. 733999.
Log. of charge for 10-inch shell, from Table D, =r . . . . 0. 301030
Lo^. of velocity for 10-inch shelly from Table D, ^ . . . . 7. 3214t>2
, • Sum.= 0.356511
Charge, in pounds, =3 • . . 5.164 Log. = .... 0.713022
Hence the charge of j^owder is 5. 164 lbs., or 5 lbs. 2i oz.
Example 2.. *
How much powder will throw a 10-inch shell 6760 feet, cm an inclined
plane which descends 7?30% the deration of the morCar being 33^ 14'? -
SoluUon. 33^41 + 7^30C qs 40?44^ is the elevation of the mortar
above the descending plane.
Inclination of the plane = 7^30^ Twice the log. co-8ine=19. 992538
Elevation above the plane=2 40. 44 ' Log. co-secant = . 10. 185393
Eleyationabovethehorizon=33. 14 Log. secant =3 . • 10.077562
One-fourth of the ranges: 1690 Log. = •• 3.227887
Impetus, in feet, = . . 3044 Log. = .3.483380
Constant log. = . '. . . . . . . . ' 1.206376.
Natural number' =
Velocity, in feet, =
221 Log. =
Divide by 2) 4. 689756
.. . , . 2.344878
442
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$88 j^jurricAL GimMBRY.
To find the Charge of Powder :—
Velocity- 442 Log. = . • • . 2.645422
Log. of charge for lO-^inqh shell, from Table D, = . . .. . 0. 301030
Log. of velocity for 10-inch shelly from Table D^ =» . . . 7« 321482
/Sum = 0.267934
Charge^ in pounds, = . • . 3.434 Log. = • . .' ; 0.53586fi
Hence the ehal-ge is 3. 434 lbs., or 3 lbs. 7 oz. nearly*.
Problbm XXXIV,
Gtoeh the Inclinatum of the Plane, the Elevation of the Piece, and the
ImpetU9 ; to find the Time of Flight.
Rule. *
To the. logarithmic secant of the inclination of the plane a^d the idge^
rithmic sine of the elevation above the plane, and half the logarithm of the
impetus : the sum (abating 20 in the index,), will be the logarithm of twice
the time of flight, in seconds.
Exfmple. ♦
In what time wilt a 10-inch shell strike an object on an inclined plane
. which ascends 7?30^, when discharged with an impetus'of 4574 feet, the
elevation of the mortar being 33?14< ; and in what, time will it strike
^ another object on a descending plane, with the same impetus and elevatkm?
Solution. 33?14:-7?30C =25 M4^ is the elevation of the mortarabove
the ascending plane ;
and, 33?14f +7?301=:40?441 is the elevation of the mortarabove
the descending plane.
To find the Time of Flight on the ascending Plane :«<-
Inclination of the plane = . . , 8?30: Log. secants 10. 004797
Elevation above the plane = . . . 25.44 Log. sine = 9.637673
Impetus = . . . . ... . 4574 Half its log. = 1.830148
Twice the time of flight =5 . . . 29.69 Log. = . . 1.472618
Time pf flight = • • • • .' . 14. 845 seconds.
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HORIZONTAL EAN6BS OP.SHBLLS. 589
To find the Time of Flighl on the descending Plane : —
Inclination of the plane = .
Elevation above the plane s
Impetus ss • • •' • •
Twice the time of flight ='
8?30^ Log. secants 10. 004797
40.44 Log. sines 9:814607
4574 Half its log. ^ 1 . 830148
44.62 Log.= , '. 1.649552
Time of flights ••••.. 22.31 seconds.
— - — - ' . --- - ■ ■ —
problbm xxxy.
CTtoai the Impetus and the Elevation ; tojind the horizontal Range.
RULB.
To the logarithm of the impetus add the logarithmic si^e of twice the
angle of elevation : the sum (abating 20 in the index^) will be the logarithm
of a natural number ; which^ being doubled, will give the required range
on the horizqnial plane.
JExample 1. ' ' *
Let a shell be discharged with an impetus of 1592 feet, at an elevation
of 34?49V; required*its range on the horizontal plane ?
Impetus := • . . 1592 Log. = . 3.201943
Twice the elevations 69^38^ Log, sines 9. 97 1964
Natural number = . 1492.5 Log. 5= • 3.173907
Horizontal range = 2985 feet*
Example 2. . :
Let a shell be discharged with an impetus of 1804 feet, at an elevation
of 25? ; required its range on the horizontal plane } '
Impetus = . . -. 1804 Log. = . 3.25^237
Twice the elevation = 50? Log. sine s 9. 884254
Natural number s . 1382 Log. =r . 3. 140491
■
Ifiorizontal range ^ 2764 feet..
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590 PRACTICAT. GUNNBRT.
Problem XXXVL
Oioen the Impetus and the Eleeatim ; to find the lime ofFBght an
the horizontal Plane.
Rule.
With the impetus and the elevation compute the horizontid range, by
the last problem ; then> with the horizontal range, thus found, and th&
elevatioijk of the piece, compute the time of flight, by Problem XXVII^
page 580. Or, the time of flight maybe computed directly, by. Problem
XXXIV., page 58&
Example.
In what time will a 13-inch shell strike an object on a horizontal pUkie,
when discharged with an impetus of 6392- feet, the elevation of the mortar
being 34 ?49:?
Impetus = • ; • . 6392 Log. =• . 8.805637
IVice the elevation = 69938^ Log. sine = 9. 971964
. Natural number = • 5992.4 Log. = •3.777601
Horizontal range s . il984. 8 feet. Log. a 4, 078631
Elevation =s . , .34749^ Log. tangent a 9. 842266
Divide by 2) 3. 920897
Fourtimesthe.flight=: 91.30 Log.s . • 1.960448|
Flights .... 22. 825 seconds.
To find the Time of Flight, by Problem XXXIV., page 588 :—
Inclination of the plane = .... 0? 0' Log. secant= 0.000000
Elevation above the plani of the ftom.=34. 49. Log. sine = 9. 756600
Imf)etus= , , . •- • - • •.6392 Jiatfiulog.s 1.90281 84
Twice the time of flight =s ^ . ,45.64 Log. a • •1.659418|
Time .of flight = 22. 82 seconds j which agrees with
the time of flight found by the last rule, as.above*
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BUMU4 AND FUZBa« 591
. Pbobum XXXVII. .
. Given the Ttme of Flight of a Shell ; to find the Length of the Fuze.
RULB.
To the logarithm of the time of $ght add the constant logarithm
9; 342423, for 13 and 10-inch shells,-!-^ 9*380211/ for 8, 5^, and 41-
inch shells :* and the sum (abating. 10 ia the index^) will be the logarithm
of the^length of the fuze^ in inches;
. "•
Example L
Let the time of flight of a 13-iiich shell be 31. 75 seconds ; required the
length of the fiize ?
•
Time of flight, in seconds, =3 . 31. 75 Log. = 1. 501744
Constant log« e • • . • . ... • .... ^ 9.342423
Length of the fuze, fai inches, = 6. 985 Log. = 0. 84416? *
Example 2.
Let the time, of flight of an 8-inch shell b^ 21.. 5 seconds ; required the
length of the fiize ?
Time of flight, in seconds, 8 • 21. 5 Log. s 1. 3S2438 '
.Constant log. = ...•.*...,.' 9.380211
Length of the fiize, in inches^ ss 5. 16 Log.'s 0. 712649
Jfotej^The fuzes for a 13 and a 10-inch shell are so constructed as to
bum • 22 of an iiich in one second $ and those £br the smaller kind, viz., 8^
5|, and 4|-inch shells, • 24 of an inch in the same Apace of time. Now,
the logarithms of these two ikcimal iiumbers, viz., 9.342423 and
9.38021 1, are therefore the constant logarithms made use of in the above
rule. • . * *•
Fuzes are generally marked off, by circular lines, into seconds and frac-
tional parts of a second, so that no time n\ay be lost in measuring and
adapting them to tlie shells for which they are intended.
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592 MBN8URAT10N OF PLANBS.
SOLUTION OF PROBLEMS IN THE MENSURATION OP
PLANES, &c.
Problem I. *
Given the Base and perpendicular Height of a Plane Triangle; to find
its Area.
RUJLE.
To the logarithm of the base add the r^gariUim of half the perpendicidar
height, and the«iim will be the logarithm of the area, or superficial content
of the triangle.
Example.
Let the base of a plane triangle be 37. 6 yards,, fmd its perpendicular
height 29. 8 yards ; required its area, or superficial content ?
Base of the triangle :s .- .• . . . 37.6yard« Log, = 1.575.188
Eerpei^. height == 29. 8 yards -h 2 = 14.9 yards • Log. ^ 1. 173186
Area, or superficial content, s^ • . 560. 24 • . . . Log. == 2. 748374
Problbm IL
Giten two ^dee and the contained Anffle of a plane TViangle; to find
. if< Jrea* ...
RULB.
To the logarithmic sine of the contained angle, add the logarithms of the
containing sides : and the sum (abating 10 in the index,} will be the loga-
rithm of twice the area of t1ie triangle.*
Let the two given sides of a triangle be 109. 5 yards and-. 168. 2 yards
respectively, and the contained angle 79? 16^ ; required the area, or super-
ficial content of that triangle?
* Or, to' the logarithmic sine of the contained angle, add the logarithm of one of the
conUining sides and the logarithm of haif ilu other emitamitig^ tidt : the turn of these tfuee
logarithms (abating 10 in tiie index,) will be'the logarithm of the area of the triangle.
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MBNSURATION OF PLANB8. 593
Contained or included angle s . • . 79? 16^ Log. sine = 9. 992335
One of the containing sides 3 . . . 109. 5 Log. = . 2.039414
The other containing side = ... 168. 2 Log. s . 2. 225826
Twice the area of the triangle ss . 18095. 7 Log. s . 4.25757S
Area of the triangle s . • • • 9047. 85 yards^ as required.
Note. — ^The above problem will be found exceedingly useful in the
practice of land-surveying.
Problem IIL
Oivm the three Sides of a THar^le) to find its Area, or superficial
. . Content.
RULB.
Add' the three sides together, and take half their sum; subtract each
side severally from that half sum, noting the remainders : then.
To the logarithm of the half sum add the logarithms of the three
remainders; now, the sum of these four logagthms, being' divided by 2^
will give the area of the triangle. *
Example.
Let the three sides of a triangle be 433, 312, and 205 yards respectively;
required its area ?
First »de :s 433 First remainder = 42 Log. s 1.623249
Second side = ^12 Second remainder zct 163 Log. r= 2.212188
Third side = 205 Third remainder* = 270 Log. = 2.431364
Sum s . . 950
Half sum = A15 \og. ^ . 2.676694
Divide by 2) 8. 943495
Area of the triangles 29631.08 Log. = ..... 4.471747i
2a
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504 MBNsumAnoN oi planes.
Problem IV. ^
Given the Diameter of a Grcle } to find its Ckcumjerence, trnd
• conwrsely.^
Rule.
To the logarithm of the diameter add the constant logarithm 0. 497150,
and the mm will be the logarithm of the circumference. And^ to the
logarithm of the circumference add the constant logarithm 9. 502850^ and
the sum will be the logarithm of the diameter.
Esfamipl^ 1. . .
The earth's diameter is 7917* 5 miles ; required its circumference?
Diameter of the earth = .... 7917.5 miles Log. =. 3.898588
Constant log, = • • • ••»...«.•••.• 0.497150
Circumference of the earth = . . 24873.miles Log. = ' 4. 395738
Example 2.
If the circumference of the earth be 2500O.mned, what is its difuheter ?
Circumference of the earth s . . ., 25000 miles Log. =? 4. 397940
Constant log. :e . . ..« • ^ ..«•.... . 9.502850
Diameter of the earth, in miles, = . 7957. 7 Log. = 3. 900790
Note. — The diameter in the first example, viz., 7917. 5 miles^ appears to
be the true diameter of the earth, on the spherical hypothesis. The con-
stant logarithms used in thia problem will be found in tl|e TMe of Miscel-
laneous Numbers, at the end of the second volume.
Probi^m V.
Given the Dicmetery or the Circumference of the Earth; to find the whole
Area of its Surface.
Rule. ' \ ^'
To twice the logarithm of the earth's diameter add the constant loga«
rithm 0.497150, and the sum vrill be the logarithm of the area of the
earth's surface, in square miles. Or,
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MBNSUftATtON OF PtAKXS. 595
To twice the logarithm of the earth's circumference add the constant
logarithm 9. 502850, and the sum will he the logarithm of the earth's
surface, in square miles.
Example I.
Required the area or superficial measure, in square miles, of the whole
of the earth's surface, allowing its diameter to be 7917< 5 English miles ?
Diameter of the earth = 7917. 5 Twice its log. =? 7. 797176
Constant log. ^ . « « g 0. 497150
Area, in square miles, =: 196936545. 5 Log. = « .8. 294326
Example 2.
Required the area or superficial measure of the whole of the earth's
surface, in square miles, allowing its circumference to be 24873 English
miles?
Circumference of theearthat 24873 Twice itslpg.aS. 791476
Constant log. » .. ^ ...*«.. . 9.502850
Area, in square miles, s 196936545. 5 Logi s . 8. 294326
■lUM mm
Problem VI.
To Jind ib$ Length of any Arc qfaCbreU,
Rule.
To the logarithm of the degrees in the given arc, considered as a natural
number, add the . logarithm of the radius of that arc, and the constant
logarithm 8. 241878; the sum will be the logarithm of the length of the
arc.
ExomvplB,
Required the length of an arc of 45 degrees, the radius being 9 inches ?
Length of the arc, in degrees, s 45 Log. s 1. 653213-
Radius of the arc, in inches, = 5 Log. = 0. 954243
Constant log, = . .* • • 8.241878
Length of the arc, in inches, ^ ^ 7. 0686 Log. a 0. 849334
2q2
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596 PRACTICAL OAtJGINO.
SOLUTION OF PROBliEMS IN GAUGING.
Gauging is the art of finding the number of gallons^ &c^ contained in
any vesseL
By a recent Act of Parliament, therie i3 to be but one general standard
gallon throughout His Majesty's dominions of Great Britain and Ireland;
which gallon is to contain 10 lbs. (avoirdupois weight) of distilled water,
each pound of which is to weigh 7000 granis (troy weight) : hence the
new standard gallon is to contain 70000 grains (troy weight) of distilled
water. Now, since a cubic inch of distilled water weighs 252. 458 grains
(troy weight), the contents of the new standard gallon may be readily
reduced to cubic measure, by the following proportion; viz.. As 252.458
grains : 1 inch H 70000 grains : 277. 27384357 inches ; which, there-
fore, is the number of cubic inches in the new stiuidard gallcm. And
because the mieasure of the present ot old standard wine gallon is 231
cubic inches, and that of the old standard ale gallon 282 sudi inches^ we
have sufficient data for obtaining proper multipliers for the reduction of
the old standard wine and ^e measure into the new general standard
measure, and conversely. Hence,
277.27384357-1-231 = 1.200319671 is tiie general multiplier for reducing
Log.ssO. 079297 r the new standard measure into the
J old Handard mne tl^easure ; and,
231h-277. 27384357=^0. 8331 1 140 "j is tiie general multiplier for reducing
Log.=9. 920703 ^ the old standard wine measure into
J the new standard measure.
277. 27384357-^282=0. 98324058 ^ is the general multiplier for reducing
Log.=s 9.992660 y the new standard measure into the
J • old standard ale measure; and>
282H-277. 27384357=1. 01704508^ is Oie general multiplier for reducing
Log.ssO. 007340 I the old standard ale measure into
j the new standard measure.
Now, the respective multipliers and their corresponding logarithms being
thus obtained, the reduction of the old standard wine and ale measure into
the new general standard* measure, and conversely, may be very readily
performed^ by means of the foUpiving problems.
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PRACTICAL 6AU6IK6. 597
Problem L
To reduce the old standard fFine Measure into thenew Imperial Measure.
Rule.
To th^ logarithm of the old standard wine gallons add the constant
logarithni 9. 920703^ and the sum will be the logarithm qf the new standard
gallons.
Example 1.
Reduce 400 gallons of the old standard wine measure into the new
general standard measure.
Given number of gallons = 400 Log. = 2.602060
Constant loe. = . . . . , 9.920703
New standard gallons s 333.245 Log. = 2.522763
ExampUQ^
Reduce 9864 gallons of the old standard wine measure into the new
general standard measure-
Given number of gallons = 9864 Log. = 3.994053
Constant log. =5 .« , , 9.920703
New standard gallons = 8217.8 Log. s: 3.914756
Probi^m II.-
To reduce the new Imperial Measure into the old stamdard wine Measure^
Rule.
To the logarithm of the new standard gallons add the constant logarithm
0. 079297^ and the sum will be the logarithm of the* old standard wine
gallons.
Example 1,
Reduce 400 gallons of the new general standard measure into the old
standard wine measure.
Given number of new standard galloris = 400 Log. s 2. 602060
Constant log. = • 0^079297
p}d standard wine gallons = , , , 480.128 Log. = 2*681357
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Example 2.
Reduce 9S64 gallons of the new general standard measure into the old
fttf^ndar4 wipe m^fwire*
Given number of new standard gallons = 9864 Log. = 3. 994053
Constant log. = 0.079297
Old standard wine gallons «... 11889.95 Log. s 4.073850
Note. — ^From the above problems it appears that the new general
standard gallon is^ very nearly^ ovae-Jifih greater than the present or old
standard wine gallon.
- I- ■ — • '
Problbm liL
To reduc0 tk$ M itandard Ale Measure into the new Iwferial iileosure.
RULB,
To the logarithm of the old standard ale gallons add the constant loga-
rithm 0.007340^ and the buiq will be the logarithm of the new general
standard gallons.
Exaimiple 1.
Reduce. 400 gallons of the old standard ale measure into the new
general standard measure.
QtvQQ mmber of ale gallons = 400 Ifig. 9 2, 602060
Constant log. :;s . 0.007340
New standard gallons = 406.82 Log. = 2.609400
. JBsampk 3^
Reduce 9864 gallons of the old standard ale measure into the new
general standard measure.
Given number of ale gallons s; 9864 Log. = 3.994053
Constant log. =; ...*.,..... Q. 007340
New standard gallons s 10032. 13 Log. s. 4. 001398
PaolitBM IV.
To reduce the new Imperial Measure into the old strndair^ 4k Measw^
Rum.
To the logarithm of the new standard gallons add the eonstaiit logarithm
9.992660, and the sum will be the logarithm of the old standard ale
gallons.
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Reduce 400 gallons of the new general standard measure into the old
standard ale measure.
Given number of new standard gallons ^ 400 Log. = 2.602060
Constant log. aa « < « • « 9.902660
Old standard ale gallons = . . • 393.296 Log. = 2.594720
Example 2.
Reduce 9864 gallons of the new' general st^dard measure into the old
standard ale measure.
Given number of new standard gallons = 9864 Log. = 3. 994053
Constant log. := 9.992660
Old standard ale gallons ac . • . . 9698.7 hog. sa 3*986713
No<e.-»From the two laat problems it appears that the new general
standard gallon is^ very nearl]|r/ one-sisiieih less than the present or old
standard ale gallon.
Problem V. •
Cfiten the DiinensionB of a drcutar-headed Cash; tojtn^ its Content in
Ale and in fTine Gallons^ and alto agreeabhf tf) the new general
standard or Imperial GaUon.
RULK.
Divide tlie bead diameter by the bung difimeter, to two places of deci-
mals in the quotient; then,
Add together the logarithm for ale or wine gallons corresponding to
this quotient in the first part of Table LVIL, the logarithm of the bun^
diameter in the second part of that table, and the common Icq^arithm of
the length of the cask; iVe .sum (abating 10 in the index,) mil be the
logarithm of the content of the cask in ale or wine gallons. Now, to the
logarithm, . thus found, add the constant logarithm 0. 007340 for ale
gallons, or 9. 920703- for wine giallons ; and the sum will be the logarithm
of the true content of the cask in gallons^ agreeably to the new general
standard or imperial measure.
Example I,
Let the bung diameter of a cask be 25 inches, its head diameter 19. 5
inches, and length 31 inches; required its content in ale and wine gallons^
and also in gallons agreeably to the new general standard measure )
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600 PAACTICAL GAUGING.
19. 50 -I- 25 =s 0. 78j quotient of the head diameter divided by the bong
diameter.
First,— For Ale Gallons :—
Quotient = 0. 78 Log. for ale gallons = 7. 362671
Bung diameter = ... 25 inches Corresponding log. = . 2. 795880
Length of the cask = • • 31 inches Common log. = . .1. 491362
-
Content in ale gallons = . ; 44.66 Log. = .... .1.649913
Constant log. = 0.007340
Content in imperial gallons =: 45'. 42 Log. <= • . . . L 657253
Second, — ^For Wine Gallons :—
Quotient = 0. 78 Log. for wine gallons =. 7. 449340
Bung diameter = • . 25 inches Corresponding log. a? 2. 795880
Length of the cask == • . 31 inches Common log. = . . 1. 491362
Content in wine gallons =3 54.52 Log. = . • . . • 1.736582
Constant log. = .....*...' 9.920703
Content in imperial gallons=45. 42 Log. = ..... 1. 657285
See the example for illustrating the use of Table LVIL, page 153, and
also page 154.
NotoiP— In gauging a cask, it is to be remembered that the dimendons
of the bung diameter, the head diameter, and the length of the cask, be-all
takjsn within the cask. In measuring these dimensions, it must be care-
fully observed that the bung-hole be in the middle of the cask, and that
the bung-stave and the stave directly opposite thereto be' both regular and
even within the cask j also, that the heads of the cask be equal and truly
circular : if so, the distance between the inside of (he chimb, to the outside
of its opposite stave %vill be the head diameter intKn the cask, very nearly.
. Example 2.
Let the bung diameter of a cask be 31. 25 inches, its head diameter
23. 75 inches, and length 39 inches ; required its content in ale and wine
gallons, and also in gallons agreeably to the new general standard or
imperial measure?
23. 75 -«- 31. 25 rs 0. 76> quotient of the head diameter divided by the
bung diameter.
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#racti6al gauging. 601
•First,— For Ale Gallons :—
Quotient =..•..• 0. 76 Log. for ale gallons = 7. 355087
Bung diameter = . . 31. 25 inches Corresponding log. = 2. 989699
Length^of the cask =s: . • 39 inches Common log. = . 1. 591065
Content in ale gallons = . 86.268 Log. = . 4 . . 1.935851
Constant log. = . 0.0073*40
Content in imperial gallons s 87. .74 Log. = . • . . 1.943191
Secqnd. — ^For Wine Gallons :-^
ftuotient = . . . . '. . 0, 76 * Log. for wine gallons = 7. 441742
Bung diameter = . . 31. 25 inches Corresponding Tog. = 2.989699
Length of the cask = ' 39 inches Common log. = . • 1.591065
Content in wine gallons = 105. 32 Log. =s ..... 2. 022506
Constant log. = . 9. 920703
Content in imperial gallons^ 87. 74 Log. = ..... 1.943209
Remark,— The above problem will be found exceedingly useful to Pursers
in the Royal Navy, to Commissaries in the Army, aiid to other officers in
charge of gpvernment stores, who may have occasion, to purchase beer,
wine, or spirits, on His Majesty's account^ in foreign countries ; because it
enables them to ascertain, in a few minutes, the absolute number of gallons
contained in any given quantity of liquor, of the old measure, agreeably to
the newly-established standatd^ or imperial measure.
Problem VI.
Given the Content of a XJask lying in a horizontal Posiiiony its Bung^
Diameter, and the Depth of the Ullage or wet Inches; to find the
Quantity oflAqaor in the. Cask.
Rule.
Conceive the. bung diameter to be represented by urfity or 1 inch, and
that it be dividedinto 10000 equal parts ; then the half of this, viz.^ . 5000,
is to be considered as a constant decitnal.
Divide the wet inches, or depth of the. ullage, by the bung diameter, to
four places of decimals in the 'quotient ; find the difference between this
quotient and the constant decimal. Now,* one-fourth of this difference
being subtracted from the quotient, if the latter be less than th6 constant
decimal, or adde<i^ thereto if it'be more than that decimal, the difference or
sum will be the multiplier.
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602 PRACTICAL OAUGlNOi
Then, to the logarithm of the mukipliefi thus founds add the logarithm
of the content of the cask, in wine measure; and the sum Mall be the loga-
rithm of the ullage, or number of gallons of liquor in the cask, in wine
measure. And if to this logarithm the constant logarithm 9. 920703 be
added, the sum will be the logarithm of the ullage^ agreeably* to the
imperial measure.
Note. — If the content of the cask be given in ale measure, the constant
logarithm will be 0. 007340.
Example 1.
Let the buAg diameter of a cask be 31. 25 inches, its content in wine
measure 105.32 gallons, and the depth. of the ullage, 11.5. inches;
required the quantity of liquor in the cask ?
Depth of the ullage, or wet inches, 1 1 • 5 -f- 31 . 25
inches (B. D.) = . 3680 quotient, . 3680, which is less than the constant
Constant decimal = . 5000 decimal.
Difference = . . . 1320 -h 4 =5 . .330, subtractivc.
Multipliers ....... .3350 Log. »: 9.52504S
Content of cask, in wine measure,s£ 105. 32 gallons Log. » 2« 02250*6
Content of ullage, in wine gallons, s= 35 . 28 . Log. ^ 1 • 54755 1
Omstant log* « 9.020703
Content of ullage, in imperial galls.:= 29. 39 Log. = 1 . 468254
Note. — If the content of the cask be given in imperial measure, let the
logarithm thereof be added to the logarithm of the multiplier ; and die
sum will be the logarithm of the ullage..
Thus, in the above example, let the content of the cask be given agree-
ably to the new general standard or imperial measure J vis., 87. 74 gallons;
then.
Multiplier, aa abcAre, 3s: 3350 Lo|;. a 9.625045
Content of the cask, in ioipl. Bieas.8s87 • 74 galldnt Lqg. » 1 • 943209
Content of ullage, in imperial galls. = 29. 39 Log. = 1.468254
Example 2. .
Let the bung diameter of a cask be 25 Inches, its content hi vrine
measure 54. 52 gallons, and the depth of the ullage 15. f$ inches ; required
the quantity of liquor in the cask ?
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PRACTICAL GAUGING.
60»
Depth of the ullage, or wet inches, 15. 75 -»- 25
inches (B. D.) ^ . 6300 quotient, • 6300, which is more than the constant
Contaot decimal = . 5000 decimal.
Difference ==
. 1300 -4- 4 = . . 325, additive. .
Multiplier = ....... .6625 Log. = 9.821186
Content of the cask, lawine meas.=54. 52 gallons Log. = 1. 736582
Content of ullage, in wine gallons,»36. 12 Log. == 1 . 557768
• Constant log. = 9. 920703
Content of ullage, in imperial gall^. ss 30. 09 Log. = 1 . 47847 1
But if the. content of the cask be given agreeably to the imperial standard
measure, yiz,^ 45.42 gallons, then the latter part o/the operation will-be
as thus :-r-
Muitiplier, as above, = , . . . . 6625 Log. = 9. 821 186
Content of the cask, in impl. meas. = 4'5. 42 gallons Log. = 1 . 657285
Content of ullage^ in imperial galls. = 30. 09
Log. = 1.478471
Remark. — If the dry inches of the bung diameter be made use ot instead
of the wet, the result of the operation will express the vacuity in the cask ;
and if thb vacuity be ^ded to the ullage, the sum will be the contest of
the cask, which will be a proof that the work is right.
Thus^ in the- last example, where the bung diameter is 25 inches, and
the. depth of the ullage 15.75 bches, the difference of these is 9.25,
which,, therefore, is the number of dry inches.
Then, dry inches 9. 25 -h 25
inohes (B. D.). sx , 3700 quotient,^ • 3700, which is less than the constant
Constant decimals .5000 decimal..
Difference = . . . 1300 •»- 4 a . .325, subtracUve. .
MultipKe?= •....• . ,3375 Log. =; 9.528274
Content of the cask, in impl. meas.^ 45. 42 gallons Log. s 1 . 657285
Vacuity ia the cask r:
Content of the ullage 9
Content of the caskss
1*
15.33
30.09
Log. s 1. 185559
45. 42 ; wbich proves the work is right.
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604 PRACTICAL GAUGING.
Problbm VII. •
Owen the Content qfa Cask standing in a vertical or upright Position, its
Lengthy and the Depth ofth^ Ullage or wet Indies s to find the Quan-
tify of Liquor in tlie Cask.
RULB,
Conceive the lengthof the cask to be represented by unity or 1 inch^ and
that it be divided into lOOOO equal parts -, then the half of this^ viz«^ • 5000^
is to be considered as a constant decimal.
Divide the wet inches, or depth of the ullage, by the length of the cask,
to four places of decimals in the quotient; find the difference betwreen this
quotient and the constant decimal : now, one-tenth of this difference being
subtracted from the quotient, if the latter be less than the constant deeimal,
or added thereto if it be more than that decimal, the difference or sum will
be the multiplier.
Then, to the logarithm of the multiplier^ thus founds add the logarithm
of the content of the cask, in. wine measure; and the sum will be the
logarithm of the ullage, or number of gallons of liquor in the cask, in wine
measure. And if to this logarithm the constant logarithm 9. 920703 be
added, the sum will be the logarithm of the ullage agreeably to the imperial
standard measure.
Note. — If die content of the cask be given in ale measure, the constant
logarithm will h^ 0. 007340.
». • *
Example L
Let the length of a cask, between the heads, be 39 inches, its content in
^vine measure 105.32 gallons, and the depth of the ullage 16.5 inches;
required the quantity of liquor in the cask ? '
Depth of ullage, or wet inches, 16. 5 -*- 39
inches (length) = .4231 quotient^ .4231, which is less than the conatant
Constant decimal 3=.. £000 decimal*
Difference = - • . . 769 -«- 10 = . . 77, subtractive.
Multiplier = 4154 Log. = 9.618467
Content of the cask, in wine meas.= 105. 32 gallons Log. s 2. 022506
Content of ullage, in wine gallons,sr 43. 75 . Log. » 1. 640973
Constant bg. s 9.920703
Content of the ullage, in imperial galls.=36. 45 Log. s 1 • 56 1 676
f.
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PRACTICAL GAUQINO. 605
Nbt^. — If the content of the cask be given agreeably to the imperial
standard measure, let the logarithm thereof be added to the logarithm of
the multiplier; and the sum will be the logarithm of the ullage. Thus, in
the above example, let the content of the cask be given in imperial
treasure ; viz., 87* 74 gallons j then,
MulUpliers:: 4154 Log. s 9*. 618467
Content of the cask, in impl. meas. = 87. 74 gallons Log. s 1. 943209
Content ofullage, in imperial galls, s 36.45 Log. = 1.561676
Example 2.
Let the length of a cask, between the heads, be 31 inches, its content
in wine measure 54. 52 gallons, and the. depth jof th6 ullage 18. 5 inches;
required the quantity of liquor in the cask ?
Depdi of ullage, or wet inches, 18. 5 -h 31
inches (length) s .5968 quotient, . 5968, which is more than the constant
Constant decimal = . 5000 decimal.
Difference s • . . . 968 -f- 10 = . .97, additive.
Multiplier =£ ........ .6065 Log. = 9.782831
Content of the cask, in wine galls.^: 54. 52 Log. =: 1 « 736582
Content of ullage, in wine galls, a 33. 07 ' Log. = 1 . 5 1 9413
: Constant log. = 9. 920703
Content of ullage, in imperial galb.=27. 55 Lo^. = 1 . 4401 16
But if the content of the cask be given in imperial measure, viz., 45, 42
gallons,, then the latter part of the operation will be as thus : —
Multiplier, as above, = . 6065 Log. = 9.782831
Content of the cask, in imperial meas.=45. 42 gallons Ldg. — I. 657285
Content of ullage, in imperial galls. = 27. 55 Log. =2 1. 4401 16
Remarks — If the dry inches of the length of the cask be made use of
instead of the wet, the result of the operation will express the vacuity in
the cask ; and if this vacuity be added to the ullage, the sum will giv^ the
content of the cask : but this, it is presumed, does not need to be elucidated
by an example.
Note.— The following Table, which is particularly adapted to the reduc-
tion of the old-established wine and ale measure into the new general
standard or Imperial measure, and the contrary, will, be found of very con-
siderable use in the event oi* purchasing' wine or spirits in places out of
His Majesty's dominions.
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606
PftAOnCAL OA.06INO.
A Tablb
For readily finding the Number of Wine or Ale OdUon$ tohieh is adualif
equivalent to any given Number of OaUcm of the newly^e8taUi$hed
general standard or Imperial Measure, and conversely.
n
r
Wine Measure
Ale Measure.
H
Imperial Meas. .
4
Imperial Meas. 1
G.
0
Q
0
p.
0
GiUs.
Ss
G.
a
p.
GUU.
^s.
G.
2
GUU.
•^s
G.
fi
P.
Gills.
1.200
Iffil)
0
0
0.983
if^ii
0
0
.0.833
T
0
0
0
1.017
0
0
0
2.401
2<o.
0
0
1.966
2do.
0
0
1.666
0
0
0
2.034
Ido.
0
0
0
3.601
3do.
0
0
2. 950
ido.
0
0
2.499
Mo.
0
0
0
3.051
Lpt.
0
0
1
0.801
lpt.
0
0
3.933
lpt.
0
0
3.332
lpt.
0
0
.1
0.068
Iqt.
2qt8.
0
0
1
2
0
0
1.603
Iqt.
0
0
-
3.866
Iqt.
0
0
1
—
2.665
iqt.
0
1
0
0.136
3.20.5
2qts.
0
1
3.732
2qts.
0
1.330
2qts.
0
2
0
0.273
'Tf
0
3
1
0.808
3qts.
0
2
3.598
%
0
2
3.1595
)qt8.
0
3
0
0.409
1
0
1
2.410
G.1
0
3
3.464
0
3
2.660
G.1
1
0
0
0.645
2
2
1
1
0.820
2
1
3
2.927
2
1
2
1.319
2
2
0
0
1.091
3
4
3
4
2
3
0
3.231
3
2
3
2.391
3
2
1
3.979
3
3
0
0
1.636
0
1.641
4
3
3
1.855
4
3
1
2.638
4
4
0
0
2.182
5
6
tf
0
0.051
5
4
3
1.318
5
4
0
1.298
5
5
0
0
2.727
6
7
0
1
2.461
6
5
3
0.782
6
4
3
3.957
, 6
6
0
0
3.273
7
8
]
1
0.872
7
6
3
0.246
7
5
3
2.617
7
7
.0
0
3.818
8
Si
9
10
2
3
0
0
3.282
8
7
3
3
3.710
8
6
2
1
1.277
«
, 8
0
0
1
0.364
1.692
9
8
3.173
9
7
3.936
9
9
1
0.909
10
12
0
0
0.102
10
9
3
2.637
10
8
1
2.596
10
TO
0
1
1.454
20
24
0
0
0.205
20
19
2
1.274
20
16
2
1
1.191
20
20
1
0"
2.909
30
36
0
0
0.307
30
29
1
.3.911
30
24
3
3.787
30
30
2
0
0.363
40
48
0
0
0.409
-
40
50
39.
"49^
0
0
—
2.548
40
33
1
2.383
40
40
2
1
1.818
50
60
0
0
0.511
1.185
50
41
2
0.978
50
5U
3
0
3.272
60
72
0
0
0.614
60
58
3
3.822
60
49
3
3.574
60
61
0
0
0.727
70
84
0
0
0.716
70
68
3
2.459
70
58
1
2. 170
70
n
0
1
3.181
80
96
0
0
0.818
80
78
2
1.096
80
66
2
0.765
80
81
1
0
X635
SO
100
108
120
0
0
0
0
0.921
90
8»
98
1
1
-
3.733
90
74
3
1
3.361
90
91
2
0
1.090
1.023
100
2.370
100
83
1.956
100
roi
2
I
'2.544
200
240
0
0
2.046
200
196
2
0.740
.200
166
2
3.91.3
200
203
1
1
1.0B9
300
360
0
0
3.069
300
294
3
.3.110
300
249
3
1.869
300
305
0
0
3.633
400
480
0
1
0.092
400
393
1
1.479
400
33S
0
3.826
400
406
3
0
2,177
500
600
600
.720
0
7
1
1.U5
500
491
2
3.849
500
«00
416
499
2
3
1.782
500
m
2
0
0
1
0.721
1
2.138
€00
589
2^219
3.739
600
610
3.166
700
840
0
1
3.161
700
688
1
0.589
700
583
0
1.695
700
711
3
1
hSlO
800
960
1
0
.0. 184
800
786
2
2.959
80^
666
1
3.652
800
813
2
1*
0.354
900
1080
1
0
1.807
900
884
3
i.329
900
749
3
1.608
900
915
1
0
2.898
1000
1200
1
0
2*229
1000
983
0
3.699
1000
833
0
0
3.5^5
1000
1017
0
0
1.443
Note. — ^In using the above TaUe, if the given number of gaHon^ csnnot
be exactly found, or if it fall without the limits of the Table, tl»e sum ^
the different quantities corresponding to the several terms wbieh mtke wp
the given number of gallons is, in such eases, to be taken ; as in tbi
following jexamples : —
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BnaCBLLANBOlTfl PROBLEMS. 607
'Example L
In 1736 gallons^ imperial measure^ how many gallons of wine measure ?
G. Q. P. Gills.
1000 galls., impl. meas., are equal to 1200. 1. 0. 2. 229 W. M.
700 ditto ditto 840.0. 1.3. 161 ditto.
30 ditto ditto 36.0.0.0. 307 ditto.
6 ditto. ' ditto 7. 0. 1. 2. 461 ditto.
Hence, 1736 galls., impl. irieas., artf equal to 2083. 3. 0. 0. 158 W. M.
• Example 2.
In 1839 gallons, \nne measure, how many gallons imperial measure ?
G. 2. P. GiUi.
1 000 galls, wine meas.^ are equal to 833. 0. Q- 3. 565 impl. meas.
800 ditto clitto 666. 1. 1.3. 652 ditto.
30 ditto ditto 24. 3. 1. 3. 787 ditto.
9 ditto . ditto 7.1.1.3.936 ditto.
Hence, 1839 galls.^ wine meas'., are equal to 1532. 0. 0. 2. 940 imjil. meas.*
SOLUTION OF MISCELLANEOUS PROBLEMS.
Problem I.
Qioen the Grcumference of a Cablcy and its Length; to find its Weight,
Rule.
To twice the logarithm of the circumference of the given cable, add the
logarithm ^f its. length, and the constant logarithm 9. -734967: the sum of
these three logarithms (abating 10 in the index,) will be the weight of the
given cable, in pounds, avoirdupois.
. Example.
Let the circumference of a cable be 21 inches, and its length 110
fathoms ; required its weight ? .
Circumference or girt of given cable=21 inches Twice its log. =2. 6^4438
Length of ditto, in fathoms, =: • 110 Log. = . , 2.041393
Ckmstant log. = • . '. . . . .. . . '. . .... 9.734967
Weight of the given cable, in pounds,=:2635 1 , Log. =: . « 4. 420798
* A general VictuaUing Table is ^ven at the end of the gecood volume, which wiU be
found of considerable utility to the Pureers of the Naval service, in mal^ing^ out their annual
accoiuits, or in contpletiBg^ the pnyvidioDs of their ships to any given time.
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608 MISCBLIJINBOUS PROBLBM8*
Remark. — It has been found, by actual experimeit, that 1 fiitlioni of a
hemp cable which measuret 9 inches in circumference weighs 44 lbs. avoir-
dupois. Now, since cylinders of equal lengths are as the squares of their
circumferences, — therefore, as the square of 9 inches (the circumference
of the experimented cable), is to the weight of 1 fathom thereof, viz^
44 lbs. ; so is. the square of the circumference of any other cable* to the
weight of 1 fathom of such cable: which, multiplied by the length of the
cable, will give ite whole weight The constont' logarithm 9.734967 is
found by adding the arithmetical complement of twice the logarithm of 9
inches to the logarithm of 44 pounds.
Problbm II.
Given the Diameter of a Circle i to find Us Circun^er€nce*
Rule.
To the logarithm of the diameter of the given circle add the constant
logarithm 0. 497150, atid the sum will be the logarithm of the circumfer-
ence of that circle.
Example 1.
Let the diameter of a circle be 78. 41 yards ; required its circumference?
Diameter of the given circle = . • 78. 41 yards Log. = 1. 894372
Constant log. = . .0.497150
Circumference of the given circle, in yards, = 246. 33 Log. = 2. 391522
No/«.^— The circumference of a circle whose (Uameter is wiity or 1, is
3. 14159265 ; and, since the circumferences of circles, are to each other, as
their diameters,, or radii,— therefore, as the diameter 1, is to its circum-
ference 3. 14159265 ; so is the diameter of any otlier circle, to its circum-
ference : and hence the above rule. The constant logarithm is expressed
by the logarithm of 3. 1 4 1 59265.
Example 2.
If the diameter of the earth be 79 17* .5 miles, what is its circumference?
Diameter of the earth q= . . . . 7917.5 miles Log. « 3.898588
Constant log. =s . • . 0.497150
Circumference of the earth, in miles,=: 24873. 5 Log. = 4. 395738
The converse of this problem, viz., deduchig the diameter from the
circumference, is obvious.— See Problem* IV., page 594. *
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MISCELLANEOUS PROBLEMS, 01151
PROBJUkM in.
Gioen the Diamei^ of a Grcle ^ to find its Jrea, or wpeftftdal Omient
Hulk.
All circles are to one another, as the squares of their diameters; and as
die area of a circle whose diameter is unity or 1, is • 7853982, the loga-
rithm of which is 9. 895090, — therefore, to twice the logaridim of the
given circle, add the constant logarithm 9. 895090 ; and X\ie sum (abating
10 in the index,) will be the logarithm, of the area^ or superficial contept
of that ciTcIe.
Example. •
If the diameter of a circle be 78. 41 yrf^ds, what is its area or superficial
content?
Diameter 6( the given circle s± 78. 41 yards IVnce its log. == 3. 788744
Constant log. = .., ....•., 9.895090
Area of the g^ven circle, in yards, s 4828. 8 Log. = . • 3. 683834
Problem IV.
Given the Area or supeificial Content of a Circle : to find Us Diameter.
Rule.
As this problem is evidently the converse of the last,— therefore, to the
logarithm of the area of the given circle, add the constant logarithm
0. 104910 (the arithmetical complement of 9. 895090). : divide the sum by
2, and the quotient will be the logarithm of the diameter of the given
circle.
Example.
Let the area of a. circle be 48^8. 8 yards; required its diameter ?
Area or superficial content of given crrcle;=4828. 8 yards Log.=s3. 683834
Constant log. = . . ............ .0.104910
Divide by 2) 3. 788744
Diameter of the given circle, in yards> s 78. 4 1 .Log. =s • • 1 • 894372
2 R
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610 VISfSLLANBOUS PAOBUUCS*
PfU>BJLBM V.
Given the Diameler tffa Grcfe; to jmd I/m Siilfi qfa Square equal m
Area to thai .CSrcfe.
RULB.
To the logarithm of the diameter of the given cifble^ add the conrtant
logarithm 9. 947545 (the logarithm of the square root of . 7853982) ; and
the sum (abating 10 in the index,) will be the logarithm of the §ide of a
square equal in area or eaperficial content to that circle.
Example* "
If the diameter of a circle be 78. 41 yards^ what is the side of.a square
equal in area to that circle ? *
Diameter of the given circle =n • • • 78.41 yards Log. == 1.894372
Constant log. =a ...'.. ^ ........ . 9.947545
Side of the required square^ in yards^ = 69. 49 Log. = 1. 841917
Problem VI.
Given the Diameter ofaGrcle; to find the Side of a Square mecribed
in that CSrcle.
RviB.
To the Ic^arithm of the diameter of the. g^ven circle^ add the constant
logarithm 9; 849485 ; and the sum (abatiiig 10 in the index,) will be the
logarithm of the side of a square inscribed in that circle.
Example.'
If the diameter of a circle be 78. 41 yards, what is the side of a square
inscribed in that circle ?
Diameter of the ^ven circle = . . . 78*41 yards Log. = 1.894372
Constant. log. s ••....... ^ «... . 9.849485
Side of the .inscribed square, in yards, = 55. 44 Log. = 1. 743857
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MUCJUXANXOns PR0BIJIM8. 611
pRomsM VII.
Gken the iran$verse and the conjugate JHameten qfon EOipsiet tofiiui
.itiJrea.
RULV.
To the logarithms of the longer and the shorter diatneters of the ellipsis,
add the constant logarithm 9, 895080 : the sum (abating 10 in the index,)
will be tlfe area of that ellipsis. ^
Example.
' Let the transverse diameter of an ellipsis be 616 yards, and its conjugate
diameter 445 yards ; requiredthe area or superficial content of that ellipsis?
Transverse diameter s «. • • 616 yards Log. =: 2*789581
Conjugate diameter :s • . , 445 yards Log. s 2. 648360
Constant log. = ..•••••,... 9.895090
Area of the given ellipsis, inyards,32l5294 Log. s 5.833031
Probjjbm VIII.
Given the transverse and the conjugate Diameters of an Ellipsis ; to find
the JMameter of a Circle equal in Area to that EOipeis*
Rule.
To the logarithm of the longer diameter, add the logarithm of the
shorter diameter; divide the sum. by 2, and the quotient will be the loga-
rithm of the diameter of a circle equal in area to the ellipsis.
'Example*
Let the transverse diameter of an ellipsis be 616 yards, and its conjugate
diameter 445 yards ; required the diameter of a circle equal in area to that
ellipsis?
Transverse diameter of the given ellipsisst 616 yards Log. s= 2.789581
Conjugate diameter of ditto =s « . • 445 yards Log. =s 2.648360
Divide by 2) 5.437941
Diameter ofa circle equal in area=:52. 356 yards Log. = 2.718970^
2b2
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612 mSCBLLAMSOUS PBOBUUIS.
Problbm DC.
Given the trantvene and the angugaie Diameten qfan EOifme^ to fied
Us drcumference.
Rule.
Square the two diameters ; add those squares together : take half the
sum, and find the logarithm correspondhig thereto. Now, the half of this
logarithm will be Uie logarithm of a natural number, which, being added
to half the sum of the two diameters, will give die cdtreeted mean diameter.
To the logarithm of the corrected mean diameter, thus found, add the
constant logarithm 0. 1915121 ; and the sum will be the logarithm of the
circumference of the ellipsb.
Example.
Let the transverse diameter of an ellipm be 616 yards, and its conjugate
diameter 445 yards ; required its circumference ?
Tr.diam.616x616i=379456, the square.
Conj.do.445 x 445= 198025, ditto.
Divide by 2 ) 577481, sum of the squares.
Halfsuinofsquarte=:288740iLg.5.4605077
Half the logarithms^ . .. 2.7302588iNtJVoJ>37.3519
Half the sum of the two diameters = ... 530.5
Corrected mean diameter = . . . . . . . 1067. 851 9Lg^. 0285 11
Constant log. = . 0.196121
Circumference of the ellipsis, in yards, = 1677.38 Log. s , 3.224632
PaoBLmtf X.
I • • • ■ •
Giventhe Diameter of a Sphere, or Globe: to find iti SoluKiy.
To thrice the logarithm of the diameter of the given sphere, add the
constant logarithm 9. 718999 j and the sum (abating 10 in the index,) will
be the solid content of such sphere.
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MISC&LLANBOUS PROBLEMS. 613
Example^ •
If the diameter of the earth be ^[9 1 /• 5 miles^ what is its solidity ? ,
Diameter of the earth =s 7917:5 miles Thrice its log* s IL 695764
Constant log. = 9.718999
Solidity of the earth, in miles, ^ 259874059701 . 5 Log. s 1 1 . 414763
Note. — It has been found that the solidity or solid content of a sphere,
whose diameter is unity or 1, is . 5235988 ; and sincje spheres are to one
another, . as the cubes of their dian)eters,-^therefore, as the cube of the
diameter 1, is to its solidity . 5235988 ; so is the cube of the diameter of
any other sphere or globe, to the solidity of such sphere or globe : and
hence the above rule. The constant logarithm is expressed by the loga-
rithm of . 5235988.
Remark. — ^For the method of finding the number of square miles con-
tained in the earth's superficies, see Problem V., page 594.
Prqblbm XI.
Given the Earih'i Diameter; to find the Height to which a Person should
be raised to see onC'third of its Surface.
RuLB. .; .
From twice the logarithm of the earth's semi-diameter, subtract the
logarithm of its one-third : the remainder will be the logarithm of the
height to which a person should be raised above the earth's centre, to see
one-third of its surface ; from which let the earth's radius or semi-diameter
be taken, and the remainder will be the required height abore its surface.
Example.
How high above the earth must a person be raised, that he may see one-
third of its surface ?
Earth's semi-diameter = . . 3958. 75 miles Twice its log.=:7. 195 1 16
One-third of ditto = . . . 1319.5833 miles Log. = 3.120437
Height above the earth's centre = 1 1876. 25 miles Log. s 4. 074679
Deduct the earth's semi-diameter s 3958. 75 miles.
Remainder = 7917^^50 miles; which is the true
height to which a person should be raised above the earth, to see one-
third of its stirface. '
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614 MISCSU^NBOUS PR0BLBM8.
Rrobubm XIL
GiveA the Etirth*s SemUDiaimetery and the SufCi fnean horizonial
ParalUuf; to find the Earth's Distance from the Sm.
RtJLE.
To the lofpurithm of the eartVa 6emi«diameter^ add the logarithi
co-tangent of the sun's mean horizontal parallax; and the sum (abating
10 in th^ index^) will \>e the logarithm of the sun's m^an distance from
the earth.
Example,
By the traosito of Veutis over the sun s disk in the yetfr^ 1761 end 17695
the aun'a me^u borisontal parallax appears to be about 8. 65 seconds of a
degree; now^ if the earth's semi- diameter be 3958. 7& mites^ its mean
distance from the sun is required ?
Semi-diameter of the earth ^ . 3953. 75 nules Log« =? 3^ 5975£81
Mean horizontal parallax of the sun=8^. 65 L(>g. cp-tang.= 14. 3730860
Earth's mean distance from the 8im^94546196 miles Log. = 7. 9756441
Problem XIIL
Gioen the Sun's mean Distance frtm the Earth, and his apparent Semi^
JHameieir, at a mean Rater tojind the true Measure ^fkis Diameter,
in EngUsk JIMles.
RULB.
"Te the logarithm of the sun's mean distanee from the earthy add Ae
logarithmic tangent of his seni-dlameter | and the sum (abating 10 tn the
index,) will be the logarithm of the sun's semi-diameter, in English miles ;
the double of which will be the measure of his whole diameter.
Example.
If th^ sun's mean disUwce from the earth be 94546196 English nules^
and his mean Apparent semi-diameter I6in^.65, the true measure of his
diameter is required ? *
Sun's mean distance from the eartK=9454619e miles Log. = 7. 9756441
Sun's apparent semi-diamct»r = 16f 1 *. 63 Lqg. tangent=:7. 6683950
Sim's tfue semi-diameter » . . 440797. 5 miles Log. a 5. 6442391
■ ' .III
True measure of the sun's.diameter= 88 1 595 English veilm*
/
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PftOBLBM XIV.
Oioen ike Diameters of the Earth and the Sun; to find the Batio af
thrir Magniiudee* .
RULB.
.Since the magnitudes of all spherical bodies are as the cubes^ or tripli-
cate ratio, of their diameters (Euclid, Boole XII., Prop. 18), — therefore,
from thrice tiie logarithm of the sun's diameter subtract thrice the lo^-
rithm of the earth's diameter, and the remainder will be the logarithm 9f
the ratio of their magnitudes.
Esampte.
If the earth's diameter be 7917.& Bnglkh miles, and that of the sun
88 15 95 such miles, required the ratio of their magnitudes ?
Sun's diameter, in English miles, = 881595 Thrice its log. =17.8358076
Earth's diameter. In ditto, = . . 7917. 5 Thrice ita log. = 11. 6957643
Ratio of the magnitudes of the earth and sua»( 1380532 Log.ss6. 1400433
Problsm XV. .
Owen the Circumference qf the Earth; to find the Rate, per HauTf at
< which the Inhabitants under the JSquator are carried, in consequence of
the Earth's diumtil Motion round Us Axis.
; Rule.
To the arithmetical ooniplement of the logarithm of 24 hours, add the
logarithm of the earth's circumference, and the logarithm of 1 hour:- the
sum of these three logarithms (abating 10 in the index,) will be the loga-
rithm of the rate per hour at .which the inhabitants under the equator are
carried by the earth's diurnal motion on its axis.
Example.
' Let the circumference of the earth be 24873. 5 miles ; required the rate
per hour at which the inhabitants under the equator are carried, in con-
sequence of the earth's diurnal motion ?
One day, or 24 hours, Arith. comp. of its log. = 8.6197888
Earth's circumference s± 24873. 5 miles Log. = 4. 3957369
Given tittle, or .... I hour Log. = 0.0000000.
Rate per hour, in miles, = 1036.396, ' Log/ ss 3.0155257
616 ICIflCBLLILNBOnS FEOBLBMS.
Pkoblbm XVI.
To find the Bate ai whieh the InhabUanis tmdtr amf gioem ParaBel of
Latitude are carKed, in consequence of the Earth's dkamal MoUtm em
its Axis.
Rule.
The circumference of the earth under the equator is 24873. 5 miles ;
and since the circumference under- any parallel of latitude decreaaes in
proportion to the co-sine of the latitude of such parallel, — therefore, to
the logarithm of the earth's circumference, under the equator, add die
logarithmic co-sine of the latitude of the given parallel ; and the sum
(abating 10 in the index,) will be the logarithm of the earth's drcumfcr-
ence under that parallel : with whifeh proceed as- directed in the last
problem*
Example.
Let the circumference of the earth be 24873. 5 miles; required the rate
per hour at which the inhabitants under the parallel of London are carried
by the earth's motion on its axis ?
Circumference of the earth = 24873. 5 miles Log. = . . 4. 3957369
Latitude of the parallel of London = 51 ?3 1 C Log. co-sine = 9. 793990?
Circumference under given parallels: 15478. 45 Log. = . . 4. 1897276
One day, or 24 hours, Arith. comp. of its log. = . . . 8. 6197888
Rate per hour, in miles, as required, = 644. 93 Log. = • • 2. 8095164
PaoBLBM XVn.
To find the Length of the tropkdl or solar Year.
Rule.
It has been found, by observation, that the sun apparentlgf advances in
the ecliptic 59^8^.33 of a degree every day at a mean rate; thatis, from
the time of his leaving any given meridian to the time of his returning to
the same meridian. Now, since the ecliptic is a great circle of 360 degrees,
— ^therefore, as the sun's apparent diurnal motion in the ecliptic, is to I
day, or 24 hours ; so is the great circle of 360 degrees, to the true length
of the tropical or solar year ; that is, to the time of the sun's periodical
revolution round the ecliptic from any equinoctial or solstitial point to the
same point again; Hence, by logarithms.
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MISGBLLANBOUS FROBLBMS., 617
Example,
The sun's daily motion in the ecliptic is 59' 8^. 33 in every natural day,
or 24 hours, at a mean rate; required the length of the tropical or solar
year?
Sun's app. diur. motion 59 'S"". 33, in 8ecs.=3548. 33 Log.ar.co.6. 4499760
One day, or 24 hours, in seconds = . . 86400 Log. = 4. 9365 1 37
Ecliptic, or great circle of 360?, in secs.= 1 296000 Log. = 6.11 26050
Length of the tropical year, in seconds=:3IS56928 Log. = 7. 4990947
Hence the tro)iical or solar year con^sts of 365 '5 ! 48T48 ' , as required.
Problbm XVIIL
To Jtnd the Rate at winch the Earth tnoves in the EcUptk during the
Time of its annual or periodical Revolution round the Sun»
Rule.
Since the earth's mean distance from the sun .is 94546196 miles
(Problem XII., page 614), the diameter of the orbit in which it moves
round that great luminary is 189092392 miles; and since the diameter of
a circle is to its circumference in the ratio of unity or 1, to 3. 14159265,
the circumference of the earth's orbit is 594031320 mWei. Now, as the
earth describes this circumference in 36if5*48T48! (last problem), or
8766 hours nearly, we have the following computation by logarithms : —
As the length of the year, in hours, =^ 8766 Log. ar. comp.=6. 0571985
Is to the circumf. of the f^atth's orbits=59405 1320 miles Log.3=8. 7738239*
Sois * Ibour Log.=:0. 0000000
To the earth's hourly motion in i|» orbit = 67768 miles Log.=4. 8310224
Problem XIX.
Given the Moon's mean Distance from the Earth, and her apparent
Semi-diameter, at a mean Rate; to find the true Measure of her
Diameter, in English MUes.
RuLBt
it b shown in page 9, under the head ^'Augmentation, of the Moon's
Semi'diameter/' that, the moon's- mean distance from the earth is
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618 MI8CBXXAKBOUS PR0^JE1C8«
236692.35 miles. Now, since her apparent semi-diameter is 15 M3? at
a mean rate, — ^therefore, to the logarithm of her mean distance from the
earth, add the logarithmic tangent of h^r apparent semi^diameter^ and
the sum (abating 10 in the index,), will be the logarithm of the moon's
semi-diaAeter in English miles : the double of which will be the measnxe
of her whole diameter. .
Example.
•
Let the moon's distance from the earth be 236692. 35 miles, and her
semi-diameter 15 M3^; required the true measure of her diameter in
English miles ? .
Moon's mean distance from the earths 236692. 35 miles Log.ft=5. 374IS42
Moon's apparent semi-diameter = 15'43? Log. tangent = 7* 6600896
Moon's true semi-diameter = • . • 1082. 1 miles Log.=3. 0342738
True measure of the moon's diameter=2164. 2 English miles.
Problem XX.
CUcen the Diameters of the Earth and the Moon ; to find the Ratio
of their Magnitudes,
Nofe.^This is performed by Problsm XTV., page 615.
* Example^ ,
If the earth's diiimeter be 7917* 5 English miles, and that of the moon
2164. 2 such miles, required the ratio of their magnitudes ?
Diameter of the earth :a . . 7917.5 Thrice its log.=r II. 6957643
Diameter of the moon ^ • . 2164.2 Thrioe its log.s 10. 0058922
Ratio of the magnitndes of the earth and moon=48. 96 LfOg.=I. 6898721
t^ROBLBM XXI*
To find haw wmeh larger the Earth appears to the hmar InhaBUasiis
than the Moon appears to the terrsstrial Inhabkanis.
RULB.
Since the distance between the earth and the moon is such as to cause
their opposing hemispheres to appear^ reci]meal)y from each other, like flat
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circles \ and Binee circles are to one another as the squares of their diameters
(Euclid^ Book XII., Prop. 2^) or, which is the same thbig, since spherical
surfaces are to each other, as the squares of their radii,-«*therefore, from
twice the logarithm of the earth's diameter, subtract twice the. logarithm
of that of the moon ; and the remainder will be the logarithm of the number
of times that the earth appears larger to the inhabitants of the moon than
the moon does to the inhabitants of the earth.
Example.
The diameter of the earth is- 79 17. 5 miles, and that of the moon 2164. 2
miles; requited how much larger the earth appears from the moon than the
moob does'from the earth ? «
Diameter of the earth == , , . 79l7. 5 Twice its. lQg.=:7. 7971762
Diameter of the moon = . . . 2164. 2 Twice its log.=6. 6705948
Number of times the earth is larger than the D =« 13. 88 Log.» 1 . 1265814
PkoBLBM XXII.
Tojind the Rate at which the Mbon revolver round her Orbit.
Note^r^This is performed by Problem XVIIL, page 617 S as thus :-«
Since the moon's mean distance from the earth is 236692. 35 miles, the
diameter of her orbit must be twice that distance, or 473384. 70 miles :
hence its circumference is 1487182 miles.- And since the moon goes
through this circuit, or orbit, in 27^7*43r5!, her hourly motion may be
determined \n the Ibllowing manner ; viz.,
As one luiiation=27'7*43r5!, in secs.= 2360585 Log.ar.co.=:3. 6269804 •
Is to the circumference of the moon's orbits 1487182 Log.=6. 1723641
So is one hqur^ in seconds =.,.,. 3600 . I40g.=:3. 5563025 <
To the moon^s hourly motion in her orbit == 2268 miles LfOg.s=3. 3556470
Problbm XXIIL
To^nd ike mean Distance 4^ a Planet fiom the Sun.
Rule.
It has been demonstrated, by the celebrated Kepler, that if two or*niore
bodies move round another body as their common centre of motion, the
620 MI'8C£LLAN£0US l»ROBLBMS«
squares of their periodical times ^ill be to each other in the same* propor-
tion as thectthfs of their mean distances from the central body; and hence
the following rule ;-*-
As die square of the earth's periodical or annual motion round the sun,
is to the cube of its mean distance from that luminary ; so is the square of
any other planet's periodical revolution round tlie sun, to the cube of its
mean distance therefrom ; the root of which will be the distance sought.
Example.
Tlie earth's periodical or annual motion round the sun is completed in
365 days, 5 hours, 48 minutes, ^8 seconds, and that of Venus in 224 days,
16 hours, 49 minutes, 1 1 seconds. « Now, if the earth's mean distance from
the sun be 94546196 miles, what is Venus' distance from that luminary ?
Earth'3 periodical revolution
365f5t48T48!, in sees. 31556928 Ar. cb. of twice its log. 5.0018106
Earth's mean dist.from sun, inmiles,=94546196 Thrice its log.23. 9269323
Venus' per.rev.224M6M9ril!,insecs.l94l4l51 Twice log. 14. 5762368
Reject 20 from the index; and, to extract the root, divide by3) 23. 5049797
Venus' mean distance from sun, in miles, « 68390098 Log.=7. 8349932^
PftOBLBM XXIV.
To^nd how much more Heat and Light the Planets adjacent to'the Sun
receive from that Jjuminary than those tohich are more remote.
Rule.
Since the effects of heat and light are redprocally proportional to the
squares of the distances from the centre whence they are generated,—
therefore, from twice the logarithm of the remote planet's distance from
the sun, subtract twice the logarithm of the adjacent planet!s distance
therefrom ; and the remainder will be the logarithm of the number of times
that the planet adjacent to the sun is hotter and more luminous than that
which is more remote.
. Exainple.
If the earth's distance from the sun be 94546196 miles, and that of
Venu9 68390098 miles, required how much more heat and light the
kites planet receives from the sun than the former ? .
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JJflSCXUJkNBOUS PftOBLSMS. 621
Earth's mean di^t. from 8un=;94546196 miles Twice its Iog.= 15. 95 12882
Venus' ditto s68390098 mUes Twice its log.= 15. 8699865
Heat and light VeHus receives more than the earths 1 . 205 Log. 0. 08 1301 7
Probliim XXV.
Given the apparent Diameter of a Planet; to find the Measure of its
true Diametei\
Find the difference between the earth's and the planet's mean distances
from the sun, and it will show the planet's mean distance from the earth ;
with which and the planet's apparent semi-diameter, compute* her true
diameter, by Problem XIX., page 617-
Example*' . ' ^
Let the earth's distance from the sun4>e 94546196 English miles, that
of Venus 68390098 such miles, and her apparent diameter 58''. 79;
required the true measure of her diameter, in English miles ? '
Earth's distance from the sun = 94546 196'mile8
Venus' ditto = 68390098 miles
Venus' mean dist. from earth = 26I5609S miles Log. = . 7* 4175729
Venus' apparent semi-diameter = 29"^. 395 Log. ung.=6. 1537885
' ■ ■
Venus^ true semi-diameter =: 3727 miles Log. = . 3. 5713614
True measure of Venus' diam.= 7454 English miles.
' IVoftfw— If the ratio of the magnitudes of the earth and a planet be
required, it may be determined by Problem XIV., page 615; thus in the
case of Venus :—
Diameter of the earth s . 7917. 5 miles Thrice ito log.s 1 1 . 6957643
Diameter of Venus = . . 7454 miles Thrice its log.= 1 1 . 6171682
Ratio of the magnitudes of th^ earth and Venus=l. 198 Log. =0.0785961
The velocity or rate at which a planet moves round its orbit may be
determined by Problem XVIIL, page 617.
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622 msGBLUiiiBpvs paoBUOCs,
PaoBuuc XXVI.
Tojinidthe Tims thai the Sm iaket to titrn romd iU Jms.
RULB.
If the bright face of the sttn be carefully observed through a good tele-
scope, large black spots will be found to make their appearance on its
eastern limb ; from this they gradually advance to the middle of the diak,
and thence to the western limb) where they disi^pear. After being absent
for nearly the same period of time that they were visible, they will be
observed to appear again on the eastern' limb as at first; thus finishing
their career in 27 days, 12 hours, and 20 minutes* Call this time the
observed intervaL
Find the number of degrees and parts of a degree that the earth has
moved eastward in the ecliptic during that interval^ in the following man-
ner; viz.,
•
As the earth's annual motion
round the sun:is36$ f 5 1 48748 ! ,in secs.3 1556928 Log. ar. co.2. 5009053
Is to eclip., or great circle of 360^, in sees. 1 296000 Log, s 6. 1 1 26050
Soistheob8.interv.==27fl2^20!,insecs. 2377200 Log. =s 6.3760657
Earth's advance in ecHptic during obs. interv. = 97628r Log.= 4. 9895760
Ditto, in degrees and parts of a degree, aa 27"? 7 -8?
Now, as 360 degrees, augmented by the earth's advance in the ecliptic
during the observed interval, thua found, is to the observed interval ; so is
the great circle of 360 degrees, to the absolute time of the sun's rotatoiy
motion on its axis ; thus :«-
As 360? +27?7'8rss387?7'8r, in secs.= 1393628 Log. at. co.3. 8558532
Is to the obs. int. « 27' I2t20r, in secs.«2377200 Log. « 6« 3760657
So is the great circle of 360? in secs.s 1296000 Log..» 6. 1126050
To the time of the sun's rotatory motioh=2210670f Log. s 6. 3445239
Ditto, in days and parts of a day = . 25f 14M*30t ; which^ therefore,
is the true time that the sun takes to turn round once upon its axis^ as
required.
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PaoBLRM XXVIL
Tofnd ih$ Length of a Pendulum far Vibra^g Seconds in the Latitude
of London. ;
RUUB.
It has been found by actual experiment that a heavy body let fall in the
latitude of London^ will descend, by the force of gravity, 16tV ^^^ ^^ ^
second of time ;-^now, since the circumference of a circle, whose diameter
is unity or 1, is found by computation to be 3. 14.159265 ; and since the
pendulum vibrates in the arc of a cicele, or cycloid, the radius of which is
equal to the length of the pendulum from the centre of oscillation ; there^
fore if twice the space piass^d through by a falling body in one second of
time, be divided by the square of the computed circumference of a circle,
as above, the quotient mU be the length of the pendulum for vibrating
seconds in the parallel of London.
Thus. 16^ feet = 193 inches x 2 = 386 inches, Log. = 2. 5865878
Circumf. of circle to diam. I =3. 14159265, twice its Log.= 0. 9942998
Length of thepend. in inches = 89. 1 1 Logarithm =:1. 5922875
Note. — By actual experiment the length of a pendulum for vibrating
seconds in London is 39^ inches, or 39. 125,
Problem XXVIII.
Tojind the Length qfa Pendulum for vibrating Sdlf-Seconds.
Rule. *
To the arithmetical complement of twice the logarithm of 120 (the
number of vibrations in a minute for the half-seconds' pendulum), add
twice the logarithm of 60 (the number of vibrations in a minute for the
seconds' pendulum), and -the logarithm of the length of the latter pendu-
lum : the sum of these three logarithms (abating 10 in the index,} will be
the logarithm of the length of the pendulum for vibrating half-seconds.
Example.
Let the length of a pendulum for vibrating seconds be 39. 1-25 inches ;
required the length of a pendulum that will vibrate half- seconds ?
Vibrations for i-secs.' pendulum = 120 Ar. co. of twice its log.=5. 841638.
Ditto for the seconds' pendulum = 60 Twice its log. = . .3. 556302
Length of the pendulum for sees. ^ 39. 125 inches Log* r= 1* 592454
Length of half-seconds' pendulum » 9. 781 inches Log. ^ 0. 990394
Hence the length of a pendulum for vibrating half-seconds^ is 9} mches.
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624 A COMPBMDIUM' OF PBACnCAL NAVIGATION.
A. COMPENDIUM OP PRACTICAL NAVIGATION; including the
direct maimev of making out a Day*s fFork at sea ; intended for the use
of persons unacquainted with the elements of Geometry and Trigonometry.
Problem I.
To reduce the Sun's DecUnatioriy as given in the Nautical Jlmanac, to the
Thne of apparent Nom under any known Meridian.
Rule.
From page II. of the month in the Nautical Almanac take out the son's
declination for noon of die given day, and note whether it is increasing or
decreasing ; and, at the same time, take out ^the variation of the aun's de-
clination between the noons of the given and preceding days if the longi-
tude be east, but between those of the given and following days if the lon-
gitude be west. Then, with this variation, or difference of declination,
enter Table XV, at top, and the longitude of the given meridian in the
right hand column i — in the angle of meeting will be found a correction,
which being applied to the declination, taken from the Nautical Almanac,
agreeably to the directions expressed at the bottom of the Table^ will give
the sun's correct declination at the noon of the given place. * •
Note. — ^When the longitude of the given meridian, and the variation of
declination cannot be exactly found in the. Table ; then, the sum of the
proportional parts, corresponding to the several terms which make up the
whole longitude and the whole variation, will be the correction of declina-
tion required.
Ej^ample I.
Required the sun's declination at noon, August lOth., 1825% in longi-
tude 100?30C East ?
Variation of declination between given And preceding noons
• (the longitude being east) is 17-26^
Sun's declination at noon of the given day per Nau-
tical Almanac (decreasing) :=: . . . / . . 15?36'30TN.
Pro. pt. to Ion. 90? 0^ and var. 17^ Or = 4^15^ OT
Ditto . . . 90.. 0 ditto 0.26 =0. 6.30
Ditto . . . 10.30 ditto IJ. 0 ^0.29.45
Ditto • . . 10.30 ditto 0.26 n 6. 0.45.30
Correction of dec. additive = . . • . 4152r OTSOV + 4C52r
Sun's reduced, or corrected declination = ;15'^4l!22rN.
* It is the nautical or tea day that is made use of in this and the foUowin; Examples :—
this day, like the civil, hegios at midnifht, and ends at the foUowini; miiifaiii^ht:— Itis
divided into two parts, of 12 hours' each ; the first pitrt or that contained between nidai^t
and noon is called A.M. or ante taieridiemy and the other part, or that between noon aod
midaiffhty P.M. or post meridiem.
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A COMPBNBIUll OF PRACTICAL NAVIGATION. 625
JSxample 2.
Required the sun's declination at noon, April 3d., 1825 *, in longitude
75°45^ West?
Variation of declination between given and foUcnving noons
(the longitude being west) is 22'54f
Sun's declination at noon of the given day, per Nau-
tical Almanac, (increamg) = S?18U0C' N.
Pro.pt. tolon. 75? OC andvar. 22: 0r = 4^35r (T,
Ditto 0. 45 ditto 22. 0 = 0. 2. 45
Ditto 75. 0 ditto 0.54 =0.11.15
Ditto 0.45 ditto 0.54 =0. 0. 6.45
Correction of declination, additive zr . 4:49r 6r45V + 4M9r
Sun'a reduced, or corrected declination =: . . . . . 5?23:29f N,
Problem II.
Given the Sun's Meridian Altitude, tojini the Latitude of the Place of
Observation*
RULB.
Reduce the sun's declination to the meridian of the given place by the
preceding Problem.
Then, to the observed altitude of the sun's lower limb add the difference
between its semi-diameter (page III. of the month in the Nautical Alma-
nac,) and the dip of the horizon, (Table II.) and the sum will be the appa-
rent altitude of the sun's centre ; .from which, let the difference between
the parallax and refraction answering thereto (Tables VII. and VIII.) be
subtracted, and the remainder will be the sun's true central altitude ; which
being taken from 90 degrees will leave the sun's meridional zenith distance
of a contrary denomination to that of its observed altitude. Now,
If the sun's meridian zenith distance and its reduced declination are
both north, or both south, their sum will be the latitude of the place of
observation : but if one be north and the other south, their difference will
be the latitude, and always of the same name with tiie greater term.
JExample I.
April 10th, 1825, in longitude 75? west, the meridian altitude of tlie
8un*s lower limb wa» 57?40'30^ south, and the height of the observer's
See Note, page 624.
2 S
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626 A COMPENDIUM OF PRACTICAL NAVIGATION.
eye above the level of the sea 22 feet ; required the latitude of the place
of observation i
Variation of the sun's declination between the given and ft^lowing
noons, (the longitude being west,) is 22'6r .
Sun's declination at noon of the given day per Nau-
tical Almanac (increcwin^) =: 7?56U2^ N.
Propl.parttolong.75?0^ andvar. 22^ 0^=4^35^ OT
Ditto . . . .75.0 ditto 0. 6 =0. 1.15
Correction of declination, additive = • . 4^86^15^ + 4^36r
Sun's reduced, or corrected declination = . . . . 8? I'lS'T N.
Observed altitude ofthesun's lower limb = . . . 57*?40'SO? S.
8nn*8semi.diameter=: • * *^^59r-i ^^^^^^ ^^^ jj,29r
Dip of the hor. for 22 feet = 4. 30 J
Dip
Apparent altitude of the sun's centre =
Parallax 0^5? refrac. 0^35 r, difF. = 0'.30
Sim'fl true central altitude =.••••
Sun's meridional zenith distance =: • •
Sun'^s reduced declination =: « • • •
Latitude of the place of observation = .
, . . 57^51C59r S.
subtractive =: 0'30^
57^5r.29f S.
32? 8^3ir N.
8. 1.18 N.
40? 9<49r N.
Example 2«
September 21st., 1825, in longitude 60? cast, the meridian altitude of
the sun's lower limb was 56?26^ north, and the height of the obaerver's
eye above the level of the sea 26 feet ; required the latitude of the place of
observation ?
Variation of the sun's declination between the given and preceding
noons, the longitude being east, is 2^^.22^.
Sun's declination at noon of the given day per Nau'
tied Almanac (decreflwing) = 0?43C34r N.
* Propl.pirttolong.60?0:«advar,23'. 0r=3^50T Cf
Ditto . • . . 60.0 ditto 0.22 =0. 3.40
Correction.of declination, additive = . . 3'53r407 + 3:54*
Sun's reduced or corrected declination = • . . . 0?47'28. N.
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A COMPtiIlt>iUM OF PRACTICAL MAVIQATION. 627
Observed altitude of the sun's lower limb = . • . 56? 26^ 0? N.
Sun's semi-diameter 15 '58^ dip of the horizon for 26
feet = 4'52r difiference r= + 11? 6lf
Apparent altitude of the sun's centre = .... 56?37' 6f N.
Parallax O'.SI refrac. 0:37^ diff. = 0'32r subtractive =s 0^327
Sun's true central altitude =: . .
Sun's meridional zenith distance =:
Sun's reduced declination =
Latitude of the place of observation =
66?36:34f N.
88^33^26? 8.
• 0.47.28 N.
• • •
32?35:58f S.
pROBLBBf III.
Given the difference of Longitude between two Places^ both under the
same Parallel of Latitude, tojind their Distance^
RCLB.
To the logarithmic co-sine of the latitude, add the logarithm of the diff-
erence of iQngitude, in miles ; and the sum, abating 10 in the index, will be
the logarithm of the distance.
Example.
Required the distance between Portsmouth, in longitude 1?6' wes{,^and
Green Island, Newfoundland, in longitude 55 ?35^ west, their common
latitude being 50?47' north ?
Long, of Portsmouth =s . . • * 1? 6i W.
Long, of Green Island = . • • 55.35 W.
Difference of Longitude = . . 54?29^ == 3269 ms. Log. 3. 514415
Latitude of the parallel = 50? 47? N. Log. co-sine tz . . 9. 800892
Distance, in miles = • - • 2(966.8 Log.- . .3.315307
PaoBUSM IV.
* Gitrm the Distance between two Places, both under the same Parallel of
Latitude, tojind their Difference ofLomgitude.
Rui^.
To the logarithmic secant of the latitude, add the logarithm of the dis- *
tance, and the sum, abating 10 in the index, will be the logarithm of the
difference of longitude*
282
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628 A COMFBMBIUM OF PRACTICAL NAVIGATION.
Example.
A ship from Cape Gear, in latitude 51?25^ north, and longitude 9?29!
west, sailed due west 1040 miles ; required the longitude at which «he
then arrived ?
Lut. of the parallel =s . . 5 1 ?25 ' Log. secant = ... 10. 205037
Distance sailed = ,. . • 1040 miles, Liog. = • * . 3.017033
Difference of long. = . . 27?48^ W.=: 1667. 6 mUes, Log. 3. 222090
Longitude sailed from = 9. 29 W.
Longitude arrived at = 37? 17' W.
No/e.— The ahove two Problems are essentially Useful when a ship sails
upon a parallel of latitude ; that is, when she steers either due east, or due
west.
PROBJ.BM V.
Given the LatUudei and Longiludes of two Places, to find the Omne
and Distance.
RULK.
From the logarithm of the difference of longitude, the index being aug-
mented by 10, subtract the logarithm of tlie meridional difference of lati-
tude ; the remainder will be the logarithmic tangent of the course : — then,
to the logarithmic secant of the course, thus found, add the logarithm of
the difference of latitude, and the sum, abating 10 in the index, will be the
logarithm of the distance.
£rampZe.
Required the course and distance between CapeBajdli, in latitude 40?3'
north, longitude 3?52r east, and Cape Sicie, in latitude 43?2' north, and
longitude 5?58' east ?
Lat. of C. Bajoli 40? Z'. N. Merid. pts. 2626. 6, Longitude 3. 52 E.
Lat. of C. Sicie =43. 2 N. Merid. pts. 2865. 8, Longitude 5. 58 E.
Diff. of latitude 2?59^ Merid. diff lat. 239: 2. Diff. long. 2? 6^
= 179 miles. = 126 miles.
To find the Course :—
Difference longitude 126 miles, . . Logarithm = 2. 100371
Merid. difference of latitude 239 miles Logarithm = 2. 378398
Course N. 27^47 :53rE.= ... Log, tang, = 9.721973
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A COMPENDIUM OF PRACTICAL NAVIGATION. 629
To find the Distance : —
Course 27 M7^ 53 r .... Log. secant =. 10.053254
Difference latitude = 179 miles. Logarithm = . 2. 252853
Distance in miles = 202. 3 • . Logarithm = . . 2.306107
Hence the true course is N. 27^47 '•53r R, or N. N, E;i E. nearly, and
the distance 202^ miles.
Problem VL
Given the Latitude and Longitude of the Place sailed fiom, with the
Course and Distance ; to find the Jjatitude and Longitude of the Place
come to.
Rule.
To the logarithmic co-sine of the course, add tlie logarithm of the dis-
tance ; the sum, abating 10 in the index, will be the logarithm of the differ-
ence of latitude 5 which being applied to the latitude left by addition or
subtraction, according as the latter is increasing or decreasing, the sum, or
difference will be the latitude come to. Now, to the logarithmic tangent
of the course, add the logarithm of the meridional difference of latitude;
the sum, abating 10 in the index, will be the logarithm of the difference of
longitude ; which being applied by addition or subtraction to the longitude
left, according as the latter is increasing or decreasing, the sum or differ-
ence will be the longitude come to.
Example 1. .
A ship from Cape Ortegal, in latitude 43947' N. and longitude 7?49^
W., sailed N. W. i N. 560 miles; required the latitude and longitude of
the place come to ?
To find the Difference of Latitude :—
Course steered =: 31 points . . . . • Log. co-sine = 9. 888185
Distance sailed 560 miles Logarithm = 2. 748188
Difference of latitude 432. 8 miles . . JiOgarithm = 2. 636378
«
To find the Latitude come to :—
Latitude of Cape Ortegal 43^47' N. Meridional parts . . 2927.8
Diff. of lat. 432. 8 N. = 7. 13 N.
Latitude come to . . .51? 0' N. Meridional parts . 35B8. 8
* _
Meridional difference of latitude = . .641.0
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6ao
A COMPBNDIUM Of PRACTICAL MAVIOATION,
To find the Difference of Longitude, and hence the Longitude come to :—
Course steered = 3| points. Log, tang. • . 9.914173
Meridional difference of lat, =: , 641 miles. Logarithm • • 2. 806S58
Diflferencc of long. = 8?46' W.= 526 miles. Logarithm . . 2. 721031
Long.of C. Ortegal = 7. 49 W.
Long, come to sx . 16?35^ W.
Remarks. — ^When a ship decreases her latitude ; that is, when the diflFer-
ence of latitude made good is pf a different name to the latitude sailed
from ; then, if the difference of latitude, expressed in degrees, be greater
than the latitude left, their difference will be the latitude come to ; which
will be of a contrary denomination to that sailed from ; beeaute, in this
case it is evident that the ship must have crossed the Equator.
And, when a ship decreases her longitude ; that is, when the difference
of longitude made good is of a contrary name to the longitude sailed from ;
then, if the difference of longitude^ expressed in degrees, be greater than
the longitude left, their difference will be the longitude eome to ; which
will be of a eontrary name to that sailed from ; because^ in this case the ship
will have crossed the meridian whence the longitude is reckoned.
Agaln.-^When a ship increases her longitude ; that is, when the differ^
ence of longitude made good, expressed in degrees, is of the same name
with the longitude sailed from, their sum will be the longitude .come ta ;
but, if this sum exceeds 180 degrees, then, its differance to 360 degrees wUl
express the longitude cojne to, which will be of a contrary denomioalioii to
that sailed from ; for, in this case, also, the ship will have crossed the me-
ridian that the longitude was reckoned fi-oni :— see Problems, Rules, and
Remarks, between pi^s 211 and 217* .
Example 2.
A ship from the Island of Annabona, in latitude l?23f S., and longitufle
5?34^'%., sailed W.N, W. 546 miies; i^wed the latitude and longitude
of the place at which she arrived ?
To find the Difference of Latitude ; —
Course steered =: 6 points . • . . L(^. co-sine . . 9. 582840
Distance sailed S4Q mka .... Logarithm « . 2.737198
Difference of latitude 208, 9 miles zz Logarithm .
T(> find the Li^titude come to :^^
Latitude sailed from = . . 1?23' S. Merid. piMfts
Piff.tet. = 208. 9 miles = 3.29 N.
2.320033
83,0
Latitude come ta = . . 2? 6; N. Merid. pvts= • .126.0
Meridional difference of latitude = ..,.., ^ , 209.0
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A COMPENDIUM Of PRACTICAL NAVIGATION. 631
To find the Difference of Longttude^ and hence the Longitude come to:-—
Course steered = . • • . 6 points. Log. t»ng. = 10. S82776
Meridional diff. of lat. = . .209 miles. Logarithm = 2. 320146
Diff. of long. = . 8? 25 C W. = 504. 6 miles Log. = 2. 702922
Long, sailed JErom 5.34 E.
Long, come to = . 2?5l< W»
Hence^ the latitude come to is 2?6^ N. and the longitude 2?5I C W.
Example 3.
A ship from Pitt's Island, in latitude 2?541 N. and longitude 174?30C
E. sailed S. £. by E. ^ E. 760 miles ; required the latitude and longitude of
the place at which she arrived?
To find the Difference of Latitude : —
Course steered =: . • 5^ points. Log. co-sine :=: • , 9* 673387
Distance sailed 760 miles. Logarithm = , , 2* 880814
Diff. of lat. = 358. 4 miles. Logarithm ... 2. 5S4201
To find the Latitude come to :-—
Latitude sailed from = « . 2?54C N. Mend, parts =: • 174. )
Diff. of lat. 358. 4 miles = 5. 58 S.
Lat. come to = . • • • 3? 4^ S* Merid. parts x • 184.1
Meridional difference of latitude = 858. 2
To find the Difference of Longitude, and hence the Longitude come to :-—
Course steered =: « . . 5| points. Log. tangent = 10. 272043 .
Meridional diff. of lat. = 358. 2 miles. Logarithm := . 2. 554126
Diff. of long, made good = II ? 10^ E. = 670 miles Log. 2. 826169
Longitu4e sailed from ::: 174.30 &
Sum = 185?40'. E.
Longitude come to = . 174°20'. W.
Hence, the latitude come to is 3 ?4^ S., and the longitude 174?20C
west.
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632
A COMFENBiUM OF PRACTICAL KAVIGATION.
Problem VIL
Given both Latitudei and the Course; to fnd the Dutance Sailed and
the Longitude come to.
RULB.
To the logarithmic secant of the course, add the logarithm of the differ-
ence of latitude ; the sum, abating 10 in the index, will be the logarithm
of the distance.— Then,
To the logarithmic tangent of the course, add the logarithm of the me-
ridional difference of latitude; the sum, abating .10 in the index,, will be
the logarithm of the difference of longitude ; which being applied to the
longitude left by addition or subteaction, according as it is increasing or
decreasing, the sum or difference will be the longitude come to.
Example.
A ship, from a place in latitude 3?4' S., and longitude 174?20t W^
esuled N. W, by W. ^ W. until she was found, by observation, to beiu
latitude 2?54C N.; required the distance sailed, and the longitude at which
the ship arrived ?
Lat- sailed from iz 3? 4'. S.
Lat. come to = . 2. 54 N.
Mer. parts = . 184. 1 miles.
Mer. parts == . 174. i miles.
Diff. oflat. = .
Course = .
Diff. of lat.
. 5?58<= 358 ms. Mer. diff. lat. = 358. 2 miles.
To find the Distance Sailed : —
. 5| points. Log. secant =: •
. . 358 miles. Logarithm = . .
Distance sailed =: 759. 4 miles. Logarithm = . .
To find the Difference of Longitude :—
Course = . . .
Merid. diff. lat .
Diff. of long. = •
51 points. Log. tang. =
358. 2 Logarithm =
670. 1 ras. Logarithm r=
10.326613
2.553883
2.880496
10. 272043
2.554126
2.826169
Long, sailed from = •.•••••
Difference of long, made good 670 miles =
174?20^ W.
11.10 W.
Sum = .
185?30^ W.
Longitude come to zz
174?30t E.
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A COMPENDIUM OF PRACTICAL NAVIGATION. 633
Note.— The three last Problems comprehend all the cases that usually
occur in the practical part of Mercator*8 sailing ; — for the speculative
cases^ see pages from 236 to 248, inclusive.
Problbm VIIL .
To find the Course^ Disiancey Difference of Latitudej and Difference of
Longitude made good upon compound Courses, and also the Bearing
and Distance from a Ship to the Place to which she is ^^ound, viz : —
To make out a Day's Work at Sea,
Rule.
Make a Table of any convenient size, and divide it into six columns :-*
in the first of these place the several courses, taken from the log board
(corrected for lee-way, if any, and also for variation), and in the second
place their corresponding distances.— The third and fourth columns are to
contain the differences of latitude, and, therefore, to* be marked N. S. at
top 5 and the fifth and sixth the departures, or meridian distances, which
are to be marked at top, also, with the letters B. W. — Now,
Enter the general Traverse Table, and take out the difference of latitude
and departure answering to each corrected course and distance, and place
them in their respective columns :— then, the difference between the sums
of the N. and S. columns will be the whole difference of latitude made
good, of the same name with the greater ; and the difference between the
sums of the E. and W. columns will be the whole departure made good, of
the same name with the grieater term. '
Remark. — ^The courses, taken from the log board, are to be corrected
for variation, and lee-way, if atiy, in the following manner, viz.
If the variation be easterly, it is to be allowed to the right hand of
the course steered by compass ; but to the left hand if it be westerly: —
And,
If the larboard tacks be aboard^ the lee-way is to be- allowed to the
right hand of the course steered by compass } but, to the left hand if the
starboard tacks be aboard.
To find the Course and Distance made good :-—
From the logarithm of the departure, the index being increased by 10/
subtract the logarithm of the difference of latitude } the remainder will be
the logarithmic tangent of thie course. — Then,
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634 A COMPBNDIUU Of PRACTICAL NAVIGATION,
To the logarithmic secant of the course^ thus found, add the logarithm
of the difference of latitude, and the eum, abating 10 in the indexj will be
the logarithm of the distance.
To find the Latitude in, by Account, or Dead Reckoning: —
If the difference of latitude, and the latitude of the place from which
the ship's departure was taken, or the yesterday *s latitude, be of the same
name ; their sum will be the latitude in, by account : but if of contrary
names, their difference will be the latitude in, of the same name with the
greater term.
To find the Difference of Longitude ; and thence the Longitude
come to :
To the logarithmic tangent of the course made good, add the logarithm of
the meridional difference of latitude (by observation) 3 the sum, abating 10
in the index, will be the logarithm of the difference of longitude. — ^Now,
if the difference of longitude, and the longitude of the place from which
the ship's departure was taken, -or the yesterday's longitude be of the same
name ; their sum will be the longitude in, by account, when it does not
exceed 180 degrees ; otherwise it is to be taken from 360 degrees, and the
remainder will be the longitude in, of a contrary name to that left : — but, if
the difference of longitude, and the longitude left be of contrary names,
their difference will be the longitude come to, of the same native with the
greater term.
To find the Bearing and Distance of the Ship to the Port, or Place to
which she isi Bound :^
From the logarithm of the difference of bngitude between the ship and
the place to which she is bound, the' index being increased by 10, subtract
the logarithm of the meridional difference of latitude ; the remainder will
be the logarithmic tangent of the course. Then,«-To the logarithmic secant
of the course, thus found, add the logarithm of the difference of latitude, and
the sum, rejecting radius, will be the logarithm of the distance.
Note. — ^l^he true bearings or course thus found, may be reduced to the
magnetic, or compass course, if necessary, by allowing the value of the
variation to the right hand thereof if it be westeriy j but» to the left hand,
if easterly : — this being the converse of reducing the course steered by com-
pass to the true course.
And this rule comprises the substance of that nautical operation, which
is generally termed making oat a day'a work at sea.
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Example 1,
A ship from Cape Espichell, in latitude 38925^ N. and longitude 9? 13'
W. bound for Porto Santo, in ktitude 33?3^ N, and longitude 16? 17 • W.,
by i'eason of contrary winds was- obliged to sail upon the following courses :
viz., (with the larboard tacks aboard,) W. by S. 56 miles j N. W, by W,
110 miles; W. N. W. 95 miles; (and then with the starboard tacks
aboard,) S. by E. i K 50 miles ; S. by W. i W. 103 miles ; and S. S. W.
116 miles, when she was found- by observation to be in latitude 34? 1 7 '* N ,
and longitude 13?42 • W. ; the lee- way on each of the courses was about
half a pcHnt ; the variation was two points westerly on the three first
courses, and If point on the three last ; required the true course and dis«
tance made good; the latitude and longitude at which the ship arrived by
account ; and the direct course and distance between her true place, by
observation, and the port to which she is bound ?
TRA.VEKSK TaBLB.
Corrected
Cottnee.
Dis-
tances,
. Difference of Latitude.
Departure.
N.
S.
E.
W.
S.W.b.W,4W.
W.bvN.iN.
•
86
110
95
50
103
116
31.9
9.3
26.4
37.0
102,5.
115.9
33.6
10.1
5.7
49.4
105.8
94.5
41.2
Diff. lat.
281,8
41.2
49.4
Departure.
249.2
49.4
240.6
199.8
To find Iho Cdurse made good i^
DqMrturd = . . . . 199« 8 miles. « . Logarithm = 2. 300596
Difference of lat. = 240. 6 miles. . . Logarithm = 2.381296
Course made ghoi S. 39?42:25^' W. a . Log. tang. = 9. 919300
"to find the Distance made good : —
Course made good » S. 39?42^25? W. . Log. see. s: 10. 1 13892
.Difference of lat. = 240. 6 miles. . . Logarithm =: 2.381296
Distance made good an 312« 7 miles.
Logarithms 2.495188
636 A COMPENDIUM OF PJtACriCAL NAVIGATION.
Hence, the course made good is S. 39?42'.25r W. orS. W. i S. nearly,
and the distance 313 miles nearly.
To find the Latitude and Longitude come to by Account, or Dead
Reckoning : —
Latitude sailed from 38^25' N 38?25^ N.M.pts.— 2500. 1 ms.
Diff. oflat. made
gocfJi = 240. 6 ms.=4? 1 ^ S. .
Lat.cometobyacc.=s34?24^ N.Byob.= 34?17< N.Nf.pts.= 2192.0 ms.
Meridional difference of latitude by observation = • • • . 308. 1 ms.
Meridional difference of lat. = 308. 1 miles. Logarithm = 2. 488692
Course made good = . . S. 39M2'25r W. Log. tang. =i 9. 919300
Diff. of long, made good = 4? 16' W. s 255. 8 ms. Log. s 2. 407992
Longitude sailed from = 9. 13 W.
Long, come tobyacct. = 13? 29' W. . .
To find the Course and Distance from the Ship to her intended Port : —
Lat. of ship by ob.=34?17' N. Mer. pts. = 2192.0 Long. =13?42: W.
Lat- of Porto Santo=33. 3 N. Mer. pU. = 2103. 1 Long. =16. 17 W.
Diff. oflat. = . . 1?14' Mer. diff. lat. 88.9 Diff. long. 2?35^
= 74 miles. = 155 miles.
Difference of longitude = 155 miles. Logarithm = . . . 2.190332
Mer. diff. latitude = . 88,9 miles. Logarithms. • . 1.948902
Courses . . . S. 60?9M9':' W. Log.Ung. . . .* 10.241430
Course = . . . . S. 60?9M9r W. Log. secant = • . 10.303185
Diff^srence of latitude 74 miles. . . Logarithm s . . 1.869232
Distances . . • 148. 7 miles. Logarithms . . 2. I724I7
Hence,— ITie course made good is S. 39?42' 25r W. or &;. W. J S. nearly.
Distance made good s 313 miles.
Latitude come to by account =; '• • . 34?24^ N.
Latitude by observation s 34?17- N.
Longitude come to by account s . ... 13? 29^ W.
Longitude by observation = . . . . 13?42' W.
Porto Santo bears S. 60?9M9l' W. or S. W, by W. J W. nearly.
Distant. . . . 149 miles.
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A VQtAnfiViVM %}¥ I'KACTIUAL. FIAVIQATION*
uo/
Note.'^lf the variation be one point and three-quarters west, the ship
roust steer W. b. S., by compass.
Example 2.
A ship from Port Royal, Jamaica, in latitude 17^58^ N., and longitude
76^53'. W., got under weigh for HayU, St Domingo, in latitude 18?30?N.,
and longitude 69?49'W., and sailed upon the following courses; viz.,—
S, 40 miles, 8.E.b.S. 97 miles, N.b:E. 72 miles, S.E.iS. 108 miles,
N.b.E.|E. 114 miles, S.E. 126 miles, N.N.E. 86 miles; and then by
observation was found to be in latitude 16?55'N., and longitude 72?30C
W. ; the lee-way on each of those courses was a quarter of a point (the
wind being between S.E. b. E. | E. and E. b. N. i N.), and the variation of
the compass half a point easterly ; required the true course and distance
made good, the latitude and longitude at which the ship arrived by account,
with the direct course and distance between her true place by observation
and the port to which she is bound I
Traverse Table.
Corrected
Courses.
Distances.
Difference of Latitude.
Departure.
N..
s.
E.
w,
S.iW.
btSaE^XE.
N.b.E.|E.
s.s.e.|e.
N.b.E.|E.
S.E.iS..
N.N.E.iE.
40
97
72
108
114
126
86
69.8
107.3
■ 77.7
39.6
87.7
92.6
101.2
• ■
41.5
17.5
55.5
38.4
75.1
36.8
5.9
254.8
Diff.ofLat.
321.1
254.8
264.8
5.9
5.9
Departure.
66.3 ■
258.9 =
To find the Course made good :—
Departure s • • • • .
Difference of latitude = •
Course ss
. 258.9milesvLog. = . . . '2.413132
. 66.3 miles Log. ^ . . . 1.821514
S.75?38n0?E. Log, tangents 10.591618
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638 A OOMPXKOIUM OP PRACTICAL NAVIOATION.
To find the Distance made good : —
Course made good = . , S. 75°38'. IOC E. Log. secant = 10/605409
DifTerence of latitude = . . 66. 3 miles Log. = . . 1.821514
Distance = 267.3 miles Log. = . . 2.426923
To find the Latitude and Longitude come to by Account, or Dead
Reckoning :—
Lat. sailed from = 17?58;N. . . 17?58'.N. Mer. pts=1096. 1 miles.
Diif. of lat. made
good 66.3miIes=:I. 6 S.
Lat.cometobyacc.l6?52?N, Byobs. 16?55rN. Mer. pts=cl030. 1 miles.
Meridional difference of latitude^ by observation, =r • • • 66. 0 miles
Meridional difference of latitude => 66 miles Log. = . . 1. 819544
Course made good = . . S. 75 ?38< lOr E. Log. tangent^:: 10. 5916 1 8
Difference of long, made good84? 18^ E.»257. 7 miles Log.B:2. 411 162
Longitude sailed from = . 76. 53 W.
Longitude come to by account=72?35 ' W.
To find the Course and Distance from the Ship to her intended Pent :—
Lat.ofshipbyobs.l6^55'.N. Merid. pts=: 1030. 1 Long. = 72?30fW.
Lat. ofHayti= 18.30 N. Merid.ptsr=1129.8Long. = 69.49 W.
Diff. of latitude 1?35^ Mer.diff.lat. 99.7 Diff. long. 2?4 11
= 95 miles. s 161 miles.
Difference of longitude — . . . 161 miles Log. = • 2.206826
Meridional difference of latitude = 99. 7 miles Log. = . 1. 998695
Course = N. 58n3C 55? E. Log. tongrs 10.208131
Course :a' ..... • 'N. 58? 13l55rE. Log.secant^lO. 278617
Difference of latitude «= . . . . 95 miles Log, as , 1*977724
. _
Distance :« 180.4 miles ljOg.sa . 2.256341
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Hence,--'nie course made good is S, 75^38'. lOrfi., or E.b.S. JS. nearly.
Distance made good =a 267 j^ miles.
Latitude come to, by account, s= • 16?52'. north-
Latitude by observation s • . • • 16^55^ north.
Longitude come to, by account, = 72?35' west.
Longitude, by observation, ss . . 72?30C west.
The true course from the ship to Hayti is N.58?i8<65TE., 6t
N.E.b.E.j(B. nearly.
The course, by compass, is N.E. ^ E. •
And the distance 1801 miles nearly.
Note. —For the method of making out a day's work by inspectum, see
Problem IX., page 249.
OP THE LOG-BOOK.
A Log-Book is a true and correct register of all the various transactions
which happen on board of a ship, whether at sea or in harbour :. such as,
coming to an anchor, getting under weigh, loosing or furling sails, mOoring
or unmooring, making or shortening sail, mustering at quarters or by
divisions, exercising great guns and small arms, &c. &c. &c. This book
should be a faithful transcript of the log-board.
The sea day, like the civil, begins at midnight in the Royal Navy, and
ends at the midnight following : it is, however, divided into two parts,
each consisting of 12 hours. The first 12 hours, or those contained
between midnight and noon, are denoted by a.m., which signifies ante
meridiem, or iefore mid -day; and the other 12 hours, or those from noon
to midnight, are denoted by p.it., which signifies post meridiem, or ajler
mid-day. The reckoning, however, is kept from nooft to noon, the same
as in the merchant service.
When a ship is bound to a distant port or place, the bearing and
distance of that port or place must be previously computed, by Problem
V^ page 628. The bearing or true course, thus determined, must be
reduced to the compass course, by applying the variation to the fight hand
thereof if it be westerly, but to the left hand if easterly : — (see Problem V., .
page 496). If islands, capes, or headlands intervene, it will be necessary
to find the several courses and distances between each successively; making
proper allowance for the variation.
At the time of leaving the land, the bearing of some point or place is to
be carefully observed, whose latitude and longitude are known ; which,
together vrith the estimated distance of the ship from such point or place,
is to be noted down on the log-board. This is called taking a departure^
640 A COMPENDIUM OF PRACTICAL NAVIGATION.
As the distance inferred from estimation is very susceptible of error,
particularly in hazy weather, or when that dis,tance is considerable, it will
be advisable to make use of the following method in taking a departure ;
viz., Let the bearing of some well-known place be observed, and, when the
ship has run a convenient distance, on a direct course, let the bearing of
the same well-known place be again observed; then there will be a triangle
formed, in which there is one side given : that is, the distance sailed between
the times of observation, and all the angles, to find the distance between
the ship and the place observed. This may be done by Problem I.,
Oblique Sailing, page 256 ; or it may be very readily determined by means
of a good chart. In like manner may a departure be taken from a light-
house at night.
In making out the first day's work after leaving the land, especial care
must be taken, in setting down the bearing and distance of the departure
in a traverse table, to make use of the opposite point of the compass to
. that bearing ; and, also, to make due allowance for the variation. Thus,
if the object from which the departure was taken bore N.E. b. E., and the
variation of the compass be 2 points westerly, then the true course for
the traverse Table is S.W. b. S. ; abreast of which, in the proper column,
is to be placed the estimated or computed dis^ce.
The course steered, is indicated by the compass ; the distance sailed, in
a given time, is determined by the log-line and the half-minute or quarter-
minute glass. In His Majesty's Royal Navy, the log is hove once in every
hour ; and so it is on board ships belonging to the East India Company.
The several courses and distances sailed during the interval of 24 hours,
or from noon to noon, together with all the remarks and occurrences that
are worthy of notice, are generally marked down with chalk on a board,
painted black, called the log-board. This board is usually divided into
six columns : the first column on the left hand contains the hours from
noon to noon, viz., •from noon to midnight, and then from midnight to
noon ; the second and third columns contain the knots and fathoms suled
every hour; the fourth contains the courses steered; the fifth the winds;
and in the sixth the various remarks are written,-^8uch as, the state of the
iveather, the sails set or taken in, the observations for ascertaining the
ship's plate, the variation of the compass, and whatever else may be
deemed necessary. The log-board is transcribed every day at noon (under
the direction of the Master,) into the log-book, which is divided into
columns exactly in the same manner.
The form of the log-book which is now made use of in the Royal Navy,
will be shown presently.
The courses steered must be corrected for the variation of the compass,
and also for lee- way, if any. .If the variation be westerly, it must be
allowed to the left hand of the course steered ; but if easterly, to the right
hand thereof, in order to obtain the true course.— See Problem VI., page497»
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The lee- way is to be allowed to the right hand of the course steered,
if the larboard tacks be on board ; btit to the left hand, if the starboard
tacks be on board.
The variation of the compass should be determined twice a day (every
moriiing and evening,) if possible. The method of doing this is «hown in
the several problems contained between pages 483 and 495.
With respect to the lee-way, its nature or effect may be thus ex-
plained:—
When a ship is close-hauled, and the wind blowing fresh, that part of
the wind which acts upon the hull and rigging, together with a consider*
able part of the force which is exerted on the sails, tends to drive her
immediately from the direction of»tlie wind, or, as it is .termed, to lee-
wards But since the bow of a ^ip exposes less surface to the water than
her side, the resistance will be leiis in the fii^t case than in the second ;
the velocity, therefore, in the direction of her head, vrill, in most cases,
be greater than in the direction of her side ; and the ship's real course will
be between those two directions. Hence the angle contained between the
line of the ship's apparent course and the line she actually describes
through the water, is termed the angle of lee^way^ or, simply, the lee-way.
The angle oflee-wfiy that a ship makes may be very readily determined
in the following manner ; viz., Draw a semi-circle on the taffrail, with its
diameter at right angles to the ship's keel, and its circumference divided
itito points and quarter-points ; then let the angle be observed which is
contained between the semi-diatneter pointing right aft, .or parallel to the .
keel, and that which points in the direction of the wake, and it will be the
lee-way required. Or, after heaving the log, if the line (before it is drawn
in) be applied to the centre of the semi-circle, the points and quarter*
po|nts contained between ita direction and the fore and lift radius of the
aemi-cirde will be the lee-way, as before. «
Many writers on navigation have given rules for ascertaining the quan-
tity of lee- way which a ship makes, independent of observation. These
are as follow 5 viz., .
1. When a ship is close-hauled, has all her sails set, the water smooth,
with a light breeze of wind, she is then supposed to make little or no
lee-way.- . .
2. Allow one point when the top-gallant -sails are handed.
3. Allow two points when under close-reefed top-sails.
4. Allow two points and a half when one top-sail is handed.
5. Allow three points and a. half when both or the three top-saila are
handed.
6. Allow four points wheii the fore-sail or fore->course is handed.
7« Allow five points when under the main^-sail or mfiin-course only,
2t
642 A courssDWU o? practical vaticatiok.
8. Allow six pmnU when under a balaaoed i
9. ADovr seven pomts when under bare poles.
Ab these rules depend entirelv upon the quantity of sail set, '
rtgkrd to the model at the stup, or to the nature of die wajr in
may. be trimmed for sailing, it is evident that they are far ttom
general, and that they are, in reality, little more than mere probttUe eoo-
jcctures. But since the accuraey of aship's reekoning depends;, in some
measure, upon the truth of the lee-way, it ought to be dedoecdl, at aB
times, from actual observation, as above directed; and then its Tahne ahonid
be carefully noted down, in a separate orfumn, on the log-board : ao that
all concerned may be thereby enabled to Correct the course* stecicd, is
making out their days' works at noon.
In very strong gales, with a contrilry wind and a high se% it is not
prudent to attempt working to wmdward: in such cases, the grand object
is, to avoid, as much as possible, losing ground, or being driv
Whh this intention, it is customary to by the ship to, under no
than may be barely sufficient to check that violent rolling which a
otherwise acquire, to the eifdangering of her masts, yards, and' rigging.
When a ship is brought to, the helm is kept about three parts aleey which
brings her head gradually round to the wind. The force of this etenDCDt hav-
ing then very little power on the sails, the ship consequently loses her way
through the water, which ctesmgtoactupon'the rudder, herheadfidls off from
the wind ; the sail which she has set fiUsj atad gives her fresh way through
the water, which, acting on the rudder, brings her head again gradually
. round to the wind ; and .thus she obtains a kind of vibratory motion,
coming up to the wind and falling off fiom it alternately.
Ships Iie*to under different sails, according to circumstancies ; and one
* vessel will lie^to considerably better under some particular sail thAn another.
But, in general, a close-reefed ihun- top-sail is, perhaps, the most eligible
sail to lie-to under ; because of its being nearly over the centre of motion,
and, also, because of its elevated position, which renders it far less suscep-
tible of being becalmed in the trough of the sea than' either the courses or
storm-stay-sails.
When a ship is lying-to, observe the points of the compass upon which
she comes up and falls off, and take the middle point for her apparent
course : to which let the variation and the lee-way be applied, and the troe
course will be obtained. Thus, suppose a ship lying-to under a close-
reefed main-top-sail, with her larboard tacks on board, comes up S.S.W.,
and falls off to S.W.b. W. ; then, allowing the variation to be 1| point
west, and the lee-way to be 21* points, the cburse made good is S.W.i W.:
for the middle point between S,S.W. andS.W.b.W. is SlW.iS.j to
which, li point westerly variation being allowed to the left, and 2j points
lee- way to the right, makes the true course S.W.iW.
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A COMPXKDIUM Or PRACTICAL NAVIOATIOX. 849
The setting and drift of currents, ynth tht heave and drift of the sea,
should be set down as courses and distances upon the log-board: these are
to be corrected for variation only.
The xsompntation made from the several corrected cofirses^ and their
co/responding distances, is called a day's work; and the ship's place^
deduced therefrom, is called her place by account, or dead reckoning.
If the course and distance made by a ship could be correctly aacertuned,
by means of the compass and the log, nothing more would be necessary in
determining her true place at sea ; for the absolute course aiid' distance
being known, the latitude^ and longitude could be readily computed, by
Problem VL, page 629. But, in co9sequence <i{ the irregularities to which
the hcfkving of the log is subject, particularly during .the night, with many
unforeseen and itnavoidable causes, such as sudden squalls, imperfect
compasses, . unequal care in the helms-man, inaccurate allowances for
variation and lee^way, &e. &c., the latitude .and longitude of the ship, as
inferred from dead reckoning, will'very seldom agree with the truth, or
with those immediately deduced from celestial observation. In conse-
quence of this discrepancy, Mveral. writers on navigation have proposed to
apply a conjectural correction to the departure or meridian distance, in
order to find the true longitude. Thus, if the course be near the meridian,
the error la wholly Attributed to the distance, sind the departure is to be
increased or diminished accordingly; if it be nedr a parallel, that is, near
the east or west point of the compass, the course only is 8Upp<»ed to be
erroneous I- and if the course be towards the middle of the quadrant, viz.,
near four points, the assumption is that both course and. distance are wrong« .
These corrections, being computed and applied according to the rules
given by different authors, will generally place the ship upon different sides
of her meridian by account :' hence, since the corrections arising from these
rules are evidently founded upon a vague kind of guess-work, they ought
to be absolutely rejected.
'When the latitude by account differs from that by observation, the log«
line and half-minute glass should be carefully examined, and, if found
erroneous, the distance sailed, as indicated thereby, should be corrected
accordingly, by the Problems given for that purpose, between pages 272 and
276. If the corrected distance, thus found, with the course, does not pro-
duce a coincidence in the latitudes by account and observation, the mariner
should then consider whether the variation has been properly determined and
allowed upon the courses steered by compass ; if not, these courses are to
be again corrected; but no other alteration whatever should be made in
them. If the latitudes by account and observation be still found to
disagree, the navigJitor should next Consider whether the ship's place has
been affected by a current or by the heave of the sea, and allow for their
course and drift to the test of his judgmenU By carefully applying thdse
2t2
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644 A COMPKNDIUM OP VRACTECAL NAVIGATfON.
corrections, a new difference of latitude and departure, and a new course
and distance, will be obtained ; which will, in general, prodace an ap-
proximation in the latitudes : beyond this, no alteration whatever should
be made in the departure with the view of finding the longitude by
account.
However, since there are many mariners who, from long-establisked
practice, are not willing to depart from the common Aystem of correcting
the dead reckoning by the rules laid down for that purpose in certain
Epitomes of Navigation } and since, these rules are exceedingly complicated,
and admit of a variety qf cases j the following general rule is given f©r the
use and guidance of such persons, which reduces those various cases into
one very concise method, and thus does away with the necessity of con-
sulting several complex rules before the desired correctioif can be obtained.
A general Rule for correcting the Dead Reckoning :—
Augment the distance sailed by iwo-ihirds of the difference between the
latitude by account and that by observation, when -the observed latitude is
before or ahead of that by account; but diminish the distance sailed in the
same proportion, wh^n the observed latitude is -astern or behind that by
account. Then,
Enter the general Traverse Table with this corrected djstance and the
difference of latitude by observation, and find the corresponding departure.
Now, with the departure, thus found, in a latitude column, and the middle
latitude as a course, find the correspjonding distance, anditwiH be the
conected difference of longitude.
Example 1.
Suppose a ship, from a place in latitude 47?49< N. and longitude 9?29^
W., sailed S. 43? W. 160 miles, and then finds her latitude by account to
be 45'?54' N., but by observation her true latitude is 46?39< N.; required
the longitude come to by account, or dead reckoning ?
iSoInrion.— -The difierehce between the latitude by account and that by
observation, is 15' miles; the two- thirds of which is 10 miles. Now, this
being added to the distance sailed; because the observed latitude is before
or ahead of that by account, makes the corrected distance =s 170 niiles :
with this corrected distance and the difference of latitude by observation,
viz., 2? 10^ or 130 miles, the corresponding departure, in the general
Traverse Table, is 109. 3 miles. Then, with this departure, in a latitude
column, and the middle latitude (between the latitude sailed fronf and that
arrived at by observation), viz., 46?44^ as a course, the difference of longi-
tude corresponding thereto, in a distance column at the top or bottom of
the page, is 159 miles, *or 2?39^ W. ; which, being added to the longitude
left, shows the longitude at which the ship arrived to be 12 ?S^ west.
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Example 2.
Suppose a ship from Perto Santonin latitude 33^3' N. and longitude
]6?17' W.^ sailed N. 47? E. 210 miles, and then finds her latitude by
account to be 35"? 26^ N., but. by observation her true latitude is only
35?8C N.; required th^ longitude come to by accouiit, or dead reckoning ?
Solution.-^The diifference between the latitude by account and that by
observation; is IS' miles; the two- thirds of which is 12 miles. Now, this
being subtracted from the distance saifed, because the observed latitude is
astern or behind that by account, makes the corrected distance =198
miles: with this corrected distance and the difference of latitude by
observation, viz., 2^5^ or 125 miles, the corresponding departure, in the
general Traverse Table, is 153.5 miles. Then, with this departure, in a
latitude column, and the middle latitude (between the latitude sailed from
and that come to by observation), viz., 34?5|t as a course, the difference
of longitude corresfiondtng thereto, in a distance column at the top or
bottom of the page, is 185 miles, or 3?5f E.; which, being subtracted
from the longitude left, shows the longitude at which the ship arrived to
be 13? 12^ west.
Remath.^-Alihongh the above gener&l rule for correcting the dead
reckoning. i3 the most simple, and, perhaps the most accurate of any that
have been as ye^ devised for that purpose, yet the author has frequently
found, on making the land after a long voyage, that the longitude deduced
therefrom d^ered several degrees Jroin the truth: hence it is evident,
notwithstanding the easy and specious feasibility of this:method, that the
prudent mariner will do well to be extremely cautious in applying it to
practice ; nor should he ever place any manner of faith in the longitude so
deduced, particularly, if he has been any considerable time from the land.
From this it is manifest that the. navigator should determine the longitude
of his ship, as often as possible, both 1^ the lunar observations and l^y a
chronometer; and from the true longitude, thus found, the reckoning of
this element is to be carried forward, in the same manner as that of the
latitude, from the last observation. A separate account, however, should
be kept of the longitude by dead reckoning : such account i^.not only very
satisfactory, but it often .proves highly useful as a reference; particularly
in comparing the computed velocity and drift of a current with those
deduced from actual experiment.
The following' is the form a( the log-book which is now used in His
Majesty's Royal Navy, and from which we will make out apracHca/ day's
work.
646
Log'Baak qfHU Majesty's Ship •
Wedneidayj June 4th, 1823.
H.
10
11
13
K. ]
?. Ccunes.
Winds.
No. of
Sisals.
S.E.
S.E.
A.M. Moderate breezes and basy ireatber.
At 4»30«» employed washing decks.
At 7*30** unmoored ship, and hnre short on tfae best
bower.
At 8 UO" weighed and made saU.
At noon light winds and hazy weather.
Coarse.
Distance.
Lat by
Account.
Latitude b}
Obserr.
Long, hv
Account.
Ob«erv. ^''™-
Variation
at Noco.
Bearinn and
Di.st.atN«on.
34°20'S.
•
25o2d'W.
cape of Gaud
HopeK b.S.
(5 miles.
r
2
3
4
6
7
9
9
10
11
12
5 4
5 4
6 0
6 2
6 2
6 2
7 6
8 0
8 6
9 0
9 4
» 6
N.W.b.W.
N,W.§W.
N.W.<W.
N.W.
SJS.
8.E.
p. M. Moderate breeses, with ihtclc basj wcatbcr.
At 2*30" shook a reef nut of the top.-sails^ and set
the top-mast and fop-gaUaat studding-sails. •
At 5* heat to quarters, and exercised great raos
and small sirms.
At 8* firesh breexes and doady weather.
At midnight fresh breeses and dear weather.
Log-Booi qfjht Majatjf'i Ship ■
't Tkurtdoff^ June 5th, 1^3.
H.
K.
9
9
9
9
9
9
10
10
9
10
9
11
9
12
9
Course.
N.69W.
Courses,
TTWT
N.W.b.N.
Distance
214 miles.
Winds.
"SET
SJS,
S.E.
No. of
Signals.
Ramaika and Osdureaces, .
▲.M. ^resh breezes and ciear weather.
Lat. by
Account.
33o6'S.
Latitude by Long, by
At 4 * ane clear weather ; emfdoyed washing decks.
At 6* 30« set the fewer stikdding-sails*
At 8*40» in lower ttaddiog-sails.
At 10*30* mottmd by diidsipBi, and Inspcettd the
eople's clothing.
At noon fresh breezes and fine ^rlearnreather.
Long, by
Observ.
33? 4' 30*^5.
Account.
Lunar
Obsenr.
14<'24'E. U^gr-R
Lottg.by Variation
Chron.
at Noon.
14«24'E.23<»10'W,
Bearings and
Disk at Noon.
jSt. Helena
N.4flP&l'W.,
dist. lias ms.
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Note. — ^The departure is taken from the Cape of Good Hope ; and as
this place bore, at "aoon, E. b. S. from the ship^ .distant 15 miles, the com-
pass bearing or course of the ship from the Cape was, therefore, W. b: N.
NdW, the variation, 2i points west, being allowed to the left-hand of
W. b. N., shows the true bearing or course to be W. b. S, i S. The • other
courses are, in like manner, to be corrected for variation ; but since the
value of this element is not the same at both noons, it is advisable to allow
25?20<, 6r 2i points west,, on the courses in the first 12 hours, of between
noon and midnight, and 23^10% or 2 points west, on the courses in the
other 12 hours, or between midnight and noon; then, these corrected
courses, with their respective distances, being inserted in a Traverse Table,
after the following manner, the difference of latitude and departure corre-
sponding thereto, \iith the course and distance made good, maybe readily
determined by calculation^ agreeably to the rule given iii Problem VIIL,
page 633 ; or, perhaps more readily, by the general Traverse Table. — See
Problem IL, page t08. ' .
T&AVBBSB TaBLS.
Corrected
Courses.
Dtttances.
■ DiflEsrence of Latitvde.
Departure.
N..
S. .
£..
w.
W.b.9.iS.
W.JN.
W.b.N.iN.
W.b.N.}N.
W.b.N.fN.
W.N.W.
N.W.b.W.
15
11
18
22
37 .
48
6d
Diff.oflat.
1.6
4.4
. 6.4
12.5
18.4
37.8 ,
3. 6
—
14.6
10.9
17.5
21.1
d4.-8
44.3
56,5
8l.l .
3,6
3.6
Departures
199.7
o.p
77.5
199.7
To find the Course and Distance made good :—
The differeiMy 6f latitude 77 '^i <^d the departure 199. 7, are found to
sfgree nearest abreast of 69 degrees, and under or over 214 miles distance.
Hence the course made good is N. 69? W% or W,b,N,f N. nearly, and
the distance 214 miles.
Lat. in by ace, = . 33? 6'. lOrS. By ob.33? 4^30? S. Mr, p8.=2104. 9
• ■^^— — — — —
Meridionaldifferenceoflatitude^ by observation^ = • . , • 95.2
[miles.
Meridional difference of latitude = 95. 2 miles Log. = . .1. 978637
Course made good = • . . . N. 69^ W. Log. tang. = 10. 415823
Diff. of long, made good = . 4° 8^ W.=248 ms. Log.=2. 394460
Long, of C. of Good Hope = 18?32: IS^T E.
Long, come to by account = 14?24! 15 ^ E.
To find the Coyrse and Distance from the Ship to St. Helena :
Lat.ofshipbyobs.33? 4^ S. Mer.pt8. = 2104.3 Long.= 14?28^£.
Lat. of St. Helena 15.55 S. Mer.pts.= 967-5 Long*= 5.44 W.
Diff.oflat =: . 17? 5^ Mer. diff. lat,= 1136.8 Diff. long. 20? 12:
=: 1029 miles. = 1212 mUes.
Difference of longitude s 1212 miles. Log. = . • . 3. 0S3503
Merid.diff. lat. = • . 1136.8 miles. Log. = • . . 3.055684
Course = . . • . N.46?51CWr Log.tang. = . 10.027819
Courses. ... N. 46?5r. W. Log. secant = 10.165001
Diff. of lat. « , . . 1029 miles. . Log. ss . .• . 3.012415
Distance =: .... 1505 miles. Log, = , . . 3. 177416
Hence,— The course made good is N. 69? W. or W. by N. } N. nearly.*
Distance made good = . . . 2 14* miles.
Latitude come to by account '== 83? 6' IQ": S.
Latitude by observation = . . 33. 4.80' S.
Longitude come to by account =: 14. 24. 15 E.
Longitude by lunar observation := 14. 28, 0 E.
' To find the Latitude and Longitude come to by Account : —
Lat. Cape GoqdHope34?23U0rS. . 34?23M0r S. Mer.pts.=220a. I |
Diff. Lat. 77. 5 ms. . 1. 17. 30 N.
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A COMPENDIUM OF PRACTICAL NAVIGATION. D4Sl
Longittide by chronometer = . .14. 24. 0 E.
Variation at noon 23. 10. 0 W. *
St. Helena bears N. 46?5 1 ^ W. or N, W. J W. nearly, inde-
pendent of variation.
Distsant • • • 1503 miles.
OF THE MEASURE OP A KNOT ON THE LOG LINE.
It ha&beeQ remarked, page 272, in the introduction to the Problems for
correcting the distance sailed on account of any errors that may be dis--
coyered in the log line and half-minute glass, that the distance between
any two adjacent knots on the log line should bear the same proportion to
a nautical mile that half a minute does to ai) hour, viz. the.oue hundred
and twentieth part ; that a nautical mile contains 6080 feet ; and that this
number divided by 120, gives the true measure of a knot, viz. 50 feet a^d
8 inches. — But, sinc^ the young- navigator may be desirous of being ;nade
acquainted with the principles upon which th|S measure has been deter-
mined, the following considerations are, therefore^ submitted to his atteii*
tion ; which, besides satisfying him in that particular, may do something
towards giving him a just idea of the true figure of the earth ;^and, with-
out which idea he can liever clearly comprehcind the principles upon which
the art of navigation is founded.
The earth is a planet, and the next, in the solar system, above Venus.—
Our senses assure us of its opacity ; — and that it is of a glo|:>ular or spheri*
cal figure will appear evident from the arguments which follow:—-
A lunar eclipse is occasioned by the moon's passing through theiearth's
shadow ; and since this shadow, when projected on the lunar disc, is ob*
seived to be always circular in every different position of the earth, it
necessarily follows that the •earth, which casts the shadow, must be sphe-
rical, since nothing but a sqphere, when turned in various positions with
respect to a luminous body, can project a circular shadow. — ^Again,
A lunar eclipse is observed sooner by those who live eastward than by
those who live westward ; the di^erence of time being, always propor*
tional to the difference of longitude between the plfu^es of observation.
Now, if the earth were an extended plane^ as the primitive fathers as-
serted, the eclipse would happen at the same instant in all places :— but
this is so far from being the case, that the inhabitants of Jamaica will not
* The variation of the compass may he very, readily determined at noon (sufficiently c<)r*
rectly for nautical purposes,) by the second part of th« Rule to Problem IV. page494« read*
ing tun instead of star or planet .«*fiee note at bottom of pa^ 495«
OW A GOIIPJEMDIDH OF PIUCTICAL NAVIGATION.
see an eclipse of the moon until about five houn after it. takes place at
Greenwich ;— therefore the figure of the earth mttst'be spherical^ or very
nearly so.
If the earth were an extended plane^ the meridian zenith distance of any
one fixed star would be the same in all parts of the world ; because the
measure of the earth's diameter bears no more propdttion to the immea-
surable distance of the nearest fixed star than an indivisible point' does to
the diameter of the earth.-r-But, since the meridian zenith distance of the
same fixed star is found to differ with the latitude, the difference in the
zenith distance being always proportional to the intercepted arch of the
meridian ; and since it is the known property of a curve that the arches are
proportional to their correspondent angles, therefore the surface of the
earth and sea is of a curvilinear form.— Menqe the earth must be of a sphe-
rical figure.
The earth has been circumnavigated by mbny .persons, at diflerent
periods, who, by sailing in a westerly direction, allowance being made for
promontories, &c. arrived at the place whence they sailed. — Henee, the
earth- must be either of a cylindrical, or a spherical .figure ;-^but that it is
not of a cylindrical figure will appear obvious by considering that the dif^
fer^nce of longitude and meridional distance between two places would, on
the cylindrical hypothesis, be equal ;*-whereas, experience and actual ob-
servation, demonstrate that the very reverse of this takes place ^-**tberefe^e
the earth must be of a spherical form from west to east.
If a ship in north latitude sails southerly, the north polai' star will be
found gradually to decrease in altitgde till the vessel reaches the Equator ;
at which place the star virill be seen immersed in the horizoh.-**After cross-
ing the Equator^ and as the ship advances in the southern hemisphere, the
stars in the neighbourhood of the south celestial pole will be seen gradu-
ally emerging from the southern/horizon, and increasing in altitude, whilst
those about the north celestis^ pole will be entirely lost sight of; being
hid below the horizon :~hence the earth is spherical from north to south ;
but it is also spherical from west to east, as appears from its circumnavi-
gation ; therefore the figure of the earth is that of a sphere.
When two distant ships are approaching each other, at sea, the royals
and top-gallant sails only of each are visible at first; the lower sails and
hulls 'being concealed by the convex surface of the waters — but as th^
draw nearer towards each other, the parts that Were so concealed by the
convexity of the sea's surface, will be seen toilse gradually above the Im^*
zon. — ^Now, if the sea were an extended plftne, the hulls or bodies af the
ships would be the first parts seen; and because they arc the largest, they
would, evidently, be seen at the greatest distance ; nor would the small
sails near the masts' heads be visible until the ajiproacb of the ships
brought them considerably nearer.
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In making the land the most elevated parts are first seen, such as moun-
tains, kc. ; then tops of light houses and steeples, and shortly afterwards the
coast, or beach : — this plainly demonstrates that the surface of the sea is
convex. •
The sun is observed sooner at rising and later at setting by a person at
the mast-head of a ship than by one on deck ; and so is the moon and all
other celestial objeets.-^These phenomena evidently arise from the sphe-
rical figure of the earth ; and are, therefore, most convincing and satisfac-
tory proofs of Its globularity.
Again.— The continual presence of the «un, above the horizon, during
the space of several months in the neighbourhood of one terrestrial pole,,
while at a place equally distant from the other, he is as long absent, affords
another convinchig proof that the earth is of a spherical figure.
The spherical figure of the earth may be also- inferred from the method
of levelUngj or the art of conveying water from one place to another ; — for,
in this operation it fs always found necessary to make an allowance
between the true and the> apparent levels on account of the rotundity of
the earth ; the true level being a curve line which falls below the straight
lirie of apparent level about 6 inches in 1 mile ; 32 inches in 2 miles $ 128
inches in 4 miles, &c., the curvature always augmenting in proportion to
the square of the distance. See Problem X., between pages 545 and 547.
Finally,-' All die planets arie observed to be of a spherical figure } and
since the earth is a planet, subject to the same laws, and revolving round
the sun in the same manner as the other planets, it must, therefore, by
analogy, be also spherical. '
The irregularities on the earth's surface, occasioned by mountains and
TalHes, are very inconsiderable compared with its magnitude ; and take off
no more from its actual rotundity than the little risings on the coat of an
orange do from the rotundity of that fruit 2— for the highest eminence or
mountain bears a less proportion to the magnitude of the earth than the
smallest grain' of sand does tor an 18-inch globe. — ^Thus*,
The summit of Chimbora$o, one of the Andes Mountains, and the high-
est in the known world, is only 20280 feet above the level of the sea,
or not quite 4 mtle^ in perpendieular height. — ^Now, the radius of the
earth is 30902200 feet> and that of an 18-inch j^obe 9 inches ;^hence, by
the rule of proportion, as 20902200 feet : 20902200 feet 4- 20280 feet
s 20922480* feet : ; 9 inches to 9. 0087 inches i from which deduct 9
inches (the radius of the artificial globe,) and the remainder 0. 0087 is the
relative elevation of Mountf Chimbora^o on an 18'inch globe ; and as this
is scarcely the one hundred andjl/ieenth part of an inch, it. is, therefore,
considerably less than 'a common grain of sand. — Hence it is evident
that the highest mountidns, and deepest vallies, take little or nothing from
the earth's rotundity.
0^^ A. COMPENDIUM OP PRACTICAL NAVIGATION.
• Although when speaking of the earth in general terms, it may be <
dered as a globe, or sphere ; yet, in strictness it is not a perfect sphere, but
rather 'an oblate spheroid ; which is a solid generated by the re?oliitionof
a semi-ellipse about its shorter axis or diameter ;— and actual admeasure-
ments, in sundry places, have clearly proved that the polar axis, or diame-
ter is about 35 miles less thati the equatorial diameter. — However, »iice
the earth* differs so very little from a globe or sphere, it may, ^erefose, be
very safely considered as being perfectly spherical in all nautical ealcnlatioos
whatever.
The spherical figure of the earth being (bus satisfactorily established, its
jnagnitude may be determined by measuring a small portion of a meridian,
and observing the zenith distances of one or more stars at the extreme sta.-
tions; then, the difference, between the zenith distances of the same star
gives the correspondent celestial arch.— :Now,
As the celestial arch, thus found, is to the measured or intercepted por-
tion of the qaeridran ; so is one degree, to its absolute length in the same
measure in which the portion of the meridian was {aken.
In this manner the celestial arch of one degree has been fovmd to con-
tain 69. 093 English miles ; and since the earth's circomference, like that
of all other spheres, contains 360 degrees; thereforie 360 degrees x 69.093
miles = 24873. 48, is the true measure of the earth's circumference in
English miles. — Hence, its diameter is 7917^ miles, English measure.
Now, since the. nautical arch is, in every respect, equal to the celestial
arch, the length of a degree in the one being precisely equal to the length
of a degree in the other, each. containing 60 geogri4>hical miles j and since
the measure of a degree of this arch in English miles, is 69. 093, or 364815
English feet;— therefore 364815 feet •+- 60 miles = 6080 feet; whiefa,
evidently, is the true length of a nautical mile, expressed in English mea-
sure.— ^And, if 6080 feet be divided by 120 (tiie numbev of half minutes in
an hour,) the quotient 50 feet and 8 inches will be the true measure of a
knot.— And, hence the principles upon which the measure of a knot upon
the log line has been determined.
But, because it is safest to have the reckoning a-head of the ship, 48
feet, or 8 fathoms are, therefore^ commonly allow^ between every two ad-
jacent knots oil the log line : — and this measure is to correspond to a g^ass
running 30 seconds ; or, rather 29| seconds, so as to make up for any time
that may be unnvoidably lost in the act of turning the gla
jRemarL— The instruments made use of for measuring angles ats^ and
for ascertaining the latitude and longitude, are quadrants, sextants, and
reflecting circles. Since, however, splu^e cannot be afforded in this weak
for giving particular descriptions of these instruments, and the man-
ner of adjusting and using them; the reader is, therefore^ reqpecCfydly
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referred to an ocular examination of the iustniments, and to a few
explanatory remarks from some person practically acquainted with the
various uses to which they may, or can be applied. — A few hours practical
•instructipn will convey more real information to a person on these sub-
jects, than if he were to spend a whole year in poring over the voluminous
descriptions which have been puUished, by different authors, relative to
the use of the quadran^ and sextant.
Instead, therefore, qf wasting time and paper with descriptions that may
be well omitted, we will here endeavour to describe that which is of far
more importance to the practical navigator, viz. .
The true Method of finding the Index Error of* a Sextant, so as to
guard against the Errors arising from the Flexibility and the
* Friction of the Index Bar.
The customary method of finding the index error of a quadrant or sex-
tant (as directed by writers on the use -of these instruments,) is by measuring
the vertical diameter of the sun to the right and left of 0 on the arch, with
a motion of the index in contrary directions (that is, by bringing the re-
flected image to touch the lower and upper limbs of the direct object alter-
nately), and theh taking half the difference of those measures for the index
error of the instrument. — ^lliis method, it must be observed^ is very far
irom being correct ; because it is the hocizontal diameter of the sun, and
not Us vertical diameter, that should be n^easured; for, while the former
remains invariably the same, the latter is subject to continual alterations
owing to tlie effects of atmospherical refraction, as will appear evident
by an inspection of the last column of Table V.— Moreover, since the index
is not an inflexible bar, and since it doe» not turn upon its centre without
suffering some slight ' degree of friction ; it is therefore evident that the
measure of the sun*s diameter taken by the progressive motion of the index
will, in most cas^s, be more than the truth.: whilst that taken by the con-
trary or retrogressive motion will, in general, be less than the truth :—
hence, the index error established upon the above principles must frequently
mislead the mariner by rendering inaccurate what, otherwise, might be a
very correct observation. . And this accounts for the result of the evening
observations, taken on shore by means of an artificial horizon, so very sel-.
dom agreeing with the result of those taken in the morning ; even though
all imaginable care be used, and though the observer keeps the same plane
and roof of the horizon directed to him during the time *of both obser-
vations.
Now, to guard against the errors arising from the bending and the fric-
tion of the index bar, as well as that proceeding from the contraction of
the sun^s vertical diameter ; let the following observations be attended to.
in finding the ind^ error of a quadrant or soxtanty and the coojmat ciect
of thoie errors will be obviated.
JFltr^.— To find the Error for a Progrestive Motioii of the Index :--
Screw the inverting telescope into its place. Slack the index. Tun
the tangent screw backward to nearly as far as it will go. Put the noniai
of the index about 1?151 to the right of 0 on the arch, and then fiwten
the index sufficiently tight for. observation.-— Hold the sextant so that its
plane may be parallel to the horixontal diameter of the sun : direct the
sight to that object, and turn the tangent screw forward imtil the limbs of
the sun seen by reflection and direct vision make a perfect contact. — ^Note
down the angle and it will express the measure of the sun's diameter to
the right of 0 on the arch.^Direct the sight ag^n to the sun, and turn die
tangent screw gtill forward until the opposite limbs are in perfect contact :
note down the angle, and it will be the measure of the sun's diameter to
the left of 0 on the arch.— Now, if both measures of the diameter are the
same, there i^ no error in the angles shown by the progressive motion of the
index ; but if those measures do not correspond, half their difference is to
be taken as the index error of the instrument, which error will be additive
when the diameter measured to the right of 0 exceeds that measured to
the left ; otherwise, subtractive.— Then, this error is to be considered as a
constant quantity (so long as- the instrument does not meet vrith any aeci^
dent), and to be applied to allwcreasifig angles, either of altitude or dis-
tance, which may be taken by th^ progressive motion of the index*
Jgainr^To ^nd the Error for a Retrogressive Motion of the Index :--
Slack' the index. Turn the tangent screw forward to nearly as &r as
it will go. Put the nonius of the index about 1°.\5' to the I^ of 0 on the
arch, and then fasten the index sufficieudy tight for observation* — ^Hold the
sextant as before ; direct the sight to the sun, and turn the tangent screw
backward until the limbs of the sun seen by reflection and direct vision
make a perfect contact : — ^note down the angle, and it will express the mea-
sure of the sun's diameter to the left of 0 on the ^rch*:— Direct the si^t
again to the sun, and turn the tangent screw still backward until the op*
posite limbs are in perfect contact ; read off the angle, and it will be the
the measure of the sun's diameter to the right of 0 on the arch* — Now, if
both measures of the diameter are the sam^ there is no error in the angles
shown by the retrogressive motion of the index : but if those measures do
not correspond, half their difference is to be taken as the index error of the
instrument ; which error will be luiditive when the diameter measured to
the right of 0 exceeds that measured to the left ; otherwise, subCractive*
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Then, this error is to be considered as a constant quantity (so long as the
instrument does not meet with any accident)^ and to be applied to all de-
creanng angla, either of altitude or distance, which may be taken by the
backward or retrogressiire motion of the index.
Hence it is very probable that two errors may be established for the
same ii^trument } the one for increasing, and the other for decreasing an->
gles* The true values of those errors should be noted down (for the future
guidacce of the observer,) with a black-lead pencil on the inside of his sex-
tant case in the following manner, viz. i-^
Error for the. forward or progressive motion of the index OMO? sub-
tractive. .
Error for the backward or retrogressive motion of the index IMO*
additive.
Or. whatever the errors may be.
And thus the correct values T)f the index error will be properly deter-
mined, whilst the errors arising from the spring and the friction of the
bar, together with that proceeding from the contraction of the sun's ver-
tical diameter will be all safely provided against. .
OP TAKING THE ALTITUDE OP A CELESTIAL OBJECT
BY MEANS OP AN ARTIFICIAL HORIZON.
Since the generality of nautical persons do not appear to be sufficiently
acquainted with the manner of applying the necessary corrections to angles
of altitude taken by means of an artificial horizon; the author is, there-
fore, induced to make a few observations' touching the direct application
of tliose corrections ; in doing which some hints wiU be thrown out for the
guidance of young observers, relative to the nature and use of the astro*-
nomical instrument now under consideration.
Cf the Artyicial Horizon.
In settling the positions of places in-land in an astronomical manner,
or- in ascertaining the error and the rate of a chronometer on shore where
there is not an open and commanding view of the sea horizon, the ob-
server must, ia all such cases, have recourse to an artificial horizon for the
purpose of taking the necessary angles of altitude.
Although there is a great variety of artificial horizons now extant, yet,
for the sake of conciseness, I shall only touch upon the two that are in my
own posscssion*-^The first of these consists of a plane spetulum, or polished
656 OF TAKING TMS ALTITUDB OF A -CBLESTIAL Q^BCT.
ptete of glass (4 inches long by 3 inches broad,) fixed in a brass frame,
and standing upon three adjusting screws : by means of these and a spirit
level, placed in different positions on its surface^ it may be made perfectly
parallel to the plane of the horizon : observing that the adjusting screws
are to be turned until the air-bubble rests in the middle of the spirit level
on the surface of the speculum. — ^The other is the common, or quicksilver
horizon ;— this simply consists of a small wooden trough, about half an
inch deep, 3i inches long, and 2^ inches broad ; — ^into this trough « few
pounds of mercury or quicksilver are poured ; the surface of which as-
sumes when settled, agreeably to the nature of fluids, an exact horizontal
plane. — To prevent the mercury from being ruffled or sigitated by the*ac-
tion of the wind, a roof is placed oyer it, in which are fixed two plates of
glass, the two sides of each plate being ground mathematically plane and
parallel to one another: — And,.of all artificial horizons an instrument of
this description is the very best that. cai\ be employed in taking the
altitudes of the heavenly bodies.
Of the Use of the Artificial Horizon; that «, to observe the Altifude of
the Stm, or other Celestial Object, with a Sextant, and an Artificial
Horizon.
In taking the altitude of the sun, or other luminary, the observer is to
place his artificial horizon betwixt him and the object selected for observa-
tion ; and at such a convenient distance as tp see the image of that object
reflected from the middle of the quicksilver as well as the real object in the
heavens : — then, having screwed the plain tube,, or the natural telescope
of the sextant into its place in the socket ; and placed one or two of the
dark screens, according to tlie brightness of the sun, to intervene on each
side of the horizon glass ; the lower limb of the reflected image of the
sun, as seen through the erect or natural telescope, is to be brought into
contact with the upper limb of the image reflected from the artificial ho-
rizon : — but, if the altitude of the upper limb of the object be irequired, it
must be brought into contact ^lith the lower limb of the image as seen in
the artificial horizon. — Now, the angle on the arch of the sextant being
read bff, and the index error, if any^ applied to it, the result will be the
double of the sun's, or other object's altitude above the horizontal plane :
to the half of which, if the object be the sun, let the semi- diameter, re-
fraction and parallax be applied, and the true central altitude will be
obtained.
Remarks.
Since neither the plain tube, nor the natural or erect telescope can be
depended upon in taking observations when rigorous exactness is required;
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BY MJEEANS OF AN ARTIFICIAL HORIZON. 657
the inverting telescope should^ therefore, be invariably made use of in all
cases where angles of altitude are to be measured with astronomical preci-
sion : — and here, jperhaps, it may not^ be unnecessary to state that when
the inverting telescope is used, the lower limb of the sun, or moon, will ap-
pear to be the upper limb, and conversely.*— Hence, in observing the alti-
tude of the lower limb of the sun or moon, the apparent upper Umb of the
object, as 'seen in the horizon glass through the inverting telescope, is to
be brought into contact with the lower limb of the image in the artifidal
ftomon.*— in this case the reflected image in the Rrtificial horizon will ap-
pear to be uppermost. — ^Again, in observing the altitude of the upper limb
of £he sun or moon, the apparent lower Umb of the object, as seen in the
horizon glass of the sextant though the inverting telescope, is to be brought
into contact with the upper Umb of the image in the artificial horizon .—in
this case the reflected image in the artificial horizon will appear to be un-
dermost.
If an observer be placed as remote from, or as near to, an artificial hori-
zon as possible, the rays of light passing from the sun or other celestial
object to his eyej and from that object to the surface of the artificial hori-
zon, will, on account of the immense distance of such object from the
earth, be physically equal and parallel in every respect to each other :— -
hence, it is- easy to perceive that it is immaterial whether the artificial hori-
zon be placed high or low, remote or near with' respect to the observer,
provided he can but ^ee the object's reflected image Iherein.
When an angle of altitude is taken by means of an artificial horizon,
its measure on the limb of the sextant will always be the double of die
true value thereof above the horizontal plane ^7-this will appear evident by
considering that if a person places himself at any distance before a plane
mirror, or common looking-glass, his reflected image will appear just as far
behind such looking-glass as he is before it : — and, upon this simple prin-
ciple it is that the reflected image of the sun, or other object, will appear
to be as. far below the surface of die artificial horizon as the real object is
above it 5— but since the limb of the real object, as reflected from the index
glass of the sextant, is to be brought into contact with that of the image
apparendy reflected below the surface of the artificial horizon, it is there-
fore manifest that die contained angle, as expressed on the arch of the sex-
tant, must be equal to twice the measure of the. observed angle of altitude
above the plane of the horizon :— and from this we may readily perceive
that angles of altitude taken in the above manner are not affected by
the angle of horizontal depression, commonly called '^ the dip of the
horizon.'' ♦
• See page 387.
2u
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Q^b - OF TAKING THB ALTITUDB OV A CBLBSTIAL OBJECT,
Now, the double angle of altitude being thus obtained, the true altHnde
of the object is to be deduced therefrom in the following manner, viz. :-—
l^r«f •— To apply the Corrections when the Sun is observed :^*
Correct the observed angle for the index error of the sextant, if any ; — to
the hsdf of which apply the sun's semi-diameter by addition if the lower
limb be observed, but by subtraction if it be the upper limb, and the sun's
apparent central altitude will be obtained ; from which let the difference
between the refraction and parallax ^corresponding thereto be subtracted,
and the remainder will be the -sun's true central altitude.
Seeond^^To apply the Corrections when the Moon is observed ^—
Find the moon's apparent central altitude in the same manner ays if it
were the sun that was under consideration ; observing to correct her eemi-
diameter by the augmentation contained in Table IV. ;— then, to the ^-
parent altitude, thus found, let the correction in Table XVIIL be added|y
and the sum will be the true altitude of the moon's centre.
l^d.— To apply the Corrections when a fixed Star is observed : —
Correct the observed angle for the index errOr of the sextant, if any;
from the half of which subtract the refraction corresponding dierelo^ and
the remainder will be*the. star's true altitude.
EmmpU U
Let the measure of the observed angle between the lower limb of the
sun reflected from the index glass of a sextant, and the upper Kmb thereof
reiected from an artificial horizon be 103° 14^40^ 5 the index error of the
sextant 3U0T additive, and the sun's semi«diameter 16C18?; required
the true altitnde of the sun's centre }
Measure of the observed angle := , • 103?14C40T
Index error = , . +3.40
Corrected observed angle = • ...;.... 103? 18^205
The half of which is the correct observed altitude of
the sun's lower limb above the plane of the hor« x: 51?39' 107
Sun's semi-diameter =•••••« 4 • p 9 + 16* 18
Apparent altitude of the sun's centre = • • , • • 5I?55^28T
Refraction answenng to ditto = 0'A4^.\ ,.^
T*^ ti 1. « r^ ^ \ diflFerence =
P&rtdlax, ditto ditto 0. 5 >
- 0.39
Sun's true central altitude = . . . . . , . ., . 5K54M97
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BT MEANS OP AN ARTlflCIAL HORIZON. 659
Example 2.
Let the measure of the observed angle between the upper limb of the
moon reflected from the index glass of a sextant and the lower limb there-
of reflected from an artificial horizon be 39^50^20? ; the index error of the
sextant l',50^ svibtractive, the moon's semi-diameter 14^46^> and her
horizontal parallax 54' 13?; required the true altitude of the moon's
centre?
Measure of the observed angle IS • 39?50'.20'r
Inde^i error ss • ^ •• • • -* 1.50
' ■ ■ ■ ^
Correctedobserved angle = ., ..,•••.,. 39?48'30i:
The half of which is the correct observed altitude of the
moon's upper limb above the plane of the horizon ^ I9?54' 15T
Moon's semi-diameter =:s . . 14'46? ) « i^ ei
Augmentation of do, 1 ab. I v , =: U. 5 j
Apparent altitude of the moon's centre = . • • • 19?39^24T
Correction of ditto, Table XVIIL = . . . • . . +48.25
Moon's true central altitude = . . * . . • • . 20?27'49r
Example 3.
Let the measare of the observed angle betweeiK the centre of a fixed
star reflected from* the index glass of a sextant and the centre thereof re*
fleeted from an artificial horizon be 71? 16 C 10?, and the index error of the
sextant 2 '.30? subtractive ; required the true altitude of the star ?
Measure of the observed angle =s * 71? 16^ 10?
Index error as ••••••••.•.'•• -* 2,30
Corrected observed angle « 71?13'40?
The half of which is the correct observed altitude of
the star's centre above the. plane of the koriaon bi , • 95. 36. 50
llefraetion conrespondlng to ditto ^ ..••..-^ 1. 19
*■ ** III ■
Tniesltitadeeftlit»tara. 95?S5;31?
Note.'^'When the altitude of a celesUai object exceeds. 60 degrees, it
cannot be taken by means of a sextant and an artificial horizon ; because,
2u2
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• 660 OF TAKING THJt ALTITUDB OF A CBLB8TIAL OBJECT.
in this case, the measure of the double angle of altitude will exceed the
limits of the graduated arch of the former instrument.
Remarks.
1. In observing equal altitudes by means of an artificial horizon, or in
taking a continued series of altitudes for the purpose of determiniDg the
error and the rate of a chronometer ; it will be essentially necessary to
keep the same plane of the glass roof of the horizon towards the observer
in each observation ; so that in the event of there being any trifling defect
in the parallelism of the surfaces of the two plates of polished glass, which
form the roof of the horizon, the error arising therefrom may equally
affect each observed altitude. To make certain of alwap having the same
side of the roof next the observer, it will be advisable to make a small
mark in the wooden part thereof': then, this mark being kept towards the
observer, in every observation, the altitudes will thus be prevented from
being unequally afiected by any want of parallelism that may chance to
be in the phme» of the glass part of the roof.
*
2.— In calm weather the altitudes may be taken by reflection from the
quicksilver without making use of the glass roof :— in like manner they
may be taken, during such weather, by reflection from a bason of water ;
or, by reflection from a cup of tar, treacle, oil, or other fluid and viscous
substance.
3.— Mariners frequently supply themselves with, what may be termed, a
home-made, or ship-built artificial horizon ; the quicksilver in which they
shelter under a roof formed by two squares of the thick glass with which
ships are usually furnished :— this is, to say the least of it, a poor substi-
tute :— it is vainly endeavouring to accomplish that, by memis of a couple
of squares of common glass, which can scarcely be effected by the most
highly-finished and parallel planes that can possibly be produced, by the
labour and ingenuity of the most eminent optician, or mathematical in-
strument maker ;-.indeed, it is a most absurd and mistaken contrivance j
and a pertinacity in its use betrays an evident deficiency of usefiil know-
ledge on the part Of the proprietor :— for, since the surfaces of those
squares are not rendered mathematically accurate by being ground
perfectly plane and parallel to one another, they cannot possibly refract
and reflect the rays of light in an exact uniform manner :— hence, the
angle of incidence will not be equal to the angle of reflection; and thus
the angles of altitude observed in such a mahe-sf^ft and defectioe horizon
must and will be always erroneous.
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A NSW MBTHOD OF TINBINO THB XX>KOrnrDB« 661
A NEW AND CORRECT METHOD OF FINDING THE LONGI-
TUDE ON SHORE, AND, IN SOME CASES, AT SEA;
Which, besides being remarkably simple, will be found equally as strict as
the common Method by the Lunar Distances, and, at the same time,
considerably more practicable than that Method : for most Mariners
must be aware of how extremely difficult it is to. take an accurate Lunar
Observation at Sea, particularly when the Ship rolls rapidly, or pitches
with toy degree of violence ; but few find any difficulty whatever in
measuring the. Altitude of a Celestial Object, provided the Horizon be
sufficiently clear;— and, these points betag premised, we will now
proceed to the Solution of the following
Problem.
Given the Latitude of a Place or Ship, the observed Altitude of the Moon's
well-defbied Limb, and the apparent Time of Observation : to find the
Longitude of that Place or Ship.
RULB.
To the apparent tinie of observation (always reckoned from the preced-
ing nooii), add the longitude by account, in time, if it be west, or subtract
it if east, and the sum or remainder will be the estimated time at Green-
wich ; to which let the sun's right ascension, at the noon preceding the
Greenwich time, be most carefully reduced ; and let the moon's declinationj
at the period preceding the Greenwich time, be also carefully reduced to
the same time. .
To the sun's reduced right ascension, add the apparent time of observa-
tion (rigidly determined) ; and the sum (abating 24 hours, if necessary,)
will be the right ascension of the meridian/ which convert into motion or
degrees.
Reduce the observed altitude of that part of the moon's well-defined
limb which is either nearest to or farthest from the horizon, according as
the lower or upper limb may be observed, to the true central altitude, by
Problem XV., page 323. Then, with the moon's true central altitude,
thus found, her corrected declination, and the given latitude, compute her
angufa&r distance from the meridian. Now, if the moon were in the eastern
hemisphere at the time of observation, let her angular distance from the
meridian be added to the right ascension of the meridian ; and the sum
(diminished by 360 degrees, should it exceed this quantity,) will be the
moon's correct right ascension ; but if she were in the western hemisphere,
her angular distance is to be subtracted from the right ascension of the
meridian (increased by 360 degrees, if necessary) : the remainder will be
the moon's correct right ascension at the time and place of observatiofn.
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662
A HSW ftfBTHO0 OP riNpiNG THS UyKOmTOB.
With the moon's correct right ascension^ thus found^ enter page VI. of
the month in the Nautical Almanac^ opposite to the given day, or to that
which immediately precedes or follows it, and take out the next less and
the next greater tabular right ascensions ; find their difference, and find^
also^ the difference between the computed and the next less tabular right
ascension. Now, from the proportional logarithm of the last difference,
subtract the proportional logarithm of the first difference : the remainder
will be the proportional logarithm of a portion of time, which being added
to the hour standing over the least tabular right ascension^ the ^m will
be the apparent time at Greenwich; the difference between which and the
apparent time of observation, will be the required longitude in dme; — east,
if the time at ship or place be the greatest; otherwise, west.
Note.— The above rule is made out on the assumption that the moon's
place in right ascension and declination is given in page VI. of tlie month
in the Nautical Almanac, agreeably to the form exhibited in the page at
the end of this problem. It is evident that this rule may be very consi-
derably contracted, but the author gives it thus in detail with the view of
making himself the more clearly understood, — so that the reader may not
mistake his meaning.
Examipk 1.
September 8th, 1826^ tn latitude 40?27'dO? south, and lonptude, by
account, 30?40C west, at IMQTl?- apparent time, the true altitude of
the moon's centre, ea^t of the meridian, was 31?55^23l'; required the
true longitude of the place of observation ?
lUutiraiUm.
In the annexed diagram
(projected stereographically
upon the plane of the meri-
dian^) let the primitive circle
Z H N O represent the plane
of the meridian of the. place
of observation; SP the earth's
axis ; £ Q the equator ; H O
the horizon; and ZN the
prime vertical, in which Z
represents the zenith. Let the
small circle a a represent the
parallel 6f the moon's alU tude,
and that expressed by 6 6 the
parallel of her declination.
H
^>
\\
k^
/
X
K 1
\ /
^y^
w
\/
"^^
W
K
^
y
N
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A KBW urnmoD or finding tbs lonoituds.
663
rPhe intersection of those small circles in J> shows the moon's place in the
Heavens at the time of observation. Draw the oblique circles S ]) P and
Z }) N. Then, in the oblique angled spherical triangle Z ]) S, there are
^ven the three sides, to find the hour angle Z S ]) ; viz., the side D Z :£■
t;he co-altitude or zenith distance of the moon, the side ]) S ^ the moon^s
cHstance from the elevated pok of the equator, and the side S Z ss the co-
latitude of the place; to find the moon's distance from the meridian
expressed by the angle at S ; which angle is. easily determined by calcula-
tion, and )vhich being compared with the right ascension of the meridian
(by addition -or subtraction, according as the object may have been in the
eastern or western hemisphere at the time of observation,) the moon's
correct right ascension will thus be obtained, and, hence, the apparent
time of observation at Greenwich.
Ckmputaiwn*
Apparent time of observation ss . . . • . i?10?17'
Longitude by account 80?40' W., in time = +2. 2. 40
Estimated time at Greenwich
3M2T57:
Sun's reduced right ascension « 11^ 6?11'.2
Apparent time of observation a 1.10.17
Right ascension of the meridian m 12n6?28\2 s 184?7'3?
Moon's reduced declination s 20?49'3r south.
Moon's tnie central altitude =k 3 1 ?55 '. 23f
Moon's south polar distance = 69. 10.67 Log. co-secant = 0.029319
Latitude of the place of obs. = 40. 27. 30 Log. secant = . 0. 1 18685
Sumsss . 141^33^50?
Half sum = 70M6^55r Log. co-sine = 9.517413
Remainder = . . . • . 38.51.32 Log. sine = . 9.797547
Sum= 19.462964
Half the hour angk= . . 32°36^24r Log. sine = . 9.731482
i '8 angular dist. east of mend. 65^2^ 48?
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DIM A NEW MSTHOD OP FINDING TBB IjONGmUXB.
Moon's angular dist. E. of merid.^? 12M8r
Right ascension of the inerid.= 184. 7. 3
J'sILA.attime&placcofob8.249?19:5iri|^-^, q^ i;^ k_ , ^.^t
rsRJLperNant.Al.atIIlhrs 249. 11.46 i^"' ^^ ^^^^-J-^'?
J's ditto ditto atVIhr5251. 4.53 Jl>iff.l.53. 7 Rlog. .201/
Portion of time = 0M2?52! Pn^.log.= iTufiO
Time corresponding to least tabular R.A.=3. 0. 0
Apparent time of obsenr. at Greenwich = 3M2?52!
Apparent time at the place of obsenr. = 1 . 10. 17
Long, of the place of obsenr., in time = 2? 2?35 ! = 30?38!45f wtaL
iVbte.— The latitude and the altitude are made use of directly^ in the
above calculation, instead of their complements, so that the hour a&ele
may come out a logarithmic sine.
Example 2.
September I4th, 1826, in latitude 47?12^20r north, and loiigitnd«^ by
account, 38?47' east, at 14*58r38! apparent time, the tnie altitude of
the moon's centre, west of the meridian, was 16?24'18r; rapured the
true longitude of the place of observation ?
Apparent time of observation = .... 1 4 1 58?38 !
Longitude by account 38M7 ' cast, in time = — 2. 35. 8
Estimated time at Greenwich = .... 12?23T30!
Sun's reduced right ascensions 11 ^297 8'. 3
Apparent time of observations 14.58.38 .0
Right ascension of the mend. == 2? 27r46'.3 = 36?56^34if
Moon's reduced declination = 4?24'0f south.
Moon's true central altitude = 16?24 ^Sr
Moon's north polar distance = 94. 24. 0 Log. co-secant = 0. 001282
Lat. of the place of observ. = 47. 12. 20 Log. secant = .0, 167S93
Sum = 158? 0^38r
Half sums 79? 0^9? Lpg. co-sine =s . 9.280393
Remainders 62.36. 1 Log. sine = . .9.948324
Sum s 19.397892
Half the hour angles • . 29?59(53r Log. sine s= , . JJ. 698946
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Pom.- ■*-XV,|, ,
^•»? taken „,f eh °'''''»^^o^ '
«»o»„. '^^"^'^nt time of oi
^e will heft. . '
^^^ to eve
f"**"' centre r^«^S*2f
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I
6M A NSW MBTHO0 Of KNOINO THB IX>NGITUDB.
Moon'8 tnie central altitude = 17*41 UO^
Moon's north polar distance = 76.22.22 Log. co-secant = 0. 012401
Lat of the place of observ. = 50. 4 S. 15 Log. secant ^ . 0. 199301
Sumr= 144?52'.17^
Half sum = 72? 26 C SirLog. co-sine «B . 9.479683
Remainder = 54« 44. 28^ Log. sine = . . 9. 91 1984
Sum:s 19.603371
Half the hour angle = . . 39^8^ 8r Log. sine = * . 9.80I6JS5|
J 's angular dist. west of merid. 78?36( 16r
Right ascension of the merid. 106. 8. 10§
])'8R.A.attime&placeofobs.=27?31'54^^ I T\-irn«^i / jrt#ri>t /m»^/«
r.iUV.perNautAL.tXVIII.h.26.50. 9 |l>'ff.0?41.45irPJog.,6346
Moon's ditto ditto atXXI.h»28. 19. 16 JDiff.1.29. 7 P.log..8a53
Portion'of time = . l?24r20:Prop.log.=.8298
Time corresponding to least tabalar R. A. = 18. 0. 0
Apparent time of observation at Greenwich3Ll9i24T20!
Apparent time at the place of observation b 19. 20. 0
Longitude of the place of observ., in time = 0^ 4^20: = l?5'0r west;
which is the correct longitude of the given place*
Remarks.
In finding the longitude by the proposed method (which is evidently
founded on the most natural and unerring principles,) the moon's ri^t
ascension becomes the principal element in the calculation; aud since this
must be deduced from her true central altitude, the observed altitude of
her limb must be taken with all imaginable care : for which purpose the
observation should necessarily be made with a sextant or circular instru-
ment, and the inverting telescope ought invariably to be used, particularly
if the altitude be observed at night *, moreover, \t is that part of the moon's
round or well-defined limb (upper or lower, as the case may be>} which is
perpendicular to the plane of the horizon, that must be brought donin^ to
the surface of the water or sea.
The most favourable time for observing th« moon's altitude, ao as to
obtain the longitude to the greatest possible degree of exactness, is when
her change of altitude is the quickest ; and this always happens when she
is in or near tq the prime vertical ; that is, the east or weH point of the
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horizon. The altitude^ however^ should not be less than 5 degrees^ on
account of the uncertainty of the atmospherical refraction near the horizon;
nor should the object's distance from the meridian be less than 3 hours^ or
45 degrees. •
Since an error of 1 minute of a degree in the moon's computed right
aaeension will have . the same effect upon the deduced longitude that an
error of the same value has in the computed distance by the method of
the lunar observations,* it is'' essentially requisite that several altitudes of
the moon's round or well-defined limb be most carefully observed, and the
corresponding apparent times per watch noted down : the sums of these,
divided by their number, will give the mean altitude and the mean corre-
sponding apparent time.; which should be depended upon in preference to
any single observation* If there be an assistant observer to take the altitude
of the sun, a planet, or a fixed star, at the same moment that the principal
observer takea the altitude of the moon's well-defined limb, the apparent
timci may then* be deduced from the altitude of such object ; and thus any
imperceptible irregularity in the going of the watch, since the last time of
ascertaining its error, will be provided against.
In a very rough sea, and when the ship rolls or pitches considerably, it
wiH be most highly advisable to multiply the observations, and to take the
mean as the true result.
It is to be noted, however, that the sextant must be properly acyusted,
07 the value of its index error very carefully determined, by the method
in page 653 ; the instrument must then be held in a direct vertical posi-
tion, so that its plane, if produced, would meet with an imaginary plumb-
line passing throujfh the moon's centre, and let fall from that gart of
her well-defined limb which is exactly perpendicular to the plane of the
horizon^ and which is either the nearest to or the most remote from the
said plane s any deviation from this position, either to the right or left,
will make the angle of altitude something more than the truth.
As it may be rather difficult to make a perfect contact between that part
of the moon's well-defined limb, thus indicated, and the horizon imme-
diately under, when she is so posited in the heavens that an imaginary
straight line joining the cusps of her horns is at right angles, or perpen-
dicular to the plane of the horizon, (a position in which she is at times
during the two or three days preceding and following her conjunctive
Sy^ygia,. or change,t) it will therefore be advisable to take her altitude at
* This I St a mean rate, will b« ftbout 27§ milet in places under the equatpr ; but since
the error decreases in proportion to the co-sine of the latitude, it will only amount to about
17i miles in tbe parallel of Portsmouth.
f The moon is never exactly in this position, except when she arrives at the nona^esimal
degrSe ; that it, the 90th degree of tbe ecliptic above the borison i and then she is too near
to the meridian for observation.
668 A NSW MSTHOD OF FINDING THB LONGITUDE.
some convenient moment before or after the time of her being ia this
mifavourable position ; observing, however, that the moment so chosen be
sufficiently far from the period of her passing over the meridian of the
given place.
It will appear very perceptible, from what has been thus adduced, that
the proposed method of finding the longitude possesses a most decided
advantage over that by the lunar distances ; because, while most marinerB
are found competent to take a very correct altitude of a celestial object,
few are found sufficiently qualified to m^easure the angular distance between
the moon and sun, or a fixed star, to that degree of precision which is so
indispensably necessary to the obtaining of the true longitude ; particularly
when the object seen by reflection is to the left-hand of, and considerably
lower than, that seen by direct vision. Besides, there is, at times, a very
considerable degree of uncertainty attendant on the admeasurement of a
lunar distance from the sun or a fixed star ; for the objects approach or
recede from each other so very slowly, that the eye of the observer is very
frequently deceived, unless aided by a high magnifying power ; and this
cannot always be used, when the ship suffers a violent degree of agitation :
hence it commonly happens, after making the contact between the limbs
of the objects apparently perfect, that, on directing the sight to them
again, their limbs will appear to be separated, or, perhaps, entered upon
each other ; and this separation, or entering, in direct opposition to the
absolute motion of the moon. But if the horizon be clear, and the moon
at a proper distance from the meridia$i^ all uncertainties vanish in bringing
her well-defined limb in contact with the visible horizon expressed by the
convex surface of the sea j because she then rises or falls so] very rapidly^
that a careful observer may take the altitude of her limb to the sixth part
of a minute ; and this is a degree of exactness that can very rarely, if at
all, be introduced into the measured lunar distances taken at sea.
Though what has been said here is strictly true in theory, yet it may often
fail in'practice, owing to the uncertainty of the sea horizon: hence it is
evident that the method in question is onit/ properly adapted to the deter'-
mination of the hiigU^tde of places on shore, where the altitudes can* be
correctly taken by mearu of an artificial horizon; and, certainly, .with this
view, it will be found to be one of the very best methods for settling the
true positions of places inland or along the coast.
Were the moon's place in right ascension and declination computed in
the Nautical Almanac agreeably to the form in page 670, it would very
considerably facilitate the finding of the longitude by the metiiod now
under consideration ; for then the mariner would be provided with all the
necessary elements that enter into the. calculation, with the bare exception
of the moon's correct central altitude and the apparent time ; but these, it
is supposed, his own diligence will always furmsh. Moreover, the loxigi«
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^ude could then be as readily inferred from an altitude of the moon's limb
and the corresponding time indicated by a chronometer, as it is now by
that of the sun and the same delicate piece of mechanism.
Note. — Page 670 shows the manner in which the moon's right ascension
and declination should be computed and arranged, so as to answer the
intention of the present problem ; these right ascensions and declinations
may be readily deduced from the elements given, at noon and midnight,
in page* VI. of the month in the Nauticfd Almanac, by means of the
problem for illustrating the use of Table XVIL, page 34. Thus, to find
the moon's right ascension at III., VI., and IX. hours, September 1st, 1826:
First Second Mean
Diff. Diflf. 2dDiff.
Moon'aR.A.atmidnt.,Aug.31st,142M3a2r [5032' 4^
Moon's do^at noon, Sept. 1st, = 149. 15. 16 1 ( 0. 38 .
• '^ J6. 32 42 fl'DiT
Moon's do. at midnt., Sept. l8t,= 155; 47. 58 ^ * ' i 1. 23 ^
^ > 6« 34. 5 )
Moon's do. at noon, Sept. 2d, = 162. 22. 3 3
The variation of right ascension betwee*n noon and midnight of the
given day, is 6?32C42T ; the proportional part of which corresponding to
III. hours, is . 1^38: 10^ 5
Equation of second difference == • . ^ 5. 6
Proportional part corrected = • • • 1?38^ 4^.9
Moon's R, A. at noon, 1st September, =» 149. 1 5. 16
Moon's R. A. at III. hours = . . . 150?53 '. 20^. 9
The propor. part qf 6?32M2r corresponding to VI. hours, is 3?16:2K. 0
Equation of second difference = .^ ...... • — ?• 5
Proportional part corrected = 3?16'13''. 5
Moon's R. A. at noon, 1st September, = 149. 15. 16
Moon's R- A. at VI. bourse 152?3^.29^5
The propor. part of 6?32M2f corresponding to IX. hours, is 4?54^3ir. 5
Equation of second difference = ^5.6
Proporticmal part corrected = 4?54^25''.9
Moon's R, A. at noon, 1st September, s 149.15.16
Moon's R. A. at IX. hours = 154? 9'41\9
The' moon's declination is to be determined in the same manner pre-
cisely ; observing, however, to apply the corrected proportional part by
addition or*subtraction, according as the declination at the preceding noon
OT midnight is increasing or decreasing.
t&
217.
232.
247.
262,
277.
291
305.
318.
330.
342.
6
17.
29.
41.
NoOD,orO* III.*
U9«l&'lb'
162.22. 3
175.38. 2
189. 10.4:1
203. 6.46
19.26
16.25
18.36
21.17
7.59
219.19. 6
234. 8.38
249.11.46
264.13.20
278. 56. 58
25.14
5.50
9. 8
39.44
45.13
14.27
16.18
58.48
48.34
50.19
I. 9.53
306.45.43
319.44.36
332.11.36
344. 14. 34
150»53'2r
164. 0.53
177.18.33
190. 53. 52
204. 53. 8
356. 2.24
7. 43. .55
19. 26. 57
31.18. 4
43.21.31
VI.*
I52«31'30'
165.39.53
178.59.22
192. .37. 22
206. 39. 56
IX.*
221. 9. 8
236. 1. 3
251. 4.53
266. 5. 4
280. 45. 27
222.59.
237. 53.
252.57.
267.56.
282. 33.
294.53.56
308.25. 2
321. 19. 35
333.43. 7
345. 43. 42
357.30.16
9.11.35
20. 55. 17
32. 47. 45
44. .^2. 57
154° 9'
167. 19.
180.40.
194.21.
208. 27.
42*
4
28
16
8
168.58.25
182.21.51
196. 5.32
210.14.46
224.50.17
239. 46. 26
254. 50. 56
269.47.36
284.20.54
296.37.
310. 3.
322. 54.
335. 14.
347. 12.
358.58.
10.39.
22.23.
34. 17.
46.24.
XII.*
157«26' 19"
170.37.58
184. 3.34
197.50.13
212. 2.50
296. 20. 18
311.41.57
324.28. 6
336.45. 4
348.41.18
0. 25. 46
12. 7. 1
23.52.26
35. 47. 45
47.56.30
XV.* XVIII.»
2.^6.4l.2.i
241.39.22
256.43.48
271.38.20
f. 7.49
\b^ A' AT"
172. 17. 46
185. 45. 37
199.35.19
213. 51. 18
228.32.47
243. 32. 22
258. 36. 29
273.28.38
287. 54. 1 1
300. 2.34
313.19.33
326. 1.39
338.15.38
350. 9.49
3OI.44.T5
314.56.37
327.34.46
339.45.43
351.38.10
XXI
173.57.4:
187.28.
2f»1.20.5(>
215.48. lu
i30.24.2:
245.25.2:
260.28,^.^
275.18.31
289.39.59
9113.25.20
316.33. t
329. 7.38
1.15.;
6.23
1.53.26
13.34.50
25.21.12
37.18. 4
49.28.39
3.21. 4
15. 2.44
26.50. 9
38.48.36
51. 0.59
I341I]
J353.
4.441.42
16.30.43
28. 19. 16
40.19.21
52.33.35
8i. 2. 7
77.42.53
90.32L 4
103.25.27
116.19.45
54.
^.
79.
92.
105.
117,
130.
143.
156.
17.0.
6.23
36.37
18.40
8.35
2.14
55.39.25
68.11.18
80.54.S4
93. 45. 10
106.39. I
57.12.41
69.46. 9
82. 30. 34
95.21.48
108. 15. 49
58. 46.
71.21.
84. 6.
96.58.
109.52.
56.31
50.42
46.47
49.38
6.16
119.33.17
132. 27. 32
145.24. 8
158.28.17
171.47.12
121.10. 2
134. 4.23
147. 1.37
160. 7.11
173.28.31
J22. 46.
1135.41.
i48.39.
161.46.
175. 10.
60.
72.
85.
98.
IIL
124.
137.
150.
163.
176.
19.50
56.23
42.54
35.11
29.24
61.53.44
74.31.45
87. 19. 13
100.11.55
113. 6.11
, 63.27.50
I 76. 7.15
I 88.55.36
1101.48. 40
114.42.58
23.34
18.15
16.58
25.44
52.14
126. 0.20
138. 55. 16
151.54.51
165. 5.24
178.34.41
127.37. 6
140.32.21
153.32.56
• 166.45.23
180.17.34
129.13.54
142. 9.32
155.11.11
168.25.40
1S2. 0.52
The Moon's Declination, September, 1826.
Noon,orO* 111.*
70 10'
2.16.
2.50.
7.50.
12.25.
16. 16.
19. 6.
20.42.
21. 0,
20. 0.
0"N
2n.
17 8,
39
42
26
9
5
23
32
17.51,
14.44.
10.55.
6.37.
2. 7.
5
25
4
56 s.
8 s.
2.24.
6.45.
10.45,
14. 15.
17. 7,
a7w
18
13
13
24
19.15.
20.32.
20.55.
20.21.
18.49.
16.22.
1.3. 2.
8.58.
4.18.
0.4.3.
0
23
20
18
3
"10
41
1
2n
.508.
6«34'34"n
1.38. 3n
3.28.348.
8.26.50
12.57.21
16.41.18
19.22.26
20.48.40
20.57. 1
19.47.56
17. 5. 9
19.37.32
20.54. 6
20.52.28
19.34.16
17.30.34
14. 17. 46
10.24. 8
6. 4.32 s.
1.32. .58 s.
2.57.53N.
7.16.38
11.13. 17
14.38.i^8
17.25.55
19.27.31
20.38.18
20. 53. 49
20. 13. I
18.3:1.45
15.59..'>U
12.34.23
8.24.42
3.41. IOn.
1.22.198.
VI.*
5058' 45"n
0.59.56N
4. 6.416.
9. 2.33
13.28.14
17.27.59
19.51.25
20.58.23
20.46.43
19. 19. .34
17. 9.13
13.50.30
9.52.49
5.30.59 8,
0. 58. 50 s.
3.31. 6n
7.47.36
11.40.50
15. 2. 5
17.43.44
19.39.14
20 43.21
20.51.40
20. 3.49
18.17.35
15. 3b. 58
12. 5.25
7. 50. 52
3. 4. On.
2. 0.488.
IX.»
5»ir2'33"w
0.21.4111
4.44.388.
9.37.48
13.58.22
16.47. 2
13.22.37
9.21. 8
4.57.168
0.24.45 8.
4. 4. 6n.
8.18.12
12, 7.54
15.24.36
18. 0.48
19.50. 9
20.47.32
20.48.54
19.53.42
18. 0.35
15.13.10
11.3.5.46
7.16.32
2.26.34N.
2. .39. 17 8,
XII.*
4045' 38"N
0.16.418.
5.22.258.
10.12.34
14.27.43
49
7
31
47
49
1
8
4
24 s,
19n.
54n.
27
28
29
10
16
52
31
40
45_
.35
27
41
5lN.
46 ft.
XV.*
4« 8'37"N
0.55. 38,
5.59.578.
10.46.47
14.56.14
1». 10. 34
20.15.25
21. 1.19
20.31.40
^8^l7. 3
i6.~ 0. 12
12.25. 6
8.16.41
3.49.25 s.
0.43. 17n.
5. 9.26N.
9. 18. 18
13. 0.30
16. 7.43
18.33. 0
20. 9.33
20.59.29
20.40.49
19.30.34
17.23.53
14.28.13
10.34.29
6. 6.22
1.10.54N.
.3.56.118.
xvm.*
XXI.*
353F2PN »i3tAS^s
1.33.27 8. 2.11.51s.
6.37.10 s.
11.20.22
15.23.52
18.30.12
20.25.30
21. 1. 4
20.22.35
18.29.31
18.48.44
20.34.24
21. 0.45
20.12. 2
18.10.41
15.35.40
11.55.35
7.44. 2
3.15.238.
1.17. 8n
6. 18. 41 N.
9.47.42
13.25.58
16.28.17
18.47.37
20.18. 0
20.58.38
20.35.13
19.17.35
17. 4.10
13.5*. i
10. 2.56
5.30.40
0.32.48N
4.34.278.
7.14, 4 s.
11.53.21
15.50.36
15.10.24
11.25.34
7.11. 7
2.41.17$.
I.50.51N.
6.13.38N.
9.56.41
13.50.52
16.48.11
19. 2. I
20.25.37
L 54.36
20.28.52
19. 3.45
16.43.35
13. 30. 16
9.30.46
4.54.33
0. S.26S.
5.1231>.
Tbe author is of opmioB,- that, for the purpose of determinijig the true longitude of plans oq
shore, the method in question will be found prefer»ble to any otbWf pwlicolarly wbtn Hk altlMes
are taken by ineann of an artificial horizon.
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J. u jneauce ine luoon 9 nxgn^ JMMcetwvan ona x/rQimono/i \q any gwen
Time under a known Meridian.
RuLB.— To the apparent time of observatioa (always reckoned from the
preceding noon,) add the longitude, in time, if it be west, but subtract it
if east ; the sum, or remainder will be the time at Greenwich.
With this time enter page VL of the month in the Nautical Almanac,
opposite to the given day, or to that which immediately precedes or follows
it, and take out the right ascensions and declinations at Uie periods directly
preceding aqd following the said time $ find their difference, and, also, the
difference between the said or Greenwich time and the preceding period : then.
To the proportional logarithm of the last difference, add the propor-
tional logarithm of the first difference, and the sum will be the propor-
tional, logarithm of a correction which is always to.be added to the right
ascension at thft preceding period ; but» to be applied by addition, or sub-
traction to the declination at that period, according as it may be increas-
ing or decreasing. — And, thus, the right ascension and declination will be
obtained, sufficiently near the truth for all nautical operations, indepen-
dent of interpolations, or second differences.
This rule is founded on the assumption that the moon's right ascension and
declination are given in theNautical Almanac agreeably to the form in page670.
Examfie.
Required the moon's right ascension and declination, Sept. 18th, 1826^
^t 19^20?0! apparent time, in long. 1 ?5 ^ W. of the meridian of Greenwich ?
Apparent time 19*20? 0!
Longitude 1?5^ west, in time = . . . + 4. 20
Greenwich time 19*24720!
To find the Moon's Right A/M:ension 1 —
Ap.timeatGr.= 19*24'r20'. ) difference = . . 1»24T20: P.L.x=.3293
D 's RA.Sep.l8, at IShours 5 =26?50^ 9T. j.^^ ^039^ 7^ p. L.=. 3058
]> 's ditto ditto at 21 hours s 28. 19. 16 ) .. .
Correction of right ascension = . . .. . . . 0^4 1 US ^ P, L.= . 6346
R.A. at the period preced. the Greenwich times 26.50. 9
Moon'^ correct R.A. at time and place of obs. = 27?31 '541^
To find the Moon's Declination : —
Ap.timeat A.= 19*24?20! 1 difference » . . 1*24*20! P.L.=.S293
D'8dec.Sep.l8,atl8hoursJ =13?25!58rN.|^^Q024/54i^ PL- 8591
D'sdo. do. at 21 hours « 13.50.52 N.i
Correction of declination = + 0?ll!40^P.L.= l. 1884
Declln. at. period preced. the Greenwich times 13. 25. 58 north.
Moon's correct dec. at time and place of obs. s IS'^S? "SST north.
vr rixvj/tnv
J<1J> IjAllAUATASy
Hence, it is evident, that if the moon's right ascension and declination
be given in the Nautical Almanac to every third hour, the reduction of
those elements, to a given time under a known meridian, will become a
more simple operation than the reduction of the sun's right ascension and
declination to a given time and place.
SOLUTION OF PROBLEMS relative to finding the Latitudes and
Longitudes, Right Ascensions and Declinations of the Heavenly Bodies,
and to the computing of the Central Distances between the fdoon and
Sun, a fixed Star, or a Planet
Problem L
Given the Bight Ascension and Declinaiion of a Celestial Objed, tojimd
iU LaHiude and Longitude.
Example.
The apparent right ascension of a Arietis, August 1st. 18^5, was
1*57*23M76, audits declination 22938^4-'. 44 north; required its
apparent latitude and longitude, the obliquity of the eclipdc being
23?27^42-'.875?
a Arietis, right ascen, = 1*57^23', 176, in degrees = 29?20'.47'. 64,
and north polar distance = 67^21 155 ""• 56.
Construction.
Describe the pri-
mitive circle PESQ,
with the chord of
60? on the plane of ^
the solstitial colure :
— draw the Equator
EQ, and, at right
angles thereto,' the
axis PS; in which
P, represents the N.
celestial pole.
Make £ a, Q £ =
22?38U-'.44, the
star's north declin.
adon ; and with the
tangent of its com-
plement, viz. 67?21 C55''. 56, describe the parallel of declination a £.— Take
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. OF THB HBAVBNLY BODIES. 673
the star's right ascension^ viz., 29?20U7'^. 64^ in the compasses, from the
scale of semi-tangents, and lay it off on the Equator from <y> to R, and with
the secant of its complement, viz. 60?39' 12*'. 36, describe the circle of
right ascension PRS; the intersection of which with the parallel circle
a by at 9|c , will be the apparent place of the star in the heaveQs.-^Make
P ]V, S 0=s 23?27 -42". 875 the obliquity of the ecliptic :-.draw the polar
line NO, and, at right angles thereto, the ecliptic line yf <r ® ' — ^through
the intersecting point i^, draw the circle of longitude N ^te O, cutting the
ecliptic in L; — then, the arc t L> will be the longitude of the star, and
the arc L 9|e , its latitude ) the former being taken in the compasses, and
applied to the scale of semi-tangents, will be found to measure about 35{
degrees : — the angle FN* (measured by the arc L o = 54} degrees,)
represents the co-longitude of the star, and the arc N ^ its co-latitude;
the latter being reduced to the primitive circle will be found to measure
about 80 degrees.
Now, in the oblique angled spherical triangle FN 3|e, given the side
PN = 23?27M2''.875, the obliquity of the ecliptic ; the side F * =
67?21 ' 55^. 56, the star's north polar distance, and the included angle N F
4c == 1 19?20<47^. 24 ; tofind the sideN i^ = the co-latitude of the star,
or its distance from the north pole of the ecfiptic, and the angle FN i|c :±=
its co*longitude.
Note, — ^The circle of right ascension which passes through yp, viz. PQS,
is always equal to, or expressed by 270 degrees ; and that which passes
through npj or atv, by 0, or 360 degrees ; the difference, therefore, between
yf and nn, or, which is the same, between Q and <r , Is 90 degrees ; which
being added to the arc v R, 29?20'47''. 64, makes the whole arc Q R=
1 19?20'47''. 64, which is the true measure of the angle R F Q ; that is,
the angle N P :|c , comprehended between the two given sides.
Hence, by spherical trigonometry. Problem IIL, Remark 1, page 203^
An.NP*=sll9?20M7''.64^-2=59?40^23^82.tw.L.sin. 19. 8721827-68
Side FN = obi. of the ecliptic = 23.27.42 .875 L.sin. 9.6000350.88
SideP :|c = star's N. polar dis.= 67.21.55 .56 L.8in. 9.9651914.48
Sum= 39.4374093.04
Difference of the two sides = . . 43?54n2^685 Hf.S. 19.7187046.52
2x
Digitized by VjOOQ IC
Diff. of the two 8ide8=43?54'. 12". 685 Half S. of logs. 19. 7187046. 52 +
Half difference . . 21?57'- 6''.3425 Log. aine » 9.5726693. 15
Archss , , . , 54?27'.23''.53 Log.tang.=s 10.1460353.37
I
\
Log, sine of arch^ subtract from half sum of logs. =? 9. 9104508. 95 —
Half the side P * = 40» I'. 14". 26 = Log. sine = 9. 8082537. 57
Die whole side P 4^ sSO^'. 2 '. 28''. 52 ; which is the co^latitude of the gtren
star, or its distance from the north pole of the ecliptic; heuce the
latitude of a Arietis is 9?57 '31''. 48 north.
Now, in the oblique angled spherical triangle PN 4k > the three sides aie
given, and the angle P ; to find the angle N =:: the co-longitude of the
star.— Hence, by spherical trigonometry, Problem I., page 198,
As the side N * = . 80? 2128^^.52 Log. co-secant = 10.0065934.92
Is to the ang.NP 4k » 119*20.47 .64 Log.sine«= . . 9.9403527.32
SoistheudeP^s » 67.21.55 .56 Log: sines • . 9.9651914.48
TotheanglePN 4c=::54?46nr.S8 Log. sine a . . 9.9121376.72
The angle N, thus foiind»54?46; 1 1 ". 38, which is the co-longitude of
the star, and which is measured by the arc of the ecliptic L o, being taken
from 90 degrees ; that is, from <r ®, leaves the arc cy» L=35?13'48'. 62 ;
which, therefore, is the apparent .longitude of the star a Arietis. — Hence,
the apparent longitude of the givenstar is 1 'S't ISMS'". 62j and its apparent
latitude 9?57'3r.48 north.
Now, from the above Problem we obtain the following
General Rule
for computing the latitude and longitude of a celestial object, vis»
Find the difference between the object's right ascension, expressed in de-
grees, and 270 degrees 3 ^ — to twice the logarithmic sine of half this diff-
erence, add the logarithmic sines of the object's north polar distance, and
of the obliquity of the ecliptic : from half the sum of these three logarithms
subtract the logarithmic sine of half the difference between the object's
north polar distance and the obliquity of the ecliptic ; the remainder will
be the logarithmic tangent of an arch, the logarithmic sine of which being
subtracted from the half sum of the three logarithms will leave the logar-
ithmic sine of an arc, which being doubled will give the object's distance
from the north pole of the ecliptic ) the difference between which and 90
degrees will be the latitude of the object, which will be north when the
ecliptic polar distance is the least 5 otherwise,. south.
• In all caseSy whenever this difference exceeds 180 degreei, it must be subuacled fimn
360 degrees.
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OF THB HBAVSNLY BODIES. 6/5
To find the Longitude : —
To the logarithmic secant of the object's latitude^ add the logarithmic
sine of the difference between its righl ascension and 270 degrees^* and
the logarithmic sine of its north polar distance ; the sum of these three
logu-ithms, abating 20 in the index^ will be the logarithmic sine of an archi
virhich being subtracted from 90 degrees, will leave the object's true longi-
tude when its right ascensio|i is less than 6 hours or 90 degrees ; but which
is to be increased by 6 signs when the right ascension b between 12 and
18 hours^ that is, between 180 and 270 degrees.
Again. — If tlie right ascension lies between 6 and 12 hours; that is, be-
tween 90 and 180 degrees, the arch, so found, is to be augmented by 3
sig^ns ; but if the right ascensioq is belweeii 18 and 24 hours, viz. between
270 and 360 degrees, it is to be augniented by 9 signs ; in either case
the true longitude of the object will be obtained.
Example I.
The apparent right ascehsioh of Aldebaran, August Srd., 1825, was
4^ 25T56' . 1 15,and its declination 16?9'5''.35 north; required k» apparent
latitude and longitude, the obliquity of the ecliptic being 23?27'42'[. 875?
Aldeb's.RJV.=4*25T56M15=66?29^ 1^.725
Diff. between R.A.and 270? =^ 203?80'.58'^, 275
And 360? -.a03?301 58^275= 156929C 1^725
The half of which is == . . . 78?14'*80". 862 Tw.L.»in.l9.d8l580
Aldebaran's.N. polar distance =f 73.50. 54 .650 Log. sii).=:fl.fi8.25.U
Obliquity of ecliptic = ... 23. 27. 42 . 875 Log. sin.=9. 600035
Sum= 39.564126
Diff. bet. P. dist. and ob. of eclip.=50? 23 H P. 775 Half S.= 19. 782068 +
Half ditto =x . 25. 11.85 .887 Log. sin. 9.629077
Arch=: 54?53C2ir Log. teng.= 10. 152986
Log. fiine of drch, subtract from . '
balfsumoflogs. = . .-......•,, 9.912775-
Arc= . • 47?44<22'\5 Log. sine = 9.869288
Twice the arc = . . . . . 95^28 f 45'. 0; which is Aldebaran's dis-
tance from the north pole of the ecliptic j its latitude, therefore, is
5^28'.45r south.
tm^i^^m^^mmm I'^p m m n i n man ■ i ■ ■ ■— * ■'■; n ■ . - ; -ii^ ■■■! imi n i h i ii i i i i ^ '
* See Note, page 674.
2x2
676 OF FINDING THE LATJTUBBS^ &C.
To find Aldebaran's Longitude :— -
Lat. of Aldebaran = . . . . 5?28U5r S. Log. 8ec.= 10. 001989
Diff. bet\v. R. A. and 270 deg.= 156. 29. 1 . 725 Log. 8ine= 9. 600982
Aldebaran'fl N. polar distance = 73. 50. 54 • 65 Log. sine = 9. 9825 1 1
Aldebaran's co-longitude = . . 22?38M1 ' . 6 l«g. sine = 9. 585482
Aldebaran's longitude =s . . 67?2la8^4 5 or2!7?2ia8^4.
Example 2
The apparent right ascension of Pollux, September 3rd., 1825, was
7*34r38'.74, and its declination 28?26n9^.48 north; required it»
apparent latitude and longitude^ the dbliquity of the ecliptic being
23?27M2^875?
Pollux's R.A.=7*34r38'. 74=113?39M1^ 1
DifiF.bet. R.A.and270? s . 156?20:l8^9
The half of which is = ... 789 10^ 9^. 45, Tw. L.S. 19. 981350
Pollux's N. polar distance = . 61 . 33. 40 . 520 L. sin. = 9. 944150
Obliquity of ecliptic s . . . 23. 27. 42 . 875 L. sine = 9. 600035
Sum = . 39. 525535
Difr.bet.P.dis,andob.of ecl.= 389 5(57^.645 Half sum 19. 762767 i+
Half dittos ...... 199 2C58^ 822 Log. sine 9. 513733 J
Arch= 60935(4ir L. tang. = 10.249034
Log. sine of arch, subtract from
half sum of Logs. = 9.940102 —
Arcs ....... 4l939^50^5Log.si^c= 9. 822665 i
Twice the arc 839 1 9U P. 0; which is Pollux's distance
from the north pole of the ecliptic : — hence, its latitude is 6940' 19.'
north.
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y
To find Pollux's Longitude : —
Latitude of Pollux = • . . 6^40^ 19r N. Log. secant = 10. 00295 1
Diff. bet. R. A. and 270 deg.= 156. 20. 18 .9 Log. sine = . . 9. p03503
Pollux's north polar distances 61.33.40 .52 Log. sine = • . 9.944150
Arch = . 0!20?48U4^2 Log, sine =.. 9. 550604
Add . ..,•.. 3. 0. 0. 0 .0.
Sum = 3 !20'? 48 U4^ 2; which is the true longitude
of Politic.— Hence, the latitude of Pollux is 6?40'19f north, and its lon-
gitude 3 ! 20?48 1 44 ''. 2, as required.
In like manner may the latitudes and longitudes of the moon and planets,
be deduced from their respective right ascensions and declinations.
Problem IL
Qivmi the Latitude and LmgUude of a Celesiial Object ; to find its Right
Jscension and Declination.
Example. .
The apparent long, of a Arietis, August 1 st,. 1 825, was! 1 1 5? 13 M8''. 62,
and its latitude 9?57 -3^.48 norths required its right ascension and de-
clination, the obliquity of the ecliptic being 23?27'42''. 875 ?
The construction of this Problem is exactly like that of the preceding ;
thus, lay the longitude of the given star off on the ecliptic line from on to
L, and draw the circle of longitude N L O. — ^Make yf d, © c = the star's
latitude, and draw the parallel circle cd; the intersection of which with
the circle of longitude, at ^y will represent the apparent place of the star
in the heavens.-— See the last projection.
Now, in the oblique angled spherical trifmgle NP a|e ; given the side
PN = 23?27U2''.875, the obliquity of the ecliptic; the sideN « =
80°2'28''. 52, the star's distance from the north pole of the ecliptic, and
the included angle P N * = 54?46' 1 1 ". 38, the complement of the. star's
longitude (measured by the arc L ©) ;^ to find the side P * = the star's
north polar distance, and the angle N P * (measured by the arc R Q) ;
the difference between which and r Q, viz. 90 degrees, expressed by the
arc Y H = will be the star's right ascension. «
Pence, by spherical trigonometry, Problem IIL Remark 1, pi^e 203^
O/O OP M Mill mi THS KIUIIT A9(;jinaiUPl9y (»i;«
Angle PN*= .. . . 54?46:il".38
Half ditto 5= . . . • 27?23^ 5^69 Tw.L.sm.l9. 32545 14. 58
Side P N = obliquity of the
ecliptic = 23.27.42 . 875 Log.sin.=9. 6000350. 88
Side N :f: = star's ecliptic
polar distance = • . 80. 2.28 .250 Log.8in.= 9. 9934065. 08
Sum =s38. 9188930. 54
Difference of the two sides =56?34^45^ 645 HalfS. 19.4594465.27 +
Half difference = ... 28^7 ^ 22*. 822 Log. sin. 9. 675713/. 08
Archtt 4 3l?17;22^56 Log.T. 9.7837328.19
Logarithmic sine of arch = 9.7154718.60 —
Half the side P * = . . 33^40^57''. 76 Log. sine 9. 7439746. 6?
The whole side P ]|c = . . 67^21 ^55^^. 52 ; which is the co-declination
of the ^ven star, or its north polar distance; hence^ the declination of a
Arietis is 22?38U''. 48, north.
Now, in the oblique angled sphisrical triangle N P 4c ; the three sides
are given, and the angle N; to find the angle P = the arc RQ; or the
sum of the starts right ascension and 90 degrees.
Hence, by spherical trigonometry. Problem L, page 198,
As the side P % s: . . 67?2H55''; 52 Log. co-sec. «10. 0348085. 84
IstotheanglePN* = 54.46.11 .38 Log. sine « . 9.9121S76.70
So is the side N )fe = . 80. 2. 26 . 52 Log. sine s . 9. 9934065. 08
To the sup. of at^. NP^e s60. 39. 12 . 38 Log. sine ±s . 9. 9403527. 62
Hence, theang. NPsjc isal 19. 20M7'. 62 =:the arc RQ ; from which take
the are tQ, 90 degrees 5 and the remaining arc t R =: 29?20'47*'. 62, is
the star's right ascension.— The apparent right ascension of a Arieds,
on the given day, was therefore, 29?20U7''.62, and its dedination
22?38M^ 48 north.
Now, from the above Problem the following general rule is dedwced for
computing the right ascension, and declination of a celestial-object^ vis.
Find the difference between thei ol^ect's {on^tude wd three signs (e)
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ence^ add the logarithmic sines of the object's distance from the north
pole of the ecliptic, and of the obliquity of the ecliptic ; from half the sum
of these three logarithms subtract the logarithmic sine pf half tlie difference
between the obliquity of the ecliptic and the pbject's ecliptic polar distance^
and the remainder Ivill be the logarithmic tang* of an arch ; the log. sine of
which being subtracted from the half sum of the three logarithms, will leave
the logarithmic sine of an arc^ the double of which will be the object's dis-
tance from the north pole of the equator :— now, the difference between
this distance and 90 degrees will be the declination of the object ; which
will b^ north when the first term is less than the latter ; otherwise south.
To find the Right Ascension :—
To the logarithmic secant of the object's declination, add the logarithm
mic sine of the difference between its longitude and 90 degrees, * and the
logaTithmic sine of its distance from the north pole of the ecliptic ; the sum
of these three logarithms; abating 20 in the index, will be the logarithmic
sine of an arch, which being subtracted from 90 degrees will leave the ob-
jeet'a correct right ascension when its longitude is less than 3 signs or 90
degrees; but which is to be increased by 180 degrees when the longitude
is between 6 and 9 signs, or between 180 and 370 degrees. Again, if the
longitude is between 3 signs and 6 signs, that is between 90 and 180 de-
grees, the arch, so found, is to be augmented, by. 90 d^roes; but if the
longitude lies between 9 and 12 signs, vis. between 270 and 360 degrees,
it is to be augmented by 270 degreei ; in eithier case the correct right
ascension of the object wjll.be obtained, which may be converted into time,
if necessary, by Problem I., page 296.
Example K
The apparent longitude of Aldebaran, Aug. 3rd,1825,was 2* 7°21 : 18". 4
audits latitude 5? 28! 45?, south; recpiired its apparent right ascension
and declination, the obliquity of the ecliptic being 23'?27'42''. 875 ?
Aldebaran's longitude =s . • . 67?2la8^4
Difference to 90 degrees » , , 22?38M1''.6
The halfof which is = . . . Hei9'20\ 8 TwJi. sine 18.585974
Aldebaran's ecliptic polar dist. es 95. 28, 45 . 0 Log. sine 9. 998011
Obliquity of ecliptic = . . » 23. 27. 42 . 875 Log. sine 9. 600035
Sum = 38. 184020
Difference betw. ob. of edip. and
star's eclip. polar distance = . 72? 1 ' 2". 125 Half S. 19. 0920 10 +
• In an cues, whenever tbit difference exceeds }89 deprees^ it must be subtrscted from
JV jrsi^«^jit^^« «-«
Difference betw. ob. of eclip. and
star's eclip. polar distance = . 72? 1 \ 2\ 125 Half S. 19. 092010 +
Half ditto = 36? 0C31\062 Log sine 9. 769309
Arch = . 11.52.21.27 L. tang. 9. 322701
Log, sine of arch, subtract from ■
half sum of logs. = 9.313310-
Arc= 36?55^27*'.3 Log.sine 9.778700
Twice the arc= 73?50'.54*'.65 which is Aldebaran's
.distance from the north pole of the equator \ hence its declination is
16?'9'.5%4 north.
To find the Right Ascension :—
Declination of Aldebaran= 16? 9\ 5\4 Log. secant = 10.017490
Diff. bet. long, and 90 deg.=: 22. 38. 4 1 .6 Log. sine = 9. 58548 1
Aldebaran's ecliptic P. dis.=:95. 28.45 . 0 Log. sine s 9. 99801 1
Arch 5= ..... 23?30:58^.3 Log.sine = 9.600982
Aldebaran's right asc. = 66?29'. r. 7,or4i25?56M
Example 2.
The apparent longitude of Pollux, Sep. 3rd, 1825, was 3!20?48M4'. 2,
and its latitude 6? 40^ 19^ north; required its apparent right ascension and
declination, the obliquity of the ecliptic being 23? 27 '42^.875 ?
Pollux's longitude = .... 110?48M4^2
Difference to 90 degrees = . , .20?48<44^.2
The half of which is = . . . 10?24^22M Tw.L^in.l8. 513353
Pollux's ecliptic polar distance = 83.19. 41 .0 Log. sin. 9.997049
Obliquity of the ecliptic = . . 23. 27. 42 . 875 L. sine 9. 600035
Sum 38.110637
Difference between ob. of ecliptic _.«...— ...^
and sUr's eclip. pol. distance s 59?5K58'', 125 Hf, S. 19.055318it
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OF THE UBAVBNLY BODIES. 681
Difference between ob. of ecliptic * • *
and star's eclip. pol. distance = 59?51 ^58^. 125 Hf. S. 19. 055318^+
Half ditto = ... . . ... 29?55^59^ 062 L. sin. 9.698090
Arch= 12?49^25''.32 L. tang. 9.3572281
Log. sine of arch, subtract from
half sum of Logs. = 9.346258*-
Arc= ......... 30M6^50^.03 Log. sin. 9. 709060
TVicethearc= .... . 61^33^0'', 06 ; which is Polhix^s dis-
tance from the north pole of the equator ;— hence, its declination is
28?26n9^. 94 north.
To find the Kght Ascension :-—
Declination of Pollux = . $8?26n9^.94 Log. secant =10.055850
Diff^betw. Ion. and 90 deg.=20. 48. 44 .20 Log. sine = 9. 550603
Pollux's ecliptic polar dis. =83. 19. 41 . 00 Log. sine = 9. 997049
Arch= 23?39'.4ir Log. sine = 9.603502
Add ....... 90. 0. 0
Right ascension of PolInx=l 13?39M1 = 7*34r38^ .73
Note. — In the same manner may the right ascensions and declinations of
the moon and planets be deduced from their respective latitudes and lon-
gitudes.
PROBLSM in. *
Given the Latitudes and Longitudes of the Moon and Sun, Moon and
fixed Star, or. Moon and Planet; to find the tnie Central Distance
between them.
Note. — If this Problem be projected stereographically on the plane of
the circle of longitude passing through one of the objects, it will be found,
in every respect, like Problem XXIV., page 273, of " The Young Navi-
gator's Guide to the sidercfd and planetary parts of Nautical Astronomy ;*'
reading, however, difference of longitude for difference of right ascension,
and ecliptic polar distances for polar distances : — henc^, it is evident that
there is a spherical triangle to work in,, where two sides and the included
angle are given to find the third side, or central distance between the ob-
jects, ^d which may be computed by the following
General Euk.
.To twice the logarithmic sine of half the difference of longitude between
thf two objects, add the logarithmic co-sines of their latitudes ; from half
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the sum of these three logarithms subtract the logaridimie rioe trf half the
difference or half the sum of the latitudes^ according as they are of die tame
or of contrary names ; the remainder will be the logarithmic tangent of an
arch, the logarithmic sine of which being subtracted from the half sura of
the three logarithms, will leave the logarithmic sine of half the true central
distance between the two given objects.
Example. L
September 3rd^ 1825, the moon's apparent longitude, at noon^ was
1 '. 16? 19^29r, aud her latitude 2?33^S0? north; at the same time the ap-
parent longitude of Pollux was 3f20?48^3S:, and its latitude 6?40U9?
north ; required the true central distance between those two objects ?
Longitude of Pollux s= •
Longitude of the moqn c
U0'?48^38r
46.19.29
Difference of longitude = 64?29' 9^
The half of which is = .
Latitude of Pollux =s
Latitude of the moon s
32?l4^34ir Tw. the log. sJn.l9. 454285
. 6. 40. 19 N. Log. co.sine=9. 997049
. 2. 33. 30 N. Log, co-sine=9. 999567
Sum = 39. 450901
DilTerenee of latitude . . # 4? eU9? Half som « I9.7254ftO| +
Half differences .... 2? 3:24ir Log. sine . 8. 5549771
Arch= 86? 8air Log.tang.= 11.170473
Log. sine of arch, subtract ■
from half sum of logs, s . . . . ^ . . « . 9.999012-
Half true distances . . 32?11^ 4^ Log. sine = 9.726438|
Hence, the true central distance between the moon and Pollux, at the
g^ven time, was 64?22!8T • which corresponds exactly with the computed
distance in the Nautical Almanac.
Note.^^lt is evident that the same result will be obtdned by mdiing
use of the right aseensions and declinatlona of the objects.
JS^HnpIe 2.
August 22, 1825, the moon's apparent longitude, at noon, was
8! 19?6' 15r, and her latitude 0?d^2K north; at the same time the appa-
rent longitude of Spica Virginis was 6121?24l32?, and its latitude
2^3:^5? apttth ) rei|[aiitdtbe tnie^entria dtaUaeeUttmo Hmc ob|^t»^
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uw i;uMruimu thjs lukak uiSTAjncjS9*
ooo
Longitude of the moon &= • 259? 6^ IS'T
Ifongitude of Spica Virginks 201. 24. 32.
Difference of longitude = • 57?41'43'r this divided
. by 2 gives 28?50'.51ir Twice L. sin.= 19. 366962
Lat.ofmoon= 0. 3.21 N. Lo^.co-siue 10.000000
IfatofSpicaVirg. 2. 2.23 S. Log.co-sine 9.999725
Sum =
.39.366687
Sum of lata. = • 2? 5M6r Half sum = .19. 683343 i . . 19. 683343 ^
Half ditlosi
Arch » •
1? 2^53r Log. sine = . 8. 2622374
87?49M2^ Log. tang.
n. 421106 L.sin. 9. 99968S
Half true dist. = 28?5 1 ^37^ Log. sine = .9. 683655 J
Hence, the true central distance between the moon and Spica Virgtnia
is 57^43 H 4^ ; which is 1 1 more than that given in the Nautical Almanac.
Example S.
August. 4th, 1825, the moon's apparent longitude, at noon, was
0'. 14?l3C32r, tod hiSr latitude 4?d8Uirnorth ; at the same time the sun's
longitude was 4 ! 1 1 ?4d ' 46 ; required the true central distance 7
Nol^.— Since the sun apparendy moves in the ecliptic, he has, therefore,
no latihide.
Mvkm's longitude =« 14?13^82?
Sun's tongitude aa 131.43.46
Difference pf long.= 1 1 7. 30. 1 4 this divided
iy 2 gives 5^?45 1 7' Twice the L. sines m 863660
Moon's lat. 4. 38. 4 1 N. Log. co-sine = 9. 99857 1
Sun's Ut. S5 0. 0. 0 Log. co-sine s . lO.OOOOOO
Sum as k
Diff.oflat.r=4?38!4lr Hdfium:
Half dittx>= 2? 19^2Qir Log. sine a
Arch = . 87?16'55r Log.tang..=
Hf. req. dis. 58^42: lOjr Log. sine =
39.862431
19.9312151
S. 607688
19.931215^
11. 323527iL.8in.9. 999511
9,9317044
Hence^ the true central distance between moon and sun is 1 17?24^2ir j
wW^h corresponds with that in th^ Nawtjc^ Almwuc^
684 . OP couPvriVG the lunar di^ancbs. >
Rem(7r%'.— Since the co-latitudes of the sun and moon and the compre-
hended angle (expressed by their difference of longitude,) form a quadran-
tal spherical triangle; therefore the true central distance between these
particular objects may be more readily determined by the following con-
cise method than by the above general Rule, viz.
To the logarithmic co-sine of the difference of longitude, add the logar-
ithmic co-sine of the moon's latitude ; the sum of these two logarithms,
abating 10 in the index, will be the logarithmic co-sine of the true central
distance between the sun and moon.
Example 1.
August 6th, 1825, the moon's longitude, at noon, was l!d?0'34'y and
her latitude 3? 23 ^20? north; at the same time, the sun's longitude was
4113?38!46T ; required the true central distance ?
Moon's longitude = 38^ 0'34r
Sun's longitude = .. 133.38.46
Difference of long. =: 95?38n2r Log. co-sine = . , . . 8.992199
Moon's latitude = . . 3.23,20 Log. co-sine =;= . . . .9.999240
True central distance =95?37'36r Log. co-sine = . .. .. *. 8.991439
which is precisely the same as that given in the Nautical Almanac.
Example 2.
August 7th, 1825, the moon's longitude, at noon, was l!20?4M2f, and
her latitude 2?30^42? north; at the same time the swi's longitude waa
4 : 14?36 ' 1 8^ ; required the true central distance ?
Moon's longitude = 50? 4M2^ .
Sun's longitude = . 134.36. 18
Difference of long. = 84 ?3 1 ^ 36r Log . co-sine = .... 8. 979468
Moon's latitude = . 2. 30. 42 Log. co-sine = • . . . 9. 999583
True central dist = 84?31^55r Log. co-sine = ... 8.979051
which exactly corresponds with the computed distance in the Nautical
Almanac.
Example 3.
Required the true central distance between the moon and sun at noon,
August 8th ; at midnight, August 8th; at noon, August 9th^ and at ovd-
night, August 9th, 1825 ?
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Moon's long, noon Aug. 8th = 62?22^38T
Sun's longitude ditto = • . 135. 33. 5 1
JJifTerence of longitude = . . 73? 1 1 : 13f Log. co-sine = 9. 461273
^Moon's latitude = . . • . 1.30. 9 Log. co-sine = 9.999851
Distance at noon, Aug. 8th, 73? 1 1 '. ZV. Log, co-sine s 9. 46 1 1 24
A^ooii's long. mid. Aug. 8th = 68?38'20r
Sun^s longitude ditto s . *. 136. 2.38}
Difference of longitude =
IVloon's latitude =s . •
67?24n8jr Log. co-sine
► 0. 57. 33 Log. co-sine
9. 584572
9. 999939
Distance at midnight, Aug. 8th=67? 24 '. 3 K Log. co-sine s 9. 5845 1 1
Moon*s long, noon, Aug! 9th =3 74?59' 18r
Sun's longitude ditto = . . • 136.31.26
Difference of longitude s . . 61?32^ 8r Log. co-sine = 9. 678166
Moon's latitude = 0. 23. 49 Log. co-sine = 9. 999990
Distance at noon, Aug. 9th, = 6ie32' llr Log, co-sine = 9. 678156
Moon's long, at mid. Aug. 9th =81 925 1 59^
Sun's longitude at ditto = . . 137* 0. 14
Difference of longitude = . . 55934M5r Log. co-sine = 9.752346
Moon's latitude = • . ; • . 0.10.42 Log. co-sine = 9.999998
Distance at midnight Aug. 9th = 55934' 16^ Log. co-sine = 9. 752344
Now, from these four consecutive lunar distances, the distances at the
intermediate periods, oi> every third hour may be readily determined in the
following manner, viz.
Find' the proportional parts of the difference at the middle interval be-
tween the four distances (that is, between the second and third distances,)
answering to 3 hours, 6 hours, and 9 hours : correct these proportional
parts by the equation of second differences agreeably to the rule given, for
that purpose, in page 34 ;-^then, these corrected proportional parts being
applied to the second lunar distance by addition or subtraction, according
as the distances are increasing or decreasing, the sum or difference will be
the true distances at the given periods :^thus.
OF COMPUTING THE LUNAR mSTANGBS*
Aug.8,1825,di8.atN. =73nH34
Ditto • . . do. at M. =67.24.31
Aug. 9» • . do.atN. =6L82. )K:
Ditto . • . do.atM.s=55.84. 16
UtdUt SddUff.
5?47' 3^^
McwMCC
c c9 on }0?5n7^i ,
The proportional parts of 5?52'20.T (the middle first difference) anawer^
ing to the intermediate periods, viz.
To 3 hours it is =s 1*?28'. 5^}
Equation of second difference
Proportional part corrected, a .
Distance at midnighty Au^« 8th =
Distance at 15 hours, Aug. 8th ss
To 6 hours it is =;= • . • •
Equation of second difference =
Proportional part corrected = .
Distance at midnight Aug. 8th =
Distance at 18 hours Aug. 8th =s.
And to 9 hours it is = ' • • •
Equation of second difference =
Proportional part corrected ss' .
Distance at midnight Aug. 8th =
Distance at 21 hours Aug. 8th =s .
- 31
1?27'34
67.24.31
.6S?d6^57f
2?56U0r
• - 41
2?55i29r
87.^4.31
> f. I II ■■ I I
64°29C 2r
4?24'15r
- 31
4^23:447
67.24,31
63? 0^477
The distances for the intermediate periods corresponding to the first
and to the last 12 hours; that is, for every third hour between the first
and second distances, and between the third and fourth distance, may be
also very readily determined by means of the Formulae which are given in
page 1 17 of the Nautical Almanac for 1825.
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APPENDIX.
Showing the direct application of logarithms to the solution of pro-
blems connected with the ctoctrine of compound interest ; which deve-
lops the extraordinary powers of logarithmical arithmetic more than
imy other department of science which has been touched upon in this
work.
D^miion.— .Compound Intbrest is that which is deduced not only
from the* sum of money lent as the principal^ but also from the interest
ariring thereon ; which interest, as it becomes due at the stated times of
payment, is added^ or supposed to be added^ to the principal/
Although it is illegal to lend money at compound interest, yet hi
purchasing annuities, pensions, or leases in reversion, it is usual to allow
the purchaser compoutid intel'est for the use of his ready money. — ^And^
these points being premised, we will proceed to the solution of the most
interesting problems relating to this department of science] — ^for which
purpose the following Tables have been computed.
Note. — The rates, or .ratios, o( £ 1 sterling contained in these tables
were computed by .the rule of proportion in the following manner, viz.—
As £ 100 I £3 : : £ 1 to £. 0300 ; — hence, the amount of £ 1 for one yedr is
£1.0300; which, therefore, is the ratio; and so on for the rest.— The
respective numbers annexed to these ratios are expressed by the common
logarithms corresponding thereto :-*-thus, the logarithm of 1.0300 is
0.0128372; and the logarithm of .0300 is 8.4771213, and so on of
others. — In this part of the work it has been deemed advisable to take
out the logarithm^ to seven places of decimals ; though for ordinary
purposes six places of decimals will be found amply sufficient.
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Problem I.
Given the Princgtaly Bate of Interest, and Time; to find the Amount.
RULB.
Multiply the logarithm of the ratio, Table A, by the time, or number of
payments, according as the instalments may be reckoned in years, half
years, or quarters ; to the product of those two numbers, add the logar-
ithm of the principal, and the sum will be the logarithm of the amountt
l?emarfc.-«-The logarithm of the ratio, contained in Table A, must always
be taken out so as to correspond with the nature of the instalments ;— -
thus, if the payments are to be made yearly, the logarithm is to be taken
out of the first column ; if half yearly, it is to be taken out of the second
column ; but if quarterly, it must be taken out of the third column.-— The
same is to be observed of liable B ; — and, in order to avoid repetition, it
must be remembered that this remark is applicable to the various cases of
compound interest, &c. which may be given.
Example L
What will a principal of £240000 sterling amount to in 45 years, by
annual, payments, at the rate of 5 per cent, per annum, compound
interest?
Kate, 5 per cent. ; ratio = 1. 0500, logarithm^ Table A. s .0. 02 U 893
Multiply by the number of payments = .•••••• 45
1059465
0847572
Product = . 0.9535185
Principal = £ 240000, the logarithm of which is = • . • 5 . 38021 1 2
Amount, or improved principal = . £2i 56402 Log. =; 6.3337297
Example 2.
What will a principal of £240000 sterling amount to in 45 years, by
half-yearly payments, at the rate of 5 per cent, per annum, compound
interest ?
Rate, 5 per cent. ; ratio = 1. 02500; logarithm, Table A, = 0. 0107239
Multiply by the number of payments = 45 years x 2 = . . 90
Products .......: 0.9651510
Principals £240000, the logarithm of which is = . . . 5.3802112
Amount, or improved principal, = £2214941 Log. = . 6. 3453622
2 Y
690 COMPOUND IKTBRXST.
Example 3.
What will a principal of £240000 sterling amount to in 45 years, by
quarteriy payments, at the rate of 5 per cent, per annum, compound
interest ?
Rate, 5 per cent. ; ratio = 1 . 012500 ; log., Table A, = .0. 0053950
Multiply by the number of payments = 45 years x 4 = • • 180
4316000
053950
Products 6.9711000
Principal = £240000, the log. of which is s 5.3802112
Amount, or improved principal, = £2245490 Log. = • 6.3513112
Note. — ^The above examples will show the reader the great evil which is
attendant upon other than yearly payments of interest. Thus, in the
present instance, while thie principalis augmented to nine times its original
value, the excess of the half-yearly above the yearly instalments is £58539 ;
but by quarterly instalments, it is full £89088 : which excess is evidently
to the manifest injury of the borrower or debtor. This will appear still
more evident from the following
Example.
Admitting that the national debt of Great Britain was £206590000
sterling in the year 1786, what would its probable amount be in the year
1814 (being a lapse of 28 years), by yearly, half-yearly, and quarterly
payments, at the rate of 5 per cent, per annum, compound interest ?
Pirst,-*For Yearly Payments : —
Rate, 5 per cent ; ratio » 1 . 0500 ; log. Table A, == . . 0^ 021 1893
Multiply by the number of payments = 25
1695144
0423786
Product = 0.5938004
Amounf of debt in 1786 = £206590000, the log. of which is=8. 3151093
Amount of ditto in 1814 ss £809859533, very nearly Log.»8. 9084097
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COMPOUND IKTSASST^ 691
Second,— For Half-yearly Payments :—
Rate, 5 per cent. ; ratio = 1 . 02500 j log.. Table A, = « . 0. 01 07239 .
Multiply by the number of payments a 26 years x 2 oa « 56
0648434
0536195
Products 0.6005884
Amount of debt in 1786 = £206590000, the log. of wMch i8»8. 8151098
I ■!■ Mil ^
Amount of debt in 1814 a £823469720^ very nearly Log*a8. 9156477
Third,— For Quarterly Payments t-^
Rate, 5 per cent. ; ratio = 1 . 01 2500 $ log., Table A, = .0. 0053950
Multiply by the number of payments = 28 years x 4 =s • 112
0107900
0593450
Products . ,0.6042400
Amount of debt in 1786= £206590000, the log. of which i8=8. 3151093
Amount of ditto in 1814 ^ £8305 18462, very nearly Log.aS. 9193493
Hence it appears, that if the national debt were £206590000 in the
year 1786, and that it were allowed to multiply by the accumulation of
interest upon interest, it would amount, at the end of 28 years, rht^ in
1814, by annual payments, to the sum of £809859533 3 by half-yearly
payments, to the sum of £823469720; but, by quarterly payments, to the
enormous sum of £8305 18462 sterling ! And this is a sum of such mag-
nitude, that it could not be liquidated by all the gold and silver now in ,
circulation amongst the different kingdoms, states, and empires, in the
known world.
We- will only adduce one more example in order to apprize the reader
of the almost incredible manner in which a smn of money may be imp? oveA
by the accumulation of interest; for which purpose, let us suppose that six-
pence sterling is the sum put out at compound interest, that the rate is &
per cent, per annum, payable by half-yearly instalments, and that the time
is 450 years. — Then,
2y2
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Rate, 6 per cent ; ratio = 1 . 0300 ; log., Table A, = . . 0. 0128372
Multiply by number of payments = 450 years x 2 == • . 900
Products .11.5534800
Principal = 6d. sterling, or . 025 £, the Ic^. of which is = . 8. 3979400
Amount, or improved principal, = £8941700000 Log. = 9. 9514200
Hence it is evident, that sixpence sterling put out at mterest vpon interest j
agreeably to the given rate and time, would amount to the amazing sum
of eight thousand nine hundred and forty-one millions and seven hundred
thousand pounds sterling; which sum could not be made up by all the
gold and silver that have been dug out of the bowels of the earth from the
creation of the world to the present day !
Problem II. -
Given the Amount or improved PrindpaU Rate of Interest, and Ttme;
to find the original PrindpaU
Rdlk.
Multiply the logarithm of the ratio. Table. A, by the time or number of
payments ; subtract the product from the logarithm of the amount, and the
remainder will be the logarithm of the original principal or sum put out at
interest. ^
Example.
What principal put to interest for 31 years, and payable half-yearly, will
amount to £29876 sterling, at 5 per cent, per annum, compound interest?
Rate, 5 per cent. ; ratio = 1 . 02500 ; log.. Table A> = ... 0, 0107239
Multiply by number of payments = 31 years x 2 = . • . 62
0214478
0643434
Products 0.6648818
Amount, or improved principal,=: £29876, the log. of which is=4. 4753225
Original principal, or sum put to interest, = £6463 Log. =3. 8104407
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COMPOUND INTEREST. 693
Problem • IIIl
Given the original Principal or Sum lent, the 7«m^, and Hie Amount or
improved Principal; to find the Rate of Interest.
' Rule.
From the logaiithtn of the amount or improved principal, subtract the
logarithm of the original principal) divide the remainder by the number
of payments, and the quotient will be the logarithm of the ratio ; with
which enter the proper column of Table A, according to the modes of
payment^ and opposite thereto, in the left-h&nd column^ will be found the
required rate of interest.
Example.
At what rate per cent, per annum will £2360 amount to the sum of
£4792 in 15 years, the payments being made quarterly ?
Amount, or improved principal, = £4792, log. of which is=: 3. 6805168
Original principal or sum = £2360, the log. of which is = 3.3729120
Divide by the number of payments = 15 years x 4 = 60)0.3076048
The ratio to which, in Table A, is 1. 01 1875, = quotient = 0, 0051267. 5
Hence, the rate of interest is 4| per cent, per annum.
Problem IV.
Gftt^ the oriffnal Principal or. Sum lent, the Rate oflnteregt, and the
Amount or improved Principal ; to find the Time.
Rule.
From the logarithm of the improved principal, subtract the logarithm of
the original principal | divide the remainder by the logarithm of the ratio.
Table A, and the quotient will be the time or number of payments in years,,
halves, or quarters,* as the case may be.
Example,
In what time will the deposits or funds in the savings' banks, which are
now (June, 1826,) estimated at £14500000 sterling, amoutft, by half-
yearly payments, to the sum of £107415024 sterling, at the rate of 4^ per
cent per annum, compound interest ?
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Q94 AAnUlllAB in AKJUUlKSy
Amount^ or improved pTincipaI,£l074l5024, the log, of which is 8. 0310457
Principal, or sum funded in June, 1826,=:f 14500000 Log.=:7. 1613680
Rate, 4| per cent. ; ratio = 1. 02250 ; divide by log, of this,
TableA,= '. 0.0096633)0.8696777
= 90 half-yearly payments : hence, the required time is 45 years.
From the above result, an abstract reaaoner would be apt to imagine
that the savings' banks are more of individual than of national uUlity«
ANNUITIES IN ARREARS, AT COMPOUND INTEREST,
Definition. — An annuity is said to be in arrears when the debtor keeps
it in his hands for any certain time after the term or period of payment
becomes due. The sum of all the single payments, together with the
interest due upon each payment, from the time of its becoming payable to
the time that the whole is paid off, is called the amount of such annuity.
Problem I.
Given an Annuity ^ the Tirtke or Number ofPaymenU, and the Rate fer
• Cent, per Annum ; iofind the Amount.
RULK.
Multiply the logarithm of the ratio, Table A, by the number of pay-
ments ; find the natural number answering to the product, and diminish it
by the integral part of (he ratio, viz., I ; then, to the logarithm of this
diminished natural number, add the logarithm of the annuity (proportioned
to the modes of payment) : from the sum of these two logarithms, subtract
the logarithm of the decimal part of the ratio contained in Table B; and
the remainder will be the logarithm of the amount of the annuity.
Note.'-U the payments be yearly, take the logarithm of the whole
annuity ; if half-yearly, take the logarithm of half the annuity ; and if
quarteriy, take the logarithm of the one-fourth part of the annuity.
Example
B560 per ai
^^ ^ _r cent, com
quarterly payments ?
If an auteity of £560 per annum be unpaid for 9 years, what will it
amount to, at 4i per cent, compound interest, by yeariy, half-yearly^ and
arterlv uavments ?
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Rate, 4J per cent. 5 ratio= 1 . 0425 ;
log,, Table A = 0.0180761
MuLbyno.ofpajTi.z: 9
Product = . . 0. 1626849 Nat. no.= 1 . 4544033
Subtract the integral part of the ratio, viz., 1
I>imiRished natural number rz . . . 0.4544033 Log.=:9. 6574415
Annuity = £560, the log. of which i« =: ^.7481880
Sum of the two logarithms = ;....,.•.. 12.4056295
Decimal part of the ratio=. 0425 j log., Table B, = . . ,8. 6283889
Amount of annuity by yearly payments = £5987. 432 Log.=:3. 7772406
Second, — For Half-yearly Payments : —
Rate, 4^ per cent. ; ratio = 1 . 021 25 ;
log.. Table A, == 0,0091321
Multiply by no. of
paym.::=9 years X 2:= 18
0730568
0091321
Produets . . 0.1643778 Nat^Bo^K 4600838
Subtract the integral part of the ratio, viz., 1
Diminished natural number =: ... 0. 4600838 Log.=9. 6628370
Annuity, £560 ; one half of which is £280 : log. of this = . 2. 447 1580
Sum of the two logarithTns ^ . , 12.1099950
Decimal pait of the ratio = . 02125 5 log., Table B, — . . 8. 3273589
Amount of the annuity by half-yearly paym.=£6062. 382 Log. 3. 7826361
Thii;d,— For Quarterly Payments :—
Rate, 4i per cent. ; ratio=: 1 . 010625 ;
log.. Table A, = 0. 0045900
Multiply by no. of
paym.::;:9yrsx4= 36
0275400
0137700
Product ;= . • 0, 1652400 Nat. no.= 1.46?9855
ANNUITIES IN ARRBARS,
Product = . • 0. 1652400 Nat. no.r: 1 . 4629855
Si^btract the integral part of the raiio^ viz.^ 1
Diminished natural number = . . • 0» 4629855 Log.=:9. 6655674
Annuity, £560 ; oae fourth of which is £140 : log. of this = 2. 1461280
Sum of the two logarithms == 11.8116954
Decimal part of the ratio =: .010625 ; log., Table B, =: . . 8. 0263289
Amount of the annuity by quarterly payments=£6100. 515 Log.3. 7853665
iVb^^.-^Instead of subtracting the logarithm of the decimal part of the
ratio (Table B) from the sum of the logarithms of the diminished natural
number and the annuity, its arithmetical complement may be added to
these two logarithms ; which, perhaps,* will render the operation a little
more concise, or, at least, apparently so. Thus : —
Diminished natural number x: 0. 4629855 Log. == • . 9. 6655674
Annuity, £560i one fourth ofwhich is £140 Log. = . . 2.146^1280
Decimal part of ration . 0 1 0625 ; log., Table B, arith. comp. = 1 . 97367 1 1
Amount of the annuity by quarterly paym. = £6100. 5 15 Log. 3. 7853665
Problem II,
Ct!>en the Time or Number of Payments^ the Rate per Cent, per Anrnm^
and the Amo\mt ; to find the Jnnvity.
RUJJB,
Multiply the logarithm of the ratio (Table A) by the number of pay-
ments 5 find the natural number answering to the product, and diminish it
by the integral part of the ratio. Then,
To the arithmetical complement of. the logarithm of the diminished
natural number, add the logarithm of the amount and the logarithm of the
decimal part of the ratio (Table B): the sum of these three logarithms (abat-
inglO in the index,) wJU be the logarithm of the. annuity for yearly payments,
but of half the annuity for half-yearly payments, or of one fourth thereof
for quarterly payments.
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AT COMPOUND INTBRB8T. 697
Example.
A certain annuity, at the rate of 4i per cent, per annum, amounted, at
the end of 9 years, hy annual payments, to the sum of £5987. 432; by
half-yearly payments, it amounted to the sum of £6062. 382 ; but by
quarterly payments, to the sura of £6100. 5 15 ; required the yearly value of
that annuity ?
First,— For Annual Payments ;—
Rate, 4i per cent. ; ratio= 1. 0425 ;
log.. Table A,= 0.0180761
Mult.byno.ofpaym.= 9
Products . . 0.1626849 Nat.no. 1.4544033
Subt. the integral part of the ratio, viz., 1
Diminished natural number = . • 0. 4544033 Log. ar.co.0. 3425585
Amount of annuity by annual payments == £5987. 432 Log.=3. 7772406
Decimal part of the ratio =: . 0425 ; log., Table B, = . . 8. 6283889
Yearly value of the annuity =: £560 Log. = . . . . 2. 748 1 880
Second,— For Half-yearly Payments :—
Rate, 4i per cent.; ratio =: 1 . 02125 ;
log., Table A, =: 0. 009132 1
Multbyno.ofpaym.
=:9yearsx2=: 18
0730568
0091321
Products . • 0.1643778 Nat. no. 1.4600838
Subt. the integral part of the ratio, viz., I
Diminished natural number =: . ; 0. 4600838 Log. ar. co. 0. 337 1 630
Amount of annuity by half-yearly payments=£6062. 382 Log.3. 7826361
Decimal part of the ratio = . 02125 ; log.. Table B, = • . 8. 3273589
Half the value of the annuity = . , , . £280 Log. = 2. 447 1580
Yearly value of the annuity =••••• £560, as required.
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Third, — For Quarterly Payments : —
Rate, 4^ per cent ; ratio = 1. 010625 ;
log., Table A, n 0. 0045900
Mult, by no. of paym .
= 9year8x4= 36
00275400
00137700
Product = . . 0. 1 652400 Nat. no. 1 . 4629855 Lpg. ar. co. 0. 3344326
Amount of annuity by quarterly payments= £6 100. 515 Log.=3. 7S5366S
Decimal part of the ratio =: , 010625 ; log., Table B, =z ,8. 0263289
One fourth the value of the annuity = * . . £140 Log.=2. 1461 280
Yearly value of the annuity =t • • • • . • £560, as required.
. Problem III.
Qiven the Jnnuity^ the Rate peg Cent ^ per Jnnum^ and the Amount^ tott&
tlie Modes of Payment; to find the Time, or Number of PaymeHts.
To the logarithm of the amount, add the logarithm of the decimal part
of the ratio (Table B) ; find the natural number answering to the sum of
these two logarithms, and increase it by the whole, the half^ or the fourth
of the annuity, according as the instalments may be annual, half-yeariy, or
quarterly. From the logarithm of this increased natural number, subtract
the logarithm of the whole, the half, or the fourth of the annuity, according
to the manner in which the instalments may be made payable : divide the
remainder by the logarithm of the ratio (Table A), and the quotient will
express the time, or number of payments } which will be in years, if those
be annual; otherwise, in half years or quarters, as the case .may be. The
latter expressions iv e to be divided by 2 or 4, to find the time.
Example.
An annuity of £560^ at 4^ per cent, per annum, amounted, by annual
pa}'ments, to the sum of £5987* 432 ; by half-yearly payments, to the sum
of £6062.382; but, by quarterly payments, to the sum of £6100. 5 15;
required the time corresponding to each mode of payment }
.]itizedbyV^OO*^l^
AT COMPOUND INTBEBST. • 699
First,— For Yearly Payments : —
Amount byyearly payments
=£5987- 432 Log. 3. 7772406
Dec. part of the ratio,
Tab.B, . 0425 Log.8. 6283889
Sum of the two Iogs.=:2. 4056295 Nat. no. 254. 46585
Whole annuity for yearly payments =: £560
Increased natural number = ... 814. 46585 Log.2. 9108729
Whole annuity = £660, the log. of which is = . . . 2. 748 1 880
Rate, 4^ per cent. ; ratio= 1 . 0425 ; divide by log. of this,
Table A, = 0.0180761)0.1626849=9
hence the time, by annual payments, is 9 years.
Second, — For Half-yearly Payments :—
Amoui(t by half-yearly payments
=£6062. 382 Log.3. 7826361
Dec. part of the ratio.
Tab. B, . 02125 Log.8. 3273589
Sum of the two logs. 2. 1099950 Nat.no.l28. 82347
Half the giTMi annuity s; • . • . £2(90
Increased natural number = . . . 408. 82347 Log.2. 61 15358
Half the given annuity = £280, the log. of which is = 2.4471580
Rate, H per cent. ; ratio= 1 . 02 1 25 ; divide by log. of this.
Table A, = 0.0091321)0.1643778=18
Now, 18 half-yearly payments, divided by 2, shovf the time to be 9
years.
Third, — ^Por Quarterly Paym^ts ;—
Amount by quarterly payments
=£6100. 515 Log.3. 7853665
Dec. part of the ratio.
Tab. B,. 010625Log.8. 0263289
Sum of (be twologs.= 1. 81 16954 N«t.po.64, 81797
/Google
Digitized by '
Sum of the two logs.r: 1 . 81 16954 Nat.no. 64. 81797
One fourth of the given ahnuity = « £140
Increased natural number = . . . 204. 81797. Log. 2. 31 13680
One fourth of the annuity = £140, the log. of which is = 2. 14612S0
Rate, H per cent ; ratio =: 1 . 010623 ; divide by log. of
this. Table A, = 0.0045900)0.1652400=36
Now, 36 quarterly payments, divided by 4^ give 9 years 5 which, there-
fore, is the required time.
PRESENT WORTH OP ANNUITIES IN ARREARS, AT
COMPOUND INTEREST.
Definiiim.^^When an anittu/y, to be entered on immediately, is sold
for ready money, the price which ought to be paid for it is called its
present worthy
Problem I.
Gioen an Jnnuiiy, the Time qf its Continuance, and the Rate per Cent,
per Annum ; tojtnd the present Worth of thai Anmtky.
RULB. .
Multiply the logarithm of the ratio fTable A) by the number of payments;
subtract the product from the logarithm of tlie whole, the half, or the fourth
of the annuity, according to the mode of payment : find the natural num-
ber of the remainder, and subtract it from the whole, the half, or the fourth
of the ahnuity, according as the instalments may be yearly, half-yearly, or
quarterly ; find the logarithm of this difference, from which let the loga-
rithm of the decimal part of the ratio (Table B) be subtracted, and the
remainder will be the lo^rithm of the present worUi of the annuity.
Example.
An annuity of £365 is to be continued 7 years; required the present value
thereof, by yearly, half-yearly, and quarterly payments ; allowing the pur-
chaser 5 per cent, per annum, compound interest, for the use of his ready
money?
Digitized by
Google
AT COMPOUND INTBRBST, 701
First, — ^For Yearly Payments : —
Rate, 5 per cent. ; ratio s= 1. 0500 ;
log., Table A, = . 0.0211893
Mult, by no. of payments= 7
Products .... 0.1483251
Whole annuity,£365Log.2. 5622929 . . £365
Remainders . . . 2.4139678 Nat.no. 259. 39869
Difference between nat. number and annuity=105. 60131 Log. 2. 0236693
Decimal part of the ratio. Table B, = . 0500 Log. = . 8. 6989700
Present worth of given annuity by yearly paym. £2112. 0267 Lg.3. 3246993
Second, — ^For Half-yearly Payments :-*
Rate, 5 per cent; ratio = 1. 02500;
log.iTableA,= . 0.0107239
Multiply by number of
payments =7x2= 14
0428956
0107239
Product = , . . . 0. 1501346
Halfannuity,£182. 5 Log.2. 2612629 . . £182.5
Remainder = . ... 2. 11 1 1283 Nat.no. 129. 16009
Diff. between nat. no. and half the annuity = 53. 33991 Log. 1 . 7270523
Decimal part of the ratio = . 02500 Log., Table B, = . 8. 39/9400
Present worth of annuity by half-yearly paym. £21331 5966 Log.3. 3291 123
Third, — For Quarterly Payments : —
Rate, 5 per cent. ; ratio =1.01 2500 ;
log., Table A, = .0. 0053950
Multiply by number of
payments=7yearsx4= 28
0431600
0107900
Products . . . . 0.1510600
Digitized by
Google
Products .... 0.1510600
One4thofann.£91.25Lg.l. 9602329 . . £91.25
Remainder = . . . 1 . 809 1 729 Nat. no. 64. 442582
Diff.betw.nat. num. and one 4 th of annuity = 26.807418 Log. 1.4282550
Decimal part of the ratio = .012500 Log.^ Table B, s 8.0969100
Present worth of the ann. by quarterly paym. £2144. 5936 Log.3. 3313450
Pboulbm II.
Given the present Worth qfan Annuity, the Time qfUs Cqntibiuaneef emd
the Rate of Interest ; to find the yearly Value of that Jnmdty.
Rule.
, Multiply the logarithm of the ratio (Table A) by the number of pay-
ments ; take the arithmetical complement of the product, find the natural
number corresponding thereto, and let its arithmetical complement or
difference to zero be noted. Then, to the arithmetical complement of die
logarithm of this difference to zero, add the logarithm of the purchase-
money, or present worth of the annuity, and the logarithm of the decimal
part of the ratio (Table B) : the sum of these three logarithms (abating 10
in the index,) will be the logarithm of the annuity, or of its half, or quarter,
according as^the mode of payment may be by yearly, half-yearly, or quar-
terly instalments. . .
Example,
The purchase-money, or present worth of an annuity, which is to con-
tinue 7 years, amounts, by yearly payments at 5 per cent, per annum, to
the sum of £2 112. 0267 ; by half-yeariy payments, to the sum o]r£2133. 5966 ;
but by quarterly payments, to the sum of £2144. 5986 } required the yearly
value of the annuity, agreeably to each mode of payment ?
First,— For Yearly Payments :—
Rate, 5 per cent. ; ratio = 1 . 0500 ;
log.. Table A, = 0.0211893
Mult, by no. of paym. 7
Products . . 0.1483251
AT COIIMUNO IKTBRI
Product = . . 0.1483251
Arith. comp, s 9. 85 16749 Nat.no. 0. 7 106
Ar4 CO. of nat. niim«, or diff. to zero s 0. 2893
Purchase-money^ or present worth of the annuitj
payments^ a £2112. 0267 . Log. = .
Decimal, part of the ralio^ Table B, a .0500
Yearly value of the annuity as £365 IiOg« i
Second, — For Half-yearly Pa |
Rate, 5 per cent. ^ ratio =: 1 . 02500 5
log.. Table A, = 0. 0107239
Multiply by no. of
paym.=7yr8x2=i 14
0428956
0107239
Products: . . 0^501346
Arith, comp. = 9. 8498654 Nat.no. 0. 70!
Ar. CO. of nat. num., or diff. to zero, = 0. 29
Purchase- money, or present worth of the anm
yearly payments, £2133. 5966 Log. =
Decimal part of the ratio. Table B, =: • O250
Half the annuity =: . . £182.5 Log
Yearly value of the annuity=:£365, as require<
Third,~For Quarterly Pa I
Rate 5 per cent 3 ratio = 1. 012500 ;
log., Table A., = 0. 0053950
Multiply by no. of
paym.= 7yrs x 4 = 28
U431600
0107900
Products . . • 0.151060
Digitized by
Google
Products . . 0,151060
Arith. comp. = 9. 8489400 Nat. no. 0. 7062200
Ar. CO. of nat. num., or diff. to zero^ =r 0/2937800 Log. ar. ^o. 0. 5319778
Purchase money or present worth of the annuity by quar-
terly payments, £ 2144. 5936 Log. = ' 3.3313450
Decimal part of the ratio, Table^B. = . 012500 Log. . . 8. 0969100
One fourth of the annuity :s £ 91.25 Log. = . . . 1.960232S
Yearly value of the annuity =£365, as required.
Probijim III.
Gioefi an AnnuiSy^ the present worth of that Annuity, and the Mate per
Cent, per Annum ; to find the Time of Us Continuance.
RlTLB.
To the logarithm of the present worth of the annuity, add the logarithm
of the ratio^ Table A., and find the natural number corresponding to their
sum ;— augment tlie present worth of the annuity by the yearly value of
the annuity^ its half, or quarter, according to the modes of payment ; and
find the difference between it and the natural number found as above : —
find the logarithm of this difference, and subtract it from the logarithm of
the given annuity, its half or quarter, as the case may be :-^ivide the re-
mainder by the logarithm of the ratio. Table A, and the quotient will be
the number of payments ; which will be in years if those payments be an-
nual ; otherwise in halves, or quarters according to the modes of instal-
ment.
Example.
An annuity of £365, can be purchased for the sum of £ 21 12. 0267, by
annual payments; for £2133.5966, by half-yearly payments, or for
£2144.5936, by quarteriy payments ; the purchaser is to be allowed 5
per cent, per annum for the use of his ready money ; required the time of
the continuance of that annuity agreeably to each mode of payment ?
First, — For Yearly Payments :—
Present worth of the ann. = £2112. 0267 Log. = 3. 3246993
Rate, 5 per cent ; ratio. Tab. A. = 1 . 0500 Log. = 0, 02 1 1 893
Sum of the two logs. = 3. 345886
iQQgjk
Sum of the two logs. =3. 3458886 Nat.N.=2217. 6031
Present worth of the annuity =
£2112,0267 + whole ann. £365 = 2477.0267
DiflFerence= ..•.,.... 259. 4236 Log.=2. 4140095
Given annuity £ 365, the log. of which is =...... 2. 5622929
Rate 5 per cent; ratio =3 1. 0500; divide ,
bylogarithmof this. Table A. = . . . 0.0211893) 0.1482834
^ 7 payments; — hence the time, or ' Continuance of the annuity is 7
years.
Sccond^Por Half Yearly Payments :—
Present worth of the ann. =
JE 21S3. 5966 Log.=3. 3291 123
Rate 5 per cent ;
ratio. Tab. A. =
1. 02500 Log.=0. 0107239
Sum of the two logii.=3. 3398362 Nat.N.=2I86. 9367
Present worth of the annuity =
£2133.6966 -h half ann. £ 182.5 =2316.0966
Differences 129. 1599 Log.= 2. 1111277
Halfgivenann.=£I82.5thelog. ofwhichis 2.2612629
Rate 5 per cent; ratio = 1,02500; divide
by logarithm of this. Table A. = ... 0. 0107239 ) 0. 1501352
= 14 payments; and 14 -s-' 2 = 7 years, is the time or continuance of
the annuity, as required.
Third,— For Quarterly Payments :«^
Present worth of the ann. =
£2144. 5936 Log.=3. 3313450 .
Rate 5 per cent ; ratio
1. 012500
Log.Tab.A.= 0.0053950
Sum of the two Iog8,=3. 3367400 Nat.N.=217I . 4010
Present worth of the annuity =
^ 2144. 5936+one fourth
of the ann. £91.25 = .... 2235.8436
Differences 64.4426 Log.= 1.8091730
One fourth of the given ann. = £ 91. 25, the log. of which is=: 1. 9602329
Rate 5 per cent ; ratio = 1 . 012500 : divide •
by logarithm ofthis, Table A. = . . . 0.0053950) 0.1510599
= 28 payments ; now 28 -*- 4 = 7 years, which, therefore, is the time
required.
22
PRESENT WORTH OF ANNUITIES IN REVERSION, AT
COMPOUND INTEREST.
•
D^ithfi.'^An annuity is said to be in retyermn when it is not to b^
entered upon until some particular etent has happened^ or until some oer*
tain period has elapsed after the .time of its sale :-^the ready money which
should be paid down for an anntiity of this dcscriptioh is called its present
worth.
pROmjftM I.
Given an Annuity in Reversion, the Time of iU Continuance^ the Period
at which it is to be entered upon, with tlw Rate of Interest; to find Us
present Worth.
RULB.
Find the logarithm of the present worth of the annuity agreeably to the
time of its continuance^ as if it were to be entered upon immediately, by
Problem I., page 700 } from this logarithm subtraet the logarithm of the
ratio. Table A., multiplied by the time which is to elapse before the pur-
chaser enteni tipon the annuity, and the remainder will be the logarithm of
the present worth of the annuity in reversion.
Noiet^f the annuity be made payaUe by half-yearly, or quarteriy m-
stalmentSi the ratio, &c. &c* are to be proportioned accordingly, the same
as in the preceding Problems.
Example,
What is the present worth of the reversion of a lease of £475 per an-
num, by yearly payments, to continue 25 years, but not to be entered upon
till the end of 7 years after the time of sale ; allowing the purchaser 5| per
cent, for present payment ?
Rate 5f per cent. ; ratio, Tab« A. && 1. 0550 Log.a « . .0. 0232525
Multiply by number of payments =•...••••.•• 25
1162625
0465050
Product == 0.5813125
Annuity, or lease = £ 475, Logarithm = 2. 6766936
Natural Number sea 124.5607 . . • .Logarithms 2.0953811
Differences . . 350.4393
and annuity^ or lease = . . 350.4393 Log. = 2.5446128
Decimal part of the ratio, Tab. B=. 0550 Log. = 8. 7403627
Logarithm of the present worth of the lease, sup-
posing it were to be entered upon immediately s=3. 8042501ss:£6371«6
Rate 5^ per eent. ; ratio, Tab. A.ss 1. 0550 Log.a
Ol 0232525 X 7 years = ...... .0.1627675
Present worth of the rev. « £4380.0848 Log.sS. 6414826
Note. — The latter part of the operation may be abridged in the follow-
ing manner, viz : —
To the logarithm of the difference between the natural number and the
gjyen annuity or lease, add the arithmetical complement of the logarithm of
the decimal part of the ratio. Table B., and the arithmetical complement
of the product of the logarithm of the ratio, Table A., by the time which is
to elapse before the purchaser enters upon the annuity; the sum of these
three logarithms, abating 10 in the index, will be the logarithm of the pre-
sent worth of the annuity or lease in reversion.
Thus. Diff. betw.nat. numb, and ann. or lease=350. 4393 L.=2. 5446128
Pec. part of ratio. Tab. B.=. 0550 Log. arith. comp.= 1. 2596373
Ratio,Tab. A.=l . 0550 L,=0. 0232525 x 7 ys. Ar. com. 9. 8372325
Present worth of the reversion =£4380. 0848 Log.s 3« 64 14826
PaoBLEM II.
Given the present Worth of an Annuity in Rev'ersion, the time of its con*
tinuancej and the Period at which it ia to be entered upon, with tlie
Rate of Interest ; to find the yearly value of the Annuity.
RuL?.
Multiply the logarithm of the ratio. Table A., by the number of pay-
ments or time of continuance : take the arithmetical complement of the
product, and find the natural number corresponding thereto, and let its
arithmetical complement, or difference to zero be noted :-^theii, to the
arithmetical complement of the logarithm of this difference to zero, add the
logarithm of the decimal part of the ratio. Table B., the product of the log"
arithm of the r&tio. Table A. by the time that is to elapse before entering
upon the reversion, and the logarithm of the present worth of the reversion ;
the sum of these four logarithms, abating 10 in the index, will be the log-
arithm of the yearly value of the annuity.
Note. — If the annuity be made payable by half-yearly, or quarterly in-
stalments, the ratio, &q. &c, are to be proportioned accordingly, the same
as in the preceding Problems. -
2z 2
Example.
What is the yearly value of an annuity or lease, to be entered upou 7
years hence, and then to continue 18 years, that is 25 years continuance,
and which may be bought for the sum of £ 4380. 0648, ready money,
allowing the purchaser 5i per cent, compound interest ?
Rate5§ percent; ratio,Tab. A. = 1.0550 Log. = . . .0.0235525
Multiply by number of payments = 25
1162625
0465050
Products • • 0-5813125
Arithmetical complement = 9.4186875
The natural number answering to arithmetical complement,
considered as a Logarithm, is =0. 2622331
Difference to zero =^ - . . 0. 7377669 Log. ar. comp. 0. 13^0808
Dec. part of the ratio. Tab. B. = . 0550 Logarithm= 8. 7403627
Ratio, Table A.=l. 0550 Log.= 0. 0232525 x 7 years = 0. 1627675
Present worth of the reversion = £4380.0848 Logarithm = 3.6414826
Yearly value of the annuity or lease = £ 475 Logaritlim = 2. 6766936
Problem III.
Gftt?^ an Annrnty in Reversio?}, its present Wofth^ the Period at txMch it
is to be entered ttpon, and the Rate of Interest ; to find the Time of its
Continuance.
RULB.
To the logarithm of the present worth of the reversion, add the product
of the logarithm of the ratio, Table A., by the time that is to elapse before
entering upon the annuity, and the logarithm of the decimal part of the
ratio, Table B ; the sum of these three logarithms, abating 10 in the index,
will be the logarithm of a natural number. Take the diflference between
the natural number, thus found, and the yearly, half-yearly, or quarterly
vijue of the annuity, according to the nature of the instalments ; subtract
the logarithm of this difference from the logarithm of the yeariy, half-
yearly, or quarterly value of the annuity, as the case may be : divide the
remainder by the logarithm of the ratio. Table A., and the quotient will be
the whole time of the continuance of the given annuity or lease ; which
will be in years, hdves, or quarters, according to the mode of payment.
mmgm
An annuity or lease worth £ 475 per annum (to be entered upon at the
end of 7 years after the time of sale,) may be purchased for the sum of
^4380.0848; required the time of the continuance of that annuity,
allowing the purchaser 5^ per cent, for the use of his ready money }
Present worth of the annuity or lease=£4380. 0848 Log. = 3. 6414826
Ratio, Tab. A.= 1. 0550 Log.=0. 023252&X 7 yrs. (Rate
Si per cent.) = 0. 1627675
Decimal part of the ratio, Tab. B. = • 0550 Logarithm = 8. 7403627
JNTatural number =
350. 4393 . Logarithm = 2. 5446128
Yearly value of given ann.s • 475.
Logarithm = 2. 6766936
Difference = 124.5607 Logarithms 2.0953811
Rate 5 i percent; ratio. Table A.= 1 . 0550 ; divide by
logarithm of this = 0.0232525) 0*5813125 =
' 25 payments ; — hence the tinie» or continuance of the annuity is 25
years, as required*
JN'ote, — Should it be required to find the time that should elapse between
the perio(l3 of purchasing and entering upon the annuity or lease, it may
be very readily determined by an indirect solution of Problem I., page
706 ; — as thus :
From the logarithm of the present worth of the amiuity determined as
if it were to be entered upon immediately, subtract the logarithm of the
present worth thereof in reversion : — divide the remainder by the logarithm
of the ratio, Table A., and the quotient will be the time required ; — ^this^
it is presumedi is so very obvious as not to require the illustration of an
exannple.
PRESENT WORTH OF FREEHOLD ESTATES, OR PERPETUAL
ANNUITIES,
At an Assigned Rate per Cent. Compound Interest, to be entered on
immediately.
DefinUion.^Freehold Estates, or perpetual jinnuitiessignify any interest
of money, rents, pensions, grants, &c. payable yearly, half yearly, or quar-
terly, and to continue for ever:— in buying these, the purchaser is allowed
a certain per centage for his ready money ; which money is called the pre^
sent worth of the perpetual annuity, to be entered on immediately.
Paoblbm L
Given the Yearly Bent qf a freeliold Estate ^ or perpetual Annuity, Ofid
the Rate per Cent.; to find the present fVarth thereof.
RULB.
From the logarithm of the yearly, half-yearly, or quarterly ralue of the
f given perpetual annuity, according to the mode of payment (the index
being increased by 10), subtract the . logarithm of the decimal part of the
ratio corresponding to such payment. Table B., and the remainder will be
the logarithm of the present worth of the given perpetual annuity, &c«
Example.
What is the present worth of a freehold estate, or perpetual annuity of
£360, per annum, payable half yearly, allowing the purchaser 4| per cent,
compound interest, for his ready money ?
Given perpetual annuity=£360 ; half of which is £180 Log.s2. 2552725
Rate, 4} per cent.; dec. part of ratio. Tab. B^. 02375 Log.^8« 3756636
Present worth of the given freehold estate=£7578. 93 Log.=3. 8796089
Pr6bl|im II.
Gtioen the present Jf^orth of a fteehM EsMe, or perpetual jhmuUyj and
the Bate per Cent.; to find. the yearhf Value of that Estate, or per*
petual Annuity.
Rule,
To the logarithm of the present worth of the given perpetual annuity,
add the logarithm of the decimal part of the ratio (Table B), according to
the mode of payment ; and the sum will be the logarithm of the yearfy
value of the annuity, or of its half or quarter, as the case nay be.
Example.
If a freehold estate, or perpetual annuity, be bought for £7578. 93, what
ought the yearly value thereof to be, allowing the pureheaer 4} per cent,
compound interest, for his ready money ; the rent being payable half-
yearly ?
Present worth of the given perpetual annuity=£7578. 93 Log. 3. 87960S9
Rate, 4f per cent. ; decimal part of the ratio. Table B, =
• 02375 (half-yearly payments) Log. =: . * • . • 8.3756636
Half-yearly value of the annuity s £180 Log.= . . . 2.2552725
Yearly rent of the freehold e8tate=:£360^ as required.
OR FERPBTUAL ANN!
Problbm III
Oilmen 4he yearly Value of a freehold Etta
the presefit tVorth thereof; to Jim
. RULR.
From the logarithm of the yearly, half-y
given perpetual annuity, according to the
being increased by 10), subtract the logari
that annuity, and the remainder will be the
of the ratio : with thia enter Table B., and th
thereto will be found standing abreast there
Table A.j which will be the ratte per cent, re
Example.
If a freehold estate, or perpetual annuity
half-yearly, be sold for £ 7576. 93, what i:
interest, allowed to the purchaser ?
Yearly Value of the annuity, £860, half of wl
Present worth of ditto ss £7578. 93. the lo
Rate, 4| per cent.. Table A, = decimal pan
Table B, « .02375 Log. a , .
PRESENT WORTH OF FREEHOLD E
ANNUITIES IN REA
At an assigned Rate per Cent., '
D^fmiion^^Freehold estates, or perpefuc
any interest of moneys rents, pensions, gra
yearly, or quarterly, and to continue for e
entered upon until some particular event
specified time has elapsed after the time o(
purchaser holds the freehold, &c., after he
the reversion ; and the money which he
called its' present worth.
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712 PRBSBNT WORTH OF FRSEHOLD BSTATBS,
Problem I.
Given the yearly Rent of a freehold EsiatCy or perpetual Amnaiy^ the
Time at which it is to be entered upon, and the RaJLe per Cent. ; to find
the present Worth of the Reversion of that Estate.
Ruu.
To the product of the logarithm of the ratio (Table A) by the time that
is to elapse before entering upon the reversion, add the logarithm of the
decimal part of the Tatio (Table B) : subtract the sum of these two loga-
rithms from the logarithm of the yearly value of the given perpetual
annuity (the index being increased by 10), and the remupder will be the
logarithm of the present worth of that perpetual annuity, or freehold
estate.
Example.
The reversion of a freehold estate^ or perpetual annuity of £490 per
annum, to be entered upon 7 years hence, is to be sold ; what is its present
worth, allowing the purchaser 5 per cent*, compound interest, for his ready
money ?
Rate, 5 per cent. ; ratio. Table A, = 1 . 0500 Log. = • 0. 021 1893
Mult, by given time, viz., the time before entering upon the revers. 7
Products 0.1483251
Decimal part of the ratio. Table B, = . 0500 Log. = . 8. 6989700
Sum of the two logarithms 8.8472951
Yearly value of the estate, or perpetual annuity,=:£490 Log.=2. 6901961
Present worth of the reversion r^ £6964. 6774 Log. = • 3. 8429010
Problbm IL
Given the present Worth of a freehold Estate^ or perpetual Annuity in
Reversion^ the Time at which it is to be entered upon, and the Rate per
Cent.; to find its yearly Value.
RULB.
To the product of the logarithm of the ratio (Table A) by the time that
is to elapse before entering upon the reversion, add the logarithm of the
decimal part of the ratio (Table B), and the logarithm of the present worth
of the reversion of the given perpetual annuity : the sum of these three
logarithms (abating 10 in the index^} will be the logarithm of the }'early
value of that perpetual annuity.
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If a freehold estate, or perpetual annuity, to be entered upon 7 years
hence, be sold for £ 6964, 6774, what is the yearly value thereof, allowing
tlie purchaser 5 per cent., compound interest, for his ready money ?
Rate, 5 per cent. ; ratio. Table A, =s ] . 0500 Log. = . 0. 02 1 1 893
JVf ult. by given time, viz., the time before entering upon the revers. 7
Product = 0.1483251
Decimal part of the ratio. Table B, = . 0500 Log. = . 8. 6989700
Present worth of the reversion=jE;6964. 6774 Log. = . 3. 8429010
Yearly value of the given perpetual annuity ^ £490 Log.=2. 6901961
Problbm IIL
Given the yearly Value of a freehold Estate^ or perpetual Amuity in
Reversion, ilie Rate per Cent,, and its present Worth; to find the Time
that must elapse before entering upon the Reversion,
Rule.
To the logarithm of the present worth of the reversion, add the loga-
rithm pf the decimal part of the ratio (Table B) ; subtract the sum of
these two logarithms from the logarithm of the yearly value of the given
perpetual annuity : now, the remainder being divided by the logarithm of
the ratio (Table A), the quotient will express the time that must elapse
before entering upon the reversion of that perpetual annuity.
Example,
If a freehold estate, or perpetual annuity of £490 per annum, be sold for
£6964. 6774, in what time hence will the purchaser be. entitled to enter
thereon, allowing him 5 per cent., compound interest^ for the use of his
ready money ?
Pres. wprth of the revers. of given annuity=6964. 6774 Log. 3. 8429010
Rate, 5 per cent ; dec. part of ratio, Table B, 0. 500 Log. 8. 6989/00
Sum of the two logarithms = 2.5418710
Yearly value of the given perpetual annuity £ 490 Log. = 2. 6901 96 1
Rate, 5 per cent j ratio, Table A,= 1 . 0500; divide by log,
of this == . 0.0211893)0.1483251=7
. Hence, the purchaser will be entitled to enter upon the reversion at the
end of 7 years.
PftOBLBM.
To find in hcuo many Years any Principal or Sum of Money wiU dauUe
itself, at compound Interest, by yearly, half-yearly, or quarterfy
Payments.
Let the logarithm of the ratio (Table A) be considered as the decimal
part of a natural number \^ fiod the logarithip corr^ponding thereto^ and
subtract it from the constant logarithm 9* 4786093 : the remainder will be
the logarithm pf the time in which a given sum of money will double iUelf
at any proposed rate of interest within the limits of Table A : if the pay-
ments of interest be annual^ the time will be expressed in years ; odierwise,
in half years^ or quarters of years^ as the case may be.
Note. —The constant logarithm is thus determined :— Let the drnMe of
any given sum of money be represented by the number 2, the logarithm of
which is 0. 3010300; consider this as the decimal part of a natural num-
ber: then^ the logarithm corresponding thereto is 9.47860983 which,
therefore, becomes a constant expression for all modes of payment and
ratos of interest.
Example*
Required the number of years in which any given sum of money will
double itself, at compound interest, by yearly^ half-yearly, and quarterly
payments ; the rate being 5 per cent, per annum ?
First,— For Yearly Payments :—
Constant log. = 9.4786098
Rate, 5 per cent ; log*, Tsj^lt A, S7 0» 021 1893 ; consider this
as the decimal part of a natural number, the log» of which \s^%» 3261 167
Number of years, as required, = 14. 2067 Log. ss . . 1. 15249S1
Second, — For Half*yearly Payments :—
Constant log. =3 9.4786098
Rate, 5 per cent.; log. Table A, = 0. 0107239 ; consider this as
the decimal part of a natural number, the log. of which is = 8. 0303528
Time, in half years, cs . . . 28. 07094 Log. s: . h 4482570
Number of years, as required, = 14.03547; which, therefore, is the
time in which a sum of money will double itself, at 5 per cent* ccnnpoimd
interest, by half-yearly payments.
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OR PBRPETUAL ANVUmX3 IN
Third,— For Quarterly Pa
Constant log, = . • •
Rate, 5 per cent.; log., Table A,=0. 005395(
the decimal part of a natural number, the log
Time, in quarters of years, ps 55. 79797
Number of years; as required, 13^ 94949 ; whi
which a sum of money will double itsel
ioterest^ by quarterly payments*
It is after this manner that the followin
excepting, however, the last column, or that
merely expressed by the quotient of £100, d
cent. : thus, £100 -+- £5 = 20 years ; which,
a given sum of money will double itself, at 5
interest.
A Tablb,
Exhibiting tlie Time ia which any Sum of
several given Rates per Cent, per Annum,
at Simple Interest.
Rates
Crat.
Time, for compouod Interes
Yearly
Payineots.
Half.yearly
Payments.
£
3
H
3i
4
44
4i
4|
5
f
?
8
9
10
.Years.
23. 4498
21. 6723
20. 1488
18. 8284
17. 6730
16. 6535
15.7473
14.9365
14.2067
13. 5464
12. 9461
12.3981
11.8957
11.0067
10. 2448
9. 0065
8.0432
7.2725
Years.
23.2779
21. 5003
19. 9770
18. 6567
17.5013
16.4820
15.5759
14. 7652
14.0355
13.3753
12. 7752
12.2273
11.7249
10.8361
10.0744
8. 8365
7. 8736
7. 1033
By the above table it is evident, that if £
compound interest, at the rate of 5 per ce
itself, by yearly payments of the interest,
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716 COMPOUND INTERBST.
payments^ in 14 years j and by quarterly payments, in 13^ years ; whilst
at simple interest, it will not double itself in less than 20 years. Hence,
if the given sum be £1250, it will amount, in 13-^^ years, by quarterly
payments, to the sum of £2500 ; in 21 -fc years, to the sum of £5000 ; in
41-/V years, to the sum of £10000 ; in 55/^ years, to the sum of £20000 ;
and so on in geometrical progression : while at simple interest, the same
sum would only amount, in the same space of time, viz., 55-^ years, to the
sum of £4737. 7s. 6d.
jRcmarfc.— The preceding problems and examples contun all that is
essentially necessary to be known, independently of theory, in the doctrine
of compound interest. The author is not aware that the direct loga-
rithmical solutions of the various complex cases connected with this
subject have been given by any other writer : many, indeed, have published
theorems for this purpose ; but these theorems (such as those given by
the late ingenious Dr. Maskelyne, in his very learned Introduction to
Taylor's Logarithms, under the head Compound Interest,) are expressed in
such a scientific manner as to be of little use, in a mere practical point
of view ; being much better adapted for employing the minds of the cu-
rious in mathematical researches and investigations, than for abridging
the labour attendant on arithmetical computations.
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THE USE OF THE GENERAL VICTUALLING TABLE,
Contained in Vol. II., Page 661.
As this Table contains the exact daily proportion of sea provisions for
any given number of men within the ordinary limits of victualling, it will
"be found of considerable utility to the Pursers of the Royal Navy, in
closing their annual accounts, and in completing the ship's provisions to
any specified time : it will also be of great use to officers serving as
Commanders and Pursers ;-— and, perhaps, to those gentlemen in the
Victualling Department of His Majesty's service who are employed in the
examining and auditing of the Naval Victualling accounts of the above-
mentioned officers*
Remarks.
1 . — ^As the size of the page would not admit of separate columns being
employed for the salt beef and salt pork, and as the allowance of each of
these species is precisely the same ; one column only has been introduced
into each page of the Table on account of those articles of victualling 5 —
which column contains the exact proportion of each. — Hence, in taking out
the proportions from this column, corresponding to any given number of
men, care must be taken to put down such proportions ixvice ; that is, first
for salt beef, and then for salt pork ; or, otherwise, to double those propor-
tionsy at once, for salt meat generally.
2. — In like manner, as the size of the page would not admit of separate
columns being employed for bread and beer, and for oatmeal and vinegar ;
it will be necessary, in taking out the proportions of those species, corres-
ponding to any given number of men, to put down as many gallons of beer
as the gecoJid column expresses pounds of bread ; and as many gallons of
vinegar as the last column expresses gallons of oatmeal.
The following problems, will illustrate the principal uses to which this
Table may be applied.
PaofiLEM I. ^
Given the Number of Men victualled for one Day, to find tlie corresponding
Proportion of each Species of Provisions.
Rule*
If the given number can be found in the left-hand column of the table,
the corresponding proportion of each species of provisions will be found
abreast of it in the same horizontal line j but if it cannot be exactly found,
which in general will be the case, write down any two or more of the
tabular numbers that will make up the given one, opposite to which pat
down the corresponding quantities of provisions : then, the sums of these
quantities will be the true proportion of each species of provisions.
Example.
Let the number of men victualled for one day be 45685 ; reqiured the
true proportion of each species of provisions corresponding thereto ?
Men. Bread. Beet' Salt Meat. Flour. Peaae. Sagar. Coeoa. Tea. OatmatL ViMgar.
45000 ^vc 46600 45000 33750.0 16875.0 1406.2 4218.12 2S12.8 703.2 401.6.4 401. 6L 4
600 do. 600 600 450. 0 225. 0 18. 6 56. 4 37. 8 9. 6 6. 2. 12 5. 2. 12
85 do. 85 85 63.12 31.14 2.5^ 7.15} 5.5 1.5(0.6.1 0.6.1
45685 give 45685 45685 34263.12 17131.14 1427.5^ 4282.15} 2855.5 713.13^ 407.73 407. 7.3
Note. — In making out the Purser's annual victualling account, the pro-
portions of salt beef and pork, as given in the table, are to be doubled, or
thrown into one sum under the head of salt meat, as above.
If there be miy fresh meat issued during the period of the account, sub-
tract the amount thereof from the number victualled, and then take out
the proportion of salt meat, flour, and pease^ corresponding to the remain-
der : thus, suppose the quantity of fresh meat issued to be 22238 pounds.
No. vie. for one day is 45685
Lbs. fr. meat issued^: 22238
— — — Salt meat. Flour. Pease.
Remainder = . 23447. Now, 23000 gives 17250.0 8625. 0 718.6
400 do. 300.0 150. 0 12.4
47 do. 35.4 17.10 1.3|
23447 gives 17585. 4 8792.10 732. 5i
See Pursers' Instructions (Appendix), No. 21, page 11 7*
Problem II.
Given the Complement of Men, and the Number of Days for which they
are to be victualled; to find the Proportion of each Spedes of Pro-
visions.
Rule.
Multiply the complement of men by the given number of days for which
they are to be provisioned, and the product will be the number to be
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erwgr
directed in the last problem*
Example.
Let the complement of a ship be 275 men, and the time for which they
a.re to be victualled 4 lunar months or 112 days; required the propprtion
of each species of sea provisions ?
Given complement of mens 275
M ultlply by time, in days^ss 1 1 2
Product = • . « • 30800, which tsjthe number to be victualled for
one day.
M«B. Bread. Spitits. Salt Beef. Salt pAfk. Flour. Peaia. Sugar. Cncoa. Tea. Oatmeal. Vinega'
3OOOO^ve30000 937.4 11250 11250 11250 937.4 2812.8 1875 468.12 267.6.12 267.6.12
800 do. 800 25.0 300 300 300 25.0 76.0. 50 12.8 7.1. 2f 7.1.2
30800 g^ive 30800 962.4 11550 11550 11550 962.4 2887.8 1925 481.4 275.0.0 275.0.0
And these are the exact proportions of the different species of provisions
for the given complement and time. *
RenfKirks.
1. Since the salt beef is generally cut up in 8 lbs. pieces, and the pork
in 4 lbs. pieces, the pounds of salt beef are to be divided by 8, and the
pounds of salt pork by 4 : the respective quotients will be the number of
pieces of each species. Thus, in the above case, tlie number of pieces of
salt beef is 1443^, and of pork 2887^.
2. It being customary to substitute a proportion of raisins and suet for a
part of the flour, a deduction is to be made from the full allowance of
the latter, on account of such quantities of those substitutes as it may be
deemed advisable to demand from the victualling stores ; observing that
one pound of raisins is equal to one pound of flour, and that half a pound
of currants or half a pqund of suet is to be considered as being equal to one
pound of raisins or one pound of floun
3. As tobacco and soap are directed to be issued to the ship's company,
in the proportion of two pounds of the former and qhc pound of the latter
per man per lunar month, the Purser Is to include those articles in his
demand for provisions from the Victualling Agents ; but it is to be ob-
served, that he must only demand as much tobaccco and soap as will
answer for the two-thirds of the complement, agreeably to the time for
which the ship is ordered to be provisioned.
If the complement of men be multiplied by 2, and the product divided
by 3, the quotient will be the two-thirds of the complement : this, being
multiplied by the number of months for which the ship is ordered to be
victualled^ will give the number of pounds of soap ; the double of which
will be the number of pounds of tobacco. Thus^ in the above case^ where
the complement is 275 men^ and the time 4 lunar months ; —
Now, 275 Baen x 2-^3= 183 J, which is the f of the complement.
Multiply by no. of months = 4
Product ; 733^ = pounds of soap.
Double of ditto • • • • 146^f s pounds of tobacco.
I'HB END OF VOLUME T.
Printed by MUU, Jowctt,aad MUlt, Bolt-coait, Flect-ttrcet
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ERRATA
Page 141, Note, line 4, for "kt, 50? Si
Page 463, line 6 from the bottom,^ '^ii
2. 5L I '' ready moon's corrected ri(;
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Just publisfiedf tlie Second EdUiony in royal 8vo.y price lis. boards.
OF
To the Sidereal and Planetary parts of NAUTICAL ASTRONOMY :
being the theory and practice of. finding the LATITUDE, the LON-
GITUDE, and the Variation of the Compass by the FIXED STARS
and PLANETS. To which is prefixed the description and use of the
NEfF CELESTIAL PLANISPHERE.
By THOMAS KERIGAN. R. N.
LONDON '. — PRINTED FOR BALDWIN AND CRADOCK.
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