This is a digital copy of a book that was preserved for generations on library shelves before it was carefully scanned by Google as part of a project to make the world's books discoverable online. It has survived long enough for the copyright to expire and the book to enter the public domain. A public domain book is one that was never subject to copyright or whose legal copyright term has expired. Whether a book is in the public domain may vary country to country. Public domain books are our gateways to the past, representing a wealth of history, culture and knowledge that's often difficult to discover. Marks, notations and other marginalia present in the original volume will appear in this file - a reminder of this book's long journey from the publisher to a library and finally to you. Usage guidelines Google is proud to partner with libraries to digitize public domain materials and make them widely accessible. Public domain books belong to the public and we are merely their custodians. Nevertheless, this work is expensive, so in order to keep providing this resource, we have taken steps to prevent abuse by commercial parties, including placing technical restrictions on automated querying. We also ask that you: + Make non-commercial use of the files We designed Google Book Search for use by individuals, and we request that you use these files for personal, non-commercial purposes. + Refrain from automated querying Do not send automated queries of any sort to Google's system: If you are conducting research on machine translation, optical character recognition or other areas where access to a large amount of text is helpful, please contact us. We encourage the use of public domain materials for these purposes and may be able to help. + Maintain attribution The Google "watermark" you see on each file is essential for informing people about this project and helping them find additional materials through Google Book Search. Please do not remove it. + Keep it legal Whatever your use, remember that you are responsible for ensuring that what you are doing is legal. Do not assume that just because we believe a book is in the public domain for users in the United States, that the work is also in the public domain for users in other countries. Whether a book is still in copyright varies from country to country, and we can't offer guidance on whether any specific use of any specific book is allowed. Please do not assume that a book's appearance in Google Book Search means it can be used in any manner anywhere in the world. Copyright infringement liability can be quite severe. About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http : //books . google . com/ I i n Digitized by Google Digitized by Google Digitized by Google Digitized by Google Digitized by Google Digitized by Google ^i^^irnpht^n ,md f,^. t,,,,, ^gy Digitized by Google Digitized by Google LONDON : PRINTED BY MILLS, JOWETT AND MILLS, (late bensley) BOLT-COURT, FLEET-STREET. • » • Digitized by Google 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. cyAQA QQ Digitized by VjOOQ IC Digitized by Google 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* Digitized by Google 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. Digitized by Google 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 Digitized by Google 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. Digitized by Google 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 Digitized by VjOOQ IC CO|iTKNT$. XXVU Takle. Page. 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 Digitized by VjOOQ IC 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 Digitized by Google 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 Digitized by VjOOQ IC inp^ COMTBNTS. 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 Digitized by VjOOQ IC PfMem. Pmg§. 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 Digitized by Google XXxii CONTENTS. Problem. Page. 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 Digitized by Google CONTENTS. XXXIU Problem. Page. 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 Digitized by Google 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 Digitized by Google CONTENDS. xxxr FroMem. Page. 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 Digitized by Google 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 Digitized by Google 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. Digitized by Google Digitized by Google 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 Digitized by Google 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. Digitized by VjOOQ IC 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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« Digitized by Google 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. Digitized by Google 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 Digitized by Google 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. Digitized by Google 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. Digitized by Google 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. Digitized by Google 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. • .47 seconds, answering to which is • ' • . 0. 0. 3. 8 Urn. 47?47'47^ the time corresponding to which is . 3M1?U!8! B Digitized by VjOOQ IC .• . 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. « . 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. Digitized by Google 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. Digitized by Google 4 BMCftlPTION AN0 USB OF THB TAKLBS. • 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 Digitized by Google 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 ^ Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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* Digitized by Google 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 Digitized by Google 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 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 Digitized by Google 12 DBSCRIPTION AND USB OF THB TABLB8. 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 . . . ' 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 ? . Digitized by Google 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 Digitized by Google 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. Digitized by Google 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. Digitized by Google 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. Digitized by Google 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 /Google Digitized by ' 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 Digitized by Google 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. 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' w ill 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. c2 Google Digitized by 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 Digitized by VjOOQ IC 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. • . • 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^^ Digitized by Google 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. 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 /Google Digitized by ' 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 . . 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 corre s po n ding 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 ? Digitized by Google 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? Digitized by VjOOQ IC 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 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^ Digitized by Google 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. 3. 37. 30 Do. to 9. and 0.50 = 0.18.45. Do. to 0.29 and 0.50 = 0. 1. 0.25 bo. to 9. and 0. 2 = 0. 0.45. 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> Digitized by VjOOQ IC Df iCRIPTI^N M«9 USB QV Ttt« TA1IJUB9* 27 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. = 0.19.20. Do. to 9, and 0.30 = O.ll.lS.H) Do, to 0.29 and 0.30 = 0. 0.36.15 Do. to 0. and 0. 8 = 0. 3. 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. » (K 4.50 Do. to 9. 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* Digitized by VjOOQ IC ^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. 5.52.30 Do.^ to 12. and 0.50 = 0.25. 0. Do. ^ to 0.47 and 0.50 ,= 0. 1.37.55 Do. to 12. and 0. 2 =0. I. 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. = 0.29.22.30 Do. to 12, and 0.30 =: 0.15. 0. Do. to 0. 47 and 0. 30 =0. 0. 58. 45 Do. to 12. and 0. 6 =0. 3. 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. 7.50 Do. to 12. and 0.30 s 0.15. 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! Digitized by VjOOQ IC 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. and 7. = 2.37.30. Do. to 0.50 and 50. s 1.44.10. Do. to 0.50 and 7. s 0.14.35. Do. to 9. and 0.40 = 0.15. 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. Digitized by Google 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? /Google Digitized by ' 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. =z 0. 6. 0. Do. to 7. and 0.30. = 0.17'.30. Do. to 0.12 and 0.30. =0. 0.30. • Do. to 7. and 0. 1. =0. 0.35. Do. to 0.12 and 0. 1. =s 0. 0. 1. Do. to 7. 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. = 0.20. Do. to 7. and 3. = 1.45. Do. to 0. 12 and 3. = 0. 3. Do. to 7. 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. =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. Digitized by Google 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. 6. 0. Do. to 7. and 0.50. =: 0.29.10. Do.. to 0.12 and 0.50. = 0. 0.50. Do. to 7. and 0. 1. 0- = 0. 0.35. Do. to 0.12 and 0. 1. = 0. 0. .1. Do. to 7. and 0. 0.40 rs 0. 0.23.20 Do. to 0.12 and 0. 0.40 =0. 0. 0.40 Do. to 7. 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. = 1. 0, Do. to 7. and 0. 20. ^z 11.40. Do. to 0.12 and 0.20. zz 0.20. Do. to 7. and 0. 3. ZZ 1.45. Do. to 0. 12 and 0. 3. = 0. 3. Do. to 7. and 0. 0.40 "" 0.23.20 Do. to Oi.12 andO. 0.40 ™* 0. 0.40 Do. to 7. 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. Digitized by Google » 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. 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 Digitized by VjOOQ IC 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 Digitized by Google 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^ and0.20 =2.4 andO. 7J= .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. Moon's correct longitude, at the given time • • . . 7 ' 2 1 ? 1 2 : 1 7 742r j)2 Digitized by VjOOQ IC ' SQ, 2>BSCR1FTI0N AND USB OF THB TABLBS. 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. 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 * Digitized by Google 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. =0.14 .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 Digitized by Google 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 Digitized by Google ]»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. Digitized by VjOOQ IC 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 ? Digitized by Google 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 Digitized by VjOOQ IC 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 Digitized by Google 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 M l l l l i M Log. . . , 9.999)586 CotMtantli^, 0.S0091Q i j I n w inyn 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« Digitized by Google 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« Digitized by Google 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^ Digitized by Google 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. 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 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. Digitized by Google 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. Digitized by Google 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 Digitized by Google 52 BBSCRIPTION AND USB OF THE TABUS* Star's apparent altitude 10? ' 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 /Google Digitized by ' 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. Digitized by Google 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^ Digitized by Google 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^ Digitized by VjOOQK 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, Digitized by Google 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 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 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 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* Digitized by Google 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 /Google Digitized by ' 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 Digitized by VjOOQ IC ,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 ? Digitized by VjOOff IC 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. Digitized by Google 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. Digitized by Google ]>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 ; 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 Digitized by Google 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 + 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/ . Digitized by Google 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 Digitized by Google 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 ; 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 at top, or over 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 Digitized by Google ' PBSCRIPnOM ANB USB OF THB TABLBSf 07 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 : 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 Digitized by Google 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 ; 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. 'Digitized by Google 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 Digitized by VjOOQ IC 70 0BtCllIFTION AN0 VMM OF TMB tAJUJU. 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, Digitized by Google 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 Digitized by Google 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. • • . - , Digitized by Google 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» Digitized by Google 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; Digitized by VjOOQ IC 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 Digitized by Google 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. 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, /Google Digitized by ' DBSCttlPTlON AND VSB OF THE TABLES. 77 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. Digitized by VjOOQ IC 78 PSBCRIPTION AND USB OF TOB TABUfl^ 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? /Google Digitized by ' J>BiCEIFnOM AKO VBM OF THB TABLES. 79 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! Digitized by VjOOQ IC 80 BBSCRIPTIOK AND USE OF THB TABLES. 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^, Digitized by VjOOQ IC DBSORIPTION AND USB OF THB TABJLBS. 81 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 Digitized by Google 82 HBSeaiPTlOK and USB of tub TABtlS. 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« Digitized by Google JDBSOSIPTIOV AVD USB OF TBB T1BLB8« BS 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 Digitized by Google 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. Digitized by Google DESCRIPTION AND USB OF THE TABLES. B5 ■ 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. Digitized by VjOOQ IC 86 MMCRlPTIdK AN0 U8B OF tBB tA&lM. 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. Digitized by Google mscamioN and vn of thb tabu8# $7 • 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. • • Digitized by Google 88 BKSCRIPTIOK AND -USE OP THB TABLSS*. 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. Digitized by Google 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. Digitized by Google 80 usomipnoif avo vsb or thb TAVMi 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 to' 180 de- grees ; as thus t the arcs corresponding to the log.' secants are giv^n in regular succession at top from 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 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. Digitized by Google '»88CfttPtlO!f ANI> VM OF THS tABUEI. 01 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? } y Google Digitized by ' 92 . PBSCRIPTION AND USB OF THE TABUBS. 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. Digitized by Google DSSCRIFTION AND USB 01^ THB TAfiJLBS. , 93 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. Digitized by_ Gqogle 94 ^ 2>pscEiFnofir and usb of thb tabu|8. 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 to 180 degrees ; as thus : the arcs corresponding to the log. sines are given in regular succession at top, from 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 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. Digitized by Google jmSCtlPTIOM AMD USB OF TUB TlBLBt* 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 « li m n ■ '■ ^» 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. Digitized by Google 96 DBSCaiFTION AND USE OF THB TABLES. 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. • Digitized by Google BBSCRIfTION AND USB OP THB TABLES. 97 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 /Google H Digitized by ' 98 DBSCRIPnOV AffP USB 09 m TABUS. 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 Digitized by Google DMCAIFnON AM]> USB OF THB TABLB8. 9Q 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 Digitized by Google 100 BBSC&IFTION ANJ> USB OF THB TABLB8. 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. Digitized by Google 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 ? Digitized by Google 102 ' PBSCfi.lH'ld^r AND U8B OF THE TABLB8* 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 Digitized by Google 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. Corrected time of transit 4^9^23! Time of high water at Ci^ Florida on full and change days ?• 30. 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? ? Digitized by VjOOQ IC 104 DBSCKIFTION AND USB OF THE TABtBS. 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. Corrected time of transit 10i26?31! Time of high water at given place on full and change days • 9. 0. Trnie of high water at Queen Charlotte's Sound, past noon of the given day . • . . 19?26T8l! Subtract 12.24. 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, Digitized by Google 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. Digitized by Google 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. Digitized by Google 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. 1 heiie« the dlffefenee of latitude is 155^ 2^ and the departure SS. 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. and 508. 4 1 hence (he differeuce pf Mtode is 182. 0| and the departure 508* 4 milest Digitized by Google lOS DBSCRIPtlON AND USB OP THE TABLES. 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. 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* Digitized by Google 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. Digitized by Google 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. Digitized by Google uacftimoii ▲>» 088 w thb yablsi. Ill 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 Digitized by Google 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. Digitized by Google DfiSCRIFTION AND USB OF THfi TABLES. 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 Digitized by Google 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«. Digitized by Google 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 Digitized by Google 116 DESCRIPTION AND USB OF THB TABtSS. 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. Digitized by Google 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 Digitized by Google 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, Digitized by Google 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, . 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^ /Google Digitized by ' 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.^ Digitized by VjOOQ IC ]>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. Digitized by Google 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. Digitized by Google 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 Digitized by Google 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* Digitized by Google 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 = Tab.diff. to 1? of dec.= 3: 5 now ^^ = 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. Digitized by Google 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. Digitized by Google 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 Digitized by Google 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. Digitized by Google 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! /Google Digitized by ' 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 Digitized by Google 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 -^ Digitized by Google 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. 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 ? Digitized by Google 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. 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. . Digitized by Google 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 ? Digitized by Google 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. 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. 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. 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. 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: Digitized by VjOOQ IC 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. 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. 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. Greenwich time = 16*15T28! Moon's dec. reduced to this time, is 7° 17^52^8, Digitized by Google ]>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. 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. Digitized by VjOOQ IC 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 Digitized by Google 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. Digitized by Google 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? Digitized by Google 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^= . . - .2 Corr. to lat 60<»22' and dec. 23*^7'45'' a .... 0"53'".8 /Google Digitized by ' 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. Digitized by VjOOQ IC BXSGBIPTION AND USB OF TUB TABLBI. 148 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. Digitized by Google 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. Digitized by Google 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 Digitized by Google 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, 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 Digitized by Google BSSCRIPTION ANB USB OF TBB TABLBS. 147 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 Digitized by VjOOQ IC 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. Digitized by Google 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. Digitized by Google ISO 2IB9CRlFT|eN AND V9E OF THB TABUI8« 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. 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 Digitized by VjOOQ IC 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^ &€. Digitized by Google 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 Digitized by Google 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 Digitized by VjOOQ IC 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 Digitized by Google 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* Digitized by Google 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. Digitized by Google 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. Digitized by Google 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- Digitized by Google 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. Digitized by Google 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 ' Digitized by Google 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 Digitized by Google 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 Digitized by Google .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? Digitized by Google 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. Digitized by Google 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- Digitized by Google .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. Digitized by Google 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* Digitized by Google 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* Digitized by Google 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* • , : Digitized by Google 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 . /Google Digitized by ' 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« Digitized by Google 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 /Google Digitized by ' 174 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^ Digitized by Google 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 Digitized by Google 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.. Digitized by Google 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 Digitized by Google 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 Digitized by VjOOQ IC 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 Digitized by Google 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? ^ . Digitized by Google 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. Digitized by Google 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 Digitized by Google 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* /Google 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, Digitized by Google 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. 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. ' 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, Digitized by Google 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. 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. 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. Digitized by Google 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. 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, Digitized by Google 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. 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. 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, Digitized by Google 190 $?BBaiCAL TaiGQVOMnftT* Now, the leg AC, being connected with the required part, Is therefore to be the first term in the proportion. Hence, As the leg A C = . . 68?29'45r Log, tang, ar, comp. = 9. 595490 Is to radius = ... 90. 0, Log. sine = . . • 10.000000 So is the angle A = . 74. 45. 15 Log. co-sine = . • 9. 419891 To the hypothenuse A Cs 84? 5i 6r Log. co-tangent a 9. 015381 Note.'^^The hypothentise v acute, because the given leg and angle are of the same affection. . Peoblbm V. Gwen the two Legiy to find the Aagle$ 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 Digitized by Google 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. 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 Digitized by Google 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. 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, * . * Digitized by Google SPHSmCAL TRt GONOIIBTRY. 1 93 As the angle B = 64?20^45r Log. sine ar. comp* = . 10. 045091 Is to radius = . 90. 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. Digitized by Google 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. 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. 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 Digitized by Google 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. 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« Digitized by Google 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. 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. Digitized by Google 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. Digitized by Google 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 ■ Digitized by Google 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. Digitized by Google 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 Digitized by VjOOQ IC 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 ? Digitized by Google 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? Digitized by Google 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. Digitized by Google 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. Digitized by VjOOQ IC 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. Digitized by Google 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? Digitized by Google 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. Digitized by Google 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? Digitized by Google 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. Digitized by VjOOQ IC 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 Digitized by Google 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* Digitized by Google ^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. Digitized by Google 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 Digitized by Google 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* Digitized by Google 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. Digitized by Google 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. Digitized by Google ^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^ Digitized by Google 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 ^ • Digitized by "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, Digitized by Google 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 :— * * . Digitized by Google 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» Digitized by Google 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, Digitized by Google 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 ? Digitized by Google 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 Digitized by VjOOQ IC 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* Digitized by Google 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 Digitized by VjOOQ IC 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. Digitized by 2.327430 Google 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 Digitized by 9.823855. Google 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. Digitized by Google 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 /Google Digitized by ' 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$e, and Distance. Example. A ship from the Snares, New Zealand, in lati- tude 48?3C S., and longitude 166?20: E., sailed upon a direct course between the south and west, till she was found byjobservation to be in longitude 151?27- E., and to have made 546 miles of departure; required the latitude come to, the course steered, and the distance sailed ? Digitized by Google 284 NAVIQATfON. Longitude of the Snares = . . 166?20fE. Long, of the ship by observation = 151. 27 E. Difference of longitude s • • • 1 4 ? 53 ' :s 893 miles. To find the Middle Latitude s the Angle BCD:— With the departure = B C^ and the difference of longitude s= C D^ the angle of the middle latitude may be found by trigonometry, Problem IIL, page 174; as thus: As the diff. of long. = 893 miles. Log. ar. comp. = 7* 049148 Is to radius s: . 90? O: Or Log. sines . 10.000000 So is the departure = 546 miles, Log. = . « 2. 737193 To the mid. lat. = 52? 1 8 '. 28r Log. co-sine sa 9. 78634 1 Twice mid. lat = 104?37' Of nearly. Lat.ofthe Snares=48. 3. OS. Latitude come to =56?34f O^S. Diff. of latitude » 8?3l! 0?s 511 miles. To find the Course =c the Angle A :— With the difference of latitude A B, and the departure B C, the eourat may be found by trigonometry, Problem IV., page 175 ; as thus : As the diff. of lat. = 511 miles. Log. ar. comp. = 7* 291579 Is to radius s • . 90? 0' Or Log.sine^ . 10.000000 So is the departure = 546 miles, Log. = . • 2. 73/193 To the course as . . 46?5aM8r Log. tangent at 10.026772 . To find the Distance ss A C :~ Wiih the angle of the course, thus found, and the difference of lati^de A B, the distance may be computed by trigonometry. Problem IV., page 175 : hence. As radius « . . 90? 0^ Or Log. co-secant s 10. 000000 Is to diff. of lat = 511 miles. Log. &r . . . 2.708421 So is the course = 46?53M8r Log.secant = . 10. 165378 To the distance s • 747. Smiles, Log. =: • . • 2.873799 Digitized by VjOOQ IC MIDDLB LATITODJI SAILING. 23$ 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 Digitized by VjOOQ IC 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. 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^ Digitized by VjOOQ IC 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 Digitized by Google 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 Digitized by VjOOQ IC- 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 . 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." ^^^ ^^ Digitized by Google 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 Digitized by Google 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. uigitizea oy Google 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 Digitized by Google 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 Digitized by Google 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, Digitized by Google 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 ? Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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. Digitized by Google 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 Digitized by Google 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. Digitized by Google 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- Digitized by Google 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 Digitized by Google 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., Digitized by Google 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 :— . Digitized by Google 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 Digitized by VjOOQ IC 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- '-. >-■ ^ ^■^'-'' Google Digitized by 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 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 Digitized ^by LjOOQ IC 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»a p proye d IcD^hof aknot. • • Digitized by Google 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 Digitized by Google 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 Digitized by oy Google 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 Digitized by Google 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 ^ Digitized by Google 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. Digitized by Google 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. Log. sine = , 9.911405 So.is the co-lat. - BS . 50.23. 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. Log. sine = . 9.911405 . So is the co-lat. AS « 55.36. Log. sine = . 9.916514 To the ang. of pos. SB A 429l7:20irLog,rine = . 9.827932 Google Digitized by 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. 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-^ Digitized by Google 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. 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. 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? Digitized by Google 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, Digitized by Google 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. 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. . 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. Digitized by Google ts I I B u II n V u b'b-h B y y .|t y-tt II H 11 u n II H H' It <9« Off rfh> *. cou lo ^ ^ •*• »^>-*ioioo»CA9.uiik.c;«c;iAcn k?ki2S:2bJ22bifibkp feOMNfedlOMfeOfeOfeCMlOMOOOOOQOaOCOaDaDOOQOaDODQDoig ^kOIOfeOfdfcCtOMfeOfSlOUpMkOtdfedkdtO^IOIOlCttvtO^ (Ik • oooeooeoooeooooooooe o-o o o o o o w c*» 1^ »« M c;« o» ^ >r r* e>* »d 1^ ilk. CO 10 c;i 0*5010^50 ^ ko vivivi*N«ppaD^c;«p»c;i;^pp»«p^^top^<fiOa^piU >000000©00vji0 000© ^^^ Coppppppppppk^UOpppppppppppp oooeoeeoooo^oDooooooooeeeo iooo>WoD*-oow<ccnGoigo«o>-*.coto«;»'vijccj'^iowg oc&0Ot2o)oci^*^^u*u^uc;^poio<n~^5i)«5'<p)pv^*<^S oaDO»C7*ieo»u«c;«otUMCi9iutc^»-Uii^ac^09ar«^^^^ 5I^»-<yt5o0dM>-u<<C0dp^wdQ0^ppkCaD*-.p9)C#s'^ D0C©rf^*^W<O^O»^<C^€»«ilfc.<OWO4^*t c»i»«j^2j-c»wwo>o*»*opvij-v*tj»cnQow»c;'7-7-pp If fii s. ft r 5 ? 8 ?2 Digitized by Google 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^ Digitized by Google 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. Digitized by VjOOQ IC 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 ; Digitized by Google 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. Log. sine = . . 9.834865 So is the co-latitude B S= 56. 59. 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 Digitized by VjOOQ IC 290 NAVIOATIOH. As the distance AB« 101?47^52r Log. co-secant s 10.009273 Istodiff.long.ASBa 136. &2. log. sine a . . 9.834865 Soi8theco-lat.ASs: 56. 8. 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. 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- Digitized by Google 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. 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. 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 Digitized by Google 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. 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, 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.' Digitized by Google 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. Digitized by Google ^J! I I. s m 1|3 N 1} i H ■9 55 3KSi!!3S£:S8!!3a3!!iSS^Si!>&S!3Sii^£SS38&S -^U9C^9»«A| 3'§SSSSSSSS.SSS8SS;SSSSSSSSSSSSSS l^^;;ss:^:ss$s3$^S3^8^ssss:ss^!$;$*^$i s § M s H n H fl n R fl n B I n I R B IB B B B I! Q I n B I n I B .n GO (« w oa on en o) CO w cfi (ft eft «n w « w CO w w GO com CO w CO en Qo » ao «) Digitized by Google 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. Digitized by Google !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 Digitized by Google 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! Digitized by VjOOQ IC 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 Digitized by Google 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. Digitized by Google 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 Digitized'by LjOOQ IC 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^ Digitized by VjOOQ IC 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 Digitized by Google 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 ? ! 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. 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' Digitized by VjOOQ IC 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. 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 Digitized by Google 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 Digitized by Google 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. Digitized by •Google 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 /Google Digitized by ' 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^ Digitized by Google 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' « 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' • 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 id by Goog lj 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. Digitized by Google 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 Digitized by Google 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 Digitized by Google 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. 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 as cen s i o n, 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. /Google Digitized by ' 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. Digitized by VjOOQ IC 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. 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 Digitized by VjOOQ IC 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 ? Digitized by Google 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 Digitized by Google 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 /Google Digitized by ' 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^ Appar e nt 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. Digitized by Google 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. Digitized by Google 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. Digitized by Google 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, /Google Digitized by ' 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* Digitized by Google 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. 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 obeei vati on ^ .. 58t50^2»*S. . + 28. 12 . 5»?I8UirS. . 30?4in9rN. . 19.35.25 N. • 50?16:44rN. Digitized by VjOOQ IC 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. 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. /Google Digitized by ' 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. Greenwich time =5 ....... 3fl6*48r30! Digitized by VjOOQ IC 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^ (}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. /Google Digitized by ' 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 Digitized by Google 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« Digitized by VjOOQ IC 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. Digitized by Google 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 ? Digitized by Google 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 Digitized by Google 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. Digitized by Google 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 Digitized by Google 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. Digitized by Google 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 ! 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 : I I ■ 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. Digitized by VjOOQ IC 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. Digitized by Google 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. Digitized by VjOOQ IC 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 /Google Digitized by ' 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- Digitized by Google 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. Digitized by Google / 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. Digitized by Google 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. Digitized by VjOOQ IC 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. Digitized by Google 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. 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. Digitized by Google 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 Digitized by Google 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 Digitized by Google 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 Digitized by Google 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. 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^ Digitized by Google 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; = + ,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. Digitized by Google 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. /Google Digitized by ' LATITUDE BV AN ALTtTtoB TAXKN NBAR THB MERIDIAN. 859 Apparent time of obserration =: . .... 3*55T17! Longitude45? W., intima =: 3. 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 + .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? 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. Digitized by VjOOQ IC 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* 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. Digitized by Google 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. Digitized by Google $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. Digitized by VjOOQ IC 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- 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 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* Digitized by Google 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. 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. Digitized by Google LATITUDE BY AN ALTITUDE TAKKN NEAR THE MERIDIAN^ 365 Cor. in Tab. LII., ans. to lat 52?N. and dec. 0^ = ... 1^32*'. Diff.to2?onat.= -6''.45now,6^4xl2^-Hl20'=: . . . - .6 Diff.toI?ofdec.=:-l-'.2;now,1^2x24n9r^60C= . . . - ,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. Digitized by Google 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. 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 • Digitized by VjOOQ IC 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*'. Difference to 2? of lat.= -7^ 0; now, 7^ 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' Digitized by VjOOQ IC 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-». 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 Digitized by Google 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 Digitized by Google 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. 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 • - •! 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 Digitized by VjOOQ IC 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 Digitized by Google 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. 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. 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- / Google Digitized by ' 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, Digitized by Google 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. 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. 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, Greenwich time pa^ noon, September 30th =; « • • • 11' 57 7|! Digitized by VjOOQ IC 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. Digitized by Google 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. Digitized by Google 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? Digitized by Gbogle 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. Digitized by Google 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. 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|! Digitized by Google 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. Digitized by Google 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 ' . Longitude 60?45 : W., in time = • -f 4, 3. 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 Digitized by Google 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. 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 Digitized by Google 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. . 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. Digitized by Google 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 /Google Digitized by ' OF FINDING THB APPARBNT TIMB. 885 Time, pe^ Watch. Altitude of San'g Lower Limb, 3* 0r30! ' 13?49M0r 3. 1.15 . 13.44. 3. 2. 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. 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. 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 Digitized by Google 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. 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. 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* Digitized by Google 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 Digitized by Google 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! /Google Digitized by ' 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. 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" Digitized by VjOOQ IC 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.. Greenwich time s= • . # . , ♦ 16*44?45* Digitized by VjOOQ IC 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"^ /Google Digitized by ' 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' Digitized by VjOOQ IC 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. 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 Digitized by VjOOQ IC 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 } Digitized by Google 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. 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 Digitized by VjOOQ IC 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. • Digitized by Google 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. 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* /Google Digitized by ' 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. 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. 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! /Google Digitized by ' 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. Digitized by Google 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, Digitized by Google OF FINDING THB APPARENT TIME. 401 Time of observation, per watch, = . 7*28ri8! Longitude 60?W., in time = . -f 4. 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 Digitized by Google 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 /Google Digitized by ' 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 Digitized by Google 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! /Google Digitized by ' 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. 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. 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, Digitized by VjOOQ IC 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. 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. 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! /Google Digitized by ' 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. 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* /Google Digitized by ' 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- Digitized by Google 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. 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 . Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. Digitized by Google 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 Digitized by Google 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 Digitized by Google 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! . 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 * . Apparent time at Greeniwchs • . . 10? 9?36'.3 /Google Digitized by ' 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 Digitized by Google 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. 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. Digitized by VjOOQ IC 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 Digitized by VjOOQ IC 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. Digitized by LjOOQ IC 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. 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. Digitized by Google 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. 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 Digitized by Google 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 Digitized by VjOOQ IC 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 ? /Google Digitized by ' 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 '. 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. Digitized by Google 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. Digitized by VjOOQ IC 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. Digitized by Google 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. /Google Digitized by ' 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. .5 Sun's reduced right ascension = . . • . 10*45 T54 ' . 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 Digitized by VjOOQ IC 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. Digitized by Google 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 Digitized by Google 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. Digitized by Google 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. =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 ' . Digitiz'ed by Google 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 Digitized by Google 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 Digitized by Google 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. . Digitized by Google 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, . Digitized by Google 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. Digitized by Google 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. Digitized by Google 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 /Google Digitized by ' 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. -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 ? Digitized by Google 442 VAxmcih Awnmour. Sun's apparent alt. = 39?25^ 0^— Correc. 1^ 3r=trucalt,=39?23:57^ MooD'sapparentaIt.= 19.56. +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 /Google Digitized by ' 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. Digitized by Google 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? Digitized by Google 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. +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. 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 /Google Digitized by ' 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* Digitized by Google 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 ? Digitized by VjOOQ IC 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 ? Digitized by Google 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. sin e 9. 944902 Halflhe true distances •, . . . 45?36:38i7 Log, 8ine=9. 854065 True central distance a .'v . . 91?13''17' 2 o Digitized by Google 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 Digitized by VjOOQ IC 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. +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 Digitized by Google 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 Digitized by VjOOQ IC 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? Digitized by Google 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 Digitized by Google 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. 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. Prop. log. = 1984 Portion of time = .• l*24rlS! Prop. log. = 3299 Time corresponding to first distance = 12. 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. Digitized by VjOOQ IC ^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 Digitized by Google 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. .15. 5 1070 48' 30'' . 47. 50 . 47. 15 . 46. 40 . 46. 70 9' 0* 7.22. 30 7.36. 7.49.30 8. 3. 530 13' 30" 53. 2. 52. 50. 30 52. 39. 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: Digitized by Google 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. 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. 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" Digitized by Google 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. 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 ? Digitized by Google 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. Apparent distance = 108*! 0'.25r Alt. J 'slower limb = 39?30'45r Moon's 8emi-diameter= +16. 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. 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 declination s 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 ! Digitized by Google 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. 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. Apparent time of obs. at Greenwich =: 7 J24r 1 7 ! Digitized by Google 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' Digitized by Google 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. 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. /Google Digitized by ' 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. 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. Apparent time of observ. at Greenwich = 19^ 16T24! Digitized by Google 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. 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 Digitized by Google 466 NAUTICAL ASTILONOlfT. Apparent time of observing the sun's altitudes: 1 ^ 5S?35 * Time of observing dittOj per watch, s . • 1 • 53. Watch slow for apparent time c= . . « « 0T3$'. Timei per watch, of gbs. the lunar distances 21. 30. 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? Digitized by Google 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. 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. S21f 9T32? 47? 7^ 21.10.33 Ditto 47.16. J 2h2 i^ Digitized by Google 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. 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. J>21. 21.43 20. 0. 3 91.22.43 Ditto 19.50.30 J 21.23.48 Alt.of8un'8low.limb=49. 14. ^ 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 Digitized by VjOOQ IC 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 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. 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. Digitized by Google 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 } Digitized by Google 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. Moon's apparent alt.^ 45? 3! 157 Digitized by Google 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 su