UNIVERSITY OF CALIFORNIA
ANDREW
SMITH
HALLIDIL:
1868 3SteT 19O1
Return fr
A COMMONPLACE-BOOK of THOUGHTS, MEMORIES,
A FANCIED Original and Selected. PART I. Ethics a"d Character ; PART Literature and Art? .SW - revised and corrected ; vrith Etchings and
\Vood Engravings. Crown 8vo. 18*.
the
TWO LECTURES on the sOCIAL EMPLOYMENTS of
1 y. xters f>f Charity and the • »f Labour. New Edition
^ith a Prefatory Letter on the Present Condition and Requirements of th Women of England. Fcp. 8vo. 2*.
6
QKETCHES in CANADA, and RAMBLES among the
O ^ ,11 Two Parts, Is. each.
„ • ,iy a record of her mvn ^
J. i
London : LONGMAN, GREEN, and CO. Paternoster Row.
3
'
NAUTICAL ASTRONOMY
LONDOX
PRINTED BY SPOT TIS WOO »E AND CO. NEW-STEEET SQUAKB
THE
PKOJECTION AND CALCULATION
or
THE SPHEEE
FOR YOUNG SEA OFFICERS
BEING
A COMPLETE INITIATION INTO NAUTICAL ASTRONOMY
BY S. M. SAXBY, RN.
PRINCIPAL INSTRUCTOR OF NATAL ENGINEERS, HEB MAJESTY'S STEAM KESEBVE LATE OF CAIUS COLLEGE, CAMBRIDGE
LONDON
LONGMAN, GREEN, LONGMAN, AND ROBERTS 1861
VK,
53
PREFACE.
SPHERIC TRIGONOMETRY is so little understood by those rvho are not professionally mathematicians, while the sub- ject is so important to a maritime nation, that the question whether there be not defect in the present mode of educa- ting nautical men demands earnest attention.
For more than a century the method of teaching what is called the "Art of Navigation" has been gradually changing, and the knowledge of principles has in England been consequently decreasing, while other nations, pos- sessed of a navy and commercial marine vastly inferior to those of England, have maintained a standard which actu- ally originated in the labours and experience of our own early nautical teachers.
It is remarkable that, while in the year 1700 no one who was ignorant of Spherics, and the principles of Nau- tical Astronomy, was considered a competent navigator, in the year 1860, the "epitomes" of navigation used both in the Eoyal Navy and the Merchant Service of this great country, not only make no pretension to the teaching of Spherics, either by calculation or construction, but one of
A 4
107275
VI PREFACE.
them (which on all other naval subjects is elaborate and accurate) even repudiates the propriety of mixing theory with practice, while certain other works insist on theory as the only medium by which to obtain a correct practical knowledge of the subject.
The true system of teaching lies between these extremes ; and this little work is an attempt to facilitate the study by sufficiently explaining general principles while adapting them to immediate practice.
A few recent works could be referred to which have already, by an advance in the right direction, whetted the mental appetite for really digestible and nourishing suste- nance ; but the young naval officer and the young aspiring merchant commander of the present generation have ab- solutely (beyond these) no available means of improving their acquaintance with what they feel to be indisputably essential to their self-respect as navigators, inasmuch as works in publication either disregard theory altogether as unimportant to a beginner, or bewilder the student with a phalanx of formula, which in their very aspect so commonly suggest a series of difficulties insurmountable, that relapse into indifference is the natural consequence.
And yet naval instructors are expected to impart a knowledge of Nautical Astronomy adequate to the needs of a sailor. Such is undoubtedly done to the very limit of their powers, but few besides nautical teachers can ap- preciate the labour of the work of elucidating a subject like Nautical Astronomy, in the absence of well-digested plans of instruction in a printed form, in which every- thing would be explained so clearly as to encourage the student, and lead him step by step up the ladder of
PEEFACE. Vll
knowledge, when the ascent is to be attempted away from professional assistance.
The subject of Spherics is one of such profound research and extent, that it seems to have been expected that a certain conciseness of language and solemnity of tone would best become the author who would enter publicly upon its teaching. But familiar description, divested of " hard words," has been so successfully attempted by an excellent mathematician of our own -times while treating of sister branches of science, that the example of such men as the Kev. Harvey Goodwin (whose illustrations of general mathematics have so much benefited the Cam- bridge student) may safely be imitated ; certainly by one, it is presumed, whose experience as a teacher of adults extends over a period of some three and thirty years.
There may be about 100,000 navigators in Great Britain, very likely there are considerably more, inasmuch as in the merchant service alone there are about 45,000 vessels afloat. It is scarcely, however, to be expected that the " wary mariner," trained as he has been to vigilance, and ever alive as he consequently is to necessity for precautions in his sea voyage, will not apply the same to his voyage of life. It is contrary to his professional habits to venture heedlessly among the rocks and shoals which may beset a coast to him unknown ; and in like manner such are his precautions with regard to any science to him unknown ; but give him even a tolerably clear outline " chart," and general " sailing directions," or even furnish him with a " hand lead " wherewith he may feel the bottom if he can- not see it, and there are in his occupations days and hours of thoughtful leisure in which he may be tempted to vary
Vlll PKEFACE.
his " cruise," and at least examine for himself the creeks and lagoons of science. At present these are obscured from his vision by characteristic prejudices and distrust. To remove these is then the object of this little book. Whatever may be the beauties of a science so captivating as Nautical Astronomy, the very approaches have been made toilsome from the decay of the landmarks which once guided the young seaman on his path. And when a straggling, casual, and mere wayfaring adventurer has accidentally gained a a peep " within the barriers, what has he seen but a frowning array of sines, secants, tangents, &c., twisted into every imaginable equational and fractional form, and distorted into a thousand labyrinthine channels, leading into deeps to him unfathomable. It is in vain to say such is a befitting initiation for a young aspirant in Nautical Science. It appals him ; it drives him back to his ordinary level, discourages from other attempts to ad- vance himself in the scale of proficiency, and throws him upon the dangerous and debilitating " consolation " that among his own class he can still " pass muster." Many indefatigable navigators do however brave the difficulties which present themselves, and attain a very fair footing : the object of this book is to vastly increase their number.
It may be asked, could one in a thousand even illustrate geometrically or by fe projection" the simple question of latitude from a meridian altitude, or answer it by the use of scale and dividers ? The writer would be very sorry to be compelled to publish, even from his individual ex- perience, details of positive danger which this state of things has entailed upon the commercial world ; but one thing is certain, viz. the importance of the interests of
PREFACE. IX
maritime commerce is involved in the safe transit of its vast treasures ; such interests are paramount, and demand our best efforts in their support, — a sufficient apology for this attempt.
It is not enough that naval officers have peculiar ad- vantages in nautical training, or that certain inducements are proffered, by the Board of Trade to men capable of acquiring, or willing to contend for, " extra certificates " in the merchant service ; for not even with the latter is a knowledge of Spherics expected beyond its one simple application to great circle-sailing, and this without any requirements whatever as to calculation. Hence we may safely affirm that the subject of teaching Spherics seems to demand complete revision.
Nor is it sufficient that the nautical tutors of the pre- sent day perform their duties in a manner which has al- ready loosened, as it were, the stability of an erroneous system of teaching, and which would in time restore that system to its previous sturdy and useful basis. But it is desirable to shorten such time of changeful probation, and an endeavour, in order to be successful, must be bold and radical. Navigation in these days is not as formerly required for the slow hulls of the beginning of the last century, which averaged from four to six knots an hour, but for steamers, swiftly flying against wind and tide at the rate of — (it is scarcely prudent to say what!). A more ready system of navigation is therefore called for ; and if years since it were important to the venerable structures to which allusion is made, that they should be navigated in the best possible method, by so much the more is it necessary that in this improved age of "clippers,"
X PREFACE.
all the efficacy of a sound knowledge of principles should be called into operation.
If Spherics as a basis of all oversea navigation was con- sidered and found to be indispensable to the safe conduct of ships in 1680, by so much the more must a knowledge of it be essential to shipmasters in 1860.
In tracing the cause of the deterioration which the system of nautical training has undergone during the last century and a half, we shall find it shown in the preface to the still admirable and elaborate " Practice of Naviga- tion," written by the lamented and unrequited Lieut. Raper, R.N., wherein he states that the theory and the practice are kept "purposely distinct;" and he adds " it is the custom generally to teach the theory first ; the im- pression forced on me is on the contrary, that the practice is itself the best foundation for sound and rapid advance- ment in the theory. For he who has acquired the practice knows the nature and extent of the subject, and in pro- ceeding to the theory he has a distinct perception of the object to be attained."
The author of this little work concurs fully in the views of Raper, wherein he advocates the all importance of practical knowledge to the student in theory ; but the ex- clusion of theory is merely the lesser evil, and has led to the " rule of thumb " system, to the final exclusion of elucidation of principles from our best existing works on navigation on the one hand, or has engrossed it too exclusively on the other, in works intended for young sea officers; its inevitable consequence is therefore the state of things which it is the object of the writer of this to combat.
PEEFACE. XI
The mind of a seafaring man is so much occupied with responsibilities of duty, or as regards passengers, the crew, &c., that he has in general too little leisure for deep studies, and when on shore his necessary and reasonable repose admits of little interruption; therefore to benefit the whole class a work is wanted which, in familiar lan- guage, and in a proper blending of theory with practice can lead to a solid and respectable acquirement. The more easy the steps of knowledge can be rendered to so peculiarly situated a class as sea officers, the faster pro- gress will he make who aims at a higher intellectual level ; and, indeed, a very gently inclined plane of science is more necessary for the navigator than for any other class of individuals.
Such, then, this book is intended to be; such, indeed, as a young seaman may take up with a firm conviction that all essentials are fully explained; and with an in- terest in believing that if it be read even with the atten- tion usually paid to a work of fiction, as a mere pastime, sound and valuable information will infallibly take root in his mind.
The oft-quoted assertion attributed to Euclid in his reply to one of the Ptolemies, viz. " that there is no royal road to learning," must no longer obtain among us. Not every seaman aspires to become an Archimedes, and if he did, an Alexandrian school might not be the only one and the best in which to form him.
What Sir John Herschel has done for astronomy, Fara- day for chemistry, Sedgwick for geology, Arnold for 'classics, Hutton and Colenso for arithmetic and algebra, and Goodwin, Snowball, &c., for the study of mechanics,
Xll PEEFACE.
may surely be done for the navigator. Hence the object of this book.
Ill understood, isolated and forbidding formulae thrust imperiously upon an already burdened memory are a load and an oppression, while pleasant, easy illustration of principles is an enticement, and wins upon the mind until its powers are secured by the silken bonds of a willing captivity.
The British Grovernment are seconding the community in their struggles for a better middle class education. Shall then those whose perilous avocations demand espe- cial intelligence, and to whose hands and heads we trust our lives and properties, and above all our national honour and the defence of our hearths, shall these alone be ex- cluded from the social arena of progress, and as mere spectators see the palm of proficiency in the grasp of a
foreigner ?
S. M. S.
CONTENTS.
Page PROJECTION: —
Orthographic 2
Mercator's 7
Stereographic 8
GENERAL REMARKS ON TRIGONOMETRY .... 10
Definition of an angle 15
Geometrical theorems 16
Ratios . , . . .21
Natural sines 22
LOGARITHMS, the nature of . .23
Computation of 29
NAUTICAL ASTRONOMY 31
Lines of the sphere . 32
Spheric triangle 34
General diagram of daily astronomical phenomena . . .36
SPHERIC PROJECTION 38
Plane scale 39
PEOB. I. To make an angle so that the angular point shall be
at the centre of the primitive . . . .39 II. To lay off any number of degrees on a right circle . 40 III. To draw an oblique circle through any point lying
within the primitive circle . . . .40
IY. To draw an oblique circle through any two points . 41 V. To draw a small circle parallel to the primitive at
any distance from it 41
VI. To draw a parallel circle at a given distance from a right circle, or at any distance about a given point at the primitive . . . . .42 VII. To find the pole of an oblique circle . . .43
XIV CONTENTS.
Page SPHERIC PROJECTION (continued} : —
PROB. VIII. To draw a great circle through any given point so
as to make any desired angle at the primitive . 43 IX. To draw a small circle through a given point which
shall be at a given distance from a right circle . 44 X. To draw an oblique circle perpendicular to a given
oblique circle 45
XI. To draw a great circle which shall make any angle
with the primitive 45
XII. To measure any part of an oblique circle . . 46
XIII. To measure an angle at the primitive . . .47
XIV. To measure an angle which is not at the primitive . 47
CONSTRUCTION OF SPHERIC TRIANGLES .... 48
To find the latitude of a place 51
To find the apparent time . . . .... . . .56
To find an azimuth 58
CALCULATION OF SPHERIC TRIANGLES . ... . 60
Right angled -. . 60
The five circular parts . . . . . . . .62
Quadrantal spheric triangles . ' . 69
Oblique spheric triangles 69
CASE 1. Two sides and an opposite angle . . . .71
2. Two angles and an opposite side . . . .74
3. Two sides and an included angle . . . .76
4. Two angles and an included side . . . .80
5. Three sides • « .83
6. Three angles 85
NAUTICAL ASTRONOMY:—
Ex. 1, 2, 3, 4, 5. To find latitude 87
6, 7, 8. To find apparent time 96
9, 10. To find an azimuth 100
11. To find a great circle course .... 102
12. To find a great circle distance . . . .104
13. To find the latitude of a great circle vertex . 105
14. To find an altitude 107
15. To find an amplitude . . . . .108
16. To find the time of daybreak . . . .109
17. To find the time of rising of any celestial body . Ill
Erratum. Page 24, line 12, for 1 + 1 + ^ read 1 + 1 + —
TRIGONOMETRY
AND
NAUTICAL ASTRONOMY.
WHILE plane trigonometry in its application to the practice of navigation, is so well set forth in the usual epitomes and other nautical works, that little need be said herein upon the subject, beyond a few remarks on certain of its fundamental principles, no one can properly comprehend even the very elementary principles of nautical astronomy, without a better insight into spherical trigonometry than is to be obtained from any work at present accessible to the sea officer.
PROJECTION.
The study of spherics is too often undertaken in igno- rance of a method of constructing a spheric angle by scale. Such methods (for there are several) are called " projections," a general term in spherics, signifying the transferring of spaces from rounded surfaces to flat or " plane " ones, in the forms in which the eye would trace them, according to their assumed relative position.
Distortion must of necessity accompany the projection
B
2 NAUTICAL ASTRONOMY.
of a spheric surface upon a plane, and all methods used in accomplishing this are not equally good.
1. For example, the gnomonic projection presents peculiar advantages in dialling, and also in great circle or tangent sailing; while the orthographic projection is used by the astronomer in the delineation of eclipses, the transits of heavenly bodies, &c. Mercator's is ex- ceedingly useful in the construction of sea charts ; while the stereographic is more applicable than all others to the purposes of nautical astronomy, in consequence of all its parts being either arcs of circles or straight lines. The globular is used by map engravers, the scenographic for perspective, and need not occupy the attention of the nautical astronomer.
2. Our consideration, then, will be restricted to the orthographic, Mercator's and stereographic projections ; merely first noting that in the gnomonic projection, the eye is supposed to be at the centre of the sphere, viewing the meridians as straight lines; and in this projection the shortest distance on the globe between two places is re- presented by the shortest distance between the two cor- responding points on the flat surface.
ORTHOGRAPHIC.
3. In the orthographic projection, we, for the occasion, must suppose the eye to be placed at an immense distance,
Fig. 1.
.,n
\\s»
ORTHOGRAPHIC PROJECTION. 3
and as viewing" a sphere as if it were a mere disc, such as the sun or moon appears to be, the visual rays, as 1, 2, 3, &c., in Fig. 1, being supposed to be parallel to each other ; in such case parallels of latitude in a right sphere would appear (as c, d, x, in Fig. 2) to be straight lines, and se- parated in the proportion of the divisions on what we call the line of sines upon the scale ; while the meridians lying near the diameter would appear to be at nearly their actual distances asunder, but would be crowded as they lie nearer to the periphery as at a (Fig. 1).*
4. Supposing, further, that a globe of immense size had the usual lines of the sphere marked upon it, it would, as
Fig. 2.
seen by the eye at (if possible) an extreme distance, have the appearance of Fig. 2, in which all the meridians are ellipses and not arcs of circles ; this would readily be un- derstood by holding a small toy globe at arm's length, with the polar diameter in a vertical direction.
5. There is, perhaps, at the present time, no work in pub-
* Many illustrations might be given from one crowded diagram ; but as plainness is so desirable in a work like this, it is better to give separate diagrams with the text therewith connected.
B 2
4 NAUTICAL ASTRONOMY.
lication which would much assist the ordinary student in projecting the orthographic sphere. The methods given, and the explanations accompanying them, are beyond the comprehension of a beginner ; and are, moreover, so suffi- ciently troublesome to the practised draughtsman as to cause him in most cases merely to mark off the ortho- graphic distances of the meridians on the equator only, and draw them as arcs of circles. This may even be seen in the diagrams illustrating the works of our greatest philosophers. Hence orthographic plates will seldom bear the inspection of the mathematician. It is not therefore extraordinary that the orthographic projection has been so little used for nautical purposes, until the writer sug- gested its introduction (as it will be further explained), to relieve the stereographic projection of certain disabilities in its application to the purposes of navigation.
6. The following directions for constructing an ortho- graphic hemisphere, in what is called a " right sphere " (having the poles in the circumference as distinguished from a " parallel sphere " in which the poles are at the centre, and from an " oblique sphere " in which the poles are neither at the circumference nor at the centre) are_ given in detail, the method not having been yet published, and being the one long used by the writer. Its utility may at once be inferred from the circumstance of its requiring neither tangents nor secants, but simply the line of sines and scale of chords. It should be borne in mind that in Fig. 2 the parallels all appear as straight lines, like the equator ; and if the line, a b, in its divisions forms the line of sines as se^n at the diameter of Fig. 1, any corresponding por- tions; of the other parallels must do the same, because they are supposed to be viewed from the same point; hence, c d, and e f, Fig. 2, &c., are each crossed by meridians, at distances proportional to the natural sines of their respective latitudes. In other words, d'2 would
ORTHOGRAPHIC PROJECTION. 5
be a semi-minor axis of the meridian z 2 n, or the cosine of its meridian. Every nautical astronomer, therefore, will do well to have a scale prepared for general use, similar to the one at Fig. 3. It is formed by making the lines B C and B A meet at any angle (the nearer 90° the more convenient), and laying off on B A the distances B 15, B 30, &c., to B 90, from any scale of sines ; then join C 15, C 30, &c., and parallel to B A draw any number of lines through the figure, and the scale is complete.
Fig. 2 may be drawn from this scale in the following manner : —
Take A B from the scale (Fig. 3), as a radius, and de- scribe a circle, drawing two diameters right-angled at the centre. Take a straight edge of paper and lay it at B A, Fig. 3; transfer on to it the distances, B 15, B 30, &c., up
Fig. 3.
Scale of sines for orthographic projection.
to 90 ; lay these off on Fig. 2, as 6 1, 6 2, &c., up to 6 a, and also from b towards 2, from b to y, and from b to n. Through b z and b n draw parallels as in Fig. 2 ; then take the length of any half parallels, as c d, on a straight edge of paper, find its distance on the scale (Fig. 3), as at c d, and having copied the divisions (as was done at B A in dividing a 6, Fig. 2), lay off these on Fig. 2, as dl9 d2, d3, &c. : proceed thus with each half parallel, and points will be accurately obtained through which the whole of the desired meridians may be constructed.
B 3
6 NAUTICAL ASTKOXOMY.
7. If it be required to project the sphere on the plane of the equator, or the " parallel sphere," proceed as fol- lows. From the scale of sines (Fig. 3) take B A as radius, and with it draw a circle abed, quartering it, as before, at right angles, as in Fig. 4. If it be desirable to draw meridians through every 10°, or every 15° (suppose the latter) take the division of the line of sines, Fig. 3 (as before), apply them from e towards b, in Fig. 4 ; and with the centre e, and distances el, e2y &c., describe circles ; then lay off from a to b, b to c, c to d, and d to a, every 15th degree from a scale of chords corresponding
Fig. 4.
to the radius (or by dividing either quadrant into six equal parts); join these several points with the centre, and these form the meridians, the poles of the primitive circle being at the centre.
8. This mode of construction is particularly useful in projecting places in the polar regions, or places of heavenly bodies when near the poles of the ecliptic or equator ; in such cases the least distortions being at and near the centre of the figure; but is most especially valuable in general questions of nautical astronomy, where the data
MERCATOR'S PROJECTION. 7
lie near the prime vertical, such as altitudes, &c., at hours which are near 6 A.M. or 6 P.M., while for other periods of the day or night nearer to noon or midnight, the stereo- graphic projection has its advantages.
MERCATOR'S.
9. Mercator's projection is too well known by nautical men to require much mention here, but its properties may be thus very briefly stated. On a- globe it will be noticed that the actual measured length of degrees of longitude diminishes as they recede from the equator towards either pole.
In about the year 1590, a Mr. Wright, of Caius College, Cambridge, conceived the notion that he could on paper conveniently compensate this contraction of the degrees of longitude (in placing the round upon a flat surface), by expanding each degree of latitude in proportion to its corresponding degree of longitude in that latitude; by this means he projected the meridians and parallels as straight lines; hence, in this all rhumbs (or compass bearings) cross the meridians at equal angles, and a ship's course is laid down as a straight line. But this projection greatly distorts the outline and figures of places lying far from the equator. The miles in a degree of latitude are to the miles in a degree of longitude as radius is to cosine of latitude in which the degree of longitude is situated. For example : — Eequired the length in measured nautical miles of a degree of longitude in latitude 50°.
In Fig. 5, making hypothenuse radius, we have (by plane trigonometry)
rad AC : 60 : : cos 50° : AB = 38' -57 miles of long;
B 4
NAUTICAL ASTRONOMY.
or in Fig. 6, making base radius
sec 50° : 60 : : rad A B : 38'-57 miles of long.
Fig. 5.
Fig. 6
STEREOGRAPHIC.
10. The stereographic projection (which every navi- gator will feel the greatest pleasure and advantage in comprehending if he read the following with ordinary attention), enters largely into the daily work and interest of every one who has command of a ship.
In this projection the eye is supposed to be at some part of the earth's surface, say, upon the equator, and it would in such case see the meridians of the further hemisphere, as they would appear if produced on to a flat surface tangential to the opposite point of the equator which would be exactly under the eye. The following (Fig. 7) will best illustrate this ; in which we suppose the globe to be transparent, and resting on a point of its equator upon a table ; the eye being at a, and viewing the meridians, c 5, c 4, &c., as they would appear to it if their intersection of the equator, a b D, were produced to the surface of the table.
STEEEOGEAPHIC PEOJECTION. 9
In this projection we require the use of the scale of chords, semi-tangents, tangents, sines, and secants. The visual rays from a passing and cutting the radius b c, form on it what is called the scale of semi-tangents ; and the two triangles, acb and "aDB," being similar, the divisions
Fig. 7.
on b c and B D are proportional. This needs special remem- brance.
In this projection, spaces lying near the centre are con- tracted in size, the largest degrees being near the primitive circle.
Meridians, as drawn obliquely, may be imitated by hold- ing a ring or coin in various positions, the circumference at one time appearing as a circle, or at others as a straight line or slightly curved, &c.
10
NAUTICAL ASTRONOMY.
GENERAL REMARKS ON TRIGONOMETRY.
11. A few elementary hints, which, although essential to the proper understanding of the subject, are not always suf- ficiently explained to a beginner, and thereby materially retard his progress, will properly precede the consideration of spheric construction.
The following (Fig. 8) is called the trigonometrical " canon " (a word merely signifying a collection of mathe- matical truths or formulae), and from it are derived all the terms and rules used and practised in trigonometry.
Fig. 8
In the above,
AB is called the radius of the circle (of course A K, A I, AH, and AE, are also radii. Euc. I. def. 15.)
B C is called a tangent (always of the opposite angle A), and is so called from Latin, tangere to touch, be- cause it only touches but does not cut the circle.
A C is called a secant [always of the angle between it
PLANE TKIGONOMETRY. 11
and radius, therefore of L (angle) A], so called from Latin, secare to cut, because it cuts through the circle. D E is called a sine, because it lies in the hollow, or
bosom, of the curve E B L (Latin, sinus). FE is called a cosine, or sine of an l_ which is com- plementary to another, or required to make up 90°. Thus DE is the sine of the arc EB, and FE is the sine of arc E H, which is the complement of E B (for H E + E B = 90°). Therefore F E is called the cosine of E B, and is equal to A D, because F E and A D are drawn parallel, and E D is perpendicular to both. GrH is, in like manner, the tangent of HE, or cotan- gent of EB. AGr is, in like manner, the secant of HE, or cosecant
ofEB. HI is called a line of chords, from its serving, as it
were, to tie or confine the ends of the arc Unl. D B is called a versed sine, and is the " height of the
segment " E B L D E.
12. Every circle is supposed to be divided into 360 de- grees (marked 360°). If, there- fore, in the following figure, (9) FlS- 9- PZ equals 90° or a quadrant, it is plain that if D a? also equals 90°, the word "degree" refers to no measure of length,but merely signi- fies the 360th part of a circle, what- ever the size of that circle may be ; and, therefore, a degree may be of any length. As, however, degrees D r enter into calculations, some de- finite value of them must evidently be necessary ; and, consequently, geometers express the value of degrees by taking any two lines from those given in the trigo-
12
NAUTICAL ASTRONOMY.
nometrical canon (Fig. 8), and consider the length of one as compared with the length of another in the same triangle : so that we use the terms sine, tangent, secant, &c., as referred generally to the radius of the circle, con- sidering the length of radius to be 1 or 10, &c. (inches, feet, miles, leagues, &c., at will) : this will be further illustrated.
13. It has long been customary to call a line, as E D (Fig. 8), a sine, or as BC, a tangent; but such is only correct when the length of a radius of a circle is known or understood. It is generally useful to describe the sine, &c.,
~p"T)
as fractions, thus, — — (Fig. 8), as they express fairly the AJl
value referred to. Anticipating by a few pages the question of proportion, it may here be noted that a vulgar fraction is a " ratio " or proportion in itself, and is deduced from a triangle. Thus, when we speak of -fths of anything we refer to some magnitude which can only be appreciated by considering the fraction -J in relation to its integer or whole number " one." As this whole number itself, expressed as a fraction, is -Jths, fths, fths, &c. ; and when, therefore, we speak of |ths, we express a " ratio," meaning as 3 is to 4 (or symbolically 3 : 4), so then we express the value of degrees by using ratios, and com- paring them with the radius of the circle, which it is convenient to do by a number capable of decimal division, for obvious reasons (such as 1, 10, 100, &c.); and as at least one side of a plane triangle is al- ways given, we are at liberty to compare this with the length of what is thus called the sine, tan- gent, secant, &c., of an angle, and hence the length of the arc itself.
14. For further example, in Fig. 10, let BD be what is
Fig. 10.
PLANE TRIGONOMETRY. 13
commonly called the sine of the angle A. We describe its value by saying it is as the perpendicular is to radius, and
write it thus : £^? or from the figure -— • rad 015
In like manner the other fundamental trigonometrical ratios are represented by fractions thus : —
= -— - is an expression for the tangent of L A
rad CN
rad CN cotang
perp PN
«"
From the above it will be seen that certain ratios are reciprocals ; for instance : —
PN C P
sine =— — and cosec = — —
and cotang =
CP CN
sec =7^: and cosine = -— —
CN C P
15. Therefore, in works on logarithms, when we want the secant of an angle we can find it by subtracting its cosine from 20 (the diameter of a circle whose radius is 10), and to find the log sine we subtract the log cosec from 20 ; and to find the log tang we subtract the log cotang from 20, &c. Other useful, deductions maybe made, such as to find the log tang : add 10 to the log sine, and from the
14 NAUTICAL ASTRONOMY.
sum subtract the log cos (or log tang = log sme+1Q).
log cos
and to find the log cotang add 10 to the log cos, and from
the sum subtract the log sinefor log cotang= log COS+1Q)
log sine
&c. ; so that the values of the six fundamental ratios may be expressed thus : —
sine = cosine =
cosec sec
tang = cot =
cot tang
sec = • — : — cosec = — —
cosine sine
N.B. — The unit here meaning 1 diameter =2 radii each of 10, or diameter =20.
16. This, however, which forms the elementary base of a proper knowledge of trigonometry, is not absolutely essen- tial to the navigator, whose practical operations in plane trigonometry may be performed in total ignorance of prin- ciples, by dint of mere intelligence and skill from repetition; but in like manner does the blacksmith strike with the face of the hammer, and not with the handle, and would not probably perform his work more effectually if he were, previous to every blow, to calculate the force required to fashion his heated iron. But this must be remembered : a mathematical " blacksmith " would probably give fewer blows, because he would better know how to make each stroke tell, from bringing the face of the hammer to bear on the iron in the best direction with the greatest effect. In like manner the mathematical navigator will obtain his result in the shortest method.
It is beneath the dignity of a British sea officer to be content with mere knowledge of the use of formulse. After reading this little book he may be safely advised to take up
ANGLES. 15
" Jeans's Trigonometry," or some such small work on the subject.
DEFINITION OF AN ANGLE.
17. Before proceeding it may be well to explain what is really meant by an angle : that such explanation is necessary cannot be denied. A work of this description, which is designed as a mere stepping-stone to study, must needs adopt assertions without proofs, for fear of alarming the timid who desire improvement, but who yet doubt their own powers. It may, however,' be safely asserted, that since our proofs are deduced mainly from the Books of Euclid a knowledge of his system of proving should be imparted at the earliest opportunity. In works upon navi- gation generally more extracts from Euclid are given than the sea officer thinks necessary for his satisfactory working, and too few to satisfy his after-desire of research ; while the Books of Euclid themselves are supposed to be too heavy an undertaking for any but a schoolboy having no other em- ployment than study. These are delusions. A ground- work in mathematics well laid is a continual source of mental profit and amusement. There is no limit (but the powers of mortal intellects) to the structure which may be raised upon it. A very long acquaintance with the subject of teaching can only lead to a belief that whenever mathe- matical study is to any mind found to be repulsive, it may be suspected that the individual student has not had its details sufficiently explained. Thousands upon thousands can work an equation by logarithms who have but an in- distinct notion of what is really meant by an angle ; and it may cheer the student when he sees that we may go even to the " dreaded " Euclid to obtain a full and clear com- prehension even of this trifle.
For instance, he says among his definitions (Book I. def.
16 NAUTICAL ASTRONOMY.
8) : ft A plane angle is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction." And (Book I. def. 9), "A plane recti- lineal angle is the inclination of two right (or straight) lines to one another, which meet together, but are not in the same right line ;" so that, in the following figure the right
rig. 11.
line AB meets the right line AC at the point A, and the " angle " is the inclination of these two lines as measured in degrees upon any circle drawn from A as a centre cutting these two lines. For instance, de, or fg, or hi, is each the measure of the (( angle A " in degrees, 20 degrees meaning -f/^ of any circle drawn round the point A.
GEOMETRICAL THEOREMS.
18. A few of Euclid's theorems may here be introduced with advantage.
Book I. XIII. — The angles which one right line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Depart- ing from the precise language of absolute and complete proof (for obvious reasons), we may say that the semicircle HEB contains 180 degrees. If EGr be perpendicular to AB, the arcs HE and EB being equal will each contain 90°, but BC is less than 90°, being, say, 55°; then EC must be 35°, and EH being 90°, CH will be 125°. Now, EC is
THEOREMS.
17
called the complement of CB, and HC is called the supple- ment of C B.
•,55°
19. Book I. XV. tells us that if two right lines cut one another the vertical or opposite angles shall be equal. Thus, the angles CEA and BED are equal to each other, as are . also C E B and A E D ; the angle CEA means the angle at the point E formed by the two lines CE EA. (We always put the letter indicating the point between the others.) Now, from the XHIth Proposition, Book I., as above, it is evident that the two contiguous angles HGrC and C Gr B equal 180 degrees; so in Fig. 13 ac + cb equal 180 degrees, and bd + d a form the other 180 degrees.
Fig. 13.
20. Euc. I. Prop. XXIX. — If a right line falls upon two parallel right lines, it makes the alternate angles equal to one another, &c. &c. ; so that, indeed, A GrE, C H E, FHD,
C
18 NAUTICAL ASTKONOMY.
FGrB, are equal to each other, being in this case about 28 degrees, while EGrB, EHD, FHC, and FGrA are also equal, being about 152°, — the circles render this apparent.
To satisfy this it needs only that AB and CD be pre- cisely parallel.
21. Euc. I. Prop. XIX. — The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.
The arcs of the circles drawn about each of the angular
points with an equal radius show at once that, for in- stance, the small angle B 30° is opposite to the smaller side AC, &c.
22. Euc. VI. Prop. VIII. — In a right-angled triangle if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another (that is, triangle ADC, triangle BDC, and triangle ACB are similar).
Now, by similar triangles we mean triangles which have
THEOEEMS. 19
the three angles in the one equal to three angles respectively in the other ; and although their opposite sides may be of different lengths, they are nevertheless proportional : thus, by the figure, if we lay off the distance C D at B c, and draw ac parallel to CD, we shall find the triangle Be a equal in its angles to CD A, and its sides proportional to tri- angle BCD, that is, Be will be to ac as BD is to DC (Be : ac :: BD : DC), &c. &c., and here (by L, XXIX.), because BC falls across the two parallel lines ac9 CD, the angles Bac and BCD are equal. (And this is the
Fig. 16.
way in which one proposition of Euclid rests for proof upon others which precede. As an example, proof of the above proposition could only be made with mathe- matical accuracy by reference to 34 propositions of Book L, 10 of Book V., and 3 of Book VI.; in all 47 pro- positions, besides definitions, axioms, and postulates, repetitions, &c.).
23. Euc. III. Prop. XX. — The angle at the centre of a circle is double the angle at the circumference upon the same base, that is, upon the same part of the circum- ference.
In the triangle ADC, the side AD equals DC, there- fore, as equal sides are opposite to equal angles (as de- duced from Bookl. Prop. XVIIL), the angle DAC equals D C A. But Prop. XXXII. Book I. implies that the angle EDA equals ^D AC and DC A together; let ACB and
c 2
20
NAUTICAL ASTKONOMY.
A D B be two angles standing upon the same base A B ; therefore, as ADE is the double of A CD, and by like reasoning E D B would be the double of E C B, so must the whole angle A D B be the double of the angle A C B.
Fig. 17.
Euc. I. Prop. XXXII. — If any side of a triangle be produced, the exterior angle is equal to the two in- terior and opposite angles; and the three interior angles of any triangle are equal to two right angles.
Fig. 18.
Reference to Prop. XXIX. Book I. will show that the angle CAB will equal the angle DBE, because the right line C B falls on the two parallels A C and B D ; also, that the angle A C B will equal C B D. Therefore the angle CBE, which is exterior to the triangle, and is made up of the two angles DBE and D B C, is equal to the two in- terior and opposite angles of the triangle (the angle ABC being the adjacent angle). And it is also evident that the two 6( opposite " angles, together with the " adjacent " angle form the interior angles of the triangle) are, equal to the
RATIOS. 21
exterior and adjacent angles together, therefore (by I. XIII.) are equal to 180° or two right angles.
The above will be sufficient to give a general notion of the importance of Euclid.
25. Every triangle has three sides, and in lettering a right-angled triangle it is customary to place B at the right angle and to make AC designate the hypothenuse.
KATIOS.
26. The rules of Plane Trigonometry do not fall within the present compass of this work ; they are to be found in all works on navigation. It may, however, be remarked that the rules given therein for the working of questions in navigation, take for instance —
as diff lat is to rad so is dep to tang course,
are in the form of a proportion or ratio (proportion is the equality of ratios) — three things, as in what is called the Eule of Three, being given to find a fourth. Every con- ceivable arithmetical calculation is the working of a pro- portion, or the comparison of ratios. When we say 5 times 8 make 40, we mean to say that 1 : 5 :: 8 : 40, or, fractionally, ^-, £$ ; and these fractions form what is called an "equation," for £ = TV But Euclid, Book VI. 16, demonstrates that, if four straight lines (or quantities) be proportionals, the product of the means (or middle terms) of such proportion shall be equal to the product of the extremes (or first and last terms), as from the above 1 x 40 = 5 x 8 ; and, as another example, if 3 coils of rope cost 20 shillings, 6 coils will cost 40, or 3 : 20 :: 6 : 40; that is 3 x 40 = 20 x 6.
c 3
22 NAUTICAL ASTRONOMY.
2i7. Euclid, further, in Book V. (Def. 13 to 17), shows that such proportions may be varied by division, conver- sion, inversion, alternation, &c., so that the last example admits of being varied in form; for
As 20 : 40 :: 3 : 6 6 : 40 :: 3 : 20 40 : 6 :: 20 : 3, &c.
So far, therefore, as the calculation of straight lines is con- cerned (which may be put to represent by their propor- tionate lengths any proportionate quantities), the work of calculation becomes easy; but in entering upon the calcu- lation of angles, 'it should be shown that common arith- metic fails altogether in its powers to readily solve all the parts of a trigonometrical figure. Nor is the difficulty properly explained in elementary works of the present day.
NATUEAL SINES.
28. If we examine Fig. 19, in which the circumference is divided into equal parts of ten degrees each, we shall see in the line C D a scale of what are called " Natural Sines." C 30 being equal to a 30, &c. The divisions on
Fig. 19.
D
on '90
|
60 ^ e |
|||
|
jn /" 5 |
O >. .Bft |
||
|
/_ o 4 |
0 ™ \ 40 |
||
|
/_ |
" 3 |
5 \ 0 . \ Qn |
|
|
..? |
2, |
2 |
Ui |
|
„/ |
Q z |
I \ |
|
|
1 |
C D, moreover, are not equal; for C 10 is much larger than the distance 40 to 50 ; and, indeed, the sine of 30° is exactly in length half the sine of 90°, while the natural
LOGARITHMS. 23
number 30° is only one third of 90°; and we shall also find that the secant of 60° is equal to twice the sine of 90°, while the natural number 60 is two thirds of 90 ; that the tangent of 45° is equal to the sine of 90°, &c.
LOGARITHMS.
29. To obviate the above inconvenience, an artificial system of numbers was sought for by mathematicians ; and, happily, Lord Napier, about 250 years since, gave the world his System of Logarithms; and our astonishment is excited when we learn that this great discovery was made nearly half a century before what has since been called the " Logarithmic Series " was invented.
NATURE OF LOGARITHMS.
30. The very word Logarithm is a stumbling-block to many; but it is easy to show that, although the ground- work is so very little known to many who use logarithms, a few hints put in familiar language will not merely gra- tify a laudable curiosity, but pleasantly assist in further investigation.
31. Lord Napier based his system of logarithms upon the following infinite series, in which it will be seen that values of fractions are systematically diminished by adding in- creasing multipliers to the denominators.
1 + T + F2 + 1^3 +P2:^4 + 1-2-3-4-5 + &C* Any one who understands vulgar fractions can resolve these into the following : —
1 + 1 + 1 + i + A + ITS + &c. Of course the series might be extended to great length
C 4
24 NAUTICAL ASTRONOMY.
and increased accuracy, but the above is enough for our purpose. Now, if we reduce the above fractions to a com- mon denominator, we get
and adding numerators we get 2T8-^-, which, reduced to decimals, becomes =2-71666, an approximate base of the Napierian logarithms, which, if extended to further terms, and the division be made as usually shown in works on logarithms, becomes (as under) 2-7182818, &c., which, so far as it extends, is the true base of the Napierian log- arithmic system ; thus : —
1 + 1
|
+ 1-2 |
*i =: |
4-0 |
|
1 1^3 |
i § |
•1666666666 |
|
1 1-2-3-4 |
i 24 |
•0416666666 |
|
1 1-2-3 -4-5 |
1 120 |
•0083333333 |
|
1 |
1 720 |
•0013888888 |
|
1-2-3-4-5-6 |
||
|
1 |
1 |
•0001984126 |
|
1-2-3 -4-5 -6 -7 |
5040 |
|
|
1 |
1 |
•0000248015 •0000027557 |
|
1-2-3-4-5-6-7-8 1 |
40320 1 |
|
|
1-2-3-4-5-6-7-8-9 |
362880 |
|
|
1 |
1 |
•0000002755 |
|
1-2-3-4-5-6-7-8-9-10 |
3628800 |
2-7182818 &c. Now, the Napierian system, arising, as above shown, from
LOGARITHMS. 25
so simple an infinite series, and one which is so easily re- membered, is made the basis of all other systems.
32. Any number may be taken as a base; let us at random take the number 3. Then, as logarithms are de- nned to be a series of numbers in arithmetical progres- sion, placed opposite to and corresponding with another series in geometrical progression, and so placed that 0 in the logarithmic stands opposite 1 in the geometric — we can easily form a skeleton system based on the number 3, as under : —
Natural Numbers. Geometrical. Logarithms.
1 = 1 . . v • 0-000000
31 = 3 . V . 1-000000
32 = 9 (3 x 3) . '..... . 2-000000
33 = 27 (3 x 3 x 3) . .; 3-000000
34 = 81 (3 x3 x 3 x3) :. 4-000000
35 = 243 &c. &c. . . . 5-000000
36 = 729 '., ; . . . 6-000000
37 = 2187 . .. >. . .. * 7-000000
38 = 6561 . , . \ 8-000000
39 = 19683 '. „ . . 9-000000 310 = 59049 ; . . . 10-000000
33. To prove that the above is really a table of loga- rithms, let us attempt calculation l)y it as we would by the table in common use : remembering the rules, that —
In logarithms we multiply numbers by adding their logarithms, and we divide numbers by subtracting their logarithms. Suppose, for example, we desire to multiply 729 by 81.
In the above table
the log of 729 is 6 log of 81 is 4
the sum 10 = log of the answer, 59049
26 NAUTICAL ASTRONOMY.
By arithmetic.
729
81
729
5832
59049
And again, to divide 2187 by 243,
By arithmetic. By logarithms.
243)2187(9 log of 2187 is 7
2187 log of 243 is 5
the difference 2 = log of 9.
And further, to extract the cube root of 19683. This is done by dividing the logarithm of the number by the index of the power.
9
logof 19683 = 9,and 3 tliein(j x = 3 which is the log of 27 ;
or, as written V 19683 = 27
and again, to raise the number 9 to the fourth power : multiply the log by the number of the index of the power ; thus —
log of 9 is 2 index 4
8 = log of the number 6561 ; or 94 = 6561.
34b. A table of the above description has, however, serious defects ; the greatest is apparently the want of analogy between the number of figures in the whole number and the index of the logarithm, as will be shown immediately. The index or characteristic of the logarithm, is the in-
LOGARITHMS. :>7
teger, or whole number, and the decimal is called the mantissa. To facilitate calculation by logarithms, Mr. Henry Briggs, a contemporary of Lord Napier, published at Cambridge, in 1615, a system having for its base the number 10, the root of our decimal scale of notation, in which the powers of the number 10 are shown by merely adding to unity as many figures as are " indicated," by what are therefore aptly called the indices of different powers, as we see in the following skeleton table to the base 10. Here, again, the logarithm is merely the index of the poiver, while it indicates absolutely the number of figures, less one, in the whole number to which it cor- responds. This is the common system of logarithms.
35. The above table (32) only gives the logarithms of numbers which are multiples or powers of 10, but we might, for instance, require to know the log of 270 — which would evidently lie between the log of 100 and the log of 1000; its index, however, would be 2 (because the number contains three figures), together with a decimal or mantissa, and we see in a more extended table it would be as represented by the log 2-23044.
Natural Numbers. Logarithms.
10° = 1 ,,. , ,...-.'• . 0-000000 &c.
101 = 10 . . .... . 1-000000
10 2 = 100 . . . . 2-000000
10 3 = 1000 . . . . 3-000000
10 4 = 10000. . . . 4-000000
10 5 =100000 . . . 5-000000
10 6 = 1000000 . , . 6-000000
10 7 =10000000 . . . 7-000000
10 8 = 100000000 . . . 8-000000
10 9 = 1000000000 . . 9-000000 10 10 = 10000000000 . 10-000000
28 NAUTICAL ASTEONOMY.
36. It is important to add, that if the natural number be a vulgar fraction, such as -|, we may (because it means 5 divided by 8), subtract the log of the denominator from that of the numerator (increased by unity if necessary) thus —
log of 5 = 0-698970 p f 8)5000
„ „ 8 = 0-903090 -7625
1-795880 = dec. fraction -625.
It is obvious, therefore, that it would have been as simple to have reduced the vulgar fraction to its decimal at once, and then taken its logarithm.
37. But we borrowed an unit in subtracting, there- fore the resulting 9 was absolutely -9 (decimal 9) or minus 1 (written 1). (A word here to those who are not " well up " in decimal arithmetic ; be advised and lose not a day in " brushing up " a little. It is not, however, likely, that any one having sufficient interest in the sub- ject to enable him to read this little book thus far, will do otherwise.) It will then be easily seen that it would, in the above example, have been better to borrow 10 than 1, writing the resulting index 9 as minus 9 (— 9). An- other example : multiply 100-6 by -1006.
log of 100-6 = 2-002598 •1006 = -9-002598
1-005196
casting off the borrowed ten it will be 1-005196, equal to the number 10*12; this is a more simple plan than writing the log of -1006, as 1*002598, and operating algebraically.
The following abstract will have its uses, and illustrate the above. (The number 1006 is taken at random, any number may be substituted.)
LOGARITHMS. 29
Natural Numbers. Logarithms.
1006 3-002598
100-6 2-002598
10-06 . ,' . . . . 1-002598 1-006 . ,.-; . . . 0-002598 •1006 . . . . —9-002598 •01006 . , . \. —8-002528 ' •001006 . , . .- —7-002598 &c.
N.B. — All works on logarithms have rules attached, for taking out numbers, whether representing linear or angular quantities.
COMPUTATION OF LOGAEITHMS.
38. We have seen (34) that only logarithms which have a certain base are conveniently applicable to practical pur poses, and that the Napierian system is the most simply obtained from a series, which gives its base 2-718281828, &c. This is commonly called the natural or hyperbolic system, and is written thus —
Loge 2-718281828, &c., while the decimal or Brigg's or the common system is written log 10 (read, log to the base 10, or log to the base e). .
We use the Napierian system as a foundation of our common system in the following deduced formula : —
The common log \ _ Nap. log of number Nap. log. of number of any number J ' the Nap. log of 10 = 2-3025851
1, = -434:29448 x Nap. log of the number. 2'302o8ol
Hence, to construct the common logarithm of any number, we use a number which maybe called a Napierian constant ; it is the double of the above equals -43429448 (which is the modulus of the common system), and equals -86858896.
30 NAUTICAL ASTROXOMY.
It is here quite unnecesary to give the algebraic reasons why we use the following infinite series : it is enough for our purpose to say that in it we have a series, by which we may compute the logs of all natural numbers, and this without knowing the log of any previous num- ber. It is this : —
Now, if we let P represent the number whose log we re- quire, say the number 2, and M the modulus, '43429448, the above will, in figures, be as under: —
,og , .
(or, as reduced)
log 2 - -86858896 ( 1 + ! + _L_ + 1 + &c.\ V 3 81 1215 15309 )
(or decimally) log 2 = -86858896 (-333333 + -012347 + '000823 + -000065 + &c.)
(or, after addition) log 2 = -86858896 x -346568
(or, after multiplication)
log 2 = -301025 approximately, but if the series be extended log 2 = '3010300 as we find in our tables of common logarithms.
31
NAUTICAL ASTRONOMY.
39. Before entering upon precise rules for construction and calculation of spheric angles, a few remarks upon the subject of nautical astronomy itself will lighten the labour of the student.
The general complaint of those who have " looked into " spherics, is, that although they have been taught to work spheric angles, they do not understand the principles sufficiently to be able to apply their knowledge with con- fidence to ordinary or rather extraordinary work.
If, say they, we could always see the figures illustrating our questions, were it only in the mind's eye, it would assist us in obtaining solutions, and increase our interest in the study; it would also very materially help our memories.
Too generally, however, a spheric triangle is drawn by hand, without any reference whatever to its adaptation to , the data of the question under consideration ; hence much theoretical difficulty arises from the consideration of angles, as acute or obtuse, and of complements, supplements, &c. If in the study of spherics, the maturing of the reasoning powers is to be our main object (as in teaching Euclid for the mere logic of its reasonings), present works on the subject are abundant. But if the study is to be undertaken by those circumstanced like sea officers in general, whose object is to master enough of the principles to enable them to pursue their professional avocations in mathematics, with confidence and success, an abridged arrangement of scientific facts suitable to their purpose will doubly benefit them, inasmuch as not only will their nautical work be more accurately and more readily per- formed, but the having once obtained a well-grounded acquirement in principles, will render it less difficult for
32 NAUTICAL ASTKONOMY.
them to employ their few hours of leisure, which some services permit, in pleasant advancement. . 4O. The special object of the following explanations is then, to lead the young seaman to the most agreeable part of a navigator's study, viz., "construction." Not that in practice he will be required to actually draw his diagrams, but a knowledge of construction will greatly aid him in calculation.
LINES OF THE SPHERE.
To give a general notion of the imaginary lines of the sphere, we will suppose that I am standing on the coast, looking seaward towards the west. The north would in such case evidently be on my right hand, the south on my left hand, and the east would be behind me. I might imagine a point over my head called the zenith, and a point below my feet called the nadir. The distance of the zenith and nadir would, of course, be unlimited ; but let us for the sake of precise illustration limit it to any distance, say, 1000 yards ; in such case I must con- sider the horizon to the north and south also limited to the same distance. Now, suppose further, the meridian of the place on which I am standing to be a line drawn from the north point of the horizon, up over my head through the zenith point, and down precisely to the south point of the horizon, this would give me a semicircle. And again, imagine lines drawn from the point over my head (zenith) downwards, so as to cut each point of the com- pass at the horizon ; these would be called azimuth circles. Let me now further suppose that I walk backwards in a line due east, until I see the figure my imagination has been constructing with, say, a radius of 1000 yards; it
LINES OP THE SPHERE.
33
would, if visible, appear precisely as the following figure, 20 (if drawn on the stereographic projection, which alone will be used, with a slight exception, in the following de- monstration of spherics), the centre marked west (W). being the point on which I had been previously standing.
Fig. 20.
Suppose, again, that the centre of this figure was situated in latitude 50° N., and for illustration, that the polar star is exactly at the north polar point of the heavens, or exactly over the north pole of the earth (which it is not by about 1£ degrees), let me now, as in Fig. 21, divide my figure from north to Z, and from Z to south into degrees, 90 in each quadrant ; and through the 50th degree from north or N. I place the pole of the heavens, being at about the spot at which I should see the polar star at night in the heavens, while standing at the centre of the figure. I would next imagine a line connecting this pole with the- centre of the figure, and drawing another line at right angles to it from W to Q, the latter would represent the part of the equator above the horizon, S N, because every part of it would be 90° from the pole. It has already been shown that the great circle, NZS, is a meridian ; it is also to an observer at W, the 12 o'clock <e hour circle ; " because, supposing the sun to be in north declination (or distance north of the equator), and rising at the point of the horizon marked c, it would, between
D
34 NAUTICAL ASTRONOMY.
its rising and noon, seem to describe the small arc, cd, until^ being at d, on the meridian of the place at noon, it would descend from d towards c, where it would " set " below the horizon. Thus we see that the point W answers either for east or west. In Figs. 20 and 21, the azimuths are drawn to every point of the compass, and whichever azimuth circle cuts the horizon at the point of the sun's setting would be its true bearing at such sunset; in Fig. 21, it would set at about W.N.W. But if P Z S is an hour circle, so P W would be another (viz., the 6 o'clock hour circle), and we may conceive others to be drawn intermediate. P y is therefore one, and represents the hour circle of about half- past 2 P.M. or i past 9 A.M., while 0 the intersection of d c and PO^/, would be the sun's place at that time ; and it is, for example, the work of " spherics " to calculate the pro- portions of the spheric triangle Z 0 P, P 0 being evidently the " polar distance," and Z 0 the zenith distance, Z P the co-latitude, &c.
Fig. 21.
SPHEEIC TRIANGLE.
4:2* We may now, by another figure, 22, show that what is meant by a spheric triangle is really a part of the surface of a solid globe, bounded by three arcs of great circles, as the triangle ABC. The angles being the in-
SPHEEIC TRIANGLE.
35
clinations of the planes of the great circle to each other, and the lengths of the sides have always reference to the angles they make at the centre of the solid figure, al- though only the triangle itself is usually shown in dia- grams.
Having now, it is presumed, given a correct idea of what is to be understood by " spherics," it will be plea- sant to see how this very interesting branch of science is illustrative of daily ordinary phenomena. If we inquire as to what extent of information might be obtained from very simple data, the reply will be highly encouraging, because we shall find that the drawing of a few curves by very easy rules (which are about to be fully explained further on) open to us a vast and satisfactory insight into astro- nomy itself.
D 2
36
NAUTICAL ASTRONOMY.
GENERAL DIAGRAM OF DAILY ASTRONOMICAL PHENOMENA.
43. Let it be supposed that we are in latitude 50° N., that the date is the 20th May, and the time of day 10 in the morning.
N.B. — On the 20th May the sun's declination or distance from the equator would be about 20° N.
Figure 23 is drawn from these data, and gives us the following information accordingly : —
Fig. 23.
oo
S N would be the horizon.
OB „ the sun's altitude.
0 Z j, „ zenith distance.
OP „ „ polar distance.
DIAGRAM OF ASTRONOMICAL PHENOMENA.
37
|
E p would be a pai |
||
|
Z |
55 |
the z |
|
K |
55 |
55 I |
|
N |
55 |
55 E |
|
S |
55 |
55 & |
|
C |
55 |
55 e |
|
Od |
55 |
5, S |
|
ZOB |
55 |
55 |
|
SB |
55 |
55 |
|
BN |
55 |
99 |
|
CB |
3J |
59 |
|
Zd |
55 |
95 |
|
Sd |
55 |
55 |
|
pan |
>5 |
59 |
|
POH |
55 |
59 |
|
zc |
55 |
99 |
|
dy |
55 |
55 |
dA
y
A
d 0
F
Z.QPR
AT
dOy wEx Ew Ex
NZ
a parallel of altitude, zenith, nadir.
north part of the horizon, south.
east or west part of the horizon, sun's distance from the meridian, azimuth circle, azimuth from south. „ north.
amplitude from C (east or west), meridional zenith distance, altitude at noon (meridional alti- tude).
six o'clock, hour circle, ten o'clock, or 2 P.M. hour circle, prime vertical, declination, or apparent path in
the heavens for the day. half the length of the day.
night, sun's place at midnight.
„ rising or setting. „ noon.
„ ,,10 A.M., or 2 P.M. „ „ 6 o'clock. „ „ when on the prime vertical, time of the sun's rising or setting, the limit of duration of twilight, parallel of declination, 20th May.
„ „ 21st December,
half the length of the shortest winter's day. „ „ longest winter's night,
the sun's distance below the horizon on 21st December at midnight.
D 3
38 NAUTICAL ASTRONOMY.
S w would be the sun's meridian altitude on 21st December. N P „ the latitude of the place (or height of the pole). Z P „ complement of latitude. A Gr ascensional difference.
SPHERIC PROJECTION.
The general terms used in nautical astronomy being thus understood, it remains to illustrate <e spheric projection ; " but as we mean to explain as we go, let us advance warily. Only those problems which are abso- lutely essential to sea officers will be at first given. The remainder may possibly follow in a supplementary volume for the assistance of those who desire a more extensive acquaintance with the subject.
In the following figures the eye is supposed to be op- posite the centre, which point is called the (< pole " of the primitive or boundary circle ; but the word pole will not henceforward in this book signify anything more than a point exactly 90° from some great circle.
Circles are either great or small, not so much from their dimensions as from their position on the sphere. None but great circles can divide a sphere into two equal parts, their planes cutting the centre. Small circles are those whose planes do not cut the centre, but divide the sphere into two unequal parts. Small circles are also called parallel circles, because their planes are parallel to the plane of the equator (of this kind are parallels of latitude, of declination, of altitude, &c,).
Before explaining the problems, the following is to be specially remembered, viz., all the measures used in spherics, such as sines, tangents, &c., are taken from parts of the circle ; and in case the student may have forgotten the construction of the plane scale, its formation may be
PEOBLEMS.
39
seen in the accompanying diagram, from which all mea- sures in the succeeding figures are drawn.
45. PLANE SCALE.
Fig. 24.
12 30 40 50 6D SCALE 70 OF SECANTS
PROBLEMS.
46. PROB. I. — To make an angle so that the angular point
shall be at the centre of the primitive (say 50°). From centre 0 through 50° on the primitive draw 0 a if the arc be already divided. If not divided, make the angle a OH equal 50° by the use of a scale of chords, lay- ing 50° from H to a (first drawing the circle with a radius of 60° from the scale, as in all cases).
Fig. 25.
40
NAUTICAL ASTKONOMY.
47. PROS. II. — To lay off any number of degrees on a right circle (say 60°).
1st (say upon OZ from the centre). Take 60° from the scale of semi-tangents and lay off from the centre to a.
2nd (say upon X 0 from X towards 0). On the scale of semi-tangents count 60° backwards from 90°, and lay it off from X towards 0, at b.
48. PROB. III. — To draw an oblique circle through any
point lying within the primitive circle.
Fig. 27.
PROBLEMS.
41
Connect the point (as at A) with the primitive HZ EX, and from the point (say H) at the primitive draw a diame- ter H 0 K ; draw also another diameter at right angles to the first, as ZOX. Bisect the line AH, and produce (or lengthen) the bisecting line till it cuts 0 X at x ; then x will be the centre of an oblique great circle which will cut the point A.
49. PKOB. IV. — To draw an oblique circle through any two points, say through point I and point A.
Through either point (say A) draw a diameter as H AO E» Draw ZX at right angles to it at the centre. Join AZ. Make AZy a right angle at Z. The intersection of Zy on
Fig. 28.
HE produced will give a third point, y. Join ly and A I, and bisect Zy and A I, the lines of intersection will meet at x, which will be the centre of the great circle A I B.
50. PKOB. V. — To draw a small circle parallel to the pri- mitive at any distance from it (say 40°).
With its complement 50° from the scale of semi-tangents
42
NAUTICAL ASTRONOMY.
and centre 0, draw a circle as abc, and it will be 40° from the primitive.
Fig. 29.
51. PROB. VI. — To draw a parallel circle at any given distance from a right circle (say at 50° from HR\ or at any distance about a given point at the primitive (say at 40° from Z.)
1st. Lay off 50° from the scale of chords from H to a,
PROBLEMS.
43
also from E to c, and 50° from the scale of semi-tangents from the centre to b. Then through the three points abc describe a circle as in Prob. IV. Or, take the secant of (90 -50°) = 40°. Lay off this distance from 0 to d, and then, with the tangent of 40° and d as a centre, draw abc, the small circle required. Whether the point be 6, or e, or /, &c., use the same means.
52. PKOB. VII. — To find the pole of an oblique circle (say ofZcX).
Draw and produce Xc till it cuts the primitive. From a lay off the distance of 90° (or HZ) from a past Z to 6.
Fig. 31.
Join 6X ; and where this line cuts the radius OE (or at p) will be the pole of ZcX. Or thus: Measure CO on the scale of semitangents, and lay off its complement from 0 towards E, as at p, .the pole.
S3. PROB. VIII. — To draw a great circle through any given point so as to make any desired angle at the primitive (say 40° and through the point d).
With centre 0 and tangent 40° describe an arc as ab;
44
NAUTICAL ASTRONOMY.
with centre d and secant 40° describe another arc which cuts the first at x ; then x is the centre of the oblique circle Zc£X, and it is drawn accordingly.
Note. — Diameters are always at right angles to the primitive.
54. PKOB. IX. — To draw a small circle through a given point (x), which shall be at a given distance from a \ right circle (say at 40°).
Take the secant of 50° (the complement of the given Fig. 33.
PROBLEMS.
45
distance) and from centre 0 sweep an arc as p q ; cross this with the tangent of 50° laid off from the point x. Then will n be the centre of the small circle adb. Join On, and draw the diameters ce and fg. Then g b and fa will each measure 40° on the scale of chords, and 0 d will measure 40° on the scale of semi-tangents.
55. PROB. X. — To draw an oblique circle perpendicular to a given oblique circle.
Find the pole p of the given circle ZaX (Prob. VIIL). Draw any diameter at pleasure, say 6 0 c ; through cp b draw
Fig. 34.
a great circle (Prob. IV.), and it shall be perpendicular to Z a X. If the perpendicular be required to pass through any point, as a, through the two points a and p, describe an oblique circle by Prob. IV.
56- PROB. XI. — To draw a great circle which shall make an angle of, say 30°, with the primitive.
Draw a diameter from Z at right angles to H E from the point Z as a centre, and with the chord of 60° from the plane scale by which the circle was drawn describe the arc 0 E Of. From the centre 0 make OE (on OEGr) equal to 30° on
46
NAUTICAL ASTRONOMY.
the scale of chords. Join Z E and produce to x, and x will be the centre of the oblique circle ZYX, and it measures 30° on the scale of semi-tangents, counting from 90° on the
Fig. 35.
scale, or from H towards 0. Thus any oblique circle may be drawn by taking the number of degrees from the scale of secants ; for example, the radius of the oblique circle 30° (according to the problem) is the secant of 30°.
57. PKOB. XII. — To measure any part of an oblique circle (as a b in Ha&K).
PROBLEMS.
47
Find the pole of the oblique circle as at p (by Prob. VII.). Join pa and produce to c, and join pb and produce to d, and the measure cd on the scale of chords will be the measure of a b.
Note. — By this problem any number of degrees may be laid on an oblique circle.
58* PKOB. XIII. — To measure an angle at the primitive, as HZ a.
N.B. — The angle at the primitive is always measured on a right circle which lies 90° distant, or which passes through the pole p of the oblique circle. Apply the distance H a
Fig. 37.
in the dividers to the scale of semi-tangents, counting back- wards from 90°. Then Ha is the measure of HZ a and = 50°.
N.B. — This problem is very useful to the navigator.
k59. PROB. XIV. — To measure the angle ZaR. Having found the pole of ZaX (by Prob. VII.) to be at p, and the pole of HaR to be akp, join ap and produce
48
NAUTICAL ASTRONOMY.
it to b, and join apf and produce it to c, and the distance be measured on the scale of chords (or 68°) will be the measure of the angle ZaE or HaX.
60. It must be remembered that the pole of a right circle is always at the primitive : thus the pole of Z X is at H or E (being at a distance of 90°), and the pole of HE is at Z or X.
N.B. — The right circle appears as a diameter in the stereographic projection.
CONSTRUCTION OP SPHERIC TRIANGLES.
61. In proceeding to the construction of spheric triangles the navigator must bear in mind that it is convenient to make arcs of latitude, declination, altitude, &c., to occupy the same relative positions in the sphere — that is to say, latitude is always on the primitive, &c., or in other words, he had better use a right sphere, as above, instead of an oblique sphere.
Declination is always a small circle parallel to the equator.
CONSTRUCTION OF SPHEEIC TRIANGLES.
49
Altitude is always a small circle parallel to the horizon.
Indeed, the lines, as they are drawn on Fig. 23, the illustrative diagram, ought to be perfectly understood and remembered.
In all the figures following the same letters will desig- nate the same parts. Thus —
H K will always represent the horizon. PS the polar axis of the earth as prolonged or pro- duced to the heavens (of which the primitive is the imaginary limit, the earth itself being now supposed to be the very small point at the centre of each figure).
ZN the zenith and nadir. EQ the equator. dc a parallel of declination. ab a parallel of altitude.
x the position of the heavenly body.
While the answers to the problems will be indicated by a thick line at the part of the figure where the answer is to be measured.
Fig. 39.
The line HE will consequently exactly coincide in posi-
50 NAUTICAL ASTRONOMY.
tion with, and represent the wooden horizon of, the artifi- cial globe (Fig. 39), while the other lines of the sphere correspond also with those on the globe, but seen as slightly distorted by the nature of the projection.
62. Those who possess a globe will do well to compare the following problems with the lines on it.
It is necessary to remember that a mere spheric triangle may be formed from any three parts given, and either a side or angle may be placed according to convenience in construction. But the limiting of certain data to certain parts of the projection is a mere conventional rule, in order to simplify the study to the minds of those who have neither time nor inclination to perfectly master the whole doc- trine of spherics, but who desire a mere knowledge of its principles and practice as applicable to the wants of the navigator.
MEMOEANDA.
63* All azimuth circles meet at the zenith and cut the horizon at right angles, and are measured along it.
All hour circles meet at the poles of the world, which are points on the primitive 90° from the equator.
A parallel of declination is a small circle which the sun or a heavenly body seems to describe round the pole.
The spheric figure in general use, although a hemisphere, really represents the whole sphere, inasmuch as the hour circles merely imply time from noon, A.M. or P.M. ; and, consequently, the hour circle for 10 A.M. answers for 2 P.M., each being two hours from noon. In like manner with azimuth circles, the point next to south may be either S. by E. or S. by W. according as we consider the centre as the east or west point.
Amplitude is distance of an azimuth circle from W. or E., as measured on the horizon.
TO FIND THE LATITUDE OF A PLACE.
51
Azimuth is distance of an azimuth circle from N. or S., as measured on the horizon.
Latitude is distance from the equator.
Longitude is distance from the meridian which passes through Greenwich Observatory.
64. As the numbers used in the following constructions are merely intended to serve the purpose of illustration, answers are given to the nearest degree only.
The usually-occurring questions in nautical astronomy will first be answered by projection, and afterwards (107) the same figure will be repeated when working by calcula- tion, such additions and arrangements being made to them as the process of calculation requires, in order to adapt them to it.
TO FIND THE LATITUDE OF A PLACE.
65. I. — Given, meridian altitude sun's centre, 60°.
„ declination . . . 20° N. (Observer north of the sun.)
Draw the circle with the chord of 60°.
Draw two diameters, H K and Z N, at right angles to
52
NAUTICAL ASTKONOMY.
each other. (In the following questions the circle is sup- posed to be drawn and quartered.)
Lay off the given altitude 60° taken from the line of chords, from H on the horizon towards Z the zenith, say to d, and d will represent the sun's place on the meridian. The sun being in 20° N. declination the equator will be 20° southward of d, or at E. Join E 0, and produce it to Q, and make the polar axis, P S, at right angles to it ; then P E will be the height of the pole P, which is equal to the latitude 50°, as measured on the line of chords (45).
66. II. — Given, sun's altitude . 50°.
declination 20° N. azimuth, S. 45° E.
Draw a I, the parallel of altitude 50° (parallel to H E), by Prob. IX. Draw the azimuth circle Z x N, 45°, from H,
Fig. 41.
the south point of the horizon, by Prob. XL, making it 45° from the primitive : where these intersect will be the sun's place x.
Then with the secant of the complement of the decli- nation, .or 70°, intersect the tangent of 70° laid off in the
TO FIND THE LATITUDE OF A PLACE.
53
same direction northward of the equator (because decli- nation is north), where these intersect will be the centre from which with the tangent of 70° describe the small circle d c (by Prob. IX.) ; through this centre and the centre of the primitive draw the polar axis, P S, and the measure, R P, on the scale of chords will be the lati- tude.
67. III.— Given, declination, 10° S. altitude . 50°. time . 10A.M.
N.B. — In south latitude (being south of the sun) draw the primitive with the chord of 60°.
In the primitive assume a point S (above what is in- tended to be the south point of the horizon), and quarter the circle.
Fig. 42.
Lat 40° S.
Draw S#P, two hours or 30° from the primitive (Prob. XL).
Draw d c = 10° on the south side of the equator by Prob. IX. 10° from E Q.
Then with the secant from centre 0, and tangent from
54
NAUTICAL ASTRONOMY.
centre x of the co-altitude = 40°, find the centre of a I (Prob. IX.), and draw it with tangent 40° as a radius; through the centre of this parallel and the centre of circle draw a diameter, Z 0 N, and another, HOE, at right angles to it, and the distance, H S, will be the lati- tude, 40° south.
68. IV. — Given, altitude of a celestial body on the meridian, below the pole, say, —
Altitude of a Lyrae . Declination of a Lyrae
. 20° below the pole.
. 381° N.
(Observer south of the star.)
Let x be the star's place, the decimation being 38-*-°, the equator E Q, will be 38£° south of it, as measured on
a Lyrae
(below pole)
the scale of chords, and the altitude, 20°, will give the horizon at E 20° below x. Draw HE and E Q, and diameters at right angles to each, and the height of the pole P above E will be the latitude, 71-J-0. The parallels of declination and altitude may be drawn by Prob. IX.
TO FIND THE LATITUDE OF A PLACE.
55
69. V.— Griven, time . 9A.M. „ declination 20° K „ azimuth S. 60° E.
(Observer north of the sun.)
Draw the circle with the chord of 60°.
Assume the polar point P, and draw the polar diameter, P 0 S, and also E Q, the equator, at right angles to it.
Draw the hour circle, 9 A.M. = 45° from the primitive (by Prob. XI.). Draw d c9 the declination (by Prob. IX.), then the intersection, x, is the sun's place.
Fig. 44.
Time, 9 A.M. Dec. 200 N. Az. S. 60° E.
Lat 58° N.
The given azimuth circle is 60° from south, and the azimuth circle passing through x must be drawn by Prob. VIII., as Z x N ; lay off 90° on the scale of chords, from Z to E, and E P will measure the height of the pole or be the latitude = 58° N.
£ 4
56
NAUTICAL ASTRONOMY.
TO FIND THE TIME.
70. VI.— Given, latitude . 45° 40' N. „ declination . 10° N. „ azimuth . S. 45° E.
Draw H E and Z N at right angles to each other.
From the scale of chords lay off 45° 40' from E to P. Draw P S and E Q.
Draw the azimuth circle, Z x N, by Prob. XL, and the parallel of declination, 10° N., by Prob. IX., then the point of intersection, x9 will be the sun's place.
Fig. 45.
Lat 45° 40'. Dec 10° N. Az. S 45° E.
Time, 10 A.M. H
Through P&S draw an oblique circle by Prob. IV., and the angle Z P a? will be the hour angle, equal to two hours from noon (or the meridian), or 30°, or 10A.M., measured on E y, from semi-tangents, backwards from 90°.
71. VII. — Given, declination „ altitude azimuth
. 20° S. . 20°. S. 45° W.
Draw H E and Z N.
Draw the azimuth circle by Prob. XL, as Z x
TO FIM) THE TIME.
57
Draw the parallel of altitude, 20°, by Prob. IX., then x will be the sun's place.
Draw the parallel of declination, d c, through the point x, by Prob. IX.
By Prob. IV. draw P x S through the three points, and the angle Z P x will be the hour circle, and is measured on E y, a diameter at right angles to P S, and is equal to 45°, or three hours from noon, westerly, by the equator, or 3 P.M.
Fig. 46.
Time, 3 P.M.
7Z. VIII.— Given, latitude . 21° N. „ declination 20° S. „ altitude . 30°.
Draw H K and Z N.
Make E P equal to 21° from the scale of chords.
Draw P S and E Q.
Draw the parallel of declination, d C9 by Prob. IX.
Draw the parallel of altitude, a I, by Prob. IX., and the intersection x will be the sun's place ; through the points Pa; and S draw the oblique circle (by Prob. IV-)? and the angle Z P x will be the hour anlej-dineasured on
58
NAUTICAL ASTEONOMY.
E y = 45°, or three hours from noon, being 9 A.M. or 3 P.M. V
Fig. 47. z
Lat 21° N. Dec 20° S. Alt 30°.
Time, 9 A.M. u
TO FIND AN AZIMUTH.
73. IX.— Given, latitude . 36° S.
„ declination 20° N. „ altitude . 20°.
Draw H R and Z N.
Lay off the latitude from the scale of chords = 36° from the south point of the horizon at H to S.
Fig. 48.
Lat 36° S. Dec 20° N. Alt 20°.
Az N. 45° E.
TO FIND AN AZIMUTH.
59
Draw S P and E Q.
Draw declination north by Prob. IX. (as d c\ and
Draw a I, the parallel of altitude, 20° by Prob. IX., then x is the sun's place.
Through the three points, Z x and N (by Prob. IV), draw an oblique circle, Z y N, and the angle y Z E will be the azimuthal angle = 45°, as measured at K to y, on the semi-tangents backwards from 90° on the scale.
74. X.- Given, latitude . 21° N.
„ time . . 9 A.M.
„ declination 20° S. Draw H K and Z N. Make K P equal to the latitude 21°. Draw P 0 S and E Q.
S. W E.
Draw the hour circle, three hours or 45° from noon (or the primitive), by Prob. XI., as E y.
Draw the parallel of declination, d c, south (by Prob. IX.), the point of intersection, x will be the sun's place ; through the points Z x N draw an oblique azimuth circle by Prob. IV., and the angle x Z R will be the azimuth from north, or N. 130° E., and the angle H Z y will be the azimuth from south, or S. 50° E.
60 NAUTICAL ASTRONOMY.
CALCULATION OP SPHERIC TRIANGLES.
75. In proceeding to the calculation of spheric triangles, we notice that such are either right angled (i.e. having one angle equal to 90°), quadrantal (i.e. having one side equal to 90°), or oblique (i.e. having neither an angle nor a side equal to 90°).
1. RIGHT-ANGLED SPHERIC TRIANGLES.
76. Every triangle, as in plane triangles, has " six parts," viz., three sides and three angles ; and any three of these being given, the rest may be found by proportion. But in a right-angled spheric triangle two parts only need be given besides the right angle.
77. In calculating parts of a triangle, whether plane or spherical, we shall often save much trouble if we consider, first, whether of the three things or parts given any tivo of them are a side and an opposite angle. When such is the case the " rule of sines," as it is called, founded on the fundamental theorem that " the sides of a triangle are in proportion to the sines of their opposite angles " is peculi- arly simple.
To find a Side.
78. EULE. — As the sine of any given angle is to the sine of its opposite side, so is the sine of any other given angle to the sine of its opposite side.
To find an Angle.
79. EULE. — As any given side is to the sine of its op- posite angle, so is any other given side to the sine of its opposite angle.
It is not, as already declared, the purpose of this book to do more than give a plain but substantially practical
CALCULATION OF SPHEEIC TEI ANGLES. 61
introduction to the study of spherics, leaving the mentative proofs of various theorems to the few works on the subject which are already before the public, or to a succeeding volume.
80. But that which above has been called a "funda- mental " rule deserves, in passing, a little attention, be it only to encourage the student towards further, research ; as in this he will see the simplicity of the study of geometry when it is approached by a proper path.
In the following plane triangle ABC bisect each side and erect perpendiculars ; they will meet in 0, the centre
Fig. 50.
(N.B.— Lengths and angles are marked in order that the student may easily veriiy by logarithms).
of the circumscribing circle. Join 0 A, 0 B, and 0 C. Now, by reference to page 19, we shall find that Euclid, in Book III., Prob. XX,, proves that " an angle at the centre of a circle is double the angle at the circumference upon the same base ;" therefore, in the above figure, the angle A 0 B is double the angle A C B ; but by construction A D is the half of A B ; similarly, the angle A 0 D is the half of the angle A 0 B ; therefore, the angle A 0 D equals the angle A C B. Now, A B, the base of angles A 0 B and A C B, is a chord of the arc ACB, and AD being half of AB (being by
62 NAUTICAL ASTRONOMY.
definition called a "sine"), subtends the angle AOD or ACB.
Hence we find A D equals the sine of the angle A 0 D — equals the sine of the angle ACB.
By taking another base as A C, and another as C B, we shall, by the same method of demonstration, find that E C is equal to the sine of angle CBA, and also that FB is equal to sine of the angle BAG, and putting a for the side B C, and b for the side A C, and c for the side A B, we shall have the following equations : —
•J- a = sine A (i.e. sine of £ A)
\ 6 = sine B
\ c = sine C
And by combination : —
£ a : ± b : : sine A : sine B, Or, a : b :: sine A : sine B, &c.
Or, a : sine A : : b : sine B, &c.
Thus the sides are in proportion to the sines of these opposite angles.
A number of useful formulae, which seem to wear so forbidding an aspect in ordinary works upon Trigonometry, are really nothing more than easily obtained deductions from the above.
THE FIVE CIRCULAR PARTS.
81- When, however, the calculation of right-angled spheric triangles cannot fall under the rule above given (from having no angle and opposite side given), a method invented by Lord Napier and published in 1614, and which is called the " Circular Parts " (or because the right angle is never considered one of them, is called also the " Five Circular Parts ") is singularly applicable to all cases which can occur.
THE FIVE CIRCULAR PARTS. 63
82. Anyone of these five circular parts maybe considered the middle part, the parts joining thereto being called extremes conjunct ; but the parts which are separated by an angle or a side are called extremes disjunct.
N.B. The right angle does not separate its two contain- ing sides.
83. In every case, then, as three things must enter into every consideration of proportion, viz. the two given parts (excluding the right angle) and the part required, one must be called a middle, while the others are considered as con- junct or disjunct, as the case may be, but in using the part in computation it must be remembered that —
" When angles or hypothenuse Among the parts you trace, Their complements Or supplements Must always take their place."
(A quaint rhyme or two may aid the memory.)
84. The manner in which the equations are formed from which we derive the proportions may be thus explained : —
Fig. 51.
Draw any spheric triangle ABC.
Then, because the right circle AB passes through 0 the
64 NAUTICAL ASTRONOMY.
pole of the primitive of which one side B C of the triangle is an arc, the angle B is a right angle. The " Five Parts" are, therefore (83) :—
1. The complement of the hypothemise AC.
2. The side A B
3. The side CB
4. The complement of angle A
5. The complement of angle C.
Now, suppose, in the above triangle, the sides AC and AB are given to find the L A.
The middle part must be so selected as to make the other parts either disjunct or conjunct (not one conjunct and the other disjunct). The middle part in this triangle will evidently be the Z. A, as the two given sides include it, and because they join it they will be extremes conjunct.
N.B. The hypothenuse is always the side opposite the right angle. Calculation then depends on the following universal, or as it has been long called the " catholic pro- position," viz. : —
35* The sine of the middle part multiplied into radius is reciprocally proportional with the tangents of extremes conjunct, and with the cosines of extremes disjunct.
Expressed as an equation it would stand thus : —
sine of middle x radius = tan extr conjunct x other extr conjunct.
Eemembering that of four numbers in proportion, the product of the means always equals the product of the ex- tremes (see page 21), we may vary the above as follows, viz.: —
radius : tan extr conjunct : : tan other extr conjunct : sine of middle, or, tan extr conjunct : radius : : side of middle : tan other extr conj.
If the extremes are disjunct we have as follows : —
sine of middle x radius = cos extr disjunct x cos other extr disjunct.
THE FIVE CIRCULAR PARTS. 65
or, radius : cos extr disjunct : : cos other extr disjunct : sine of middle ; or, cos extr disjunct : radius : : sine of middle : cos other extr disjunct.
It follows, then, that as we can only want, in any case of right-angled spheric trigonometry, to find either a middle part or an extreme, the following four simple formulae are all-sufficient : —
If extremes are conjunct
RULE A. Sine of middle = tang extr conjunct x tang other extr conjunct
radius.
RULE B. Tang extr conjunct = radius x sine middle part
tang other extr. conjunct.
If extremes are disjunct.
RULE C. Sine of middle = cos extr disjunct x cos other extr disjunct
RULE D. Cos. extr disjunct = radius x middle part
cos other extr disjunct.
Or, as adapted at once to logarithmic calculation, we can put the above still more plainly, thus : —
When extremes are conjunct.
86. RULE a. To find log sine middle /Fromlogtangextr conjunct + logtang
\other extr conjunct subtr rad (or 10)
87. RuLE&.Tofindlogtangextrconjunct / From log rad + log sine of middle
\ subtr log tang other extr conjunct
When extremes are disjunct.
To findlogco^ disjunct
care being taken to use complements of angles and hypo- thenuse. We will take an example, and at once apply the above formulae. *
66 NAUTICAL ASTRONOMY.
9 0. In the right-angled triangle suppose the following : ' sile BC I IT] re1uired> Vpothemise AB and L A. Fig. 52.
In constructing a spheric figure, it is, moreover, always convenient to place a given side on the primitive : we do so by making A C = 30° (by scale of chords) join C 0.
By Problem V. draw a small circle parallel to the primi- tive — 40° distance ; and where this cuts 0 C will be the third angular point B. Through the points A B D draw a great circle (by Prob. IV.), and ABC will be the triangle, of which the L C will be a right angle (because the right circle C 0 passes through 0, the pole of the primitive, and therefore is perpendicular to the arc A C), and its opposite side will be A B the hypothenuse.
N.B. It is convenient to mark the given parts as in the figure, and the parts required with an o.
To find the hypothenuse A B.
It will be seen that in order to get two " parts " alike (that is, two parts which shall be either conjunct or disjunct}, the middle part should in the above figure be the hypo- thenuse itself, for it is separated from the parts given by
THE FIVE CIRCULAR PARTS. 67
L A at one end, and by L B at the other. The parts A C and B C are therefore called extremes disjunct.
91. The following rhymes may help the memory in applying tangents or cosines, &c. : —
Tangents join the middle
(Put the middle where you please) ;
Cosines afar>
From middle are (Fire parts you have in these).
We have, therefore, in this example, to use cosines with the extremes (subject to the correction for hypothenuse and angles), and we want to find the middle part, A B. Rule C gives the following equation : — -
Sine of middle part -cos ext ^ x cos of other ext disJ-
radius.
Now, before proceeding, let us consider what is meant by this equation, and why it was further altered into rule c (88). We have already shown that multiplication is per- formed in logarithms by addition ; and division by subtrac- tion ; and in the fraction standing on the right side of the sign of equation, there are two quantities to be multiplied, and a quantity which is to divide their product. We there- fore add the logarithms of the two factors in the nume- rator from the sum and subtract the logarithm of the denominator.
92. In a proportion worked by logarithms it is better to place the terms vertically (putting the divisor as the first term), thus : —
As radius
is to cos extreme disjunct,
so is cos of the other extreme disjunct
to the sine of the middle part.
(Remember old Dr. Kelly's Hibernian rhyme: — )
" Now the product of radius and middle part sine, Equals that of the tangents of parts that combine, And also the cosines of those that disjoin") F 2
68 NAUTICAL ASTRONOMY.
93. But we have first (83) to correct the above propor- tion if the hypothenuse or an angle form part of it. It should therefore appear accurately thus : —
as radius . . . co ar 10-000000 is to cos side A C 30° , ". 9-937531 so is cos side B C 40° 9-884254
to cos hyp. AB . . . =9-821785 « 48° 26' 21
To find/. A: — Use the rule of sines (79), (having now opposite sides and angles).
as sine A|B 48° 26' 21" . . co ar 0-125952
to sine of opp. L (radius) . •„ 10-
so is sine of B C 40° . \' _. . 9-808067
to sine of L A . , - , . 9-934019 = 59° 12' 37"
But as it is better to thoroughly master one question in all its bearings, we will take a different view of the same question, and determine on finding the angle A, before we find the hypothenuse.
We now evidently call A C the middle, and then the L A and the side B C will be extremes conjunct (the right angle is not one of, and does not separate the parts re- member), and Eule 87 gives us as an equation : —
To find log tang «* conjunct {*«*$>££%£«
tang of the other ext. conj. B C 40° . co. ar. 0-076186
rad '•>.• 10-
sine of middle A C 30° % . . & 9-698970
co tang L A . . ., . . . 9*775156 = 59° 12' 37"
The hypothenuse can now be found by the rule of sines (78), thus :—
as sine L A 59° 12' 37" . . co ar 0-065981 to sine side B C 40° . '. -''M . 9-808067 sine 90° 10-
sine hyp. •}> . . . ' . >, . 9-874048 = 48° 26' 21"
QUADRANTAL SPHERIC TRIANGLES. 69
94. The co. ar. (read arithmetical complement) of an arc is what the logarithm wants of radius, and is readily formed by subtracting each figure of the logarithm (be- ginning at the left hand) from 9 and the last from 10 : thus the logarithmic co. ar. of -333333 is -666667. This saves subtraction as the three logs may then be added.
Every right-angled spheric triangle may in like manner be worked b}' the four equations A, B, C, D, or a, b, c, d (page 65). But in such triangles as have a side for a right angle, a modification of the above rules is necessary, for we have in such cases what are called
QTJADRANTAL SPHERIC TRIANGLES.
95. The only difference in the mode of working arises from an apparently whimsical perversion of terms.
For now the merry quadrant
Its pranks with us to play, Transforms itself to radius,
And laughs our rules away. It calls legs, angles ! — angles, legs !
Our notions to confuse ; "While its opposite angle's supplement,
It calls hypothenuse !
So that, considering the quadrantal side as radius, and the supplement of its opposite angle as hypothenuse, the solution of quadrantal angles is performed by rules already explained ; viz., the rule of sines, and the four rules for the five circular parts.
OBLIQUE SPHERIC TRIGONOMETRY.
96. The majority of spheric questions which occur in practice fall under the denomination " oblique," i. e. having neither an angle nor a side equal to a right angle.
F 3
70 NAUTICAL ASTKONOMY.
97. Oblique spheric trigonometry admits of the six following cases, viz. : —
The given parts will be either
1. Two sides and an opposite angle.
2. Two angles and an opposite side.
3. Two sides and an included angle.
4. Two angles and an included side.
5. Three sides.
6. Three angles.
98. The solution of the first four cases may either be effected by means of a perpendicular let fall from one of the angles to its opposite side, or by special rules not requiring the perpendicular. There is a little difficulty with beginners in constructing the triangle so as to admit of a perpendicular being drawn in such 'manner as to retain two of the given parts in one of two new right- angled triangles thus formed. This may often be avoided by attending to the following directions, viz., describe the usual circle and quarter it. Then lay off a given side, A C, on the primitive • (A being the angular point of the left hand of the side of the triangle, which lies on the primitive, and C the other end of it. It is merely convenient to have one method), and at C lay off the given angle (Prob. XI). Consider how you can secure two of the given parts in a new right angle you are about to form, and (by Prob. X.) let fall the perpendicular accordingly.
99. When the two angles which lie upon the side which is to be crossed by a perpendicular are of like affection, i. e., both greater or both less than a right angle, the perpendicular will fall within the triangle, but when un- like, i. e. when one is acute and one obtuse, the perpen- dicular will fall without the triangle.
100. N.B. In explaining the nature of " construction " from observation, it was recommended that the various
OBLIQUE TRIGONOMETRY,
71
circles of the sphere should always be made to represent their respective parts in the general astronomic diagram, but for calculation it is better to take the parts merely as sides or angles.
The following will show the method of "projecting" and calculating any oblique spheric triangle which, can possibly occur.
101. CASE I.
Two sides and an opposite angle.
Given, a side 60°
a side 100° j-To find a side and the other angles,
an opposite angle 130°
By construction :—
N.B. In nearly all cases we suppose a circle to be already drawn with a chord of 60°, and two diameters at right angles within it.
Fig. 53.
Lay off from C to A the one side, 60° (from scale of chords). From A draw A Gr, with the supplement of 130°, and draw (by Prob. VIII.) the great circle A B Gr. About E
72 NAUTICAL ASTRONOMY.
describe the small circle v y, at a distance of the supple- ment of 100° (or 80°) from it (by Prob. VI.) ; and where v y cuts A B G- is the angular point B.
With the three points E B C draw an oblique circle (by Prob. IV.), and A B C is the triangle.
To measure the parts required, viz., AB, L C, and ZLB.
A B is measured by Prob. XII. L C is measured by Prob. XIII. L B is measured by Prob. XIV.
By calculation:—
1st. By means of a Perpendicular.
Having a side on the primitive with an adjacent L A given, let fall a perpendicular from the L C upon AB (produced if necessary) to D (Prob. X.).
To find the other opposite /.ABC by rule of sines (79).
as sine BC 100° . * . " co ar 0-006649 to sine L A 130° '. . . 9-884254
so is sine AC 60° . A .. ._..;.. 9-937531 to sine / ABC ,\ .; . 9-828434 = 42° 21'
180 L CBD = 137 39
To find AD (in triangle ADC) : —
The L A will be the middle part, and the hypothenuse AC and the side AD will be extremes conjunct (87).
as cotang hyp AC 60° co ar 0-238561
to rad 10-
cos middled A 130° . -. 9-808067
to tang . . . r . . 10-046628 = 48° 4'
180 tang AD = 131 56
OBLIQUE TRIGONOMETRY. 73
To find DB (in triangle CBD) :—
The L B will be middle, and hyp BC and the side DB will be extremes conjunct (87).
as tang BC 100° co ar 0753681
to radius . . .10'
cos £B 137° 39' . 9-868670
tangDB . . . 10-622351 = 76° 35'
To find L C by rule of sines (79).
|
AD = 131° 56' DB= 76 37 |
as sine AC 60° . to sine zB 42° 21' so is sine B A 55 19 to sine L C . |
. 0-062469 . 9-828434 . 9-915035 |
|
BA= 55 19 |
||
|
. 9-805938 |
2nd. Without a Perpendicular.
When the solving of a spheric triangle presents any diffi- culties to the unpractised as to where to let fall the per- pendicular, if time is precious, recourse may be had to the following rule : —
When two sides and an opposite angle are given.
First find the angle opposite to the other of the two given sides, and the third angle of the triangle may then be found thus : —
RULE. — As the sine of half the difference of the two sides is to the sine of half their sum
so is the tangent of half the difference of the two angles to the cotangent of half the contained angle.
Or, as applied to the preceding question, viz. : —
Given, aside AC = 60° asideBC = 100° L A =130° / B =42° 21'
74
NAUTICAL ASTRONOMY.
To find the angle C : —
side AC 60° as sine half diff 20° . . coar 0-465948
side EC 100 to sine half sum 80° . . . 9-993351
"J^Q" tang half diff angles 43° 49^' . 9-982182
— Q cotanghalf contd£Cl9° 53' = 10-441481
20~
half sum half diff Z.A1300 ZB 42
whole L C is 39° 46'
21
87 39 half diff 43 49£
Then the side AB may be found by the rule of
102. CASE II.
Two angles and an opposite side,
Given, an angle 60° ") To find a side angle 70° I a side
opp. side 50°J an angle
By construction : —
At the point C make an angle of 60° with the primitive (Prob. VIII.) by drawing C BE ; about C, at the distance of
Pig. 54.
50° (the given side), draw the small circle Ay (by Prob. IX.) ; at B, where these intersect, draw (by Prob. XI.) the great circle AB, making an angle of 70° with the primitive, and ABC will be the spheric triangle.
OBLIQUE TEIGONOMETRY. 75
To measure the parts required, viz. AC, AB, andZ_B: —
1. measure AC on the scale of chords = 49° 37'
2. AB is measured by Prob. XII. . = 44 55
3. L B is measured by Prob. XIV. . =110 52
By calculation : —
1st. By means of a Perpendicular.
In order to preserve the two parts B C and L C in the same rectangle, let fall a perpendicular on side AC. This is done by drawing a line from the centre of the circle, which is always the pole of the primitive, through the Z_ B till it cuts AC at D. (Prob. X.)
To find the other opposite side AB by rule of sines : —
; as sine L A 70° . co ar 0-027014
to sine side BC 50° . . 9-884254
so is sine L C 60 ! .*. » 9'937531
to sine side AB . . 9-848799 = 44° 54' 35"
To find the segment CD in ABCD :—
The angle C will be "middle," and the hypothenuse BC and CD will be extremes conjunct (87).
as cotang hyp BC 50° . co ar 0-076186
to rad .... 10-
so is cosine L C 60° . . 9-698970
to tang side CD . . . 9-775156 = 30° 47' 23"
To find segment AD in triangle ABD : —
The angle will be middle, and the side A D and hyp. AB will be extremes conjunct (87).
as cotang AB 44° 54' 35" co ar 9-998631
to radius ... 10-
so is cos /A 70° . .•£' 9-534052
to tang segment AD . . 9-532683 = 18° 49' 36"
segCD = 30° 47' 23" segAD = 18 49 36
49 36 59 = sideAC
76 NAUTICAL ASTKONOMY.
To findZ. B in A ABC by rule of sines (79).
as sine side BC = 50°. coar 0-116746 to sine/. A = 70 . . 9*972986
so is sine side AC = 4937 , 9-881799 to sine ZB 9-970531= 69° 7' 51"
180
/. /B = 110 52 9
N.B. We take the supplement of 69° 1' 51" because the construction shows the L B to be obtuse.
2nd. Without a Perpendicular.
The side AB opposite the other given angle can be found by rule of sines as before, and equals 44° 54' 35". Then find L B by the special rule given in Case I. thus : —
a side AB 44° 54' 35" L A 70°
asideBC 50 LC 60
2)94 54 35 2)10
half sum of sides 47 27 18 half diff two angles 5 diff two sides ~5 5 25 half diff two sides 2 32 42
as sine | diff 2 sides 2° 32' 42" eo ar 1-352577 to sine | their sum 47 27 18 \ 9*867318 so is tang £ diff 2 angles 5 00 .. 8-941952
tocotang^ contained L~B 10-161847 = 34 33 46
2_
69 7 32 180 ZB=110 52 28
Find AC by rule of sines (78).
as sine ZA70° . . . . co ar 0-027014 .
to sine side BC 50° . ,V . 9'884254
so is sine^B 110° 52' 28" . . 9-970516
to sine side AC . . . = 9-881784 = 49° 36' 51"
103. CASE IIL Two sides and an included angle,
Given, a side A C = 60° r To find side B C.
„ asideAB = 110° 1 „ L B
„ the included angle A 45° L „ / C
OBLIQUE TRIGONOMETRY.
77
By construction : —
Lay off A C = 60° from the scale of chords.
Draw the great circle AD B (by Prob. XI.), at the distance of 45° from the primitive; about Gr draw the parallel circle vy, distance of the supplement of 110 (Prob. V.) ; where the two circles cross is the point B, through the points C B H draw a great circle (Prob. IV: ), and ABC will be the triangle.
To measure the parts required : —
1. The L C, ax is measured on the scale of semi-tangents (Prob. XIII.).
2. measure / B by Prob. XIV.
3. measure side CB by Prob. XII.
Fig. 55.
By calculation : —
1. By means of a Perpendicular.
Draw a perpendicular from the L C upon the side A B to D (by Prob. X.).
Find the segment AD in the A A CD (78).
78 NAUTICAL ASTRONOMY.
Then the z. A is the middle, and hyp A C and side A D are extremes conjunct (87).
as cotang hyp A C 60° , ... . . co ar 10-238561
to radius f ..... 10*
so is cos L A 45° . .' ... . 9-849485
to tang of segment AD . . 50° 46' 17" = 10-088046 subtr from side AB ' .' . 110
segment. . '' . . . 59 13 43 = segment AD
Find D C in A A D C by rule of sines (78) :—
as rad. . . ^ . co ar 10*000000 sine side (hyp.) A C 60° . f 9-937531 sine L A 45° . 9*849485
to sine side DC . . r 9*787016 *? 37° 46'
Find L B in A B D C.
Here B D is middle, and D C and L B are extremes conjunct.
as tang D C 37° 46' . co ar 0-110839
to rad. . . . .f \ • 10'
so is sine side B D 59° 14' 9-934123
to cotang L B 42° 2' • ." . . 10-044962
Find side B C by rule of sines (78) : — as sine L B 42° 2' . co ar 0-174209
to sine A C 60° ..'••-, . 9-937531 so is sine L A 45° . ;..' . 9-849485
to sine side B C , . . 9-961225 = 66° 9'
Find L C by rule of sines (79) :—
as sine side A C 60° . ' '. . 0-062469 to sine Z B 42° 2' . . . 9-825791 so is sine side AB 110° . . 9-972986*
to sine L C 46° 36' . . = 9-861246 180
L C = 133° 24' N.B. We take the supplement because the Z C is obtuse (by construction).
• OBLIQUE TKIGONOMETRY. 79
2. Without a Perpendicular.
KULE. — When two sides and an included angle are given :—
1. As the sine of half the sum of the two sides is to the sine of half their difference
so is the cotangent of half their contained angle
to the tangent of half the difference of the other angles ;
and again,
2. As the cosine of half the sum of the two sides is to the cosine of half their difference,
so is the cotangent of half the contained angle
to the tangent of half the sum of the other two angles.
And half the difference thus found added to half the sum gives the greater angle ; and half the difference sub- tracted from the half sum gives the smaller angle,
side A C ^ 60° side A B = 110
2)170 half sum 80 Z A (contained angle) = 45°
2)50 halfdiff 25
Then by the above rules : —
as sine ^ sine of 2 sides 85° „ .co ar 0-001656
to sine £ diff ditto 25° , 9-625948
so is cotang £ contained L A 22° 30' . 10-382776
to tang \ diff of other angles . , . 10-010380 = 45° 41'
as cos \ sum 85° .,- v; rr • . co ar 1-059704
to cos. \ diff. 25° ... . . 9-957276
so is cotang of \ the cont. L A 22° 30' . 10*382776
to tang \ sum of other angles . . . 11-399756= 87° 43'
greater LC = 133 24 less / B = ~42 2
80
NAUTICAL ASTRONOMY.
Find B C by rule of sines (78).
as sine L C 133° 24' . to sine side AB 110° . so is sine L A 45°
co ar 0-138720
. 9-972986
9-849485
9-961191
66° 8'
104. CASE IV. Two angles and an included side.
Given, an angle 60° "J To find an angle „ an angle 36° I a side
„ included side 70° J a side.
Fig. 56.
By construction : —
Lay off A C equal to 70° from the scale of chords.
Draw the great circle A B (Prob. XI.), at a distance of 36° .from the primitive, and in like manner C B at a dis- tance of 60° ; where these two intersect will be the point B, and ABC will be the triangle.
To measure the parts required : —
1. measure side AB by Prob. XII.
2. measure side B C „ XII.
3. measure L B XIV.
OBLIQUE TRIGONOMETRY. 81
By calculation : —
1. By means of a Perpendicular.
Let fall a perpendicular from L A on side B C produced (Prob. X.) (or from L C on A B produced, suppose the former).
Then will ADC be a right angle, as will also A D B right angled at D.
Find the Z. D A C in the A D A C.
The side A C will be middle, and the angle C, and Z. D A C will be extremes conjunct (82) : then by (87): —
|
as cotang L C 60° . » |
. co ar 0-238561 , <* . 10* |
|
so is cosine side AC 70° . ••• #? |
. •> . 9-534052 |
|
9-772613 — 59° 21' |
|
|
L BAG - 36 |
L DAB = 23 21
Find' AD in A A CD.
The angle D A C is middle, and extremes are conjunct.
as cotang side A C = 70° . * . co ar 0-438934
to rad . .10-
so is cos L D A C 59° 21' . . . '. 9707393
to tang AD , «,'' • , -« • .... 10-146327 = 54° 28'
Find AB in A ADB.
The Z. D A B is middle, and sides A D and A B are extremes conjunct.
*as tang side AD 54° 28' . . . co ar 9*853673
to rad 10-
so is cos L DAB 23° 21' . 9-962890
to tang side A B 9-816563 = 56° 45'
* It promotes accuracy to use the logarithm itself, from which the tang AD was taken on the previous calculation. 9-853673 is the ar. co. of 10-146327.
a
82 NAUTICAL ASTRONOMY.
Find BC by rule of sines (78).
as sine L C 60° . . co ar 0*062469
to sine side AB 56° 45' . 9*922355
so is sine L A 36° , , 97G9219
to sine side BC . . , 9*754043 = 34° 35'
Find L A B C by rule of sines (79).
as sine side AB 56° 45' co ar 0*077645 to sine L C 60° . . . 9*937531 so is sine side AC 70° . 9*972986 to sine L ABC ;' *. ,' . . =9*988162= 76° 41'
180 0 Z.B 103 19
N.B. — The supplement is used because construction shows the angle to be obtuse.
2. Without a Perpendicular.
KULE. — When two angles and the included side are given: —
As the sine of half the sum of the two angles
is to the sum of half the difference
so is the tangent of half the contained side
to the tangent of half the difference of the other two sides.
And again: —
As the cosine of half the sum of the two angles
is to the cosine of half their difference,
so is the tangent of half the included side
to the tangent of half the sum of the other sides ;
and half the difference added to half the sum will give the greater side, and half the difference subtracted from half the sum will give the smaller side.
an angle 36° side 70° _ggc
an angle 60 ~2
2)96 half sum 48
5)24"
halfdiff 1?
•D
OBLIQUE TRIGONOMETRY.
83
Find sides A Band BC.
as sine ^ sum 2 angles 48°
to sine | diff 12°
so is tang £ included side 35°
to tang £ diff other sides .
as cos | sum of 2 angles 48°
to cos I diff 12°
sine tang | included side 35°
to tang ^ sum other sides
co ar 0-128927
9-317879
9-845227
9-292033 = 11° 5' co ar 0-174489
9-990404
9-845227
45° 40'
'.'' 10-010120 greater side AB smaller side B C
Find L B by rule of sines (79).
as sine side EC 34° 35' to sine L 36° . so is sine side AC 70° to sine L B
co ar 0-245954 9-769219 9-972986 9-988159= 76° 180 /B "
130 19
N.B. — The construction of the figure shows the L B to be obtuse ; therefore we use the supplement.
105. CASE V.
Three sides given : —
side AC 60° sideAB 70° sideBC 100°
Pig. 57.
G2
84
NAUTICAL ASTRONOMY.
By construction : —
Lay off one side (say 60°) on the primitive, as at AC, from chord of 60°. About E draw (Prob. VI.) a small circle ay at the distance of the supplement of BC. About A draw a small circle xz. at the distance of 70° (Prob. VI.). Through point B, where the small circles intersect, draw (by Prob. IV.) an oblique circle AB, and ABC will be the triangle.
To measure the parts required : —
1. rmn on the scale of semi-tangents measures Z. A.
2. The Z. C is measured from Z to o (semi-tangent back-
wards from 90°).
3. The Z.B by Prob. XIV.
By calculation : —
KTJLE. — Find half the sum of the three sides. Subtract from this half sum each of the two sides which, together, contain the required angle. Then add the sines of these two remainders to the sines of the two sides which contain the angle (using the co arcs of the latter). Half the sum of these four logarithms will give the sine of half the re- quired angle.
To find the angle A : —
side AC = 60° side AB = 70° side BC = 100° 2)230 half sum of sides 115
sine sine
co ar co ar
0-062469 0-027014
side AC
side AB
60 ~55~
115 70 45
first remainder 9-913365
second remainder 9-849485 2)19-852333
sine 57° 32' 2
= 115 4
9-926166
OBLIQUE TRIGONOMETRY. Find/. B by rule of sines (79).
as sine side B C 100° . . eoar 0-006649 to sine /A 115° 4' .... 9*957040 so is sine side AC 60° . 9-937631
to sine L B 52° 48' . . . . =9 '90 1220
Find/. C by rule of sines (79).
as sine side BC 100° . . coar 0-006649 to sine L A 115° 4' .... 9-957040 so is sine side AB 70° ... 9'972986 tosine^C 59° 48' .... =9'936675
106. CASE VI.
Three angles given : —
Angle A 130° Angle B 50° Angle G 4£°
Fig. 58.
85
By construction: —
Make the L A = 1 30° (by Prob. XL) by drawing the oblique circle ABZ. Take the measure of the angle which is to be at the primitive =45° from the scale of semi tangents, and sweep it round the pole of the primitive (from the centre) as s t.
Find the pole, x, of the oblique circle (Prob. VII.), and round x draw a small circle ^qual to the other given angle,
Q 3
$6 . NAUTICAL ASTRONOMY.
or 50°, as ef, from scale of semi-tangents. Where the two small circles cut, as at n, will be the pole of the oblique circle, which shall make an angle of 45° with the primitive and 50° with the other oblique circle.
Through on draw a diameter ; measure on on the scale of semi-tangents, and lay off its complement beyond o ; say to v.
Through the three points D v C (D C being at right angles to GrH) draw the oblique circle DvC (Prob. IV.), and ABC shall be the triangle.
To measure the parts required : —
1. AC is measured on the scale of chords.
2. AB „ by Prob. XII.
3. BC „ by Prob. XII. By calculation : —
RULE. — From half the sum of the three angles take each of the angles next to the side required. Add the co arcs of the sines of the two angles which adjoin the re- quired sides, together with the cosines of the two remain- ders. Then half the sum of these logarithms will equal half the cosine of the side required.
Find side BC.
LA. = 130°
L~B = 50 sine * , coar 0-115746
ZC = 45 sine . * co ar 0-150515
2)225
ZB = 50
624- 1st remainder co sine 9-664406
2£ 45
67^ 2nd remainder co sine 9-582840 2)19-513507
cosine 34° 50* = 9756753 2_
69 40 180
110 20 = sideBG
EXAMPLES IN NAUTICAL ASTRONOMY. 87
Find side AC by rule of sines (78).
as sine/ A 130° . . co ar O'l 15746
to sine BC 110° 20' . . 9*972986
so is sine / B 50° . . . 9-884254
to sine side AC x. . . 9-972986 = ZAC 70°
Find side AB (78).
as sine L A 130° . . co ar 0-115746 to sine BC 110° 20' . . . 9 972986 so is sine Z C 45° . . . 9-849485
to side AB .. ; * « . 9-938217= ZAB 60° 9'
107. All the rules necessary for calculating a spheric triangle have been explained. Other formulae might have been added, but where the application of the subject to practice is the main object sufficient has been given. But in order to obviate any possible difficulties which a student might at first encounter in the application of what has been said, we shall now return to the examples which were con- structed from supposed observation (page 65), and show the manner of calculating the results ; and the more espe- cially is this necessary, because, in spherics, the triangle drawn as a problem in nautical astronomy differs from that which is more adapted to calculation ; and again this, if an oblique triangle, requires some management in order to so construct the triangle as to render it convenient for letting fall an available perpendicular to be used with the "5 circular parts," or Napier's rules. An example (No. 2) will be given in illustration of this difference, while the others will be worked without a perpendicular, leaving it to the student to exercise his ingenuity in further con- struction or calculation.
108. EXAMPLE 1. — To find the latitude of a place : —
Given, meridian altitude 60° (The observer north of the sun) „ declination . 20° N.
This question needs no other figure than that already
G 4
88
NAUTICAL ASTRONOMY.
given (fig. 40), because the sun being on the meridian or primitive, its solution is a mere question of length of an arc, no triangle is formed. The sun's altitude being known, together with its declination, fixes the position of the equator, and as the pole is always 90° distant from it, and as latitude is the height of the pole above the horizon, lati- tude will be the complement of the height of the equator above the horizon, and is measured on the primitive.
Thus, in Fig. 59, if H d be the altitude = 60°
and E dthe declination = 20° K
height of the equator = 40 90
' /. E Z or P R = the latitude = 50° Fig. 59.
109. EXAMPLE 2.— To find the latitude:—
Given, sun's altitude 50° „ declination 20° N.
„ azimuth . S. 45° E.
1. By special rule (page 73). It is generally more con- venient to use this rule, because it applies to the astro- nomical figure at one, while to use a perpendicular and the circular parts requires a reconstruction of the figure, (which, see onward).
EXAMPLES IN NAUTICAL ASTKONOMY.
89
Find L C by rule of sines (79).
as sine side B C 70° . . . co ar 9-027014
to sine L A 135° 9'849485
so is sine side A B 40° .... 9-808067
to sine L C 9-684566 = 28° 56'
sum
0-587004
9-913365 10-123411
J.0-623780 = 13°-23' 2
B = 26 46
|
side 70 side 40 |
L 135°-0f / 28-56 |
|
|
2)TlO~ ~55 |
2)106-4 53-2 ~ |
fcdiff |
2)30
as sine $ diff of sides 15° . . co ar
to sine \ sum 55°
so is tang \ diff 5 3° 2' .
to tang \ contained L . . . .
Find side A C by rule of sines.
as sine L C 28° 56' to sine side A B 40° so is sine L to side A C
Fig. 60.
|
°56'. |
. coar 0-315342 |
|
|
340° |
9-808067 |
|
|
26° 46' . |
9-653558 |
|
|
9776967 |
= 36° 45' |
|
|
90 |
||
|
latitude |
= 53° 15' |
N.B. — Notice that while C R will be the. latitude, we, in forming the triangle ABC, use the complement of the two given sides, and that the side required in the triangle A B C is therefore the co latitude.
90
NAUTICAL ASTRONOMY.
2. By means of a Perpendicular.
As previously recommended (90), it is well, as a general rule, to put a given angle at C (by Prob. XI.), and a given side upon A C by scale of chords, letting fall the perpen- dicular upon B C (or B C produced if necessary) from the angle A by Prob. X.
In this question we have two sides, and their two op- posite angles, one of the latter having been found by rule of sines.
The following will be the triangle as constructed for the perpendicular let fall from A : —
o /
side A C being 70 side A B „ 40
L C ,,28 56
L B „ 135
Fig. 61.
Find D C in A A D C.
The L C is middle and extremes are conjunct (82). Then correcting for complements or supplements (83), we have by (87)
as cotang A C 70°
rad .
cosine Z C 28° 56'
co ar 0-438934 . 10-
9-942099
to tang side DC . » * « . 10-381033 = 67° 25'
EXAMPLES IN NAUTICAL ASTRONOMY. 91
Find A D in A A D C.
AD is middle, and extremes are disjunct, then by (88).
as rad .' » » 10-
to sine A C 70° 9-972986
so is sine L C 28° 56' , 9*684658
to sine AD ,.' 9-657644 = 27° 2'
Find D B in A A D B.
AB is middle and disjunct, then by (89).
as cos A D 27° 2 . . . . . 0-050248
to rad . ' . \ " . . . . 10-
so is cos A B 40° , 9-884254
to side DB . . - •'. . - . . 9-934502 = 30° 41
side DC 67 25
co lat or side B C = 36 44 90
latitude required = 53° 16'
It would have been shorter (perhaps not so obvious to a beginner) to have found D B by using the A A D B, and using the complement of ABC (= 45°) as the /.DBA, thus :—
The L B 45° is middle and conjunct, then by (87).
as cotang 40° . '».'*" , . 9-923814 .*
to rad . . . . . .10-
so is cos L B 45° 9*849485
totangDB . -> ... « 9*773299 = 30° 41' co lat
110. EXAMPLE 3.— To find the latitude of a place :
Given altitude 50° south of the sun dec 10 S. time 10 A.M.
In this example, the sun being in south declination and the observer being south of the sun, it is evident that the observer must be in south latitude (remember that lati- tude is the height of the pole of the sphere above the horizon).
92
NAUTICAL ASTRONOMY.
By construction : —
Being in south latitude the south pole will be above the south part of the horizon. We generally, in construction, consider the north part of the horizon to lie on the right of the centre of the figure, and the south part on the left part.
Assume S the south pole, and draw 3 0 P, and E Q, the equator at right angles to it.
Draw the great circle S x P (by Prob. XI.), making an angle of two hours or 30° with primitive.
By Prob. VI., draw the parallel of declination d c at a distance of 10° south of E Q, or 80° from the primitive.
Fig. 62.
To draw the great circle Zx N, so that Z % — 40° (the co of 50° the altitude), take the secant of 40 from the centre o, and sweep an arc as g n, and the tangent of 40° from the point x, and sweep another arc hm across it; the inter- section will be the zenith line OZa (Prob. VI.). With the three points Z x N, describe the great circle Z x N (Prob. IV.), and S Z a? is the triangle, and S Z the co latitude, S H the latitude required.
EXAMPLES IN NAUTICAL ASTKONOMY.
93
By calculation : —
Find L Z, by rule of sines (79).
as sine of side Z X = 40° . is to sine side of L S = 30 so is sine side S x — 80
to sine Z Z =
(obtuse by construction)
. co ar
0-191933 9-698970 9-993351
9-884254 = 50° 180°
130° = L Z
To find L
a side Z a? = a side S x —
40° . 80 .
40° 80
2)120 2)40
half sum sides = 60 £ diff ~20~
By rule (page 73) : —
as sine J diff 2 sides 20° . is to sine |- their sum 60° . so is tang £ diff 2 angles 50°
to co tang £ then contained L
an angle 130° an angle 30 2)100
50 half diff 2 angles
. co ar 0-465948
9-937531
, . 10-076186
.' • . 10-479665 = I8tf20'
36° 40'
To find co lat S 2, by rule of sines (78).
as sine L S = 30° . is to sine side Z x = 40° so is sine L x = 36° 40' to sine side S Z
. co ar
0-301030 9-808067 9-776090
9-885187
co lat 50° 9' S. 90
lat 39° 51'
111. EXAMPLE 4. — To find the latitude by a celestial body on the meridian below the pole :
Given, altitude of a Lyrse on meridian below the north pole = 20° declination of do. ...*..
This forms no triangle, the star being on the primitive.
94
NAUTICAL ASTRONOMY.
Let HE be the horizon. Lay off the alt 20° to a, the star's place. The star being 38-1- N. of the equator (in declina- tion), lay off 381- from a to Q. Draw QE and P S, and PK will be the latitude, and will equal 90° — (38-1-0 — 20°) = 90° ^=71£ north.
Fig. 63.
a Lyra
112. EXAMPLE 5. — To find the latitude of a place : —
Given, app. time 9 a.m.
declination, 20° K azimuth, S. 60° E. Observer N. of sun.
Fig. 64.
EXAMPLES IN NAUTICAL ASTKONOMY. 95
By construction : —
Assume a point P on the primitive, and draw P S and E Q. Draw the ohlique circle P x9 making 45° with the primitive (Prob.XL). About P draw the parallel circle dc 70° (the co declination) distant from P (Prob. V.). Through the intersection x draw Zx9 making the angle at the pri- mitive = 60, the azimuth from noon, or south (Prob. XL). Then Z#P will be the spheric triangle, and ZP the co latitude.
By calculation : —
To find Z x by rule of sines (78).
60° as sine L Z 120° . . co ar 0-062469
186 is to sine side x P 70° . y 9-972986
Z#.= 120 so is sine L P 45° , . . 9*849485
to sine side Z x . . , 9 -884940 -=50° 6'
To find t_x (page 73).
sine 50° 6' L 120°
side 70 L 45
2)120 6 2)75
half sum 60 3 37 30 half diff
2)19 54
half diff 957
as sine \ diff 2 sides 9° 57' . co ar 0762485 to sine \ sum 2 sides 30° 3' . . 9-937749 so is tang \ diff 2 angles 37° 30' . 9-884980 to cotang \ cont L x . . . 10-585214 = 14 34'
2
Z ar 29 8
Find side ZP by rule of sines (78).
as sine L Z 120° , . co ar 0-062469
is to sine side Pa: 70° . . . 9'972986 so is sine L x 29° 8' , . . 9-687389 to sine of side ZP , '. \ 9-722844 = 31° 53'
90 latitude ZP 58 7 N
96
NAUTICAL ASTRONOMY.
113. EXAMPLE 6.— To find the time (or hour angle),
Given, latitude 45° 40'
declination 10° N. azimuth S. 45 E.
Fig. 65.
By construction :—
Lay off the lat 45° 40' from Eto P (scale of chords), and draw diameters at right angles from P (the pole of the world) draw the parallel d, c with the co declination =80° (Prob. V.). Draw Zx, making the angle 45° with the pri- mitive (Prob. XI.); and through the intersection at a? draw (Prob. IV.) #P, and ZxP will be the triangle and £P the hour L required.
By calculation : —
FindZ_# by rule of sines (79).
as sine side #P 80° . tosine^Z 135° so is sine side ZP 44° 20' . to sine L x
co ar 0-006649 9-849485 9-844372
9700506=30° 7'
EXAMPLES IN NAUTICAL ASTKONOMY.
97
|
side . side . half sum . halfdiff . |
. 80° 0 . 44 20 |
L 135° 0 0 L 30 7 0 |
|
2)124 20 |
2)104 53 0 |
|
|
. 62 10 2)35 40 17 50 |
52 26 30 half diff Zs |
Find L P (page 73).
as sine of half diff sides 17° 50' to sine of half sum 62° 10' so is tang half diff L s 52° 26^' to cotang half cont L P . .'
coar 0-513925
9-946604
10-114104
. 10-574633 = 14° 55' 2_
hour angle £P = 29 50~
N.B. — To convert space into time and the reverse, use the following rules : —
KULE.— Multiply space by 4 and divide the degrees by 60, thus :—
the arc 29° 50' 4_
60)119 20 Ih 50m 2 Os = the above hour angle
To convert time into space : —
RULE. — Eeduce hours to minutes and divide by 4, thus: —
the time Ih 59m 20s
60
4)119 20 29° 50'
114. EXAMPLE 7. — To find time.
Given, declination, 20° S. (being in N. lat.) altitude 20 azimuth S. 45° W.
By construction : —
From Z draw Z#N, making an angle 45° with the pri- mitive (Prob. VIII.). About Z draw the parallel e'f at a
98
NAUTICAL ASTKONOMY.
distance of co alt 70° from it (Prob. VI.). Through their intersection x draw a parallel of declination cfo=to polar distance 110 ( = 90° + dec. 20°) (Prob. VI); and draw an oblique circle through P x S. (Prob. X.) Then Z x P will be the triangle and the L P the hour angle required.
By calculation : — ' iso°
Find Z. P by rule of sines. ^ DZ* 45
as sine side 110° co ar 0-027014 z arZP 135°
is to sine L 135° . . 9'849485
so is sine side 70° . 9-972986
to sine /P , . . 9-849485 =hourZP 45D
_ 4 (page 97) 60)180
3 hours
N.B. — 3h. p.m., because sun was west of meridian.
Fig. 66.
115. EXAMPLE 8. — To find time.
Given, latitude 21° N. declination 20° S. altitude 30
By construction : —
off PK from the scale of chords = the latitude 21°
EXAMPLES IN NAUTICAL ASTRONOMY. 99
(considering P, as usual, the north pole of the world). About S, the south pole, draw a parallel equal to sun's north polar distance, or 110° (Prob VI.). About Z, the zenith, draw a parallel / g equal to co alt (or zenith distance) (Prob. VI.). Through Z#N draw an oblique circle (Prob. IV.) and ZxP will be the triangle and/.P the hour angle.
Fig. 67.
By calculation : —
N.B. — Three sides are given ; find L P (page 84).
|
a side ZiX a side ZP a side # P half sum ZP |
. = 60° = 69 co ar sine = 110 co ar sine |
. 0-029848 . 0-027014 . 9-887406 . 9-217609 |
|
2)239 119 30' 69 50 30 = 119 30 110 |
1st remr |
|
|
9 30 = 2ndremr sine 22° 24 2 |
||
|
2)19-161877 |
||
|
'= 9-580939 |
||
|
time from noon H 2 |
44 48 4 |
|
|
60)179 12 |
||
|
2h 59m 12s |
100
NAUTICAL ASTEONOMY.
116. EXAMPLE 9. — To find an azimuth.
Given, latitude 36° S. declination 20° N. altitude 20°
By construction : —
Place S} the south pole of the heavens, 36° above H, the
assumed south part of hori- Fig. 68. zon Z. Draw diameters as
usual. About P, the north pole, draw the parallel d c at a distance equal to the co declination 70°(Prob.VL). About Z draw the parallel ef equal to the co alt 70° (Prob.VL). Through the point x, where these cut, draw ZzN and Sx~P (by Prob. IV.) and ZxS is the triangle, and angle Z the required azimuth.
By calculation : —
N.B. — Three sides are given.
a side SZ 54° a side Sx 110 a side Zx 70 2)234
117
54
Find/.Z (page 84).
* / coar 0-092042 co ar 0-027014
SZ
Zx
63 1st remainder sine 9-949881 117 70
47 2nd remainder sine 9-864127 2)19-933064
sine 67° 48'
9*966532
_
azimuth L Z 135 36~
N.B.— SZ and Zx are the two sides which make the required angle, therefore their co ar sines are used.
EXAMPLES IN NAUTICAL ASTRONOMY.
101
117. EXAMPLE 10. — To find an azimuth.
Given, latitude
time
declination
21° N. 9 a.m. 20° S.
declination
20 S 90
110
polar distance from N
O'jo
By construction : —
Place the north pole at P, 21° above the hori- zon K. Draw diameters. Draw the oblique circle P#S with the Z_45° from the primitive(45° = 3 hrs.) (Prob. VIII.). About S, the south pole, draw a parallel equal to the co de- clination (by Prob. VI.), and through the intersec- tion x draw (Prob. VI.) Z#N and Zx~P will be the triangle, and /_ Z the azimuth required.
By calculation : —
We have two sides given and an included angle.
a side # P a side ZP
half sum
lialf sum
110 69
2)179
included z2)45°
22 30 half contained Z
89 30
2)41
20 30
To find the other angles (page 79).
as sine i sum 2 sides 89° 30' to sine | diff 2 sides 20 30 so is cot £ contd Z 22 30 to tang ^ diff other angles
as cos £ sum 2 sides 89° 30' to cot | diff 2 sides 20° 30' so is cot £ contd L 22 30
to tang \ sum other angles
co ar 0-000017 . 9-544325 . 10-382776
9-927119 = 40°
co ar 2-059158 * 9-971588 . 10-382776
. 12-413522 = 89° 47
sum is the azimuth greater L Z 130 00 less L x ~49 34 H 3
102 NAtJTICAL ASTRONOMY.
118. EXAMPLE 11. — To find a ship's course when sailing on a great circle, and the distance between port and port.
Given, ship's latitude in . . . . . 50° K
latitude of place bound to . . . 10° N.
the difference of longitude between the two places 60° W.
The term great circle " course " is deceiving, inasmuch as no part of a circle is a straight line. A ship could not sail upon a great circle without constantly changing her course by compass ; and, therefore, " great circle sailing " is positively impracticable, because a great circle tf track " cuts no two meridians at an equal angle. The passage of a ship along a great circle track is evidently a series of courses tangential to it.
Fig. 70.
Suppose, in the above figure, AB to be a given great circle track ; upon a ship's starting at a for the point B a compass course, if long continued, would take her con- siderably away from her proper track, and, we will say, place her at af. It is evident, therefore, that the shorter these compass courses are made, the more will the ship keep to her true course, and the shorter will be the distance required to be sailed over. As an instance : A ship was 12 hours in going from b to &', and 24 hours in going (at the same rate) from c to c' ; we find that in these cases she would be leaving her great circle track altogether. It remains, therefore, to provide a ready method of finding how to steer by compass so as to depart as little as possible from
EXAMPLES IN NAUTICAL ASTRONOMY.
103
our proper track, which, of course, would be the nearest distance between the port left and port bound to.
The term tangent sailing is the only correct designation of this method, and was first suggested to the Astronomer Royal by the author in 1857.
From the above example to find the first or initial tan- gent course, we proceed thus —
By construction :—
Consider the diagram as a hemisphere drawn on the plane of a ship's meridian (the ship being somewhere on the meridian). PS would represent the poles, and EQ the equator. Let L represent the latitude, or the
Fig. 71.
ship's place, and I the latitude of port bound to, or LQ= 50° and £Q=10°. Draw diameters to point L, and draw Po;S, making an angle of 60° (the difference of longitude) with the primitive (Prob. VIIL). Draw the parallel circle 1 1', and through the point of intersection x draw the oblique circle Lie N, Prob. X., and the /_ #LQ will be the tangent course required ; measured on r n (being 90° distant) it equals 63° 17'.
H 4
104 NAUTICAL ASTRONOMY.
By calculation : —
We have here two sides and an included angle.
a side 80 included / 2)60°
a side 40 30 = half contained L
2)120
half sum 60
2)40
half diff 20
To find the other angles (page 79.)
as sine \ sum of sides 60° . . co ar 0*062469 to sine \ diff 20° .... 9*534052 so is co tang \ contained angles 30° . 10-238561 to tang \ diff angles .... 9-835082 = 34° 22'
cos \ sum of sides 60° . . . co ar 0-301030 to cosine \ diff 20° .... 9*972986 so is cotang \ contained angle 30° . 10-238561 to tang \ sum other angles f ••'. -/. . 10-512577 = 72° 55'
L L 107 17
the tangent course
or S 72 43 W
119. EXAMPLE 12. — -To find the distance between port and port upon a great circle (in the above example),
Find side La? by rule of sines (78).
as sine L L (as above) 107° 17' . co ar 0-020066 is to sine side 80° . . . . 9'993351 so is sine L P 60° . . . . 9*937531
to sine side La? 9-950948 = 63° 17'
60
distance to make good L# 3797 miles.
Suppose that, three days afterwards, the ship vas in lat, not 50° but 45° N, and diff of long between the two places was not 60°, but 50°, what would be her altered course ?
Working as above would give the course about 67°, and the distance about 55°, or 3300 geographical miles.
EXAMPLES IN NAUTICAL ASTRONOMY.
105
N.B. — By Saxby's spherograph (a simple instrument which thoroughly illustrates nautical astronomy, and works any spheric triangle without calculation) a great circle9 or tangent course, is easily obtained in five to ten seconds ; being thus rendered more simple than even a Mercator's course.
120. EXAMPLE 13. — To find the "latitude of vertex" of a great circle track.
Given, latitude of the ship 36° S. " bound to 40° S. difference of long. 110°
Fig. 73.
Lot. 36° S.
Lot. 40° S.
106 NAUTICAL ASTKONOMY.
By construction : —
Make EL equal to lat 36° (from the scale of chords) south of the equator, and L will be the ship's place on her meridian. Draw diameters L I and mn. About the south |K>le, S, draw a parallel circle Lo, distant 50° the co lat (Prob. V.). Draw the oblique circle S#N, making 110° with the primitive (Prob. VIII. ). Through the intersection x draw L# I, and L&S will be the triangle. Let fall a perpendicular Sy upon ILx (the great circle track) (by Prob. X.) from the pole S, and y will be the part of the track nearest to the south pole, or the vertex. Measure Sy (by Prob. XIL), and its complement will be the latitude of vertex.
By calculation : —
To find the two angles.
We have two sides and the included angle given.
a side 54°
a side 50_ included / 2)110°
2)104 55 £ contained L
half sum 52 half diff 2)4 ~2
To find the other angles (page 79).
as sine i sum of sides 52° . . co ar 0-103468
to sine § diff of sides 2° . . . S'542819
so is cotang \ contained angles 55° . 9-845227
to tang £ diff other angles . r « • . 8-491514 = 1° 46' 30"
as cos \ sum of 2 sides 52° . . co ar 0-210658 to cos | diff 2 sides 2° . . -.. 9'999735 cotang £ contained angles 55° . "•. 9'845227 to tang \ sum of other angles . . 10-055620 = 48° 40' 00"
sum= L x 50 26 30 diff= ZL 46 53 30
Find S y by rule of sines (78).
asrad . . . lO'OOOOOO to sine side 50° . . 9-884254 so is sine L x 50° 26' 30" 9-887041
to sine side Sy . . 9771295 = 36° 12'
90
latitude of vertex "" 53 48~
EXAMPLES IN NAUTICAL ASTRONOMY. 107
121. EXAMPLE 14. — To find an altitude of a celestial body.
Given latitude 30° N. „ time 11 A.M. declination 20° S.
By construction : —
Lay off the lat from K to P by scale of chords.
Draw the hour circle 1 1 A.M. = 15° by Prob. VIII. =P x S.
Draw the parallel of declination d e by Prob. VI. ; where these intersect will be x the altitude sought.
Through Z x N draw a great circle (Prob. IV.), and x y will be the measure of the altitude (Prob. XII. ), and P a? Z is the triangle.
By calculation : —
With the co lat Z P = 60°, the pole distance x P = 110, and the hour /_ P = 15°, we have two sides and an in- cluded angle.
By rule page 79 : —
a side 110°
a side 60 2)15°
2)170 sum 7° 30' half contained L
85 \ sum 2)50 the diff
108
NAUTICAL ASTRONOMY.
To find L Z.
as sine of \ sum of 2 sides 85°
is to sine \ diff 2 sides 25°
so is cotang \ contained L 7° 30'
to tang \ diff 2 angles
as cos \ sum 2 sides 85° .
is to cos £ diff 2 sides 25°
so is cotang \ contained L 7° 30'
to tang \ diff 2 angles
so. ar. 0-001656 9-625948 . 10-880571
. 10-508175 = 72° 45' 30" 1-059704 9-257276 . 10-880571
. 11-897551 = 89° 16' 30'' L Z 162 2~~
To find Z x the zenith distance (78).
as sine L Z 162° 2' . is to sine side 110° . so is sine L P 15° .
to sine side Z x
0-510796 9-972986 9-412996
9-896778 = 52° 30' 90
altitude = scy
122. EXAMPLE 15. — To find an amplitude.
Given latitude 50° N. „ declination 20° K.
Fig. 75.
37° 30'
By construction : —
Lay off the lat E P = 50° (scale of chords).
Draw P S and E Q.
EXAMPLES IN NAUTICAL ASTRONOMY. 109
Draw a parallel of decimation d c = 20 N. from E Q (Prob. VI.).
Through point of intersection x draw the great circle PxS (Prob. IV.), and Px K will be the triangle, and x o the amplitude.
By calculation : —
P x will be a middle part (83), and the extremes are dis- junct.
To find x E (89).
as cos 50° co ar. 0-191933
is to rad 10'
so is cosine 70° 9-534052
to cosine a? E ..... 9725985 = 57° 51' azim
90
the amplitude x o = 32° 9' N.
N.B. — The ordinary rule is, add the secant of the lati- tude to the sine of the declination, and the sine of the sum is the amplitude.
Thus,— sec 50° . . . . 0-191933 sine 20 , 9-534052
9725985 = 32° 9'
Eemember the secant is the reciprocal of the cosine, and the cosine of 70° = the sine of 20°; hence the rule.
123. EXAMPLE 16. — To find the time of daybreak.
N.B. — Daybreak is the time at which the sun's centre is just 18° below the horizon of the place.
Given latitude . . 50° N. „ sun's declination . 9° 43' S.
By construction : —
Lay off the lat 50° N. from E to P from scale of chords. Draw the parallel of declination d c by Prob. VI.
110
NAUTICAL ASTRONOMY.
Draw the parallel tw of 18° below HE the horizon
by Prob. VI. ; where the parallels cut at x is the sun's place at daybreak.
Draw the great circle P x S (the hour circle), and Z&N the azimuth by Prob. IV., and Z^P will be the triangle, and the Z_ # P E the time from midnight.
By calculation : — Here are three sides given, viz., the co lat 40°,
the polar distance 99° 43' and the side Z# = 90° + 18°
= 108°.
By rule, page 84. To find L P.
a side 108° Co arc sine .
99 43' co arc sine .
40 2)247 43
123 52 \ sum 9943
24 9
12352 40
83 52
1st remainder sine
0-006275 0-191933
9-611858
|
2nd remainder sine 2 |
. 9-997507 |
|
2)19-807573 9-903786 |
|
|
ZPar = 106 30 180 |
L RP# = 73 30 in space
4 by rule page
60)294 00 B 4" 54"
apparent time of daybreak «= 4" 54m in time N.B. — To be corrected by the equation of time for the day as given in almanacks, so as to get mean time.
EXAMPLES IN NAUTICAL ASTKONOMY.
Ill
. EXAMPLE 17. — To find the time of rising of a celestial body.
Given latitude of place . . . 51° 27' N. „ declination of the object . 10 15 S.
Fig. 77.
By construction :—
Lay off the latitude at R P.
Draw the parallel of declination d c by Prob. IX.
Where this cuts the horizon will be the place of the sun's rising, as at x.
Draw Pa; S, Prob. IV., and #P R will be the triangle, and the /_ # P R will be the time of rising. P R Q, &c., being the midnight meridian.
By calculation : —
To find/. P.
L P is middle and extremes are conjunct (83).
Then by (86):—
As rad 10'
istocotangPtf 100° 15'. . . . 9-257269 so is tang 51° 27' . 10-098617
to cosine / P .... . 9-355886 = 76° 53'
4
60)307 42 time of sunrise . = 5h 7m 42 S.
112 NAUTICAL ASTEONOMY.
If we use the triangle Z x P, we have a quadrantal triangle for Z x = 90°.
By rule (page 69). Find/.
asrad ....... 10*
is to cotang Z P 38° 33/ . . . . 10-098617
soiscotangPar 100° 15' .... 9-257269
to cosine / ZPx 9-355886 = 76° 53' &c.
The preceding examples comprise all ordinary questions to which the attention of the navigator is likely to be called. When a student has read this little book he will be better able to comprehend works written upon the subject of spherics, in which the theory is explained. It is more than probable that a small volume may shortly follow the pub- lication of this, adapted to those who, not content with an ( ' initiation " into nautical astronomy, desire to become further acquainted with so beautiful, enticing, and useful a branch of study.
THE END.
LONDON
PRINTED BY SPOTTISWOODE AND CO. NEW-STREET SQUARE
UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY
THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW
Books not returned on time are subject to a fine of 50c per volume after the third day overdue, increasing to $1.00 per volume after the sixth day. Books not in demand may be renewed if application is made before expiration of loan period.
MAR 1 1918
APR 08 1988
must MAR
50m-7,'16
Feb 20 1913 Lyon
107275
AULAY,
lisrht Hon. Lo
rterly Magazin Historical Essai Poems, &c. 2 vc
n of James t sed and correct'
LAND from t
and II. 8vo. price 32i ontributed to tl
. Library Edition (I
RICAL ESSA1
Volumes for the Pock
RICAL ESSA1
ition, complete in O ; calf, by HAYDAY, 3
CAL ESSA1
complete in 2 vo
:>ry of "England' s Life of John-
BEECHES of the P
3 HIMSELF. 8vo. 12s.
[ ORD MACAULA
i RE FORM in 1831 and 1832.
[ AYS of ANCIENT RUMU. JtJy tne M
MACAULAY. With Illustrations, original and from the Antique, o» HAP
.A., engraved on Wood by S. WILLIAMS. New Edition. Pep. 410.^^62 morocco, by HAYDAY, 42s.
LORD MACAULAY'S LAYS™ ANCIENT ROME, with IVR and the ARMADA. ICmo. 4s. 6d. cloth ; morocco, by HAYDAY, 10s. 6d.
London: LONGMAN, GREEN, and CO. Paternoster Row.