IN MEMORIAM FLOR1AN CAJORI >mBb ■'•■• ■■ - KJ: : ■S.Vi:_.a3-- *° A %;3? CONTENTS. PLANE TRIGONOMETRY. Chap. Page - I. General Principles of Plane Trigonometry ... 3 II. Sines, Tangents, and Secants ..... 6 III. Right Triangles .20 IV. General Formulas 27 V. Values of the Sines, Cosines, Tangents, Cotangents, Secants, and Cosecants of certain Angles ... 37 VI. Oblique Triangles 47 NAVIGATION AND SURVEYING. I. Plane Sailing 67 II. Traverse Sailing 76 III. Parallel Sailing 80 IV. Middle Latitude Sailing 83 V. Mercator's Sailing ...*... 90 VI. Surveying 103 VII. Heights and Distances 115 SPHERICAL TRIGONOMETRY. I. Definitions . 129 II. Right Triangles 134 III. Oblique Triangles 156 o?is>€a»3*»ii IV CONTENTS. SPHERICAL ASTRONOMY. CHAP. Page. I. The Celestial Sphere and its Circles .... 191 II. TJhe Diurnal Motion 196 III. The Meridian 210 IV. Latitude 223 V. The Ecliptic . . . . . . . .261 VI. Precession and Nutation 275 VII. Time 293 VIII. Longitude 309 IX. Aberration 340 X. Refraction 356 XI. Parallax 367 XII. Eclipses 383 ERRATA. \ 11, ] 13, • -2, for 2, 6.7914 read 6.7935 BCT ACT P. 307, I. 21, for 9 h 17 m Q s r> ^ 42 7n 52 15 13, 0.00354 0.00364 325, —3, fiar.43 fig. 40. 56, 13, AB AC 338, 4, 51° 2 51^43 '39 186, -5, tangent sine 8& £ , dele R. A. 206, 15, fig. 2 fig. 35 356, -5, for dec. read upper an, 7, fig. 37 fig. 38 369, — 1, attain obtain 235, 15, ZSD ZSP 370, 6, AE AL 264, 8, long. 90° long. —90- 14, ABR BAR 277, 15 & 16, PZ h& -3, RAB RB 278, 10, APA X *I 384, 7, OPO' opo 1 — 1, NAA X SPS' 4 A 36 w Q s 390, 3, JVg Mg 315, 307, -6, 17, SS'D 7 / *23 w 51 s 401, 4, 15, MUG 61 JVMg 63 PLANE TRIGONOMETRY. Digitized by the Internet Archive in 2008 with funding from Microsoft Corporation http://www.archive.org/details/elementarytreatiOOpeirrich PLANE TRIGONOMETRY. CHAPTER I. GENERAL PRINCIPLES OP PLANE TRIGONOMETRY. 1. Trigonometry is the science which treats of angles and triangles. 2. Plane Trigonometry treats of plane triangles. [B. p. 36.]* 3. To solve a Triangle is to calculate certain of its sides and angles when the others are known. It has been proved in Geometry that, when three of the six parts of a triangle are given, the triangle can be constructed, provided one at least of the given parts is a side. In these cases, then, the unknown parts of the triangle can be deter- mined geometrically, and it may readily be inferred that they can also be determined algebraically. But a great difficulty is met with on the very threshold of the attempt to apply the calculus to triangles. It arises from the circumstance, that two kinds of quantities are to be intro- * References between brackets, preceded by the letter B., refer to the pages in the stereotype edition of Bowditch's Navigator. 4 PLANE TRIGONOMETRY. [CH. I. Solution of all triangles reduced to that of right triangles. duced into the same formulas, sides, and angles. These quan- tities are not only of an entirely different species, but the law of their relative increase and decrease is so complicated, that they cannot be determined from each other by any of the com- mon operations of Algebra. 4. To diminish the difficulty of solving triangles as much as possible, every method has been taken to com- pare triangles with each other, and the solution of all triangles has been reduced to that of a Limited Series of Right Triangles. a. It is a well known proposition of Geometry, that, in all triangles, which are equiangular with respect to each other, the ratios of the homologous sides are also equal. [B. p. 12.] If, then, a series of dissimilar triangles were constructed con- taining every possible variety of angles : and, if the angles and the ratios of the sides were all known, we should find it easy to calculate every case of triangles. Suppose, for instance, that in the triangle ABC (fig. 1.), the sides of which we shall denote by the small letters a, b, c, respectively opposite to the angles A, B f C, there are given the two sides b and c and the included angle A, to find the side a and the angles B and C. We are to look through the series of calcu- lated triangles, till we find one which has an angle equal to A, and the ratio of the^ including sides equal to that of b and c. As this triangle is similar to ABC, its angles and the ratio of its sides must also be those of the triangle ABC, which is therefore completely determined. For, to find the side a, we have only to multiply the ratio which we have found of b to a, that is, the fraction - by the side b or the ratio - by the side c. § 4.] GENERAL PRINCIPLES. Solution of all triangles reduced to that of right triangles. 6. A series of calculated triangles is not, however, needed for any other than right triangles. For every oblique triangle is either the sum or the difference of two right triangles ; and the sides and angles of the oblique triangle are the same with those of the right triangles, or may be obtained from them by addition or by subtraction. Thus the triangle ABC is the sum (fig. 2.) or the difference (fig. 3.) of the two right trian- gles ABP and BpC. In both figures the sides AB, BC 9 and the angle A belong at once to the oblique and the right triangles, and so does the angle BCA (fig. 2.) or its supple- ment (fig. 3.) ; while the angle ABC is the sum (fig. 2.), or, the difference (fig. 3.) of ABP and PBC; and the side AC is the sum (fig. 2.), or the difference (fig. 3.) of AP and PC. c. But, as even a series of right triangles, whiqh should con- tain every variety of angle, would be unlimited, it could never be constructed or calculated. Fortunately, such a series is not required ; and it is sufficient for all practical purposes to calculate a series in which the successive angles differ only by a minute, or, at the least, b ya second. The other triangles can be obtained, when needed, by that simple principle of inter- polation made use of to obtain the intermediate logarithms from those given in the tables. 1* PLANE TRIGONOMETRY. [cH. II. Sine, tangent, secant. CHAPTER II. SINES, TANGENTS, AND SECANTS. 5. Confining ourselves, for the present, to right tri- angles, we now proceed to introduce some terms, for the purpose of giving simplicity and brevity to our language. The Sine of an angle is the quotient obtained by dividing the leg opposite it in a right triangle by the hypothenuse. Thus, if we denote (fig. 4.) the legs BC and AC by the letters a and 6, and the hypothenuse AB by the letter A, we have. sin. A = -, sin. B = T . (1) h li 6. The Tangent of an angle is the quotient obtained by dividing the leg opposite it in a right triangle, by the adjacent leg. Thus, (fig. 4.), tang. A = -, tang. B= -. (2) 7. The Secant of an angle is the quotient obtained by dividing the hypothenuse by the leg adjacent to the angle. § 10.] SINES, TANGENTS, AND SECANTS. Cosine, cotangent, cosecant. Thus, (fig. 4.) sec. A = T , sec. B = -. b a (3) 8. The Cosine, Cotangent, and Cosecant of an angle are respectively the sine, tangent, and secant of its complement. 9. Corollary. Since the two acute angles of a right triangle are complements of each other, the sine, tan- gent, and secant of the one must be the cosine, cotan- gent, and cosecant of the other. Thus, (fig. 4.), sin. cos. A = cos. sin. tang. A = cotan.I2 == a cotan. A = tang. B = - A — cosec.2? = cosec. A = sec. B = (4) 10. Corollary. By inspecting the preceding equations (4), we perceive that the sine and cosecant of an angle are reciprocals of each other ; as are also the cosine and secant, and also the tangent and cotangent. PLANE TRIGONOMETRY. [CH. II. To find the tangent. So that cosec. A X sin. A = - X t — — 7 = 1 a A a A sec. .4 X cos. ^4 tang. A X cotan. -4 A b_bh_ b* h-~ bh~ a b ab a ab whence cosec. A =22 - 7 , or sin. A == T sin. .4 cosec. A sec. 1 A 1 cotan. A — -, or tang. A = tan. ^4 cotan. A (5) : --, or cos. A i= j- ^ (6) cos. ^4 sec. A { v ' As soon, then, as the sine, cosine, and tangent of an angle are known, their reciprocals the cosecant, secant, and cotan- gent may easily be obtained. 11. Problem. To find the tangent when the sine and cosine of an angle are known. Solution. The quotient of sin. A divided by cos. A is, by equations (4), sin. A a b ah a cos. A ~~ h ' k bh~~ b' But by (4) hence tang. 4 = -; tang. A = sin. A cos. A* (?) § 4.] SINES, TANGENTS, AND SECANTS. 9 Sum of squares of sine and cosine. 12. Corollary. Since the cotangent is the reciprocal of the tangent, we have cotan. A =. — — . (8) sin. A v ' 13. Problem. To find the cosine of an angle when its sine is known. Solution. We have, by the Pythagorean proposition, in the right triangle ABC (fig. 4.) a 2 + b 2 =z h 2 . But by (4) / • -V- , / A- « 2 , ° 2 a 2 +b 2 h 2 • or (sin. 4) 2 + (cos. 4) 2 = 1 • (9) that is, the sum of the squares of the sine and cosine is equal to unity. Hence (cos. A) 2 = 1 — (sin. A) 2 , cos. A = VI — (sin. A) 2 . (10) I 14. Corollary. Since h 2 — a 2 z= b 2 , we have by (4) h 2 a 2 h 2 — a 2 b 2 t { sec.A) 2 -(t^g.A) 2 = -^-- = —^— = - = h or (sec. A) 2 — (tang, ^l) 2 = 1 ; (11) whence (sec. A) 2 == 1 + (tang. A) 2 . 10 PLANE TRIGONOMETRY. [CH. II. Calculation of cosine, &c. 15. Corollary. Since h 2 — b 2 =za 2 we have by (4) axo / * A h2 b2 h 2 —l> 2 a 2 , (cosec-4) 2 — (cotan.^4) 2 = — - = — = — _ 1, v ' v ' a 2 a 2 a 2 a 2 — ' or (cosec. A) 2 — (cotan. A) 2 = I ; (12) whence (cosec. A) 2 = 1 -f- (cotan. A) 2 . 16. Scholium. The whole difficulty of calculating the trigonometric tables of sines and cosines, tangents and cotangents, secants and cosecants is, by the pre- ceding propositions, reduced to that of calculating the sines alone. 17. Examples. I. Given the sine of the angle A, equal to 0.4568, calculate its cosine, tangent, cotangent, secant, and cosecant. Solution. By equation (10) cos. A = s/l — (sin. A ) 2 = V(l + sin..4)(l — sin.^l). 1 -+- sin. A = 1.4568 0.16340 1 — sin. A = 0.5432 9.73496 (cos. A) 2 2|9.89836 cos. A = 0.8896 9.94918. By (7) and (8) - sin. A . cos. A tang. A = cotan. A = -: -r. cos. A sin. A $ 17.] SINES, TANGENTS, AND SECANTS. 11 Calculation of cosine, &c. sin. A = 0.4568 9.65973 (ar. co.) 10.34027 cos. A = 0.8896 (ar. co.) 10.05082 9.94918 tang. A = 0.5135 9.71055 (ar. co.) 10.28945 cotan. A = 1.9474. By (6) sec. A = r, cosec. A cos. A sm. A log. sec. A = — log. cos. A = 0.05082, sec. A = 1.1241. log. cosec. A = — log. sin. ,4 = 0.34027, cosec. it = 2.1891. 2. Given sin. 4. =0.1111 ; find the cosine, tangent, cotan- gent, secant, and cosecant of A, Ans. cos. A = 0.9938 tang. .4=0.1118 cotan. A = 8.9452 sec. A = 1.0062 cosec. 4 f= 9.0010. 3. Given sin. J. = 0.9891 ; find the cosine, tangent, cotan- gent, secant, and cosecant of A, A?is. cos. A = 0.1472 tang. A = 6.7173 cotan. 4 = 0.1489 sec. A = 6.7914 cosec. A= 1.0110. 12 PLANE TRIGONOMETRY. [CH. II. Sine, &c. in the circle whose radius is unity. 18. Theorem. The sine of an angle is equal to the perpendicular let fall from one extremity of the arc, which measures it in the circle, whose radius is unity, upon the radius passing through the other extremity. Proof. Let BCA (fig. 5.) be the angle, and let the radius of the circle AB A 1 A be AC = unity = 1. Let fall, on the radius AC, the perpendicular BP, and we have by § 5, in the right triangle BCP, sin .B C P = |J = ^ = BF. 19. Theorem. In the circle of which the radius is unity, the cosine of an angle is equal to the portion of the radius, which is drawn perpendicular to the sine, included between the sine and the centre. Proof. For if BCA (fig. 5.) is the angle, we have, by § 9, C P CT> cos. BCA = ~ = ZZL = CP. 20. Theorem. In the circle of which the radius is unity, the secant is equal to the length of the radius drawn through one extremity of the arc which measures the angle, and produced till it meets the tangent drawn through the other extremity. The trigonometric tangent is equal to that portion of the tangent, drawn through one extremity of the arc, which is intercepted between the two radii which termi- nate the arc. § 22.] SINES, TANGENTS, AND SECANTS. 13 Sine, &c. in this and other systems. Proof. If CB (fig. 5.) is produced to meet the tangent A at T, we have, by (2) and (3), in the right triangle BCT, sec. BCA = ~ - -^- =Cr tang. BCA = -^ = — = ^T. 21. Scholium. The preceding theorems (18-20), have been adopted by most writers upon trigonometry, as the definitions of sine, cosine, tangent, and secant, except that the radius of the circle has not been limited to unity. [B. p. 6.] By not limiting the radius to unity the sines, &c. have not been fixed values, but have varied with the length of the radius ; whereas their values, in the system here adopted, are the fixed ratios of their values as ordinarily given to the radius of the circle in which they are measured. Thus, if R is the radius, we have sin., cos., &c. in the common system == R X sin., cos., &c. in this system. 22. Corollary. If the angle is very small, as C (fig. 6.), the arc AB will be sensibly a straight line, perpendicular to the two radii CA and CB, drawn to its extremities, and will sen- sibly coincide with the sine and tangent ; while the cosine will sensibly coincide with the radius CA 9 and the secant with the radius CB. Hence, the sine and tangent of a very small angle are nearly equal to the arc ivhich measures the angle, 2 14 PLANE TRIGONOMETRY. [CH. II. Sine, &c. of very small angles. in the circle the radius of which is unity ; and its cosine and secant are nearly equal to unity. 23. Problem. To find the sine of a very small angle. Solution. Let the angle C (fig. 6.) be the given angle, and suppose it to be exactly one minute. The arc AB must in this case be ^jsirjir of the semicircumference, of which unity or CA is radius. But the value of the semicircumference, of which unity is radius, has been found in Geometry to be 3.1415926. Therefore, by §22, ™,.r=^ = Hi^ = o.o<« 9 . m In the same way we might find the sine of any other small angle, or we might, in preference, find it by the following proposition. 24. Theorem. The sines of very small angles are proportional to the angles themselves. Proof. Let there be the two small angles, BCA and B'CA (fig. 7.) Draw the arc ABB 1 with the centre C, and the radius unity. Then, as angles are proportional to the arcs which measure them, BCA : BCA = BA : B'A. But, by § 22, sin. BCA =f BA, sin. BCA = B' A ; whence BCA : BCA = sin. BCA : sin. BCA. a. This proposition is limited to angles so small, that their arcs may be considered as straight lines. It is found in prac- § 26.] SINES, TANGENTS, AND SECANTS. 15 Sines of small angles. tice, that the angles may be as large as two degrees, provided the approximations are not carried beyond five places of deci- mals. The investigation of the sines of larger angles requires the introduction of some new formulas. 25. Examples. 1. Find the sine of 12' 13", knowing that sin. 1' = 0.00029. Solution. By (28) 1': 12' 13": : sin. V : sin. 12' 13", or 60" : 733" : : 0.00029 : sin. 12' 13". Hence sin. 12' 13" = ** X 0-00029 = Qm . Ans bO 2. Find the sine of 7' 15", knowing that sin. V = 0.00029. Ans. sin. 7' 15" = 0.00210. 3. Find the sine of 2' 31", knowing that sin. 1' == 0.00029. Ans. sin. 2' 31" = 0.00073. 26. Problem. Given the sine of any angle, to find the sine of another angle which exceeds it by a very small quantity. 16 PLANE TRIGONOMETRY. [CH. II. Sine of an angle differing very little from a given angle. Solution. Let the given angle be BCA (fig. 8), which we will denote by the letter 31; and let the angle whose sine is required be B'CA, exceeding the former by the small angle B'CB, which we will denote by the letter m ; so that M= BCA, m =B'CB, 3f+m = B'CA. From the vertex C as a^ centre, with the radius unity, de- scribe the arc ABB'. From the points B and B' let fall BP and B'P' perpendicular to AC. We have, by § 18 and 19, Sine. M=BP sin. BCA = sin. (31 -f m) = B' P' cos. 31= PC; Draw BR perpendicular to B' P', and B> P< =BP + BR, or sin. (31+ in) = sin. 31 + BR. The triangles BCP and BBR } having their sides perpen- dicular each to each, are similar, and give the proportion BC: BB=CP : BR. But, by § 22, BB' === sin. m. Hence 1 : sin. m = cos. 31 : BR ; and BR = sin. m. cos. 3T, which gives, by substitution, § 28.] SINES, TANGENTS, AND SECANTS. 17 Cosine of an angle differing very little from a given angle, sin. (M + m) = sin. M + sin. m. cos. M. (13) 27. Corollary. If m were 1', (13) would become sin. (M -f- l'J = sin. iff -f- sin. 1'. cos. M, = sin. Jf + 0.00029 cos. M. (14) We may, by this formula, find the sine of 2' from that of 1', thence that of 3', then of 4', of 5', &,c, to the sine of angle of any number of degrees and minutes. 28. Corollary. We can, in a similar way, deduce the value of cos. (M+ m). For, by § 19, cos. (if + to) = P'C= PC— PP', = cos. M—BR. But the similar triangles BB'R and BCP give the proportion BC:BB' = BP: BR, or Hence 1 : sin. m sss sin. Jf : .RR. BR =z sin. m. sin. M, whence cos. (Jf -f- m) = cos. ifcT — sin. m. sin. i!f, (15) and, if we make m = 1', this equation becomes cos. (1/ + I') = cos « M — sin. T # sin M, ss cos. if— 0.00029 sin. ilif. (16) 2* 18 PLANE TRIGONOMETRY. [cH. If. Sine and cosine of angles. 29. Examples. 1. Given the sine of 23° 28' equal to 0.39822, to find the sine of 23° 29'. Solution. We find the cosine of 23° 28' by (10) to be cos. 23° 28' = 0.91729. Hence, by (14), making M = 23° 28' sin. 23° 29' = sin. 23° 28' + 0.00029 cos. 23° 28', = 0.39822 + 0.00026, = 0.39848. Ans. sin. 23° 29' — 0.39S48. 2. Given the sine and cosine of 46° 58' as follows, sin. 46° 58' == 0.73096, cos. 46° 58' = 0.68242, find the sine of 46° 59'. Ans. sin. 46° 59' = 0.73116. 3. Given the sine and cosine of 11° 10' as follows, sin. 11° 10' m 0.19366, cos. 11° 10' fc= 0.98107, find the cosine of 11° ll 7 . Ans. cos. 11° 11 = 0.98101. 30. By the formulas here given a complete table of sines and cosines might be calculated. Such tables have been actually calculated ; and table XXIV. of the Navigator is such a table ; their logarithms are given in table XXVII. of the Navigator. § 30.] SINES, TANGENTS, AND SECANTS. 19 Natural and artificial sines. Radius of table. The sines, cosines, &c. of table XXIV. are called natural, to distinguish them from their logarithms, which are some- times called their artificial sines, cosines, &c. The radius of table XXIV. is 10 5 == 100000, so that this table is, by § 21, reduced to the present system by dividing each number by 100000, that is, by prefixing the decimal point to each of the numbers of the table. The radius of table XXVII. is 10 10 z=z 10000000000, so that this table is reduced to the present system by subtract- ing from each number the logarithm of this radius, which is 10, that is by subtracting 10 from each characteristic. The method of using these two tables is fully explained in pp. 33 - 35, and p. 390, of the Navigator. 20 PLANE TRIGONOMETRY. [CH. III. Hypothenuse and an angle given. CHAPTER III. RIGHT TRIANGLES. 31. Problem. To solve a right triangle^ when the hypothenuse and one of the angles are known. [B. p. 38.] Solution. Given (fig. 4) the hypothenuse h and the angle A, to solve the triangle. First. To find the other acute angle B, subtract the given angle from 90°. Secondly. To find the opposite side a } we have by (1) a sin. A z= T , h which, multiplied by h, gives a = h sin. A ; (17) or, by logarithms, log. a = log. h + log. sin. A. Thirdly. To find the side 6, we have by (4) cos. A = -r, a which, multiplied by h, gives b = 7i cos. .4 ; (18) or, by logarithms, log. b =z log. 7* + l°g* cos. ii. § 33.] RIGHT TRIANGLES. 21 Leg and an angle given. 32. Problem. To solve a right triangle, when a leg and the opposite angle are known. [B. p. 39.] Solution. Given (fig. 4.) the leg a, and the opposite angle A, to solve the triangle. First. The angle B is the complement of A. Secondly. To find the hypothenuse h, we have by (17) a zz: h sin. A, which, divided by sin. A, gives by (6) h = ~ — — a cosec. A ; (19) sin. A v or, by logarithms, log. h = log. a + (ar. co.) log. sin. A s= log. a -J- log. cosec. A. Thirdly. To find the other leg 6, we have by (4) b cotan. .4 = -, a which, multiplied by a, gives b =i a cotan. A ; (20) or, by logarithms, log. b = log. a -j- log. cotan. ^4. 33. Problem. To solve a right triangle, when a leg and the adjacent angle are known. [B. p. 39.] Solution. Given (fig. 4.) the leg a and the angle JB, to solve the triangle. First. The angle A is the complement of B. 22 PLANE TRIGONOMETRY. [CH. III. Hypothenuse and a leg given. Secondly. The other parts may be found by (19) and (20), or from the following equations, which are readily deduced from equations (4) and (6), ** h z= - = a sec. B, cos. B (21) b =z a tang. B ; (22) or, by logarithms, log. h =z log. a -f- log. sec. B, log. b = log. a + log. tang. B. 36. Problem. To solve a right triangle^ when the hypothenuse and a leg are known. [B. p. 40.] Solution. Given (fig. 4.) the hypothenuse h and the leg a, to solve the triangle. First. The angles A and B are obtained from equation (4), sin. A = cos. B = j ; (23) or, by logarithms, log. sin. A = log. cos. B = log. a -j- (ar. co.) log. h. Secondly. The leg b is deduced from the Pythagorean prop- erty of the right triangle, which gives a 2 + 6 2 = h*, (24) whence &2 p- 7,2 _ a 2 = (A + a) (A — a), 6 = V (7,2 _ a 2) — ^ [(^ + fl ) ^ _ a)] j (25) by logarithms, log. b. = i log. (A 2 — a*) = J [log. (h + a) + log. (A — a)]. <§> 36.] RIGHT TRIANGLES. 23 The legs given. 35. Problem. To solve a right triangle, when the two legs are known, [B. p. 40.] Solution. Given (fig. 4.) the legs a and 6, to solve the triangle. First. The angles are obtained from (4) ^ a tang. A =p cotan. B = - ; or, by logarithms, log. tang. A = log. cotan. B = log. a + (ar. co.) log. b. Secondly, To find the hypothenuse, we have by (24) h = V (a 2 +& 2 ). Thirdly. An easier way of finding the hypothenuse is to make use of (19) or (20) h = a cosec. A = a sec. B ; or, by logarithms, log. h = log. a. + l°g- cosec. ^L = log. a -|- log. sec. Z?. 36. Examples. 1. Given the hypothenuse of a right triangle equal to 49.58, and one of the acute angles equal to 54° 44' ; to solve the triangle. Solution. The other angle = 90° — 54° 44' = 35° 16'. Then making h = 49.58, and A = 54° 44'; we have, by (17) and (18), 24 PLANE TRIGONOMETRY. [CH. III. Examples of right triangles. h = 49. 58 1.69531 1.69531 A = 54° 44' * sin. 9.91194 cos. 9.76146 a = 40. 481 1.60725; b = 28. 627 1.45677. Am. The other angle = 35° Iff; The less -f 40 ' 481 > me legs _| 28.627. 2. Given the hypothenuse of a right triangle equal to 54.571, and one of the legs equal to 23.479 ; to solve the triangle. Solution. Making h = 54.571, a == 23.479; we have, by (23), % a = 23.479 1.37068 h = 54.571 (ar. co.) 8.26304 A = 25° 29' sin. B = 64° 31' cos. | 9.63372. By (25), h + a = 78.050 1.89237 h — a = 31.092 1.49265 b 2 2 |3.38502 b = 49.262 1.69251 Ans. The other leg = 49.262 The angles-} f° JJ 3. Given the two legs of a right triangle equal to 44.375, and 22.165; to solve the triangle. * To avoid negative characteristics the logarithms are retained as in the tables, according to the usual practice, with the logarithms of decimals, as in B. p. 29. <§> 36.] RIGHT TRIANGLES. 25 Examples of right triangles. Solution. Making a — 44.375, b — 22.165 ; we have, a — 44.375 1.64714 1.64714 b = 22.165 (ar. co.) 8.65433 A z= 63° 27' 28" tang. ) 1 o m 47 . cosec. > - n 04ft o 7 J5z=z26°32 / 32 / cotan. r 030147 ' sec. j 1004837 h == 49.603 1.69551 Ans. The hypothenuse == 49.603 {63° °7 / 28 // 26^3232" 4. Given the hypothenuse of a right triangle equal to 37.364, and one of the acute angles equal to 12° 30' ; to solve the triangle. Ans. The other angle = 77° 30' rp, , f 8.087 The legs = | 3a47g 5. Given one of the legs of a right triangle equal to 14.548, and the opposite angle equal to 54° 24' ; to solve the triangle. Ans. The hypothenuse == 17.892 The other leg— 10.415 The other angle = 35° 36^ 6. Given one of the legs of a right triangle equal to 11.111, and the adjacent angle equal to 11° ll 7 , to solve the triangle. Ans. The hypothenuse =n 11.326 The other leg =z 2.197 The other angle = 78° 49 ; . 3 26 PLANE TRIGONOMETRY. [CH. III. Examples of right triangles. 7. Given the hypothenuse of a right triangle equal to 100, and one of the legs equal to 1, to solve the triangle. Arts. The other leg = 99.995 0° 34' 23" The angles = { £ ^ 23" 8. Given the two legs of a right triangle equal to 8.148, and 10.864, to solve the triangle. Ans. The hypothenuse = 13.58 The angle, ={S: 5 ?:$ § 38.] GENERAL FORMULAS. 27 Sine of the sum of two angles. CHAPTER IV. GENERAL FORMULAS. 37. The solution of oblique triangles requires the introduction of several trigonometrical formulas, which it is convenient to bring together and investigate all at once. 38. Problem* To find the sine of the sum of two angles. Solution. Let the two angles be BAC and B' AC (fig. 9), represented by the letters M and N. At any point C, in the line AC, erect the perpendicular BB 1 . From B let fall on AB' the perpendicular BP. Then represent the several lines, as follows, a — BC, a 1 = BC, b = AC h = AB, h 1 = AB', x — BP M=BAC, NzzzBAC. Then, by (4), sin. BAC = sin. M = T , sin. N z=z %- ii h; «r b b COS. M = T , COS. N = -77 ft /*' sin. JBAP = sin. (if + N) =^-=|. 28 PLANE TRIGONOMETRY. [CH. IV. Sine of the difference of two angles. Now the triangles BPB 1 and B'AC, being right-angled, and having the angle B 1 common, are equiangular and similar. Whence we derive the proportion AB< : BB< — ACBP, whence and h! : a + « ' = & • x ? _ab + a'b X ~ h' ' ,v, . , T \ x ab4-a'b The second member of this equation may be separated into factors, as follows, /,r . , T x ab , ba' a b b a' whence, by substitution, we obtain sin. (i*f + N) — sin. M cos. iV+ cos. 1/ sin. iV. (26) 39. Problem. To find the sine of the difference of two angles. Solution. Let the two angles be BAC and B 'AC (fig. 10), represented by M and N. At any point C in the line AC erect the perpendicular BBC. From B let fall on AB' the perpendicular JBP. Then, using the notation of § 38, we have sin. BAP == sin. (M— N) = ^ — | 7 A B h $ 40.] GENERAL FORMULAS. 29 Cosine of the sum of two angles. The triangles B' AC and BB'P are similar, because they are right-angled, and the angles at B' are vertical and equal. Whence AB 1 : BB' — AC: BP, or h' : a — a 1 = b : x ; whence ab — a'b X = A'"' and sin. , ,. _ mr% x « b — ba' ( *~^=;A= ft - = p' — TV a b b a! and by substitution, sin. (M — N) = sin. M cos. iV — cos. M sin. iV. (27) 40. Problem. To find the cosine of the sum of two angles. Solution. Making use of (fig. 9), with the notation of § 38, and also the following !/ = AP,z=PB>; we have co,(^ + iV)=^|=f- But y = AB' — PB< =zh' — z. 3* 30 PLANE TRIGONOMETRY. [cH. it. Cosine of the sum of two angles. The similar triangles BPB' and B'AC, give the propor- tion AB' : BB' sb BC . B'P } h 1 : a -f- a' = a 1 : z; whence « a' + a' 2 * = — ¥~ ; and y — h' — zzzzh' j± _ &/2 __ a /2 rt ^ ~~ ^ " But, from the right triangle AB'C, h' 2 — a' 2 ss (4JB») 2 — (^'C) 2 =b (4C) 2 bb & 2 ; whence and y _ b 2 h> « «' n + iV) = .y ' h b 2 — ad hh' _ h2 ~hh' — aa! W b ~ h b a a' whence by substitution, cos. (M+ N) =z cos. M. cos. iV — sin. M. sin. JV. (28) 41. Problem. To find the cosine of the difference of tioo angles. § 41.] GENERAL FORMULAS. 31 Cosine of the difference of two angles. Solution. Making use of (fig. 10.) with the notation of the preceding section, we have cos. BAB' == cos. (M— N) = ^-= = f. v ' AB h But y — AB' + PB' = h' + ft The similar triangles BB'P and B AC give the proportion .4J3' : BB' — B'C : J3P, or h' : a — d = d : z ; whence and a d — a' 2 _h' 2 — a' 2 + ad h' But h' 2 — a 12 = b 2 . Hence b 2 -f- ad and cos. <*T? ( H=^ 6 2 , aa' b b a d or, by substitution, cos. (if — N) = cos. if cos. iV+ sin. if sin. iV. (29) 32 PLANE TRIGONOMETRY. [CH. IV. Sum and difference of sines and cosines. 42. Corollary. The similarity, in all but the signs, of the formulas (26) and (27) is such, that they may both be written in the same form, as follows, sin. (Jf dc N) = sin. M cos. N =b cos. M sin. N. (30) in which the upper signs correspond with each other, and also the lower ones. In the same way, by the comparison of (28) and (29), we are led to the form cos. (Jf dc N) =s cos. Jf. cos. N ^F sin. Jf. sin. N, (31) in which the upper signs correspond with each other, and also the lower ones. 43. Corollary. The sum of the equations (26) and (27) is sin. (Jf + N) + sin. (M — N) = 2 sin. 31 cos. N. (32) Their difference is sin. (Jf + N) — sin. (Jf — N) = 2 cos. M sin. N. (33) 44. Corollary. The sum of (28) and (29) is cos. (Jf + N) + cos. (Jf — N) = 2 cos. Jf cos. iV. (34) Their difference is cos. (Jf — N) — cos. (Jf + 2V) = 2 sin. Jf sin. JV. (35) 45. Corollary. If, in (32-25), we make M+N= A, andM—N=:B; that is, 3r=i(A + B), N = i (A — B) ; they become, as follows, <§> 48.] GENERAL FORMULAS. 33 Sum and difference of sines and cosines. sin. A + sin. B = 2 sin. £ (A + #) cos. J (^L — JB) (36) sin. A — sin. JB = 2 cos. £ (it -f B) sin. £ (4 — B) (37) cos. ^4 + cos. B = 2 cos. £ (4 + JB) cos. | (.4 — J3) (38) cos. B — cos. A =2 sin. J (4 + 5) sin. | (4 — jB). (39) 46. Corollary. The quotient, obtained by dividing (36) by (37), is sin. A + sin. B __ sin. £ (A -{- B) cos. ± (A — B) sin. ^4 — sin. B cos. ± (A -\- B) sin. £ (A — B)' Reducing the second member by means of equations (6), (7), (8), we have sin. A 4- sin. B «-«-*»* , - « sTnTZ^smTl* * tang ' * (^ + *) COtan ' i( A ~ B ) j tang.^+#) = cotan. j. (.4 — _2J) tang. £ (4 — #) cotan. $ (A + £)' * ' 47. Corollary. The quotient of (39) divided by (38) is, by reduction, cos. B — cos. A .j ,\ \ ■ ' _ % „ ^ • • = tang.^ + I?) __ tang.j(il — g cotan. £(A—B) cot*n.£(A+ B' { ' 48. Corollary. Putting in (26) and (28), M and iV both equal to A i we obtain sin. 2 A= sin. ^4 cos. A -\- sin. A cos. A = 2 sin. ^4 cos. A (42) cos. 2 ^4 = cos. A cos. ^4 — sin. A sin. ^4 = (cos. A)* — (sin. A)*. (43) 34 PLANE TRIGONOMETRY. [CH. IV. Sine &c. of double, and half of an angle. 49. Corollary. The sum of (43), and of the following equation, which is the same as (9), 1 = (cos. A) 2 -f (sin. A) 2 , is 1 + cos. 2 A = 2 (cos. A) 2 . (44) Their difference is 1 — cos. 2 A = 2 (sin. A) 2 . (45) 50. Corollary. Making 2 A = C, or C= £ A, in (42-49), we obtain sin. C = 2 sin £ C cos. J C (46) cos. C = (cos. J C) 2 — (sin. J C)* (47) 1 + cos. C = 2 (cos. J C)2 (48) 1 _ cos. C = 2 (sin. J C) 2 . (49) The equations (48) and (49) give cos. i C == V[i (1 + cos. C)] (50) sin. J C = V[J(1 — cos. C)] (51) -. ^ #/l COS. C\ ,__, 51. Problem. To find the tangent of the sum a?id of the difference of two angles. Solution. First. To find the tangent of the sum of two angles, which we will suppose to be M and N 9 we have from (*% tmm , ™ s\n. (M+N) tang. (M + N) = W -T »n '- b v ~ ' cos. (if + iV) § 51,] GENERAL FORMULAS. 35 Tangent of sum and difference of angles. Substituting (26) and (28), sin. Mcos. N-\- cos. M sin. N tang. (M-{~ N) = cos. Mcos. N — sin. M sin. N' Divide every term of both numerator and denominator of the second member by cos. 31 cos. N ; sin. M cos. N cos. ill" sin. N I ht . 7%rv cos - M cos. iV ■" COS. M cos. N tang. (M-h- N) = — , & v ' ' cos. M cos. iV sin. if sin. N cos. Jf cos. iV cos. M cos. iV sin. M sin. iV _ cos. M ' cos. iV " ~ sin. M sin. iV' cos. iH ' cos. iV which, reduced by means of (7), becomes mr \ 7»rv tang. Jf 4- tang. iV tang. (Jf+ 2V) = -^—t-l-. (68) Secondly. To find the tangent of the difference of M and N, since by (7) tang. (Jf-jy) = 8in - <*-* [) v ' cos. (M— JV) 9 a bare inspection of (30) and (31) shows that we have only to change the signs, which connect the terms in the value of tang. (M + N) to obtain that of tang. ( M — N). This change, being made in (53), produces •tang. (M - N) = . **ng. if- tang. * T S V ; 1 + tang. M tang. JV l > 36 PLANE TRIGONOMETRY. [CH. IV. Tangent and cotangent of double an angle. 52. Corollary, As the cotangent is merely the reciprocal of the tangent, we have, by inverting the fractions, from (53) and (54), cotan. (M+N) = i#-r~: ^-*r, (55) v ' ' tang. M -f- tang. N ' v ' 1 + tan £- ^"tano*. N , " cotan. (M—N)z= —X — =1 2—. (56) v t tang. JIT — tang. N v ' 53. Corollary. Make M = X = A, in (53) and (55). They become _ . 2 tang. A ,^ v cotan^^ 1 -^"^. (58) 2 tang, vl v ' § 54.] PARTICULAR VALUES OF SINES, &C. 37 Sine, &c. of 0° and 90°. CHAPTER V. VALUES OF THE SINES, COSINES, TANGENTS, COTANGENTS, SECANTS, AND COSECANTS OF CERTAIN ANGLES. 54. Problem. To find the sine, fyc. of0° and 90°. Solution. Supposing M = N, in (27) and (29), we have sin. (M — M) = sin. 0° == sin. M cos. M — cos. iJfsin. M cos. (M—M) = cos.0° — (cos. M) 2 + (s'm.M) 2 ; whence, by (9), and the consideration that 0° and 90° are complements of each other, sin. 0° — cos. 90° = (59) cos. 0° = sin. 90° = 1. (60) From (6) and (7), we have tang. 0° sa cotan. 90° = Sm ' ° no = ° = 0, (61 ) cos. * v y cotan. 0° = tang. 90° = — -— b = 4- = oo (62) tang. x ' sec. 0° = cosec. 90° = — = 4. = I (63) cos. cosec. 0° = sec. 90° = - — -*■ = i = oo. (64) sm. u 38 PLANE TRIGONOMETRY. [CH. V. Sine, &c. of 180° and 270". 55. Problem. To find the sine, $*c. of 180°. Solution. Make A == 90°, in (42) and (43), they become, by means of (59) and (60), sin. 180° = 2 sin. 90° cos. 90° = (65) cos. 180° === (cos. 90°) 2 — (sin. 90°) 2 = — 1. (66) Hence from (6) and (7), n 1ftO° n = (67) = — oo (68) = - 1 (69) tang. TSO o _ sin. 180° 180 ~ cos. 180° — 1 cotan. 180 o = cos. 180' sin. 180° — 1 sec. 180° — i ~~ cos. 180° — i cosec. 180° - Sin ' 18 °° = IT = (70) 56. Problem. To find the sine, Sfc. of 270°. Solution. Make M == 180° and N = 90° in (26) and (28). They become, by means of (59, 60, 6o } 66), sin. 270° = sin. 180° cos. 90°+cos. 180° sin. 90°= — 1 (71) cos. 270° = cos. 180° cos. 90° — sin. 180° sin. 90° = 0. (72) Hence, from (6) and (7), _ sin. 270° — I - tang - 270 = c^27(F == -(r = ~ w (73) «^o cos. 270° COtan " 27 ° = »1F = —1 = ° < 74 > § 58.] PARTICULAR VALUES OF SINES, &C. 39 Sine, &c. of 360° and 45°. : .» o - 2w =s^W=*=- < 75 > 57. Problem. To find the sine, fyc. of 360°. Solution. Make A = 180° in (42) and (43) ; and they be- come by (65, 66, 59, 60) sin. 3G0° = = sin. 0° (77) cos. 360° cs= 1 = cos. 0°. (78) Hence the sine, fyc. of 360° are the same as those of0°. 58. Problem. To find the sine, fyc. of 45°. Solution. Make C = 90° in (50) and (51). They become, by means of (59), cos. 45° = V [J (1 + cos. 90°)] z=z «/i (79) sin. 45° — s/ [J (1 — cos. 90°)] = +/i = cos. 45°. (80) Hence, from (6) and (7), tang. 45> == " = 1 (81) cos. 45 v ' cotan. 45° = -_ = 1 = tang. 45° (82) tang. 45 to v / sec. 45° = X — = -±--^2 (83) cos. 45 \/ £ v ' cosec. 45 = -r— - - = — - - = s/2 = sec. 45°. (84) sin. 45° s/ £ v ' 40 PLANE TRIGONOMETRY. [CH. V. Sine, &c. of 30°, 60^, and the supplement. 59. Problem. To find the sine, fyc. of 30° and 60°. Solution. Make A = 30° in (42). It becomes, from the consideration that 30° and 60° are complements of each other, sin. 60° = cos. 30° =- 2 sin. 30° cos. 30°. Dividing by cos. 30°, we have 1 = 2 sin. 30°, or sin. 30° = J = cos. 60° (85) whence, from (6), (7), and (10), cos. 30° = sin. 60° =± s/ (1 — \) = \ \/3 (86) tang. 30° =£ cotan. 60° == j-^r- = ^- = Vi (87) cotan. 30° = tang. 60° =z -j— = \/3 (88) 1 2 sec. 30° = cosec. 60° = - — — = -7-- (89) ^ V o \/ o cosec. 30° == sec. 60° = - = 2. (90) 60. Problem. To find the sine, Sfc. of the supplement of an angle. Solution. Make M = 180° in (27) and (29). They become, by means of (65) and (66), sin. (180° — 2V}== sin. 180° cos. N — cos. 180° sin. N = sin. iV (91) cos. (180° — N)z=z cos. 180° cos. JY + sin. 180° sin. iV == — cos. JV (92) $ 62.] PARTICULAR VALUES OF SINES, &C. 41 Sine, <&c. of obtuse angle. whence, from (6) and (7), tang. (180° — N) = — tang. N (93) cotan. (180° — N) = — cotan. N (94) sec. (180° — N) — — sec. N (95) cosec. (180° — N) = cosec. N; (96) that is, the sine and cosecant of the supplement of an angle are the same with those of the angle itself and the cosine, tangent, cotangent, and secant of the supple- ment are the negative of those of the angle. 61. Corollary. Since, when an angle is acute its supplement is obtuse, it follows from the preceding proposition, that the sine and cosecant of an obtuse angle are positive, while its cosine, tangent, cotangent, and secant are negative. This proposition must be carefully borne in mind in using the trigonometric tables, as it affords the means of discrimi- nating between the two angles which are given in B. Table XXVII, and of deciding which of these two angles is the required one. 62. Corollary. The preceding corollary might also have been obtained from (26) and (28). For by making M — 90°, we have by (59) and (60) sin, (90° + N) = cos. N (97) cos. (90° + N) = — sin. N\ (98) whence, by (6) and (7), tang. (90° + N) = — cotan. N (99) 4* 42 PLANE TRIGONOMETRY. [CH. V. Sine, &c. of negative angle. cotan. (90° + N) = — tang. N (100) sec. (90° + N) = — cosec. iV (101) cosec. (90° + N) = sec. iV; (102) that is, the sine and cosecant of an angle, which exceeds 90°, are equal to the cosine and secant of its excess above 90° 7 while its cosine, tangent, cotangent, and se- cant are equal to the negative of the sine, cotangent, tangent, and cosecant of this excess. 63. Problem. To find the sine, &c. of a negative angle. Solution, Make N = 0° in (27) and (29). They become, by means of (59) and (60), sin. ( 11 N) = — sin. N (103) cos. (-N) —cos.N (104) whence, from (6) and (7), tang. ( — N) — — tang. N (105) cotan. ( — - N) = — cotan. N (106) sec. ( — N) = sec. N (107) cosec. ( — N) — — cosec. N; (108) so that the cosine and secant of the negative of an angle are the same with those of the angle itself ; and the sine, tangent, cotangent, arid cosecant of the nega- tive of the angle are the negative of those of the angle. § 66.] PARTICULAR VALUES OF SINES, &C. 43 Sine, &c. of an angle greater than 180°. 64. Problem, To find the sine, Sfc. of an angle lohich exceeds 180°. Solution. Make M — 180° in (26) and (28). They be- come, by means of (65) and (66), sin. (180° + N) = — sin. N (109) cos. (180° + N) =z — cos. N (110) whence, from (6) and (7), tang. (180° + N) = tang. N (111) cotan. (180° + N) = cotan. N (112) sec. (180° + N) = — sec. N (113) cosec. (180° + N) = — cosec. 2V; (114) that is, the tangent and cotangent of an angle, which exceeds 180°, are equal to those of its excess above 180° ; and the sine, cosine, secant, and cosecant of this angle are the negative of those of its excess. 65. Corollary. If the excess of the angle above 180° is less than 90°, the angle is contained between 180° and 270° ; so that the tangent and cotangent of an angle which exceeds 180°, and is less than 270°, are positive ; while its sine, cosine, secant, and cosecant are negative. 66. Corollary* If the excess of the angle above 180° is greater than 90° and less than 180°, the angle is contained between 270° and 360°; so that, by § 64 and 61, the cosine and secant of an angle, which exceeds 44 PLANE TRIGONOMETRY. [CH. V. 270° and is less than 360°, is positive ; while its sine, tangent, cotangent, and cosecant are negative. 67. Corollary. The results of the two preceding corollaries might have been obtained from (27) and (29). For by mak- ing M = 360°, we have, by § 57, sin. (360° — N) = — sin. N (115) cos. (360° — N)— cos. N (116) whence, by (6) and (7), tang. (360° — N) = — tang. N (117) cotan.(360° — N) = — cotan. N (118) sec. (360° — N)=z sec. N (119) cosec. (360° — N) = — cosec. N (120) that is, the cosine and secant of an angle are the same with those of the remainder after subtracting the angle from 360° ; while its sine, tangent, cotangent, and co- secant are the negative of those of this remainder* 68. Problem. To find the sine, fyc. of an angle which exceeds 360°. Solution. Make M = 360° in (26) and (28;. They be- come, by means of (77) and (78), sin. (360° -f N) = sin. N (121) cos. (360° + N) = cos. JY (122) that is, the sine, fyc. of an angle which exceeds 360° are equal to those of its excess above 360°. <§> 70.] PARTICULAR VALUES OF SINES, &C. 45 Increase of sine, &c. of an acute angle. 69. Theorem. The sine, tangent, and secant of an acute angle increase with the increase of the angle ; the cosine, cotangent, and cosecant decrease. Proof. I. The excess of the sine of M -|- m over the sine of M is, by (13), equal to sin. m cos. M } which is a positive quantity when 31 is acute ; and, therefore, the sine of the acute angle increases with the increase of the angle. II. The excess of cos. M over cos. (M -{- m) is, by (15), equal to sin. m sin. M, which is a positive quantity ; and, therefore, the cosine of the acute angle decreases with the increase of the angle. III. The tangent of an angle is, by (7), the quotient of its sine divided by its cosine. It is, therefore, a fraction whose numerator increases with the increase of the angle, while its denominator decreases. Either of these changes in the terms of the fraction would increase its value ; and, therefore, the tangent of an acute angle increases with the increase of the angle. IV. The cosecant, secant, and cotangent of an angle are, by (6), the respective reciprocals of the sine, cosine, and tan- gent. But the reciprocal of a quantity increases with the decrease of the quantity, and the reverse. It follows, then, from the preceding demonstrations, that its secant increases with the increase of the acute angle, while its cosecant and cotangent decrease. 70. Theorem. The absolute values (neglecting their signs) of the sine, tangent, and secant of an obtuse 46 PLANE TRIGONOMETRY. [CH. V. Increase of sine, &c. of obtuse angle. angle decrease with the increase of the angle; while those of the cosine, cotangent, and cosecant increase. Proof. The supplement of an obtuse angle is an acute angle, of which the absolute values of the sine, &,c. are, by § 60, the same as those of the angle itself. But this acute angle decreases with the increase of the obtuse angle, and at the same time its sine, tangent, and secant decrease, while its cosine, cotangent, and cosecant increase. § 71.] OBLIQUE TRIANGLES. 47 Sides proportional to sines of opposite angles. CHAPTER VI. OBLIQUE TRIANGLES. 71. Theorem. The sides of a triangle are directly proportional to the sines of the opposite angles. [B. p. 13] Proof. In the triangle ABC (figs. 2 and 3), denote the sides opposite the angles A, B, C, respectively, by the letters a, b y c. We are to prove that sin. A : sin. B : sin. C — a :b : c. (123) From the vertex B, let fall on the opposite side the perpendic- ular BP, which we will denote by the letter p. Then, in the triangle BAP, we have by (1) Sm ' A = AB : =-c' or p =zc sin. A. 0^4) Also, in the triangle BPC, we have, by (1) and (91), and from the consideration that BCP is the angle C (fig. 2.), and its supplement (fig. 3.), BCP = BP p ~~ BC~ a' or p =za sin. C. (125) Comparing (124) and (125), we have c sin. A = a sin. C, 48 PLANE TRIGONOMETRY. [CH. VI. A side and two angles given. which may be converted into the following proportion sin. A : sin. C z= a : c. In the same way, it may be proved that sin. A : sin. B — a : b ; and these two proportions may be written in one as in (123). 72. Problem. To solve a triangle when one of its sides and two of its angles are known. [B. p. 41.] Solution. First. The third angle may be found by subtract- ing the sum of the two given angles from 180°. Secondly. To find either of the other sides, we have only to make use of a proportion, derived from § 71. As the sine of the angle opposite the given side is to the sine of the angle opposite the required side, so is the given side to the required side. Thus, if a (fig. 1.) were the given and b the required side, we should have the proportion sin. A : sin. B — a : b ; whence by (6) a sin. B . _ «-^l» b — — : -- ±= a sin. B cosec. A. (126) sin. A x ' 73. Examples. 1. Given one side of a triangle equal to 22.791, and the adjacent angles equal to 32° 41/ and 47° 54' ; to solve the triangle. Solution. The other angle •= 180° — (32° 41 7 + 47° 54') = 99° 25 7 . *74.] OBLIQUE TRIANGLES. 49 Given two sides and an angle opposite one of them. By (126) 99° 25' cosec. 10.00589 10.00589 32° 41' sin. 9.73239 47° 54' sin. 9.87039 22.791 1.35776 1.35776 12.475 *1.09604; 17.141 *i. 23404. Ans. The other angle == 99° 25' The other sides = { 12.475 17.141 2. Given one side of a triangle equal to 327.06, and the adjacent angles equal to 154° 22' and 17° 35' ; to solve the triangle. Ans. The other angle = 8° 3' The other sides -{ 1010.4 705.5 74. Problem. To solve a triangle when two of its sides and an angle opposite one of the given sides are known. [B. p. 42.] Solution. First. The angle opposite the other given side is found by the proportion of § 71. As the side opposite the given angle is to the other given side, so is the sine of the given angle to the sine of the required angle. Thus, if (fig. 1.) a and b are the given sides and A the given angle, the angle B is found, by the proportion * 20 is subtracted from each of these characteristics, because the two sines and cosecant were taken from the tables without any di- minution, as required by § 30. 5 50 PLANE TRIGONOMETRY. [CH. VI. Given two sides and an angle opposite one of them. a \b =. sin. A : sin. B ; whence S \n.B= bsm a A . (127) Secondly. The third angle is found by subtracting the sum of the two known angles from 180°. Thirdly. The third side is found by the proportion. As the sine of the given angle is to the sine of the angle opposite the required side, so is the side opposite the given angle to the required side. That is, in the present case, sin. A : sin. C = a : c ; whence , — a sin. c cosec. A. (128) sin. A 75. Scholium. Two angles are given in the tables corre- sponding to the same sine, which are supplements of each other, one being acute and the other obtuse. Two values of B (127) are then given in the tables, and both these values may be possible, when the given value of b is greater than that of a> and the given value of A is less than 90° ; for, in this case, there may be two triangles, ABC (fig. 11.) and AB'C, which satisfy the data. 76. Scholium. The problem is impossible, when the given value of b is greater than that of a, and the given value of A is obtuse. For the greatest side of an obtuse-angled triangle must always be opposite the obtuse angle. 77. Scholium. The problem is impossible, when the given value of b is so much greater than that of a, that we have § 79.] OBLIQ,UE TRIANGLES. 51 Given two sides and an angte opposite one of them. b sin. A > a; for, in this case, the given value of a is less than that of the perpendicular CP (fig. 11.), from C upon AP. 78. Scholium. The obtuse value of B does not satisfy the problem, when b is less than a; for the obtuse angle of a triangle cannot be opposite a smaller side. In this case, therefore, the problem admits of only one solution. 79. Examples. 1. Given two sides of a triangle equal to 77.245 and 92.341, and the angle opposite the first side equal to 55° 28' 12", to solve the triangle. Solution. Making b — 92-341, a = 77-245, A = 55° 28' 12", we have, by (127), a = 77-245 (ar. co.) 8-11213 b = 92-341 1-96540 A = 55° 28 ; 12" sin. 991584 B = 80 ; 1' or = 99° 59' sin. 9-99337 A + B = 135° 29' 12" or == 155° 27 ; 12" C = 44° 30' 48" or = 24° 32' 48" Then, by (128), 52 PLANE TRIGONOMETRY. [CH. VI. Given two sides and an angle opposite one of them. a = 77-245 1-88787 1-88787 C = 44° 30' 48" sin. 9-84576 or === 24° 32' 48" sin. 961850 A == 55°28 12"cosec. 1008416 1008416 C= 65-734 1-81779 or = 38 952 1-59053 Ans. The third side = 65-734 or = 38-952 {80° V i 99° 59' 44° 30' 48" ov — { 24° 32' 48" 2. Given two sides of a triangle equal to 77-245 and 92-341, and the angle opposite the second side equal to 55° 28' 12" ; to solve the triangle. Ans. The third side = 110-7 T , 4 , , ( 43° 3344" The other angles = j g()0 5g , 4 „ 3. Given two sides of a triangle equal to 40 and 50, and the angle opposite the first side equal to 45°, to solve the tri- angle. Ans. The third side = 54.061 or = 16-65 {62° 7' ( 72° W or = i 117° 53' 17° r 4. Given two sides of a triangle equal to 77-245 and 92*341, and the angle opposite the second side equal to 124° 31' 48", to solve the triangle. Ans. The third side == 23-129 43° 33' 44" The other angles = \ i i° ^4/ 28" $ 80.] OBLIQJTE TRIANGLES. 53 Ratio of the sum of the two sides to their difference. 5. Given two sides of a triangle equal to 77*245 and 92*341, and the angle opposite the first side equal to 124° 31' 48", to solve the triangle. Ans. The question is impossible. 6. Given two sides of a triangle equal to 75*486 and 92*341, and the angle opposite the first side equal to 55° 28' 12", to solve the triangle. Ans. The question is impossible. 80. Theorem. The sum of two sides of a triangle is to their difference, as the- tangent of half the sum of the opposite angles is to the tangent of half their difference. [B. p. 13.] Proof. We have (fig. 1.) a : b = sin. A : sin. B ; whence, by the theory of proportions, a -j- b : a — b = sin. A -f- sin. B : sin. A — sin. B t which, expressed fractionally, is a -f- b sin. A ~(- sin. B a — b ' sin. A — sin. B But, by (40), sin. A -f- sin. B tang. J (A -f- B) sin. A — sin. B tang. % (A — B) ' whence a + b __ tang. J (A + B) a — b tang, i (A — B) 5* (129) 54 PLANE TRIGONOMETRY. [cH. ft. Given two sides and the included angle. or a + b : a — 6 = tang. J (4 + B) : tang. £ (^4 — B). 81. Problem. To solve [a triangle when two of its sides and the included angle are given. [B. p. 43.] Solution. Let the two sides a and b (fig. 1.) be given, and the included angle C, to solve the triangle. First. To find the other two angles. Subtract the given angle C from 180°, and the remainder is the sum of A and B, for the sum of the three angles of a triangle is 180°, that is, A -f B = 180° — C, and £ (A + B) — 90° — £ C = complement of J C. The difference of A and 2? is then found by (129) a + 6:fl — b = tang. J (4 -f jB) : tang. % (A — B). But we have tang. J (^4 -|- B) z= cotan. J C; whence tang. J (.4— JB) = ^=|tang4(4+B)=^cotan.iG(180) The greater angle, which must be opposite the greater side, is then found by adding their half sum to their half difference ; and the smaller angle by subtracting the half difference from the half sum. Secondly. The third side is found by the proportion sin. A : sin. C =z a : c ; § 82.] OBLIQUE TRIANGLES. 55 Given two sides and the included angle. whence a sin. C sin. A 82. Examples. 1. Given two sides of a triangle equal to 99*341 and 1.234, and their included angle equal to 169° 58', to solve the tri- angle. Solution. Making a == 99.341, b = 1.234 ; and C =z 169° 58', J- C = 84° 59' ; we have, by (130), a + b — 100.575 (ar. co.) 7.99751 a — b — 98.107 ^1.99170 i (A + B) = 5° 1' tang. 8.94340 £ (A — B) = 4° 53' 39" tang. 8.93261 ^4=9° 54' 39" B =z 0° 7 / 21' a = 99.341 1.99712 C— 169° 58' sin. 9.24110 A = 9° 54' 39" cosec. 10.76416 c = 100.55 2.00238 Ans. The third side =: 100.55 T , .. , f 9° 54' 39' The other angles z= | QO ^ 21 „ 56 PLANE TRIGONOMETRY. [CH. VI. Segments of base made by perpendicular from opposite vertex. 2. Given two sides of a triangle equal to 0.121 and 5.421 and the included angle equal to 1° 2' ; to solve the triangle. Ans. The other side = 5.294 T , ,. . i 178° 56' 35" lhe other angles = t ~ 1/05// 83. Theorem. One side of a triangle is to the sum of the other two, as their difference is to the difference of the segments of the first side made by a perpendicu- lar from the opposite vertex, if the perpendicular fall within the triangle ; or to the sum of the distances of the extremities of the base from the foot of the perpen- dicular, if it fall without the triangle. [B. p. 14.] Proof Let AB (figs. 12 and 13) be the side of triangle ABC on which the perpendicular is let fall, and BP the per- pendicular. From 5asa centre with a radius equal to BC, the shorter of the other two sides, describe the circumference CC E'E. Produce AB to E 1 and ACto*C, if necessary. Then, since AC and AB are secants, we have, AC AE — AE : AC. But and AE' = AB + BE —AB + BC AE = AB — BE = AB — BC (fig. 12.) AC = AP — PC = AP — PC (fig. 13.) AC = AP + PC — AP + PC $ 85.] OBLIQ.UE TRIANGLES. 57 Given the three sides. whence (fig. 12.) AC : AB + BC = AB — BC: AP — PC (fig. 13.) AC : AB + BC = AB — BC : AP + PC 84. Problem. To solve a triangle when its three sides are given. [B. p. 43.] Solution. On the side b (figs. 2 and 3.) let fall the perpen- dicular BP. Then, by § 83, (fig. 2.) b : c + a = c — a : PA — PC (fig. 3.) b : c + a = c — a : P^l + PC. These proportions give the difference of the segments (fig. 2.), or their sum (fig. 3.). Then, adding the half difference to the half sum, we obtain the larger segment corresponding to the larger of the two sides a and c. And, subtracting the half difference from the half sum, we obtain the smaller segment. Then, in triangles BCP and ABP, we have, by (4) and (92), A AP cos. A = ; c PC and (fig. 2.) cos. C s= , P C (fig. 3.) cos. C = — cos. BCP — . The third angle B is found by subtracting the sum of A and C from 180°. 85. Corollary. From the preceding section, we have 58 PLANE TRIGONOMETRY. [CH. VI. Given the three sides. (fig. 2.) pa-pc= (« + «H«-g) = £lzi_«f (fig. 3.) pa + pc = (c + ")(c- zA = £l^_«! which, added to (% 2.) PA + PC: -AC = b (fig. 3.) PA — PC: -AC: - b gives 2Pi = c 2 — a* - + 6 = b 2 + c 2 — a 2 b b Hence PA= b2 + c2 - 26 and . PA &2 + c 2_ a 2 cos. A = — = -^_ (131) 86. Corollary. If (131) is cleared from fractions it becomes by transposition a 2 — b 2 + c 2 — 2 6c cos. ^4. (132) 87. Corollary. Add unity to both sides of (131) and we have 1 + cos. it * ^ mW - + 1 - 26 7 - ¥b~c (133) Since the numerator of (133) is the difference of two squares, it may be separated into two factors, and we have § 87.] OBLIQ.UE TRIANGLES. 59 Given the three sides. 14 . cosi _(H c + a)(b + c - a) 1 + cos. A _ 2^ • Now, representing half the sura of the three sides of a tri- angle by 5, we have 2 sz= a + b + c, (134) and 2s-2a = 2(5-«)=:fl + 6 +c— 2 a— b +c—a. (135) If we substitute these values in the above equation, it be- comes ii a 4 s (s — a) 2s (s — a, 1 + cos. A = — - — . (136) 2 6c be But, by (48), 1 + cos. A = 2 (cos. %A) 2 . Hence or (cos.J^)2=iiip^ (137) be cos.^=v( S(5 ~ a) )> (138) which corresponds to proposition LXI. of B. p. 14. In the same way, we have cos.iB = */(tilzd)\ (139) co M In the same way, we have si„.* B = v(^=4f-^) (146) riMC=V( ^ a ^'-* ) ). ,(147) $ 91.] OBLIQUE TRIANGLES. 61 Given the three sides. 89. Corollary. The quotients of (145, 146, and 147), di- vided by (138, 139, and 140), are by (7) m »»b = v( ( 't ( ;'.1V) 13.] PLANE SAILING. 71 Given distance and course. (fig. 15.) be the distance. Draw through A the meridian AC, and let fall on it the perpendicular BC. The angle A is the course, AC is the difference of latitude, and BC is the depart- ure. Then, by (17, and 18.) Diff. of lat. = dist. X cos. course. 0^2) Departure = dist. X sin. course. (153) Secondly. When the distance is great, as A B (fig. 16), then divide it into smaller portions, as A a, a b, b c, &c. Through the points of division, draw the meridians AN, an, bp, &c. Let fall the perpendiculars am, b n, cp, &c. Then, as the course is every where the same, each of the angles m A a, n a b, p b c, &c. is equal to the angle A, or the course. Moreover, the distances, A m, a n, b p, &,c. are the differences of latitude respectively of A and a, a and b, b and c, &c. Also a m, b n, c p, &c. are the departures of the points A and a, a and b, b and c, &,c. Therefore, as the difference of latitude of A and B is evidently equal to the sum of these partial differences of latitude ; and as the departure of A and B is by § 10 equal to the sum of the partial depar- tures, we have Diff. of lat. = Am-\-an-\-bp-\- &c. Departure = am-\-bn-{-cp-\- &c. But the right triangles m A a, n ab,p b c, &c. give by (152, and 153.) A m — A a X cos. course, a m = A a X sin. course ; a n = a b X cos. course, b n = a b X sin. course ; b p = b c X cos. course, cp = b c X sin. course. &c. &c. 72 NAVIGATION AND SURVEYING. [CH. I. Given course and departure. The sums of these equations give Diff. of lat. = i?a-)-«n + &c + &c = (Aa-\-ab-\-bc-\- &c.) X cos. course, Departure = a m -\- b n -\- c p ~\- &c. = (Aa-\-ab-{-bc-\- &c.) X sin. course. But Aa-\-ab-\-bc-\- &c. = AB = distance. Hence, Diff. of lat. = dist. X cos. course, Departure = dist. X sin. course ; precisely the same with (152) and (153). This shows that the method of calculating the dif- ference of latitude and departure is the same for all distances, and that all the problems of Plane Sailing may be solved by the right triangle (fig. 15.) [B. p. 52.] A table of difference and latitude and departure are given in pages 1-6, Tables I. and II. of the Navigator, which might be calculated by (152 and 153.) 14. Problem. To find the distance and difference of latitude, when the course and departure are known. [B. p. 55.] Solution. There are given (fig. 15.) the angle A and the side B C. Hence, by (19, and 20), Distance == departure X cosec. course. (154) Diff. of lat. = departure X cotan. course. (*55) § 18.] PLANE SAILING. 73 Cases of plane sailing. 15. Problem. To find the distance and departure, when the course and difference of latitude are known. [B. p. 55.] Solution. There are given (fig. 15.) the angle A and the side AC. Then, by (21, and 22). Distance = diff. of lat. X sec - course. (156*) Departure = diff. of lat. X tang, course. (157) 16. Problem. To find the course and difference of latitude, when the distance and departure are known. [B. p. 57.] Solution. There are given (fig. 15.) the hypothenuse AB and the side EC. Then, by (23, and 25), departure sin. course z= — -? , (15o) distance Diff. of lat. z= \/ [(dist.2) — (departure) 2 ]. (159) 17. Problem. To find the course and departure, ivhen the distance and difference of latitude are known. [B. p. 56.] Solution. There are given (fig. 15.) the hypothenuse AB and the leg AC. Then, by (23 and 25), diff of lat. cos. course = — , (160) distance. ' Departure = s/ [(dist.) 2 _ (diff. of lat.) 2 ]. (161) 18. Problem. To find the course and distance, when the departure and difference of latitude are known. [B. p. 57.] 7 74 NAVIGATION AND SURVEYING. [CH. I. Examples. Solution. There are given (fig. 15.) the legs AC and BC. Then, departure ,-tLL* tang, course ^ 5I ^___ (162) Dist. = diff. of lat. X sec. course. (163) 19. Examples. 1. A ship sails from latitude 3° 45' S., upon a course N. by E., a distance of 2345 miles ; to find the latitude at which it arrives, and the departure which it makes. Ans. Latitude — 34° 35' N. Departure =z 458 miles. 2. A ship sails from latitude 62° 19' N., upon a course W. N. W., till it makes a departure of 1000 miles ; to find the latitude at which it arrives, and the distance sailed. Ans. Latitude == 69° 13' N. Distance = 1082 miles. 3. The bearing of Paris from Athens is N. 54° 56' W. ; find the distance and departure of these two places from each other. Ans. Distance = 1135 miles. Departure = 929 miles. 4. A ship sails from latitude 72° 3' S. a distance of 2000 miles, upon a course between the north and the west, that is, northwesterly, until it makes a departure of 1000 miles ; find the latitude at which it arrives, and the course. § 19.] PLANE SAILING. 75 Examples. Ans. Latitude = 43° IT S. Course = N. 30° W. 5. The distance from New Orleans to Portland is 958 miles; find the bearing and departure. Ans. Bearing = N. 49° 24' E. Departure as 1257 miles. 6. The departure of Boston from Canton is 8790 miles; find the bearing and distance. Ans. Bearing as N. 82° 31' E. Distance = 8865 miles. 76 NAVIGATION AND SURVEYING. [CH. II. Traverse. CHAPTER II. TRAVERSE SAILING. 20. A traverse is an irregular track made by a ship when sailing on several different courses. The object of Traverse Sailing is to reduce a trav- erse to a single course, where the distances sailed are so small that the earth's surface may be considered as a plane. [B. p. 59.] 21. Problem. To reduce several successive tracks of a ship to one ; that is, to find the single track, leading to the place, which the ship leas actually reached, by sailing on a traverse. [B. p. 59.] Solution. Suppose the ship, to start from the point A (fig. 17.) and to sail, first from A to B, then from B to C, then from C to JEJ, and lastly from E to F ; to find the bearing and distance of F from A. Call the differences of latitude, cor- responding to the 1st, 2d, 3d, and 4th tracks, the 1st, 2d, 3d, and 4th differences of latitude ; and call the corresponding departures the 1st, 2d, 3d, and 4th departures. Then we need no demonstration to prove that, Diff. of lat. of A and P= 1st diff. of lat. — 2d diff. of lat. + 3d diff. of lat. — 4th diff. &c. ; or that the difference of latitude of A and F is found § 22.] TRAVERSE SAILING. 77 To reduce a traverse to a single course. by taking the sum of the differences of latitude corre- sponding to the northerly courses, and also the sum of those corresponding to the southerly courses, and the difference of these sums is the required difference of latitude. By neglecting the earth's curvature, we also have, Dep. of A and F = 1st dep. — 2d dep. —3d dep. + 4th dep. or the departure of A and F is found by taking the sum of the departures corresponding to the easterly courses, also the sum of those corresponding to the west- erly courses ; and the difference of these sums is the required departure. Having thus found the difference of latitude and de- parture of A and F, their distance and bearing are found by <§> 18. 22. The calculations of traverse sailing are usually put into a tabular form, as in the following example. In the first column of the table are the numbers of the courses ; in the second and third columns are the courses and distances ; in the fourth and fifth columns are the differences of latitude, the column, headed N, corresponding to the northerly courses, and that headed S, to the southerly courses; in the sixth and seventh columns are the departures, the column, headed E, cor- responding to the easterly courses, and that, headed W ; to the westerly courses. [B. p. 59.] 7* 78 NAVIGATION AND SURVEYING. [CH. 11. Examples. 23. Examples. 1. A ship sails on several successive tracks, in the order and with the courses and distances of the first three columns of the following table ; find the bearing and distance of the place at which the ship arrives, from that from which it started. No. Course. Dist. N. S. E. W. 1 N. N. E. 30 27.7 115 2 N. W. 80 56.6 56.6 3 West. 60 60.0 4 S. E. by S. 55 45.7 30.6 5 North. 43 43.0 6 S. by W. 152 149.1 194.8 42 1 29.7 Sum of col umns, 127.3 146.3 127.3 42.1 Diff. lat. = 67.5S.dep.=104.2W. Dep. = 104.2 2.01787 Diff. of lat = 67.5 (ar. co.) 8.17070 1.82930 Bearing = 57° 4' tang. 0.18857 sec. 0.26467 Dist. = 124 1 2.09397 Ans. Bearing = S. 57° 4' W. * Distance = 124.1 miles. 2. A ship sails on the following successive tracks, South 10 miles, W. S. W. 25 miles, S. W. 30 miles, and West 20 miles ; it is bound to a port which is at a distance of 100 miles from the place of starting, and its bearing is W. by N. § 23,] PLANE SAILING. 79 Examples. Required the bearing and distance of the port to which the ship is bound, from the place at which it has arrived. Ans. Bearing =5 S. 51° 47' W. Distance = 239 miles. 80 NAVIGATION AND SURVEYING. [CH. III. Differences of longitude in parallel sailing. CHAPTER III. PARALLEL SAILING. 24. Parallel Sailing considers only the case where the ship sails exactly east or west, and therefore re- mains constantly on the same parallel of latitude. Its object is to find the change in longitude corresponding to the ship's track. [B. p. 63.] 25. Problem. To find the difference of longitude in parallel sailing. [B. p. 65.] Solution. Let AB (fig. 18.) be the distance sailed by the ship on the parallel of latitude A B. As the course is exactly east or west, the distance sailed must be itself equal to its de- parture. The latitude of the parallel is AD A' or A A 1 . The angle AEB = A'D B 1 , or the arc A' B', is the difference of lon- gitude. Denote the radius of the earth A'D = B'D = A D by JR, and the radius of the parallel AE = BE by r ; also the circumference of the earth by C, and that of the parallel by c. Since AB and A'B' correspond to the equal angles AE B and A'D B\ they must be similar arcs, and give the propor- tion, AB :AB' = c : C, or Dep. : diff. long. = c : G <§> 27.] PARALLEL SAILING. 81 Difference of longitude. But, as circumferences are proportional to their radii, c : C=r :R. Hence, leaving out the common ratio, Dep. : cliff, long. = r : JR. Putting the product of the extremes equal to that of the means, r. diff. of long. = R. departure. But, in the triangle AD E, since DAE=ADA= latitude, we have, from (17), r = R X cos. lat. which, substituted in the above equation, gives, if the result is divided by R, Diff. long. X cos. lat. = departure. (1°*4) Hence, by (8), Diff. long, = ^R^™ = dep. X sec. lat. (165) cos. lat. r x ' 26. Corollary. Since the distance is the same as the depar- ture in parallel sailing. The word distance may be substituted for departure in (164) and (165). 27. Corollary. It appears, from (164) and (165), that if a right triangle (fig. 18.) is constructed, the hypothe- nuse of which is the difference of longitude, and one of the acute angles the latitude, the leg adjacent to this angle is the departure. All the cases of parallel sailing may , then, be reduced to the solution of this triangle. 82 NAVIGATION AND SURVEYING. [CH. III. Differences of places on the same parallel. 28. Problem. To find the distance between two places which are upon the same parallel of latitude. Solution. This problem is solved at once by (164). 29. The Table, p. 64, of the Navigator, which " shows for every degree of latitude how many miles distant two meridians are, whose difference of longitude is one degree," is readily calculated by this problem. 30. Examples. 1. A ship sails from Boston 1000 miles exactly east ; find the longitude at which it arrives. Ans. Longitude sought = 51° 48' W. 2. Find the distance of Barcelona (Spain) from Nantucket (Massachusetts). Ans. Distance = 3252 miles. 3. Find the distance between two meridians, whose differ- ence of longitude is one degree in the latitude of 45°. Ans. Distance = 42.43 miles. § 32.] MIDDLE LATITUDE SAILING. 83 Middle latitude. CHAPTER IV. MIDDLE LATITUDE SAILING. 31. The object of Middle Latitude Sailing is to give an approximative method of calculating the difference of longitude, when the difference of latitude is small. [B. p. 66.] 32. Problem. To find the difference of longitude by Middle Latitude Sailing, when the distance and course are known, and also the latitude of either extremity of the ship's track. [B. p. 71.] Solution. The difference of latitude and departure are found by (152) and (153), Diff. lat. = dist. X cos. course Departure = dist. X sin. course. The difference of longitude may then be found by means of (165). But there is a difficulty with regard to the latitude to be used in (165); for, of the two extremities of the ship's track, the latitude of one is smaller, while the latitude of the other extremity is larger than the latitude of the rest of the track. Navigators have evaded this difficulty by using the Middle Latitude between the two, as sufficiently accurate, when the difference of latitude is small. Now the middle lat- itude is the arithmetical mean between the latitudes of the extremities, so that we have, 84 NAVIGATION AND SURVEYING. [CH. IV. To find the difference of longitude. Middle lat. = J sum of the lats. of the extremities of the track ; (166) and, by (165), departure DifF. long, b E-— — == dep. X sec. mid. lat. (167) & cos. mid. lat. ' or, by substituting (153), DifF. long. = dist. X sin. course X sec. mid. lat. (168) " This method of calculating the difference of longitude may be rendered perfectly accurate by applying to the middle lat- itude a correction," which is given in the Navigator, and the method of computing, which will be explained in the suc- ceeding chapter. [B. p. 76,] 33. By combining the triangle (fig. 16.) of Plane sailing with that (fig. 18.) of Parallel sailing a triangle (fig. 19.) is obtained, by which all the cases of Middle Latitude sailing may be solved. 34. Problem. To find the distance and bearing of two places from each other, when their latitudes and longitudes are known. [B. p. 68.] Solution. From (fig. 19) we have Departure = diff. long. X cos. mid. lat. (169) . . . departure tang, bearing = ^^^- (170) dist. == diff. lat. X sec. bearing. (J 71) § 38.] MIDDLE LATITUDE SAILING. 85 Cases of middle latitude sailing. 35. Problem. To find the course, distance, and dif- ference of longitude, when both latitudes and the depar- ture are given. [B. p. 70.] Solution. The difference of longitude is found by (167), the course by (170), and the distance by (171). 36. Problem. To find the departure, distance, and difference of longitude, when both latitudes and the course are given, [B. p. 72.] Solution. The departure is found by the formula departure — diff. lat. X tang, course; (172) the distance by (171); and the difference of longitude may be found by (167), or by substituting (172) in (167) diff. long. z=z diff. lat. X tang, course X sec. mid. lat. (173) 37. Problem. To find the course, departure, and dif- ference of longitude, when both latitudes and the distance are given. [B. p. 73.] Solution. The course is found by the formula diff. lat. /t _ cos. course == — ; (174) dist. v ' the departure by departure zsz dist. X sin. course; (175) and the difference of longitude by (167). 38. Problem. To find the difference of latitude, dis- 8 86 NAVIGATION AND SURVEYING. [CH. IV. Examples. tance, and difference of longitude, when one latitude, course, and departure are given. [B. p. 74.] Solution. The difference of latitude is found by the formula diff. lat, = dep. X cotan. course; (176) the distance by the formula dist. = dep. X cosec. course; (177) and the difference of longitude by (167). 39. Problem. To find the course, difference of lati- tude, and difference of longitude, when one latitude, the distance, and departure are given.'] B. p. 75.] Solution. The course is found by the formula sin. course ±z ,. ; (178) dist. v the difference of latitude by the formula diff. lat. = dist. X cos. course; (179) and the difference of longitude by (167). 40. Examples. 1. A ship sailed from Halifax (Nova Scotia) a distance of 2509 miles, upon a course S. 79° 34' E. ; find the place at which it arrived. Solution. By § 32, <§> 40.] MIDDLE LATITUDE SAILING. 87 Examples. dist. = 2509 3-39950 3.39950 course == 79° 34' cos. 9.25790 sin. 9.99276 diff. lat. = 454' = 7° 34' S. 2.65740 given lat. — 44° 36' N. mid. lat. = 40° 49' required lat. =37° 2'N. cor. — 7' cor. mid. lat. =40° 56' sin. 10.12178 diff. long. == 3266' — 54° 26' E. 3.51404 given long. = 63° 28' W. required long. tk 9° 2' W. Ans. The place arrived at is one mile south of Cape St. Vincent in Portugal. 2. Find the bearing and distance of Canton from Washing- ton. Solution. By § 34, lat.ofWashington^SS^'N. long. = 77° 3 W. lat. of Canton =23° 7'N. long. =: 113° 14' E. diff. lat. = 946' — 15° 46, sum of longs. = 190° 17' mid. lat. = 31° 0' diff. long. = 169° 43=10183 / cor. = 31' cor. mid. lat. = 31° 31' cos. 9.93069 diff. long. = 10183' 4.00788 diff. lat. = 946' ar.co. 7.02411 2.97589 bearing = S. 83° 47' W. tang. 10.96268 sec. 10.96526 dist. — 8733 miles 3.94115 NAVIGATION AND SURVEYING. [CH. 1V« Examples. 3. A ship sails from New York a distance of 675J miles, upon a course S. E. £ S. ; find the place at which it arrives. Ans. Three miles to the west of Georgetown in Bermuda. 4. Find the bearing and distance of Portland (Maine) from New Orleans. Ans. The bearing — N. 49° 25' E. The distance sc 1257 miles. 5. A ship from the Cape of Good Hope sails northwesterly until its latitude is 22° 3' S., and its departure 3110 miles I find its course, distance sailed, longitude, and its distance from Cape St. Thomas (Brazil). Ans. Course = N. 76° 38' W. Distance — 3197 miles. Longitude r= 18° W. Distance to the Cape St. Thomas z= 22 miles. 6. A ship sails from Boston upon a course E. by N. until it arrives in latitude 45° 21' N. ; find the distance, its longitude, and its distance and bearing from Liverpool. Ans. Distance sailed = 920 miles. Longitude — 50° 10' W. Distance from Liverpool = 1905 miles Bearing from Liverpool =: S. 75° 22' W. 7. A ship sails southwesterly from Gibraltar a distance of 1500 miles, when it is in latitude 14° 44' N. ; find its course and longitude and distance from Cape Verde. $ 40.] MIDDLE LATITUDE SAILING. 89 Examples. Ans. Course = S. 37° 21' W. Longitude = 18° 3' W. Dist. from Cape Verde =r 339 miles. 8. A ship sails from Nantucket upon a course S. 62° ll 7 E., until its departure is 2274 miles ; find the distance sailed, and the place arrived at. Ans. Distance s= 2571 miles. The place arrived at is 261 miles north of Santa Cruz. 9. A ship sails southwesterly from Land's End (England), a distance of 3461 miles, when its departure is 3300 miles ; find the course and the place arrived at. Ans. The course = S. 72° 27' W. The place arrived at is Charleston (South Carolina) Light House. 8* 90 NAVIGATION AND SURVEYING. [CH. V. To find the difference of longitude. CHAPTER V. MERCATOR S SAILING. 41. The object of M creator's Sailing is to give an accurate method of calculating the difference of longi- tude. [B. p. 78.] 42. Problem. To find the difference of longitude, when the distance, the course, and one latitude are known. Solution. Let A B (fig. 16) be the ship's track. Divide it into the small portions A a, ab, be, &c, which are such that the difference of longitude is the same for each of them, and let d — this small difference in longitude. Let also L — the latitude of A, L 1 — the latitude of B, I — the latitude of one of the points of division as b, I' = the latitude of the next point c, C = the course. The distance b c may then be supposed so small, that the for- mulas of middle latitude sailing may be applied to it ; and (173) gives § 42.] mercator's sailing. 91 Difference of longitude. dz=(l'—l)X tang. C X sec. J (/' + l) t (180) or i d cotan. C= — ^i r ~? A ' < 181 ) cos, J (/ + If, now, the mile is adopted as the unit of length, and if R = the earth's radius in miles, (182) J7~ — is the leugth of the arc J (V — Z), expressed in terms of the radius as unity ; and since this arc is very small, its length is by § 22 equal to its sine or i{l '~ l) = sin. i(l'-l); ( 183) which substituted in (181) gives d cotan. C _ sin. £ {I 1 — Z) (184) (185) 2JR cos. \ (/' + *) Let now d cotan. C sin. J (Z' — Z ) ~~2BT = co¥. £(/' + Z) ; and (185) may be written in the usual form of a proportion sin. i (I' — Z) : cos. £ (Z' + Z) = m : 1 ; (186) whence, by the theory of proportions cos, j (V + I) + sin. £ (/' — /) __ 1 + i» cos. £ (/' + Z) — sin. i (Z' — I) ~~ \—m (187) But if in (40) we put 4 = 90°- l(l' + l),B = £ (I' — I); (188) we have 92 NAVIGATION AND SURVEYING. [CH. V. Difference of longitude. A + B = 90° — I, A — B = 90° — /', (189) and (46) becomes cos, i (f + /) + bid, j (/' — /) = cotan. (45° — jf) . cos. i (/' + /) — sin. J (/' — I) cotan.(45° — £1)' K ' whence, if we put ar=I±A (i9i) 1 — m colzn.JW.-in =M (1W) cotan. (45° — i / ) Hence, the successive values of cotan. (45° — £ I) at the points A y a, 6, &c, form a geometric progression; and if D = the difference of longitude of A and B, n = the number of portions of AB ; we have by (185) n a. 5. ==* g s ^ -, (193) d 2 Rm tang. C v ' and by the theory of geometric progression cotan. (45° — £ L 1 ) == cotan. (45° — £ L) W, (194) and by logarithms log. cotan. (45°— ££.')— lo S- cotan - ( 45 ° — lL)=\°g- M n . ( 195) If, lastly, we put e = M&% (196) we have M n s*e**H& (197) § 43.] mercator's sailing. • 93 Meridional difference of latitude. which substituted in (195) gives by a simple reduction \j£i 1 °g- cotan -( 45 °-i^ / )- 1 ^ log.cotan.(45°-jZ)] X tang. C = D (199) Now the value of— «— log. cotan. (45° — J L) has been calculated for every mile of latitude, and inserted in tables. [B. Table III.] It is called the Meridional Parts of the Latitude, and the method of computing it- is given in the following section. The difference between the meridional parts of the two latitudes, when the latitudes are both north or both south, is called the Meridional Difference of Latitude ; but when one of the latitudes is north and the other south, the sum of the meridional parts is the meridional difference of latitude. Hence (199) gives D z= diff. long. = mer. diff. lat. X tang course. (200) 43. Corollary. The difference of longitude is as in (fig. 20.) the leg DE of a right triangle, of which AD is the meridional difference of latitude, and the angle A the course ; and by combining this triangle with the triangle ABC of plane sailing, all the cases of Mer ca- tor's Sailing are reduced to the solution of these two similar right triangles. 94 • NAVIGATION AND SURVEYING. [CH. V. Table of meridional parts. 44. Problem. To calculate the Table of Meridional Parts. Solution. I. The value of R is found from the considera- tion that if n = the ratio of a circumference to its diameter = 3.1416, (201) we have, 2 TV R — the circumference of the earth = 360° = 21600' and R = 3^- m 3437.7. (202) II. In finding the value of e, the portions of the distance are supposed to be infinitely small, hence m is by (185) also infinitely small, and its reciprocal is infinitely great. If I -\- mis divided by 1 — m, as follows, l_ w )l4-m(l+2m + 2m 2 + &c. 1 — m + 2m 2 m — 2 m 2 + 2m 2 2 m 2 —2 m 3 + 2m 3 we have by (191) M= 1 + 2 m + 2 m 2 + &c. (203) But since m is infinitely small, m 2 , m 3 , &c. are infinitely M4.] MERCATOR S SAILING. 95 Table of meridional parts. smaller, and the error of rejecting them in (203) is less than any assignable quantity ; whence if/=l+2m and by the binomial theorem, and (196), i e = (1 +2ffl) s (204) (2 m) 2 2 m \ 2 m / \ 2 m / 1 . 2 . 3 ' v ' 1 But -— 2 hi ble error, 1 2m is infinite and gives therefore, without any assigna- l=-^-,J- -2= A-.&c- (206) 2 m ' 2 m which, substituted in (205), gives 6=1 2m .2> I (2m) s (2 m) s + &c. "(2m) 2 1.2 ' (2m) 3 1.2.3 + &c. (207) = i + i + _i_ j L 1 1 ^ ^1.2^1.2.3^1.2.3.4 so that e is the sum of a series of terms, the first of which is unity ) and each succeeding term is obtained by dividing the 'preceding term by the place of this pre- ceding term. 96 NAVIGATION AND SURVEYING. [CH. Neperian logarithms. Table of meridional parts. The value of e is thus computed. i) 1 . 000000 2) 1 . 000000 3) . 500000 4) . 166667 5) . 041667 6) . 008333 7) .001389 8) . 000198 9) . 000025 . 000003 e — 2.71828 (208) The sixth place of the value of e is neglected as inaccurate. This value of e is remarkable, as being the base of the sys- tem of logarithms invented by Neper, and which are called the Neperian or hyperbolic logarithms. III. The values of R (202) and e (208) give R 3437.7 _ 3437.7 _ ToiTe ~~ log. (2.71828) - 0.43429 ~ ' y15 ' 7 ' (Z ™> so that we have by (199) Mer. parts of L = 7915.7 log. co-tan. (45° — | L), (210) which agrees with the explanation of Table III. given in the Preface to the Navigator. <§> 46.] mercator's sailing. 97 Correction for middle latitude. 45. Examples. 1. Calculate the meridional parts for latitude^45° 48'. Solution. 2)45° 48' 45° — i L = 45° -- 22° 54' = 22° 6' 22° 6' log. cotan. 0.39141 log 9.59263 7915.7 3.89849 mer. parts of 45° 48' := 3098 3.49112 2. Calculate the meridional parts of latitude 28° 14'. Am. 1767. 3. Calculate the meridional parts of latitude 83° 59 / . Am. 10127. 46. Problem. To calculate the correction for middle latitude sailing. Solution. If the angle DBC (fig. 19.) were exactly what it should be in order that the hypothenuse BD should be the difference of longitude, and the leg JBCthe departure, it would be the corrected middle latitude, and we should have diff. long. =sec. cor. mid. lat. X departure z= sec. cor. mid. Iat. X diff. lat. X tang, course, (211) which, compared with (200) gives, by dividing by tang, course, mer. diff. lat. = sec. cor. mid. lat. X diff. lat. (212) 9 98 NAVIGATION AND SURVEYING. [CH. V. Correction for middle latitude. • i » m er. cliff, lat. «*.«* whence sec. cor. mid. lat. = — — — . (213) diff. lat. ' If, from the corrected middle latitude, calculated by this formula,® the actual middle latitude is subtracted, the correction of the middle latitude is obtained, as in the table on p. 76 of the Navigator. The meridional difference of latitude should be obtained for these cal- culations, not from the tables of meridional parts, but directly from the tables of logarithmic sines, &c. by means of (209) ; and when the difference of latitude is less than 14°, tables should be used in which the loga- rithrris are given to seven places of decimals. The following examples are, for the convenience of the learner, limited to cases for which the common tables are sufficiently accurate. 47. Examples. 1. Find the correction for middle latitude sailing, when the middle latitude is 35°, and the difference of latitude 14°. Solution. Greater lat. — 35° + 7° — 42° Less lat. — 35° — 7° = 28° 45° _ £ gr . lat. = 24° cotan. 0.35142 45° — J less lat. = 31° cotan. 0.22123 0.13019 log. 9.11458 7915.7 3.89849 diff. lat. = 840' ar. co. 7.07572 corrected mid. lat. = 35° 24' sec. 10.08879 correction = 35° 24' — 35° = 24'. ^ 51.] mercator's sailing. 99 Cases of Mercator's sailing. 2. Find the correction for middle latitude sailing, when the middle latitude is 60°, and the difference of latitude 16°. Ans. 46'. 3. Find the correction for middle latitude sailing, when the middle latitude is 72°, and the difference of latitude 20°. Ans. 124'. 48. Problem. To find the bearing and distance of two given places. [B. p. 79.] Solution. We have by (fig. 20.) for the bearing, diff. long. ..,.. tang, bearing = ,._ , - , (214) 5 B mer. diff. lat. v ' and the distance is found by (171). 49. Problem. To find the course, distance, and dif- ference of longitude, when both latitudes and the depar- ture are given. [B. p. 80.] Solution. The course is found by (170), the difference of longitude by (200), and the distance by (171). 50. Problem. To find the distance and difference of longitude, when both latitudes and the course are given. [B. p. 82.] Solution. The distance is found by (171), and the differ- ence of longitude by (200). 51. Problem. To find the course and difference of longitude, when both latitudes and the distance are given. [B. p. 83.] Solution. The course is found by (174), and the difference of longitude by (200). 100 NAVIGATION AND SURVEYING. [cH. V. Examples. 52. Problem. To find the distance, the difference of latitude, and the difference of longitude, when one lati- tude, the course, and departure are given, [B. p. 84.] Solution The distance is found by (177), the difference of latitude by (176), and the difference of longitude by (200). 53. Problem, To find the course, the difference of lati- tude, and the difference of longitude, when one latitude, the distance, and the departure are given. [B. p. 85.] Solution, The course is found by (178), the difference of latitude by (179), and the difference of longitude by (200), or by the following proportion deduced from the similar triangles of (fig. 20.) Diff. lat. : dep. =z mer. diff. lat. : diff. long, (215) 54. Examples. 1. A ship sails from Boston a distance of 6747 miles, upon a course S. 46° 59J' E. ; to find the place at which it arrives. Solution. Dist. z=z 6747 3.82911 Course = 46° 59£' cos. 9.83385 tang. 10.03022 Diff. lat.=76° 42' S.=4602 / , 3.66296 mer. d. I. =5007, 3.69958 Lat. Ieft = 42 -21'N. mer. p. 2810 Lat in = 34° 21' S. 2197 diff: long. = 5368' 3.72980 =89°28 / E. mer. diff. lat. — 5007 long, left .= 71° 5' W. long, in 1829' W. § 54.] mercator's sailing. 101 Examples. Ans. The place arrived at is the Cape of Good Hope. 2. Find the bearing and distance from Moscow to St. Helena. Solution. Moscow, lat. 55° 46' N. mer. parts 4049 long. 37.° 33' E. St. Helena, lat. 1 5° 55' S. mer. parts 968 long. 5° 36' W. Diff. lat. — 71° 41/ mer. diff. lat. 5017 diff.l. — 43° 9' af 4301' = 2589' Mer. diff. lat. = 5017 (ar.co.) 6.29956 diff. lonn c 2 sin. ^4 sin. JB #««*% area of ABC = -___-. (217) 57. Problem. To find the area of a triangular field, when two of its sides and the included angle are known. Solution. Let ABC (fig. 2.) be the triangle to be measured, 6 and c the given sides, and A the given angle. Then, by (216), area of ABC — £ b p, and, by (124), p zz: c sin. A. Hence area of ABC = } b c sin. A. (218) or, the area of a triangle is equal to half the continued product of two of its sides and the sine of the included angle. 58. Problem. To find the area of a triangular field, when its three sides are known. Solution. Let ABC (fig. 1.) be the given triangle. Then, by (218), area of ABC z=z £ b c sin. A ; but, by (151), sin. A = i W['(«-*)('-*) ('-«)] b c in which s denotes the half sum of the three sides of the tri- angle. § 59.] SURVEYING. 105 Area of triangle. Hence b c sin. A z= 2 a/[s (s — a) (s — b) (s — c)] ; and area of ABC = s/[s (s — a) (s — b) (s — c)] ; (219) or, to find the area of a triangular field, subtract each side separately from the half stun of the sides ; and the square 7 % oot of the continued 'product of the half sum and the three remainders is the required area. 59. Examples. 1. Given the three sides of a triangular field, equal to 45.56 ch., 52.98 ch., and 61.22 ch. ; to find its area. Solution. In (fig. 1.) let a = 45.56 ch., b =1 52.98 ch., c = 61.22 ch. 2 5= 159.76 ch. s z= 79.88 ch. 1.90244 s — a = 34.32 ch. 1.53555 s — b= 26.90 ch. 1.42975 5 — c=z 18.66 ch. 1.27091 6.13865 Area of ABC— 1 173. 1 sq. ch. 3.06932 Ans. The area = 117 A. 1 R. 9 r. 2. Given the three sides of a triangular field equal to 32.56 ch., 57.84 ch., and 44.44 ch. ; to find its area. Ans. The area = 71 A. 3 R. 29 r. 106 NAVIGATION AND SURVEYING. [CH. VI. Area of rectilinear field. 3. Given one side of a triangular field equal to 17.35 ch., and the adjacent angles equal to 100° and 70° ; to find its area. Ans. The area = 85 A. 3 R. 16 r. 4. Given two sides of a triangular field equal to 12.34 ch. and 17.97 ch., and the included angle equal to 44° 56' ; to find its area. Ans. The area = 7 A. 3 R. 13 r. 60. Problem. To find the area of an irregular field bounded by straight lines. First Method of Solution. Divide the field into triangles in any manner best suited to the nature of the ground. Measure all those sides and angles which can be measured conveniently, remembering that three parts of each triangle, one of which is a side, must be known to determine it. But it is desirable to measure more than three parts of each triangle, when it can be done ; because the comparison of them with each other will often serve to correct the errors of observation. Thus, if the three angles were measured, and their sum found to differ from 180°, it would show there was an error ; and the error, if small, might be divided between the angles; but if the error was large, it would show the ob- servations were inaccurate, and must be taken again. The area of each triangle is to be calculated by one of the preceding formulas, and the sum of the areas of the triangles is the area of the whole field. This method of solution is general, and may be ap- plied to surfaces of any extent, provided each triangle <§> 60.] SURVEYING. 107 Rectangular surveying. is so small as not to be affected by the earth's curva- ture. Second Method of Solution. Let ABCEFH (fig. 21.) be the field to be measured. Starting from its most easterly or its most westerly point, the point A for instance, measure succes- sively round the field the bearings and lengths of all its sides. Through A draw the meridian NS, on which let fall the perpendiculars BB, CO, EE ", FF', and HH'. Also draw CB'E', EF", and HF" parallel to NS. Then the area of the required field is ABCEFH =z AC'CEFF' — [AC'CB + AHFF 1 ]. But AC'CEFF' = C'CEE' + E'EFF' ; and AC'CB + AHFF' = C'CBB' + BBA + AHH' + HHFF 1 . Hence ABCEFH = [C'CEE' + EEFF] — [C'CBB' + BBA + AHH' + H'HFF] ; or doubling and changing a very little the order of the terms, 2 ABCEFH — [2 C'CEE + 2 E'EFF'] — £ [2 BBA + 2 CC'BB' + 2 HHFF' + 2 AHH']. > (22 ° } Again, 2 5 , jB4 = BB' X -4B' 2 CC'BB' =z {BB' + CC) X B'C 2 C'CjEjET = (EE' + CC) X EC" . 2 EEFF' = (EE' + JLF'J X JB'*" ^ 2 HHFF' = (Hi/' + JFF") X i? 7 ^' 2 jUTtf' = i?i/' X -4H 7 . 108 NAVIGATION AND SURVEYING. [CH. VI. Rectangular surveying. So that the determination of the required area is now reduced to the calculation of the several lines in the second members of (221.) But the rest of the solution may be more easily comprehended by means of the following table, which*is pre- cisely similar in its arrangement to the table actually used by surveyors, when calculating areas by this process. Sides AB N. s. E. W. Dep. Sum. N. Areas. S. Areas. AB BB' BB' BB' BB'A BC BC BB' CC BB'+CC CC'BB' CE CE' EE' EE' CC' + EE' CCEE' EF EF' pp// FF' EE'+FF' EEFF' FH FH' FF 7 " HH' FF'-f- HH' H'HFF' HA HA Hir O HH' AHH' In the first column of the table are the successive sides of the field. In the second and third columns are the differences of latitude of the several sides, the column headed N, corresponding to the sides running in a northerly direc- tion, and that headed S, corresponding to those running in a southerly direction. These two columns are calculated by the formula Diff. lat. — dist. X cos. bearing. In the fourth and fifth columns are the departures of the several sides ; the column headed E, corresponding to the sides running in an easterly direction, and that headed W, to those running in a westerly direction. § 60.] SURVEYING. 109 Rectangular survey. These two columns are calculated by the formula Departure s= dist. X sin. bearing. In the sixth column, headed Departure, are the depar- tures of the several vertices, which end each side of the field from the vertex A. This column is calculated from the two columns E and W, in the following man- ner. The first number in column Departure is the same as the first in the two columns E and W; and every other number in column Departure is obtained by adding the corresponding number in columns E and W, if it is of the same column with the first number in those two columns, to the previous number in column Departure ; and by subtracting it, if it is of a different column. Thus, BB' == BB> CC == BB" =2 BB' — BB" EE 1 3= E'E" + EE" =. CC + EE" FF' = F'F" + FF" ' = EE + FF" HH =: F'F'" = FF' FF'" O =: HH' — HH'. In the seventh column, headed Sum, are the first factors of the second members of (221). This column is calculated from column Departure in the following manner. The first number in column Sum is the same as the first in column Departure ; every other number in column Sum is the sum of the corresponding num- 10 110 NAVIGATION AND SURVEYING. [CH. VI. Rectangular survey. ber in column Departure added to the previous number in column Departure, as is evident from simple inspec- tion. In the eighth and ninth columns are the values of the areas, which compose the first members of (221). These columns are calculated by multiplying the mem- bers in column Sum by the corresponding numbers in columns N and S, which contain the second factors of the second members of (221). The products are writ- ten in the column of North Areas, when the second fac- tors are taken from column N, and in that of South Areas, when the second factors are taken from col- umn S. If we compare the columns of North and South Areas with (220), we find that all those areas, which are preceded by the negative sign, are the same with those in the column of North Areas ; while all those, which are connected with the positive sign, belong to the column of South Areas. To obtain, therefore, the value of the second member of (220), that is, of double the required area, we have only to find the difference between the sums of the columns of North and South Areas. [B. p. 107.] 61. Corollary. The columns N, S, E, and W, are those which would be calculated in Traverse Sailing, if a ship wa s supposed to start from the point A, and proceed round the sides of the field till it returned to the point A. The differ- ence of the sums of columns N and S is, then, by traverse § 62.] SURVEYING. ill. Correction of errors. sailing, the difference of latitude between the point from which the ship starts, and the point at which it arrives; and the difference of columns E and W is the departure of the same two points. But as both the points are here the same, their difference of latitude and their departure must be nothing, or Sum of column N = sum of column S ; Sum of column E = sum of column W. But when, as is almost always the case, the sums of these columns differ from each other, the difference must arise from errors of observation. If the error is great, new observations must be taken ; but if it is small, it may be divided among the sides by the following proportion. The sum of the sides : each side = whole error : error corresponding to each side. The errors corresponding to the sides are then to be subtracted from the differences of latitude, or the de- partures which are in the larger column, and added to to those which are in the smaller column. 62. Examples. 1. Given the bearings and lengths of the sides of a field, as in the three first columns of the following table ; to find its area. Solution. The table is computed by § 60. 112 NAVIGATION AND SURVEYING. [CH. 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CJ 5 o 9 o o 6 w 5 i— < CM T}< © l-H OS CO CM CM CO XO rH '- , OS i> bb 1 2 CM CO CO O Q. © a © CM E m W o CM CM n s? £ co* GO CO* CO* fc d rH CM CO -* 1- • t- CO *63.] SURVEYING. 113 Area of irregular field. 2. Given the lengths and bearings of the sides of a field, as in the following table ; to find its area. No. Bearings. Dist. 1 N. 17° E. 25 ch. 2 East. 28 ch. 3 South. 54 ch. 4 S. 4° W. 22 ch. 5 N. 33° W. 62 ch. Ans. The area = 173 A. R. 36 r. 63. Problem. To find the area of a field bounded by sides, irregularly curved. Solution. Let ABCEFHIKL (fig. 22.) be the field to be measured, the boundary ABCEFHIKL being irregularly curved. Take any points C and F, so that by joining AC, CF, and FL, the field ACFL y bounded by straight lines, may not differ much from the given field. Find the area of ACFL, by either of the preceding meth- ods, and then measure the parts included between the curved and the straight sides by the following method of offsetts. Take the points a, 6, c, d, so that the lines A a, a b, b c, cdy dC may be sensibly straight. Let fall on AC the per- pendiculars a a 1 , bb 1 , cc' t dd'. Measure these perpendiculars, and also the distances Aa' 9 b'c', b'c' 9 c'd', d'C. 10* 114 NAVIGATION AND SURVEYING. [cH. VI. Area of irregular field. The triangles Acta', Cdd 1 , and the trapeziums aba'b', bcb'c', cdc'd are then easily calculated, and their sum is the area of ABC. In the same way may the areas of CEF, FHI, and IKL be calculated ; and then the required area is found by the equation ABCEFH1KL =s ACFL — ABC + CEF + FHI — IKL. Example. Given (fig. 22.) A a' = 5 ch., a'b' = 2 ch., b 1 c' = 6 ch., e'tf = 1 ch., d'C = 4 ch. ; also a a' £s 3 ch., bb' = 2 ch. } cc' = 25 ch., rfrf' z= 1 ch. ; to find the area of ABC. Arts. Required area z:2A. 3 R. 36 r. § 67.] HEIGHTS AND DISTANCES. 115 Horizon. Bearing. CHAPTER VII. HEIGHTS AND DISTANCES. 64. The plane of the sensible horizon at any place, is the tangent plane to the earth's surface at that place. [B. p. 48.] The horizontal plane coincides with that of the surface of tranquil waters, when this surface is so small that its curvature may be neglected ; and it is perpendicular to the plumb line. 65. The angle of elevation of an object is the angle which the line drawn to it makes with the horizontal plane, when the object is above the horizon ; the angle of depression is the same angle when the object is be- low the horizon. 66. The bearing of one object from another is the angle included by the two lines which are drawn from the observer to these two objects. 67. Problem. To determine the height of a vertical tower, situated on a horizontal plane. [B. p. 94.] Solution. Observation. Let AB (fig. 23.) be the tower, whose height is to be determined. Measure off the distance BC on the horizontal plane of any convenient length. At the point C observe the angle of elevation ACB. 116 NAVIGATION AND SURVEYING. [CH. VII. Height of vertical tower. Calculation. We have, then, given in the right triangle ACB the angle C and the base BC, as in problem, § 33 of PL Trig., and the leg AB is found by (22). Example. At the distance of 95 feet from a tower, the angle of eleva- tion of the tower is found to be 48° 19'. Required the height of the tower. Arts. 106.69 feet. 68. Problem. To find the height of a vertical tower situated on an inclined plane. Solution. Observation. Let AB (fig. 24.) be the tower situated on the inclined plane BC. Observe the angle B, which the tower makes with the plane. Measure off the dis- tance BC of any convenient length. Observe the angle C, made by a line drawn to the top of the tower with BC. Calcidation. In the oblique triangle ABC, there are given the side BC and the two adjacent angles B and C, as in § 72 of PL Trig. Example. Given (fig. 24.) BC '= 89 feet, B = 113° 12', C "= 23° 27' ; to find AB. Ans. AB = 51.595 feet. 69. Problem. To find the distance of an inaccessible object. [B. p. 89 and 95.] Solution, Observation. Let B (fig. 2.) be the point, the dis- tance of which is to be determined, and A the place of the <§> 72.] HEIGHTS AND DISTANCES. 117 Distance of inaccessible object. observer. Measure off the distance AC of any convenient length, and observe the angles A and C. Calculation. AB and BC are found by § 72 of PL Trig. 70. Corollary. The perpendicular distance BP of the point B from the line AC, and the distances AP and PC are found in the triangle ABP and BPC, by § 31 of PI. Trig. 71. Corollary. Instead of directly observing the angles A and C, the bearings of the lines AB y AC, and BC may be observed, when the plane ABC is horizontal, and the angles A and C are easily determined. 72. Examples. 1. An observer sees a cape, which bears N. by E. ; after sailing 30 miles N. W. he sees the same cape bearing east ; find the distance of the cape from the the two points of obser- vation. Arts. The first distance ~ 21.63 miles. The second dist. — 25.43 miles. 2. Two observers stationed on opposite sides of a cloud ob- serve the angles of elevation to be 44° 56', and 36° 4', their distance apart being 700 feet ; find the distance of the cloud from each observer and its perpendicular altitude. Ans. Distances from observers =: 417.2 feet, and = 500.6 ft. Height =z 294.7 feet. 3. The angle of elevation of the top of a tower at one sta- tion is observed to be 68° 19', and at another station 546 feet 118 NAVIGATION AND SURVEYING. [CH. VII. Height of inaccessible object. farther from the tower, the angle of elevation is 32° 34'; find the height and distance of the tower, the two points of obser- vation being supposed to be in the same horizontal plane with the foot of the tower. Ans. The height ; . . — 234.28 ft. The dist. from the nearest point of observ. = 135.86 ft. 73. Problem* To find the distance of an object from the foot of a tower of known height, the observer being at the top of the tower. Solution. Observation. Let the tower be AB (fig. 23.), and the object C. Measure the angle of depression HAC. Calculation. Since ACB — HAC, we know in the triangle ACB the leg AB and the opposite angle C, as in § 32 of PL Trig. Example. Given the height of the tower — 150 feet, and the angle of depression ==17° 25' ; to find the distance from the foot of the tower. Ans. 478=16 feet. 74. Problem. To find the height of an inaccessible object above a horizontal plane, and its distance from the observer. [B. p. 96.] Solution. Observation. Let A (fig. 25.) be the object. At two different stations, B and C, whose distance apart and § 75.] HEIGHTS AND DISTANCES. 119 Distance of two objects. bearing from each other are known, observe the bearings of the object, and also the angle of elevation at one of the stations, as B. Calculation. In the triangle BCD, the side BC and its adjacent angles are known, so that BD is found by § 72 of PI. Trig. In the right triangle ABD, the height AD is, then, computed by § 33 of PI. Trig. Example. At one station the bearing of a cloud is N.N.W., and its angle of elevation 50° 35'. At a second station, whose bearing from the first station is N. by E., and distance 5000 feet, the bearing of the cloud is W. by N. ; find the height of the cloud. Ans. 7316.5 feet. 75. Problem. To find the distance of two objects, ivhose relative position is known. [B. p. 90.] Solution. Observation. Let B and C (fig. 1.) be the two known objects, and A the position of the observer. Observe the bearings of B and C from A, Calculation. In the triangle ABC, the side BC and the two angles are known. The sides of AB and AC are found by § 72 of PI. Trig. Example. The bearings of the two objects are, of the first N. E. by E., and of the second E. by^S. ; the known distance of the first object from the second is 23.25 miles, and the bearing N. W. ; find their distance from the observer. 120 NAVIGATION AND SURVEYING. [CH. VII. Distance apart of two objects. Ans. The distance of the first object is zn 18.27 miles. That of the second object = 32.25 miles. 76. Problem. To find the distance apart of two ob- jects separated by an impassable barrier. [B. p. 91.] Solution, Observation. Let A and B (fig. 1.) be the ob- jects ; the distance of which from each other is sought. Measure the distances and bearings from any point C to both A and B. Calculation. In the triangle ABC the two sides A C and BC and the included angle C are known. The sides AB and BC may be found by § 81 of PI. Trig. Example. Two ships sail from the same port, the one N. 10° E. a dis- tance of 200 miles, the second N. 70° E. a distance of 150 miles; find their bearing and distance. Ans. The distance = 180.3 miles. The bearing of the first ship from the second = N. 36° 6' W. 77. Problem. To find the distance apart of two in- accessible objects situated in the same plane with the observer, and their bearing from each other. [B. p. 92.] Solution. Observation. Let A and B (fig. 26.) be the two inaccessible objects. At two stations, C and E, observe the bearings of A and B and the bearing and distance of C from E. Calculation. In the triangle AEC we have the side CE, and the angles ACE and AEC 9 so that AC is found by § 72 of PL Trig. § 78.] HEIGHTS AND DISTANCES. 121 Distance apart of two objects. In the same way B C is calculated fronvthe triangle BCE. Lastly, in triangle ABC, we know the two sides AC and . BC, and the included angle for ACB = ACE — BCE. Hence AB and the angles BAC and CBA are found by §81. Example. An observer from a ship saw two headlands ; the first bore E. N. E., and the second N. W. by N. After he had sailed N. by W. 16.25 miles, the first headland bore E. and the second N. W. by W. ; find the bearing and distance of the first headland from the second. Ans. Distance = 55.9 miles. Bearing — N. 65° 33' W. 78. Problem. To find the distance of an object of known height, which is just seen in the horizon. Solution. I. If light moved in a straight line, and if A (fig. 27.) were the eye of the observer, and B the object, the the straight line APB would be that of the visual ray. The point P, at which the ray touches the curved surface CPD of the earth, is the point of the visible horizon at which the ob- ject is seen. The distances PA and PB may be calculated separately, when the heights AC and BD are known. For this purpose, let O be the earth's centre, let BD be produced to E, and let h =i AC, H=z BD, I = PA, L = PB, R z=z the earth's radius 11 122 NAVIGATION AND SURVEYING. [CH* VII. Distance of an object seen in the horizon. Since BP is a tangent, and BOE a secant to the earth, we have BE: BP = BP : BD; and BD is so small in comparison with the radius, that we may take BE= DE = 2R, and the above proportion becomes 2 R : L = L : H ; whence 2,2 — 2 RH, L & V(2 RH)> (222) H - L2 H -21V (223) and in the same way 72 = 2 R h, I— \/(2 R h), (224) 7 l2 (225) II. Light does not, however, move in a straight line near the earth's surface, but in a line curved towards the other centre , which is nearly an arc of a circle, whose radius is seven times the earth's radius ; so that for the point of con- tact P and the distances I and L, the positions of the eye and of the object are A' and B'. Now if we put BB' = H>, BD = H x — H— H A!C—h x , we can find the value of H 1 with sufficient accuracy by changing in (223) R into 7 R, which gives § 79.] HEIGHTS AND DISTANCES. 123 Distance of an object seen in the horizon. L 2 H' — — 1 H H x —H — H=^H= SL 2 " 7R' (226) L^^dRH,). (227) whence III. In calculating the value of L by (227), it is usually desired in statute miles, while the height H 1 is given in feet. Now we have in the Preface to the Navigator, page v, R z= 20911790 feet, (228) whence £ R = 48794177 feet, log. */(i R) = i log. i R = 3.84418, and log. (L in feet) = 3.84418 + % log. (H 1 in feet). „ - . ., L in feet But L m miles = ^ , so that log. L in miles = log. L in feet — 3.72263 = 0.12155 + £ log. H x in feet, (229) which agrees with the formula given in the preface to the Navigator for calculating table X. IV. The Table may be used for finding L or 7, when H 1 and h 1 are given, and then the required distance is the sum of L and I. 79. Corollary. Table X gives the correction for the error which is committed in § 67/- by neglecting the earth's curva- 124 NAVIGATION AND SURVEYING. [CH. VII. Table X of the Navigator. ture, for it is evident that to the height BP (fig. 28.) of the object above the visible level must be added the height PC of the level above the curved surface of the earth, as in B. p. 95. 80. Examples. 1. Calculate the distance in table X at which an object can be seen from the surface of the earth, when its height is 5000 feet. Solution. J log. 5000 ~ £ (3.69897) = 1.84948 constant log. =: 0.12155 dist. = 93.5 (as in table X) 1.97103 2. Being on a hill 200 feet above the sea, I see just appear- ing in the horizon the top of a mast, which I know to be 150 feet above the water ; how far distant is it ? Solution. By table X, 200 feet corresponds to 18.71 miles. 150 feet corresponds to 16.20 The distance is 34.91 miles. 3. At the distance of 7J statute miles from a hill the angle of elevation of its top is 2° 13' ; find its height in feet, the observer being 20 feet above the sea. § 80.] HEIGHTS AND DISTANCES. 125 Distance of an object seen in the horizon. Solution. 2° 13 tang. 8.58779 7£ miles = 39600 4.59770 By table X. 1533 feet 3.18549 7.50 1 foot correction, height 20 5.12 height = 1534 feet, height 1 1.58 4. Calculate the distance in table X, when the height is 450 feet. Ans. 28.06 miles. 5. Upon a height of 5000 feet, the top of a hill, one mile high,J r is just visible in the horizon ; how far distant is the hill? Ans. 189.6 miles. 6. At the distance of 25 miles from a mountain the angle of elevation of its top is 3° ; find its height, the observer being 60 feet above the intervening sea. Ans. 7033 feet. 11* SPHERICAL TRIGONOMETRY. SPHERICAL TRIGONOMETRY. CHAPTER I. DEFINITIONS. 1. Spherical Trigonometry treats of the solution of spherical triangles. A Spherical Triangle is a portion of the surface of a sphere included between three arcs of great circles. In the present treatise those spherical triangles only are treated of, in which the sides and angles are less than 180°. 2. The angle, formed by two sides of a spherical tri- angle, is the same as the angle formed by their planes. 3. Besides the usual method of denoting sides and angles by degrees, minutes, &c. ; another method of denoting them is so often used in Spherical Astronomy, that it will be found convenient to explain it here. The circumference is supposed to be divided into 24 equal arcs, called hours ; each hour is divided into 60 minutes of time, each minute into 60 seconds of time, and so on. Hours, minutes, seconds, &c. of time are denoted by h, ??i, s, &c. 130 SPHERICAL TRIGONOMETRY. [CH. I. Arcs expressed in time. 4. Problem. To convert degrees, minutes, fyc. into hours, minutes, fyc. of time. Solution. Since 360° = 24 h we have 15° = l\ and 1° = T ^ h = 4 W , and 15' = l m , and V =4% t&'= I s , and I'm 4'. Hence a° = 4 a m , a! = 4 a% a" = 4a'; so that to convert degrees, minutes, fyc. into time, mul- tiply by 4, and change the marks ° ' " respectively, into 5. Corollary. To convert time into degrees, minutes, fyc, ?miltiply the the hours by 15 for degrees, and di- vide the minutes, seconds, fyc. of time by 4, changing the marks m s \ into ° ' ". The turning of degrees, minutes, &c. into time, and the reverse, may be at once performed by table XXI of the Navi- gator. 6. Examples. 1. Convert 225° 47' 38" into time. Solution. By § 4. By Table XXI. 225° = 900™ = 15 h 15* 47' = 188 s — 3™ 8 s 3 m Qs 38"= 152' = 2 s 32' 2*32' 225° 47' 38" = 15 /l 3™ 10' 32' 15 h 3- 10*32' § 7.] DEFINITIONS. 131 Arcs expressed in time. 2. Convert 17* 19™ 13 s into degrees, minutes, &,c. Solution. By § 5. By Table XXI. 17* = 255° 17* 18™ £= 259° 19- 13* = 4° 48' 15" 3- 12 s = 48' 17* 19™ 13* = 259° 48' 15" I s dti 15" 17* 19™ 13 s = 259° 48' 15" 3. Convert 12° 34' 56" into time. Ans. 50™ 19 s 44*. 4. Convert 99° 59' 59" into time. Ans. 6 ft 39™ 59 s 56'. 5. Convert 3 /l 2™ 12 s into degrees, minutes, &c. Ans. 45° 33'. 6. Convert 11* 59™ 59 s into degrees, minutes, &c. Ans. 179° 59' 45". 7. When an arc is given in time, its log., sine, &c. can be found directly from table XXVII, by means of the column headed Hour P. M., in which twice the time is given, so that the double of the angle must be found in this column. The use of the table of proportional parts for these columns is explained upon page 35 of the Navigator. When the time exceeds 6*, the difference between it and 12* or 24* must be used. 132 SPHERICAL TRIGONOMETRY. [CH. Arcs expressed in time. Examples. 1. Find the log. cosine of 19* 33- IV. Solution. 24* — 19* 2S m IV == 4* 26- 49* 2 X (4* 26- 49 s ) as 8* 53- 38* 8* 53" 36 s P. M. cos. 9.59720 prop, parts of 2* 7 8* 53- 38 s P. M. cos. 9.59713 2. Find the angle in time of which the log. tang, is 10.12049. T 2- 40 s P.M. tang. 10.12026 7* prop, parts 23 2) T 2- 47 s P.M. 10.12049 Ans. 3* 31- 23J* 3. Find the log. sine of 3* 12- 2 s . Ans. 9.87113. 4. Find the log. cosine of IP 3- 13 s . Ans. 9.98653. 5. Find the log. tang, of 15* m 9*. Ans. 10.00057. 6. Find the log. cotan. of 22* 59- 59 s . Ans. 10.57183. 7. Find the angle in time whose log. secant is 10.23456. Ans. 3* 37- 26*. 8. Find the angle in time whose log. cosecant is 10.12346* Ans. 3* 15- 15*. § 9.] DEFINITIONS. 133 Right and oblique triangles. 8. An isosceles spherical triangle is one, which has two of its sides equal. An equilateral spherical triangle is one, which has all its sides equal. 9. A spherical right triangle is one, which has a right angle ; all other spherical triangles are called oblique. We shall in spherical trigonometry, as we did in plane trigonometry, attend first to the solution of right triangles. id 134 SPHERICAL TRIGONOMETRY. [CH. II. Investigation of Neper's Rules. CHAPTER II. SPHERICAL RIGHT TRIANGLES. 10. Problem. To investigate some relations between the sides and angles of a spherical right triangle. Solution. The importance of this problem is obvious ; for, unless some relations were known between the sides and the angles, they could not be determined from each other, and there could be no such thing as the solution of a spherical triangle. Let, then, ABC (fig. 29.) be a spherical right triangle, right-angled at C. Call the hypothenuse AB, h ; and call the legs BC and AC, opposite the angles A and B, respectively a and 6. Let O be the centre of the sphere. Join OA, OB, OC. The angle A is, by art. 2, equal to the angle of the planes BOA and CO A. The angle B is equal to the angle of the planes BOC and BOA. The angle of the planes BOC and AOC is equal to the angle C, that is, to a right angle; these two planes are, therefore, perpendicular to each other. Moreover, the angle BOA, measured by BA, is equal to BA or A; BOC is equal to its measure BC ox a, and AOC is equal to its measure ^4 Cor b. Through any point A' of the line OA, suppose a plane to pass perpendicular to OA. Its intersections A'C and A'B' § 10.] SPHERICAL RIGHT TRIANGLES. 135 Investigation of Neper's Rules. with the planes CO A and BOA must be perpendicular to OA'y because they are drawn through the foot of this perpen- dicular. As the plane B'A'C is perpendicular to OA, it must be perpendicular to AOC \ and its intersection B'C with the plane BOC, which is also perpendicular to AOC, must like- wise be perpendicular to AOC. Hence B'C must be per- pendicular to A'C and OC, which pass through its foot in the plane AOC. All the triangles A' OB', AOC, BOC, and ABC are then right-angled ; and the comparison of them leads to the desired equations, as follows : First. We have from triangle A' OB' by (4) OA 1 cos. A 1 OB' z=z cos. h — ; and from triangles AOC and BOC OA' cos. A'OC == cos. b = yr-^-, cos. B'OC = cos. a = oc> oc OB r The product of the two last equations is OA! OC OA! cos. a cos. b^ — x-^^^r; hence, from the equality of the second members of these equa- tions, cos. h =z cos. a cos. b. (230) 136 SPHERICAL TRIGONOMETRY. [CH. II. Investigation of Neper's Rules. Secondly. From triangle A'B C we have by (4), and the fact that the angle B'A'C is equal to the inclination of the two planes BOC and BOA, A'C cos. B'A'C = cos. A =z — — ; A'B 1 ' and, from triangles A'OC and A' OB', by (4), tang. CO A' = tang. 6= — , cotan. B'OA' = cotan. ^ = -rrrrr A'B' The product of these equations is _ A'C v A'O A'C tang, b cotan. A = _ X ^ = ^; hence cos. A — tang. 6 cotan. h. (231) Thirdly. Corresponding to the preceding equation between the hypothenuse h, the angle A, and the adjacent side b, there must be a precisely similar equation between the hypothenuse h, the angle B, and the adjacent side a ; which is cos. B z=z tang, a cotan. h. (232) Fourthly. From triangles BOC, BOA', and B'A'C, by (4), B'C sin. ^OC' = sin. a sin. B'OA' = sin. A = sin. B'A'C M sin. 4 = OB 1 ' B'A' ~OB' 9 B'C B'A' § 10.] SPHERICAL RIGHT TRIANGLES. 137 Investigation of Neper's Rules. The product of these last two equations is . _ BA 1 B'C B'C sin. h sin. ^^X^ = M/ J hence sin. a 5=3 sin. h sin. A. (233) Fifthly. The preceding equation between h, the angle A, and the opposite side a, leads to the following corresponding one between h, the angle B, and the opposite side b ; sin. b ±z sin. h sin. B. (234) Sixthly. From triangles C'OA' 9 BAC, and BOC, by (4), sin. CO A 1 =. sin. 6 == cotan. B'A'C = cotan. -4. = '//^/> B'C B'C tang. B'OC == tang, a =7777/- The product of these last two equations is 4'C , B'C A'C cotan. ^ tang, a = — X ^ = ^; hence sin. 6 = cotan. J. tang. a. (235) Seventhly. The preceding equation between the angle A y the opposite side a, and the adjacent side b, leads to the fol- lowing corresponding one between the angle B, the opposite side b, and the adjacent side a ; sin. a = cotan. 5 tang. b. (236) 12* 138 SPHERICAL TRIGONOMETRY. [CIJ. II. Investigation of Neper's Rules. Eighthly. From (7) sin. a tang, a — , cos. a sin. b tanp*. b =. -\ & cos. 6 which, substituted in (235) and (236), give cotan. B sin. b sin. a s= sin. b : cos. b cotan. A sin. a cos. a Multiplying the first of these equations by cos. b and the second by cos. a, we have sin. a cos. b = cotan. B sin. 6, sin. b cos. a z= cotan .4 sin. a. The product of these equations is sin. a sin. b cos. a cos. 6 ss cotan. ^4 cotan. 2? sin. a sin. 6 ; which, divided by sin. a sin. b, becomes cos. a cos. b sg cotan. ^4 cotan. I?. But, by (230), cos. h zz: cos, a cos. b ; hence cos. h = cotan. ^4 cotan. A (237) Ninthly. We have, by (230) and (234), cos. h COS. a :zz -, cos. 6 _ sin. b sin. B=z— — r , sin. h § 11.] SPHERICAL RIGHT TRIANGLES. 139 Neper's Rules. the product of which is, by (7) and (8), _ sin. b cos. h sin. 6 cos.h COS. a Sin. B = : : = r. - 7 cos. 6 sin. a cos. 6 sin. ft r= tang, b cotan. h. But, by (231), cos. A = tang, b cotan. It ; hence cos. A = cos. a sin. B. (238) Tenthly. The preceding equation between the side a, the opposite angle A, and the adjacent angle B, leads to the fol- lowing similar one between the side b 9 the opposite angle B , and the adjacent angle A ; cos. B ±s cos. 6* sin. ^4. (239) 11. Corollary. The ten equations, [230-239], have, by a most happy artifice, been reduced to two very simple theorems, called, from their celebrated inventor, Neper's Mules. In these rules, the complements of the hypothenuse and the angles are used instead of the hypothenuse and the angles themselves, and the right angle is neglected. Of the five parts, then, the legs, the complement of the hypothenuse, and the complements of the angles ; either part may be called the middle part. The two parts, including the middle part on each side, are called the adjacent parts ; and the other two parts are called the opposite parts. The two theorems are as follows. I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. 140 SPHERICAL TRIGONOMETRY. [CH. II. Neper's Rules. II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. [B. p. 436.] Proof. To demonstrate the preceding rules, it is only neces- sary to compare all the equations which can be deduced from them, with those previously obtained. [230 - 239.] Let there be the spherical right triangle ABC (fig. 30.) right-angled at C. First. If co. h were made the middle part, then, by the above rule, co. A and co. B would be adjacent parts, and a and b opposite parts ; and we should have sin. (co. h) = tang. (co. A) tang. (co. B) sin. (co. h) = cos. a cos. b ; or cos. h = cotan. A cotan. J5, cos. h ±3 cos. a cos. b ; which are the same as (237) and (230). Secondly. If co. A were made the middle part ; then co. h and b would be adjacent parts, and co. B and a opposite parts ; and we should have sin. (co. A) = tang. (co. h) tang, b, sin. (co. A) =s? cos. (co. jB) cos. a; or cos. A === cotan. h tang, h, cos. A = sin. B cos. a ; which are the same as (231) and (238). In like manner, if co. B were made the middle part, we should have § 12.] SPHERICAL RIGHT TRIANGLES. 141 Sides, when acute or obtuse. cos. B = cotan. 7i tang. a f cos. B = sin. A cos. b ; which are the same as (232) and (239). Thirdly. If a were made the middle part, then co. B and b would be the adjacent parts, and co. A and co. h the opposite parts ; and we should have sin. a == tang. (co. B) tang 6, sin. a = cos. (co. ^1) cos. (co. h) ; or sin. a = cotan. 2£ tang. 6, sin. a = sin. A sin. 7^ ; which are the same as (236) and (233). In like manner, if b were made the middle part, we should have sin. b = cotan. A tang, a, sin. b = sin. B sin. h ; which are the same as (235) and (234). Having thus made each part successively the middle part, the ten equations, which we have obtained, must be all the equations included in Neper's Rules; and we perceive that they are identical with the ten equations [230-239]. 12. Theorem. The three sides of a spherical right triangle are either all less than 90° ; or else, one is less while the other two are greater than 90°, unless one of them is equal to 90°, as in § 16. Proof. When h is less than 90°, the first member of (230) is positive ; and therefore the factors of its second member 142 SPHERICAL TRIGONOMETRY. [CH. II. Angle and opposite leg both acute or obtuse. must either be both positive or both negative ; that is, the two legs a and b must, by PL Trig. § 61, be both acute or both obtuse. But when h is obtuse, the first member of (230) is negative ; and therefore one of the factors of the second member must be positive, while the other negative ; that is, of the two legs a and 6, one must be acute, while the other is obtuse. 13. Theorem. The hypothenuse differs less from 90° than does either of the legs, the case of either side equal to 90° being excepted. Proof. The factors cos. a and cos. b of the second member of the equation (230) are, by (4), fractions whose numerators are less than their denominators. Their product, neglecting the signs, must then be less than either of them, as cos a for instance, or cos. h < cos. a ; and therefore, by PL Trig. § 69 and 70, h must differ less from 90° than a does. 14. Theorem. An angle and its opposite leg in a spherical right triangle must be both acute, or both ob- tuse, or, by § 16, both equal to 90°. Proof When A is acute, the first member of (238) is posi- tive, and therefore the factor cos. a of the second member, being multiplied by the positive factor sin. JB must be positive ; that is, a must be acute. But if A is obtuse, the first member of (238) is negative, and therefore the factor cos. a of the second member must be negative ; that is, a must be obtuse. <§> 16.] SPHERICAL RIGHT TRIANGLES. 143 One side equal to 90°. 15. Theorem. An angle differs less from 90° than its opposite leg, the case of either side, equal to 90°, being excepted. Proof. Since the second member of (238) is the product of the two fractions cos. a and sin. B, the first member must be less than either of them. Thus, neglecting the sines, cos. A < cos. a ; hence A differs less from 90° than a does. 16. Theorem. When in a spherical right triangle either side is equal to 90°, one of the other two sides is also equal to 90° ; and each side is equal to its opposite angle. Proof. First. If either of the legs is equal to 90°, the cor- responding factor of the second member of (230) is, by (59), equal to zero ; which gives cos. h = 0, or, by (59), h=z90°. .'j Again, if we have h =r 90°, it follows, from (59) and (230), that = cos. a cos. b, and therefore either cos. a or cos. b must be zero ; that is, either a or b must be equal to 90°. 144 SPHERICAL TRIGONOMETRY. [CH. II. Sides equal to 90°. Secondly. When either side is equal to 90°, it follows, from the preceding proof, that h 5= 90° ; which substituted in (233) produces, by (60), sin. a =. sin* A ; which gives a = A ; because, from § 14, a could not be equal to the supplement 17. Corollary. When both the legs of a spherical right triangle are equal to 90°, all the sides and angles are equal to 90°. 18. Theorem. When two of the angles of a spherical triangle are equal to 90°, the opposite sides are also equal to 90°. Proof. For in this case, one of the factors of the second member of the equation (237) must, by (61), be equal to zero, since either A or B is 90° ; hence cos. h = ; or, by (59), k = 90° ; and the remainder of the proposition follows from § 16. 19. Corollary. When all the angles of a spherical right triangle are equal to 90°, all the sides are also equal to 90°. § 20.] SPHERICAL RIGHT TRIANGLES. 145 Limits of the angles. 20. Theorem. The sum of the angles of a spherical right triangle is greater than 180°, and less than 360° ; and each angle is less than the sum of the other two. Proof. I. It is proved in Geometry, that the sum of the angles of any spherical triangle is greater than 180°. II. It is proved in Geometry, that each angle of any spheri- cal triangle is greater than the difference between two right angles and the sum of the other two angles. Hence if the sum of the two angles A and B is greater than 180°, we have 90° > A + B — 180°, or A + B < 270°, or A + B + 90° < 360° ; that is, the sum of the three angles is less than 360° ; for in case the sum of the angles A and B is less than 180°, the sum of the three angles is obviously less than 360°. III. When the right angle is greatest of the three angles, we have 90° + A + B > 180°, or A + B > 90° ; that is, the greater angle is in this case less than the sum of the other two. But if one of the other angles A is the greatest of the three angles, we have, by the proposition of Geometry last referred to, B > 90° + A — 180°, or B > A — 90°, or A < B + 90° ; 13 146 SPHERICAL TRIGONOMETRY. [CH. II. Hypothenuse and an angle given. so that in every case one angle is less than the sum of the other two. 21. To solve a spherical right triangle, two parts must be known in addition to the right angle. From the two known parts, the other three parts are to be determined, separately, by equations derived from Ne- per's Rules. The, two given parts, with the one to be determined are, in each case, to enter into the same equation These three parts are either all adjacent to each other, in which case the middle one is taken as the middle part, and the other two are, by <§> 11, adjacent parts ; or one is separated from the other two, and then the part, which stands by itself, is the middle part, and the other two are, by § 11, oppotite parts. 22. Problem. To solvo a spherical right triangle, when the hypothenuse and one of the angles are given. Solution. Let A B C (fig. 30.) be the right triangle, right- angled at C; and let the sides be denoted as in § 10. Let h and A be given ; to solve the triangle. First. To find the other angle B. The three parts, which are to enter into the same equation, are co. /*, co. A, and co. B; and, by § 21, as they are all adjacent to each other, co. h is the middle part, and co. A and co. B are adjacent parts. Hence, by Neper's Rules, sin. (co. h) =z tang. (co. A) tang. (co. B), or cos. h == cotan. A cotan. B ; and, by (6), § 24] SPHERICAL RIGHT TRIANGLES. 147 Hypothenuse and an angle given. _ cos. h cotan. B = = cos. /* tang. A cotan. A Secondly. To find the opposite leg a. The three parts are co. A, co. k, and a; of which, by § 21, a is the middle part, and co. h and co. A are the opposite parts. Hence, by Neper's Rules, sin. a = cos. (co. h) cos. (co. A) y or sin. a — sin. h sin. A. Thirdly. To find the adjacent leg b. The three parts are co. A y co. h, and 6 ; of which co. A is the middle part, and co. h and b are the adjacent parts. Hence, by Neper's Rules, sin. (co. A) z=z tang. (co. h) tang. 6, or cos. A =z cotan. h tang, b ; and, by (6), cos. J. tang. 6 = j- z= tang, h cos. -4. a cotan. h ° 23. Scholium. The tables always give two angles, which are supplements of each other, corresponding to each sine, cosine, &/C But it is easy to choose the proper angle for the particu- lar case, by referring to § 12 and 14 ; or by having regard to the signs of the different terms of the equation, as determined by PL Trig. § 61. 24. Scholium. When h and A are both equal to 90°, the values of cotan. B and tang, b are indeterminate ; for the nu- merators and denominators of the fractional values are, by (59) and (61), equal to zero ; and in this case there are an 148 SPHERICAL TRIGONOMETRY* [CH. II. Hypothenuse and an angle given. infinite number of triangles which satisfy the given values of h and A. The problem is impossible by § 18, if the given value of h differs from 90°, while that of A is equal to 90°. 25. Examples. 1. Given in the spherical right triangle (fig. 30.), 1i = 145° and A = 23° 28' ; to solve the triangle. Solution, h, cos. 9.91336 *n, sin. 9.75859, tang. 9.84523 n, 4, tang. 9.63761, sin. 9.60012, cos. 9.96251 jB,cotan. 9.55097 n; asm. 9.35871; b tang. 9.80774 n. Ans. B = 109° 34' 33", a = 13° 12' 12", b = 147° 17 1 5". 2. Given in the spherical right triangle, (fig. 30.), h = 32° 34', and A =* 44° 44' ; to solve the triangle. Ans. ^ — 50° 8' 21", a ~ 22° 15' 43", b ■=. 24° 24' 19". 26. Problem. To solve a spherical right triangle, when its hypothenuse and one of its legs are given. * The letter n placed after a logarithm indicates it to be the logarithm of a negative quantity, and it is plain that, when the number of such logarithms to be added together is even, the sum is the logarithm of a positive quantity ; but if odd, the sum is the logarithm of a negative quantity. <§> 28.] SPHERICAL RIGHT TRIANGLES. 149 Hypothenuse and a leg given. Solution, Let ABC (fig. 30.) be the triangle; h the given hypothenuse, and a the given leg. First. To find the opposite angle A ; a is the middle part, and co. A and co. h are the opposite parts. Hence sin. a z=i cos. (co. h) cos. (co. A) j or sin. a at sin. h sin. A ; and, by (6), sin. a sin. A = — — T = sin. a cosec. h. sin. A Secondly. To find the adjacent angle B ; co. JB is the mid- dle part, and co. h and a are the adjacent parts. Hence sin. (co. B) == tang, a tang. (co. h), or cos. J5 := tang, a cotan. ^. Thirdly. To find the other leg b; co.h is the middle part, and a and b are the opposite parts. Hence cos. h ss cos. a cos. 6 ; and, by (6), cos. h _ cos. o = = sec. a cos. h. cos. a 27. Scholium. The question is impossible by § 13, when the given value of the hypothenuse differs more from 90° than that of the leg. 28. Solution. When h and a are both equal to 90°, it may be shown, as in § 24, that the values of B and b are indeter- minate. 13* 150 SPHERICAL TRIGONOMETRY. [CH. II. A leg and the opposite angle given. 29. Example. Given in the spherical right triangle (fig. 30.), az= 141° \l> and h z=z 127° 12' ; to solve the triangle. Ans. A = 128° 6f 54", B z= 52° 22' 24', b =z 39° 6' 23". 30. Problem. To solve a spherical right triangle, when one of its legs and the opposite angle are given. Solution. Let ABC (fig. 30.) be the triangle, a the given leg, and A the given angle. First. To find the hypothenuse h ; a is the middle part, and co. h and co. A are the opposite parts. Hence sin. a =z sin. h sin. A ; and, by (6), . sin. a . ; sin. h z= z= sin. a cosec. A. sin. A Secondly. Te find the other angle B ; co. A is the middle part, and a and co. B are the opposite parts. Hence cos. A — cos. a sin. B ; and, by (6), cos. 4 sin. 5 zzz z= sec. a cos. A. cos. a Thirdly. To find the other leg. b ; 6 is the middle part, and a and co. A are the adjacent parts. Hence sin. b z= tang, a cotan. ^4. <§> 34.] SPHERICAL RIGHT TRIANGLES. 151 A leg and the opposite angle given. 31. Scholium. There are two triangles ABC and A'BC (fig. 31.) formed by producing the sides AB and AC, to the point of meeting A 1 , both of which satisfy the conditions of the problem. For the side BC ox a, and the angle A, or by § 2 its equal A', belong to both the triangles. Now ABA' and AC A' are semicircumferences. Hence h', the hypothenuse of A'BC, is the supplement of A; b 1 is the supplement of b ; and A'BC is the supplement of ABC. One set of values, then, of the unknown quantities, given by the tables, as in § 23, corresponds to the triangle ABC, and the other set to ABC. 32. Corollary. When the given values of a and A are equal, the values of h, B, and b become sin. li — 1, sin. B z= 1, sin. b =. 1 ; or, by (60), h = 90°, B = 90°, b == 90° ; as in § 16. 33. Corollary. When a and A are equal to 90°, the values of b and B are indeterminate, as in § 24. 34. Scholium. The problem is, by § 14, impossible, when the given values of the leg and its opposite angle are such, that one is obtuse, while the other is acute, or that one is equal to 90°, while the other differs from 90° ; and, by § 15, it is im- possible, when the given value of the angle differs more from 90° than that of the leg. 152 SPHERICAL TRIGONOMETRY. [CH. II. A leg and the opposite angle given. 35. Example. Given in the spherical right triangle, (fig. 30.), a =± 35° 44', and A = 37° 28' ; to solve the triangle. Ans. h = 73° 45' 15" ) ( h = 106° 14' 45" B — IT 54' \ or { B = 102° 6' b c= 69° 50' 24" j ( b = 110° 9' 36". 36. Problem. To solve a spherical right triangle, when one of its legs and the adjacent angle are given. Solution. Let ^4jBC(fig. 30.) be the triangle, a the given leg, and B the given angle. First. To find the hypothenuse h ; co. B is the middle part, and co. h and a are adjacent parts. Hence cos. B = tang, a cotan. h ; and, by (6), cos. B _ cotan. h z= =s cotan. a cos. 2*. tang, a Secondly. To find the other angle A ; co. J. is the middle part, and co. B and a are opposite parts. Hence cos. A = cos. a sin. J3. Thirdly. To find the other leg b ; a is the middle part, and co. B and 6 are adjacent parts. Hence sin. a = tang, b cotan. J5 ; and, by (6), sin. « . tang. 6 = =; = sin. a tang. J5. 6 cotan. ii & §38.] SPHERICAL RIGHT TRIANGLES, 153 The legs given. 37. Example. Given in the spherical right triangle, (fig. 30.), a= 118°54 / , and B — 12° 19' ; to solve the triangle. Ans. h =. 118° 20' 20", A =- 95° 5# 2", b — 10° 49' 17". 38. Problem. To solve a spherieal right triangle, when its two legs are given. Solution. Let ^.BC (fig. 30.) be the triangle, a and b the given legs. First. To find the hypothenuse h ; co. h is the middle part, a and b are opposite parts. Hence cos. h =z cos. a cos. b. Secondly. To find one of the angles, as A ; b is the middle part, and co. A and a are adjacent parts. Hence sin. b ±b tang, a cotan. ^4 ; and, by (6), sin. cotan. A = = cotan. « sin. 6. tang, a In the same way, cotan. B == cotan. b sin. a. 154 SPHERICAL TRIGONOMETRY. [CH. II. The angles given. 39. Example. Given in the spherical right triangle, (fig. 30.), a =i 1°, and b = 100° ; to solve the triangle. Ans. h — 99° 59' 52 /r , i= 1° 0' 56", J5z=90° 11' 24". 40. Problem. To solve a spherical right triangle, when the two angles are given. Solution. Let ABC (fig. 30.) be the triangle, A and B the given angles. First. To find the hypothenuse h ; co. h is the middle part, and co. A and co. B are adjacent parts. Hence cos. h = cotan A cotan. B. Secondly. To find one of the legs, as a ; co. A is the mid- dle part, and co. B and a are the opposite parts. Hence cos. A 33: cos. a sin. B \ and, by (6), cos. A cos. a = -^ — =r = cos. J. cosec. B. sin. X5 In the same way, cos. b — cosec. A cos. Z?. 41. Scholium. The problem is, by § 20, impossible, when the sum of the given values of A and B is less than 90°, or $ 42.] SPHERICAL RIGHT TRIANGLES. 155 The angles given. greater than 270°, or when their difference is greater than 90.° .42. Example. Given in the spherical right triangle, (fig. 30.), A = 91° 1 1', and B = 111 IV, to solve the triangle. [jins. h = 89° 32' 28", a = 91° 16' 8", b = 109° 52 EH / o * 156 SPHERICAL TRIGONOMETRY. [CH. Ill Sines of sides proportional to sines of opposite angles. CHAPTER III. SPHERICAL OBLIQUE TRIANGLES. 42. Theorem. The sines of the sides in any spherical triangle are proportional to the sines of the opposite angles. [B. p. 437.] Proof. Let ABC (figs. 32 and 33.) be the given triangle. Denote by a, b, c, the sides respectively opposite to the angles A, By C. From either of the vertices let fall the perpendicu- lar BP upon the opposite side AC. Then, in the right triangle ABP y making BP the middle part, co. c and co. BAP are the opposite parts. Hence, by Neper's Rules, sin. BP =± sin. c sin. BAP == sin. c sin. A. For BAP is either the same as A, or it is its supplement, and in either case has the same sine, by (91). Again, in triangle BPC, making BP the middle part, co. a and co. C are the opposite parts. Hence, by Neper's Rules, sin. BP =s= sin. a sin. C; and, from the two preceding equations, sin. c sin. A = sin. a sin. C, which may be written as a proportion, as follows ; sin. a : sin. A = sin c : sin. C. In the same way, sin. a : sin. A = sin. b . sin. B. § 43.] SPHERICAL OBLIQ.UE TRIANGLES. 157 Bowditch's Rules. 43. Theorem. BowditcK's Rules for Oblique Tri- angles. If, in a spherical triangle, two right triangles are formed by a perpendicular let fall from one of its verticles upon the opposite side ; and if, in the two right triangles, the middle parts are so taken that the perpendicular is an adjacent part in both of them ; then The sines of the middle parts in the two triangles are proportional to the tangents of the adjacent parts. But, if the perpendicular is an opposite part in both the triangles, then The sines of the middle parts are proportional to the cosines of the opposite parts. [B. p. 437.] Proof. Let M denote the middle part in one of the right triangles, A an adjacent part, and O an opposite part. Also let m denote the middle part in the other right triangle, a an adjacent part, and o an opposite part ; and let p denote the perpendicular. First. If the perpendicular is an adjacent part in both tri- angles, we have, by Neper's Rules, sin. M ==; tang. A tang, p, sin. m — tang, a tang, p ; whence sin. M tang. A tang, p tang. A sin. m tang, a tang, p ~~ tang, a * or sin. M : sin. m =z tang. A : tang. a. Secondly. If the perpendicular is an opposite part in both the triangles, we have, by Neper's Rules, 14 whence 158 SPHERICAL TRIGONOMETRY. [CH. III. Two sides and the included angle given. sin. M z=z cos. O cos. p t sin. m z= cos. o cos. p ; sin. 31 cos. O cos. p cos. O sin. m cos. o cos. p cos. o ' sin. M : sin. m — cos. O : cos o. 44. Problem. To solve a spherical triangle, when two of its sides and, the included angle are given. [B. p. 438.] Solution. Let ABC (figs. 32 and 33.) be the triangle, a and b the given sides, and C the given angle. From B let fall on .AC the perpendicular BP. First. To find PC, we know, in the right triangle BI£C, the hypothenuse a and the angle C. Hence, by means of Neper's Rules, tang. PC = cos. C tang. a. (239) Secondly. AP is the difference between AC and PC, that is, (fig. 32.) AP — b — PC ot (fig 33.) AP = PC— b. (240) Thirdly. To find the side c. If, in the triangle BPC f co. a is the middle part, PC and PjB are opposite parts ; and if, in the triangle ABP t co. c is the middle part, BP and AP are the opposite parts. Hence, by Bowditch's Rules, cos. PC : cos. AP = sin. (co. a) : sin. (co. c), or cos. PC : cos. -4P z= cos. a : cos. c. (241) <§> 45.] SPHERICAL OBLIQUE TRIANGLES. 159 Rules for acute or obtuse angles and sides. Fourthly, To find the angle A. If, in the triangle BPC, PC is the middle part, co. C and BP are adjacent parts; and if, in the triangle ABP, AP is the middle part, co. BAP and BP are adjacent parts. Hence, by Bowditch's Rules, sin. PC : sin. PA = cotan. C : cotan. BAP ; (242) and BAP is the angle A (fig. 32.), when the perpendicular falls within the triangle ; or it is the supplement of A (fig. 33.), when the perpendicular falls without the triangle. Fifthly. B is found by means of (344), sin. c : sin. C = sin. b : sin. B. (243) 45. Scholium. In determining PC, c, and BAP, by (239), (241), and (242), the signs of the several terms must be care- fully attended to; by means of PI. Trig. § 61. But to determine which value of B, determined by (243), is the true value, regard must be had to the following rules, which are proved in Geometry. I. The greater side of a spherical triangle is always opposite to the greater angle. II. Each side is less than the sum of the other two. III. The sum of the sides is less than 360°. IV. Each angle is greater than the difference between 180°, and the sum of the other two angles. There are, however, cases in which these conditions are all satisfied by each of the values of B. In any such case this angle can be determined in the same way in which the angle A was determined, by letting fall a perpendicular from the 160 SPHERICAL TRIGONOMETRY. [CH. III. Fundamental equation. vertex A on the side BC. But this difficulty can always be avoided, by letting fall the perpendicular upon that of the two given sides which differs the most from 90°. 46. Corollary. By (240), (104), and (29), we have cos. AP = cos. (b — PC) s= cos. (PC— b) = cos. b cos. PC + sin. b sin. PC, (244) which, substituted in (241), gives cos. PC: cos. b cos. PC-\- sin. b sin. PC=z cos. a : cos. c. Dividing the two terms of the first ratio by cos. PC, we have by (7), 1 : cos. b -|- sin. b tang. PC r= cos. a : cos. c. (245) The product of the means being equal to that of the extremes, we have cos. c z=z cos. a cos. b -|- sin. b cos. a tang. PC. (246) But by (239) tans. PC = cos. C tang, a = , s 5 cos. a ' or cos. a tang. PC = cos. C sin. a ; (247) which, substituted in (246), gives cos. c —= cos. a cos. 6 -f- sm - a sm « ^ cos « C> (248) which is owe o/* the fundamental equations of Spherical Trigonometry. 47. Corollary. We have, by (48), cos. C=zl-+2(cos. £C) 2 , § 49.] SPHERICAL OBLIQ.UE TRIANGLES. 161 Column of Log. Rising of Table XXIII. which, substituted in (248), gives, by (28), cos. c == cos. (a -f- b) + 2 sin. a sin. b (cos. £ C) 2 , (249) from which the value of the side c can readily be found by using the table of Natural Sines. 48. Corollary. We have, by (49), cos. C— 1 — 2 (sin.JC) 2 , which, substituted in (248), gives, by (29), cos. c — cos. (a — b) — 2 sin. a sin. b (sin. J C) 2 , (250) which can be used like formula (249). 49. Corollary. The use of formula (250) is much facilitated by means of the column of Rising in Table XXIII of the Navigator. This column contains the values of log. 2 (sin. £ C) 2 — 2 log. sin. £ C + log. 2 = 2 log. sin. iC+ 0.30103. (251) But the decimal point is supposed to be changed so as to correspond to the table of Natural Sines, that is, 5 is added to the logarithm ; and 20 is to be subtracted from the value of 2 log. sin. J C, which is given by table XXVII, as is evident from PI. Trig. § 30. So that the coltimn Rising of Table XXIII is constructed by the formula log. Ris. C = 2 log. sin. J C + 5.30102 — 20 — 2 log. sin. J C — 14.69897, (252) which agrees with the explanation in the Preface to the Navigator. 14* 162 SPHERICAL TRIGONOMETRY. [CH. III. Two sides and the included angle given. 50. By using table XXIII, the following rule is ob- tained for finding the third side, when two sides and the included angle are given. Add together the log. Rising of the given angle, and the log. msines of the two given sides. The sum is the logarithm of a number, which is to be subtracted from the natural cosine of the difference of tioo given sides {regard being had to the sign of this cosine). The dif- ference is the natural cosine of the required side. 51. Examples. 1. Si Calculate the value of log. )lution> £ (4*28 w ) = 2 h U m log. Ris. 4*28" Ris. of 4 A sin 28-. 9.74189 2 19.48378 14.69897 4.78481 2. Given in the spherical triangle two sides equal to 45° 54', and 138° 32, and the included angle 98° 44'; to solve the triangle. Solution. I. by (239), €=: 98° 44 / cos. 9.18137 n. a == 45° 54' tang. 0.01365 PC =171° 6' 16" tang. 9.19502 n. § 51.] SPHERICAL OBLIQUE TRIANGLES. 163 Two sides and the included angle given. « By (940), AP = 171° 6' 16" — 138° 32' == 32° 34' 16". By (241), PC = 171° 6' 16" cos. (ar.co.) 10.00525 n. AP = 32° 34' 16" cos. 9.92569 a = 45° 54' cos. 9.84255 c = 126° 24' 45" cos. 9.77349 w. By (242), PC = 171° 6' 16" sin. (ar. co.) 10.81071 AP = 32° 34' 16" sin. 9.73106 C = 98° 44' cotan. 9.18644 ». BAP = 118° 8' 19" cotan. 9.72821 n. A = 180° — 118° 8' 19" =3 61° 51' 41". By (243), c = 126° 24' 45" sin. (ar. co.) 10.09433 C = 90° 44' sin. 9.99494 b = 138° 32' sin. 9.82098 B = 125° 34' 48"' sin. 9.91025 Ans. c = 126° 24' 45" A = 61° 61' 41" B = 125° 34' 48". 164 SPHERICAL TRIGONOMETRY. [CH. III. • Two sides and the included angle given. II. The third side is thus caloulated by means of (249), 2 log. 0.30103 45° 54' sin. 9.85620 138° 32' sin. 9.82098 £(98° 44') = 49° 22' 2 cos. 19.62744 0.40332 9.60565 — 0.99683 = Nat. cos. (13S° 32' + 45° 54')= N. cos. 18426 _ 0.59351 = Nat. cos 126° 24' 23" = c. III. The third side is thus calculated by § 41. 98° 44' « 6* 34™ 56' ' log. Ris. 5.06139 45° 54' sin. 9.85620 138° 32' sin. 9.82098 54774 4.73857 92° 38' N. cos. — 4594 c = 126° 25' 8" N. cos. — 59368 3. Calculate the log. Ris. of HM2 m 20*. Ans. 5.29632. 4. Given in a spherical triangle two sides equal to 100°, and 125°, and the included angle equal to 45° ; to solve the tri- angle. Ans. The third side = 47° 55' 52" The other two angles — 69° 43' 48", and = 128° 42' 48". § 52.] SPHERICAL OBLIQUE TRIANGLES. 165 A side and the two adjacent angles given. 52. Problem. To solve a spherical triangle, when one of its sides and the two adjacent angles are given. [B. p. 438.] Solution. Let ABC (figs. 32 and 33.) be the triangle ; a the given side, and B and C the given angles. From B let fall on AC the perpendicular BP. Fir$t. To find PBC f we know, in the right triangle BPC y the hypothenuse a and the angle C. Hence, by Neper's Rules, cotan. PBC = cos. a tang. C. (253) Secondly. ABP is the difference between ABC and PBC, that is, (fig. 32.) ABP =B — PBC, or (fig. 33.) ABP = PBC — B. (254) Thirdly. To find the angle A. If, in the triangle PBC, co. C is the middle part, PB and co. PBC are the opposite parts; and if, in the triangle ABP, co. BAP is the middle part, PB and co. ABP are the opposite parts. Hence, by Bowditch's Rules, cos. (co. PBC) : cos. (co. ABP) — sin. (co. C) : sm.(co.BAP), or sin. PBC : sin. ABP = cos. C : cos. BAP ; (255) and BAP is either the angle A or its supplement. Fourthly. To find the side c. If, in the triangle PBC, co PBC is the middle part, PB and co. a are the adjacent parts ; and if, in the triangle ABP, co. ABP is the middle part, PB and co. c are the adjacent parts. Hence, by Bow- ditch's Rules, cos. PBC : cos. ABP s= cotan. a : cotan. c. (256) 166 SPHERICAL TRIGONOMETRY. [CH. III. A side and the two adjacent angles given. Fifthly, b is found by the proportion sin. C : sin. c s= sin. B : sin. b. (257) 53. Scholium. In determining PBC, BAP, and c by (253), (255), and (256), the signs of the several terms must be care- fully attended to, by means of PI. Trig. § 61. To determine which value of b, obtained from (257), is the true value, regard must be had to the rules of § 45. # But if all these conditions are satisfied by both values of b, then b may be calculated by letting fall a perpendicular from C on the side c in the same way in which c has been obtained in the preceding solution. But this case can be avoided by let- ting fall the perpendicular from the vertex of that one of the two given angles, which differs the most from 90°. 54. Corollary. Since 180° — a, 180° — b, and 180° — c are the angles of the polar triangle, and 180° — A, 180° — B, and 180° — C are its sides ; we have given in the polar tri- angle the two sides 180° — jB, and 180° — - C, and the in- cluded 180° — a ; so that the polar triangle might be solved by § 44. 55. Corollary, If formula (248) is applied to the polar tri- angle of the preceding section, it becomes by PI. Trig. § 60, — cos. A = cos. B cos. C — sin. B sin. C cos. a, or cos. A*— — cos. jB cos. C+sin.jBsin. Ccos. a. (258) 56. Corollary. In the same way (249) becomes by (92) and (116), cos. A = — cos. (B + C ) —2 sin. B sin. C(sin. £ a)*, (259) §58.] SPHERICAL OBLIQUE TRIANGLES. 167 A side and the two adjacent angles given. from which the value of the third angle may be found by means of table XXIII. 57. Corollary. In the same way (250) becomes by (92), cos. A=z — cos.(J3 — C)+2sin.jBsin.C(cos.£a) 2 , (260) from which the value of the third side may be found. 58. Examples. 1. Given in a spherical triangle one side equal to 175° 27', and the two adjacent angles equal to 126° 12', and 109° 16'; to solve the triangle. Solution. I. By (253), a = 175° 27' COS. 9.99863 n. C = 109° 16' tang. 0.45650 n. PBC = 19° 19' 24" cotan. 0.45513 By (254), ABP = 126° 12' — 19° 19' 24" = 106° 52 36". By (255), PBC— 19° 19' 24" ABP = 106° 52' 36" C = 109° 16' BAP = 162° 36' sin. (ar. co.) 10.48031 sin. 9.98088 cos. 9.51847 n. COS. 9.97966 n. 168 SPHERICAL TRIGONOMETRY. [CH. III. A side and the two adjacent angles given. By (256), PBC = 19° 19' 24" cos. (ar. co.) 10.02518 ABP === 106° 52' 36" cos. 9.46288 n. a = 175° 27' cotan. 1.09920 n. c = 14° 30' 9" cotan. 0.58726 .4 = .R4P = 162° 36'. • By (257), • C = 109° 16' sin. (ar. co.) 10.02503 c = 14° 30' 9" sin. 9.39867 B = 126° 12' sin. 9.90685 b = 167° 38' 21" sin. 9.33055 Ans. A = 162° 36' b = 167° 38' 21" cz=z 14° 30' 9". II. The third angle is thus calculated by means of (259). 175° 27' = 1 1 h 41 m 48* log. Ris. 5.30035 126° 12' . sin. 9.90685 109° 16' sin. 9.97497 — 152115 5.18217 235° 28' — N. cos. 56689 162° 36 12" N.cos. — 95426 § 59.] SPHERICAL OBLIQUE TRIANGLES. 169 Two sides and an opposite angle given. III. The third angle is thus calculated by means of (260), 2 log. 0.30103 J (175° 27) = 87° 43' 30" 2 cos. 17.19750 126° 12' sin. 9.90685 109° 16' sin. 9.97497 0.00240 7.38035 16° 56' — N. cos. — 0.95664 A = 162° 36' N. cos. — 0.95424 2. Given in a spherical triangle one side = 45° 54', and the two adjacent angles =z 125° 37', and = 98° 44' ; to solve the triangle. Ans. The third angle = 61° 55' 2", The other two sides = 138° 34' 22", and = 126° 26' 11". 59. Problem. To solve a spherical triangle, token two sides and an angle opposite one of them are given. [B. p. 437.] Solution. Let ABC (figs. 32 and 33.) be the triangle, a and c the given sides, and C the given angle. From B let fall on AC the perpendicular BP. First. To find PC. We know, in the right triangle PBC, the side a and the angle C. Hence, by Neper's Rules, tang. PC z=l cos. Ctang. a. (261) Secondly. To find AP. If, in the triangle PBC, co. a is the middle part, CP and PB are the opposite parts; and if, 15 170 SPHERICAL TRIGONOMETRY. [CH. III. Two sides and an opposite angle given. in the triangle ABP, co. c is the middle part, AP and PB are the opposite parts. Hence, by Bowditch's Rules, cos. a : cos. c — cos. PC : cos. AP. (262) Thirdly. To find b. There are, in general, two triangles which resolve the problem, in one of which (fig. 32.) b = PC + AP, (263) and in the other (fig. 33.) b = PC— AP. (264) But, if AP is greater than PC, there is but one triangle, as in (fig. 32.), and b is obtained by (263) ; or, if the sum of AP and PC is greater than 180°, there is but one triangle, as in (fig. 33.), and b is obtained by (264). Fourthly. A and B are found by the proportion sin. c : sin. C =. sin. a : sin. A (265) sin. c : sin. C =. sin. b : sin. B. (266) 60. Scholium. In determining PC and AP by (261) and (262), the signs of the several terms must be carefully at- tended to by means of PL Trig. § 61. The two values of A, given by (265), correspond respective- ly to the two triangles which satisfy the problem ; and the one, which belongs to each triangle, is to be selected, so that the angle BAP, which is the same as A in (fig. 32.), and the supplement of A in (fig. 33.), may be obtuse if C is ob- tuse, and acute if C is acute. For BP is the side opposite BAP in the right triangle ABP, and the side opposite C in the triangle BCP ; and therefore, by § 14, BP, BAP, and C are all at the same time acute, or all obtuse. § 62.] SPHERICAL OBLIQUE TRIANGLES. 171 Two sides and an opposite angle given. Of the two values of B, given by (266), the one which be- longs to each triangle is to be determined by means of the rules of § 45. 61. Scholium. The problem is, by a proposition of Geome- try, impossible, when the given value of c differs more from 90° than that of a ; if, at the same time, the value of one of the two quantities, c and C, is acute, while that of the other is obtuse. And in this case we should find that AP was larger than PC, and at the same time that the sum of AP and PC was more than 180°. 62. Examples. 1. Given in the spherical triangle, one side z= 35°, a second side = 142°, the angle opposite the second side = 176° ; to solve the triangle. Solution. By (261), C — 176° cos. 9.99894 n. a=L 35° tang, tang. 9.84523 PC = 145° 3' 56" 9.84417 n. By (262), a— 35° cos. (ar. co.) 10.08664 PC —. 145° 3' 56" cos. 9.91371 n. c == 142° cos. 9.89653 n. AP = 37° 56' 30" cos. 9.89688 172 SPHERICAL TRIGONOMETRY. [CH. III. Two sides and an opposite angle given. By (264), b = 145° 3' 56" — 37° 56' 30" = 107° 7 / 26". By (265), c = 142° sin. (ar.co.) 10.21066 C =z 176° sin. 8.84358 a— 35° 43' 34" sin. 9.75859 A= 3° sin. 8.81283 By (266), c ±s 142° sin. (ar.co.) 10.21066 C= 176° sin. 8.84358 b == 107° r 26" 12' 58" sin. 9.98030 J5= 6° sin. 9.03454 Am r. b = 107° 7' 26" A = 3° 43' 34" B = 6° 12' 58". 2. Given in a spherical triangle, one side == 54°, a second side rz= 22°, the angle opposite the second side = 12° ; to solve the triangle. Am. The third side — 73° 14' 29", or = 33° 32' 59". One angle = 26° 40' 49", or = 153° 19' 11". The third angle = 147° 53' 51", or = 17° 51' 43". 63. Problem. To solve a spherical triangle, when two angles and a side opposite one of them are given. [B. p. 438.] § 64.] SPHERICAL OBLIQ.UE TRIANGLES. 173 Two angles and an opposite side given. Solution, Let ABC (figs. 32 and 33.) be the triangle, A and C the given angles, and a the given side. From B let fall on AC the perpendicular BP. This per- pendicular must fall within the triangle if A and C are either both obtuse or both acute ; but it falls without, if one is obtuse and the other acute. First PC may be found by (261). Secondly, To find AP. If, in the triangle PBC, PC is the middle part, co. C and PB are the adjacent parts ; and if, in the triangle ABP, AP is the middle part, co. BAP and BP are the adjacent parts. Hence, by Bowditch's Rules, cotan. C : cotan. BAP = sin. PC : sin. AP. (267) Thirdly. To find b. We have (fig. 32.) b = PC+ AP, ' ' (268) (fig. 33.) b = PC— AP. (269) Fourthly, c and B are found by the proportion sin. A : sin. a = sin. C : sin. c, (270) sin. a : sin. A = sin. b : sin.jB. (271) 64. Scholium. Either value of AP, given by (267), may be used, and there will be two different triangles solving the problem, except when AP -\- PC (fig. 32.) is greater than 180°, or PC (fig. 33.) is less than AP. It may be that both values of AP satisfy the conditions of the problem, or that only one value satisfies them, or that neither value does ; in which last case the problem is impossible. 15* 174 SPHERICAL TRIGONOMETRY. [CH. III. Two angles and an opposite side given. Of the values of c, determined by (270), the true value must be ascertained from the right triangle ABP by§ 12; or since PB and C are both acute or both obtuse at the same time ; it follows, from § 12, that when C and AP are both acute or both obtuse, that c is acute ; but when one of them is obtuse and the other acute, c is obtuse. From the two values of B (271), the true value must be selected by means of the rules of § 45. 65. Scholium. The problem is impossible, by Geometry, when A differs more from 90° than does C, and when at the same time one of the two quantities a and A is acute, while the other is obtuse. This case is precisely the same as the impossible case of § 61. 66. Examples. \, 1. Given in a spherical triangle, one angle ±= 95°, a second angle — 104°, and the side opposite the first angle z= 138° ; to solve the triangle. Solution. By (261), C = 104° cos. 9.38368 n. a 5= 138° tang. 9.95444 n. PC z= 12° 17' 20" tang. 9.33812 By (267), C = 104° cotan. (ar.co.) 0.60323 n, PC =± 12° 17' 20 " sin. 9.32802 BAP- 95° cotan. 8.94195 n, AP s= 4° 16' 59" sin. 8.87320 § 66.] SPHERICAL OBLIQUE TRIANGLES. 175 Two angles and an opposite side given. By (268), b = 12° 17' 20" + 4° 16' 59' = 16° 34' 19". By (270), A = 95° sin. (ar.co.) 10.00166 a = 138° sin. 9.82551 C = 104° sin. 9.98690 c = 139° 19' 40" sin. 9.81407 By (271), a = 138° sin. (ar.co.) 10.17449 A= 95° sin. 9.99834 b = 16° 34' 19" sin. 9.45518 B = 25° 7' 38" sin. 9.62801 Again, by (269), b = 12° 17' 20" — 4° 16' 59" = 7° 0' 21" c = 180° — 139° 19' 40" = 40° 40' 20". By (271), a = 138° sin. (ar. co.) 10.17449 A— 95° sin. 9.99834 b= 7° 021" sin. 9.08623 B=z 10° 27' 42" sin. 9.25906 Ans. b ss 16° 34' 19" or = 7° 0' 21" c = 139° 19' 40" or = 40° 40' 20" B = 25° 7' 38" or = 10° 27' 42". 176 SPHERICAL TRIGONOMETRY. [CH. 111. The three sides given. 2. Given in a spherical triangle, one angle = 104°, a second angle == 95°, and the side opposite the first angle = 138° ; to solve the triangle. Ans. The two sides are 17° 22' 13", and 136° 36' 27". The other angle is 25° 39' 9". 67. Problem. To solve a spherical triangle, when its three sides are given. # Solution. Equation (248) gives, by transposition and di- vision, -, cos. c — cos. a cos. b igj*^ COS. C s= : : - , (272) sin. a sin. b ' whence the value of the angle C may be calculated, and in the same way either of the other angles. 68. Corollary. An equation, more easy for calculation by logarithms, may be obtained from (249), which gives, by transposition and division, 3 (cos. $ C Y = «~ '-<"■; <«+*). (273) v ' sm. a sin b ' Now, letting s denote half the sum of the sides, or s = i(a + b + c); (274) if we make in (35) M=i(a + b + c)=,s, N= % (a-\-b — c) = s — c; we have M + N = a + b, M—N= c; § 70.] SPHERICAL OBLIQUE TRIANGLES. 177 The three sides given. and (35) becomes cos. c — cos. (a -j- b) = 2 sin. s sin. (s — c) ; which, substituted in (273), gives 2 (cos. i Cy = »*»;»«M' --**?; (275) 7 sin. a sin. o and cos ^n ^£&% ^ 69. Corollary. The angles .4 and JB may be found by the two following equations, which are easily deduced from (276), /sin. 5 sin. (s — a)\ co,^ = V( sm.6sin. T^)' %& \ sin. a sin. c / v cos, 70. Corollary. Another equation, equally simple in calcula- tion, can be obtained from (250), which gives, by transposition and division, cos. (a — 6) — cos. a cos. b -j- sin. a sin 6, which, substituted in the numerator of (250), gives 2 (sin . j cy - cos. (.-6) -cos., v 2 y sin. a sin. 6 ' v y whence C can be found by Table XXIII. 71. Corollary. If, in (35), we make M = | (a — 6 + c) = s — 6, iV=£( — a -|- & -{- c ) = * — a, 178 SPHERICAL TRIGONOMETRY. [CH. III. The three sides given. we have M + N = c, M—N= a — b, and (35) becomes cos. (a — b) — cos. c = 2 sin. (5 — a) sin. (5 — b) ; which, substituted in (279), gives 2(sin.£C)3= 2sin.(s- fl )sin(5-6j v 2 ' sin. a sin. 6 v ' and sin. |C=V P' (f f* ) Tf^* ) fr (281) 2 ^ \ sin. a sin. b ) 72. Corollary. In the same way we might deduce the fol- lowing equations ; sin. i A = v (^.(s-b) sin, (s-c)^ 2 ™ \ sin. 6 sin. c / sin. X B = v /!in_(^-«)sin.(s- C )V \ sin. a sin. c / 73. Corollary. The quotient of (282), divided by (277), is °y (7), tang. J4*= S -N4 = V / sin -^- 6 ) s ; n -( s -i)\, (284) cos. £ A \ sm. 5 sm. (s — a) / In the same way, n ./sin. (s — a) sin. (s — c)\ /nDn tang. A B =z \/( ^ J--, — ^-r — '- I , (285) 5 2 ^\ sin. s sin. (s — b) / v J *74] SPHERICAL OBLIQUE TRIANGLES. 179 The three sides given. 74. Examples. I. Given in the spherical triangle ABC the three sides equal to 46°, 72°, and 68° ; to solve the triangle. Solution. I. By (277), by (278), by (279), a=46°sin. (ar.co.)10.14307(ar.co.)10.14307 6=72°sin. (ar.co.) 10.02 179 (ar.co.)10.02179 c=68° sin. (ar.co.) 10.03283 (ar.co.) 10.03283 5=93° sin. 9.99940 9.99940 9.99940 s-a=47°sin. 9.86413 s-6=21°sin. 9.55433 s-c=25° sin. 2) 9.62595 19.91815 2)19.72963 2)19.79021 cos. 9.95908 9.86482 9.89511 } A = 24° 29', £ B = 42° 54', $ C = 38° 14 / 18" ; Ans. A = 48° 58', B = 85° 48', C = 76° 28' 56". II. By Table XXIII and equation (279), a — b = 26° N. cos. 89879 c = 68° N. cos. ,37461 52418 a = 46° . b = 72° C=z 5*5*55* = 76°28 / 45 // log. 4.71948 sin. (ar.co.) 0.14307 sin. (ar.co.) 0.02179 log. Ris. 4.88434 180 SPHERICAL TRIGONOMETRY. [CH. III. Neper's Analogies. 2. Given in a spherical triangle the three sides equal to 3°, 4°, and 5° ; to solve the triangle. Ans. The three angles are 36° 54', 53° 10', and 90° 2'. 75. Neper obtained two theorems for the solution of a spherical triangle, when a side and the two adjacent angles are given, by which the two sides can be calcu- lated without the necessity of calculating the third angle. These theorems, which are given in § 78 and 79, can be obtained from equations (284-286) by the assistance of the following lemmas. 76. Lemma. If we have the equation tang. M x tang. N y' we can deduce from it the following equation, sin. (M + N) __ x + y sin. (M— N) T x — y' Proof. We have from (7) sin. M . ,_ sin. N XdLiciv.Mz=. — f and tang. iV = — : 5 - cos. if' s cos.^V' which, substituted in (287), give sin. M cos. N x cos. M sin. N y This equation is the same as the proportion sin. Mcos. N : cos. if sin. N = x : y ; hence, by the theory of proportions, (287) (288) (290) <§> 78.] SPHERICAL OBLIQUE TRIANGLES. 181 Neper's Analogies. sin. M cos. N + cos. M sin. N : sin. M cos. N — - cos. M sin. iV =± x -\- y : a; — y, or, by (26) and (27), sin. {M-\- N) : sin. (IT— 2V) z= x + y : a; — y ; which may be written in the form of an equation, as in (288). 77. Lemma. If we have the equation tang. i*f tang. iV = -; (289) we can deduce from it the equation cos. ( M -f- N) __ y — x cos. (M — N) *~" y-\- x Proof. We have, by (289) and (7), sin. M sin. N x cos. M cos. N y This equation is the same as the proportion cos. M cos. N : sin. M sin. N =. y : x \ hence, by the theory of proportions, cos. M cos. N — sin. M sin. N : cos. M cos. N -\- sin. M sin. iV = y — x \x -\- y, or, by (28) and (29), cos. (ilf + iV) : cos. (iKf— N) =z y — x\y + x% which may be written as in (290). 78. Theorem, The sine of half the sum of two angles of a spherical triangle is to the sine of half their differ - ence 7 as the tangent of half the side to which they are 16 (292) 93) 182 SPHERICAL TRIGONOMETRY. [CH. III. INeper's Analogies. both adjacent is to the tangent of half the difference of the other two sides ; that is ; in the spherical triangle AB C (figs. 32 and 33.), sin.J (A+C) : sin. £ (A— C)= tang. J b : tang. \ {a—c). (291) Proof The quotient of (284), divided by (286) is, by an easy reduction, tang, i A sin. (s — c) tang. J C sin. (s — a) Hence, by § 76, sin. £ (A + C) sin. (5 — c) -f- sin. (s — a) sin. %(A — C) sin. (s — c) — sin. (s — a)' If we make in equation (40) A =z s — c =z J (a -f- b — c), B = s — a = i(— a + b + c); we have A + B = b, A — B — a — c; and (40) becomes sin. (s — c) 4- sin. (s — a) _ tang. £ b sin. (s — c) — sin. (s — a) tang.£(a — c)' This equation, substituted in the second member of (293), gives s in, j (A + C) _ tang. | 6 sin. J (A — C) ~ tang. J (a — c)' V ; which is the same as (291). 79. Theorem. The cosine of half the sum of two angles of a spherical triangle is to the cosine of half <§> 79.] SPHERICAL OBLIQUE TRIANGLES. 183 Neper's Analogies. their difference, as the tangent of half the side to which they are both adjacent is to the tangent of half the sum of the other two sides ; that is, in the spherical triangle iiflC(figs. 32 and 33.), cos.i(A+C):coa.£(A— C)— tang.^-6 :tang. J(« + c). (295) Proof. The product of (284) and (286) is, by a simple re- duction, „ sin. (s — b) tang, i 4;tang. £ C = - ^ g J hence, by § 77, cos. £ (A -f- C ) sin. s — sin. (s — b) cos. £ (A — C) sin. 5 -\- sin. (s — b) If in equation (40) inverted we make A = s = i (a + b + c), we have A + JB r= a + c, 4 — J3 = 6 ; and (40) becomes sin. s — sin. (s — b) tang. J b sin. 5 -)- sin. (s — b) tang. £ (a-(-c)" This equation, substituted in (296), gives cos.^(A + C) __ tang, j b cos. £ (A — C) tang, ^-(a-f-c)' which is the same as (295). (296) (297) 184 SPHERICAL TRIGONOMETRY. [CH. III. Neper's Analogies. 80. Scholium. In using (291) and (295), the signs of the terms must be attended to by means- of PL Trig. § 61. 81. Examples. 1. Given in a spherical triangle two angles z= 158°, and : 98°, and the included side ar 144° ; to find the other sides. Solution. By (291), } (A + C) = 128° sin. (ar.co.) 10.10347 t(A — C)= 30° sin. 9-69897 i b =12° tang. 0.48822 £ (a — c) == 62° 53' 1" tang. 0.29066 By (295), £ (A + C) == 128° cos. (ar. co.) 10.21066 n. %(A—C)=z 30° cos. 9.93753 i b = 72° tang. 0.48822 i (a + c) =z 103° 0' 25" tang. 0.63641 n. Ans. a = 165° 53' 26", c = 40° 7' 24". 2. Given in a spherical triangle two angles zs 170°, and : 2°, and the included side — 92° ; to find the other sides, Ans. a == 103° 6' 30", c = 11° 1730". § 85,] SPHERICAL OBLIQUE TRIANGLES. 185 Three angles given. 82. Problem. To solve a spherical triangle, when its three angles are given. Solution. If A, B, C are the angles of the given triangle, and a, b, c its sides 180° — A, 180° — B y 180° — C are the sides of the polar triangle, and 180° — a, 180° — 6, 180° — c the angles of the polar triangle, the sides are then given in the polar triangle ; to find the angles. For this purpose we may use the formulas of the preceding problem. 83. Corollary. Applying (272) to the polar triangle gives cos. C+ cos. A and B /nA0 . COS. C — r J — / : : ^ . (298) sin. A sin. B 84. Corollary. Equations (276-278) give, for the polar tri- angle, if we put 8 = i (A + B + C), (299) if we use (71 and 72), ,/ — cos. S cos. (S — A)\ /rt/w , v sin. i a - s/ ( r n . -^ L 1 , (300 2 \ sin. B sin. C J • i i // — cos- S cos. (S—B) \ /qm , sin. i b =V( : 7— — ?t I? ( 301 ) 2 \ sin. A sin. C / sin. ic = v| : t-^td I- ( 302 ) 2 \ sin. A sin. I* / 85. Corollary. Equations (281-283), applied to the polar triangle, give ,/cos. (S— B)cos.(S— C)\ cos. i a = \/( Wl — ET^T^ ' I > ( 303 ) 2 \ sm. B sin. C / ,/cos. (# — yl)cos.(#— • C)\ /OAy4X cos. J 6 = s/ 1 i^ /-.— ^ - 1 , (304) 2 \ sin. A sin. C / 16* 186 SPHERICAL TRIGONOMETRY. [CH. III. Three angles given. COS. i c = vC° 5 - {8 - A l C0S ^- B) ). (305) 2 v \ sm. A sm. B t 86. Corollary. Equations (284-286), applied to the polar triangle, give // — cos.tfcos. (S— A) \ ,/ — cos. S cos. (S — B) \ , OAW v taQ g- * h = fi^J^^^rcg • < 307 > ,/ — cos. #cos. (S — C) \ /OAO v tang, i c = V( C0S- is _ A T< tis^ B) y ( 308 ) 87. Corollary. Equation (273), applied to the polar tri- angle, is ~/. vo — cos. C — cos. (A4-B) /r »™v 2 sin. £ c)* H: r— A . v p ^ ' , (309 v * ' sin. ^1 sin. B which may be used like equation (279). 88. Example. Given in the spherical triangle ABC, the three angles equal to 89°, 5°, and 88° ; to solve the triangle. Ans. The three sides are 53° 10', 4°, and 53° 8'. 89. Theorem. The sine of half the sum of two sides of a spherical triangle is to the tangent of half their difference, as the cotangent of half the included angle is to the tangent of half the difference of the other two angles, that is, in ABC (figs. 32 and 33.), sin.|(a -f- c) : sin. £(a—c)=z cotan. J B : tang. £(A — C) (310) <§> 92.] SPHERICAL OBLIQUE TRIANGLES. 187 Neper's Analogies. Proof. This theorem is at once obtained by applying § 78 to the polar triangle. 90. Theorem. The cosine of half the sum of two sides of a triangle is to the cosine of half their difference, as the cotangent of half the included angle is to the tangent of half the sum of the other two angles , or in (figs. 32 and 33.), cos.£(a-\-c( : cos.4-(ez — c) — cota.n.£ B :tzng.£(A-\-C). (311) Proof. This theorem is at once obtained by applying § 79 to the polar triangle 91. Corollary. These two theorems, similar to <§> 78 and 79, were given by Neper for the solution of the case, in which two sides and the included angle are given. By means of them the other two angles can be found without the necessity of calculating the third side. In using them regard must be had to the signs of the terms by means of PI. Trig. § 61. 92. Examples. 1. Given in a spherical triangle two sides z= 149°, and — 49°, and the included angle = 88° ; to find the other an- gles. Solution. By § 89, £(a+ c) = 99° sin. (ar. co.) 10.00538 £ (a — c) = 50° sin. 9.88425 £B z=U° cotan. 0.01516 £ {A — C) — 38° 46' 10" tang. 9.90479 188 SPHERICAL TRIGONOMETRY. [CH. III. Neper's Analogies. By § 90, £ (a + c ) = "° cos. ar. co.) 10.80567 n. i(a— c) = 50° cos. 9.80807 £B = 44° cotan. 0.01516 i (A + C) = 103° 12' 31" tang. 0.62890 n. Arts. A ~ 141° 59' 41", Cz: 64° 27' 21". 2. Given in a spherical triangle two sides ca 13°, and = 9°, and the included angle a: 176° ; to find the other angles. Arts. 2° 24' 7", and 1° 40' 13". SPHERICAL ASTRONOMY. SPHERICAL ASTRONOMY. CHAPTER I. THE CELESTIAL SPHERE AND ITS CIRCLES. 1. Astronomy is the science which treats of the heavenly bodies. 2. Mathematical Astronomy is the science which treats of the positions and motions of the heavenly bodies. The elements of position of a heavenly body are (Geo. § 8) distance and direction. 3. Spherical Astronomy regards only one of the ele- ments of position, namely, direction, and usually refers all directions to the centre of the earth. 4. In spherical astronomy all the stars may, then, be regarded as at the same distance from the earth's centre upon the surface of a sphere, which is called the celes- tial sphere. Upon this imaginary sphere are supposed to be drawn vari- ous circles, which are divided into the well known classes of great and small circles. [B. p. 47.] 192 SPHERICAL ASTRONOMY. [CH. I Secondaries. Declination. Hour circles. " All angular distances on the surface of the sphere, to an eye at the centre, are measured by arcs of great circles." [B. p. 48.] 5. " Secondaries to a great circle are great circles which pass through its poles, and are consequently per- pendicular to it." [B. p. 48.] 6. "If the plane of the terrestrial equator be pro- duced to the celestial sphere, it marks out a circle called the celestial equator ; and if the axis of the earth be produced in like manner, it becomes the axis of the celestial sphere ; and the points of the heavens, to which it is produced, are called the poles, being the poles of the celestial equator." " The star nearest to each pole is called the pole star." [B. p. 48.] 7. " Secondaries to the celestial equator are called circles of declination; of these 24, which divide the equator into equal parts of 15° each, are called hour circles" " Small circles, parallel to the celestial equator, are called parallels of declination." [B. p. 48.] The parallels of declination correspond, therefore, to the terrestrial parallels of latitude, and the circles of declination to the terrestrial meridians. A certain point of the celestial equator has been fixed by astronomers, and is called the vernal equinox. The circle of declination, which passes through the vernal equinox, bears the same relation to other circles of $ 11.] CELESTIAL SPHERE AND ITS CIRCLES. 193 Right ascension. Horizon. declination, which the first meridian does to other terrestrial meridians. 8. " The declination of a star is its angular distance from the celestial equator," measured upon its circle of declination. [B. p. 49.] 9. The right ascension of a star is the arc of the equator intercepted between its circle of declination and the vernal equinox. [B. p. 49.] Right ascension is either estimated in degrees, minutes, &c. from 0° to 360° ; or in hours, minutes, &,c. of time, 15 de- grees being allowed for each hour, as in Sph. Trig. § 3. The positions of the stars are completely determined upon the celestial sphere, when their right ascensions and declina- tions are known. Catalogues of the stars have accordingly been given, containing their right ascensions and declina- tions. [B. Table viii. p. 80.] 10. " The sensible horizon is that circle in the heav- ens, whose plane touches the earth at the spectator." " The rational horizon is a great circle of the celes- tial sphere parallel to the sensible horizon." [B. p. 48.] 11. The radius, which is drawn to the observer, is called the vertical line. The point, where the vertical line meets the celestial sphere above the observer, is called the zenith ; the op- posite point, where this line meets the sphere below the observer, is called the nadir. 17 194 SPHERICAL ASTRONOMY. [CH. I. Prime vertical. Cardinal points. Hence the vertical line is a radius of the celestial sphere perpendicular to the horizon ; and the zenith and nadir are the poles of the horizon. [B. p. 48.] 12. Circles whose planes pass through the vertical line are called vertical circles. [B. p. 48.] The vertical circles are secondaries to the horizon. 13. The vertical circle at any place, which is also a circle of declination, is called the celestial meridian of that place. [B. p. 48.] The plane of the celestial meridian of a place is the same with that of the terrestrial meridian. 14. The points, where the celestial meridian cuts the horizon, are called the north and south points. [B. p. 48.] The north point corresponds to the north pole, and the south point to the south pole. 15. The vertical circle, which is perpendicular to the meridian, is called the prime vertical. [B. p. 48.] 16. The points, where the prime vertical cuts the horizon, are called the east and west points. [B. p. 48.] "To an observer, whose face is directed towards the south, the east point is to his left hand, and the west to his right hand. Hence the east and west points are 90° distant from the north and south. These four are called the cardinal points." "§> 18.] CELESTIAL SPHERE AND ITS CIRCLES. 195 Altitude. Azimuth. H The meridian of any place divides the heavens into two hemispheres, lying to the east and west; that lying to the east is called the eastern hemisphere, and the other the western hemisphere." 17. The altitude of a star is its angular distance from the horizon, measured upon the vertical circle passing through the star. [B, p. 48.] 18. The azimuth of a star is the arc of the horizon intercepted between its vertical circle and the north or south point. [B. p. 48.] A star may be found without difficulty, when its altitude and azimuth are known. But these elements of position are constantly varying. 196 ■■*- , Al, -VSTRONOMY. [CH II. rixrtl stars. Planets. Constellations. OllAPTKR 1L TIIK niruNvi, MOTION. 19. " Stars are distinguished into two kinds, fixed and wandering." [H. p. 15. J Most of the stars are fixed, thai is, retain constantly almost the same relative position ; so that the same celestial globes and maps continue to be accurate representations of the fir- mament for many years. This is a fact of fundamental impor- tance, and furnishes the fixed points for arriving at a complete knowledge of the celestial motions. Small changes of position have, indeed, been detected even in the fixed stars, as will be shown in tho course of this treatise; but these changes are too small to disturb the general fact ; they are, indeed, too small ever to have been detected, if the positions of the stars had been subject to great variations. 20. Of the wandering stars there are eleven, which are called planets. They are Mercury ( g )> Venus ( ? ), the Earth (0), Mars ( $ ), Vesta (g), Juno ($ ), Pal- las ($), Ceres (?), Jupiter (#), Saturn (h), and Cranus ( * ). [B. p. 45.] 21. For the sake of remembering the stars with greater ease, they have been divided into groups called constellations ; and to uive distinctness to the constella- tions, they have been supposed to be circumscribed by $ 22.] DIURNAL MOTION. 197 Diurnal motion. the outlines of some figure which they were imagined to resemble. [B. p. 45.J The stars have also been distinguished according to their brilliancy, as of the^rs^, second, &c. magnitude. Proper names have been given to the constellations and to the most remarkable stars. The catalogues and the maps of the stars are now so accu- rate, that no new star could appear without being detected ; and any change in the place of any of the larger stars would be immediately discovered. 22. All the stars appear to have a common motion, by which they are carried round the earth from east to west in 24 hours. This rotation of the heavens, or of the celestial sphere, is called the diurnal motion. By its diurnal motion, the celestial sphere rotates, with the most perfect uniformity, about its axis. The pole star would, therefore, if it were exactly at the pole, remain stationary ; but since it is not exactly at the pole, it revolves in a very small parallel of declination about the stationary pole. Any star in the equator revolves in the plane of the equator, and all other stars revolve in the planes of the parallels of declination in which they are situated. If O (fig. 34.) is the place of the observer, NES W his horizon, Z his zenith, P and P' the poles, the star which is at the distance from P, PM = PM! will appear to describe the circumference MH' M' H. It will rise in the east at H and set at H', if the distance PM' 17* 198 SPHERICAL ASTRONOMY. [CH. II. Sideral time. from the pole is greater than the altitude PN of the pole. But if its distance from the pole PL == PL' is less than PN, the star will not set, but will describe a circle above the horizon ; and if its distance from the pole PG = PG< is greater than the greatest distance PS from the pole to the horizon, the star will never rise so as to be seen by the ob- server at O, but will describe a circle below the horizon. 23. The time which it takes a star to pass from any position round again to the same position is called a sideral day, that is, literally, a star-day. This day is divided into 24 hours, and clocks regulated to this time are said to denote sideral time. [B. p. 147.] 24. Each point of the celestial equator passes the meridian once in a sideral day ; and the arc contained between two hour circles passes it in a sideral hour. The sideral time, therefore, which has elapsed since the vernal equinox was upon the equator, is equal to the right ascension of the meridian expressed in time. [B. p. 208.] The meridian changes its right ascension at each instant, precisely as if the celestial sphere were stationary, while the observer, with his meridian and zenith, is carried uniformly round the earth's centre from west to east once in a sideral day. <§> 28.] DIURNAL MOTION. 199 Hour angle. Amplitude. 25. The angle MPB (fig. 35.), which the circle of declination of the star makes with the meridian, is called its hour angle. While the star moves from the point M in the meridian to the point B with an uniform motion, the arc MP is carried to the position PB, and the angle MPB is described with an uniform motion. This angle converted into time is, then, the sideral time since the passage of the star over the meridian. 26. Corollary. The difference of the right ascensions of the star and of the meridian is the hour angle of the star. 27. The distance of a star from the east or west points of the meridian, at the time of its rising or setting, is the amplitude of the star. [B. p. 48.] 28. ^Problem. To find the altitude and azimuth of a star , when its declination and hour angle are known, and also the latitude of the place. Solution. If P (fig. 35.) is the pole, Z the zenith, and 3 the star ; we have PZ — polar dist. of zenith == co. latitude = 90° — L, PN = 90° — PZ—L, PB = polar dist. of star = p, = co. declination of star, when it is on the same side of the equator with the pole. = 90° -)- declination of star, when it is on the differ- ent side of the equator from the pole. = 90° q= D, 200 SPHERICAL ASTRONOMY. [CH. II. To find a star's altitude and azimuth. ZB = zenith dist. of star = z, == co. altitude of star, when it is above the horizon. = 90° -f- depression of star, when it is below the horizon. ZPB = #'s hour angle = h, PZB = azimuth of star counted from the direction of the elevated pole. = a = azimuth, when less than 90°, = 180° — azimuth, when greater than 90°. There are, then, given in the spherical triangle PZB, the two sides PZ and PB, and the included angle ZPB ; so that the side BZ and the angle PZB can be calculated by Sph. Trig. § 44. If we let fall the perpendicular BC upon PZ> tang. PC— cos. h tang. (90° =pD)==p cos. A cotan. D (312) CZ= PZ— PC— 90° — (L + PC) or —PC— PZ=(L + PC)— 90°. (313) Hence, by (241), cos. PC : sin. (L + PC) z= d= sin. Z> : cos. z ; (314) in which formulas the upper sign is used when the star is upon the same side of the equator with the elevated pole, that is, when D and L are of the same name ; and, by (242), sin. PC : cos. (L + PC) — cotan. h : cotan. a. (315) 29. Corollary, When the altitude and azimuth are both to be found, the calculation by the above method is as short as <§> 30.] DIURNAL MOTION. 201 To find a star's altitude. by any other; but when, as is usually the case, the altitude only is required, the following method is preferable. We have PZ + PB = 180° — L =f D = 180° — (L ± D) PB — PZ=^D + L = L^D)*, whence, by (249) and (250), cos.z = — cos. (L^zD)-{-2cos.Dcos.L(cos.^h) 2 (316) cos.z = cos.(Z=pZ>) — 2cos. ZJcos. Z,(sin. J/*) 2 , (317) which may be used at once, and {317) may be calculated by the aid of the column of Rising in Table XXIII. The rule obtained from (317) is the same with that on p. 250 of the Navigator, remembering that when the star is above the hori- zon cos. z s= sin. ^c's alt. (318) But when the star is below the horizon cos. z z= — sin. sjc's depression. (310) 30. Corollary. If the given hour angle is 6 h = 90°, the problem is at once reduced to the solution of a right triangle. We in this case have, by Napier's Rules, cos, z == sin. L cos. p, or sin. ^c's alt. z= ± sin. L sin. D (320) cotan. a z= cos. L cotan. p cotan. sfc's azimuth == dc= cos. L tang. D. (321) The upper sign is to be used in formulas (320) and (321), when the declination is of the same name with the latitude; otherwise the lower sign. In the former case, therefore, the 202 SPHRICAL ASTRONOMY. [CH. II. To find a star's altitude and azimuth. star is above the horizon when its hour angle is six hours, and on the same side of the prime vertical with the elevated pole ; but, in the latter case, it is below the horizon, and on the same side of the prime vertical with the depressed pole. 31. Corollary, If the star is in the celestial equator, as in (fig. 36.), we have in the right triangle BZQ } BQ == BPQ — h ZQ — L QZB = 180° — a, whence cos. z t= cos. L cos. h, or sin. ^c's alt. = cos. L cos. h (322) cotan. (180° — a) s± sin. L cotan. ft, or cotan. a = — sin. L cotan. h, (323) Hence, if the hour angle is less than six hours, the star which moves in the celestial equator is above the horizon, and on the same side of the prime vertical with the depressed pole ; but if the hour angle is greater than six hours, this star is be- low the horizon, and on the same side of the prime vertical with the elevated pole. 32. Corollary. If the place is at the equator, as in (fig. 37.), the celestial equator ZE is the prime vertical, so that if the hour circle PB is produced to C, we have in the right tri- angle ZBC ZC— ZPBz= h BZC = 90° — a BC— D, § 33.] DIURNAL MOTION. 203 To find a star's altitude and azimuth. whence cos. z = cos. D cos. h, or sin. 2(c's alt. — cos. D cos. h (324) cotan. (90° — a) txz sin. h cotan. D, or tang, a == sin. h cotan. D ; (325) so that the star is above the horizon when the hour angle is less than six hours, and below the horizon when the hour angle is greater than six hours. 33. Examples. 1. Find the altitude and azimuth of Aldebaran to an ob- server at Boston, in the year 1830, when the hour angle of this star is 3* 25 m 12 s . Solution. We find by tables VIII and LIV D— 16° 11' N. L — 42°21'N. Hence h = 3* 25 m 12* log. col. Ris. 4.57375 L ts 42° 21' cos. 9.86867 D=i 16° 11 ; cos. 9.98244 26599 4.42486 ■ D = 26° 10 ; nat. cos. 89752 alt. — 39° 10' nat. sin. 63153 sec. 10.11052 h — 51° 18' sin. 9.89233 D. cos. 9.98244 azimuth from South = 75° 10' sin. 9.98529 204 SPHERICAL ASTRONOMY. [CH. II, To find a star's altitude and azimuth. 2. Find the altitude and azimuth of Aldebaran at Boston* in the year 1830, six hours after it has passed the meridian. Solution. By formulas (320) and (321), L = 42° 21' sin. 9.82844 cos. 9.86867 D = 16° IV sin. 9.44516 tang. 9.46271 alt. = 10° 49' sin. 9.27360 azimuth from North = 77° 54' cotan. 9.33138 3. Find the altitude and azimuth of a star in the celestial equator, to an observer at Boston, when the hour angle of the star is 3*25™ 12 s . Solution. By formulas (322) and (323), L = 42° 21' cos. 9.86867 sin. 9.82844 h z= 51° 18' cos. 9.79605 cotan. 9.90371 alt. = 27° 31 7 sin. 9.66472 azimuth from South = 61° 39' cotan. 9.73215 4. Find the altitude and azimuth of Aldebaran to an ob- server at the equator, in the year 1830, when the hour angle of the star is 3*25™ 12 s . Solution. By formulas (324) and (325), Z>z=16°ll' cos. 9.98244 cotan. 10.53729 h = 51° 18' cos. 9.79605 cotan. 9.90371 alt. = 36° 54' sin. 9.77849 azimuth from North = 70° 5' tang. 10.44100 § 33.] DIURNAL MOTION. 205 Altitude and azimuth. 5. Find the altitude and azimuth of Fomalhaut to an ob- server at Boston, in the year 1840, when its hour angle is 2*3^20*. Ans. Its altitude . . =11° 51'. Its azimuth from the South == 15° 24'. 6. Find the altitude and azimuth of Dubhe to an observer at Boston, in the year 1840, when its hour angle is 9 h 30 m . Ans. Its altitude . . — 19° 14'. Its azimuth from the North =z 17° 15'. 7. Find the altitude and azimuth of Fomalhaut to an ob- server at Boston, in the year 1840, when its hour angle is 6\ Ans. Its depression below the horizon t=z 19° 58'. Its azimuth from the South = 38° 31'. 8. Find the altitude and azimuth of Dubhe to an observer at Boston, in the year 1840, when its hour angle is 6\ Ans. Its altitude . . = 36° 44'. Its azimuth from the North = 69° 3'. 9. Find the altitude and azimuth of a star in the celestial equator to an observer at Stockholm, when its hour angle is 2*3"' 20*. Ans. Its altitude . . == 25° 58'. Its azimuth from the South = 34° 45'. 10. Find the altitude and azimuth of a star in the celestial 18 206 SPHERICAL ASTRONOMY. [CH. II. Altitude of a star in the prime vertical. equator, to an observer at Stockholm, when the hour angle is 9* 30 m . Ans. Its depression below the horizon ■= 23° 51'. Its azimuth from the North = 41° 44'. 11. Find the altitude and azimuth of Fomalhaut, to an ob- server at the equator, in the year 1840, when its hour angle is2*3 w 20*. Ans. Its altitude . . — 47° 45'. Its azimuth from the South = 41° 4'. 12. Find the altitude and azimuth [of Dubhe, to an observ- er at the equator, in the year 1840, when its hour angle is 9*30 m . Ans. Its depression below the horizon = 21° 24'. Its azimuth from the North — 17° 30'. 34. In the triangle ZPB (fig. 2.) other parts might be given instead of the two sides ZP, PB, and the included angle P, and the triangle might be resolved. Of the problems thus derived, we shall only, for the present, consider two cases. 35. Problem. To find a given star's hour angle and altitude, when it is upon the prime vertical. Solution. The angle PZB is, in this case, a right angle, and if we use the preceding notation, we have cos. 7i zc cotan. L cotan. p — ± cotan. L tang. D (326) cos. z z=. cos. p cosec. L, or sin. &'s alt. = db sin. D cosec. L ; (327) m § 38.] DIURNAL MOTION. 207 Altitude of a star in the prime vertical. so that when the declination and latitude are of the same name, the hour angle is less than 6 hours, and the star is above the horizon ; but when the declination and latitude are of differ- ent names, the hour angle is greater than 6 hours, and the star is below the horizon. 36. Scholium. The problem is, by Sph. Trig. § 27, impos- sible, when the declination is greater than the latitude; so that, in this case, the star is never exactly east or west of the observer. 37. Scholium. The problem is, by Sph. Trig. § 28, indeter- minate, when the latitude and declination are both equal to zero ; so that, in this case, the star is always upon the prime vertical. 38. Examples. 1. Find the hour angle and altitude of Aldebaran, when it is exactly east or west of an observer at Boston, in the year 1840. Ans. The hour angle = 4 A 45 w 44*. The altitude m 24° 26'. 2. Find the hour angle and altitude of Fomalhaut, when it is exactly east or west of an observer at Boston, in the year 1840. Ans. The hour angle . . s= 8 h 40 m 50'. The depression below the horizon = 48° 49'. 3. Find the hour angle and altitude of Dubhe, when it is 208 SPHERICAL ASTRONOMY. [CH. II. Time of a star's rising. exactly east or west of an observer at Boston, in the year 1840. Arts. Dubhe is never upon the prime vertical of Boston. 4. Find the hour angle and altitude of Canopus, when it is exactly east or west of an observer at Boston, in the year 1840. Ans. Canopus is never upon the prime vertical of Boston. 39. Problem. To find the hour angle and amplitude of a star, when it is in the horizon. Solution. In this case the side ZB (fig. 35.) of the triangle ZPB is 90°. The corresponding angle of the polar triangle is, therefore, a right angle, and the polar triangle is a right triangle, of which the other two angles are 180° — PZ — 180° — (90° — L) — 90° + L y and 180° — PB = 180° — (90° =p D) — 90° d= D. The hypothenuse of the polar triangle is 180° — h, and the leg, opposite the angle, 90° ± D, is 180° — a. Hence, by Sph. Trig. § 40, and PI. Trig. § 60 and 62, — cos. 7i = zh tang. L tang. Z>, or cos. h =. =F tang. L tang. D (328) — cos. a = ^p sin. D sec. L, or cos. a = d= sin. D sec. L ; (329) in which the upper sign is used when the latitude and declina- tion have the same name, and the lower sign when they have different names ; so that in the former case the hour angle is greater than 6 hours, and the azimuth is counted from the <§> 41.] DIURNAL MOTION. 209 Time of a star's rising. direction of the elevated pole ; but in the latter case, the hour angle is less than 6 hours, and the azimuth is counted from the direction of the depressed pole. The amplitude is the difference between the azimuth a and 90°. Hence cos. ^c's azim. = sin. ^c's amp. = sin. D sec. L. (330) 40. Scholium. The problem is, by Sph. Trig. § 41, impos- sible, when the sum of the declination and latitude is greater than 90° ; so that, in this case, the star does not rise or set. 41. Examples. 1. Find the hour angle and amplitude of Aldebaran, when it rises or sets, to an observer at Boston, in the year 1840. Ans. The hour angle == 7 h l m 21*. The amplitude = 22° 9' N. 2. Find the hour angle and amplitude of Fomalhaut, when it rises or sets, to an observer at Boston, in the year 1840. Ans. The hour angle = S h 5l m 18*. The amplitude — 43° 19' S. 3. Find the hour angle and amplitude of Dubhe, when it rises or sets, to an observer at Boston, in the year 1840. Ans. Dubhe neither rises nor sets at Boston. 4. Find the hour angle and amplitude of Canopus, when it rises or sets, to an observer at Boston, in the year 1840. Ans. Canopus neither rises nor sets at Boston. 18* 210 SPHERICAL ASTRONOMY. [CH. III. Determination of the meridian line. CHAPTER III. THE MERIDIAN. 42. The intersection of the plane of the meridian with that of the horizon is called the meridian line. 43. Problem. To determine the meridian line. Solution. First Method. Stars obviously rise to their great- est altitude in the plane of the meridian ; so that if their progress could be traced with perfect accuracy, and the instant of their rising to their greatest height be observed, the direc- tion of the meridian line could be exactly determined. But stars, when they are at their greatest height, change their altitude so slowly, that this method is of but little practical value. Second Method. A star is evidently at equal altitudes when it is at equal distances from the meridian on opposite sides of it. If, therefore, the direction and altitude of a star are ob- served before it comes to the meridian ; and if its direction is also observed, when it has descended again to the same alti- tude, after passing the meridian ; the horizontal line, which bisects the angle of the two horizontal lines drawn in the directions thus determined, is the meridian line. Third Method. [B.p. 147.] The time which elapses between the superior and inferior passage of a star over the meridian is just half of a sideral day. If, then, a telescope were placed so as <§> 43.] THE MERIDIAN. 211 Meridian determined by circumpolar stars. to revolve on a horizontal axis in the plane of the meridian, the two intervals of time between three successive passages of a star over the central wire, must be exactly equal. But if the vertical plane of the telescope is not that of the meridian, these two intervals will not be equal, and the position of the telescope must be changed until they become equal. Thus, if ZMmN (fig. 37.) is the plane of the meridian, Z S s T that of the vertical circle described by the telescope, MS WsmE the circle of declination described by the star about the pole P; this star will be observed at the points S and s instead of at the points M and m. Now the star de- scribes the circle of declination with an uniform motion, and therefore the arc SP moves uniformly with the star around the pole, so that the angle SP M is proportional to the time of its description; that is, the angle SP31, reduced to time, denotes the sideral time of its description. Let then T = the sideral time of describing the arc SM } t = the sideral time of describing the arc 5 m, I = interval from the observation at £ to that at s, i == interval from the observation at s to that at S } <5 i = the difference of these two intervals ; we have then, in sideral time, I = 12 h — T — t = 12 h — ( T + t) i = 12* + T + t = 12* + ( T + t ) di=ri — I=:2(T+t); (331) so that if T and t were equal to each other, and they are nearly so in the case of the pole-star, we should have 2L2 SPHERICAL ASTRONOMY. [CH. III. Meridian determined by circumpolar stars. di = 4 T= 4t that is, the time of describing the arc MS or m s is nearly one quarter part of the difference between the intervals. But the error of this result can be calculated without much difficulty. For this purpose, let L == the latitude of the place, = 90° — PZ, p s=s the polar distance of the star == PS — P s, a = the azimuth of ZS T = TN = TZN. The arcs MS and m s are so small, that they do not differ sensibly from the arcs of great circles drawn from S and s perpendicular to ZPN. If, then, in the two right triangles PSM and ZSM f PM and ZM are the middle parts, SM, co. SZM, and co. SPM are the adjacent parts, so that sin. PM : sin. ZM = cotan. SPM : cotan. SZM 1 1 tang. SPM ' tang. SZM = tang. SZM : tang. SPM. But Z3I= ZP — PM = 90° — L —p and the angles SZM and SPM are so small, that they are sensibly proportional to their tangents, whence sin. p : cos. (p + L) = a : SPM, (332) or a : SPM =? sin. p : cos. p cos. L — sin. p sin. L fats 1 : cotan. p cos. Z< — sin. L 9 § 43.] THE MERIDIAN. 213 Meridian determined by circumpolar stars. Table A [B. p. 151.] and if T is expressed in sideral hours T. 15° ±= SPM — a cotan. p cos. L — a sin. L. In like manner, we find t . 15° = 5 Pm = a cotan. p cos. L -f- a sin. L. Hence, by (331) (T+ t) 15° = JJi.15° — 2« cotan.pcos. i a cotan. p cos. L = £$i. 15° T. 15° = ^ (52.15° — a sin. Z, £.15° ±= | &\t# -f a sin. L a = J (5 i . 15° tang. j9 sec. Z. (333) T = £ 45.] THE MERIDIAN. 219 Determination of the meridian line. so that the error in the time of the upper transit is £.300* — 25* ~ 75 s — 25* M 50% and the error in the time of the lower transit is £.300* + 25 s ±- 75 s + 25 s = 100 s — l m 40*. The times of the star's passing the meridian the second day were, then, 7 h 41™ + l m 40 s = T 42 w 40* A. M. and T 41- 32 s — 50 s = l h 40 w 42 s P. M. The error in the azimuth of the instrument was 9' 19" to the west of north. 5. An observer at Boston, wishing to determine his meridian line, on the morning of January 1, 1840, observed, by means of a clock regulated to solar time, the superior transit of Y UrssB Majoris at 5 h 6 m 54 s A. M,, and the inferior transit of Polaris at 6' 1 12 w 23 s A. M. What was the azimuth error in the position of the transit instrument? Solution. The interval between these two transits is 6 h 12™ 23 s — 5 h 6 m 54 s = I* 5 m 29*. But, by the Nautical Almanac, 12* + R. A. of Polaris = 13* \ m 59 s R. A. of y Ursae Majoris = 1 i h 45 m 25 s Sideral Interval = 1* 16™ 34 s Solar Interval = 1* 16™ 22 s Observed Interval = 1* 5 m 29 s Error of Interval = 10 m 53* = 653*. 220 SPHERICAL ASTRONOMY. [CH. III. Determination of the meridian line. Now for 1000" of azimuth error, and the latitude of Boston, Table C gives, since Dec. of y Ursae Majoris . . = 54° 35' Error of lower trans, of Polaris . — 1866 s Error of upper trans, of y Ursae Majoris =; 25 s Sum of errors . . . . — 1891 s Then the proportion 1891* : 653 s = 1000" : azimuth error, gives azimuth error — 345" = 5' 45'' W. 6. An observer, at Boston, wishing to determine his merid- ian line, in the evening of December 17, 1839, observed by means of a clock regulated to solar time, the superior transit of « Cassiopeae at 6 h 4S m 35 s P. M., and that of Polaris at 6 h 53"* 15 s P. M. What was the azimuth error in the posi- tion of the transit instrument ? Solution, By the Nautical Almanac, R. A. of Polaris = I* 2 m 26 s R. A. of a Cassiopeae '== h 31 m 28 s Sideral Interval = 0* 30 m 58 s Solar Interval — 0* 30™ 53 s Observed Interval — h 4 m 40 s Error of Interval = h 26 m 13 s = 1573 s . Now Table C gives', for 1000" of azimuth error and the lati- tude of Boston, since § 45.] THE MERIDIAN. 221 Determination of the meridian line. Dec. of a Cassiopeae = 55° 40' Error of trans, of Polaris = 1777* Error of trans, of a Cassiopeae s= 26* DifF. of errors — 1751* Then, the proportion 1751 s : 1573 s = 1000" : azimuth error gives azimuth error z* 900" : = V 30" E. 7. Calculate the proportional logarithm of 0° 2' 33". Ans. 1.8487. 8. Calculate the proportional logarithm of 2° 59' 12". Ans. 0.0019. 9. Calculate the corrections of tables A and B, when the latitude is 54°, and the star's polar distance 20°. Ans. Corr. A = 125 s . Corr. B = 38' 48". 10. Calculate the corrections of table C, when the latitude is 20°, and the polar distance 5°. Ans. For the upper transit, corr. C = 091 s . For the lower transit, corr. C = 737 s . 11. An observer at Boston, in the year 1840, wishing to determine his meridian line, observed three successive transits of Polaris, by means of a clock regulated to solar time. The first lower transit was observed at 6 h A. M., the next transit at 19* 222 SPHERICAL ASTRONOMY. [ctt. III. Determination of the meridian line. 7 h 2 CT 11 s P. M., and the second lower transit at 5* 56 m 4 s A. M. What was the time of the star's passing the meridian the second morning? and what was the azimuth error in the po- sition of the instrument ? Ans. The time of third merid. trans, was & 1 58 m 11 s A. M. The azimuth error = V 8" W. 12. An observer at Boston, wishing to determine his me- ridian line by means of a clock regulated to solar time, ob- served the inferior tfansit of Polaris on April 4, 1839, at h A. M., and the superior transit of n Ursse Majoris at h 53 m 59* A. M. What was the azimuth error in the position of his transit instrument? The R. A. of Polaris is V 1 m 50*, that of n Ursse Majoris is 13* 4i» 14% and the declination of n Ursae Majoris is 50° 7' N. Ans. The azimuth error = T 18" W. 13. An observer at Boston, wishing to determine his me- ridian line, in the evening of May 1, 1839, observed by means of a clock regulated to solar time, the lower transit of Polaris at 9* 49 71 22-" P. M., and that of « Cassiopeae at 9* 52 m P. M. What was the azimuth error of the instrument? The R. A. of Polaris = t* m 56 s . The R. A. of « Cassiopese == 0* 31 771 22 s . The Dec. of « Cassiopese = 55° 39' N. Ans. The azimuth error = 18' 23" W. <$> 46.] LATITUDE. 223 Latitude found by meridian altitudes. CHAPTER IV. LATITUDE. 46. Problem. To find the latitude of a place. Solution. The latitude of the place is evidently, from (fig. 34.), equal to the altitude of the pole ; so that this problem is the same as to find the altitude of the pole, which would be done without difficulty if the pole were a visible point of the celestial sphere. First Method. By Meridian Altitudes. [B. p. 166-175.] Observe the altitude of a star at its transit over the meridian, and let A z= the altitude of the star, A 1 =z >fc's dist. from point of horizon below the pole ; then, if the notation of § 28 is used, it is evident, from (fig. 34.), that L = A f ^pp; (344) the upper sign being used when the transit is a superior one, and the lower sign when it is an inferior one. I. Suppose the observed transit to be a superior one ; then, if it passes upon the side of the zenith opposite to the pole, we have A' = 180° — A, p = 90° =p 2>, 224 SPHERICAL ASTRONOMY. [CH. IV. Latitude found by meridian altitudes, arid (344) becomes L=z90° —{A±D) = (90 — A)±D — z±D , (345) the upper sign being used when the declination and latitude are of the same name, and the lower sign when they are of different names. But if the star passes upon the same side of the zenith with the pole, we have A' = A, p = 90° — D } and (344) becomes L — (A -f D) — 90° = D — (90° — A) = D — z. (346) II. If the transit is an inferior one, we have A' — A y p — 90° — D y and (345) becomes L =x (A — D) + 90° = A + (90° — D). (347) Equations (345) and (346) agree with the rule of Case I, [B. p. 166.], and (347) with Case II, [B. p. 167.] III. If both transits are observed, and if A* and A are re- ferred to the upper transits, and A 2 = the altitude at the lower transit, we have, by (344), Lz=z A —p the sum of which is L = i{A' + A 1 ); (348) <§> 46.] LATITUDE. 225 Latitude found by a single altitude. so that the latitude is determined in this case without knowing the star's declination. Second Method. By a Single Altitude. Observe the altitude and the time of the observation. I. If the star is considerably distant from the meridian, we have given in the triangle PBZ (fig. 35.), PB, BZ, and BPZ to find PZ, which may be solved by Sph. Trig. § 59, and gives, by the notation of § 28, tang. PC — cos. h tang, p = ± cos. h cotan. D (349) cos. ZC z= cos. PC. cos. z sec. p — ± cos. PC . cos. z cosec. D, (350) in which the upper sign is used if the declination and latitude are of the same name, otherwise the lower sign. 90° — L = PZ = PC ± ZC L = 90° — (PC do ZC) ; (351) in which both signs may be used if they give values of L contained between 0° and 90°, and in this case other data must be resorted to, in order to determine which is the true value of L. Scholium. The problem is, by Sph. Trig. § 61, impossible, if the altitude is greater than the declination, when the hour angle is more than six hours. II. If the latitude is known within a few miles, it may be exactly calculated by means of (317), or cos.z:= cos. [90°— (L+p)]— 2cos.£cos.Z>(sin.p) 2 . (352) 226 SPHERICAL ASTRONOMY. [CH. IV. Latitude found by a single altitude. But if A is the star's observed altitude, and^4 2 its meridian altitude at its upper transit, (344) gives A x — L + p, or = 180° — (L+p), and (352) becomes, by transposition, sin. A 2 = sin. A -f- 2 cos. L cos. D (sin. J A) 2 ; (353) from which the meridian altitude may be calculated by means of table XXIII, as in the Rule. [B. p. 200.] III. A formula can also be obtained from (281), which is particularly valuable when the star is, as it always should be in these observations, near the meridian. In this case we have in (281) applied to PBZ 2sz=90° — L-\-p+z — 180° — L + p — A 2s—2PZ—L+p — A =zA 1 —Aov = 180° — (A x + A) (354) 25— 2PBz= 180° — L—p — A — 180° — (A 1 + A)ov=A 1 — A; (355) and if these values' are substituted in (281), after it is squared and freed from fractions, they give (sin. J h) 2 cos. L cos. Z>:=sin. J (^1 x — ^4) cos. J (A l -\-A) i (356) or sin.J(J. 1 — ^l)=(sin.J/«) 2 cos. J Lcos.Z>sec.J(-4 1 4-^); (357) and if, in the second member of this equation, the value of A t is used, which is obtained from the approximate value of the latitude, the difference between the observed and the meridian altitudes may be found at once ; and this difference is to be added to the observed altitude to obtain the meridian altitude. § 46.] LATITUDE. 227 Single altitude near the meridian. IV. If the star is very near the meridian, %(A 1 — A) and J h will be so small, that we may put mn.i{A 1 — A) _ i(A l —A) _ j(A' — A) sm.jh _ sin. 1" - I" — 1 '" ' sin.F — * ' or sin. i (A 1 — A) = i (A' — A) sin. \" sin. J li — J h sin. 1* = l 5 h sin. 1" ; which, substituted in (357), give, by supposing A x equal to A in the second member, which is very nearly the case, A , — A = y> h 2 sin. 1* cos. L cos. Z> sec. Jl 1 . (358) This value of A 1 — A is proportional to h 2 , so that if it were calculated for A= I s , any other value might be calculated by multiplying byA 2 .% Now Table XXXII, of the Navigator, contains the values of A 1 — A for all latitudes and for all declinations less than 24°, excepting a few latitudes in which the meridian transit of the observed body is too near the zenith for this observation to be accurate ; and Table XXXIII contains all the values of h 2 , where h is less than 13 m . V. If the observed star is very near the pole, we have in (349) tang. PC =: cos. h tang, p ; (359) so that as p is very small, PC must be likewise small, and we have tang. PC PC cos. h = = tang, p p PC =p cos. h; (360) 228 SPHERICAL ASTRONOMY. [CH. IV. ; _ , — Altitude of the pole star. and, by PI. Trig. § 22, cos. PC == 1, sin. D = cos. p = 1, whence, by (350), and (351), cos. ZC == cos. z, ,ZC = z 9 L=90 o — PC—ZC=90° — z—PC = A — p cos. h ; (361) so that p cos. 7i may be regarded as a correction to be sub- tracted from A when it is positive, that is, when the hour angle is less than 6 hours, or greater than 18 hours; and it is to be added when the hour angle is greater than 6 hours and less than 18 hours. The table [B. p. 206.] for the pole star was calculated for the year 1840, when its R. A. = l h 2 m ; its dec. == 88° 27' nearly. Third Method. By Circummcridian Altitudes. I. If several altitudes are observed near the meridian, each observation may be reduced separately by (357) and (358), and the mean of the resulting latitudes is the correct latitude. II. But if (358) is used, the mean of the values of A x — A is evidently obtained by multiplying the mean of the values of h 2 by the constant factor ; and if to the mean of the values of A x — A, the mean of the values of A is added, the sum is the mean of the values of A lt whence precisely the same mean of resulting latitude is obtained as by the former method, but with much less calculation. <§> 46.] LATITUDE. 229 By circummeridian altitudes. III. If the star is changing its declination in the course of the observations, this change may, in all cases which can occur if the hour angle is small, be neglected in the value of cos. D. But the value of A x will not, in this case, be at each observation equal to the meridian altitude, but will differ from it by the difference of the star's declination, Let the change of the star's declination in one minute be denoted by , which is positive when the star is approaching the elevated pole; and if h is J;he star's hour angle at the time of observation, which is negative before the star arrives at the meridian and afterwards positive, the whole change of declination is hdD, so that the correct meridian altitude is The mean of the values of the corrected meridian altitude is, therefore, equal to the mean of the values of A 1 diminished by the mean of the values of h $ D ; and, if H denotes the mean of the hour angles h (regard being had to their signs), the correct meridian altitude is the mean of the values of A x diminished by H$D. Fourth Method. By Double Altitudes. I. Let two altitudes of a star, which does not change its xleclination, be observed, and the intervening time. Then (fig. 39.) let ^be the zenith, P the pole, 8 and 8' the po- sitions of the star; join ZS, Z8\ PS, PS', and SS'M; draw PI 1 to the middle Tof 88', join ZT, and draw ZV perpendicular to PT. Let P=lPS—PS — 90° — D, SPS'=i elapsed time == h 8T=A ~8'T, PT— 90° — B 20 230 SPHERICAL ASTRONOMY. [CH. IV. By double altitudes. A x b£ 90° — ZS, A\ = 90° — ZS 1 ZTP =z T, ZT=F, ZV=C TV= Z, PV=90° — E; in which D and B are positive, when the latitude and decli- nation are of the same name, but negative, if they are of con- trary names ; Z is positive, if the zenith is nearer the elevated pole than the point M. Now the triangle TPS gives sin. A == sin. PS sin. SP T — cos. D sin. £ k cos. PS =: cos. P T cos. A s or sin. D = sin. B cos. A, (362) or cosec. A = sec. D cosec. £ h (363) cosec, B = cos. A cosec. D. (364) The triangles ZTS and ZTS' give sin. A 1 = cos. Jr cos. J. — sin. J 1 , sin. .4 sin. T, (365) sin. A ', saa cos. .F cos. ^4 -{- sin. .F. sin. -4 sin. T, (366) The sum and difference of which is, by (36) and (37), sin. i{A x + A\) cos. £ (A\ — A x ) = cos. F cos. ^4, (367) sin. J (■*', — 4,) cos. | (A| + A\) z^sin.Fsin.^sin.T 7 . (368) But triangle ZTV gives sin. C = sin. F sin. T, (369) cos. F — cos. C cos. ^; (370) which, substituted in (367) and (368), give sin. C= sin. £(A\ — A J cos. £ (A 2 + -4 i) cosec. A f (371) sec.JZT=: cos. A cos.Csec. £ (A 2 + A \ ) cosec. ^(^i — A t ). (372) § 46.] LATITUDE. 231 By double altitudes. But PV= PT— TV, or 90° — E = 90° — B — Z E = B + Z. (373) Lastly, triangle ZPV gives cos. PZ z= cos. ZV cos PV sin. L = cos. C sin. E. (374) Equations (363, 364, 371 - 374) correspond to the rule and formula given in the Navigator. [B. p. 180.] II. Another method of calculating the values of B, C, and Z has been given, which dispenses with A and one opening of the tables, and may therefore be preferred by some calcu- lators, although it requires one more logarithm. Triangle TPS gives tang. PT — cos. J h tang. PS, or cotan. B = cos. J h cotan. D. (375) The substitution of (364) in (371) gives sin. Cz= cos. %(A x -\-A \ ) sin. %(A\ — A x ) sec. D cosec. %h. (376) Triangle PTS gives cos. A == sin. D cosec. B ; (377) which, substituted in (372), gives (378) sec.Z= cos.Csin. Z> cosec. B cosec. ±(A 1 -f-^i;)sec.^(^4 1 ' — A ± ). Corollary, The hour angle ZP T is the mean between the hour angles ZPS and ZPS' t and if we put ZPTz=H, 232 SPHERICAL ASTRONOMY. [CH. IV. By double altitudes. Douwes's method. the triangle ZP V gives tang. II ±= tang. C sec. E, (379) as in B. p. 181. III. Douwes's Method. When the latitude is known within a few miles. In this case let U = the assumed latitude, and the triangles ZSP and ZSP give sin. A x = sin. L 1 sin. D -f cos. U cos. D cos. ZPS } (380) sin. A\ = sin. L' sin. D -f- cos. L 1 cos. D cos. ZPS 1 ; (381 ) whence, and by (39), sin. A\ — sin. A x = cos. Z/cos. D (cos. ZPS 1 — cos. ZPS) = 2cos.L'cos.Dsm.i(ZPS'-{-ZPS)sm.i(ZPS'--ZPS) = 2 cos. Z/ cos. Z) sin. Z/ sin. ^- /a 2 sin. Zf= (sin. 4 ' x — sin. .4 , ) sec. L 1 sec. Z> cosec. ^ /«, (382) ZPS = H — l-h] (383) whence the hour angle ZPS corresponding to the observation at S' is known, and the latitude may be found by the method of a single altitude. The combination of the formulas (380, 381), and the method of computing the latitude by a single altitude, corresponds exactly to the rule given in the Naviga- tor. [B. p. 185.] The log. cosec. £ h is not only given in table XXVII, but also in table XXIII, where it is called the log. £ elapsed time off*. § 46.] LATITUDE. 233 Table XXIII. The value of log. 2 sin. H — 5 = log. sin. H -f- log. 2 = log. sin. H — ar. co. log. 2 -J- 6 = log. sin. H— 4.69897 ±= 5.30103 = log. elapsed time of H (384) is inserted in table XXIII, and is called the log. middle time of H. The 5 is subtracted from log. 2 sin. if, on account of the different values of the radius in tables XXIV and XXVII. Scholium. When the calculated latitude differs much from the assumed latitude, the calculation must be gone over again, with the calculated latitude instead of the assumed latitude. This labor may be avoided by noticing, in the course of the original calculation, the difference which would arise from a change of 10' in the value of the assumed latitude, and calcu- lating the correction of the latitude by the rule of double position. The error of the hypothesis is in each case the ex- cess of the calculated above the assumed latitude, and the proportion is diff. of errors : diff. of hyp. = least error : corr. of hyp. (385) IV. If the star has changed its declination a little, during the interval between the observations, the second altitude will correspond to a declination D', a little different from D. If D 1 is put instead of D in (381), and if A 2 denotes the second observed altitude, A x being retained to denote what this second altitude would have been, if the declination had remained unchanged, (381) becomes sin. A' 2 = sin. L' sin. D' + cos. L' cos. D' cos. ZPS'. (385) 20* 234 SPHERICAL ASTRONOMY. [cH. IV. Table XLVI. Now, if (381) multiplied by cos. D' is subtracted from (385) multiplied by cos. D, the remainder is cos. Dsin.il ' 2 — cos. 2> 7 sin.il j zz: sin.Z/sin.(Z>'— D). (386) But if we put D' — D = dD, A' 2 — A' 1 =dA 1 (387) we have, by (13) and (15), cos. D h± cos. (D + 3 D) = cos. D — sin. a 2> . sin. Z> (388) sin. ilgzzisin. (^4^ + *A x )=sin. A\-\-&\n. *A X . cos.il j, (389) which, substituted in (386), give sin.il'jSin.Dsin.^Z). -}-cos. A\ cos.Z?sin. ^iljZzzsin.Z/sin. §D sin. (j^ __ SA X _ sin. L 1 — sin. A\ sin. D . . sin. d D Z ~~ TJj ~~ cos. A\ cos. D * ' and, by (34) and (35), 2si n.Ii 7 — cos.(^ , 1 --.D) + cos.(^ , 1 + J9)^ **p ^(A\ -D)+ cos. (A> 1 + D) — D > (391) in which D is to be negative, when the latitude and decli- nation are of contrary names. Hence the value of d A 1 can be computed by this formula, and thence A\ — A' 2 — 3 A 19 and in calculating t A t , A' 2 may be substituted fov A\. Since the value of = 90° — PS' — the declination of star at S', H = SPS' ±= hour angle = interv. of sideral time. Then, in the triangle PSS', PS, PS', and H are given to find SS' = C, and S'SP = 90° — F. Next, in the triangle ZSS', the three sides are known, to find the angle ZSS' == z. Hence ZSD = 90° — # = 90° — jP— Z G = F+ Z. Lastly, in the triangle ZSP, ZS, SP, and the included angle ZSP are given to find ZP 4 90° — L. This solution is precisely similar to the Rule in B. p. 193 ; and it is easy to prove the rules for the s gns which are there given. VI. If the distance SS' were observed, the angles ZSS' and S'SP might be found from the triangle ZSS' and S'SP, in which the sides are all known, and the rest of the calcula- tion would be as in the last method, and this method corre- sponds exactly to the Rule in B. p. 197. 236 SPHERICAL ASTRONOMY. [CH. IV. Meridian altitudes. 47. Examples. 1. The correct meridian altitude of Aldebaran was found by observation, in the year 1838, to be 55° 45', when its bear- ing was south ; what was the latitude ? Solution The zenith distance =z 34° 15' N. The declination = 16° 10' N. The latitude = 50° 25' N. 2. The correct meridian altitude of Canopus was found by observation, in the year 1839, to be 16° 25', when its bearing was south; what was the latitude? Solution. The zenith distance = 73° 35' N. The declination z= 52° 36' S. The latitude = 20° 59' N. 3. The correct meridian altitude of Dubhe was found by observation, in the year 1830, to be 50° 45', when its bearing was north ; what was the latitude ? Solution. The zenith distance = 39° 15' S. The declination = 52° 36' N. The latitude = 13° 21' N. 4. If the correct meridian altitude of Dubhe, at its greatest elevation, were found by observation, in the year 1830, to be 50° 45', when its bearing was south ; what would be the lati- tude? § 47.] LATITUDE. 237 Meridian altitudes. Solution. The zenith distance — 39° 15' N. The declination = 52° 36' N. The latitude = 91° 51' N. The problem is impossible. 5. The correct meridian altitude of Dubhe, at its least ele- vation, was found by observation, in the year 1830, to be 50° 45' ; what was the latitude ? Solution. The polar distance ±= 37° 24'. The altitude ±= 50° 45'. The latitude = 88° 09' N. 6. The correct meridian altitudes of Dubhe, at its greatest and least elevation, which were on opposite sides of the zenith, were found by observation to be 41° 56' and 53° 16'; what was the latitude ? Solution. The greatest altitude — 53° 16'. The least altitude = 41° 56'. Diff. of altitudes == 11° 20 ; . 180° — Diff. of altitudes = 168° 4(K Latitude ' ~ 84° 20' N. 7. The correct meridian altitudes of a northern star, at its greatest and least altitudes, which were on the same side of the zenith, were found by observation to be 12° 14' and 72° 14' ; what was the latitude ? 238 A SPHERICAL ASTRONOMY. [CH. IV. Single altitude. Solution. Greatest alt. — 72° 14'. Least alt. = 12° 14'. Sum of alts. = 84° 28'. Latitude = 42° 14' N. 8. In a northern latitude, the altitude of Aldebaran was found by observation, in the year 1839, to be 25° 38', when its hour angle was 4 A 12™ 20 s ; what was the latitude ? Solution. By (349, 350, 351), h = 4* 12™ 20 s cos. 9.65580 Z>=]6°11' cotan. 10.53729 cosec. 10.55484 90°— PC =32° 40' ZC — 33° 6' cotan. 10.19309 A = 25° 38' sin. 9.73215 sin. 9.63610 cos. 9.92309 L = 65° 46' N. 9. In lat. 65° 40' N. nearly, the altitude of Aldebaran was found by observation, in the year 1839, to be 25 38', when its hour angle was 4 A 12 m 20 s ; what was the true latitude? Solution. I. 65° 40' COS. 9.61494 16° 11' COS. 9.98244 A h 12 m 20* Nat. num. 21657 log. Ris. 4.73823 4.33567 25° 38' Nat. sine 43261 49° 31' N. Nat. cos. 64918 16° 11' N. 65° 42' N. = the latitude. $ 47.] LATITUDE. 239 Single altitude. Had the assumed latitude been taken 10' more, the calcu- lated latitude would have been 65° 48J' N. ; hence, by (385), 3£ : 1£ — 10' : 4' = corr. of second hypothesis, or the latitude — 65° 46' N., as in the preceding example. II. By (357), £ h = 2 k 6 m 10* 2 log. sin. 9.43720 L = 65° 40' cos. 9.61494 D — 16° IV cos. 9.98244 A x = 40° 31' A =z 25° 38' A' = 25° 38' A l —A=U 5V i(A l + A) = 33° 4f sec. 10.07678 A 1 =40°29' i(A t — A)— 7°25J' sin. 9.11136 corr. A 1 = 15° 2' =z corr. lat. = 65° 40' + 2 = 65° 42' as before. 10. Calculate the variation of a star's altitude in one minute from the meridian, when the declination is 12° N. and the latitude 5° N. Solution. If A 2 — A is required in seconds, (358) gives A j — A — 450 sin. l m cos. L cos. D sec. A x by 450 sin. l m — log. 450 + log. sin l m = 2.65321 -f 7.63982 = 0.29303 L ±t 12° cos. 9.99040 D =z 5° cos. 9.99834 i4 i: =83 sec. 0.91411 ^ — A = 15".7, as in table XXXII. 1.19588 240 SPHERICAL ASTRONOMY. [CH. IV. Single i altitude near the n leridian. 11. Calculate XXXIII. the tabular number for 11 m 48 s in table Solution. Il m 48 s - = 708 s log. 2.85003 60 s log. 1.77815 1.07188 2 139.2, as in table XXXIII. 2.14376 12. In lat. 45° 28' N. nearly, the correct altitude of Alde- baran was found by observation, in the year 1839, to be 60° 40' 20", when its hour angle was 7 m 17 s . What was the true latitude, if the declination of Aldebaran was 16° 11' 9".2 N. ? Solution. From Table XXXII 2".7 From! 'able XXXIII 2 23". 1 = 53 143". 1 60° 40' 20" Third alt. = 60° 42' 43". 1 Dec. — 16° 11' 9 '.2 Lat. = 45° 28'26".l N. 13. In lat. 40° N. nearly, the sum of ten correct central altitudes of the sun, when its declination was 20° S. were 300° 20'. The hour angles of these observations were 4" 15 s , 3", 2 W 6% l m 8 s , 30 s , 50 s , l m 12 s , 2 m 15 s , 3" 10 s , 4 OT 25 s . What is the true latitude, if the change of declination is neg- lected ? § 47.] LATITUDE. 241 Latitude by circummeridian altitudes. Solution. The numbers of Table XXXIII are 4 m 15 s gives 18.1 3 9.0 2 6 4.4 1 8 1.3 30 0.2 50 0.7 1 12 1.4 2 15 5.1 3 10 10.0 • 4 25 19.5 Sum == 69.7 Mean = 6.97 Table XXXII gives 1".6 11" Mean of observations =r 30° 0' 40" Merid. alt. = 30° 0' 51" Dec. = 20° S. Lat. z=39°59' 9" N. 14, AtGottingen, in lat. 51° 32' N. nearly, the correct central altitudes of the sun on the 11th of March, 1794, were by observation 21 Ml SPHKRICJLL *T. [CH, IT. Laftaafe fey * «*ew*s _9M1* *i 19 — 1 h — 5 16 — 1 m — 2 47 19 -2 5 3 9 4 ■ »6 S declination was ^ 30 a* S., and it at the rate of 0.9$ in a minote. What is the of the ahkodes is $4* 56 » .5 : of Table XXXm is by 1 ,5 from Ta- The Mean of the boor angles is, regard- ing their signs, — 1» 50% which, multiplied by J98, gwes by (363), for the correction of :ne MrifiMi nttSE I 5 The meridian altitude = S4 57 II The declination — K S Tiff hi ton* = 51 ? » 4 TN, ^ 47%] ok. Mtt Hide by pole star. i agrees exactly with the calculations of Littrow in his 15. Calculate the correction for the altitude of the pole star ben the right ascension of the zenith is &* ?"*. > tot©* By (301), * i w > w =i^ sec. o.oi:: j> = 1° 33* Prop. log. O.SSoS Corr. alt. = 1 \\ the table, Prop. log. 0.3045 u> Wlu n the ti^ht ascension of the zenith was 7*9^*, the altitude of the pole star was obsenod at Newburvport to be 4<. flbb/toN. By Table XXVII and (384), 3* 7* 10* cosec. 0.13735 = log. elapsed time 5.30103 5 16968 = log. mid, time. 244 SPHERICAL ASTRONOMY. [CH. IV. Variation of star's altitude. 18. Calculate the variation of the altitude of a star arising from the change of 100 seconds in the declination, when the latitude is 40°, the declination 10°, and the altitude 30°. Solution. By (391), L' z=z 40°, 2 X Nat. sin. 1.2856 1.2856 A\ — D =z 20° Nat. cos. 0.9397 — 0.9397 0,9397 A\ + D=: 40° Nat. cos. 0.7660 0.7660 —0.7660 1.7057 1.1119 1.4593 1.7057 (ar. co.) 9.7681 9.7681 100" X 1.1119 2.0661 100" X 1.4593 2.1641 65" =; var. when D is -f, 1.8142 86" =z var. when D is — , 1.9322 19 The moon's correct central altitude was found, by ob- servation, to be 53° 43', when her declination was 14° 16' N. After an interval, in which the hour angle was l h 44 m 15% her correct central altitude was 42° 29', and her declination 13° 52' N. The latitude was 48° 50' N. nearly ; what was it exactly ? Solution. Table XLVI gives, for the second alt. 83" Whole change of declination 24' Correction of second altitude 20' Corrected second alt. = 42° 49', dec. = 14° 16' N. § 47.] LATITUDE. 245 Latitude by double altitudes. I. By Bowditch's first method. l h U m 15* cosec. 0.64675 14° 16' sec. 0.01360 cosec. 0.60830 A cosec, 0.66035 cos. 9.98936 cos. 9.98936 B = 14° 38' N. cosec. 0.59766 cos. 9.82326 £ sum alts.=z48° 16' cosec. 0.12712 sin. 8.97762 % diff. alts.= 5° 27' sec. 0.00197 C sin. 9.46123 cos. 9.98103 cos. 9.98103 Z=37°19 / N. sec. 0.09948 E = 51° 57' N. sin. 9.89624 Latitude m 48° 55J'N. sin. 9.87727 II. By the method (375 - 378). l h M m 15 s cos. 9.98852 cosec. 0.64675 14° 16' cotan. 0.59469 sec. 0.01360 sin. 9.39170 B 55 14° 38' N. cotan. 0.58321 cosec. 0.59753 J sum alts. = 48° 16' cos. 9.82326 cosec. 0.12712 J diff. alts.= 5° 27' sin. 8.97762 sec. 0.00197 C cos. 9.98103 sin. 9.46123 cos. 9.98103 Z=37°18'N. sec. .09935 jEJ=z51°56'N. sin. 9.89614 Lat = 48°54£'N. sin. 9.87717 21* 246 SPHERICAL ASTRONOMY. [CH. Latitude by double altitudes. III. By Douwes's method. 48° 50' sec. 0.18161 53° 43' N. sin. 80610 14° 46' sec. 0.01360 42° 49' N. sin. 67965 log. ratio 0.19521 12645 log. 4.10192 I (l h 44 w 15') = 52 w 7£ s log. el. time 0.64689 l h U m 18* log. mid. time 4.94402 52 m ll£* log. ris. 3.41152 log. ratio 0.19521 1645 log. 3.21631 80610 34° 40' N. N. cos. 82255 14°16 / N. Lat. = 48° 56' N. Had the latitude been supposed 10' greater, the calculated latitude would have been 48° 55' N. § 47.] LATITUDE. 247 Latitude by double altitudes. IV. By Bowditch's fourth method. l h U m 15 s sec. 0.04657 tan. 9.68938 14° 16' N. tan. 9.40531 sin. 9.39170 A = 15° 48' S. tan. 9.45188 cosec. 0.56485 cos. 9.98326 13°52 N. B — 1° 56' S. cos. 9.99975 cosec. 1.47190 C 25° 16' cosec. 0.36961 cos. 9.95630 F~ 4° 6'N. cotan. 1.14454 53° 43' Z="51°38'N. G — 55°44'N. sin. 9.91720 42° 29' sec. 0.13225 sin. 9.82955 cotan. 0.03820 £ sum =60° 44' cos. 9.68920 J sec. 0.12936 tan. 9.95540 Rem. = 7° V sin. 9.08692 K sin. 9.91823 J~ 42° 4'N. 2) 19.27798 lat. sin. 9.87714 13°52'N. £ Z =i 25° 49' N. sin. 9.63899 lat. = 48° 54' N. K= 55° 56' N. 19. The correct meridian altitude of Aldebaran was, by observation, 56° 25' 40" bearing south, and its declination at the time of the observation was 16° S' 44" N. ; what was the latitude ? Arts. 49° 43' 4" N. 248 SPHERICAL ASTRONOMY. [CH. IV. Latitude by meridian altitudes. 20. The correct meridian altitude of Sirius was 70° 59' 33" bearing north, and its declination 16° 28' 9" S. ; what was the latitude ? Ans. 26° 28' 36" S. 21. The meridian altitude of the sun's centre was 25° 38' 30" bearing south, and its declination 22° 18' 14" S. ; what was the latitude ? Ans. 42° 3' 16" N. 22. The meridian altitude of the planet Jupiter was 50° 20' 8" bearing south, and its declination 18° 47' 37" N. ; what was the latitude? Ans. 58° 27' 29" N. 23. The altitude of the pole star was 30° 1' 30" below the pole, and its polar distance 1° 38' 2" ; what was the latitude? Ans. 31° 39' 32" N. 24. The altitude of Capella on the meridian below the pole was 9° 52' 42", and its polar distance 44° 11' 33" ; what was the latitude? Ans. 54° 4' 15" N. 25. The meridian altitude of the sun's centre was 7° 9' 11" below the pole, and its declination 23° 8' 17" N. ; what was the latitude ? Ans. 74° 0' 54" N. 26. The two meridian altitudes of a northern circumpolar star were 61° 49' 13" and 47° 34' 27" ; what was the latitude ? Ans. 54° 36' 50" N. § 47.] LATITUDE. 249 Latitude by single altitudes. 27. In a northern latitude, the altitude of the sun's centre was 54° 9', when its hour angle was 32 m 40 s , and its declina- tion 11° 17' N. ; what was the latitude? Ans. 4G° 27' N. 28. In latitude 49° 17' N. nearly, the altitude of the sun's centre was 14° 15', when its hour angle was l h 40™, and its declination 23° 28' S.; what was the true latitude? Ans. 48° 55' N. 29. Calculate the variation of a star's altitude in one minute from the meridian, when the declination is 3° and the lati- tude 7°. * Ans. It is 27".9 when the dec. and lat. are of the same name, and 11".2 when they are of contrary names. 30. Calculate the tabular number for I2 m 59' in Table XXXIII. Ans. 168.6. 31. In lat. 50° 30' N. nearly, the altitude of Sirius was 22° 59' 36", when its hour angle was 4 m 15% and its declina- tion 16° 29 ; ll" S. ; what was the true latitude? Ans. 50° 30' 49" N. 32. In lat. 20° 27' N. nearly, the sum of seven altitudes of Sirius was 371° 21'; the hour angles of the observations were 7 m , 5 m 3 s , 2 m 12*, 9% 3™, 4™ 6% 8 m 13'; what was the true latitude, if the declination of Sirius was 16° 29' 30" ? Ans. 20° 26' 18" N. 250 SPHERICAL ASTRONOMY. [CH. IV. Latitude by circummeridian altitudes. 33. In lat. 50° N. nearly, the sum of twelve central alti- tudes of the moon was 590° ; the hour angles of the observa- tions were — 9 m 3 s , — -7™ 40% — 6 W 12% — 5*30% — 3™ 2% — 1*, — 12% 50% \ m 59% 4°% l m 30% 10™ ; the moon's me- ridian declination was 19° 10' 58".4 N., and her change of de- clination for one minute 13".875 ; what was the true latitude ? Ans. 59° 50' 2".3 N. 34. Calculate the correction for the altitude of the pole star [B. p. 206.], when the right ascension of the zenith is 9 h 7 m . Ans. 48'. 35. The altitude of the pole star was 25° 9% when the right ascension of the zenith was 21° 47'; what was the latitude? Ans. 24° 8' N. 36. Calculate the log. elapsed time and log. middle time of Table XXIII for & 49- 50 s . Ans. Log. elapsed time = 0.00001 Log. middle time — 5.30102 37. Calculate the variation of the altitude of a star arising from the change of 100 seconds in declination, when the lati- tude is 60°, the declination 20°, the altitude 30°, and the declination and latitude of the same name. Ans. 85". 38. Calculate the variation of the altitude of a star arising from the change of 100 seconds in declination, when the lati- tude is 50°, the declination 24°, and the altitude 20°. Ans. It is 73" when the lat. and dec. are of the same name, and 105" when they are of contrary names. <§> 47.] LATITUDE. 251 Latitude by double altitudes. 39. The sun's correct central altitudes were found by ob- servation to be 30° 13' and 50° 4'; his declination was 20° 7'N., and the interval of sideral time between the observations was 2 h 55 m 32 s ; the assumed latitude was 56° 29' N. ; what was the true latitude? Ans. 56°47'N. 40. The sun's correct central altitude was 41° 33' 12", his declination 14° N. ; after an interval of l h 30 m , his correct central altitude was 50° V 12", and declination 13° 58' 38" N.; the assumed latitude was 52° 5' N. ; what was the true lati- tude ? Ans. 52° 5' N. 41. The moon's correct central altitude was 55° 38% her declination 0° 20' S. ; after an interval in which the hour angle was 5*30* 49% her correct central altitude was 29° 37% and her declination 1° 10' N. ; the assumed latitude was 23° 25' S. ; what was the true latitude ? Ans. 23° 24' S. 42. The sun's correct central altitude was 16° 6% his decli- nation 8° 18' N. ; after an interval in which the hour angle was 3\ his correct central altitude was 42° 14' 9", and his declination 8° 15' N. ; the assumed latitude was 49° N. ; what was the true latitude? Ans. 48° 50' N. 43. The moon's correct central altitude was 35° 21% and her declination 5° 31' 6" S. ; after an interval in which the hour angle was 2 h 20™, her correct central altitude was 70° 1% and her declination 5° 28' 54" S. ; the assumed latitude was 1° 30' N.; what was the true latitude ? A?is. 1° 29' N. 252 SPHERICAL ASTRONOMY. [CH. IV. Corrections for the pole star in the Nautical Almanac. 44. The altitude of Capella was 60° 45' 36", and her decli- nation 45° 48' 21" N. ; at the same instant, the altitude of Sirius was 17° 54' 12", and his declination 16° 28' 40" S. ; the hour angle was [ k 33™ 45 ? , and the latitude was about 53° 15' N. ; what was the true latitude? Ans. 53° 19' N. 45. The altitude of « Bootis was 50° 3' 39", and its decli- nation 20° 10' 56" N. ; the altitude of « Aquilje was 33° 33', and its declination 8° 22' 35" N. ; the hour angle of the ob- servations was 5^ 5 m 5h s , and the assumed latitude 38° 27' N. ; what was the true latitude? Ans. 38° 28' N. 46. The distance of the centres of the sun and moon was found, by observation, to be 75° ; the sun's central altitude was 37° 40' ; the moon's central altitude was 55° 20' ; the sun's declination was 0° 17' S. ; the moon's declination was 0° 36' N. ; what was the latitude, supposing it to be north ? Ans. 23° 24' N. 48. The method of determining the latitude by means of the pole star is so accurate in practice, that tables are given in the Nautical Almanac for correcting the observed altitude for differences of latitude, and changes in the right ascension and declination of the star. The first correction of the Nautical Almanac corresponds to that of the Navigator, and is calculated by (361) for R. A. of Polaris = 1 A 1» 48*.2. (392) Dec. of Polaris — 88° 26' 54" = 2>, (393) § 48.] LATITUDE. 253 Corrections of the Nautical Almanac for Polaris. which gives p = 1° 33' 6" ss 5586" (394) log. p = 3.74710 (395) A = R. A. of zenith — It l m 48*.2. (396) The second correction of the Nautical Almanac depends upon the latitude, and would vanish, if in (350) the values of p and PC were so small that we could put sin. D = cos. p -ss cos. PC ss 1. Now equation (350) is equivalent to the proportion cos. PC : cos. p ss cos. ZC : sin. A, but, by (360), ZC ss 90° — L — PC= 90° — L — p cos. *; whence cos. PC : cos. p ss sin. (L 4- p cos. A) : sin. ^4. Hence, by the theory of proportions, cos. PC — cos. p sin. ( L -f- /? cos. A) — sin. -4 /QQ ~ X cos. jPC-j- cos.p sin. (jL-f-p cos. A) -}- sin. .4' and, by (40), (41), and (360), tang. [Jp (1 —cos. A)] . tang. [Jp (1 + cos. A)] s= tang.£(Z,+p cos. A — it) m tang. J (X, ■ + p cos. A + ,4) ' K ' or, since p and L -f- p cos. A — .4 are very small, we may put tang. [Jp (1 — cos. A)] ss £p (1 — cos. A) tang. \" tang. [Jp (1 -|-cos. A)] =z \p (1 + cos. A) tang. \" 22 254 SPHERICAL ASTRONOMY. [CH. IV. Corrections of the Nautical Almanac for Polaris. tang. £ (L-\-p cos. h — A) = J (L-\- pcos.h — A)tang. \" tan. J {L-\-p cos.7i-\-A)—t&n.[L-\-p cos.Ji — %(L-\-pcos.h — A)] — tang. (L -\-p cos. h) y which, substituted in (390), give \p 2 (1 — cos. 2 A)tang.(Z/~|-pcos.7f) tang.l"— JL-{-pcos. h — A or L=p cos. h -{-A + Jjp 2 sin. 2 h tang.( L-{-p cos. h) tang, t" (399) so that £ p 2 sin. 2 A tang. (L + p cos. A) tang. 1 " (400) or J p 2 sin. 2 A tang L tang. 1" is the correction depending upon the latitude, and in calculat" ing it we have log. (£ p 2 tang. 1") s 7.49420 + 9.69897 + 4.68558 == 1.87875 (401) and 7i is the same as in (396). The third correction of the Nautical Almanac is the change in the value of the first correction arising from the changes in the declination and right ascension of the star. Thus if the declination is greater than that of (393) by <* Z>, the value of p must be less by d D, and the correction — p cos. h is in- creased by S D cos. h. (402) Again, if the right ascension is greater than that of (392) by $ R, the value of h must be less by J R, and the value of — p cos. h is increased by — p [cos. (h — dR) — cos. h]i (403) which, by (15), is equal to § 49.] LATITUDE. 255 Corrections of the Nautical Almanac for Polaris. — p sin. $ R sin. h = — p d R sin. I s sin. h — 5580 X 0.000075 X dR s\n.h=z0"AdRsm.h, (404) and the whole change is the sum of (402) and (404). The values of (402) and (404) are easily obtained from the tables of difference of latitude and departure. We may neglect the l m 48*.2 in the value of h (396), when we calculate these cor- rections, and take h — R. A. of zenith — 1\ (405) The third correction is sometimes positive and sometimes negative, but always less than 1', so that V -f- the third correction is always positive ; and this is the given sum in the Nautical Almanac ; that is, the third correction is given 1' greater than its real value, so that it may always be positive. The latitude, obtained by means of the table of the Nautical Almanac, would then be V greater than its true value, if V were not subtracted agreeably to the rule given in the Almanac. 49. Examples. 1. Calculate the first correction of the Nautical Almanac when the R. A. of the zenith =d 4* 20 w . Solution. log. p = 3.74710 L = ± h 20" — l h l m 48*.2 = 3 h 18- 11\8 cos. 9.81219 1st corr. == 3624" — 1° 0' 24" 3.55929 2. Calculate the second correction of the Nautical Almanac when R. A. of the zenith is l h 30 m , and the latitude 50°. 256 SPHERICAL ASTRONOMY. [CH. IV. Corrections of the Nautical Almanac for Polaris. Solution. log. (£p2 tang. 1") 2= 1.87875 L = 6 h 28 m 11».8 sin.2 9.99340 50° tang. 0.07619 2d corr. = 89" = 1' 29" 1.94834 3. Calculate the third correction of the Nautical Almanac for Dec. 31, 1839, when R. A. of zenith == 14\ Solution. h=U h —l h =i 13 A p= 12* + l h = 180° + 15°. Dec. of Polaris = 88° 27' 46".6, R. A. — l h l m 59 s .47 D =s 88° 26' 54" l h l m 48'.2 *D=: 52".6 <*22 z=z 11*.27 0'A$R = 4".5 By Table II., omitting the tenths of seconds in the result, V + 3d corr. — V — 50".8 + 1".2 — 1' — 50" — 10". 4. The correct altitude of Polaris on June 25, 1839, was 47° 28' 35", when the Right ascension of the zenith was 6 h 18 m 30 s ; what was the latitude? Solution. Cor. Alt. = 47° 28' 35" First corr. = — 13' 27" A + First corr. rs 47° 15' 8" Second corr. = 1' 28" Third corr. = 1' 6" — I 1 Lat. sr 47° 16' 42" N. <§> 50.] LATITUDE. 257 Observer in motion. 5. Calculate the three corrections of the Nautical Almanac for Sept. 1, 1839, and latitude 70°, when the R. A. of zenith is 8 h . At this time, we have Dec. of Polaris = 88° 27' 8".4, its R. A. = t* 2 m 21*.32. Ans. The first correction = 23' 23" The second correction =z 3' 15" 1' -f The third correction = 43". 6. The correct altitude of Polaris on March 6, 1839, when the R. A. of the zenith was 6* 39™ 24 s , was 46° 17' 28"; find the latitude. The following is an extract from the tables of the Nautical Almanac sufficient for the present example. 1st corr. 2d corr. Lat. =s 45° Lat.z=50 { 6 h 30 m — 12' 53" V 14" r 28" 6*40™ — 8' 51" 1' 16" V 30" Third correction -{- 1/ March 1, April 1, 6* V 27" 1' 27" 8 h V 13" V 19" Ans. 46° 10 y 3" N. 50. The observer has been supposed stationary in the pre- ceding observations, but if he is in motion his second altitude will differ from the altitude for this time at the first station by the number of minutes by which the observer has approached the star or receded from it ; so that the correction arising from this change of place is obviously computed by the method in [B. p. 183.] 22* 258 SPHERICAL ASTRONOMY. [cH. IV. Greatest altitude of a star in motion. 51. In observing the meridian altitude of a star, the position of the meridian has been supposed to be known; but if it were not known, the meridian altitude can be distinguished from any other altitude from the fact that it is the greatest or the least altitude; so that it is only necessary to observe the greatest or the least altitude of the star. 52. But if the star changes its declination, the greatest alti- tude ceases to be the meridian altitude. Let h denote the hour angle of the star at the time of observation. Then if the star did not change its declination, and if B were the number of seconds given by Table XXXII ibr the diminution of altitude in one minute from the meridian passage, h 2 B would be the diminution of altitude in h minutes. But, since h is small, the altitude, at this time, is increased by the change of declination ; so that if A is the number of minutes by which the star changes its declination in one hour, that is, the number of seconds by which it changes its declination in one minute, h A will be the increase of altitude in (he time h, so that the altitude at the time h exceeds the meridian altitude by h A — A 2 B. (406) If, then, h denotes the time of the greatest altitude, and h-\-dfi a time which differs very slightly from the greatest altitude ; the greatest altitude exceeds the altitude at the time h-\-dh by the quantity (h A — A* B) — [(/* + S h) A — (h + S ft)* B] =zdh[(—A+2Bh)-\-Bdh], (407) and 3 h can be supposed so small that B$7i may be insensible, and (407) becomes d h(—A + 2Bh). (408) $ 54.] LATITUDE. 259 Greatest altitude of a star in motion. Now — A -f- 2 B h cannot be negative, because h is sup- posed to correspond to the greatest altitude, and cannot be less than the altitude at the time li -\- § h. Neither can — A-{-2Bhbe positive, for the altitude at the time h ex- ceeds that at the time h — 8 h by the quantity _*/,(_ A + ZBh), which, in this case, would be negative, and the altitude at the time h — d li would exceed the greatest altitude. Since, then, — A -\- 2 B h can neither be greater nor less than zero, we must have — A + 2Bh=zO h = wm (409) and this value of A, substituted in (406), gives A* A^ A* 2B 42* 4B (410) for the excess of the greatest altitude above the meridian alti- tude. f 53. If the observer were not at rest, his change of latitude will affect his observed greatest altitude in the same way in which it would be affected by an equal change in the declina- tion of the star ; so that the calculation of the correction on this account may be made by means of (409) and (410) pre- cisely as in [B. p. 169.] 54. E XAMPLES. 1. An observer sailing N.N.W. 9 miles per hour found, by obserration, the greatest central altitude of the moon, bearing 60 SPHERICAL ASTRONOMY. [CH. IV. Latitude by greatest altitude. south, to be 54° 18 / ; what was the latitude, if the moon's declination was 6° 30' S., and her increase of declination per hour 16 / .52? Solution, J> 's zenith dist. =. 35° 42' N. 3>'s dec. = 6°30'S. Approx. lat. — 29° 12' N. J) 's increase of dec. per hour == 16'.52 S. Ship's change of lat. =£= 8.3 A = 24.82, A* = 616.0 By Table XXXII B — 2.9, 4 B = 11.6 Corr. of gr. alt. = corr. of lat. z= 52" = 1' nearly Lat. = 29° 12' + 1' = 29° 13' N. 2. An observer sailing south 12J miles per hour found, by observation, the greatest central altitude of the moon bearing south, to be 25° 15'; what was the latitude, if the moon's declination was 1° 12' N., and her increase of declination per hour 18'.5 ! / Ans. 66° 1' N. § 58.] THE ECLIPTIC. 26 L Obliquity. Equinoxes. Signs. CHAPTER V. THE ECLIPTIC. 55. The careful observation of the sun's motion shows this body to move nearly in the circumference of a great circle. This great circle is called the eclip- tic, [B. p. 48.] 56. The angle which the ecliptic makes with the equator is called the obliquity of the ecliptic. 57. The points, where the ecliptic intersects the equator, are called the equinoctial points ; or the equi- noxes. The point through which the sun ascends from the southern to the northern side of the equator is called the vernal equinox, and the other equinox is called the autumnal equinox. % The points 90° distant from the ecliptic are called the solstitial points, or the solstices. [B. p. 49.] 58. The circumference of the ecliptic is divided into twelve equal parts, called signs, beginning with the vernal equinox, and proceeding with the sun from west to east. The names of these signs are Aries (°f), Taurus (8), Gemini (n), Cancer (zb), Leo (g\,,) Virgo (v%), Libra (£t), 262 SPHERICAL ASTRONOMY. [CH. V. Colures. Tropics. Latitude of a star. Scorpio ( Tr l), Sagittarius (f), Capricornus (V?), Aquari* us (£#), Pisces (X)- The vernal equinox is therefore the first point, or beginning of Aries, and the autumnal equinox is the first point of Libra; the first six signs are north of the equator, and the last six south of the equator. The northern solstice is the first point of Cancer, and the southern solstice the first point of Capricorn. [B. p. 49.] 59. Secondary circles drawn perpendicular to the ecliptic are called circles of latitude. The circle of latitude drawn through the equinoxes is called the equinoctial colure. The circle of latitude drawn through the solstices is called the solstitial colure. [B. p. 49.] Corollary. The solstitial colure is also a secondary to the equator, so that it passes through the poles of both the equator and the ecliptic. 60. Small circles, drawn parallel to the equator through the solstitial points, are called tropics. The northern tropic is called the tropic of Cancer ; the southern tropic the tropic of Capricorn. Small circles, drawn at the same distance from the poles which the tropics are from the equator, are called polar circles. The northern polar circle is called the arctic circle, the southern the antartic. 61. The latitude of a star is its distance from the ecliptic measured upon the circle of latitude, which $ 64] THE ECLIPTIC. 263 Longitude of a star. Nonagesimal point. passes through the star. If the observer is supposed to be at the earth, the latitude is called geocentric latitude ; but if he is at the sun, it is heliocentric latitude. [B. p. 49.] 62. The longitude of a star is the arc of the ecliptic contained between the circle of latitude drawn through the star and the vernal equinox. [B. p. 50.] Corollary. The longitude and right ascension of the first point of Cancer are each equal to 6 A , and those of the first point of Capricorn are each equal to 18 A . 63. The nonagesimal point of the ecliptic is the highest point at any time. Corollary. The distance of the nonagesimal from the zenith is therefore equal to the distance of the zenith from the eclip- tic, that is, to the celestial latitude of the zenith; and the longitude of the nonagesimal is the celestial longitude of the zenith. 64. Problem. To find the latitude and longitude of a star, when its right ascension and declination are known. Solution. Let P (fig. 35.) be the north pole of the equator, Z the north pole of the ecliptic, and B the star. Then EQW will be the equator, NESW the ecliptic, and NPZS the solstitial colure, so that the point S is the southern solstice, and N the northern solstice. Now if the arc PB be produced to cut the equator at M, and ZB to cut the ecliptic at L ; the angle ZPB is measured by the arc QM, that is, by the 264 SPHERICAL ASTRONOMV. [CH. V. To find a star's latitude and longitude. difference of the right ascensions of Q and M, or by the dif- ference of the ^fc's right ascension and 18*; that is, ZPB — 18* — R. A. z= 24* — (G* + R. A.) (411) or = R. A. — 18* = (R. A. + 6*) — 24* or = 24* + R. A. — 18* — R. A. + 6*. In the same way PZB — NL — Long. — 90° (412) or = 360° — (Long. 90°) or = — (Long. — 90°), in which the first values of ZPB and PZD correspond to the star's being east of the solstitial colure ; the second and third values to the star's being west of the colure. We also have PB — 90° — Dec. (413) BZ = 90° — Lat. (414) + PZ == 90° — ZQ=QS = obliquity of ecliptic z= ± E, (415) in which the declination and latitude are positive when north, and neo-ative when south, and E has the same sign with R. A. — 12*. The present problem does not, then, differ from that of § 28, and if we put ±i = PC- 90°, (416) in which the upper sign is used, when R. A. — 12* is positive, and otherwise the lower sign, we have by (99, 105, and 98) tang. PC=n =F cotan. A = cos. (R. A. -\- 6*) cotan. Dec. b= — sin. R. A. cotan. Dec. (417) § 65.] THE ECLIPTIC. 265 To find the latitude and longitude of a star. in which the signs are used as in (416) ; so that A and Dec. are always positive or negative at the same time. Instead of (417), its reciprocal may be used, which is =p tang. A — — cosec. R. A. tang. Dec. (418) If, then, B = E + A, (419) we have AP z= =F E — 90° =|= A =z =p B — 90° (420) or z= 90° ± A ± -E = 90° ± #, in which the upper or lower signs are used, as in (415). Hence cos. PC : cos. AP = ^p sin. A : =p sin. B — sin. A : sin. J5 z= sin. Dec. : sin. Lat. (421) so that, since Dec. and A are both positive or both negative, B and Lat. must also be both positive or both negative. Again, sin. PC : sin. PA = cos. A : i cos. B (422) = dzcotang. (R. A + 6 A ) : ± cotan. (Long. —90°) =r zh tang. R. A. : ± tang. Long. in which the signs may be neglected, and Long, is to be found in the same quadrant with R. A., unless the foot P of the perpendicular falls within the triangle; in which case the first value of AP (420) is used, so that B is obtuse. In this case, the longitude is in the adjacent quadrant on the same side of the solsticial colure with the right ascension. These results agree with the Rule in [B. p. 435.] 65. Corollary, The latitude and longitude of the zenith, that is, the zenith distance and longitude of the nonagesimal, might be found by the same method. But another rule can be 23 266 SPHERICAL ASTRONOMF. [CH. V. To find the latitude and longitude of a star. used, which is of peculiar advantage, where these quantities are often to be calculated for the same place. We have by (310) and (311), calling B the zenith, and putting T = 24* — ZPB or — ZPB (423 ) F=£(PZB — ZBP) or == 180°— £ (PZB—ZBP) (424) G — i (PZB + ZBP) or = 180°— J (PZB+ ZBP) (425) tang.P^-- cosec.J(PJ5+PZ)sin. J(PjB— PZ)cotan.JZ T = tang. (24* — P) (426) tang. 6? = — sec. £ (PP+PZ)cos.J(PP— PZ)cot. J P (427) 90° -f P+#= PZB + 90° or =360° — PZB -{-90° (428) = Long, or = 360° + Long. (429) in which the first member of (426) is used when PB is greater than PZ, and the third when PB is less than PZ, that is, within the north polar circle ; and the second members of (423, 424, 425, 428) correspond to the position of the ze- nith at the east of the solstitial colure, but the third members to the west of the colure. Again, by (295), tang. J (90° — lat.) z=r tang. J alt. nonagesimal = cos. G . sec. F tang. § (PB + PZ), (430) and the preceding formulas correspond to the rule in [B. p. 402.] 66. Scholium. The rule with regard to the value of G ap- pears to be a little different, but the difference is only apparent ; $ 68.] THE ECLIPTIC. 267 To find the altitude of the nonagesimal. for it follows from (427), that G and 12 A — J T are, at the same time, both acute or both obtuse, unless i (PB + PZ)> 90°, or PjB>180° — PZ, (431) which corresponds to the south polar circle. 67. The abridged method of calculating the altitude and longitude of the nonagesimal [B. p. 403.], only consists in the previous computation of the values A — log. [cos. (*-PB — PZ) sec. i(PB + PZ)] (432) C z= log. tang, i (PB + PZ) (433) B = log. tang. £ (PB — PZ) — C (434) =z log. [tang. £ (PB — PZ) cotan. J- (PB + PZ)] == log. [co8ec.£(PB + PZ)Bin.£(PB—PZ)]—A, whence log. [cosec.i(PB-{-PZ)sm.i(PB— PZ)] — B + A (435) and log. tang. G = A + log. (-— cotan. | T) (436) log. tang. F=A + B + log. (— cotan.£ T) (437) = log. tang. G -\- B log. tang. £ alt. non. = log. cos. (2 -f- log. sec. F+ C (438) 68. The rule in [B. p. 436.] for finding right ascension and declination, when the longitude and latitude are given, may be obtained by a process precisely similar to that for the rule before it. 268 SPHERICAL ASTRONOMY. [CH. V. To find the latitude and longitude of a star. 69. Examples. 1. Calculate the latitude and longitude of the moon, when its right ascension is 4 A 42 OT 56% and its declination 27° 21' 58" N., and the obliquity of the ecliptic 23° 27' 45". Solution. 27° 21' 58" N. tang. 9.71400 4 A 42™ 56 s tang. 0.45650 cosec. 0.02503 A = 28° 44' 12" N. sec. 0.05708 tang. 9.73903 E == 23° 27' 45" S. B = 5° 16 / 27"N. cos. 9.99816 tang. 8.96524 long. = 72° 53' 31" tang. 0.51174 sin. 9.98034 lat.= 5° 2'33"N. tang. 8.94558 2. Calculate the values of A, B, and C for the obliquity 23° 27' 40", and the reduced latitude of 42° 12' 2" N. Solution. Polar dist. == 47° 47' 58" 47° 47' 58" 23° 27' 40" . £ sum =z 35° 37' 49" sec. 0.09002 tang. 9.85536 diff. = 12° 10' 9" cos. 9.99013 tang. 9.33374 A = 0.08015, B = 9.47838 3. Calculate the altitude and longitude of the nonagesimal, when the right ascension of the meridian is 19* 50™, the lati- tude 42° 12' 2" N., and the obliquity 23° 27' 40". <§> 69.] THE ECLIPTIC. 269 To find the latitude and longitude of a star. Solution. T = 19* 50™ + 6* — 24* = 1* 50- £(l*50 m ) cotan. 0.61137 A ss; 0.08015 £=101° 30' 2" tang. 0.69152 cos. 9.29968 90° B = 9.4783S C= 9.85536 F= 124° 4' 5" tang. 0.16990 sec. 0.25167 long. == 315° 34' 7" 14° 18' 40" tang. 9.40671 alt. = 28° 37' 20". 4. Calculate the latitude and longitude of the moon, when its right ascension is 18* 27 m 12 s , and its declination 27° 49 38" S., and the obliquity of the ecliptic 23° 27' 45". Ans. The i) 's long. == 276° 1' 46" its lat. = 4°30 / 26"S. 5. Calculate the values of A, B, and C for Albany, and the obliquity 23° 27' 40". Ans. A =3 0.07967 B — 9.47573 C = 9.85333 6. Calculate the longitude and altitude of the nonagesimal, when the obliquity of the ecliptic is 23° 27' 40", the latitude 42° 12' 2"N., and the R. A. of the meridian 10* 10". Ans. The long. = 138° 30' 23" alt. = 61° 18' 49". 23* 270 SPHERICAL ASTRONOMY. [CH. V. To find the declination of a star. 7. Calculate the moon's right ascension and declination, when its latitude is 5° 0' 1" N., its longitude 64° 54' 1", and the obliquity of the ecliptic 23° 27' 45". Ans. Its R. A. -= 4 h 7 m 46*. Its Dec. = 26° 3' V N. 70. Problem, To find the declination of a star. Solution. I. Observe its meridian altitude, and its declina- tion is at once found by one of the equations [345 - 347.] II. If the star does not set, and both its transits are observ- ed, we have p — 90° — Dec. = J (A j — - A'). (438) 71. Problem. To find the position of the equinoctial points. Solution. Since the right ascension of all stars is counted from the vernal equinox, and since the two equinoxes are 12* apart, the present problem is the same as to find the right ascension of some one of the stars, which may afterwards serve as a fixed point for determining the right ascension of the other stars. Observe the declination of the sun for several successive noons near the equinox, until two noons are found between which its declination has changed its sign ; and observe also the instant of the sun's transit across the meridian on these days, by a clock whose rate of going is known. Then, by supposing the sun's motions in declination and right ascension to be uniform at this time, which they nearly are, the time of the equinox, that is, of the sun's being in the equator, is found by the proportion § 72.] THE ECLIPTIC. 271 To find the. right ascension of a star. the whole change of declination : either declination = the sideral interval between the transits — 24 /l : the sideral interval between the transit of the equinox and that of the sun at this declination; (439) and this interval is the difference between the right ascensions of the sun at this declination and the equinox. If the passage of a star had been observed in the same day, the right ascen- sion of the star would have been the interval of sideral time of its passage after that of the vernal equinox. t 72. Examples. 1. If the sun's declination is found at one transit to be 7' 9".5 S., and at the next transit to be 16' 3I".l N. ; what is the sun's right ascension at the second transit, if the sideral interval of the transits is 24 /l 3™ 38 s .21. Solution. T 9".5 + 16' 31". 1 = 23' 40 // .6 — 1420'.6 ar.co. 6.84753 1G' 31 .1 ae 991".l 2.99612 3 m 38 ? .21 = 218 s .21 2.33887 ©'s R. A. == h 2" 32 s .2 152\2 2.18252 2. If the sun's declination is found at one transit to be 18' 38".8S., and at the next transit to be 5' 3".2 N. j what is the sun's right ascension at the second transit, if the sideral interval of the transits is 24* 3™ 38 s .4 ? Ans. h m 46*.5. 3. If the sun's declination is found at one transit to be 5' 57".9 N., and at the next transit to be 17' 26 // .3 S. ; what is the sun's right ascension at the second transit, if the sideral interval of the transits is 24 A 3™ 35 s .71 1 Ans. 12*2™40*.8. 272 SPHERICAL ASTRONOMY. [CH. V. To find the obliquity of the ecliptic. 73. Problem. To find the obliquity of the ecliptic. Solution. I. Observe the right ascension and declination of the sun, when he is nearly at his greatest declination ; that is, when his right ascension is nearly 6 h or 18\ If he were observed at exactly his greatest declination, the observed declination would obviously be the required obliquity. But for any other time, the sun's declination and right ascension are the legs of a right triangle, of which the obliquity of the eclip- tic is the angle opposite the declination. Hence tang. ©'s Dec. =z sin. ©'s R. A. tang, obliq. (440) Now if we put h = the diff. of ©'s R. A. and R. A. of solstice, we have 7 tang, ©'s Dec. #***, cos. h = ^^4-_ — (441 tang. Obliq. v ' and by (277) and (278), sin. (obliq. — ©'s dec.) _ I — cos. h 2 sin. 2 %h sin. (obliq. -j- ©'s dec.) ~~ 1-f-cos. h 2 cos. 2 %h — tang. 2 J £ (442) sin. (obliq. — ©'s dec.) = (obliq. — ©'s dec.) sin. \" = tang. 2 £ /Vsin. (obi. + ©'s dec.) (443) obi. — ©'s dec.= cosec. I" tan. 2 £ h sin. (obl.+ ©'s dec.) (444) — -i/^cosec.l^tan. 2 I s sin. (obi. -f ©'sdec.) and the second member of (444) may be regarded as a cor- rection in seconds to be added to the ©'s dec. to obtain the obliquity, and the obliquity in the second member need only be known approximately. <§> 74.] THE ECLIPTIC. 273 To find the obliquity of the ecliplic. 74. Examples. 1. The right ascensions and declinations of the sun on sev- eral successive days were as follows : June 19, R. A. = 5* 50™ 53% Dec. — 23°26' 45".2N. 20 5 55 3 23 27 27 .3 21 5 59 12 23 27 44 .7 22 6 3 21 23 27 37 .3 23 6 7 31 23 27 4 .6 To find the obliquity of the ecliptic. Solution. Assume for the obliquity the greatest observed declination, or 23° 27' 45", and the corrections of all the ob- servations may be computed in the same way as that of the first, which is thus found, I cosec. 1" tang. 2 I s = *§* tang. 1" 6.43570 h — 9 m T = 547 s 2 log. 5.47598 23° 26' 45" + 23° 27' 45" z= 4G° 54' 30" sin. 9.86348 cor. dec. — 59".59 1.77516 23° 26' 45".2 obliquity — 23° 27' 44".8 =z 23° 27' 44".8 In the same way the 2d observation gives 23 27 44 .9 the 3d observation gives 23 27 45 .2 the 4th observation gives 23 27 45 .3 the 5th observation gives 23 27 45 .3 sumzzr 117 18 45 .5 The mean = 23° 27' 45". 1 274 SPHERICAL ASTRONOMY. [CH. V. To find the obliquity of the ecliptic.' 2. The right ascensions and declinations of the sun on sev- eral successive days, were as follows : Dec. 20 ©'s R. A. = ll h 51 m 1 4 s 23° 26' 48".4 S. 21 17 55 40 23 27 30 .0 22 18 7 23 27 44 .0 23 18 4 33 23 27 29 .5 24 IS 9 23 26 45 .5 what was the obliquity? Ans. 23° 27' 44 // .7. § 76.] PRECESSION AND NUTATION. 275 Secular and periodical motions. CHAPTER VI. PRECESSION AND NUTATION. 74. The ecliptic is not a fixed but a moving plane, and its observed position in the year 1750 has been adopted by astronomers as a fixed plane, to which its situation at any other time is referred. The motion of the ecliptic is shown by the changes in the latitudes of the stars. 75. Celestial motions are generally separated into two portions, secular and periodical. Secular motions are those portions of the celestial motions which either remain nearly unchanged, or else are subject to a nearly uniform increase or diminution which lasts for so many ages, that their limits and times of duration have not yet been determined with any accuracy. Periodical motions are those whose limits are small, and periods so short, that they have been determined with considerable accuracy. 76. The true position of a heavenly body, or of a celestial plane, is that which it actually has ; its mean 276 SPHERICAL ASTRONOMY". [CH. VI. Position of the mean ecliptic. position is that, which it would have if it were freed from the effects of its periodical motions. The mean position is, consequently, subject to all the secu- lar changes. 77. The mean ecliptic has, from the time of the earliest observations, been approaching the plane of the equator at a little less than the half of a second each year, thus causing a diminution of the obliquity of the ecliptic. Let NAA 1 (fig. 41.) be the fixed plane of 1750, and NA X the mean ecliptic for the number of years t after 1750. Let A be the vernal equinox of 1750, and AQ the equator. Let n z= NA and n = the angle ANA 1 ; then, upon the authority of Bessel, the point of intersection N of the ecliptic, which is called the node of the ecliptic, with the fixed plane, has a retrograde motion, by which it ap- proaches A at the annual rate of 5". 18, and if this point could have existed in 1750, its longitude would have been 171° 36' 10", so that § ii— 171° 36' 10" — 5". 18 t. (445) Moreover, the angle which the mean ecliptic makes with the fixed plane increases at the annual rate of 0".48892, but this rate of increase is itself decreasing at such a rate, that at the time t this angle is rr — 0",4S892 t — 0".00000307l9 t* (446) 78. Problem. To find the change of the mean lati- tude of a star, which arises from the motion of the ecliptic. $ 78.] PRECESSION AND NUTATION. 277 Change of mean latitude. Solution. Let L = the #'slat. in 1750 3 L = its change of lat. J. z= its long, in 1750—171° 36' 10" + 5". 18* (447) = its long, referred to the node of the ecliptic . Again, the triangle ZPP gives, by (295), sin. \(PZB+P) : cos.|(PZP- P)=:tan.jU : tan.£(PP+P,Z) But £(PZB + P) = 90° — %U, i(PZB—P) = J+ltj, 24 278 . SPHERICAL ASTRONOMY. [CH. VI. Mean celestial equator. whence 9 A — n cos. 4 tang. L (449) sa (0".48892 t — 0".0000030719 t 2 ) cos. a tang. L. 79. The mean celestial equator has a motion by which its node upon the fixed plane moves from the node of the ecliptic at the annual rate of about 50", while its inclination to the fixed plane has a very small increase proportioned to the square of the time from 1750. Thus, if AQ (fig. 41.) is the equator of 1750, and A 1 Q' that for the time f, so that A is the vernal equinox of 1750, and AP A x that for the time t. Let yj = AA' t to — NA'Qi, then A' moves from A at the annual rate of 50 '.340499, and this rate is diminishing so that at the time y = 50 // .340499 t — O'.OOO 121 7945 t 2 , (450) and the value of w in the year 1750 was * — 23° 28' 18", and is increasing at a rate proportioned to the square of the time, so that w = co/ -J- 0". 00000984233 t 2 . (451) 80. Problem. To find the change of the mean ob- liquity of the ecliptic and that of longitude. Solution. Let (fig. 41.) JV^Q'z^, NAA 1 = tl + Ji; § 80.] PRECESSION AND NUTATION. 279 Change of mean obliquity and longitude. then, by (310) and (311), sin. [#+£(v + Vi)] _ tang.fr(q)-f cnj (452) (453) sin. J (V — V^) tang. J n cos, [ g + ^^ + ^a)] __ tang. 1(g) ! -co) cos. £ (v — Vj) tang. £ 7i Now in calculating the parts of y 1 — %p and Wj — to, which are proportional to the time, we may, since y and x p 1 differ but little as well as w and Ml9 and since n is small, put 2r -fi(v + ^i)= ?* sin - 5-(^ — V' 1 ) = ^(V — vjsin. 1" tang. | n — £ 7T tang. 1" — £ tt sin. 1" = ^ (0".48892) t 1 sin. 1" £(« + »!)=«', tang.JK— w ) = i( w i— w )sin.l' cos. £ (v — v^) p 1, which, subtituted in (452) and (453), give yj — yj 1 z=z (T. 48892 t sin. tt cotan. »' (454) Wl — co — ".48892 * cos. n, (455) which are thus computed, 0".48S92 9.68924 9.68924 171° 36' 10" cos. 9.99532, sin. 9.16446 — 0".48368 9.68456 n 23° 28' 18" cotan. 0.36229 0". 164431 9.21599 that is, », — a, = — 0".48368 t (456) y — y 1 — 0". 164431* (457) 280 SPHERICAL ASTRONOMY. [CH. VI, Change of mean obliquity and longitude. or mi =z 23° 28' 18" — '.48368 I (458) fl =~ 50".340499 t — 0".l 64431 t — 50M76068 t. (459) But, in computing the parts of Wj — w and y — y %9 which depend upon t 2 } we need only retain the part depending upon t 2 in the value of tang. J n, and neglect these pajrts in the other terms of (452) and (453), we thus have sin. [tz+ -J- (v + vi )] = sin, (n + 45".08 1) (460) z= sin. n-\- 45". 08 £ sin. i". cos. n cos.[^4-^(^ + ^ 1 )]z=cos.(n)—45 // .08/sin. l"sin. n (461) 'tan.^=ffi sin. I"=£sin. l // (0 // .48892 If — 0".00000307 1 9 * 2 (462) cotan. £ (co -f ij) — cotan. (o> — .24184 *, (463) _ l-j-0'.24184*sin. Ftang «' Z tang. W — "34184 t sin. 1" = cotan. J + 0' .24184 t sin. 1" ( I + cotan. 2 o') = cotan. a/ -f- // .24L84 * sin. \" cosec. 2 ^ cos. J (v — ¥ t ) = l| sin. ^ (v — Y^) == £ (^ — Vjsin. I" sin. £ (Wj — w) = J (Wj — w) sin. 1" which, substituted in (452) and 453), give v ,— 1// 1 =0 // . 164431 * + // .48892* 2 sin. 1"45".08 cos. n cotan. «>' -f '.48892 * 2 sin. 1" X // .24184sin. n cosec. 2 w (464) — (K0000030719 t 2 sin. 77 cotan. to' to t — cd — _ ".48368 * — (K48892* 2 sin. 1"45".08 sin. n — / .0000030719 t 2 cos. 77, <§> 80.] PRECESSION AND NUTATION. 281 Change of mean obliquity and longitude. which are thus computed, 0".48892 9.68924 1" sin. 4.68557 45".08 1.65398 171° 36' 10" sin. 9.16446 cos. 9.99532 n — -0". 000015605 5.19325 + 0^.000003039 0". 0000030719 4.48741 — 0".000012566 4.48273 R 0".0000030719 4.48741 171° 36' 10" sin. 9.16446 cos. 9.99532 n sin. 9.16446 23° 28' 18" cotan. 0.36229 0.36229 cosec* 0.79958 — 0".000001033 4.01416 45".08 0".48892 0".24184 sin. 1 ' 4.68557 1.65398 9.68924 4.68557 9.68924 — 0".000243445 6.38640,, 9.38353 0".000000528 3.72238 —0.000243950 so that y — xp 1 = 0". 164431 1 — 0".000243950 t* « 1 — toz=— 0".48368 t — 0".000012566 W V j t =z 50".176068 * — 0".0001217945 1 2 + 0".000243950 1* — 50".176068 t -f 0".000122156 t* (465) w x zz: 23° 28' 18" — 0".48368 t — 0".000002724 * 2 (466) 24* 282 SPHERICAL ASTRONOMY. [CH. VI. Precession of the equinoxes. or more accurately, from BesseFs Fundament a Astronomic, V ', =s= 50".176068* + O'.OOO 1221483 t 2 (467) w, =± 23° 28' 18" — 0".48368 t — 0". 00000272295 t (468) These values were afterward changed by Bessel in his TabulcB RegiomontancB to V = 50".37572 * — 0' .0001217945 t 2 (469) y, z= 50".21129 t + 0'.0001221483 tf . (470) to, =z 23° 28' 18" — 0".48368f — // .00000272295 ^ (471) Bat these formulas were obtained from the physical theory, and are, as Bessel says, subject to errors, on account of the uncertainty with regard to some of the data ; so that we shall adopt Poisson's formulas, because they agree in the variation of the obliquity almost exactly with BesseFs observations, and shall change the value of »' to that determined by Bessel from observations; our formulas are, then, w ' = 23°28'17".65 (472) y = 50".37572 t — 0. // 00010905 t 2 - (473) V I = 50 // .22300 t + 0."0001 1637 t 2 (474) w ~ 23° 28' 17".65 -f 0".0000800i t 2 (475) » t = 23° 28' 17".65 — 0".45692 t — 0".000002242 1 2 (476) If, now, the value of y x is added to that of d 4 (449), the resulting value is the total change of a star's mean longitude. 81. The backward motion y t of the equinoxes is called the precession of the equinoxes. *82.] PRECESSION AND NUTATION. 283 Change of mean equator. 82. Problem. To find the intersection of the mean equator with the equator of 1750 and its inclination to it. Solution. Produce A Q and AQ' (fig. 41.) till they meet at T, and let AT— *, AT— <*>', and the triangle ATA 1 gives, by (291, 295, and 310), cos. J (to 7 — w) : cos. J (a/ -{- w)=ztang. J v : tang. £(<*>' — <£>) (477) sin. £ (ex)' — w) : sin. J (a/ -|- w)z=tang.J v : tang. J( *'-)-*) (478) sin. J (#'-(-#): sin.J ('—*):=: cot an. J J 7 : cot. $ (o'-fco) (479) so that t 2 may be neglected in all the terms but y, and we have 1 : cos. »' ±= £ V» sin. 1" : | (*' — ) sin. 1" (480) : sin. J = £ y sin. 1" : tang. £ (<£'+<*>) (481) 1 : | (** — «£) sin. 1" = tang. j'i J. T sin. 1". (482) Hence £ (*' + *) = 90° (483) i (*' — tf>) = £ * cos. co 7 (4S4) y = (<*>/ __ i) tang, co 7 , (485) which are thus computed, a,' cos. 9.96249 cos. 9.96249 25" 18786 1.40120 23". 103 1.36369 0".000054525 5.73660 o".oooo50oi3 5.69909 co> tang. 9.63771 9.63771 10 // .032 1.00140 0".000021717 5.33680 284 SPHERICAL ASTRONOMY. [cH. VI. Change of mean right ascension and declination. so that & = 90° — 23". 103 1 + 0'.000050013 t* ( 4 86) T — 20". 0640 1 — 0".000043434 t 2 . (487) 83. Problem. To Jind the variation of a star's mean right ascension and declination. I. The variation, which arises from the change of the equa- tor's inclination, may be found precisely in the same way in which the variations of latitude and longitude were found in § 78, for a similar change in the position of the ecliptic ; so that formulas (448) and (449) give, by substituting for -^, L and 7i, A = #' s R. A. — 90° + 23".103 t =: R — 90 L = #'s Dec. = D, tv~ T 3D z= — T cos. R (488) 8R~ T sin. R tang. D ; (489) or instead of counting the value of T and t from 1750, they may be reduced to the beginning of each year, and the square of t may then be neglected. II. The variation in right ascension is to be increased by the change in the position of the equinox, arising from its precession, which is thus found. Had the ecliptic remained stationary, the equinox would have removed from A to A\ so that if AP is perpendicular to the equator, we should have for the increase of right ascension by (475) and (484), AP = AA! cos. AAP = y cos. « (490) = (*' — <*>) = 46 // .206 t — G^.000100026 t 2 . <§> 84.] PRECESSION AND NUTATION. 285 Change of mean right ascension. But the equinox advances upon the equator from the motion of the ecliptic by the arc A'A lt which is thus found. We have, by (291), cos.^-(co 1 — w):cos. ] J(a) 1 + co) = tang. £ 4'^ :tang. £ (y— yj But COS. z(o) 1 w) zzz 1 cos. £ (co 1 -f w) = COS. (a;' — 0".22846 * ) — cos. a/ + 0'.22846 1 sin. 1" sin. J sec. 2- ( w j + w) = sec. «' — '.22846 1 sin. 1" sin. w' sec. 2 w' tang. £ il'J , = i yi'4 j sin. I" tang.£(v/ — Vi|= i(v — Vi) sin - j" == J sin. 1" (0".15272 * — 0".00022542* 2 ) whence A , A 1 = 0". 15272 1 sec. w — 0' .00022542 * 2 sec . */ — 0' .22846 t 2 ; 15272 t 2 sin.l" tang. w> secV which is thus computed, ".15272 9.18390 9.18390 sec. 0.03751 0.03751 0.03751 0U665 9.22141 0".00022542 6.35299 ,/ .00024575 6.39050 0".22846 9.35881 tang. 9.63771 1" sin. 4.68557 2.90350 286 SPHERICAL ASTRONOMY. [CH. VI. Nutation. so that A' A t — 0".1665 t — 0".00024575 t 2 , (491) and, by (489) and (490), d R — 46".0395 1 + 0".00016593 1 2 + Tain. R tang, D. (492) 84. By the motions of precession and of diminution of the obliquity, the mean pole of the equator is carried round the pole of the ecliptic, gradually approaching it ; but the true pole of the equator has a motion round the mean pole, which is called nutation. This motion is in an oval, at the centre of which is the mean pole, and is such that the position of the mean equinox dif- fers from that of the true equinox by the longitude d \ong.=i sin. £1 + ^ sin. 2 Sl+i 2 sin. 2j> + i 8 sin.2© (493) where £l = the mean longitude of that point of intersec- tion of the moon's orbit with the ecliptic, through which the moon ascends from the south to the north side of the ecliptic, and which is called the moon's ascend- ing node, J) = the moon's true longitude, © =. the sun's true longitude. The values of i, i lf i 2 , i 3 are given differently by different astronomers, and those which are, at present, adopted in the Nautical Almanac are i — — 17'.2985, i x = 0".2082 (494) i 2 = — 0".2074, » 8 = — 1 '.2550. § 86.] PRECESSION AND NUTATION. 287 Nutation. This nutation of the pole causes also the true obliquity of the ecliptic to change from the mean obliquity by the quantity ^ 1 =kcos.^l + k 1 cos.2£i-\-k 2 cos.2j> + k s cos.2Q (495) in which the values of k &c, at present adopted in the Nauti- cal Almanac, are k & 9".2500, k 1 =z — 0".0903 (496) k 2 = 0".0900, k 3 = 0".5447. 85. Corollary. The effect of nutation upon the right ascensions and declinations of the stars may be com- puted by § 83, and the formulas which are obtained agree with those given in the Nautical Almanac, and which depend upon the terms, called C and D in the formulas for Reduction of the Almanac ; these terms contain also the changes arising from the mean motion of the equinoxes, and the formulas are so reduced that t is counted from the beginning of each year. 86. Examples. 1. Find the mean obliquity of the ecliptic for the year 1840, and reduce the formulas for finding the variations of right as- cension and declination to the beginning of that year. Solution. In (476) let t = 1840 — 1750 = 90, and it gives m % = 23° 28' 17 ".65 — 41".12 — 0'.02 =: 23° 27' 36' .51. In (487, 488, and 492) let t hn 90 + *', and neglect the terms depending upon t' 2 t so that 288 SPHERICAL ASTRONOMY. [CH. VI. Change in right ascension and declination. T = 30' 5 ".76 — 0'.35 + 20".0640 1 1 — 0".0078 1 1 z=30' 5".41 + 20".0562t", and the mean variations, counted from the beginning of the year, are * D — 20 ".0562 1 cos. R $ R = 46 // .0693 t 1 + 20 ".0562 t 1 sin. R tang. D. Finally, the variations arising from nutation are thus found. The change in the obliquity of the ecliptic gives at once, from (448) and (449), by referring the positions to the mean eclip- tic instead of to that of 1750, * D = — a Wl sin. R d' R = — <5 Wj cos. R tang. D, and the change in the position of the equinox gives by (485, 488, 489, and 490), 2 T = — a A sin. Wl d'D= $ A sin. m 1 cos. J? d' Rz=z 3 A cos. w x -{- $ A sin. w 1 sin. 12 tang. D. Hence, if we take 46 // .0693 C = 46 ".0693 t" + c'= 20".0562 cos.JR d — cos. R tang. X) e£ = — sin R we have ° = ^ + 46^693 ^ = ^ + 2^^562 ^ — t e — 0.3448 sin. & + 0.00415 sin. 2 & — 0.00413 sin. 2 J> — 0.02502 sin. 2 O, <§> 86.] PRECESSION AND NUTATION. 289 Nutation in right ascension and declination. and the entire changes of declination and right ascension are ^ D — Cc' — da.d' V R —. Cc —du.d, which agree with the formulas in the Nautical Almanac, ex- cept in the coefficients of V , which are 46".0206 and 20".0426 instead of 46 // .0693 and 20".0562. If, again, we take / == 46".0693 C, g cos. G = 20 // .0562 C, g sin. G = — d », the above formulas become d' D= gcos. Gcos.R — gsin. 6rsin. R =: g cos. (G + R) <*' R — f-{-gsm. R cos. Gtang. D-\-g sin. Gcos. R tang. D =f+g sin. (R + G) tang. D, as in the Nautical Almanac. 2. Find the annual variations in the right ascension and declination of « Hydrae for the year 1840, and its true place for mean midnight at Greenwich, Jan. 1, 1840; its mean right ascension for Jan. 1, 1839, being 9* 19 w 40*.620, and its decli- nation — 7° 57' 49 '.50, and using the numbers of the Nautical Almanac. 25 290 SPHERICAL ASTRONOMY. [CH. VI. Nutation in right ascension and declination. Solution. 20".0426 1.30195 R=z9 h 19 CT 40*.620 cos. 9.8S374* ( jD=- 15".335 1.18569" D = — 7° 57 49".50 * R = 46".0206 — 1' .8051 = 44 ".2155 = 2'.948 1.30195 sin. 9.80872 tang. 9.14584" 0.25651* Hence its mean place for Jan. 1, 1840, is R — 9 h 19- 43 s .568 D —-T 58' 4".83. To calculate the effects of nutation, we have a = 339° 40, J) = 242° 30', © = 281° 15' —0.3448 sin. &= 0.1205, 9".25 cos. & = 8".673 0.00415 sin. 2 & =—0.0027, — 0".0903 cos. 2 & =— 0".068 —0.00413 sin. 2 J> =— 0.0034, 0".0900 cos. 2 J> =— ".032 —0.02502 sin. 2 © = 0.0096, 0".5447 cos. 2 © =— 0".504 C = f + 0.1240, <5 Wl - 8".049 Pc;=? c' f -f 20".0426 X 0.1240 cos. R = c't' — 15".335 X 0.1240 z= c < t> — 1".901 — * ».<*' = 8' .049 sin. JE -z 5". 181 Cc = c t> -f 0.1240 X 2*.948 = ct'-\- 0*.365 — I w d = — 8' .049 cos. JR tang. Dm — 0".861 = — 0*.058, § 86.] PRECESSION AND NUTATION. 291 Nutation in right ascension and declination. whence the variations arising from nutation are d< D = 3' .28, d ' R = 0*.30, and the true places are D = — 7° 58' 1'.55, R = 9* 19" 43 s .87. 3. Find the mean obliquity of the ecliptic for the year 1950, and reduce the formulas for finding the variations of mean right ascension and declination to the beginning of that year. Ans. Wl = 23° 26' 36".18. *' D = 19".8903 1 1 cos. R *' Rz=l 46". 1059 t 1 + 19 7/ .8903 t sin. R tang. 2>. 4. Find the annual variations in the right ascension and declination of /J Ursse Minoris for the year 1839, and its true place for mean midnight at Greenwich, Aug. 9, 1839; its mean right ascension for Jan. 1, 1839, being 14 A 5l m 14 s .943, its declination 74° 48 7 48 // .89 N., the longitude of the moon's ascending node for Aug. 9, 1839, being 347° 17 7 , that of the moon 144° 2 ; , and that of the sun 136° 30 7 , and using the constants of the Nautical Almanac, which give for Aug. 9, 1839, f— 32 // .33, g — 16 '.70, G — 327° 30'. Ans. Var. in R. A. —~ 0'.277 ; var. in Dec. = U'.ll ; and for Aug. 9, 1839, R a 14 A 51 CT 16*.36 D = 74° 48' 32 // .46. 292 SPHERICAL ASTRONOMY. [CH. VI. Tables XL and XLIII. 5. Calculate the values of f t g, and G for April 1, 1839, mean midnight at Greenwich, when g\, == 354° 10', © = 11° 34', and J) is neglected. Ans. f— 12'. 53, g — 1I"&*, G z= 299° 34. In Table XL of the Navigator, the decimal is neglected, and 20 used instead of 20.0562. Table XLIII is calculated from the formulas of Bessel, which differ a little from those of Bailly used in the Nautical Almanac. The construction of these two tables is sufficiently simple from the calculations already given. § 89.] time. 293 Sideral and solar day. CHAPTER VII. TIME. 87. The intervals between the successive returns of the mean place of a star to the meridian are precisely equal, and the mean daily motion of the star is perfectly uniform ; so that sideral time is adapted to all the wants of astronomy. The instant, which has been adopted as the commencement of the sideral day, is the upper transit of the vernal equinox. The length of the sideral day, which is thus adopted, differs therefore from the true sideral or star day by the daily change in the right ascension of the vernal equinox. But this change is annually about 50 7/ or 3 S .3, so that the daily change is less than O'.Ol, and is altogether insensible. 88. Corollary. The difference between the sideral time of different places is exactly equal to the differ- ence of the longitude of the places. 89. The interval between two successive upper tran- sits of the sun over the meridian is called a solar day ; and the hour angle of the sun is called solar time. This is the measure of time best fitted to the common purposes of life. 25* 294 SPHERICAL ASTRONOMY. [CH. VII. Perigee. Apogee. The intervals between the successive returns of the sun to the meridian are not exactly equal, but depend upon the vari- able motion of the sun in right ascension, and can only be determined by an accurate knowledge of this motion. 90. The want of uniformity in the sun's motion in right ascension arises from two different causes. I. The sun does not move in the equator but in the ecliptic. II. The sun's motion in the ecliptic is not uniform. The variable motion of the sun along the ecliptic, and its deviations from the plane of the mean ecliptic, cannot be dis- tinctly represented, without reference to the variations of its distance from the earth, and to the nature of the curve which it describes. This portion of the subject, therefore, which involves the determination of the sun's exact daily position, that is, the calculation of its ephcmeris, must be reserved for the Physical Astronomy. It is sufficient, for our present purpose, to know that the sun moves with the greatest velocity when it is nearest the earth, that is, in lis perigee; and that it moves most slowly when it is farthest from the earth, that is, in its apogee. 91. The sun arrives at its perigee about 8 days after the winter solstice, and at its apogee about 8 days after the summer solstice. The mean longitude of the perigee at the beginning of the year 1800 was 279° 30' 5", and it is advancing towards the eastward at the annual rate of about 11". 8, so that, by adding the pre- cession of the equinoxes, the annual increase of its longitude is about 62". $ 95.] time. 295 Mean and apparent time; equation of time. 92. To avoid the irregularity of time arising from the want of uniformity of the sun's motion, a fictitious sun, called a mean sun, is supposed to move uniformly in the ecliptic at such a rate, as to return to the perigee at the same time with the true sun. A second mean sun is also supposed to move in the equator at the same rate with the first mean sun, and to return to each equinox at the same time with the first mean sun. We shall denote the first mean sun by Q 1} and the second mean sun by © 2 . 93. Corollary. The right ascension of the second mean sun is equal to the longitude of the first mean sun. 94. The time which is denoted by the second mean sun is perfectly uniform in its increase, and is called ?nean time; while that which is denoted by the true sun is called true or apparent time; the difference be- tween mean and true time is called the equation of time. 95. The time which it takes the sun to complete a revolution about the earth is called a year. The time which it takes the mean sun to return to the same longitude is the common or tropical year. The time which it takes it to return to the same star is the sideral year ; and the time which it takes it to return to the perigee is the anomalistic year. 296 SPHERICAL ASTRONOMY. [cH. VII. Year. Leap year. The length of the mean tropical year is F = 365 d 5 h 48" 4V.808, (497) so that the daily mean motion of the sjn is found by the pro- portion Y : l d = 360° : daily motion = 59' 8 ".3302. (498) 96. The fraction of a day is necessarily neglected in the length of the year in common life, and the common year is taken equal to 365 d . By this approximation, the error in four years amounts to 23* 15- 1P.232 = l d — U m 48*. 768, (499) or nearly a day, and an additional day is consequently added to the fourth year, which is called the leap year. At the end of a century the remaining error amounts to nearly — d .75, which is noticed by the neglect of three leap years in four centuries. For practical convenience, those years are taken as leap years which are exactly divisible by 4, and the centu- rial years would thus be leap years, but only those are re- tained as leap years which are d. visible by 400. 97. When the mean sun has returned to the same mean longitude, it has not returned to the same star, because the equinox from which the longitude is counted has retrograded by 50" .223, so that the mean sun has 50 ; .223 farther to go, and the time of describing this arc is the fourth term of the proportion 59' 8'.3302 : l d = 50 // .223 : 20 m 22 a .786, (500) so that the length of the sideral year is Y t = Y + 20* 22 s .786 = 365 d 6 h 9 m 10'.594. (501) <§> 98.] time. 297 Tables LI and LII. Reduction of solar to sideral time. 98. The length of the mean solar day is also different from that of the sideral day, because when the © 2 , in its diurnal motion, returns to the meridian, it is 59 / 8 7/ .3302 advanced in right ascension ; so that 360° 59' 8 7/ .3302 pass the meridian in a solar day, instead of 360°, which pass in a sideral day. Hence the excess of the solar day above the sideral day, ex- pressed in solar time, is the fourth term of the proportion 360° 59' 8".3302 : 59' S'.3302 — l d : d .0027305 or Z h 55-.9094 ; (502) that is, 1 sid. day = 0.9972695 sol. day, or 24 A sid. time a= 23* 56 m 4 s .O906 of solar time ; (503) which agrees with (335) and the table for changing sideral to solar time in the Nautical Almanac, and with table LII of the Navigator. In the same way this excess expressed in sideral time is the fourth term of the proportion 360° : 59' 8V3302 = l d : d .002738 or 3 m 56 s 5554 ; that is, 1 sol. day = 1.002738 sid. day, (504) or 24* sol. time == 24* S m 56 s .5554 sid. time j (505) which agrees with the table for changing solar to sideral time in the Nautical Almanac, and with table LI of the Navigator. The remainder of tables LI and LII, as well as the corre- sponding ones given in the Nautical Almanac, are calculated by simple proportions from the numbers which are given for 24\ The sideral day begins with the transit of the true vernal 298 SPHERICAL ASTRONOMY. [CH. VII. Sun's longitude on January 1. equinox. At the time of the transit of © 2 , then, that is, at mean noon, we have the sid. time := R. A. of © 2 from the equinox zs R. A. of © 2 from mean equinox -}- Nutation of equinox in R. A. =2 sun's mean long -[-Nutation in R. A. (506) 99. The sun's mean long, for Jan. 1, 1800, at Paris, was found by Bessel to be 279° 54' 11 ".36. Its longitude for Jan. 1, of any other year t, may thus be found. Let f be the remainder after the division of t by 4, the number of days, then, by which Jan. 1 of the year t is removed from Jan. 1, 1800, is 365^ (t —f) + 365/ z= t. 365^ - \ f ss Y.t + t.ll m \2°.l92 — if (507) ss Y. t + t . d .00778 — if. But in Yt days the sun's longitude increases exactly £.360°, which is to be neglected ; and its increase in longitude is 59' 8".3302 (*+0.00778— lf)—t.21"M— f. 14' 47 ".083, (508) or more accurately from Bes el, the mean longitude E, for the the first of January of the year 1800 + ' at Paris, is E = 279° 54' 1 ".36 + 1 27".605844 + 1*. 0".0001221805 — /, 14' 47".083. (509) The mean longitude is found for the first of January, for any other meridian by the following proportion, derived from the interval of time between the © 2 >s passage over this me- ridian and that of Paris. 24* : long, from Par. ss 59 / 8 // .3302 : change in value of E. (510) $ 100.] time. 299 Sideral time converted to solar time. The sun's mean longitude for any mean noon n of the year after that of the first of Jan. is E + n.59' 8'.3302. (511) Hence the sideral time of the mean noon n is •pi &= - + ».3 W 56^.555348 + Nutation in R. A. (512) so that the solar time of the transit of the equinox from the preceding noon is 24* — S (converted into solar time). (513) 100. Examples. 1. Find the sideral interval which corresponds to 10* of solar time. Ans. 10 l m 38 s .5647. 2.' Find the solar interval which corresponds to 10* of si- deral time. Ans. 9* 58" 2K7044. 3. Find the sideral interval which corresponds to 10 w of solar time. Ans. 10 m P.6428. 4. Find the solar interval which corresponds to 10* of si- deral time. Ans. 9 m 5S'.3617. 5. Find the sideral interval which corresponds to 10* of solar time. Ans. 1C.0274. 300 SPHERICAL ASTRONOMY. [CH. VII. ' Time by observation of equal altitudes. 6. Find the solar interval which corresponds to 10* of si- deral time. Ans. 9*.9727. 7. Find the sideral interval which corresponds to 5 .85 of solar time. Ans. C.85233. 8. Find the solar interval which corresponds to S .85 of sideral time. Ans. S .84768. 9. Find the sun's mean longitude at Greenwich for the mean noon of April 4, 1839, the sideral time at this noon, and the solar time of the transit of the vernal equinox from the preceding noon; the meridian of Greenwich is 9 m 21 s .5 wes, of that of Paris. Ans. The sun's mean longitude = 12° T 3".02. The sideral time of mean noon = 48 m 3P.27. Time of tran. of ver. equi. == April 3d, 23* 1 l m 39*.6S. 101. Problem. To find the time by observation. Solution. First Method. By equal altitudes. I. If the star does not change its declination. Observe the times when the star is at equal altitudes before and after pass- ing the meridian ; the arithmetical mean between these two times is the time of the star's passing the meridian, which compared with the known time of this passage, gives the error of the clock at this time, and the correction of this error gives the time of each observation. $ 101.] TIME. 301 Equal altitudes of sun. IT. When the declination of the star is changing, the time of the star's arriving at the observed altitude A is affected ; thus if L sr the latitude, D z= the declination at the meridian, d D =i the increase of declination from the meridian, h — the hour angle, supposing no change in the decli- nation, dh — the increase of the hour angle in time, we have, by (380), sin. A — sin. L sin. D -f- cos. L cos. D cos. h (514) = sin. L s'm.(D-\- 3 D) -\- cos.L cos.(D-{-d D)cos (h-\-3h) =. s'm.L sin.D-{-dD sin. I" sin.L cos.Z>-|- cos.Z cos.ZJcos.A — 3 D sin. 1" cos.Zsin. Dcos.h — 15 cotan. A 15 cotan. Zi sin. h 15 cotan. i> tang. K (515) and since the two observations are at nearly the same distance from the meridian, the value of S7i is the same for both of them ; so that their mean is augmented by and d h is consequently to be subtracted from the mean of the observed 26 302 SPHERICAL ASTRONOMY. [CH. VII. Equal altitudes of sun. times, in order to obtain the true time of the star's passing the meridian. In calculating the value of 3 h, its two. terms may be calcu- lated separately. Now if & D is the daily variation of the star's declination, we have hi'D 2h#D /e%aK ' D = -*4T -5X24* (516) and in using proportional logarithms, the proportional logarithm of the hours and minutes of 2 h, which is the elapsed time, may be taken as if they were minutes and seconds, provided the same is done with the 2-i h in the denominator. Finally, the value of <5 h is reduced from minutes and seconds to sec- onds and thirds by multiplying by 60, so that if M is taken for the denominator of either of the parts of (515), this part P is calculated by the formula Prop. log. P = — Prop. log. g * 2 ^* 15 + log.M+Prop.log.2 h + Prop. log. 6' D, (517) which agrees with [B. p. 219.], for —Prop. log. * Q A - = — Prop. log. 12- = — 1.1761 = 8.8239. (518) III. If the altitude at the two observations had differed slightly, the mean time would require to be corrected ; for this purpose, let 3 A = the excess of the second altitude above the first, 3 h z= the increase of the hour angle, § 101.] TIME. 303 Altitudes nearly equal. Single altitude. and we easily deduce from (514) cos. A 3 A =z — 15 cos. L cos. D sin. h dh t (519) xi. , 7 cos. ^4 * A ,^™ v so that s h = — — = pr^ — r. (520) 15 cos. Lt cos. i> sin. Ii The time of the second observation being thus increased by dh, that of the mean is increased by J dh, which is, therefore, the correction to be subtracted from this mean. The corrections (515) and (520) must be both of them ap- plied when the star is changing its declination, and at the same time the observed altitudes are slightly different. Second Method. By a single altitude. [B. p. 208-218.] When a single altitude is observed, there are known in the triangle PZB (fig. 35.), the three sides, to find the hour angle ZPB, which is thus found by (277), * s = J (z + 90° — L + p) (521) cos.J^v(-^^ ( ^-V (522) 2 ^ \sin.(90° — J L)sin.p/' v ' which corresponds to [B. p. 210.] The hour angle may also be found by (282), thus if we put s< = i(A + L+p), (523) we have 5 = J (180° — A — Z,+jp)=z90°— s f +pz=90 o — A— L+s< s—p = 90° — s' > s — (90° — L)=zs' — A, whence • -.t ./'cos. s f sin. (s* — A)\ /co<\ sin. i h = \/| ^A - I, (524) 2 ^ \ cos. L sin. p ] which corresponds to [B. p. 209.] 304 SPHERICAL ASTRONOMY. [cH. VII. Distance from terrestrial object. Third Method. By the distance from a fixed terrestrial object. If the position of the terrestrial object has been before de- termined, its hour angle and polar distance may be considered as known. Hence, if T (fig. 40.) is the position of the terrestrial object projected upon the celestial sphere, P the pole, and S the star. Let the distance TS be observed, and let PT=P, PS — Pt TS=d, TPZ — H, TPS = h' t SPT— h, s=zi(P+p + d), (525) we have or . iV=V (^L^^SL=i)\ (526) ■ \ sin. P sin. p J v ' 17/ ,/sin.s sin. (s — d)\ ,•*-* s. i h' — V ( — : =-3 1 I , (527) * \ sin. P sin. y / v ' h=z H+h'. If the polar distance and hour angle of the terrestrial object is not known, but only its altitude and azimuth, the polar dis- tance and hour angle can be easily found by solving the tri- angle PZT. Fcurth Method. By a meridian transit. [B. p. 221.] If the passage of a star is observed over the different wires of a transit instrument, the mean of the observed times is the time of the meridian transit, which should agree with the <§> 101.] TIME. 305 Meridian transit. Vertical transit. known time of this transit. This method surpasses all others in accuracy and brevity. Fifth Method. By a disappearance behind a terrestrial object. If the instant of a star's disappearance behind a vertical tower has been observed repeatedly with great care, the ob- served time of this disappearance may afterwards be used for correcting the chronometer. For this purpose, the position of the observer must always be precisely the same. Any change in the right ascension of the star does not affect the star's hour angle, that is, the elapsed time from the meridian transit ; this change, consequently, affects the observed time exactly as if the observation were that of a meridian transit. A small change in the declination of the star affects the hour angle, and therefore the time of observation. Thus, if P (fig. 44.) is the pole, Z the zenith, ZSS 1 the vertical plane of the terrestrial object ; then if the polar distance PS is diminished by RS — 3 D, the hour angle ZPS is diminished by the angle SS'P — $h. But the S'R is nearly perpendicular to SP, and the sides of SS'R are so small, that their curvature may be neglected, whence RS' = <* D tang. S = 15 cos. D. * A, so that dhz=L T \dD tang. S sec. D. (528) 26* 306 SPHERICAL ASTRONOMY. [cH.'viI. To find the time. 102. Examples. 1. On July 25, 1823, in latitude 54° 20' N., the sun was at equal altitudes, the observed interval was 6 h l m 36* ; find the correction for the mean of the observed t mes. The sun's declination is 19° 48', and his daily increase of declination 12' 44 ". 8.8239 tang. 0.0030 1.4759 1.1503 Solution. 8.8239 54° 20' cotan. 9.8559, 6* l m 36' sin. 9.8510 6 m 1* P. L. 1.4759 12' 44" P.L. 1.1503 12*55 1.1570 — 2*.28 — 2*.28 1.8968 10*.3 = the required correction. 2. On September 1, 1824, in latitude 46° 50' N., the inter- val between the observations, when the sun was at equal alti- tudes, was 7* 46'" 35*; the sun's declination was 8° 14' N., and his daily increase of declination — 21' 49"; what is the correction for the mean of the observations ? Ans. 16*5. 3. On March 5, 1825, in latitude 38° 34' N., the interval between the observations, when the sun was at equal altitudes, was 8* 29™ 28'; the sun's declination was 6° 2' S., and his daily increase of declination was 23' 9"; what is the correction for the mean of the observations? Ans. 15'.4. $ 102.] time. 307 To find the time. 4. On March 27, 1794, in latitude 51° 32' N., the interval between the observations, when the sun was at equal altitudes, was 7 h 29* 55 s ; the sun's declination was 2° 47' N., and his daily increase of declination 23' 26"; what is the correction for the mean of the observations 1 Arts. — 21'.7. 5. In latitude 20° 26' N., the altitude of Aldebaran, before arriving at the meridian, was found to be 45° 20', and, after passing the meridian, to b? 45° 10'; the interval between the observations was 7 h 16 m 35% and the declination of Aldebaran was 16° 10' N. j what is the correction tor the mem of the observations ? Arts. 19 s . 6. In latitude 3G° 39' S., the sun's correct central altitude was found to be 10° 40', when his declination was 9° 27' N. ; what was the hour angle ? , Ans. 7 h 23 w 51*. 7. In latitude 13° 17' N., the sun's correct central altitude was found to be 36° 37% when his declination was 22° 10' S. ; what was the hour angle ? Ans. 9* 17 m 8 8 . 8. In latitude 50° 56' 17" N., the zenith distance of a ter- restrial object was found to be 90° 24' 28", and its azimuth 35° 47' 4" from the south ; what were its polar distance and hour angle ? Ans. Its polar distance =s 121° 6' 43" Its hour angle = 2* 52 w 16 s . 308 SPHERICAL ASTRONOMY. [CH. VII. To find the time. 9. From the preceding terrestrial object three distances of the sun were found to be 78° 9' 26", 77° 39' 26", and 77° 29' 26", when his declination was 14° 7' 13" S. ; what were the sun's hour angles, if he was on the opposite side of the meridian from the terrestrial object? Ans. % h 45™ 49 s , 2 h 43 m 27% and 2* 42 w 39'. § 103.] LONGITUDE. 309 By measurement, signals, chronometer. CHAPTER VIII. LONGITUDE. 103. Problem. To find the longitude of a place. First Method. By terrestrial measurement. If the longitude of a place is known, that of another place, which is near it, can be found by measuring the bearing and distance ; whence the difference of longitude may be calcu- lated by the rules already given in Navigation. Second Method. By signals. The stars, by their diurnal motion, pass round the earth once in 24 sideral hours ; hence they arrive at each meridian by a difference of sideral time equal to the difference of longi- tude. In the same way, the sun passes round the earth once in 24 solar hours ; so that it arrives at each meridian by a difference of solar time equal to the difference of longitude. The difference of longitude of two places is, consequently, equal to their difference of time. Now if any signal, as the bursting of a rocket, is observed at two places ; the instant of this event, as noticed by the clocks of the two places, gives their difference of time. Third Method. By a chronometer. The difference of time of two places can, obviously, be determined by carrying a chronometer, whose rate is well 310 SPHERICAL ASTRONOMY. [CH. VIII. By eclipses, moon's transit. ascertained, from one place to the other ; and if the chronom- eter did not change its rate during the passage, this method would be perfectly accurate. Fourth Method. By an eclipse of one of Jupiter's satel- lites. [B. p, 252.] The signal of the second method cannot be used, when the places are more than 20 or 30 miles apart; and, when the distance is very great, a celestial signal must be used, such as the immersion or emersion of one of Jupiter's satellites. For this purpose, the instant when any such event would happen to an observer at Greenwich is inserted in the Nautical Almanac ; and the observer at any other place has only to compare the time of his observation with that of the Almanac to obtain his longitude from Greenwich. Fifth Method. By an eclipse of the moon. [B. p. 253.] The beginning or ending of an eclipse of the moon may also be substituted for the signal of the second method to de- termine the difference of time. Sixth Method. By a meridian transit of the moon. [B. p. 431.] The motion of the moon is so rapid, that the instant of its arrival at a given place in the heavens may be used for the signal. Of the elements of its position its right ascension is changing most rapidly, and this element is easily determined at the instant of its passage over the meridian by the differ- ence of time between its passage and that of a known star. The instant of Greenwich time, when the moon's right ascen- sion is equal to the observed right ascension, might be deter- § 103.] LONGITUDE. 311 By a moon's transit. jnined from the right ascension, which is given in the Nautical Almanac for every hour. But this computation involves the observation of the solar time, whereas the observed interval gives at once the sideral time of the observation. The calculation is then more simple, by means of the table of Moon -Culminating stars given in the Nautical Almanac, in which the right ascensions of the suitable stars and of the moon's bright limb are given at the instant of their upper transits over the meridian of Greenwich, and also the right ascension of the moon's bright limb at the instant of its lower transit. Hence the difference between the right ascensions of the moon's limb, at two successive transits, is the change of its right ascension in passing from the meridian of Greenwich to that which is 12* from Greenwich ; so that if the motion in right ascension were perfectly uniform, the right ascension, which corresponded to a given meridian, or the meridian, which corresponded to a given right ascension, might be found by the following simple proportion, 12* : long, of place = diff. of right ascensions for 12* : diff. of right ascensions for long, of place, (529) in which the longitude of the place may be counted from the meridian 12* from that of Greenwich, provided the change of right ascension for an upper transit is computed from the pre- ceding right ascension, which is that of a lower transit at Greenwich, that is, if the place is in east longitude. Let then T =r: long., if west, or =12* — long, (if the long, is east) ; and let A z= diff. of right ascension for the Greenwich transits, which immediately precede and follow the required or observed transit, 312 SPHERICAL ASTRONOMY. [CH. VIII. By a moon's transit. and let d A = change of right ascension from the pre- ceding Greenwich transit to the ob- served transit, and we have, by (529), 12* : T— A :tA, (530) A T 12* a A whence 6 A — -, and T = — j—, (531) 1/Z A. and if T is reduced to seconds, we have iA = mm < 532 > log. « A = log. A + log. T + (ar. co.) log. 43200 — ]og. A + log. T + 5.36452 (533) J, 43200 M § JL-' and T =s (534) log. r = 4.63548 -f ar. co. log. 4 + log. S A, (535) and formulas (533) and (535) agree with the parts of the rules in the Naviga or, which depend upon A, and are independent of the want of uniformity in the moon's motion. The corrections which arise from the change of the moon's motion may be calculated, on the supposition that this motion is uniformly increasing or decreasing, so that the mean motion for any interval is equal to the motion which it has at the mid- dle instant of that interval. If we put, then, B zz the increase of motion in 12*, (536) A is not the mean daily motion for the interval of longitude T and the instant J T after the meridian transit at Greenwich, but for the interval 12* and the instant 6* after this transit. The mean daily motion for the instant £ T is, therefore, § 103.] LONGITUDE. 313 By moon's transit. Table XLV. {6>-j T) B (538) so that the correction for A is (6 A -- jT)B _ (21600 8 — j T)B 12* "~ ""43200 ■ and the correction of 3 A in (532) is r(2i600--jr) r(4320o-r ) '*- (43200 r B ~ 2 (43200)2~ ■ B ' (539) and the value of d B is easily calculated and put into tables, like Table XLV of the Navigator. In correcting the value of T (534), the correction of + E)> (549) the triangles SMM 1 and SZM give (sin.^M)^ sin ' Ssin ^ 5 -=^l sin. E cos. 6 _ *in.[E + i(3»E — *b)] sm.j(S"E + S b) sin. d b sin. E $ 103.] LONGITUDE. 317 By a lunar distance. ( ^-" a) . -*!£ v ' ' sin. £ cos. 6 + t 18 " — i cotan - * sin - 1" [(*" E ) 2 — ( d &)*]• (551) The triangles SS'M and SZM give, by (277) and (281), (cos. iS)2 = ^(s-^ls-a) v ' sin. E cos. a _ sin. [JE + j (J" JE — 8 a)] sin, £ ( S j n ^- a ) . ^. (553) v ' ' - sm. E cos.a v ' If now M 1 K and S'L are drawn perpendicular to MS, and £'£/ to JfcT'jSf, we have nearly S'M'= E + 9 E = SM' + SL' — E + 6"E + SL' ' 9E=9"E + SL' = 9»E + 9'E + (SL , —lfE) (554) §'E=.SL = da cos. # (555) SL'z=*a cos. (£'££') =£ 3 a cos. (S— MSM') = 8 a cos. S + ^ a sin. # sin. JBS4P = *'E + S'L sin. JHSflf' (556) #£' — d'E = S'L sin. jff&Jf 7 . (557) But from MSK, sm. MSM 1 = —. — =- = — — = — , (558) sin. E sm. E 27* 318 SPHERICAL ASTRONOMY. [CH. VIII. Lunars. Tables XVII, XVIII, XIX, XX. whence SL'-*E = 8 -' LXM>K „ S ' mA -, (559) sin. E ' and ifr^*g+Vj+* A ***'** (560) 1 ' sin. E v ' 2° + m = (60' -f * JEJ)+(60'+* // JEJ)+ g in js ( 561 ) in which 1° is added to 103.] LONGITUDE. 319 By a lunar distance. by that of table XIV, and for a planet, it is that of table XII, diminished by that of table X, A. The value of 8 b is obtained by the formula H — P cos. b — 8'b 9 (564) in which 8 1 b is the number of table XII, and P is the number taken from the Nautical Almanac, and which is called the horizontal parallax. In computing table XX, the value of P is taken at its mean of 57' 30". In the formulas for the corrections, the zenith distances may be introduced instead of the altitudes, and if we put 90° — a = Z, 90° — b = Z, s 1 =zz + Z+ E, (565) we have, by neglecting the term depending upon the correc- tion of table XX, as well as the other small quantities, sin. 5 j sin. (s 1 — z) cos.^iM — sin. E sin. Z _ sin. [E + j (8"E + 9 b)] sin. £ (8 b — 9" E) sin. E sin. 8 b 9 b — 8" E „ „ . 2 sin. s, sin. (5, — z) , d"E = 8b •-,-», ~ * & sin. E sin. Z sin.Sj sin. (*. — Z) JE + da cos. 2 J 8 = ? — =V L = — j-j sin. E sin. z 2 8 a (566) (567) 2 sin. s. sin. (s. — Z) . r — ^ \ * <*a. ^ = ~^-| ' .* ^ v , 1 L 8 a . (568) sin. E sin. 2 ' Then the second term of the value of 8' E is the first cor- rection of the Third Method of the Navigator [B. p. 242.] 320 SPHERICAL ASTRONOMY. [CH. VIII, By a lunar distance. and the second term of the value of $" E is the second cor- rection of this method ; and the computation from (564, 567, 568) agrees entirely with this method. The third correction is taken from table XX, as in the first method. V. Draw ZN perpendicular to MS, so as to make SN acute. In the right triangles ZSN and ZSM let B=z90° — SN, B'z=z90°+MN, A = i(B' + B) f (569) and we have E == 3IN+ SN=B' — B, (570) and, by Bowditch's Rules for oblique triangles, cos. ZS : cos. ZM == cos. NS : cos. MN, or sin a : sin. b = sin. B : sin. B ; (571 ) and, by the theory of proportions, sin. a -\- sin. b sin. B -f- sin. B sin. b — sin. a sin. B 1 — sin. B' that is, tang. J (a -f- b) _ tang. A tang. £ (6 — a) ~~ tang. £ E tang. A = tang. £ (a -j- b) cotan. J- (b — a) tang. J E (573) B' = A + iE, B — A — ^E, (574) and the right triangles Z#iV, MZN, SLS 1 , MKM', give cos. £ =s = cotan. ZS tang. $ZV = tang, a cotan. 2? d CI — cos. ilf= -Tjr — — cot « ZMtmg, MN = tang. & cotan. i?' # E = da tang, a cotan. JB (^75) 103.] LONGITUDE. 321 Lunars. Table XLVII. and the formulas (573-576) correspond to the Fourth Method of the Navigator. [B. p. 243.] It may be observed, that since cotan. J (6 — a) is the only term of (573) which can change its sign, A is acute when b is greater than a, and obtuse when b is less than a. VI. The most important of corrections of the distance arise from that term of 8 b (564), which depends upon the parallax. If we consider this, therefore, as the only correction of the moon's altitude, we may calculate the corrections of the dis- tance arising from it by putting 9 b =z MM> = P cos. b. (577) The triangles ZSM and M'MK, give then &'E sin. a — cos. E sin. b ,„ m ~ K cos. M=z — - == : — - - (578) x*cos. 6 sin. E cos. 6 d"E s= — P sin. a cosec. E + P cotan. E sin. b, (579) and if we put ^E' = P sin. a cosec. E (580) S 2 E — ±P cotan. E sin. b, (581) in which the signs are taken so that < 592 > 324 SPHERICAL ASTRONOMY. [CH. VIII. Lunar distance from Nautical Almanac. so that by (591) and (340) Prop. log. (T+aT) = Prop. log. T — * Q (593) log.(T+3T) = \og. T+6'Q (594) . . ~,v , m , /,* *Z\ „ _ log. But if in (196) and (204) we substitute S T -jr^Z™, (596) we have, by logarithms and (595), ; lo g - = / f >g. (l+^) - ^, (597) so that by (592) and (208) a T - ^ Q - = ( 18Qfft - r ) ^Q /iboov *~ log. 6 2X 180 -.X 0.434 V ; (180 OT — T) TSQ : 156 OT (599) and the table [B. p. 245.] for correcting by second differences may be calculated by this formula ; and, in order to obtain the value of 9 T expressed in seconds, the factor T should be expressed in seconds, while (180 771 — T) is expressed in minutes ; and it must not be forgotten, that the proportional logarithms are decimals. IX. When the distance is observed for a star, whose dis- tance is not given in the Nautical Almanac, the Greenwich time of the observation can be found approximately by adding the assumed longitude, if west to the observed time, or subtract- ing it if east ; or the time can be taken from the chronometer if it is regulated to Greenwich time. <§> 103.] LONGITUDE. 325 Lunar distance not given in the Nautical Almanac. Find, in the Nautical Almanac, the right ascension and declination of the star and the declination of the moon, for this time. Then, if T and S (fig. 43.) are supposed to be the moon and star, and P the pole of the equator, D and D' their declinations, disregarding their names, so that their polar dis- tances are 90° ± D and 90° ± D 1 , and if R' is their differ- ence of right ascensions, we have, when their declinations are of the same name, by putting 8= i(D + D' + E) (600) cos^^ = cos.^P^v(- ^?" ( ^ V (601) 2 * ^ \ cos. D cos. D 1 J v ' But if the declinations are of the same name, ^ \ cos. D cos. D' / ' v ' and the right ascension of the moon being thus found, the Greenwich time, when it has this right ascension, is easily found from the moon's hourly ephemeris in the Nautical Al- manac, and this method is the same with that in [B. p. 428.] X. The latitudes and longitudes may be used instead of the right ascensions and declinations, and the calculation will be as in [B. p. 427.] The variation of daily motion is, in this case, to be had regard to, precisely as explained in (536-541). XI. The distances of the Nautical Almanac can be calcu- lated from the right ascensions and declinations of the sun, moon, and stars, or their latitudes and longitudes, by resolving the triangles TPS (fig. 43.) by either of the methods which have been given, when two sides and the included angle are known, as in [B. p. 434.] 28 326 SPHERICAL ASTRONOMY. [CH. VIIL ^ Lunar distances. In calculating the distance of the sun and moon, the latitude of the sun may be usually neglected ; so that if SR (fig. 46.) is an arc of the ecliptic, S the sun's place, M the moon's, and MR perpendicular to SR, MR z=z L — the moon's latitude, SR z= L x =- the difF. of long, of © and J) , and cos. E = cos. SM — cos. L cos. Z x , (603) as in [B. p. 433.] It would, however, be rather more accurate to take L z= the diff. of lat. of © and J) . XII. The determination of the longitude by solar eclipses and occultations will be reserved for another chapter. 104. Examples. 1. Calculate the correction of table XLV, when T=zl h 50 w , and B = 9 m = 540 s . Solution. l h 50 m P. L. l m 50* ar. co. 8.0080 12* — 1* 50" = 10* 10" P. L. 10* 10* ar. co. 8.7519 2 P. L. 12 m 2.3522 £ B = 270' 2.4314 corr. = 34 ff .9 1.5435 2. Calculate the correction of table XLV, when T=3* 10", and B = IP*. Ans. 64M, § 104.] LONGITUDE. 327 To find the moon's right ascension. 3. Find the right ascension of the moon's bright limb, Sept. 25, 1839, at the time of the transit over the meridian of New York. The right ascensions of the moon for the two preceding and the two following transits at Greenwich are Sept. 25. Moon II. L. T. 2* m 36*.69 Moon II. U. T. 2 30 38 .08 Sept. 26. Moon II. L. T. 3 1 33 .18 Moon II. U. T. 3 33 19 .89 Ans. 2M3 W 14».4. 4. At a place in west longitude, Oct. 25, 1839, the moon's bright limb passed the meridian 10 m 6'.83 sideral time, before the star C. Tauri ; find the longitude of the place of observa- tion. The fight ascension of the star C. Tauri was 5*43 m 16'.84, and those of the moon Oct. 25. Moon II. L. T. 4*43*53 '.55 Moon II. U. T. 5 16 28 .40 Oct. 26. Moon II. L. T. 5 52 51 .91 Moon II. U. T. 6 26 40 .00 Ans. 70° 25' 30" W. 5. Find the moon's parallax in altitude, and the correction and logarithm of table XIX, when the altitude is 40° 40', and the horizontal parallax is 58'. 328 SPHERICAL ASTRONOMY. [CH. VIII. "~ Tables XVII, XVIII, XIX, XX. Solution. 58' P. L. 0.4918 40° 40' sec. 0.1200 Parallax in alt. =. 44' P. L. 0.6118 By Table XII. Refrac. = V 6" 9.6990 Corr. — 16' 48"* = 59' 42" — 42' 54" P. L. 0.6228 Log. of Table XIX. = 0.2018 6. Find the correction and logarithm of Table XVII, for a star, when the altitude is 13°. 15'. Ans. Corr. = 5G' 2", Log. — 1.3433. 7. Find the correction' and logarithm of Table XVII, for Venus or Mars, when the parallax is 20", and the altitude 24° 30. Ans. Corr. = 58 / 14", Log. =z 1.6647. 8. Find the correction and logarithm of Table XVIII, when the altitude is 56°. Ans. Corr. — 59' 26", Log. — 1.9544. 9. Find the correction and logarithm of Table XIX, when the altitude is 70°, and the horizontal parallax 54'. Ans. Corr. = 41' 34", Log. = 0.2299. * The numbers of Table XIX are so disposed in the Navigator, that the corrections of proportional parts of parallax are all additive. This is effected by placing each number opposite that parallax, which is 10" less than the one to which it belongs. There is, therefore, a correction for 0" of parallax. § 104.] LONGITUDE. 329 Auxiliary angle in lunar distances. 10. Compute the value of the auxiliary angle m, in the first and second methods of correcting the lunar distance, when the moon's apparent altitude is 40° 40', its horizontal parallax 58', and the sun's apparent altitude 70°. Solution. The values of m might be computed directly from (545), but it is more convenient to obtain it by some process of approximation. For this purpose let m =- 60° fc + 8 m, and we have 2 cos. (60° + » .) = «»(» + ' »)co..(«- J a) v ' ' cos. o cos. a =*s 2 cos. 60° cos. 9 m _ 2 sin. 60° sin. § m (604) = (cos. 8b — tang, b sin. 8 b) (cos. ^a-j-tang. a sin. 8a), in which we may put 2 cos. 60° = 1, cos. & = 53'2i" a a = • 51" £'=54' 3"+45°34'10"+31". -18"-1 26'58"=45 1'28" V. J (a + 6) = 32° 15' 30" tang. 9.80014 J(a_&) — 9 56 30 cotan. 0.75626 IB = 22 47 5 tang. 9.62330 ^1 =123 28 15 tang. 0.17970 lstang. == 100° 51' 10" tang. 0.7175 2d ang. = 146° 15' 20" tang. 9.8250 a — 42°12 / cotan. 0.0425 6z=22° 19 ; cotan. 0.3867 d a = 51" P. L. 2.3259 db=. 5321" P.L. 0.5281 1st cor. = — 7" P.L. 3.1859 2d cor. = — 32' 46" P. L. 0.7398 E' ss 45° 34' 10" — 7" — 32' 46" + 31 " — 18" = 45° V 20". 336 SPHERICAL ASTRONOMY". [CH. VIII. Clearing lunar distances. VI. 60' 19" P. L. 0.4748 0.4748 a == 42° 12' cosec. 0.1728 b ~ 22° 19' cosec. 0.4205 E == 45° 34' 10" sin. 9.8538 tang. 0.0086 1st cor. — 4° 316" 0.5014 2d cor. zz:5° 2228" 0.9039 Cor. Table XLVIII = 1' 33" JE =45° 34' 10"4-4° 3' 16"+5° 22' 28"+!' 33"— 10°— 45° V 27". VII. a— 42°12 / N.sin. 0.67172 b + E = 67° 53' 1 0", AN. sin.— 0.46322 60' 19" P. L. 0.4748 b — E ——23° 15' 10", £N. sin. 0.19739 0.40589 ar. co. 0.3916 £ = 45° 34 10" sin. 9.8538 Cor. Table XLVIII = V 33" cor.zz: — 34' 17" 0.7302 E = 45° 34' 10" + T 33" — 34' 17" = 45° 1' 26". 18. The apparent distance of the sun and moon is 95° 50' 33" ; the moon's apparent altitude is 35° 45' 4", its horizontal parallax is 54 / 24 // ; the sun's apparent altitude is 70° 48' 1" ; what is the true distance ? Ans. 95° 44' 29". 19. The apparent distance of a star from the moon is 31° 13' 26" ; the moon's apparent altitude is 8° 26' 13", its horizontal parallax is 60', the star's apparent altitude is 35° 40' ; what is the true distance ? Ans. 30° 23' ^Q". § 104.] LONGITUDE. 337 Lunar distances. 20. Find the Greenwich time, Oct. 3, 1839, when the moon's distance from the sun was 38° 12' 9". Solution. Distance 1839, Oct. 3, 15* 38° 59' 2 1" P. L. 0.3180 38 12 9 18* P. L. 3189 47' 12" P. L. 0.5813 3180 T= 1*38™ 10* P.L. 2633 9 cor.T.zn —2* Greenwich time = 16* 38™ 8*. 21. Find the Greenwich time, Jan. 2, 1839, when the moon's distance from Aldebaran was 70° 45' 13". 1839. Jan. 2, 9* Greenw. Time, Dist. — 69° 26' 29" P. L. = 0.2852 12* P. L. — 0.2863 Ans. 12* 31™ 47* 22. The correct distance of the moon from /*Corvi, 1839, April 3d, 11* 20™, in longitude 70° W by account, was 54° 8' 15" ; what was the longitude ? 29 338 SPHERICAL ASTRONOMY. [CH. VIII. Lunar distances. Solution. 54° 8' 15" Gr.T.= 11* 20™ + 4* 40™ =: 16* 3>'s Dec. s± 26 48 52 by N. A. sec. 0.04941 *'s Dec. = 22 30 11 sec. 0.03439 \ sum = 51° 27' 18" cos. 9.79198 Dist.— J sum = 2 24 36 cos. 9.99962 2)19.87540 3* 59 m 42* cos. 9.93770 ♦ 's Dec. = 12 25 56 3>'s Dec. = 16* 25 m 38* r= Greenw. Time == 16* Long. = 16*— 14* 20™ = 4*40 m = 70°, as supposed. 23. The correct distance of the moon from Castor, 1839, Nov. 29 d 19*, in longitude 45° W. by account, was 78° 3'; what was the longitude 1 Greenwich, 1839, Nov. 29* 21*, D 's R. A. = 12* 15" 16*.5, Dec. = 3° 48' 31" S. 22*, 3)'s R. A. = 12 17 2 .9, Dec. — 4, 2 39 S. Castor's R.A. = 7 24 24 .4, Dec. =32 14 2N. Ans. 44° 18' W. 24. Find the distance of the moon from the sun, 1839, August 12 d , Greenwich time at mean noon. ©'s R. A. = 9*25™51*.72, Dec. = 15° 7'51".5 N. D »s R. A. = 11 42 23 .48, Dec. = 57 27 .9 N. Ans. 36° 33' 14". § 104.] ' LONGITUDE. 339 Lunar distances. 25. Find the distance of the moon from the sun, 1839, August 14 d , Greenwich time at mean noon. ©'s R. A. = 9 h 33™ 24*.57, Dec. = 14° 31' 28".2 N. 3>'s R. A. = 13 8 27 .62, Dec. = 10 25 54 .5 S. Arts. 58° 50' 38" 340 SPHERICAL ASTRONOMY. [CH. IX. Annual and diurnal aberration. CHAPTER IX. ABERRATION. 105. The apparent position of the stars is affected by two sources of optical deception, so that they are not in the direction in which they appear to be. The first of these sources is the motion of the earth, and the corresponding correction is called aberration. Aberration, like the earth's motion, is either annual or diurnal. 106. Problem. To find the aberration of a star. Solution. The apparent direction of a star is obviously that of the telescope, through which the star is seen. Let S (fig. 47.) be the star, and O the place of the observer at the instant of observation : SO is the true direction of the star, or the path of the particle of light which proceeded from the star to the observer, and it would be the direction of the tele- scope if he were stationary. But if he is moving in the direction OP, the direction of the telescope OT must be such, that the end T was at the point R, in the line OS, at the same instant in which the particle of light was at this point. The length RT is, therefore, the distance gone by the observer while the light is describing the line OR. If, then, we put <§> 107.] ABERRATION. 341 Aberration in latitude and longitude. V=z the velocity of light, v = the earth's velocity, /= TOP = RT0 9 3I=z — ROT=z the aberration from the true place, m = apghf (607) we have, V:v=z OR: TR = sin. I : — dlsin. 1" dl=z — 7w sin. J. (608) 107. Problem. To find the annual aberration in latitude and longitude. Solution. The earth is moving in the plane of the ecliptic at nearly right angles to the direction of the sun. Hence if TP (fig. 48.) is the ecliptic, T the point towards which the earth is moving, S the true star, S 1 the apparent star, © ss= the sun's longitude, A — the star's longitude, 8 A — the aberration in long. L ss the star's latitude, d L = the aberration in lat. we have ST =1, SP = L, long, of T = © — 90°, PT = © — 90° — A = j $ PP> = d A= TP— TP', * L = SP' — SP cos. !T= cotan. /tang. ^ = cotan. ( J+ * -O tan S* ( A i — * -^)j whence tan -«-^> i ta "g-( J +/ J ) > (609) tang. J t tang, i % ' 29* 342 SPHERICAL ASTRONOMY. [CH. IX» Aberration in latitude and longitude. and, by (287 and 288), sin. S A sin. 1 1 nn.(2j 1 - w #utf) sin.(2/+ / in the denominators, and reducing by means of (608), sm. 2 / sin. I cos. I sin. j , cos. ^j (611) COS. J But cos /= cos. ^j cos. Z, (^12) whence i\ SHI. T == -. =r = . , r =r# ( 614 ) bin./ sin (/-)-<$/) v ' whence sin. Z sin. (/+ <5 /) tea sin. I sin. (Z + a Z), (615) sin. Z cos. I »nd ^Iz: i — : -- * I cos. x. sin. jc = — m tang, Z cos. I z± — jw cos. -^ t sin. Z = — m s i n . (o — j) sin. Z. (616) 108. ProTdcm. To find the annual aberration in dis- tance and direction from the vernal equinox. <§> 109.] ABERRATION. 343 Aberration from vernal equinox. Solution. Let A (fig. 48.) be the vernal equinox, and let M = SA, $Mz= aberration of M, N=z SAT, 3N=z aberration of N. Now we have * M = $ I cos. AST z=z —. — --=—. — (627) a'=i — (sin.iVcos.iY^ — cos. 2 if sin. iV^ cos.iV)sec.Dsec. w , (628) we have cos. M == cos. D cos. R (629) cotan. iV\ = sin. R cotan. Z> (630) sin. D cos.iV, . ^ sin. M cos. iV\ be ■ - *- c= sin. JD cotan. iV\ (631 ) sin. iVj i v / z=z sin. D sn. R cotan. Z> z= sin. R cos. D b' = sin. Z> cos. JR cos. X> sec. D = sin. D cos. i2 (632) a'=z— [sin.(iV— iVJ-J-sin. 2 if sin. N x cos. N] sec.D sec. « = [sin. w — sin. 2 if sin. 2 N x sin. w] sec. D sec. w — sin. 2 M sin. iV^ cos. N t cos. w sec JP sec. w <§> 109.] ABERRATION. 345 Aberration in right ascension and declination. = (1 — sin. 2 D) sin. w sec. D sec. w — sin. Ms'm.D cos. N t sec.Z) =i cos. Z> tan. w — sin. JR sin. D (633) <>Z> z=^La' + Bb'. (634) Again, we have cos. M z= cos. JR cos. D (635) cos. (itf + S M) = cos. (R-j-dR) cos. (Z> + <* D) cos. I> sin. R$R — sin. Jf a ilf — cos. Rsm.DSD z= £ (sin. 2 M—b> cos. J? sin. Z>) -f- ^4 (sin. Mcos. Mcos. iVsec. ^ — a' cos. JR sin. D), (636) and if we put a=z(s'm.Mcos. M cos. Nsec. w — a! cos.R sin. D) secD cosec. J? 6= (sin. 2 M — b' cos. R sin. X>) sec. X> cosec. R 9 we have a cos. D sin. 12 — sin. M cos. if/cos. iV 2 + sin* -K cos - -B sin. 2 2> -J- (sin. if cos. ifcT sin. N ± — cos. R sin. D cos. Z>) tan. w zzz sin. jR cos. R (cos. 2 Z> -f- sin. 2 D) -{-(sin.THsin.iV^cos.jR cos.D — cos.jR sin. M sin. N x cos.D)tan. <*> ±= sin. 12 cos. jR a zs cos. J? sec. D •- (637) 6 cos. D sin. JR z= 1 — cos. 2 M — sin. 2 D cos. 2 R = 1 — cos. 2 2> cos. 2 12 — sin. 2 D cos. 2 i2 = 1 — cos. 2 JRizzsin. 2 R b =z sin. R sec. Z> (638) *R=Aa + Bb, (639) and formulas (619, 620, 632, 633, 634, 637, 638, 639) agree 346 SPHERICAL ASTRONOMY. [CH. IX. Aberration in right ascension and declination. with those given in the Nautical Almanac for finding the an- nual aberration. 110. Corollary The value of m, which is used in the Nau- tical Almanac, is m =n 20".3600, which gives m cos. co — 20".3600 cos. 23° 27' 36".98 = 18".6768. 111. Scholium. In the values of the aberration in right ascension and declination, each term consists of two factors, one of which is the same each instant for all the stars, and the other is the same for each star, during several years. i 112. Corollary. If in (634) and (639) we put i =z A tan. « (640) B =z7i cos. H (641) -4= A sin. H; (642) they become <3Z? = I cos. D — 7i sin. Hsin. R sin. D-\-U cos. H cos. R sin. D = i cos. D-\-h cos. (H+ R) sin.Z) (643) tR = h sin. H cos. J? sec. D -\-h cos. //sin. J2 sec. D = A sin. (H+ R) sec. 2>, (644) which agree with the formulas in the Nautical Almanac. 113. We have from (619-639) d 2?=zsec. D [ — m cos. w cos. © cos. R — m sin. Q sin. R] (645) § 113.] ABERRATION. 347 ~~Tabie~XLII. ss sec. D [ — J m ( cos. to + 1 ) ( cos. © cos. R -{- sin. © sin. R ) -f- J m ( I — cos. w) (cos. O cos.R — sin. © sin. R)] :=zsec.Z>[ — m cos. 2 J w cos.(jR— © )-\-m sin. 2 £to cos. (i2-}"0 )] > and if we put Q — R— o, Q' = i2+0 (646) n = — m cos. 2 J w, rc' -= 7ft sin. 2 J a>, (647) (645) beeomes S R — sec. D (n cos. Q + ri cos. Q'), (648) and the values of n cos. Q and ?*' cos. Q' may be put in tables like Parts I and II of Table XLII of the Navigator. Again, we have $ D=z sin. D (m cos. w sin, R cos. © — m cos. R sin. ) — m sin. oi cos. © cos. D z= sin. Z> [J m (cos. w-j- l)sin. Q — \m (1 — cos. w) sin. Q'] — J m sin. (o [cos. (© + D) + (cos. © — !>)] =sin.Z>[-7wcos. 2 ^cos.)Q+90°)+Jmsin. 2 Jcucos.(Q / +90 )] — \m sin. w [cos. (0 + 2?) -f- cos. (© — D)] == sin. 2> [—?i cos. ( Q + 90°) + »' cos. ( Q' — 90°)] -Jm sin. co [cos. (©+!>) + cos. ( — !>)], (649) and the values of — Jmsin. wcos. (© + -#) and — Am sin. to cos. (@ — D) may be put in a table like Part III of Table XLII. The rules for rinding the variations in right ascension and declina- tion are then the same as in the explanation of this table. 348 SPHERICAL ASTRONOMY. [CH. IX. Table XLI.~ 114. In constructing Table XLII, the values of m and w were taken m = 20", co — 23° 27' 28", (650) whence n = — 19".173, n> == 0".827, (651) — J m sin. co — — 3".9814. - (652) 115. By putting © - J = P, (653) we have, by (613 and 615),- SL — — m cos. (P — 90°) sin. L (654) d J =i — m cos. P sec. Z», (655) so that if the values of — m cos. P are inserted in tables like Table XLI of the Navigator, the variations of latitude and longitude are found by the rule given in the explanation of this table. 116. If the star is nearly in the ecliptic, the aberration in latitude may be neglected, and the aberration in longitude will be by (655) d J z=z — m cos. P. (656) 117. Problem. To find the diurnal aberration in right ascension and declination. Solution. Let v' == the velocity of a point of the equator, arising from the earth's rotation, »' = ir^-iir- ( 657 ) V sin. 1" § 119.] ABERRATION. 349 Aberration from the motion of the star. The velocity of the observer is evidently in proportion to the circumference which he describes in a day, that is, to the radius of this circumference, or to the cosine of the latitude. The velocity of the observer z=z v' cos. lat. Now, the diurnal motion is parallel to the equator, whence the formulas (613) and (616) may be referred at once to the present case by putting Z s= the right ascension of the zenith, and changing m into m! cos. lat., © — A into Z — R, and L into D ; whence the diurnal aberrations in right ascension and declination are JR = — m' cos. (Z—R) sec. D cos. lat. (658) ti D — — m! sin. (Z — R) sin. D cos. lat. (659) 118. The value of m 1 is nearly *»>===8°27' sec. 0.0047 sin. 9.1670 a R = 5".83 a S .39 0.7654, — 2 ".72 0.4352, i cos. D =z — 6 '.06 t D = — 8",78 § 120.] ABERRATION. 353 Annual aberration. III. R — © = 255° 40' = 8 s 15° 40' P. I = 4".75 R + © = 76°+ 360 = 2 s 16° + 12* P. II = 0".20 4'.95 0.6946 D sec. 0.0047 $R — 5" = s . 33 0.6993 8* 15° 40'+ 3*z=z 11 s 15° 40' P. I. — 18".57 16°+3 s =z5*16° P. II. — D — 2".85 — 2".66 — 3 .38 0" .80 I* 1 .37 sin. 1.2871,, 9.1670 @ + Z> = 48° = l s 18° © — D=32°=l' 2° 0.4541 n $D == — 8".89 7. Find the aberration in right ascension and declination of Regulusfor May 1, 1839. Ans. By Naut. Aim. *R =z 0*.38 dD=z— 1".87 By the Navigator SR — 0*.38 *!>==— 1".91 8. Find the aberration of Regulus in latitude and longitude for May 1, 1839. Ans. * 4 == 6".5 dL=0".15 30* 354 SPHERICAL ASTRONOMY. [CH. IX. Aberration of the planets. 9. Find the aberration of Venus in longitude, when the dif- ference of longitude of Venus and the sun is 45°. Solution. r 0.0000 0.0000 r t ar. co. 0.1407 J (ar. co.) 0.0703 P = 45° sin. 9.8495 20" log. 1.3010 P t ■= sin. 9.9902 cos. 9.3214 0.6927 — 5" when P x is acute, + &" when P x is obtuse, —14" from Table XLI —14" lAzn — 19" when P x is ac, = — 9" when P x is obtuse. 10. Find the aberration of each of the planets in longitude, when the difference of longitude of the sun and planet is 15°. The value of log. r ± for each of the planets is For Mercury 9.5878 is the mean value, Venus 9.8593 The Earth 0.0000 Mars 0.1829 Jupiter 0.7161 Saturn 0.9795 Uranus 1.2829 Ans. For Mercury — 43" when P x is acute, 4" when P x is obtuse, Venus — 41" when P \ is acute, 3" when P x is obtuse, Mars 35" Jupiter 28" Saturn 26" Uranus 24" § 120.] ABERRATION. 355 Diurnal aberration. 11. Find the diurnal aberration of right ascension and declination of Polaris for Jan. 1, 1839, and latitude 45°, when the hour angle is h 30 w . Solution. 0".31 9.4914 9.4914 45° cos. 9.8495 9.8495 D = 88° 27' sec. 1.5678 sin. 9.9998 0*30"* cos. 9.9963 sin. 9.1157 *' R = — 8".04 = — 0*.53 0.9050 9 D ±2 0".03 8.4564 12. Find the diurnal aberration of ^Ursse Minoris in right ascension and declination for Jan. 1, 1839, and latitude 0°, when the star is upon the meridian. Dec. of 9 Ursse Minoris = 86° 35'. Ans. 3R = — 0\35 3'D=z 0, 356 SPHERICAL ASTRONOMY. [CH. X« Refraction of a star. CHAPTER X. REFRACTION. 121. Light proceeds in exactly straight lines ; only in the void spaces of the heavens ; but when it enters the atmosphere of a planet, it is sensibly bent from its original direction'according to known optical laws, and its path becomes curved. This change of direction is called refraction ; and the corresponding change in the position of each star is the refraction of that star* 122. Problem. To find the refraction of a star. Solution. Let O (fig. 50.) be the earth's centre, A the posi- tion of the observer, AEF the section of the surface formed by a vertical plane passing through the star. It is then a law of optics, that Astronomical Refraction takes place in vertical planes, so as to increase the altitude of each star without affect- ing its azimuth. Let, now, ZBH be the section of the upper surface of the upper atmosphere formed by the vertical plane, SB the direc- tion of the ray of light which comes to the eye of the observer. This ray begins to be bent at B, and describes the curve BA, which is such, that the direction AC'is that at which it enters the eye. Let, now, $ 122.] REFRACTION. 357 Ratio of sines in the law of refraction. tp — ZAC z= the ^fc's apparent zenith distance, r =i the refraction, == the difF. of directions of AC and BS, = SBL — S'CL n = COZ, and we have LCS' = (p — u i SBL = (p — u -f- r. Again, it is a law of optics that the ratio of the sines of the two angles LBS and ZAS' is constant for all heights, and dependent upon the refractive power of the air at the observer* Denote this ratio by w, and we have sin. ((p — u -f- r) sin. (p and if = n, (664) U and R =z the values of u and r at the horizon, we have sinJ^-u + rl = = cQs> sin. y \ / \ / whence sin.g) — sin. (9 — u-\-r) 1 — cos. (U — R) sin.y -f-sin. ( — u) 1 — cos. U sin. (p -{- sin. (y — u) 1 -\- cos. U (671) tan. Jw = tang. 2 £17 tan. ( y — J w). (672) and since u is small, £ u = tang. 2 \ U tang, (

and if we put 2iV (676) "~ N' — l p = J(^'-l), (677) $ 124.J REFRACTION. 359 Table XII. we have, by (668), J (u — r) = p r (678) r =. m tan. (9 — p r), (679) and the values of m and p must be determined by observation ; their mean values, as found by Bradley, and adopted in the Navigator, are m — 57".035, p = 3, (680) by which Table XII is calculated. 123. The variation in the values of m and p for different altitudes of the star can only be determined from a knowledge of the curve which the ray of light describes. But this curve depends upon the law of the refractive power of the air at different heights ; and this law is not known, so that the variations of m and p must be determined by observation. At altitudes greater than 12 degrees, the mean values of m and p are found to be nearly constant, and observations at lower altitudes are rarely to be used. 124. The mean values of m and p, which are given in (680), correspond to the height of the barometer nd 29.6 inches, (681) the thermometer z=z 50° Farenheit. (682) Now the refraction is proportional to the density of the air ; but, at the same temperature, the density of the air is propor- tional to its elastic power, that is, to the height of the barom- emeter. If then 360 SPHERICAL ASTRONOMY. [CH. X. Table XXXVI. h = the height of the barometer in inches, r — the refraction of Table XLI, $ r = the correction for the barometer ; we have r : r + dr = 29.6 : h (683) 29.6 9 r = (h — 29.6) r (684) (A— 29.6) .- * r = 29.6 r> (685) whence the corresponding correction of Table XXXVI is calculated, Again, the density of the air, for the same elastic force, in- creases by one four hundredth part for every depression of 1° of Fahrenheit; hence the refraction increases at the same rate, so that if <5 ; r = the correction for the thermometer, f -s the temperature in degrees of Fahrenheit, we have whence the corresponding correction of Table XXXVI is cal- culated. 125. Examples. 1. Find the refraction, when the altitude of the star is 14°, and the corrections for this altitude, when the barometer is 31.32 inches, and the thermometer 72° Fahrenheit. § 126.] REFRACTION. 361 Corrections for barometer and thermometer. Solution. 15 '.035 log. 1 75614 76° tan. 0.60323 1 st app. r =z 228".7 =2 3' 48".7 2.35937 15".035 1.75614 76°— 3 r = 75° 48' 34 " tan. 0.5971 1 2d app. r — 226" == 3' 46" 2.35325 2.353 31.32 — 29.6 *= 1.72 0.235 50— 72=— 22 1.342„ 29.6 ar.co. 8.529 400 ar. co. 7.398 6r zzzlS" 1.117 a'r= — 12" 1.093* 2. Find the refraction, when the altitude of the star is 50°, and the corrections for this altitude, when the barometer is 31.66 inches, and the thermometer 36°. Ans. The refraction =z 48" Correction for barometer r= 3" Correction for thermom. =* 2" 3. Find the refraction, when the altitude of the star is 10°, and the corrections for this altitude, when the barometer is 27.80 inches, and the thermometer 32°. Ans. The refraction — 5' 15" Correction for barometer = — 19" Correction for thermom. = 15" 126. Problem. To find the radius of curvature of the path of the ray of light in the earth's atmosphere. 31 362 SPHERICAL ASTRONOMY. [CH. X. Path of the ray of light. Solution. By the radius of curvature is meant the radius of the circular arc, which most nearly coincides with the curve. Now this radius may be found with sufficient accuracy, by re- garding the whole curve AB as the arc of a circle ; and if we put r t = the radius of curvature, R x z= OA = the earth's radius, we have AC : R x = sin. u : sin. (^ + u), (687) or, nearly, AB : R 1 z= u sin. 1" : sin. y AB= R * uaia - 1 " . (688) sin. y Again, the radii of the arc AB, which are drawn to the points A and B, are perpendicular to the tangents AS' and BS, so that the angle which they make with each other is S'AS = r 5 that is, r is the angle at the centre, which is measured by the arc AB, consequently AB = r x sin. r z= 1\ r sin. l /y , (689) whence (690) (691) (692) (693) r r sin. q> But, by (678), u = 7r, whence r i _ 7JR, sin. (p so that at the horizon as in (225, 226). r i = 7 R x $ 127.] REFRACTION. 363 Dip of the horizon. 127. Problem. To find the dip of the horizon. Solution. The dip of the horizon is the error of supposing the apparent horizon to be only 90° from the zenith, whereas it is more than 90°. If O (fig. 51.) is the centre of the earth, B the position of the observer at the height AB above the surface, O' the centre of curvature of the visual ray BT, which just touches the earth's surface at T, BH 1 perpendic- ular to O'Bj is the direction of the apparent horizon and a H — HBH 1 = OBO> — the dip. The triangle BOO' gives BO 1 : OO 1 = sin. BOO' : sin. § H=z sin. BOH' : sin * H, or, siRce BO' — 1 BO nearly, and OO' =z6BO, and d H and BOH' are small, 7:6= BOH f dH AH' 3H=% BOH' = f - , , (694) 7 7 AO sin. I" v ' But, by (227), we have, if we put R = AO, h=z AB f AH' = f- V( J R h ) = 2 V(f R h (695) whence and log. a // = log. 2 — log. (a/%R) — log. sin, 1" + £ log. A = 1.77128 + J log. 7i, (697) which is the same with the formula, given in the preface to the Navigator, for calculating Table XIII. 364 SPHERICAL ASTRONOMY. [CH. X. Dip of the sea. 128. Problem. To find the dip of the sea at different distances from the observer. Solution. Let O (fig. 52.) be the centre of the earth, B the observer at the height h — AB (in feet) above the sea, and A 1 the point of the sea which is observed at the distance d — A A 1 (in sea miles) = AOA 1 from B ; and let M == the length of a sea mile in feet. If the radius OA is produced to B' 9 so that A B 1 == AB, the point B 1 will be elevated by refraction nearly as much as the point A 1 . But the visual ray BB' will, from the equal heights of B and!?', be perpendicular to the radius OC, which is half way between B and B', so that the dip of B' is, by (694), 6 B == | BOG =2-AOA'z=fd. (698) The dip of the point A! will be greater than B' by the angle i — B'BA, which it subtends at B, and which is found with sufficient ac- curacy by the formula A I Til 7» = TO =■<"*■•*' (699) AB — Md M sin. T d But, by (228), (700) ^=w <™> § 131.] REFRACTION. 365 Twilight. 1 = J0800_ M sin. T n R sin. 1' ? v 1 so that the dip of A 1 is 9 A — f d + 0.56514 £ (703) - which is the same with the formula, given in the preface to the Navigator, for calculating Table XVI. 129. Refraction, by elevating the stars in the horizon, will affect the times of their rising and setting ; and the star will not set until its zenith distance is 90° + horizontal refraction, and the corresponding hour angle is easily found by solving the triangle PZB (fig. 35.) 130. Another astronomical phenomenon, connected with the atmosphere, and dependent upon the combi- nation of reflection and refraction is the twilight^ or the light, before and after sunset, which arises from the illuminated atmosphere in the horizon. This light be- gins and ends when the sun is about 18° below the horizon ; so that the time of its beginning or ending is easily calculated, from the triangle PZB (fig. 35.) 131. Examples. 1. Find the dip of the horizon, when the height of the eye is 20 feet. Ans. 264" = 4' 24". 31* 366 SPHERICAL ASTRONOMY. [CH. X. Dip. 2. Find the dip of the sea at the distance of 3 miles, when the height of the eye is 30 feet. Solution. fK8 = f=s 1 7 .3 0.56514 X \° = 5'.6 dip == 7'. 3. Find the dip of the sea at the distance of 2J miles, when the height of the eye is 40 feet. Ans. 10'. 4. Find the dip of the sea at the distance of J of a mile, when the height of the eye is 30 feet. Ans. 68'. § 133.] PARALLAX. 367 Parallax in altitude. CHAPTER XL PARALLAX. 132. The fixed stars are at such immense distances from the earth, that their apparent positions are the same for all observers. But this is not the case with the sun, moon, and planets ; so that, in order to compare together observations taken in different places, they must be reduced to some one point of observation. The point of observation which has been adopted for this purpose is the earth's centre ; and the difference between the apparent positions of a heavenly body, as seen from the surface or the centre of the earth, is called its parallax. 133. Problem. To find the parallax of a star. Solution. Let O (fig. 53.) be the earth's centre, A the observer, £the star, and OSA, being the difference of direc- tions of the visual rays drawn to the observer and the earth's centre, is the parallax. Now since SAZ is the apparent ze- nith distance of the star, and SOZ is its distance from the same zenith to an observer at O, the parallax OSA=p is the excess of the apparent zenith distance above the true zenith distance. If, then, 368 SPHERICAL ASTRONOMY. [CH. XI. Parallax in altitude. % =z SAZ, R = OA z=z the earth's radius, r = OS r= the distance of the star from the earth's centre, we have P : R =n sin. Z : sin. p, R sin. z or sin. p =. , r (704) or R sin. z p zz: . r sin. 1" (705) 134. Corollary, If P is the horizontal parallax, we have sin. P — — , r (706) or p- R - r sin. \" ; (707) whence sin. p = sin. P. sin. z 9 (708) or p = P. sin. z. (709) which agrees with (564) and Tables X. A., XIV, and XXIX, are computed by this formula, combined, in the last table, with the refraction of Table XII. 135. Corollary. In common cases, the value of the horizontal parallax can be taken from the Nautical Almanac ; but, in eclipses and occultations, regard must be had to the length of the earth's radius, which is different for different places. The curvature of the earth is such that the radius diminishes as we recede from the equator proportionally to the square of the sine of the latitude ; the whole diminution at the pole being about ^tj sin <> L = R sin. 9 L (1 — m sin. 2 L) (716) , r .^ BR 2ms'm. L cos. L ._„. tMng.»L = mn. t L=- [5 = l _ m , m , r , <™> $ 139.] PARALLAX. 371 Reduction of the latitude. whence sin. a X — m sin. a X sin. 2 X = 2 m sin. X cos. X (718) sin. a X z= 2 m sin. X (cos. X -f- i sm » * -£' sin. X) = 2 m sin. X (cos. X -f- sin. J a X sin. X) = 2 m sin. X cos. (X, — J a X) (719) ii a X = ■*: — -- sin. X cos. (X — J a X) sm.l" v 2 ' == „ 2 t , sin. X cos. (X — J a X), (720) 5 sin. I 7 ' from which and ZPM' give _ » sin. M p sin. I? sin. (D 4-dD) d D z= £ — = x - -p- 2 £ '- cos. Z, cos. A cos. JLi — P sin. B sec. £ sin. (D + 9 D) (731) Again, the triangles ZPM and ZPM' give __ sin. Zi — cos. JBsin. (;!+») sin.Z/ — cos.Bsm.A /m , ne% . cos. Zzz — : — — v - — L£i = : — - (732) sin. B cos. (A -f-p) sin. B cos. -A whence sin.L cos. A — sin.Z/cos. (A-\-p) zzz cos. JB sin. (A-\-p) cos. A — cos. B sin. A cos.(A-{-p) = cos. B sin. p. (733) But sin.Z* =:sin.(Z/ — s L)zzz\s'm.L'cosJL — cos.L'sm.dL izzsin.Z' — cos.Z/sin. <* L — 2 sin.Z/sin. 2 J +dI>) zrcos.JBcos. 2 !,'— sin.Bsin.Z / cos.Z / cos.(I>+^Z>),(738) and also cos. B — sin.,4 sin.X/rzcos.iW 'cos.Z'cos.^l, (739) io that, for a first approximation, 3L = — cos. M'p (740) ^ L sin. 5 L =z cos. 2 M' p sin. p (741 ) psin.pnK5.Lsin. $ L=z sin. 2 M ' p sin. p — sin. M 1 cos. U sin. (742) cos. A and (736) becomes *L=z — cos. J Bcos.Z / .P-(-sin. J Bsin. J L / .P[cos.(D + aI>) + Jsin. !><>Z)] =z — cos. B cos.Z/ .P+sin.Psin.Z/cos.(Z>-f J d ^J^, (743) and formulas (731) and (743) agree with the rule in the Navi- gator [B. p. 404]. 141. Problem. To find the parallax in right ascen- sion and declination. § 142.] PARALLAX. 375 Parallax in right ascension and declination. Solution. Formulas (731) and (743) may be applied imme- diately to this case, by putting B = the altitude of the equator = the co-latitude, L == the true declination, U = the apparent declination, D = the right ascension of the body diminished by that of the zenith = the hour angle of the body. &L = the parallax in declination, $D == the parallax in right ascension. 142. Corollary. The value of 3 L may, in this case, be found by a somewhat different process, which is quite con- venient when the altitude of the body is required to be calcu- lated. Draw PN to bisect the angle MPM', and draw MH and M 1 H 1 perpendicular to PN, and we have nearly d L == HIP = HN+H'N' = MN cos. N+M'N cos. N = {MN + M'N) cos. N = MM' cos. N = P cos. A cos. N. (744) Now, in the triangle PZN, we have PZ = co-lat. ZPN —D + i5D t and may take PN = 90° — L, ZN = 90° — A, so that PZ, ZPN, and PN are given, to find ZN and N. This method of calculating the parallax in right ascension and declination is precisely that used in [B. p. 443] for calculating from the relative parallax the corrections for right ascension and declination. 376 SPHERICAL ASTRONOMY. [CH. XI. Apparent diameter. 143. The apparent diameter of a heavenly body is the angle which its disc subtends. 144. Problem. To find the apparent semidiameter of a heavenly body. Solution. Let O 1 (fig. 56.) be the centre of the heavenly body, A the observer, and A T the tangent to the disc of the body. The angle TAO 1 is the apparent semidiameter. Let R x == OT ° = OAT r =z AO', we have sin. a OT ' AO' r (745) Hence, body, by (fig. 53.), if A is the apparent altitude of the > sin. a R, sin ■P ~'P) (746) R cos .(A a R 1 sec, .(A- -P)- (747) 145. Corollary. If 2 is the horizontal semidiameter, we have which is also the semidiameter, as seen from the earth's cen- tre. Now R x — 0.2725 R (749) R = 3.67/2, (750) whence log. -^ = 9.43537, (ar. co.) = 0.5646, (751) § 147.] PARALLAX. 377 Augmentation of semidiameter. so that formula (748) agrees with [B. p. 443. No. 10 of the Rule]. 146. Corollary. If + cos. L _ cos. jPco s. (L— 3L)SD cos.(L—3L)__ cos. Z, . sin. D cos. i _ P.cos.ff.sin .ffsin.(Z> + ) cos.(Z- = H+H.P.cos.Dsm.B + H' = H+~+H'. (763) Now we have by (761) and (762) H=zi2.p. [sin. (B + D) + sin. ( B — D)] (764) H'= -2- (tang, i .^ i + cos. d L — 1), (765) and formulas (763 to 765) agree with the method of calculat- ing the augmentation of the semidiameter given in Table <§> 148.] PARALLAX. 379 Augmentation of semidiameter. XLIV of the Navigator. The three first parts of this table are calculated for the value of 2, 2 = 16' = 960", whence £s.P = 8".18. The fourth part of the table is the correction which arises from the difference between the actual value of 2 and that as- sumed in the three former parts. If we put 9' a — the value of 9 a for 2 = 16', we have, by (755) and (748), 9a :?'•<, ='**''. (16 7 ) 2 (766) 22 d — & a 256 ^+(£- 1 )" 2 2 _ 256 . d' a - " ' 256 = ,. + £±!«>£=>S».. < 767) as in the explanation of this table. 148. Examples. 1. Find a planet's parallax in altitude, when its horizontal parallax is 25", and its altitude 30°. Am. 22". 2. Find the moon's parallax in latitude and longitude, when her horizontal parallax is 59' 10".3 ; her latitude 3° 7' 19" S., 380 SPHERICAL ASTRONOMY. [CH. XI. Parallax in latitude and longitude. her longitude 44° 36' 16"; the altitude of the nonagesimal 37° 56' 14", its longitude 25° 27' 16", the latitude of the place 43°17 / 18 // N. Solution. Reduced parallax = 59' 10".3 — 5".3 = 59' 5 // =3545" Reduced latitude = 43° 17' 18"— IT 27":= 43° 5' 51" D = 44° 36' 16" — 25° 27' 16" = 19° 9' 3545 3.54962 3.54962 3.550 37° 56' 14" sin. 9.78873 cos. 9.89691 sin. 9.789 3° 7' 19" sec. 0.00064 3° 7' 19" cos. 9.99936 * 3.33899 46' 32" 3.44589 9.99899 19° 9' sin. 9.51593 3° 53' 51" 12' 2.85492 4630" 3.44552 19° 15' 19° 21' 3" sin. 9.52027 3° 53 41" sin. 8.831 3D =12' 2.85926 — 2' 20" log. 9.975 19° 21' 3" 2.145 * Z, = 44' 10" 3. Find the moon's parallax in latitude and longitude, when her horizontal parallax is 60' £".9; her latitude 1° 30' 12" N., her longitude 130° 17', the altitude of the nonagesimal 85° 14', its longitude 125° 17', the latitude of the place 46° IF 28".4 N. Ans. Parallax in longitude == 5' 18" Parallax in latitude =s 3' 30",5. § 148.] PARALLAX. 381 Augmentation of semidiameter. 4. Calculate the parts of Table XLIV, when the argument of the first part is 3 s 19° = 109° ; that of the second 12".4, the moon's true latitude 1° 20' N., the moon's parallax in lati- tude 50', the sum of the three first parts 13", and the moon's horizontal semidiameter 14' 50". Solution. 8". 1845 sin. 109° — 7".74 = Part I. (12" 4>* Part III = 960" [sin. 50' tang. 1° 20' — 1 + cos. 50'] = 960" [sin. 50 / tang. 1° 20' — 2 sin. 2 25'] = 960" [0.00023] en 0".22. , 30 50" X 1' 10" 13"X 30.83X1.17 Part IV =— 13" X 256' ~~ 256 = — 1".83. 5. Calculate the parts of Table XLIV, when the argument of the first part is 2* 16°, that of the second 15."5, the moon's true latitude 3° S., the moon's parallax in latitude 30 7 , the sum of the three first parts 11", and the moon's horizontal semi- diameter 15' 20". A?is. Part I = 7".94 Part II = .25 Part III=:—0.48 Part IV :=— .90 6. Calculate the number of Table XV, when the altitude is 45°. Ans. 11". 382 SPHERICAL ASTRONOMY. [cH. XI. Augmentation of semidiameter. 7. Calculate the augmentation of the moon's semidiameter in Example 2; when the horizontal semidiameter is 16' 50". Solution. Part I = 6".87 + 2".58 z= 9".45 Part II = .09 Part III := —1 .02 sum =? 8".52 Part IV z= .91 augmentation = 9".43 8. Calculate the augmentation of the moon's semidiameter in Example 3, when the horizontal semidiameter is 15' 30". Ans. M?',83. <§> 151.] eclipses. 383 Solar eclipse* CHAPTER XII. ECLIPSES. 149. A solar eclipse is an obscuration of the sun, arising from the moon's coming between the sun and the earth ; and occurs therefore at the time of new moon. It is central to an observer, when the centre of the moon passes over the sun's centre. It is total, when the moon's apparent disc is larger than the sun's, and totally hides the sun. It is annular, when the moon's apparent disc is smaller than the sun's, but is wholly projected upon the sun's disc. The phase of an eclipse is its state as to magnitude. 150. An occultation of a, star or planet is an eclipse of this star or planet by the moon. A transit of Venus or Mercury is an eclipse of the sun by one of these planets. 151. Problem. To find when a solar eclipse will take place. Solution. Let O (fig. 57.) be the sun's centre, and O l the moon's centre at the time of new moon, and let I? z= the latitude of the moon at new moon == OO^ 384 SPHERICAL ASTRONOMY. [CH. XII. When a solar eclipse will happen. Let ON be the ecliptic, and N the moon's node, so that N0 1 is the moon's path. Let N = the inclination of the moon's orbit to the ecliptic ; Draw OP perpendicular to the moon's orbit, and if, when the moon arrives at P, the sun arrives at O, the least distance of the centres of sun and moon is nearly equal to O'P. Now the triangle OPO 1 gives OP — p cos. N=p — P(l — cos. N) = f* — 2 p sin. 2 I N = /» — i p sin. 2 N n = ratio of the sun's mean motion divided by the moon's = T V nearly, (768) we have OO' = n X O x P = nfi sin. N. Draw O'R perpendicular to OP, and we have nearly OR= OP — OP =i OO' sin. N = n p sin. 2 N. Hence OP — p _ ( J + n ) /» sin. 2 iV:= P-±'ftP sin. 2 iV. (769) The apparent distance of the centres of the sun and moon is affected by parallax, and the true distance is diminished as much as possible for that observer, who sees the sun and moon in the horizon, and OP vertical, in which case the diminution is equal to the difference of the horizontal paral- laxes of the sun and moon. Let, then, P z=z the moon's horizontal parallax, w = the sun's horizontal parallax, 4 =. the apparent distance of the centres, we have the least apparent dist. = OP — (P — n) ~p— &p8in.*N— P+n. (770) $ 152.] eclipses. 385 When a solar eclipse will happen. Now, an eclipse will take place, when this least apparent distance of the centres is less than the sum of the semidiame- ters of the sun and moon. Thus, let s =z the moon's semidiameter, a s= the sun's semidiameter. In case of an eclipse, we must have p _ £ /s sin. 2 N — P + 7r ,- i>, —" , v ,%/ ; let » sin. i c= ' X 3600" ('80) £ (in seconds) = c tan. i. (781) Again, let ^' = !£# =z the true dist. of centres of sun and moon, we have J' — j -\- P' (782) cos. k S= — ; J' (783) MP s= p tan. A;, (784) let t — time of describing ilffP, we have * zz= = c tan. A; (785) " a — itftfiV = — i zp * s (786) the positive values of a being reckoned towards the east, so that the upper sign corresponds to the beginning, the lower to the end of the eclipse. § 157.] eclipses. 389 Places where eclipse begins and ends. Finally, if L = the latitude = 90° — ZN h = the hour angle after noon = ZNS, the triangle ZNS gives, by § 39, sin. L = cos. d cos. a (787) tan. a #.«*•». tan. h = : — y . (788) sin. a 156. Corollary. The value of 4 is for the beginning or ending of an eclipse, 4 = s + a ; (789) for the beginning or ending of total darkness in a total eclipse, j = s — 0', (790) for the forming or breaking up of the ring in an annular eclipse, J = a — s ; (791) for the central phase, J = 0. (792) 157. Problem. To find the places for which a given phase of the eclipse is seen at sunrise or sunset. Solution. Let M (fig. 59.) be the centre of the true moon, at any time after the first formation of the phase J and before its end, S that of the sun, m that of the apparent moon af- fected by relative parallax. Since the sun and moon are in the horizon, we have Mm tea P 1 , also m S =. 4. 33* 390 SPHERICAL, ASTRONOMY. [CH. XII. Places where eclipse begins at sunrise or sunset. The zenith Z is in the line Mm, at the distance ZS — 90° from S, let N be the north pole. Join MN, and draw Ng perpendicular to NS, and the right triangle NMG gives, by putting D z= the declination of g cos R = tang. Ng cotan. MN *r tang. D cotan. 2> = ^^, (793) whence, by (287) and (52), sin.(Z> — D) 1 — cos. 22 ^ , ^ . , ° n ( 3= T -j ^ xac tang. 2 J R, (794) sm.(Z?-f Z> ) 1 + cos. 22 6 2 ' v ' or, since D — D and R are small, D — D =i±;R2 sin. 1" sin. 2 D (795) D = D + | jR2 sin. 1" sin. 2 Z>. (796) Let now, z zr g-tf, y = i^^, S=MSg, and we have z z= D — d (797) y Q — R.cos. D (798) tan. S = ^ (799) 4Mb a: sec. A (800) Now since Zilf and ZS are nearly quadrants, they are nearly parallel at their extremities If and S, so that if b z=z MSZ = m MS, (801) P'-l-j P' — J and q = _^ ql _ __^ 9 (g02) <§> 161.] ECLIPSES. 391 Places where a phase is seen at sunrise or sunset. we have sin. J b = V (g ~^2/^'~ g,) , (803) whence the triangle NZS gives ZNS —S^m (804) sin. JLr=cos. (S^fm) cos. d, tan. hz= — tan.(&=pm) cosec d. (805) 158. Corollary. Since M m may be taken on either side of MS, there are two places for each place M y except when J = J -f P' t (806) which corresponds to the beginning or to the ending of the phase, or when J cos. L sec. D sin. (h — R) (812) x=zrnhz=z D — D'— relative paral. in dec. — P' sin. z cos. M=. c± P 1 [sin. L cos. D — cos. £ sin. Z>' cos. (h— R )] (813) y z=h M= (R — R) cos. D = P' cos. L sin. (h — R') — P> sin. | sin. M (814) Now, by the diurnal motion, the angle h will increase for the instant H cos. L cos. (h — R) (816) — 15° P' d t (cos. z cos. D 1 — sin. % sin. D 1 cos. M ) m 15° P d t cos. s cos. D— 15° 2 : Jt sin. Z>'. (817) Again, if u=Sr 9 v — mr, i / =rSM } (818) we have u = ^ cos. i' = x — z , v — J sin. # = y — y, (819) and if w m' is the apparent relative orbit of the moon, it must be perpendicular to Sm, because m is the point of nearest approach. Hence if m! is the place of the moon at the end of the instant S t y we have $u z=l Sx — dx =z — mm' sin. i' = 15°dtsin.D' (y — js'in.i') — 9x (820) 394 SPHERICAL ASTRONOMY. [CH. XII. Limits of a phase upon the earth. d v 3= <$y o — dyzzzmm 1 cos. i 1 as — l5°P'd 1 cos. z cos. D 1 +15° (x + D cos. i>)*t sin. D' + ty , (821) whence m m' sin. %' — — d u cos. i 1 = dv sin. z v z= — 15° dt sin. D' (y cos. fl — ^sin. i' cos. i') +<5a: cos. i' (822) = 15°a^sin.Z> / (a: sin. I'-J-^sin.i' cos. 2 V ) + dy sin. i 1 — 15° P' $ t sin. i' cos. z cos. D 1 . Now Z)' differs so little from d, that d may be substituted for it in this equation, and we have also — 9. — the hourly motion in relative declination = D 1 d * ♦ y — - sec. D = the hourly motion in relative dec. = R lt d t and if we put __ R.cos.D D, A TW^V' B -T^^T' (823) (822) becomes, by dividing by 15° $t sin. 1", P' sin. It cos. z cos. cZ = (A -f- z sin. d) sin. t' _ ( b — # sin. d) cos. t'. (824) Let now ;. and v be so taken, that A -\- x sin. cf = i P 1 cos. z cos. d cos. * P — y Q sin. d =.* P 1 cos. z cos. e? sin. *, (826) and (824) becomes cos. z sin. i' — 2 sin. {%' — *) (827) Asin.(e v — *) . 4 .. /000 \ cos. z == r-^- ; - = a cos. r — 1 sin. y cot. 1'. (S28) sin. ^ / To find t', its value may be, at first, assumed as equal to i, as § 162.] eclipses. 395 Limits of a phase upon the earth. it is nearly, because the true relative orbit PM is nearly parallel to the apparent relative orbit m m'. Hence u and v are found by (819), and thence D 1 = d qp u (829) R 1 = ± v sec. D 1 (830) i> — D + i(R — JR) 2 sin.2D (831) y — (R — R) cos. D (832) x =D -D> (833) V tan. M— - X X (834) /QO£r\ sin. z — _ _ P' cos. M [poo J and from this value of z, i' may be found by means of (828), which gives cos Z cot. i — cot.*— — r — , (836) x sin. * x ' or if 9 is taken, so that . cos. z ;_ sin. ^ sa Vol ( 837 2 * cos. v v ' 2 x cos. r sin. 2

the two sides ZS, NS, and the included angle >S' are given to find NZ and ZNS, 167. Corollary. For one place the eclipse will be central at noon, and for this place we have, obviously, J z=z diff. dec. sm.Z=± (848) L = d + Z (849) west long, of place = app. Greenw. time of cent, eclipse. 168. Problem. To calculate the time of the beginning or ending of a given phase of a solar eclipse for a given place. Solution. Find for a supposed time near the required time, such as the time of new moon, the relative parallaxes in right ascension and declination of the sun and moon, and their rela- tive right ascension and declination. Hence their apparent relative right ascension and declination is found by simple addition or subtraction. 34 . 398 SPHERICAL ASTRONOMY. [CH. XII. Time of a solar eclipse for a given place. Let D z=l their apparent relative declination, R zzzz their apparent relative right ascension, d r= the sun's declination, W — the distance apart of their apparent centres, j = the phase, A zzzz the angle which W makes with the parallel of so that If the supposed time is that of the beginning or ending ©f the phase. But if W differ from J, find another apparent distance W' of the centres for a time a little after the former one. Then we have W — W : W — A z= diff. of supposed times : the correction which is to be added to the first sup- posed time to obtain the required time. If this correction is large, a new computation must be made, using the time just obtained as a new supposed time. 169. Corollary i The time of the phase of an eclipse or occultation might also be calculated by the following process. Let jR L zzz the relative hourly motion in right ascension, D x zzzz that in declination ; then let S (fig. 62.) be the centre of the sufij and M that of the moon at the supposed time, CS the hour circle, A the declination, and we have D =z W sin. A, R cos. d — W cos. A (850) R cos. d Jr ^„ v tan, A = — — — (851) W =± D cosec. AzzziR cos. d sec. A. (852) W z= J, § 170.] eclipses. 399 Time of a solar eclipse for a given place. moon's centre at the beginning of the phase, B at the end ; we have, then, ^o,™*- ™ CM R cos. d ,0,*™ tan. CSM = tan. S= -—= - — (853) Co U CT D tan. i — tan. CMI — tan. FSI— ~ = -f> ~3 ( 854 ) CM R 1 co$.d K ' S31 = W—y cosec. S=x sec. # (855) SP = p = Wcos. PSM=z Wcos. (S+ i) (856) cos. k\— cos. P,SU = cos. PSB z=z £ (857) a =z ASM— S + i + k (858) b =2 BSM =k — (S+ i). (859) Then let t x =1 the interval of moon's passing from A to M, t 2 = the time from M to B t and we have 4 If cos. i PPsin. ^o^itf cos, t W sin. a cos. i y x cos. A; W sin. b cos. £ 2/ x cos. /c (860) (861) 170. Corollary. This method maybe used by substituting latitude and longitude for declination and right ascension, and in this case the sun's latitude is zero, so that the formulas agree with the rule in the Navigator [B. p. 425]. 400 SPHERICAL ASTRONOMY. [CH. XII. Magnitude of an eclipse. 171. Corollary, For the beginning or end of the eclipse the phase is J — the sum of the horizontal semidiameters of the sun and moon increased by the augmentation of the moon's semidiameter. (862) For the beginning or end of total darkness in a total eclipse, J = difF. of semidiam. -f- aug. of J) 's semidiam. (863) For the formation or breaking up of the ring in an annular eclipse, J =z difF. of semidiam. — aug. of D's semidiam. (864) 172. Problem. To find the greatest magnitude of the eclipse at any place. Solution Let D sz the relative apparent declination at the beginning of the phase J f A — the angle which the line joining the centre of the apparent sun and moon makes with the circle of declination at this time, D' and A' == the values of D and A at the end of the phase J, j z=z the nearest approach of the centres, we have, by (fig. 61.), in which MNM 1 is the moon's apparent relative orbit, S the sun, SD the circle of declinations, SM z=z SM> zzz J y SN = j , D D' sin. A — — , sin. A' = — , (865) J = J cos. % (A' — A). (866) $ 174.] ECLIPSES. 401 Lunar eclipse. 173. The calculation of occupations is the same as that of solar eclipses, except that the star has no parallax, and its disc is insensible. The calculation of transits of planets over the disc of the sun is the same as that of a solar eclipse, except that the planet is to be substituted for the moon. 174. Problem. To find when a lunar eclipse will happen. Solution. The solution is the same as in § 168, except that the semidiameter of the earth's shadow at the distance of the moon is to be substituted for that of the sun ; and the change in the position and apparent magnitude of the moon from parallax may be neglected, because when the earth's shadow falls upon the moon, the moon is eclipsed to all who can see it. Now if S (fig. 61.) is the sun, E the earth, GF the semidiameter of the sun's shadow at the moon, we have the app. semi. == FEG == EFL — EIF ~ P -— EIF = P — (KES — EKI) = P — G + *j or rather, this would be the apparent semidiameter, if it were not for the earth's atmosphere, which increases the breadth of the shadow about g^th part ; so that the app. semidiam. = f^- (P — a -f- tf), and therefore, in order that an eclipse must happen, we must have, by (762), P = the latitude at the time of full moon, P nearly that of the sun's corrected for relative parallax, find the values of R s= the relative right ascension -f- parallax in R. A. D z=z the relative declination -f- correction for declin. d — the sun's declination corrected for relative paraL R x = the hourly variation of R, D ± = the hourly variation of 2>, D Q — the diff. between the values of D at the time t y and of the time of app. conjunction in right ascension, t z± the time of apparent conjunction. Then we have, by (fig. 62.), if £ is the sun's place corrected for relative parallax, and BMA the moon's true orbit. * = t + -J-, D = D + -|- D, = SI, (872) tan. CMI= tang. * = jr^j (873) <§> 184.] eclipses. 405 Longitude from solar eclipse. p = PS = D Q cos. i (874) ^~ m V D n cos, i #«**■% cos. FSA — cos. a — -^ z= — -5 (875) ^/ sin. ^L#2 ^ sin. (a -f* i) /q-7«\ ^.i = — : =z ; , l c ' D ; sin. ^17>S cos. i interval of moon's passing from A to I AI cos. i ^4 sin. (a -f- *) ~~ R % cos. d ~~ R x cos. rf (877) time of obs. phase = * + S1 "' ^ + *\ (878) J.\/ « COS. CL which agrees with [B. p. 463, from No. 12 to the end of the rule]. 183. Corollary. The diminution of the semidiameter re- quired by § 181 is, by (753), da = o.P. sin. l"zin*JL s (879) or if a and P are expressed in minutes, * 's true declination — D = 2° 43' 52".3 O's true declination = d — 1° 49' 15".5 D — d = 54'36".8z= 3276".8 3>'s hourly motion in dec. z= — 14' 10" .5 O's hourly motion in dec. — — 0' 58". 3 relative motion in dec = D x ~ — 13' 12".2 == — 792".2 J) 's motion in right ascension ±z 26' 0".5 O's motion in right ascension z= 2' 14". 7 rel. motion in right asc. = R x — 23' 45".8 =e 1425".8 408 ♦ SPHERICAL ASTRONOMY. [cH. XII. Solar eclipse of Sept. 18, 1838. D x 2.S98S3„ ar.co 7.101 17 n R 1 ar. co. 6.84594 D sec. 0.00049 D—d 3.51545 3.55630 -29° 5' 4" tan. 9.74526 n cos. 9.94147 sin. 9.68673„ p == 2863 // .7 3.45692 3.45692 c 3.80112 3.80112 * — — 3518* 3.54638 n _ __ 58^ 38 s . time of middle = 7 h 56 m 38* -— t =z 8 h 55 ?n 1 6*. Now for the phases, we have J) 's equatorial horizon, paral. =z Q's equatorial horizon, paral. Relative parallax for equator = Reduction for lat. 45° Relat. par. for lat. 45° J)'s true semidiameter Q's true semidiameter For first contact J' = P> + 5 + o = 84' 18".2=5058".2 p. 3.45692 * ar.co. 6.29600 c 3.80112 = 53' 53" .7 = 8" .5 =2 53' 45" .2 == P' 5" .3 : 3219" ZZZ = 53' 39" .9z= .9 z=z s = 14' 41/ '.2 == a z=15 7 57* .1 k = 55° 31' cos. 9.75292 tan. 0.16314 — i — 29° 5' t — 9210* z= 2 A 33™ 20* 3.96426 « — — 26° 26', m. t. of begin. = time of mid. — t — 6*21*56' app. t. = 6 h 21 w 56 s + 5 m 55* -= 6 h 27 m 51* = 96° 58' b = 84° 36', time of end = time of mid. -f t = 1 P 2S CT 36' app. t. = 11* 28" 36* + 6 W — 11* 34* 36* = 173° 39' $ 188.] eclipses. 409 Solar eclipse of Sept. 18, 1838. a cos. 9.95204 tan. 9.69647 n d cos. 9.99978 cosec. 1.49793 lat. = 63°3FN. sin. 9.95182 266° 20' tan. 1.19440 at beginning long. 335 266° 20'— 96° 58'= 169° 22 ; E. lat. =a 63° 31' + 9' = 63° 40' N. b cos. 8.97363 tan. 1.02444 d cos. 9.99978 cosec. 1.49793 lat — 5° 24 N. sin. 8.97341 90° IF tan. 2.52237, at end long. = 173° 39'— 90° 1 F=83° 28' W. lat. = 5° 24 / + 2' = 5° 26' N. For central eclipse, P 3.45692 P' ar.co. 6.49215 c 3,80112 A: — 27° 13' cos. 9.94907 tan. 9.71107 — i = 29° 5' T = 3252* = 54™ 12 s 3.51219 a == 1° 52' time of begin. == t. of mid. -r-8 1 l m 4* b = 56° 18' app. time = 8* l m 4 s -J- 5 m 57* = S h 7 m I s = 121° 45' time of end = time of middle — T z= 9* 49 771 28* app. time = 9"49 w 28 8 +5 w 59'zz:9*55 m 27*:= 148° 52' a cos. 9.99977 tan. 8.51310 d cos. 9.99978 cosec. 1.49793 lat. = 87°24 / N. sin. 9.99955 134° 17' tan. 0.01103 at beginning long. = 134° 17'— 121° 45'= 12° 32' E. 35 410 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of Sept. 18, 1838. b. cos. 9.74417 tan. 0.17593 d. cos. 9.99978 cosec. 1.49793 lat. = 33° 41' N. 9.74395 91° 13' tan. 1.67386 at end long. = 148° 52' — 91° 13' — 57°39 / W. lat. = 33° 41' + 10' = 33° 51' N. 2. Find the places where the eclipse of Sept. 18, 1838, be- gins or ends in the horizon af 8* Greenwich mean time. Solution. g = I (P 1 + J) = 4139" q> = £(p/_^) =919".l time from middle eclipse = 55 m 16* = 3316' = t t 3.52061 c ar.co. 6.19888 p. 3.45692 k = 27° 40' tan. 9.71949 sec. 05273 — t = 29° 5' J 1 = 3233" 3.50965 ar. co. 6.49035 S= 1°15' i J ' — 9'==697".4 2.84348 q — J = 2522".5 3.40183 P> ar. co. 6.49215 2)19.22781 ■== 48° 32' — — 24° 16' sin. 9.61390 S-m=— 47°17 / cos. 9.83147 tan. 0.03465„ d cos. 9.99978 cosec. 1.49793 lat. — 42°41 / N. sin. 9.83125 268° 19' tan. 1.53158 $ 188.] ECLIPSES. 411 Solar eclipse of Sept. 18, 1838. Lat. = 42° 41' + 11 ' = 42° 52' N. Greenwich app. time =: 8 h 5 m 57* =: 121° 29' long. = 268° 19' — 121° 29' = 146° 50' E. and at this place the eclipse is rising. S+ m = 49° 47' cos. 9.81002 tan. 0.07286 d cos. 9.99978 cosec. 1.49793 Lat. == 40° 12' N. 9.80980 91° 32' 1.57078„ long. = 121° 29' — 91° 32' = 29° 57' W. lat. z=z 40° 12' + 11' = 40° 23' N. and at this place the eclipse is setting. 3. Find the place on the southern limit of the eclipse of Sept. 18, 1831, which corresponds to the Greenwich mean time of 8*. Solution. Since the altitude is not known, the increase of the moon's semidiameter is not known, but it may be supposed at first to be 6 7/ , which is about its mean value. Hence j=o + s + 6" =. 30' 43 '.3 = 1844 // .3 p — ^ = 1019 "A j 3.26583 i cos. 994147 sin. sec. 3.26583 9.68673' u =z 26' 52" 3.20730 d+u = 2° 16' 7" = D' 0.00034 R' = 897".2 = 14' 57" 2.95290, 412 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of Sept. 18, 1838. P c p — » ar. co. ar. co. 49' 11" 3.45692 6.19888 P'. ar.co. 6.99166 6.49215 3.00834 E t 6.64746 W cos. 3.52061 9.50049 k' = 55° 0.16807 sec. 0.25042 — i = 29° 5' 4" % =? 34° 18' sin. 9.75091 M =— 26° 44' 7" cos. 9.95089 tan. 9.70219 n Z= tan. 9.83388 6 = 31° 21' 2" tan. 9.78477 sin. 9.71623 a + D — 33° 37' 9" tan. 9.82274 sec. 0.07947 h—R= cos. 9.97950 tan. 9.49789 n L =z 32° 23' tan. 9.80224 For a more accurate determination. P 1 =z 53' 45".2 — 3".2 = 53' 42" = 3222" s = 14' 41".2 + 11".6 = 14' 52".8 j = s + a = 30' 49".9 = 1849'.9 15° sin. 1" ar. co. 0.58204 0.58204 D x 2.89883 n R x 3.15406 B = — 3026 3.48087 n Z> cos. 9.99951 4 = 5440.1 3.73561 For 8* Gr. time we have D =z J> 's dec. = 2° 43' 5" i R . 48" 2 log. 3.362 cZ =z Q's dec. = 1° 49' 12" 1" sin. 4.686 R = rel. R. A. = 50".l 2Z> = 5°26' sin. 8.976 D = D 2> — 2> = 6.924 a; = 2> — rf == 53' 83"=3233' § 188.] ECLIPSES. 413 Solar eclipse of Sept. 18, 1838. R. 1.69984 D COS. 9.99951 x o Vo = 60" 1.69935 3.50961 d sin. 8.50187 102".7 8.50187 2" 0.20122 2.01148 B— 3026" A = 5446". 1 0.00022 ■3028" 3.481 16„ 5542".8ar. co. 6.25628 tan. 9.73744„ 2 v = 58° 40' sec. 0.28396 5542".8 3.74372 P' ar. co. 6.49187 sec. 0.05671 •'==— 46°25 / tan. 0.02137„ 2 (p = sec. < p = 29° 20' %' cos. 9.83848 J 3.26926 z 2 x 0.29252 cos. 9.91703 ar. co. 9.69897 tt=21'21" 3.10774 rf + M = 2°10 , 33" 2)19.38019 sin. 9.69004 sin. 9.85996„ 3.26926 sec. 0.00032 D j± 2°43 / 5" — R '=— 1347".5 3.12954 x = 32' 32" = 1952", R—R=z 1297" '.4 3.11307 n 1> cos. 9.99951 3.11258 ?i 35* 414 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of Sept. 18, 1838. y P> ar. 3.11258, co. 6.49187 x cosec. 0.25566 3.11258, ar. co. 6.71175 COS. tan. tan. tan. cos. tan. M tan. 9.82433 sin. 9.81834 sec. 0.13824 9.92002 z sin. 9.86011 0.02160 s+D'=z 41° 16' 24" 43° 20' — 30° 8' — 31° 30' 8" L. 122° long, : 9.94162 9.97472 k—R' — tan. 9.78091 zz 38°56'N. = 153° 20' W. 9.93250 H — 9.90722 lat. == 38° 56' + 1 1' = 39° 7' N. 4. Find where the solar eclipse of Sept. 18, 1838, is central for the Greenwich mean time of 9\ t = 4 W 44 s = P 284 s 3.45692 sec. 0.00044 c ar. co. tan. tan. cos. tan. tan. cos. tan. 2.45332 6.19888 Jc=z2° 34' 8.65220 J' P ar 3.45736 . co. 6.49215 Z s z= 31° 39' sin. 9.94951 tan. 9.78987 0.29106 9.93007 4:= 59° d + d — 60° 49' sin. 9.93304 sec. 0.31193 0.22113 0.25298 h =z 47° 18' tan. 0.03484 L =z 50° 32' 9.83140 0.08438 § 188.] eclipses. 415 Solar eclipse of Sept. 18, 1838. For a more accurate determination. P> — 53' 45".2 — 6".2 = 53' 39 " == 3219 7 D = 2° 28' 54", df = 1° 48' 13", D cos. 9.99959 R — 25' 5".4 = 1505 ".4 3.17765 x = 2> — d=40'41" — 2441" 3.38757 6.61243 £ cos. 9.93013 sec. 0.06987 tan. 9.78967 j' 3.45744 P'. ar. co. 6.49228 tan. 0.29208 sin. 9.94972 q =z59° 3' 24" tan. 0.22221 sin. 9.93332 4 -f . = 3220".6 3.50794 L. cos. 9.88090 9.88090 9.88090 15 ar. co. 8.82391 2 h = S h 19"* 45*.8 sin. 9.94782 S' 2.16057 D sec. 0.00041 2" 24'. 9 2.16098 2 h'=8 h 17 m 21% A=4 A 8 wl 40*.5, log.Ris. 4.72684 sin. 9.94661 £ — d=38°43'7"N. cos. 78022 d cos. 9.99979 40507 4.60753 22° 2' N. sin. 37515 cos. 9.96706 sec. 0.03294 M cos. 9.83794 sin. 9.86045 P'. 3.50794 dd = 34' 15".6 N. 3.31294 £' 2.16057 9 = 2° 22' 29" cos. 9.99963 sec. 0.00037 sR=z2 m 24*.9 = 36' 13" = 21 73" 2.16094 JR— *JK ;= — 668" 2.82478„ D — d< = & 26" = 385" ar. co. 7.41454 2.58546 8 = — 60° 1' tan. 0.23895„ sec. 0.30125 W 2.88671 ■§> 188.] ECLIPSES. 417 Solar eclipse of Sept. 18, 1838. D x = 792".7 2.8991 l n R L z=z 1425" ar. co. 6.84619 d sec. 0.00037 i = — 29° 6' tan. 9.74567 n W. 2.88671 iST-f- f = —89° 7' cos. 8.18798 JP- 1.07469 For beginning of eclipse. ^ = 30' 38".3 : = 1838".3 ar. co. 6.73565 P sec. 2.18966 1.07469 k = 89° 38' cos. 7.81034 W. 2.88691 3600' 3.55630 l ZZl cos. 9.94137 ^ ar. co. 6.84619 tf sec. 0.00037 a= 31' sin. 7.95508 t, — — 40 OT 3.37568 mean time at N. Y. 3 ft 24 771 , For a more accurate calculation. Gr. mean time — 8 h 20 7 * 15 P' cos. L 2.21275 2h= 6" 59 m 44*.6 sin. 9.89928 Gr. m. t. s= 8* 20 w . h = 3* 29™ 52'.3 d — 1° 48' 6SKUI R = 555".0 £' 2.11203 2>— 2°38'21".l sec. 0.00046 2 m 9 8 .6 2.11249 418 SPHERICAL ASTRONOMY. [cH. XII. Solar eclipse of Sept. 18, 1838. L cos. 9 88090 9.88090 2A' = 6*57 OT 35*/ i ' = 3 /i 28 m 47*.5 log. Ris. 4.58778 sin. 9.89770 £.— d=38 42'27" N. cos. .78033 d. cos. 9.99978 .29407 4.46846 29° 5' 42" N. sin. .48626 cos. 9.94142 sec. 0.05858 M cos. 9.86114 sin. 9.83718 P>. 3.50794 &d=. 34' 4".l N. 3.31050 S' 2.11203 d' = 2° 22' 57" cos. 9.99962 sec. 0.00038 d R = 2 m 9 s 55 = 32' 23".2 = 1943".2 2. 1 1243 R — 6 R = — 1388".2 3.14245„ D—d = 15' 24".l=924'.l ar. co. 7.03428 2.96572 S = —56° 19' 32'' tan. 0.17635„ sec. 0.25612 D x = — 792.5 2.89900„ W. 3.22184 R x = 1425.3 ar. co. 6.84610 d'. sec. 0.00038 i = — 29° 5' 48" tan. 9.74548 n 8 -f i = — 85° 25' 20" cos. 8.90207 2.12391 4 = 30' 30".8 = 1830".8 ar. co. 6.73736 & = 85° 50' cos. 8.86127 § 188.] eclipses. 419 Solar eclipse of Sept. 18, 1838. k sec. 1.13873 w 3.22184 3600" 3.55630 i cos. 9.94141 R x cos. d ar. co. 6.84648 a= 24' 40" sin. 7.85583 1 1 = — 12* 28* 2.56059 mean time at N. Y. 3* 17 m 20% Gr. time = 8 h 13 w 24*. For a still more accurate calculation. Gr. mean time S h 13 m 30* h = 3* 23 m 22*.3 15 P< cos. L 2.21279 d = 1° 48' 59".3 2 ft=6M6 m 44*.6 sin. 9.88954 R s= 401.1 8' 2.10233 D = 2° 39' 53".2 sec. 0.00046 2 W 6*.7 2.10279 L. cos. 9.88090 9.88090 2h' = &M m 37\9 h'=3 h 22 m 18> log. Ris. 4.56222 sin. 9.88790 £,— d=38° 42' 21" N. cos. 78035 d cos. 9.99978 27727 4.44290 30° 12' 7" N. sin. 50308 cos. 9.93664 sec. 0.06336 M. cos. 9.86553 sin. 9.83216 P' 3.50798 dD=: 34'2".4N. 3.31015 420 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of Sept. 18, 1838. S f 2.10233 d' ss 2° 23' 1".7 cos. 9.99962 sec. 0.00038 S R = 2* 6 ff .68 = 31' 40".2 = 1900 '.2 2.10271 R — 9 R = — 1499".l 3.17593 n D — tf ' = 16' 51 ".5 = 1011 ".5 ar. co. 6.99503 3.00497 S=z — 55° 58' 2" tan. 0.17048 sec. 0.25207 W. 3.25704 3.25704 jSf+t = — 85°3'50" cos. 8.93472 (3600 "cos.*) -± R x cos. d = 0.34419^ ar.co. 6.73753 * = 85°7'32" sec. 1.07071 cos. 8.92929 a= 3 42" sin. 7.03193 t^— 50 s . 6 1.70387 Mean Gr. time == S h 12 m 39*4., N. Y. m. t. = 3 h 16 m 34*.9. For beginning of annular phase. J = V 15'.9 =: 75".9 ar. co. 8.11976 W. 2.88671 p 1.07469 k = Sr sec. 0.80555 cos. 9.19445 W 3600" cos. z 3.49767 JR 1 cos. d ar.co. 6.84656 a= — 8° 7' sin. 9.14980 n f t = 25™ 36 s 3.18629 n Gr. mean time ac 9* 25 m 36*. § 188.] eclipses. . 421 Solar eclipse of Sept. 18, 1838. For a more accurate calculation. Gr. mean time — 9^ 30 OT , h = 4 h 39™ 53 s .4 15 P cos. L 2.21266 d = 1° 47' 44".7 2A = 9 ?l 19™46 s .8 sin. 9.97291 jR = 2217".9 £' 2.18557 2> = 2 21'48".2 sec. 0.00037 2™33 s .4 2.18594 L. cos. 9.88090 9.88090 2A / z=9 A 17 ?ra 13 8 .4 A / =4*38 w 36*.71og.Ris. 4.81444 sin. 9.97202 L— d=38° 43' 35" N.cos. 78015 d cos. 9.99979 49560 cos. COS. pi 4.69513 16° 31' 56" N. sin. 28455 9.98166 sec. 0.01834 M 9.82528 sin. 9.8712C 3.50785 3d = 34' 24 "A 3.31479 S'. 2.18557 61 = 2° 22' 19".l cos. 9.99963 sec. 0.00037 > JR =5 2301" == 13 s .4 a # = 2.18594 R — *R=z— 83". 1 1.91960 n Z> — d'zn— 30".9 ar.co. 8.45967 n l.48996 n -ST — — - 112° 40' 54" tan. 0.37890 sec. 0.41 385„ D , = — 793" 2.89927 n W 1.90381 R x = 1423".2 ar. co. 6.84674 d' sec. 0.00037 i ±2 — 29° 8' 50" tan. 9.74638 n 36 422 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of Sept. 18, 1838. W 1.90381 S+i = — 141° 49' 44" cos. 9.89551, W 1.90381 1.79932 n 3600" -f- R x cos. d' 0.40341 ^ ar.co. 8.14997 k ss 152° 51' 9" sec. 0.05071 cos. 9.94929 n a ss 17° 1' 25" sin. 9.28151 • cos. 9.94120 *! ss 38M 1.58064 Gr. m. time = 9* 30 m 38M. N. Y. m. t. = 4* 34 w 33».6. In the same way we should find for the end of the annular phase. N. Y. mean time ss 4* 38 m 12% and for the end of the eclipse, N. Y. mean time ss 5* 47 m 54% 5. If the beginning of the solar eclipse of Sept. 18, 1838, had been observed at New Orleans, in 1 at. 29° 57' 45" N., at 2 h 19 OT P.5 mean time, what would be the longitude of New Or- leans ? Solution. Let the supposed long, zs 6* Greenwich mean time =. S h 19 w P.5, h ss 2* 24 OT 58' A L ss reduced lat. = 29° 57' 45" — 9' 55" ss 29° 47' 50" P'= 53' 45".3 — 2".8 ss 53' 42 7 .5 = 3222".5 d ss 1° 48' 53".3 R=z 530".l $ 188.] eclipses. 423 Solar eclipse of Sept. 18, 1838. P' 3.50820 15. ar. co. 8.82391 L. cos, 9.93841 cos. 9.93841 cos. 9.93841 2h=z4/ l 49 m 56 s .8 sin. 9.77174 8' 2.04226 D = 2° 38' 34".8 sec. 0.00046 23h'—l m 50\3 2.04272 2h'=z4: h 48 m 6 s .5,7i< — 2 h 24 m 3 s .2l R. 4.28132 sin. 9.76936 L—d=27° 58' 57" N.cos. 89115 d. cos. 9.99978 16577 4.21951 cos. 9.84045 cos. 9.83005 P'. 3.50820 sec. sin. sec. sec, W cos. 46° 10' 3" N. sin. 72, 538 ar '. CO. sec. tan. 0.15955 M 9.86732 sd=z 25' 9" d'—2° U'2'3 3.17870 cos. 9.99967 3.05082 n . co. 6.83194 2.04226 0.00033 iR = 1654 ".5 == 1110 s .3 Rz= sR — — 1124," A D — d' = 1472 ".5 2.04259 3.16806 S = — 37° 20' 16" tan. 9.88243,, 2.89894 6.84607 0.00033 0.09960 D x =• — 792<».4 22, = 1425".4 ai d' 3.26766 i =3 — 29° 5' 20" £ -f i = — 66° 25' 36" 9.74534 960198 2.86964 424 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of Sept. 18, 1838. i cos. 9.94145 p. 2.86964 W. 3.26766 J. ar. co. 6.73824 k — 66° tit 2" sec. 0.39212 cos. 9.60788 a = —20' 34" sin. 7.77689 R x cos. d' ar. co. 6.84640 3.55630 t = 60-.37 = l m 0*.37 1.78082 long. = 6 h l m 0*.37 5= 90° 15' #'. For a more accurate calculation. Gr. mean time = 8 h 20 w df = 1° 48' 52".8, D = 2° 38' 21".l JR = 555' .4, tf ' = 2° 14' 1".8 cos. 9.99967 R — dR=z — 1099 ".1 3.04 104 n D — d= 1459".3 ar. co. 6.835S5 3.16415 £' z= — 36° 57' 52" tan. 9.87656 n sec. 0.09745 3.26160 3.26160 S + i = — 66° 3' 12" cos. 9.60840 3600 cos. i ~ J^ cos. eZ' 0.34415 2.87000 a =z 42" sin. 6.30882 ^. ar. co. 6.73821 66° 3' 54" sec. 0.39179 cos. 9.60821 * x = — 2*.02 0.30636 *ong.= 90° 14' 7" W. § 188.] eclipses. 425 Solar eclipse of Sept. 18, 1838. 6. To find the times of the beginning and end of the annu- lar eclipse of Sept. 18, 1838, at Washington, D. C, and the times of the formation and rupture of the ring. Extracts from the Nautical Almanac. Greenwich mean time, of Sept. 1838. mO* ©'s R. A. = 11" 38™ 25*.08, Dec. ■=. 2° 20' 14'.3 N. 18 .0 ©'s R. A. = 11 42 .61, Dec. = 1 56 58 .2 N. 19 .0 ©'s R. A. = 11 45 36 .16, Dec. = 1 33 39 .5 N. 20 .0 ©'s R. A. = ,11 49 11 .75, Dec. = 1 10 18 .6 N. 17 .0 equation of time sfs 5 m 28 ff .55 18 .0 equation of time = 5 49 .57 19 .0 equation of time = 6 18 .57 20 .0 equation of time = 6 31 .53 18* 6 h 3) 'i R. A. = III 39 w 49 t .64, Dec. = 3° 11' 24", 6 N. 18 7 ])'sR.A.zzll 41 33 .74, Dec. = 2 57 14 .9 N. 18 8 3) 'sR. A. = 11 43 17 .79, Dec. = 2 43 4 .6 N. 18 9 2) 's R. A. = 11 45 1 .79, Dec. = 2 28 53 .9 N. 18 10 D's R. A. =: 11 46 45 .75, Dec. = 2 14 42 .6N. 18 11 D 's R. A. = 11 48 29 .67, Dec. — 2 31 .0 N. 18 D 's semid. = 14' 4 1 ".5 Hor. Par. = 53' 54".8 18 12 3) 's semid. = 14' 41 ".1 Hor. Par. = 53' 53 .3 ©'s semid. — 15' 57". 1 Hor. Par. — 8".5 Ans. Beginning of eclipse at 3* 5™15\6 W. mean time, of ring at 4 23 46 .1 end of ring at 4 29 42 .3 of eclipse at 5 39 30 3 426 SPHERICAL ASTRONOMY. [CH. XII. Solar eclipse of May 15, 1836. 7. Calculate the time of the beginning and end upon the earth of the solar eclipse of May 15, 1836, and the places where it is first and last seen. Extracts from the Nautical Almanac. Greenwich mean time of May, 1836. U d h ©'s R. A.= 3 /l 25- 5U3 Dec. = 18° 42' 21".0N. 15 3 29 1 .93 18 56 35 .9 N. 16 3 32 59 .30 19 10 31 .6N. 17 3 36 57 .25 19 24 7 .8 N. 14 equation of time z= 3 m 56 s .30 15 equation of time = 3 56 .05 16 equation of time zz 3 55 .24 17 equation of time = 3 53 .86 14 22 3)'sR.A,z=3 20 41 .89 Dec. s= 18 40 52 .9 N. 3 22 41 .69 18 51 11 .4 N. 3 24 41 .69 19 1 24 .7 N. 3 26 41 .88 19 11 33 .0 N. 3 28 42 .26 19 21 36 .1 N. 3 30 42.84 19 31 34 .ON. 3 32 43 .62 19 41 26 .6 N. 3 34 44 .60 19 51 13 .9 N. 15 6 3 36 45 .77 20 55 .9 N. 14* 12* J) 's semid. = 14' 52' .3 Hor. Par. = 54' 34' .5 15 J) 's semid. = 14 49.9 Hor. Par. = 54 25 .6 15 12 3>'s semid. = 14 47 .7 Hor. Par. = 54 17 .7