Multiplying by cos h cos h', and reducing, we have sin2 3d + cos2 ( + h ́) .......... (B) This, if s be put (D+H+H') becomes, = cos2 + (h+ h) (! Put sin2 F= sin2d+cos2 (h + h′). cos s cos (s1) cos h cos h' COS H COS H cos scos (s-D) cos h cos h' COS A COS H' COS2 § (h + h'). cos 8 cos (SD) cos h cos h' seс н sес н' cos2 (h+h') .... (c) Then the preceding equation becomes sin2 d = cos(h+h') (1-sin2 F) = cos(h+h') cos2 F, whence sin d = cos (h+h') cos F (D) This is the formula of Borda: it has the advantages of requiring only the usual tables, and of being free from ambiguity. 28. The logarithmic formulæ deduced from equations c and D, are 2 log sin F = = log cos s + log cos (SD) + log cos h +log cos h' + log sec H + log sec H + 2 log sec (h + h') 60.... (E) log sin d= log cos (h + h) + log cos F- 10.. (F) The secants being introduced into the numerator for the cosines in the denominator, render the logarithmic computation entirely additive. Example. The apparent distance of the moon's centre from the star Regulus being 63° 35′ 13′′, when the apparent altitude of the moon's centre was 28° 29' 44", the apparent altitude of the star 45° 9' 12;" the moon's correction, or the difference between the refraction and the parallax in altitude 48′ 1′′; the star's correction, or the refraction 57". Required the true distance? 1st Method, directly from the triangles zms, ZMS. zm = 90° — 28° 29′ 44′′ = 61° 30′ 16′′, zs = 90° sm = 63° 35′ 13′′: to find the angle z. This may be effected by ch. vi. art. 38, equa. 2. which, suited to the present case, becomes tan z = sin (zm + zs zm) - ms) sin § (ms + zs + zm) and this, when performing the logarithmic computation, may be best accomplished by adding the log cosecants of the terms in the denominator, instead of suhstracting their log sines, and deducting 10 from the index, at last. sz).. 40° 7′20′′.. 9.8091692 zm).. 23° 27′ 53′′.. 9.6000842 ms).. 21° 22′ 38′′.. 10-4382946 zm).. 84° 58′ 8′′..10.0016765 The half sum - 10 is tan z. 40° 3' 7" Consequently z = 80° 6′ 14.′′ 239-84.92245 Then there are given, in the triangle zмs, 9.9246122 =60° 42′ 15′′ Zs = 90° - (45° 9′ 12′′. 57) 44° 51′ 45′′ and the included angle z = 80° 6′ 14′′. The third side may be obtained as in case 2, prob. 2, by finding a subsidiary angle @ such that tan = cos 80° 6′ 14′′ tan 44° 51′ 45′′ cos 80° 6' 14" The sum tan tan 9° 42′ 21′′ 9.2330959 COS MS = 44° 51′ 45′′ cos (60° 42′ 15′′ — 9) sec Q. cos 50° 59′ 54′′ sec 9° 42′ 21′′ ...... The sum cos MS cos 63° 5' 11".... 9.6557573 Thus the true distance between the moon and star is found to be 63° 5' 11". 2dly. By Borda's Theorem. 45° 8′ 15′′ Here we have H 45°9′ 12′′, h H′ = 28° 29′ 44′′, h' = 29° 17′ 45′′, h + h = 74°26′ s = 68° 37′ 41,", SD 5° 1′51′′ The logarithmic computation for F, therefore, is cos s...... 68° 37' cos (SD).. 5° 41". 9.5617995 9.9983236: .... 9.8773416 The sum 602 sin F... Its half is sin F.... 48° 56′ 1′′ Then to find d, and thence d, we have 9.9011062 31° 32 34".... 9.7186273 Consequently d is 63° 5′ 8′′. Note. Here the difference in the results is only 3 seconds; which is quite as little as can be expected when tables are employed in which the computer has to proportion for the seconds. Had the value of r been taken F 48° 56' instead of 48° 56′ 1′′ (and it is evident from the common tables that it lies between the two), then d would have been found about 63° 5′ 15′′. The mean between these two results is 63° 5′ 11′′. Example II. Given the apparent distance between the centres of the sun and moon 83° 57′ 33′′, the apparent altitude of the sun's centre 48° 27′ 32′′, its true altitude 48° 26′ 49′′, the apparent altitude of the moon's centre 27° 34′ 5′′, its true altitude 28° 20′ 48′′. Required the true distance of the centres of the two luminaries. Ans. 83° 20′ 55′′. Example III. The apparent distance of the moon's centre and a star was 2° 20′, when the apparent altitude of the star's centre was 11° 14', and that of the moon's 9° 39′; the moon's correction was 51′ 30′′, the star's 4' 40". Required the true distance of their centres. Ans. 1°49′. Note 2. In examples of this kind the sum of the apparent zenith distances must always be greater than the apparent distance; for if not they cannot form a spherical triangle. Thus it will be found that example 4, pa. 31, of the Requisite Tables, 2d edition, relates to an impossible case. The third side 103° 29′ 27′′ is greater than 89° 50′ 22′′ the sum of the other two. Note 3. This being an important problem in nautical astronomy, we shall here refer to other works where more compendious rules, founded principally upon subsidiary tables, are given. Such are, the Requisite Tables to be used with the Nautical Almanac; Mackay on the Longitude; Mendoza's Tables for Navigation and Nautical Astronomy; Myers's Translation of Rossel on the Longitude; Andrew's Astronomical and Nautical Tables; Kelly's Spherics. See also, Mr. Sanderson's rules in the Ladies' Diary for 1787, or in Leybourn's collection of the Diaries, vol. iii. Dr. Brinkley's in the Irish Transactions for 1808, and various others in Delambre's Astronomy, vol. iii. chap. 36. PROBLEM VII. 29. Given the longitudes and latitudes of two places upon the earth, regarded as a globe, to find their itinerary distance, that is, the arc of the great circle comprehended between them. Let A and в be the two places on the surface of the terraqueous globe, of which p is one of the poles, and conceive a spherical triangle PAB to be described, such that PA, PB, shall be the respective distances of the two places from the pole P, or their respective co-latitudes, and the angle APB the difference of longitude of those two places. Hence, L and L' being the respective latitudes, and P the difference of longitude, there are given in the triangle PAB, two sides, viz. PA = 90° ‡ 1, PB = 90° L, and P, to find the third side AB. The problem, therefore, belongs to case 2, prob. 2, of oblique spherical triangles, the appropriate formula for which, suited to the present case, become tan cos P tan PB (1.) for the subsidiary angle: .... and cos ABCOS PB sec & cos (PA — 9).......... (2.) Example I. .... Given the latitude of the observatory of Paris, 48° 50′ 14′′ N. that of the observatory of Pekin, 39° 54′ 13′′ N. and their difference of longitude 114° 7′ 30′′: to find their distance. Here since the latitudes are both of the same kind, we have PA90° L= 90° PB 90°. L'= 90° and P= diff, of long. - 48° 50′ 14′′ = 41° 9′ 46′′ |