ss' = 28° 30′, sps′ = 104° 52′ — 78° 24′ = 26° 28′, and PZS 180° 73° 36′ = 106° 24′. Hence the latitude is 32° 23′ 18′′, of the same kind with the declinations of the stars. PROBLEM VI. 24. Given the apparent altitudes of the moon and sun, or of the moon and a fixed star, and their apparent distance to determine their true distance. This is an important problem in navigation, being of essential use in one of the best methods of determining the longitude at sea. The true distance of the moon and the sun, or a star at any given instant, being ascertained, we are enabled by means of the "Nautical Almanac," and the Ephemerides of different countries, to find the time at the first meridian, which corresponds with the moment of observation: the difference between these times, reduced to degrees at the rate of 15 to an hour, shows the longitude required. Besides the error arising from the imperfection of instruments, the observed altitudes of the heavenly bodies are affected by three causes, the depression of the horizon, the refraction of rays of light in passing through the air, and the parallax, or the inclination of two visual rays passing from the celestial body, one to the earth's centre, the other to the point where the observer is placed on its surface. The principal works on nautical astronomy contain tables by which the requisite allowances may be made for these at any place and for any heavenly body: and the purpose of making such allowances is, to reduce. the observed or apparent altitude of any body at or above the surface of the earth, to the real altitude at which it would appear from the earth's centre if light were transmitted in right lines. 25. The apparent altitude of the sun always surpasses the true; for the refraction elevates the sun more than the parallax depresses that luminary. With regard to a fixed star the parallax is evanescent. Hence, the true place of the sun, or of a fixed star, will always be below the apparent place. The moon, on the contrary, is more depressed by parallax than it is elevated by refrac- M tion. Hence the true place of the moon m will be always above its apparent place. H 26. Let, then, s be the apparent place of the sun or star, s its true place, on the same vertical zs, m the apparent place of the moon, м its true place, on the same vertical zm, ms the apparent distance of the two bodies. Then, there are known н= 90 Zs, apparent altitude of the sun or star; H': 90°zm, apparent altitude of the moon; D= ms, their apparent distance: also, h = -r+p= 90° zs, true altitude of the sun or star; and h' = H r' + p′ = 90° — zм, true altitude of the moon; r and p, r' and p', being the refraction and parallax corresponding respectively to the apparent altitude of the bodies. Consequently, in the spherical triangle zms are given all the sides to find the angle z: and then in the triangle zмs are given two sides zм, zs, and their included angle z, to find the third side MS=d the true distance required. The solution of the problem thus conducted is obviously as simple and natural as can well be wished, to men acquainted with theory and accustomed to computation: yet it has been found embarrassing to mariners; on which account most writers on the subject of navigation have investigated other rules, several of which (especially those which depend upon subsidiary tables) are more direct and expeditious in operation than the original rules deduced without modification from the two triangles zms, zms. 27. The second general theorem of spherical trigono, metry, when applied to these two triangles, gives But since cos (H + H') = COS H COS H sin н sin н', we have sin i sin н'= COS H COS H' and, in like manner COS (H + H'), cos (h+h'); COS D + COS (H + H') cos d+cos (h+h') · COS H COS H = cos h cos h' Now it is evident from the formulæ in chap. iv. (A) that cos D + cos (H+H') = 2 cos (H + H+ D) X COS (H+H'— D), cos (h+h') = 2 cos2 (h + h') — 1, cos d 12 sin2 d. These values being respectively substituted for the se veral quantities in equa. A, it will be transformed into 2 cos(H+H' — D) COS (Ħ + H' + D) -- COS H COS H' 2 sin2d + 2 cos2 (h + h') — 1 cos h cos h Multiplying by cos h cos h', and reducing, we have sind+cos2} (h+h') (B) .... This, if s be put = (D + H + H) becomes, = cos2 1 (h+h) (1 Put sin2 F cos s cos (sn) cos h cos h' COS H COS н' cos scos (S-D) cos h cos h' COS H COS H' COS2 † (h + h')) cos S COS (SD) cos h cos h' sec н seс н′ ... (D) .... (c) Then the preceding equation becomes sin2 d =cos2 (h+h') (1-sin2 F) = cos2 (h+h') cos2 F, whence sin dcos (h+h') COS F 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 slog cos (SD) + log cos h + log cos h' + log sec H + log sec н' + 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 68° 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. In the triangle zms (last figure) are given zm 90° 28° 29′ 44′′ = = 61° 30′ 16′′, zs = 90° · 45° 9′12′′ = 44° 50′ 48′′ 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 sin (zm + ms · sz) sin † (ms + Zs zm) sin (zm + zs ms) sin (ms + zs + zm) : tan z = 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. Thus, sin sin cosec (zm + ms ms + zs (zm + zs cosec (ms + zs + sz).. 40° 7′ 20′′.. 9.8091692 zm).. 23° 27′ 53′′.. 9.6000842 ·ms).. 21° 22′ 38′′.. 10:4382946 zm).. 84° 58′ 8′′..10·0016765 2|39-84.92245 The half sum - 10 is tan z.. 40° 3′ 7′′ 9.9246122 Consequently z = 80° 6′ 14.′′ Then there are given, in the triangle zмs, ZM 90°. (28° 29′ 44′′ + 48′ 1′′) : 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".... 9-2351804 tan 44° 51′ 45′′ The sum tan = tan 9° 42′ 21′′ .... 9-9979155 9-2330959 |