4. On the 27th of December 1841 the sun's longitude and right ascension will be 275° 38′ 58′′, and 18h 24m 378.16; required his declination and the obliquity of the ecliptic. Ans. D 23° 20' 24"-8, and O 23° 27' 39"-9. 464. PROBLEM XXIII.-Having given the longitude and latitude of a celestial body, to find its right ascension and declination; and conversely, the obliquity of the ecliptic being supposed known in both cases. Let M be the moon or any celestial body (last fig.), and TMT' an ecliptic meridian; then ALL is its longitude, reckoning from the 7 is its latitude, L ML AR A MR A' is its right ascension, reckoned from the nearest preceding equinox, true right ascension, = angle SARO the obliquity of the ecliptic; and let AMH its distance from preceding equinox, angle MAR = E, and MAL then EOC, according as M is without or within the angle of the equator and ecliptic; and CEO. C; EXAMPLE. If the moon's longitude, on the 2d of August 1841, at noon at Greenwich, be 310° 50' 1", its latitude 0° 10′ 1′′ S., and the obliquity of the ecliptic 23° 27′ 42′′; required its right ascension and declination. Here L' 130° 50′ 1′′, 7 = 0° 10′ 1′′ S., and 0 = 23° 27′ 42′′, and M is between the equator and ecliptic. In triangle MAL, To find H. L. Rad = To find C. L. Tan / = 7.4644506 = 10. L. Sin L' = 9.8154860 9-8788730 = 12-4144224 0° 13′ 15′′. = L. Cos H L. Cot C 130° 50'. Hence also, E = 0 — C = 23° 14′ 27′′. EXERCISES. 1. On the 19th of May 1841, at noon at Greenwich, the moon's longitude was 38° 13′ 48"-4, and its latitude 5° 1' 16"-6 S.; required its right ascension and declination, the obliquity of the ecliptic being 23° 27′ 41′′-96. Ans. A 2h 16m 348·02, D = 19° 0′ 32′′·9 N. 2. On the 17th of September 1841, the registered right ascension of the moon at noon will be 13h 22m 18s, and its declination 14° 3' 11"-2 S.; what will be its longitude and latitude at the same time, the obliquity of the ecliptic being 23° 27′ 41′′-8? Ans. L 204° 15′ 55′′-7, and 7 = 4° 59′ 31′′-1. = 465. PROBLEM XXIV.-The right ascensions and declinations, or the longitudes and latitudes of two stars being given, to find their distance. P Let PNP' be the solstitial colure, A the vernal equinox, MN the equator, P its pole; D, E, two celestial bodies, of which AB, AC, are the right ascensions, and DB, EC, the declinations. Then in triangle DPE are given the codeclinations PD, PE, and angle P = the M difference of their right ascensions; hence, there are known two sides and the contained angle; and therefore the distance DE can be found. A D B 10 с E N When the latitudes and longitudes of two bodies are given, their distance is found exactly in the same way. When one of the bodies, as E', is on the opposite side of MN, then PE' 90+ CE'. Let C, C' the complements of their declinations, Pthe difference of their right ascensions, D= their distance; then (361) and R: cos Ptan C: tan 6, cos : cos (C'-6)= cos C: cos D. Or, D can be found, though not so concisely, by the method in article 355. EXAMPLE. Find the distance between Capella and Procyon on 21st January 1841, their right ascensions being 5h 4m 59s-77 and 7h 31m 08.79, and their declinations 45° 49′ 58"-2 N., and 5° 37′ 37′′-9 N. Here P 2h 26m 1s 36° 36′ 30′′. C = 44° 10′ 1′′-8, To find the arc 0. L. R L. Cos P = L. Tan= To find the distance D. L. Cos = 9.8968608 9.9045699 L. Cos (C'—0) = 9.8383839 9-9873727 L. Cos C = 9.8557070 9.8919426 and C′ = 84° 22′ 22′′. 37° 56′ 40′′. When the difference of the right ascensions exceeds 12 hours, add 24 hours to the less, and from the sum subtract the greater, and the difference will be the included angle at the pole. EXERCISES. 1. When the latitudes of Sirius and Procyon were 39° 34′ S., and 15° 58′ S., and their longitudes 101° 14', and 112° 56'; what was their distance? Ans. 25° 42'. 2. Find the distance between Capella and Procyon, when their longitudes were 78° 58′ and 112° 56′, and their latitudes 22° 52′ N., and 15° 58′ S. Ans. 51° 7'. 466. PROBLEM XXV.-Given the latitude of the place, the declination and altitude of a celestial body, to find its azimuth. Let HZR be the meridian of the place, HR the horizon, and Z the zenith; EQ the equator, and P its pole; S the body; then ZS= Z zenith distance or coal- H titude, PSP body's polar distance or codeclination, Q = = L. Sin S L. Sin (S-P) L. Cosec C D WAS' E PZC colatitude, Ζ and angle PZS=A= supplement of azimuth AZR from south. R Here P, C, Z, are given; and hence A can be found by article 349; thus, if S1 (P+C+Z), then = 2 L cos A, Lsin S+L sin (S-P) + L cosec C+ L cosec Z. If the body's declination is south, as at S', while the given latitude is north, the polar distance PS' 90° + DS'. EXAMPLE. When, in latitude 44° 12′ N., the sun's altitude was 36° 30', its declination being 15° 4′ N., what was its azimuth? Here P = 74° 56' 48 30 2)174 14 2)19.5631718 Hence S = 87 7 L. Cos A = 9-7815859 And SP = 12 11 A = 52° 47′ 16′′-6, and A = 105° 34′ 33′′-2 the azimuth from the north. Hence, 9.9994498 9-3243657 .1445350 ⚫0948213 EXERCISES. 1. When, in latitude 48° 51' N., the sun's declination is 18° 30' N., and its altitude 52° 35', what is its azimuth from the north? Ans. 134° 36'. 2. If, in latitude 51° 32′ N., the altitude of Arcturus was found to be 44° 30′, when its declination was 20° 16′ N., what was its azimuth from the north? Ans. 117° 8'. 3. When, in latitude 51° 32′ N., the sun's altitude was 25°, and its declination 4° 47' S., what was its azimuth from the north? Ans. 137° 16'. METHODS OF DETERMINING TIME. 467. PROBLEM XXVI.-Given the latitude of the place, the sun's declination and altitude, to find the hour of the day in apparent time. The angle P (last figure) in triangle SPZ is evidently the horary angle. If this angle be denoted by H, then (349) 2 L cos H = L sin S+L sin (S-Z) + L cosec P+L cosec C. EXAMPLE. On the 8th of May 1813, at 5h 30m 32s P.M. per watch, in latitude 39° 54′ N., and longitude 80° 39′ 45′′ W., the altitude of the sun's lower limb was observed to be 15° 40′ 57′′; required the error of the watch. |