6. Hence, L = 51 31 54. The following method is somewhat simpler when only one star is observed, or when the mean declination of the sun is used, as in the last example.† 9.893735 Let P, H, and L, denote the same quantities as in the preceding rule, and let 6. L. sin L. = L. cos 0 + L. cos R - 10. ~ SA+ A', the sum of the true altitudes, D A = A', the difference of the true altitudes; and let M, N, O, Q, R, denote what are called the first, second, third, fourth, and fifth arcs; then, 1. L. sin ML. sin P+L. sin H-10. 2. L. cos NL. cos P+L. sec M-10. And N is of the same species as P. 3. L. sin 0 = L. sin D + L. cos S+L. cosec M — 10. L. cos Q = 4. L. cosD+L. sin S+ L. sec M + L. sec O — 10. R=NQ. 5. When the zenith and elevated pole are on the same side of the great circle that passes through the two positions of the sun or star, R = N~ Q; otherwise, R = Ñ + Q. P = 70° 21', A = 38° 19', The preceding example, computed by this rule, is added here: S = 44° 22′. D = 6 3. *Here (a c) means the arithmetical complement of L. cos. (see art. 172, Part I.) For the demonstration of this rule, consult Riddle's useful treatise on Navigation and Nautical Astronomy. In this problem it is not necessary to know the times of observation, but merely the interval between them; from the data of the problem, however, the time of either observation can be found, and the azimuth of the object. At sea the altitudes cannot be found with sufficient precision to obtain the latitude correct within several seconds, so that five figures of logarithms are in these instances sufficient. The first three of the following exercises have been thus calculated. EXERCISES. 1. At a place in north latitude, when the sun's declination was 23° 29′ N., its true altitude at 8h 54m A.M. was 48° 42', and at 9h 46m A.M. it was 55° 48'; required the latitude. Ans. 49° 49′ 28′′. 2. At a place in north latitude, when the sun's declina tion was 2° 46' S., its true altitude was 33° 11' at 9h 20m A.M., and 42° 44′ at 1h 20m P.M.; required the latitude. Ans. 40° 50′ 10′′ N. 3. Find the time at which the first altitude was taken in the example to this problem, and the azimuth. 8h 30m, and the azimuth from N = Ans. The time 107° 47'. 4. On the 6th October 1830, at a place in north latitude, the true altitude of the sun at 7h 5m 49s A.M. mean time was 8° 37′ 42′′-6, and at 1h 2m 47s-8 it was 33° 43′ 46′′-1, and the sun's declination for the middle time was 5° 9′ 48"-1 S.; required the correct mean time of the first observation, and the latitude, the equation of time to be subtracted from apparent time being 11m 42s.1.* Ans. The latitude 48° 42′ 42′′-2 N., the time = 7h 5m 398.8. In the following example, the first method must be employed, as the polar distances are different. Also, as Atair is to the east of Arcturus, and the altitude of the former was taken some minutes later than that of the latter, this elapsed time, converted to sidereal time, must be subtracted from the difference of their right ascensions, in order to obtain the angle H. 5. On the 19th September 1830, the zenith distance of Arcturus was found to be 73° 19′ 26′′-5 at 8h 2m 47s-8 mean time, and that of Atair was 40° 53′ 56′′-3 at 8h 22m 3s; the polar distance of the former was 69° 55′ 36′′-4 and that of the latter 81° 34'; required the latitude, and the correct time of the first observation, the sun's mean right ascension at mean noon being 11h 51m 298.76. Ans. The lati tude 48° 42′ 12′′, and the time 8h 4m 21s.36. LUNAR DISTANCES. 479. PROBLEM XXXV.—To find the true angular distance between the moon and the sun or a star, having given their altitudes and apparent distances. *This example and the following are selected from Francœur's "Astronomie Pratique." Leth h' H the apparent height of the moon's centre, = the apparent height of the centre of sun or star, =the true height of the moon's centre, H' the true height of the centre of sun or star, d = the apparent distance of the centres, Dthe true distance of the centres, s=h+h', and S=H+H', also, if be an arc such that sin2 = · cos H⚫cos H' cos 1 (s + d) ⋅ cos 1 (s ~ • cos h⋅ cos h' cos2 S sin D cos cos S. then Or logarithmically, cos H+L. cos H' + L. cos (s + d) L. sin 6 = {+L. sec / + L. sech' + L. cos 3 (3~d) -(L. cos S+10)...[1] and then L. sin DL. cos + L. cos S-10...[2] EXAMPLE. On the 14th December 1818, at 12h 10m nearly, latitude 36° 7′ N., longitude by account 11h 52m W., the following observations were made, the height of the observer's eye being 19.5 feet, in order to find the distance between the moon and Regulus.* Moon's semi-diam. Refraction Ob. dis. of moon's nearest l. and Regulus, d′ = 33° 15′ 25′′ = 4 18 Hor. par. 53' 59" Par. in alt. d) 61 21 54 = 0 14 56 ; h61 36 50 61 36 18.9 L. Sec P.L. = .32274 =52301 = .84575 0 25 40.5 H = 62 2 O nearly. *This example is taken from Woodhouse's excellent treatise on Astronomy, Theoretical and Practical, Part II. p. 862. There are only ten logarithms to be taken from the tables in this computation, for L. cos S occurs twice. The calculation of the time of observation, and of the longitude of the place, is performed in the example to the next problem. Let M', S', be the apparent places of the Z centres of the moon and the sun, or a star, or a planet, and M, S, their true places, and Z the zenith; then M will be above M', because the moon's parallax exceeds the refraction due to its height, but S will be below S', in consequence of the refraction exceeding the parallax of M M |