tude is less than 3 signs or 90 degrees; but which is to be increased by 180 degrees when the longitude is between 6 and 9 signs, or between 180 and 270 degrees. Again, if the longitude is between 3 signs and 6 signs, that is, between 90 and 180 degrees, the arch, so found, is to be augmented by 90 degrees; but, if the longitude lies between 9 and 12 signs, viz., between 270 and 360 degrees, it is to be augmented by 270 degrees; in either case the correct right ascension of the object will be obtained, which may be converted into time, if necessary, by Problem I., page 342. Example. The apparent longitude of Aldebaran, August 3rd, 1825, was 2:7:21:18′′.4, and its latitude 5:28:45" south; required its apparent right ascension and declination, the obliquity of the ecliptic being 23:27:42.875? Aldebaran's ecliptic polar dis.=95.28.45 .0 Log. sine 9.998011 star's eclip. polar distance=72: 1: 2". 125Half S. 19.092010+ 16:9.5". 4 north. Are = Twice the arc= tance from the north pole of the equator; hence its declination is 36:55:27".3 Log. sine 9.778700 73:50:54",6; which is Aldebaran's dis To find the Right Ascension : Declination of Aldebaran. 16: 9 5".4 Log. secant 10.017490 Aldebaran's right ascension= 66:29: 1".7, or 4:25:56:1. Hence, the right ascension of the star Aldebaran is 4:2556′.1, and its declination 16:9.5". 4 north, as required. Note.--In the same manner may the right ascensions and declinations of the moon and planets be deduced from their respective latitudes and longitudes. PROBLEM III. Given the Latitudes and Longitudes of the Moon and Sun, Moon and fixed Star, or Moon and Planet; to find the true Central Distance between them. Note. If this Problem be projected stereographically on the plane of the circle of longitude passing through one of the objects, it will be found, in every respect, like Problem XXIV., page 273, of "The Young Navigator's Guide to the Sidereal and Planetary Parts of Nautical Astronomy;" reading, however, difference of longitude for difference of right ascension, and ecliptic polar distances for polar distances: -hence, it is evident that there is a spherical triangle to work in, where two sides and the included angle are given to find the third side, or central distance between the objects, and which may be computed by the following General Rule. To twice the logarithmic sine of half the difference of longitude between the two objects, add the logarithmic co-sines of their latitudes; from half the sum of these three logarithms subtract the logarithmic sine of half the difference or half the sum of the latitudes, according as they are of the same or of contrary names; the remainder will be the logarithmic tangent of an arch, the logarithmic sine of which being subtracted from the half sum of the three logarithms, will leave the logarithmic sine of half the true central distance between the two given objects. Example 1. September 3rd, 1825, the moon's apparent longitude, at noon, was 1:16:19:29%, and her latitude 2:33:30 north; at the same time the apparent longitude of Pollux was 3:20:48:38", and its latitude 6:40:19: north; required the true central distance between those two objects? Hence the true central distance between the moon and Pollux, at the given time, was 64:22:8; which corresponds exactly with the computed distance in the Nautical Almanac. Note. It is evident that the same result will be obtained by making use of the right ascensions and declinations of the objects. The true central distance may be readily found by the formula under Remark 2, page 204. Example 2. August 4th, 1825, the moon's apparent longitude, at noon, was 0:14:13:32%, and her latitude 4:38:41 north; at the same time the sun's longitude was 4:11:43:46%; required the true central distance? Note. Since the sun apparently moves in the ecliptic, he has, therefore, very little, or no latitude. Difference of long.=117:30:14" this divided by 2 gives 58:45 7" Twice the L.sine=19. 863860 Moon's lat. 4.38. 41 N. Log. co-sine 9.998571 = Hence, the true central distance between moon and sun is 117:24:21; which corresponds with that in the Nautical Almanac. Remark. Since the co-latitudes of the sun and moon and the comprehended angle (expressed by their difference of longitude,, form a quadrantal spherical triangle; therefore the true central distance between these particular objects may be more readily determined by the following concise method than by the above general Rule, viz., To the logarithmic co-sine of the difference of longitude, add the logarithmic co-sine of the moon's latitude; the sum of these two logarithms, abating 10 in the index, will be the logarithmic co-sine of the true central distance between the sun and moon. Example 1. August 6th, 1825, the moon's longitude, at noon, was 1:8:0:34", and her latitude 3:23:20% north; at the same time, the sun's longitude was 4:13:38:46"; required the true central distance? Difference of long. = 95:38:12" Log. co-sine = . 8.991439 True central distance=95:37:36" Log. co-sine = Example 2. August 7th, 1825, the moon's longitude, at noon, was 1:20:4:42", and her latitude 2:30:42" north; at the same time the sun's longitude was 4:14:36:18"; required the true central distance? which exactly corresponds with the computed distance in the Nautical Almanac. Example 3. Required the true central distance between the moon and sun at noon, August Sth; at midnight, August 8th; at noon, August 9th, and at midnight, August 9th, 1825 ? Moon's long. noon Aug. 8th. = 62:22:38′′ Sun's longitude ditto = . 135.33.51 Distance at noon, Aug. 8th = 73:11:34" Log. co-sine 9.461124 Dist. at midnight, Aug. 8th 67:24:31" Log, co-sine 9,584511 = |