angle M A F. The sum or difference of M A F and F A B is MAB; and in the right angled triangle MA B there are then given A M and M A B, to find A B the right ascension, and B M the declination. Or in the triangle M E P are given EP, the obliquity of the ecliptic, E M the co-latitude, and M E P the co-longitude of M, to find M P the polar distance, and E PM the supplement of MP C, the co-right ascension. EXAMPLES FOR EXERCISE. The obliquity of the ecliptic being 23° 27' 50", it is required to compute the right ascensions and declinations of the objects whose latitudes and longitudes are given in the following examples ? Note. By the two preceding problems, the right ascensions and declinations of the sun and moon, in the Nautical Almanac, are computed. PROBLEM III. From the longitude of the sun, and the latitude and longitude of any other celestial object, to find the distance of that object from the sun. Let S (see the last figure) be the sun, and M the other object, then in the spherical triangle SFM, right angled at F, are given MF, the latitude of the object, and SF, the difference of longitude of the object and the sun, to compute S M, the required distance. EXAMPLES FOR EXERCISE. It is required to compute the true distance between the sun and the moon or a star, in each of the following examples ? Note. By this problem the distances of the sun and moon, in the Nautical Almanac, are computed. PROBLEM IV. From the latitudes and longitudes of any two celestial objects, to find their distance. Let M and I (see the last figure) be the two objects, then in the spherical triangle M I E, are given the two sides ME, I E, the colatitudes of the given objects, and the included angle MEI, the difference of their longitudes, tu find I M, their distance. Note. The co-latitudes of the objects are estimated from the same pole of the ecliptic. EXAMPLES FOR EXERCISE. In each of the following examples, A and B are two stars, from whose latitudes and longitudes it is required to compute their distance ? Note. By this problem the distances of the moon and fixed stars, in the Nautical Almanac, are computed. PROBLEM V. From the right ascensions and declinations of any two celestial objects, to find their distance. Let M and I, as in the last problem, represent the two objects; then distances of the objects, and MPI, the difference of their right ascensions, to find M I, their distance. EXAMPLES FOR EXERCISE. In each of the following examples, A and B are two stars, from whose right ascensions and declinations their distance is to be computed ? The distance of an unknown star, or comet, from two known stars being given, to find its right ascension and declination. Let G (see the last figure) be the unknown star or comet, and M I, the two known stars whose distances MG, IG, from G are given; then from MP, P I, and MPI, the distance M I of the two known stars may be found, as in the last problem, and in the same triangle the angle P M I may be found. In the triangle I M G all the sides are given to find the angle G MI, the sum or difference of which and PM I is P M G. Then P M, MG, and the contained angle PMG, are give to compute P G the polar distance, and MP G the difference between the right ascensions of G and M; whence, as the right ascension of M is known, that of G is also known. Note. The observed distances should be cleared from the effects of refraction, which may easily be done by Spherics, according to the method pointed out in the “ Remark” at p. 277. EXAMPLES FOR EXERCISE. 1. If on November 14, 1836, the distance of a comet from Markab be 42° 14', and from Altair 36° 35', northward from the stars, required its right ascension and declination ? Answer, right ascension 308° 13' 19'', declination 43° 20 32'' N. 2. If on February 5, 1850, the distance of a star from Rigel be 36° 58' 20", and from Procyon 34° 15' 20'', southward from the stars, required its right ascension and declination ? Answer, right ascension 109° 47' 12'', declination 28° 31' 18" S. 3. If on March 14, 1840, the distance of an unknown star from Spica be 26° 50/50", and from Antares 39° 18' 40', southward from the stars, required its right ascension and declination ? Answer, right ascension 208° 42' 45'', declination 35° 44' 14" S. Z W А. Let the annexed figure represent the celestial sphere, A ZBNA being the meridian of an observer whose zenith is Z, and nadir N. Let P be the north, and s the south pole, and let AZ, BZ, PC, and P D be quad н rants. Then the eye in view (M ing the figure being conceived K LV B to be vertically over E, the point on the surface which is U the apparent centre of A ZBNA; all great circles, as A B, C D, PS, ZN, drawn through E will appear in the figure as straight lines; A B will represent the rational horizon, C D the equator, Z N the prime vertical, and P S the six o'clock hour circle, or the hour circle at right angles to the meridian. The great circles drawn through Z will be perpendicular to A B, and those drawn through P will be perpendicular to CD. CZ=PB = ND = A S, is the latitude, and Z P = BD = S N = A C, the co-latitude of the observer, whose zenith is Z; the arcs measuring their corresponding angles at the apparent centre E, or the spherical angles on the surface of the sphere at E. FK is the altitude of an object at F, F Z its zenith distance, F H its north declination, F P its polar distance, FPZ its meridian distance, FZP its azimuth from the north, and F Z C its azimuth from the south. GK is the altitude of an object at G, G Z its zenith distance, G I its south declination, G P its polar distance, GPZ its meridian distance, G Ž P its azimuth from the north, and GZ C its azimuth from the south. FPG is the difference of the meridian distances, or of the right ascensions of F and G; CQ, the measure of ZP O, is the semidiurnal arc.of an object which rises or sets at 0; and E Q, the complement of CQ, is the ascensional difference of O, or the measure of the time between its rising or setting and passing the six o'clock hour circle ; E O is the amplitude of O, and E L the amplitude of L; TU, a small circle parallel to A B, and 18° below it, is the twilight circle, or that on which the sun is at the beginning and end of twilight. PROBLEM VII. Given the declination of a celestial object, to find its altitude and bearing when on the six o'clock hour circle in a given latitude. In this problem the latitude and declination must be of the same denomination, otherwise the object will pass the six o'clock hour circle before it rises. In the triangle EWO (see the last figure) right angled at 0, are given E W the declination, and WE O the latitude, to find W O the altitude, and E O the bearing from the prime vertical, southward or northward, according as the latitude is south or north. EXAMPLES FOR EXERCISE. In each of the following examples it is required to compute the altitude of the object, and its bearing from the east to west, when on the six o'clock hour circle. same name. Given the declination of a celestial object, to find its altitude and meridian distance, when on the prime vertical, in a given latitude. In this problem also the latitude and declination must be of the Let X (see the last figure) be the object on the prime vertical Z E, then in the triangle E M X, right angled at M, are given M X the declination, and M E X the latitude, to find X E the altitude, and M E the complement of M C the meridian distance, whence M C is determined. Or in the triangle X P Z, right angled at Z, are given X P the object's polar distance, and Z P the co-latitude, to find ZP X the meridian distance, and Z X the zenith distance, whence the altitude E X is known. EXAMPLES FOR EXERCISE. In each of the following examples the altitude of the object when on the prime vertical, and its meridian distance at the time, are required ? |