878. The apparent diameter of the sun o inversely as the ♦ being measured from apogee, the apparent diameter ∞ 1-ecos. §. Hence the augmentation of the apparent diameter is 879. vers. exterior Vince. The visible illumined part of the disk; whole disk :: of elongation: 2. See Woodhouse, p. 552, or But in the case of the moon, the distance of the sun from the earth being incomparably greater than that of the moon, the of elongation may be considered the of elongation exterior itself. Hence visible part whole disk :: vers. 60° : 2 ::1 cos. 60° : 2 .. visible part dark part :: 1:3 880. Let D, A; L, λ be the declination, right ascension, longitude, and latitude of the star; d(D) d (A) the given aberrations in declination and right ascension. Then, if a, d' (Fig. 114,) be the two places of the star, P, the poles of the equator and ecliptic, Pon, nom referred to the parallels of the equator and ecliptic, on om will denote the co-declination and the co-latitude; and we have T Again, since the aberration is small, the A orm', o'rn may be considered rectilinear. Hence the no'r mor = angle of position (= P). And da=dD x cos. PdA cos. D sin. P See also Woodhouse's chap. on Aberrations. 881. The surface is expressed generally by s being the arc of the generating curve. = 2πr. {C + √ (r2 — y3)} Let S = 0, when y = r; then C = 0, and Let lat. 90 and 30; then the whole surface and the surface between the specified parallels are :. 2S = 4#r2 and 2S' = 4πr2 X respectively; so that 2S = 2 × 2S' Q. E. D. 882. Let D, D', x be the co-declinations of the stars and planet, a the given difference of the right-ascensions of the stars, and y the difference between the right-ascensions of the planet and that star whose declination is D. Also let d, d' be the given distance of the stars from the planet then from the A (D, d, x) D', d', x) we get which, by the Arithmetic of Sines, will give us y and x. 883. The shadow will evidently be a parallelogram deviating from a rectangle by half a right-angle. Its breadth will therefore breadth of the wall x cos. 45° = of the wall. 1 x breadth √2 The datum of the sun's altitude is unnecessary. It will give the length of the shadow. 885. Since the star is known, his declination is given by the tables. Let D be the co-declination, a its altitude when on the Prime Vertical, and L the co-latitude of the place; then from the right-angled ▲ (D, a, L) we have cos. D cos. a cos. L .. cos. L= which gives L. 886. cos. D cos. a Let ss' (Fig. 115,) be two successive positions of Then if Pz the co the sun, P being the pole and z the zenith. latitude of the plane be put = L, the co-dec. D, the hour zPs = H, and azimuth szP = a, &c., as in the figure, we have o'n da. sin zs. số = dH. sin. D. and 'n=ss' x sin. s'sn = dH sin. D cos. zsP Hence, putting zsz, we have, by the question, 887. Let L be the co-latitude, z the zenith distance, the azimuth, and D the co-declination; then from the A (z, L, D) we have Again, the declination being found, it would be easy to find the ASTRONOMY. right ascension in terms of the obliquity of the ecliptic. But we are required to express it otherwise. Since the day and hour of observation are given, let the known right-ascension and hour of the sun, be denoted by A, and H; also let A', H' be the right-ascension and meridian distance of the star; then 888. Let L be the latitude of the place, D the known co-declination of the star; then from the A (L, D, 45°) we have 889. cos. Dcos. L. cos. 45 .. cos. L= cos. D x √2. Let L denote the co-latitude, D the co-declination of the sun, and a the required azimuth; then from the rightangled ▲ (L, D) we have tan. D tan. a = sin. L 890. Since the day and hour are given, the right-ascension (A), and hour (H) of the sun are known, also since the latitude and longitude of the star are given, its right-ascension (A') and declination (90° - D) may be found (see Vince.) Hence the angular distance of the star from the meridian (H) is known from H' HA' - A, and the latitude of the place (90° - L) is given. Consequently if 90z, a denote the required altitude and azimuth, from the ▲ (z, D, L) we have |