and sin.2 / = = also '(1-cos. D. cos. A) × (m + n sin. A)* which two equations involving only two unknown quantities and A will give them when reduced and resolved. The remainder of the investigation being operose rather than intricate, we leave it, as an exercise, to the student. Having found 1, and A, he will find a from equation (4); land à will give the required place of the moon at the time of her greatest declination. 908. Let H be the hour from inidnight of rising at one place, whose latitude is L; then the hour at the other whose latitude is 90° L is, by the question Hence, if D be the given co-declination of the sun, from the right-angled A, (L, D), (90 — L, D), we have - = cos. (n × 15°) tan. L + sin. (n × 15°) ✅ (tan.' D n 15°. 909. In the vertical passing through the sun and zenith describe a circle whose radius is the length of the stick. Next draw a tangent to this circle to such a point that its inclination to the horizon may be 30°. The distance of the centre of the circle from that intersection will evidently be the longest shadow possible. Hence the stick must be inclined to the horizon at the angle 90° - 30° 60°. Again, ifl be the stick, and x that of its shadow, we have 910. Since the day is given, the sun's co-declination (D) is known from the tables, and the hour (H) being known, and the co-latitude of the place (L), from the A (z, D, L,) we get cos. z = sin. D. sin. L. cos. H+cos. L. cos. D Hence the altitude of the sun 90° - z is known. Again, since the rays of the sun are parallel, the shadow of the globe will be a cylinder whose axis is inclined to the horizontal plane at the (90°), and the radius of whose base is (r) the radius of the globe. Hence the shadow upon the horizon will be an ellipse, (the section of the cylinder) whose semi-axes may easily be shewn to be { its equation is y = cos.2 z 2r COS.z - Again, the length of the shadow is the axis-major, or which varies as the secant of the zenith distance. 2r COS.Z 911. Since the two stars are known, their right ascensions A, A', and co-declinations D, D' are given. Let also the given co-declination of the other star be D", its right ascension 6, and its distances from the two stars x, x. Then from the A (D,D", 1), (D', D', x'), we have cos. x = cos. D. cos. D' + sin. D. sin. D". cos. (A But since x + x = minimum, ➡ dx = dx′ 6) Hence, and by means of the two equations above, z z', and 8, may be found, which will more than resolve the problem. 912. On the longest and shortest day the declination of the sun is equal to the obliquity of the ecliptic, and then the meridian altitude of the sun co-latitude declination. Consequently if l, l' be the lengths of the shadows, and h the height of the rod, we have h. tan. (co-latitude - obliquity) = tan. (co-latitude + obliquity). Let L be the latitude, and o the obliquity; then cot. (L+0) cot. (L0) :: l: l::n: 1 by, the question. .. cos. (L+0). sin. (L-0): cos. (L−0) sin. (L+0)::n:1 .. cos. (L+0). sin. (L — 0) + cos. (Lo) sin. (L + 0) : cos (L+0) sin. (L — 0) · cos (L — 0) sin. (L + 0) :: n + 1: N- - 1. .. sin. (L + 0 + L-0): sin. (L+0-L-0): n+1: :: n — 1, or sin. 2×L: sin. 2×0 :: n + 1: n 913. Let D be the co-declination of the star, L the colatitude, and z, z' the observed zenith distances, h the ing the interval between the equal azimuths a. Then measur◄ 914. It is easily shewn that in the given periods the num bers of seconds are 152853, 314022 and the periods are in the sesquiplicate ratio of the mean distances, 914. ::1: (2+ of 50951 :: 1:23 (1+377)3 nearly The parallax of the moon diminishes the moon's altitude; the refraction, on the contrary, elevates both the moon and the star; and they take place wholly in vertical circles. Hence, from the A having the same vertical z at the zenith, 1vers. d + sin. a sin. b) vers. (A – B) + (vers. d 1 + cos. a−b) = vers. (A−B)+ (vers, d - vers. (a - b). Q. E. D. |