COS MS = 44° 51′ 45′′ cos (60° 42′ 15′′ – 7) sec . 9.8506086 In logs.... ...... cos 44° 51′ 45′′ .... Thus the true distance between the moon and star is found to be 63° 5' 11". 2dly. By Borda's Theorem. Here we have н= 45° 9′ 12′′, h = 45° 8′ 15′′ # 28° 29′ 44′′, h' = 29° 17′ 45′′, h + h′ = 74° 26′ s = 68° 37′ 41,", S D = 5° 1′ 51′′ The logarithmic computation for F, therefore, is 9.5617995 cos (s—D).. 5° 1′51′′. 9.9983236 9.8484403 COS S .... 9.9405687 Its half is sin F.... 48° 56′ 1′′ .... 98773416 The sum is sin d......31° 32′ 34′′ .... 97186273 Consequently d is 63° 5' 8".. Note. Here the difference in the results is only 3 seconds; which is quite as little as can be expected when tables are employed in which the computer has to proportion for the seconds. Had the value of F been taken 48° 56' instead of 48° 56′ 1′′ (and it is evident from the common tables that it lies between the two), then d would have been found about 63° 5′ 15′′. The mean between these two results is 63° 5′ 11′′. Example II. Given the apparent distance between the centres of the sun and moon 83° 57′ 33′′, the apparent altitude of the sun's centre 48° 27′ 32′′, its true altitude 48° 26′ 49′′, the apparent altitude of the moon's centre 27°34′ 5′′, its true altitude 28° 20′ 48′′. Required the true distance of the centres of the two luminaries. Ans. 83° 20′ 55′′. Example III. The apparent distance of the moon's centre and a star was 2° 20, when the apparent altitude of the star's centre was 11° 14', and that of the moon's 9° 39'; the moon's correction was 51′ 30′′, the star's 4′ 40′′. Required the true distance of their centres. Ans. 1° 49'. Note 2. In examples of this kind the sum of the apparent zenith distances must always be greater than the apparent distance; for if not they cannot form a spherical triangle. Thus it will be found that example 4, pa. 31, of the Requisite Tables, 2d edition, relates to an impossible case. The third side 103° 29′ 27′′ is greater than 89° 50′ 22′′ the sum of the other two. Note 3. This being an important problem in nautical astronomy, we shall here refer to other works where more compendious rules, founded principally upon subsidiary tables, are given. Such are, the Requisite Tables to be used with the Nautical Almanac; Mackay on the Longitude; Mendoza's Tables for Navigation and Nautical Astronomy; Myers's Translation of Rossel on the Longitude; Andrew's Astronomical and Nautical Tables; Kelly's Spherics. See also, Mr. Sanderson's rules in the Ladies' Diary for 1787, or in Leybourn's collection of the Diaries, vol. iii. Dr. Brinkley's in the Irish Transactions for 1808, and various others in Delambre's Astronomy, vol. iii. chap. 36. PROBLEM VII. 29. Given the longitudes and latitudes of two places upon the earth, regarded as a globe, to find their itinerary distance, that is, the arc of the great circle comprehended between them. Let A and B be the two places on the surface of the terraqueous globe, of which P is one of the poles, and conceive a spherical triangle PAB to be described, such that PA, PB, shall be the respective distances of the two places from the pole P, or their respective co-latitudes, and the angle APB the difference of longitude of those two places. Hence, L and L' being the respective latitudes, and P the difference of longitude, there are given in the triangle PAB, two sides, viz. PA 90° I., PB = 90° L', and P, to find the third side AB. The problem, therefore, belongs to case 2, prob. 2, of oblique spherical triangles, the appropriate formula for which, suited to the present case, become tan cos P tan PB. (1.) for the subsidiary angle: .... and cos AB = COS PB SEC COS (PA — 4)......... (2.) Example I. Given the latitude of the observatory of Paris, 48° 50′ 14′′ N. that of the observatory of Pekin, 39° 54′ 13′′ N. and their difference of longitude 114° 7′ 30′′: to find their distance. Here since the latitudes are both of the same kind, we have 48° 50′ 14′′ 41° 9′ 46′′ and P diff. of long. Hence to log cos P 114° 7′ 30′′.. 9.6114352 .... add tan PB.... 50° 5′ 47′′.. 10.0776707 10= log tan 9-26° 2′ 53′′.. 9-6891059 Now, since P exceeds 90° its cosine is negative, and consequently must be taken negatively. Hence PA - p. 41° 9′ 46′′ + 26° 2′ 53′′ = 67° 12′ 39′′, and the work indicated by equa. (2) will stand thus: Log cos PB 50°′ 5′ 47′′.... 9.8071953 9.5880938 COS (PA — 4) 67° 12′ 39′′. The sum-20, cos AB 73° 56′ 40′′.... 9·4418068 Thus the distance required is 73° 56′ 40′′ of a great circle, or in English miles, reckoning 693 to a degree, it is 5139 nearly." Example II. The latitude of St. Helena is 15° 55' S. its longitude 5° 49′ W.; the latitude of the Bermudas 32° 35′ N. longitude 63° 32′ W. Required their distance? Here since the latitudes are of different kinds, we shall have PA = 90° + L = 90° + 15° 55′ = 105° 55′′ 黛 PB 90° L'— 90° 32°35′ and P 63° 32′ · 5° 49′ 57° 43'. 57° 25' The preceding formulæ applied to these data, will give 73° 26' for the distance required. Example III. What is the distance on a great circle between St. Mary's, latitude 37° N., longitude 25° W., and Cape Henry on the same parallel of latitude, but in longitude 76° 23′ W.? Ans. 40° 31, which is about 36 miles less than the distance measured upon the parallel itself. Example IV. Required the distance on a great circle between the island of St. Thomas, latitude 0°, longitude 1o E., and Port St. Julian, in latitude 48° 51' S., and longitude 65° 10′ W. Ans. 74° 35'. Note. In the 3d example the operation becomes shortened because the triangle PAB is isosceles; in the 4th it is shortened because the triangle is quadrantal. PROBLEM VIII. 30. To determine the elongation of a planet from the sun, at the time when it appears stationary; on the supposition of concentric circular orbits, in one and the same plane. As thus restricted the problem may be solved by means of the principles of plane trigonometry, combined with those of central forces in dynamics. Let two concentric circles Eel, PpQ, be drawn, the former to represent the orbit of the earth, the latter that of the planet, the common centre s of both circles being imagined the place of the sun. While the earth moves from E to e in its orbit, let the planet move from P to p in its orbit; then, when the planet appears stationary the right lines EP, ep, drawn from the earth to the planet will be parallel. In that state of things draw the radii SE, se, SP, sp; from E and P draw tangents to their respective circles to meet each other in r; and join sr. So will the triangles SET, SPT, be right angled at E and P; and the parallel lines EP, ep, will be so near each other, that the intercepted arcs Ee, Pp, of the orbits, may be regarded as coinciding with the corresponding portions of their respective tangents. Let v be the velocity of the earth in its orbit, v' that of the planet. Then, since the arcs Ee, Pp, are described in the same interval of time, they will be to each other as those velocities, that is, |