Jupiter's declination at noon, January 1st, = 17:56 0 N. + 5. 1 18: 1 1:N. Log. rising = 5.864960 . Log. co-sine= 9.978164 50.30. 0 N. . Log. co-sine 9.803511 Planet's mer. zen. dist. = 32:28:59"Nat.V. S.= 156449 Nat. num. 443236Log.=5.646635 Jupiter's true central alt.= 23:35.52"N.co-V.S.=599685 Note.-Jupiter's horizontal parallax is assumed, in the present instance, at 2 seconds of a degree, Required the true and 16th, 1825, at 7:20*45 Example 2. apparent altitude of the planet Venus, January apparent time, in latitude 34:45? S., and longitude 80:30: E., admitting her horizontal parallax, at that time, to be 18 Sun's right ascension at noon, January 16th, 19:52 41: : = 10: 9:36 S. Log. rising 5.809810 9.993136 Log. co-sine 9.914685 Planet's hor. dist. fr. mer. = 4*36*55! Planet's mer. zen. dist. = 24:35:24"Nat.V. S.=090691 Nat.num.521953Log.=5.717631 Venus' true central alt, = 22:47:24 N.co-V.S=612644 Red.Tab. XIX.215") Par.Tab.vi.=0.16 S Venus' app. central alt. 22:49:23" Remark. In these problems, a cipher is annexed to the logarithmic rising taken from Table XXXII.: this is done with the view of reducing it to six places of decimals; so that there may be no mistake in properly applying thereto the logarithmic co-sines of the latitude and of the declination. PROBLEM IV. Given the Latitude of a Place, and the apparent Time at that Place, with the Longitude; to find the true and apparent Altitude of the Moon's Centre. RULE. Reduce the given apparent time to the meridian of Greenwich, by Problem III., page 297; to which let the sun's right ascension be reduced, by Problem V., page 298; and let the moon's right ascension, declination, and horizontal parallax be reduced to the same time, by Problem VI., page 302. To the sun's reduced right ascension let the given apparent time be added, and the sum will be the right ascension of the meridian; the differ ence between which and the moon's reduced right ascension, will be the horary distance of the latter from the meridian. Now, with the moon's horary distance from the meridian, her corrected declination, and the latitude of the place, compute her true central altitude, by Problem I., page 404. Then, From the moon's true central altitude, thus found, subtract the correction corresponding thereto and her reduced horizontal parallax, in Table XIX., and the remainder will be the apparent central altitude. Note.-The moon's right ascension and declination must be corrected by the equation of second difference contained in Table XVII., as explained between pages 33 and 37. Example 1. Required the true and apparent altitude of the moon's centre, January 4th, 1825, at 7:2818 apparent time, in latitude 50:10:N., and longitude 60: W.? Moon's horizontal parallax at noon, January 4th, = 59:17" = Moon's right ascension at noon, January 4th, 98: 6:53% 369757Log.=5.567917 Moon's mer. zen. dist.=28:44:39" Nat.V. S. = 123225 Nat. num. True alt. of D's centre 30°27:55 Nat. co-V.S.=492982 App. alt. of D's cent. 29:37'47" Example 2. Required the true and apparent altitude of the moon's centre, January 30th, 1825, at 13:33 20: apparent time, in latitude 10:20: S., and longitude 100:50: E.? Sun's right ascension at noon, January 30th, 20:51:25! Moon's horizontal parallax at noon, January 30th,=57.51% Moon's right ascension at noon, January 30th, 76:21:55" Moon's corrected R. A.=512222: = 80:35:33% Remark. The natural sines may be used in the solution of the four preceding problems, instead of the versed sines: in this case, if the natural number he subtracted from the natural co-sine of the object's meridional zenith distance, the natural sine of its true altitude will be obtained. Thus, in the last example, the moon's meridian zenith distance is 34:14.2". Now, the natural co-sine of this is 826748; from which let the natural number 680116 be subtracted, and the remainder =146632 is the natural sine of that object's true altitude; the arch corresponding to which is 8:25:54%. These problems are, evidently, the converse of those for finding the apparent time, as given in pages 383, 394, 397, and 400. SOLUTION OF PROBLEMS RELATIVE TO THE LONGITUDE. The Longitude of a given place on the earth, is that arc or portion of the equator which is intercepted between the first or principal meridian and the meridian of the given place; and is denominated east or west, according as it may be situate with respect to the first meridian. The first or principal meridian is an imaginary great circle passing through any remarkable place and the poles of the world: hence it is entirely arbitrary; and, therefore, the British reckon their first meridian to be that which passes through the Royal Observatory at Greenwich; the French esteem their first meridian to be that which passes through the Royal Observatory at Paris; the Spaniards, that which passes through Cadiz, &c. &c. &c. Every part of the terrestrial sphere may be conceived to have a meridian line passing through it, cutting the equator at right angles: hence there may be as many different meridians as there are points in the equator. |