PROBLEM V. (B) Toreduce the declination of the sun, as given in the Nautical Almanac, to any other meridian, and to any given time of the day. Rule. The corresponding time at Greenwich being ascertained (Z. 266.), find the change of the sun's declination in 24 hours from the Nautical Almanac: Then, 24 hours : this change::the time from noon at Greenwich : the variation of the sun's declination in that time. This variation must be added to the sun's declination at noon*, or subtracted from it, according as the declination is increasing or decreasing. Note. By a similar process the change of the sun's longitude, or of right ascension, may be determined for any given time, or at any given place; and also the declination of a planet. EXAMPLE 1. Required the sun's declination at noon, on the 12th of October 1822, at Glasgow, longitude 4o.15' W. First, 4°.15'=17minutes, the time by which the clocks at Glasgow are slower than at Greenwich; hence when it is noon at Glasgow, it is Oh.17' at Greenwich. O's declination at noon October 12th, Naut. Alm. is 70.171.34" Increase of declination in 24 hours 22' 34" Then 24h : 22.34"::17: 16" the increase of the sun's declination in 17 minutes of time; consequently when it is noon at Glasgow, the sun's declination is (70.17.34" +16"=)70.17.50" South. EXAMPLE II. What is the sun's right ascension, June 5th 1822, at 13h.48'. in longitude 630.10'E. ? Time at the given place 13h. 48'. Longitude 63o.10', in time = 4.12. 40" E. Time at Greenwich 9. 35. 20 O's right ascension at noon June 5th, Naut. Alm. is 4h.51'.21".7 Increase of right ascension in 24 hours 4. 65.9 * The sun's longitude, right ascension in time, and declination are given, in the IId page of the Nautical Almanac, for every day in the year, at noon, calculated for the meridian of Greenwich. Then 24h : 4.6.9 :: 9h.35'.20": 1.38".6 4h.51'.21".7 1.38.6 O's right ascension at 13h.48' in long. 630.10' E. = 46.58'. 09.8 | PRACTICAL EXAMPLES. 1. Required the sun's declination on the 25th of August 1822, at 8h.20, in longitude 48° West. The declination at Greenwich, at noon, (by the Nautical Almanac) being 10°.53'.35" N. and on the 26th, 100.32.50" N. Answer. 10°.43.37". N. 2. Required the sun's right ascension at noon, on the 25th of May 1822, in longitude 124° East. The right ascension at Greenwich (Naut. Alm.) being 45.6'. 31", and on the 24th of May, 4h.2.29". Answer. 45.5'.29". 3. Required the sun's declination January 24th, 1822, at 181.40' in longitude 132° East. The declination at Greenwich, at noon, (Naut. Alm.) being 19o.16'.13" South, and on the 25th, 19o.1'.38" S. Answer. 19o.10'.14" S. 4. Required the sun's right ascension on the 16th January 1822, at 186.48', in longitude 68° West. The right ascension at Greenwich, at noon, (Naut. Alm.) being 19h.51.30", and on the 17th, 19h.55'.47". Answer. 19h. 54'.2". PROBLEM VI. (C) To reduce the declination of the moon, as given in the Nautical Almanac, to any other meridian, ana to any given time of the day. Rule. Find the time at Greenwich corresponding to the time at the given place. (Z. 266.) Take the change of the moon's declination in 12 hours from the Nautical Almanac. Then, 12 hours : this change::the time at Greenwich : the variation of the moon's declination in that time. This variation must be added to the moon's declination (at noon or midnight) if the declination be increasing, or subtracted if the declination be decreasing. NOTE. By a similar process the change of the moon's right ascension", semidiameter, and horizontal parallax may be ascertained for any given time, or at any given place. * The moon's age and time of passage over the meridian of Greenwich are EXAMPLE 1. 1. Required the moon's declination March 17th 1822, at 7h.22' in longitude 57° W. The moon's declination at noon (Naut. Alm.) at Greenwich, being 26°.12' S. and at midnight 240.58' S. Time at the given place 7h.22' Longitude 570 W. in time = 3.48'W. D's declination at noon* 26°.12 S. l 12h : 1o. 14' : : 11h. 10% : 1o. 9'va) 's declination at midnight24°.58' S. riation of the D's declination in 116.10'. Decrease in 12 hours = 1o.14 Hence 260.12 -10.9ʻ=250.3' S. the moon's declination at the time and place required. EXAMPLE II. Required the moon's semidiameter and horizontal parallax on the 26th of January 1822, in longitude 110.45' West, at 151.45' apparent time. The semidiameter, at Greenwich, at midnight, (Naut. Alm.) being 16'.1" and at noon on the 27th, 16'.3", also the horizontal parallax at the same time 58.32", and 58.41". Time at the given place 15h.45' O 47 W. = 16h.32, or 4h.32' past midnight. D's semi-dia. at midnight, 26th, = 15'.57"; hor. paral. = 58'.32" D's semi-dia. at noon, 27th,=15'.59"; hor. paral. =58'.41" Increase in 12 hours, O'.2"; increase in 12 hours · O'. O 12h : 2':: 4h.321 : D's semi-dia. at midnight O''. 12n : 9"::4h.32 : 34 15'.571. lhor · par. at midnight = 58' 32" D's ditto in long. 11°.45 W.= 15'.57".7|hor.par.in long. 11°.45°W.=58.35".4 Hence the ) 's semi-diameter at 15h.45' in long. 11°.45' W. is 15'.571.7, and the horizontal parallax is 58',35".4. given in the VIth page of the Nautical Almanac: and her latitude, longitude, right ascension, declination, semidiameter, and horizontal parallax, are given for noon and midnight at Greenwich in pages Vth, VIth, and VI) for each month. * If the time at Greenwich had exceeded 12 hours, the moon's declination must have been taken out for midnight and the noon of the next day; and the variation applied to the midnight declination. PRACTICAL EXAMPLES. 1. Required the moon's declination on the 11th of January, 1822, at 176.47' in longitude 162° West ? The moon's declination at noon at Greenwich (Naut. Alm.) on the 12th of January, being 10.19'N. and at midnight 10.35' S. Answer. The time at Greenwich is 4h.35' on the 12th of January, and the moon's declination is 0°.13' North. 2. Required the mocn's semidiameter and horizontal parallax on the 19th of May, 1822, in longitude 38°.40' E. at 11h.15' apparent time. The moon's semidiameter at Greenwich (Naut. Alm.) at noon and midnight being 16'.38" and 16'.41"; and the horizontal parallax at the same time 61.3" and 61'.13". Answer. The time at Greenwich is 8h. 40'.20", D's semidiameter = 16'.40' and horizontal parallax=61'.10". 3. Required the moon's declination on the 13th of May, 1822, at 19h in longitude 67° East. The moon's declination at Greenwich (Naut. Alm.) at midnight being 150.5' S. and at noon on the 14th 12°.27' S. Answer. The time at Greenwich is 14h 32', 's declination 149.32 S. 4. What is the moon's declination on July 19th, 1822, at 45.49' in longitude 114° East ? The moon's declination at Greenwich (Naut. Alm.) at midnight on July 18th being 21°.12' N. and on July the 19th at noon 182.51 N. Answer. The time at Greenwich is 21h.13' July 18th, and the moon's declination 190.24' N. PROBLEM VII. (D) To find the time of a star's culminating, or coming to the meridian of Greenwich. Rule. Subtract the right ascension of the sun for the given day from the right ascension of the star, and the remainder will be the time of the star's culminating nearly.-If the sun's right ascension exceed the star's, add 24 hours to the star's before you subtract. Take the increase of the sun's right ascension in 24 hours, and add it to 24 hours. Then, This sum is to 24 hours as the star's right ascension diminished by the sun's, is to the time of the star's culminating. Note. If the time of culminating be required for any other meridian than that of Greenwich ; allow 10 seconds* of time for every 15° of longitude, which subtract from the time at Greenwich for places in west longitude, or add to that time for places in east longitude, and the result will shew the time of culminating at the given meridian. EXAMPLE I. At what time will Arcturus come to the meridian of Greenwich on the 1st of December 1822 ? The right ascension of Arcturus being 146.7'.32". *'s right ascension (1822)+24h -38h. 7.32". o's right ascension, December 1st, 1822 =169.28'.16. Time of *'s culminating nearly =216.39'.16". O's right ascension, December 1st, 1822, at noon=165.28'.16'. O's right ascension, December 2d, 1822, at noon=164.32.35". Increase of the O's right ascension in 24 hours = 0h. 4'.19". Then 24b.4'.19" : 24h:: 21h.39'. 16" : 211.35'.23" I apparent time of Arcturus's culminating, or 9h.35'.23', December 2d, in the morning Note. Arcturus will be on the meridian of Philadelphia, longitude 75°.13'W,on the sameday at 214.34.33",Philadelphia time; for 15° : 10"::75°.13' : 50 and 21h. 35' 23" – 50"216.34.33". It will also be on the meridian of Pekin, longitude 116o.24' E, on the same day at 216.36'.40”, Pekin time; for 15°:10":: 116o.24': 77'71'.17", and 216.35.23"' +1'.17"= 211.36'.40". Hence it appears that the same fixed star is on the meridian of every place, in any kingdom of moderate extent, nearly at the same hour reckoned at that place. PRACTICAL EXAMPLES. 1. At what time will Aldebaran culminate at Greenwich on the 20th November 1822? Theright'ascension of Aldebaran being 4h.25'.43" (Tab. VIII.) * For the sun's right ascension varies about four minutes of time every day, but the right ascension of a star remains nearly the same during the whole year, and 24: : 4 ::]" (=150) : 10". of See the table annexed to the Nautical Almanac for the year 1822, or Table VIII. of this work. The same fixed star is on the meridian of any place at nearly the same bour, on the same day of the month, for several years; the variation in 40 years will sekcom exceed two minutes of time. |