the sun's leaving one tropic, or equinox, till he returns to it again, and consists of 365d. 5h. 48m. 48 sec. of mean solar time. (R) Nonagesimal degree of the ecliptic, is that point which is the most elevated above the horizon; and is measured by the angle which the ecliptic makes with the horizon at any elevation of the pole; or, it is the distance between the zenith, and the pole of the ecliptic. This angle is frequently used in the calculation of solar eclipses. (S) The medium Cæli, or mid-heaven, is that point of the ecliptic which culminates, or is on the meridian at any given time. (T) The Crepusculum, or twilight, is that faint light which we perceive before the sun rises, and after he sets. It is produced by the rays of light being refracted in their passage through the earth's atmosphere, and reflected from the different particles thereof. (U) A constellation is a collection of stars on the surface of the celestial sphere, circumscribed by the outlines of some assumed figure, as a ram, a dragon, a bear, &c. This division is necessary, in order to direct a person to any part of the heavens, where any particular star is situated. (W) The diurnal and nocturnal arcs. In all places of the earth, except the two poles, the horizon cuts the equinoctial into two equal parts. In all places situated on the equator, the horizon cuts all the parallels of declination into two equal parts, and here the sun and all the stars are 12 hours above the horizon, and 12 hours below. In places between the equator and the elevated pole, the parallels of declination are unequally divided; the greater are being above the horizon, and the less arc below. In all places between the equator and the depressed pole, the parallels of declination are unequally divided; the greater arc being below the horizon, and the less arc above. In all cases, the arcs which are above the horizon are called diurnal arcs, and those below, nocturnal arcs. Or, the parallel, which the sun, moon, or stars, describe from their rising to setting, is called the diurnal arc; and that parallel which each of them describes, from the setting to the rising, is called the nocturnal arc. II. Introductory Astronomical Problems.* PROBLEM I. (X) To turn degrees, or parts of the equator into time. * These are the same as in the former editions, being extracted from the general examples, and from the notes upon them. Those which depend upon the Nautical RULE. Multiply the number of degrees by 4, and the product will be the corresponding time. NOTE. Seconds multiplied by 4 produce thirds of time Minutes multiplied by 4 produce seconds of time. EXAMPLE. 4 Answer. lh. 41'. 1". 4". Also, 770.2.10” of longitude=56.8'.8”.40'" of time, and 1249.16.30" of the equator=8h.17.6" of time. PROBLEM II. (Y) To turn time into degrees. Rule. Multiply the hours by 60, and add the odd minutes, if any, to the product, one-fourth of which will be degrees; multiply the remainder by 60, and add the odd seconds, if any, to the product, one-fourth of which will be minutes, &c. EXAMPLE. Find the number of degrees, &c. corresponding to 11.41'. ib. 41. 1". 4'". 60 Answer 250.15'.16" 16' Also, 34.4'.28" of time=46°.7' of longitude, and 8h17'.6" of time=124o.16'.30% of longitude. PROBLEM III. (Z) Given the time under any known meridian to find the corresponding time at Greenwich.* * Since the earth makes one revolution on its axis froin west to east in 24 hours, the sun must apparently make one revolution round the carth from east to west in RULE. Turn the longitude of the place under the known meridian into time (X. 265.): add this time to the time at the given place if the longitude be west, or subtract it if east, and the sum or remainder will be the time at Greenwich. If the sum exceed 24 hours, subtract 24 hours from it, the remainder will shew the time at Greenwich on the following day: if the longitude, when turned into time, cannot be subtracted from the time at the given place, add 24 hours to the time at the given place before you subtract, the remainder will shew the time on the preceding day. EXAMPLE I. Find the time at Greenwich, on the 12th of August, when it is 76.25' at a place in longitude 970.45' west. Time at the given place 79.25' = 6.31 W. Time at Greenwich 13 . 56. or 56 minutes past 1 in the morning on the 13th of August.* EXAMPLE II. Find the time at Greenwich, on the 1st of May, when it is 22h.40' at a place in longitude 160° W. Time at the given place 22h.40'. Sum 33 .20 24 Time at Greenwich 9.20, on May 2d. EXAMPLE III. Find the time at Greenwich, on the 8th of April, when it is 16 .26' at a place in longitude 98°.45' East. the same time. Now, the longitudes of all places on the earth are reckoned on the equator, which is divided into 360 degrees, and the whole of it passes the sun in 24 hours; it follows that every 15° of motion is one hour in time, every degree 4 minutes, &c. (as in Prob. I. and II.) Hence, a place one degree eastward of Greenwica will have noon, and every hour of the day, four minutes sooner than at Greenwich; and a place one degree westward of Greenwich will have noon, and every hour of the day, four minutes later. * The astronomical day begins at noon, and is counted forward to 24 hours, or the succeeding noon, when the next day begins, being 12 hours later than the civil day, which commences at the preceding midnight; thus August 12th, at 13h.56' astronomical time, is August 13th at 1h.56' in the morning, according to civil Time at the given place 16h. 26' Time at Greenwich 9.51 on the 8th of April. EXAMPLE IV. Find the time at Greenwich, on the 4th of June, when it is 5h. 26' at a place in longitude 120° East. Time at the given place + 24b. = 29h.26' = 8.- E. Time at Greenwich of June. 21 .26, on the 3d PRACTICAL EXAMPLES. 1. What Greenwich time answers to noon at a place in 60° East longitude ? Answer. 20 hours, on the preceding day. 2. What Greenwich time answers to noon at a place in longitude 60° West ? Answer. 4 hours. 3. Find the time at Greenwich when it is 19h.42' at a place in 28°.30' E. longitude. Answer. 176.48'. PROBLEM IV. (A) Given the time at Greenwich to find the corresponding time under any known meridian. RULE. Turn the longitude of the place under the known meridian into time (X. 265.): add this time to the time at Greenwich if the longitude be east, or subtract it if west, and the sum or remainder will be the time under the known meridian. If the sum exceed 24 hours, subtract 24 hours from it, the remainder will shew the time at the given meridian on the following day: if the longitude, when turned into time, cannot be subtracted from the given time at Greenwich, add 24 hours to the time at Greenwich before you subtract, the remainder will shew the time on the preceding day. EXAMPLE I. When it is 9h.51' at Greenwich, on the 8th of April, what hour is it in longitude 98°.45' East? Time at Greenwich Gh.51' Time in long. 98°.45' E. = 16.26, on April sth. EXAMPLE II. When it is 216.26' at Greenwich, on the 3d of June, what hour is it at a place in 120° East longitude ? Time at Greenwich 215.26' East. Sum 29. 26 24. Time in longitude 120° E. = 5.26, on June 4th. EXAMPLE III. When it is 13h. 56' at Greenwich, on the 12th of August, find what hour it is at a place in longitude 97°.45' West. Time at Greenwich 13h. 56' Time in longitude 970.45' W.=7.25, on the 12th of August. EXAMPLE IV. When it is gh.20' at Greenwich, on the 2d of May, whiat hour is it at a place in longitude 160° West? Time at Greenwich + 24h.= 33h, 20 Longitude 160°, in time =10.40 W. Time in longitude 160° W. = 22 . 40, on the 1st of May, or 40 minutes past 10 in the morning on the 30th of April. PRACTICAL EXAMPLES. 1. When may an emersion of the first satellite of Jupiter be observed at Bombay, in longitude 729.54.30" E. which, by the Nautical Almanac, happens at Greenwich on the fourteenth of January 1822, at 6h.23.33". Answer. 115.15'.11". 2. What is the expected time of the beginning of the Lunar eclipse, which happens on August 2d, 1822, at 10h.51'.40" at Greenwich, in longitude 76o.49.30" West? |