PROBLEM II. To convert Time into Longitude, or parts of the Equator. RULE. Reduce the hours to minutes, to which add the odd minutes, if any; then, the minutes divided by 4 give degrees; the seconds divided by 4 give minutes, and the thirds divided by 4 give seconds. The above problems may be readily solved by means of Table I:the principles upon which they are founded are set forth in the paragraph which follows the Examples in page 2. PROBLEM III. Given the Time under a known Meridian, to find the corresponding Time at Greenwich:-or, to reduce the Mean Time at any place to the Meridian of Greenwich. RULE. Let the given mean time be always reckoned from the preceding noon, to which apply the longitude of the place in time (reduced by Problem I.), by addition, if it be west, or subtraction, if east; the sum, or difference, will be the corresponding mean time at Greenwich. Note.-This Rule is general; but it is subject to certain conditions, viz. -In west longitude, when the sum is less than 24 hours, it will express the mean time at Greenwich past noon of the given day; but, when it is greater, then the sum, minus 24 hours, will be the mean time at Greenwich past noon of the day following the given one. Again. In east longitude, the difference will be the mean time at Greenwich past noon of the given day : but when the time at ship is less than the longitude in time, it is to be increased by 24 hours :—in this case, the increased time minus the longitude in time will be the mean time at Greenwich past noon of the day preceding the given one. Example 1. January 8th, 1836,-Required the corresponding times at Greenwich when it is 4:40:13: and 21:50:45, at a ship in longitude 80:53:15. West? February 10th, 1836,-Required the corresponding times at Greenand 2:40:35:, at a ship in longitude wich, when it is 20:11:41 98:14:45 East? Given the Time at Greenwich, to find the corresponding Time under a known Meridian. RULE. Let the given mean time be always reckoned from the preceding noon, to which apply the longitude of the place in time (reduced by Problem I. as above), by addition if it be east, or subtraction if west; and the sum, or difference, will be the corresponding time under the given meridian. Example 1. When may the emersion of the first satellite of Jupiter be observed at Trincomalee, in longitude 81:22 E., which, by the Nautical Almanac, happens at Greenwich, March 6th, 1836, at 7:619: ?. Mean time of emersion at Greenwich = Longitude of Trincomalee 81:22: E., in time = Mean time of emersion at Trincomalee = Example 2. 7: 6:19: When may the immersion of the first satellite of Jupiter be observed at Port Royal, Jamaica, in longitude 76:52:30" W., which, by the Nautical Almanac, happens at Greenwich, Nov. 6th, 1836, at 15:49:55:? 15:49:55: Mean time of immersion at Greenwich = Mean time of immersion at Port Royal = 10:42:25: As the above problem is the converse of Problem III., the longitude in time is therefore applied to the given mean time with a contrary sign, viz., by subtraction in west, and addition in east longitude. It is particularly useful in looking out for the approximate times of observing the eclipses of Jupiter's satellites. PROBLEM V. To Reduce the MEAN Sun's Right Ascension (viz. the "Sidereal Time," which is given in page II. of the Month in the Nautical Almanac,) to any given Meridian, and to any Time under that Meridian. RULE. Turn the longitude into time (Problem I., page 341), and add it to the mean time at ship or place, if it be west, but subtract it if east; the sum, or difference, will be the corresponding mean time at Greenwich, subject to the conditions in Problem III., page 342.-To the mean time at Greenwich, thus found, take out the corresponding equations in Table XLVI., volume II., page 597; which being added to the mean sun's right ascension at the noon preceding the Greenwich time; the result will be the correct right ascension of the mean sun at the given time and place. Should the result exceed the measure of a day, it is to be diminished by 24 hours. Example 1. Required the mean sun's right ascension, January 1st, 1836, at 20:40:36: mean time; the longitude being 120:45 East? 37 minutes, in ditto. 36 seconds, in ditto. of the given day. Mean sun's R. A. at noon of given day, per Naut. Alm. 18:40:43:04 Equation answering to 12 hours, in Table XLVI.= Ditto 1.58.28 6.08 0.10 Required the mean sun's right ascension, January 10th, 1836, at 21:10:30: mean time; the longitude being 140:35 west? Longitude 140:35 West, in time. +9.22.20 of the day following the given one, viz. of January 11th. Mean sun's R. A. at noon January 11th, per Naut. Alm. 19:20 8:62 Equation answering to 6 hours, in Table XLVI. = 59.14 5.26 0.14 19:21 13:16 Mean sun's correct right ascension, as required = Note. See Explanatory Article 37, page 313, relative to the "Sidereal Time," in page II. of the month in the Ephemeris, being called the MEAN Sun's Right Ascension. PROBLEM VI. Given the Mean Time at Ship or Place, and the Longitude, to find the Right Ascension of the Meridian; or, to Reduce Mean Solar Time to Sidereal Time. RULE. Turn the longitude into time (Problem I., page 341), and add it to the mean time at ship, if it be west, but subtract it if east; the sum, or difference will be the mean time at Greenwich, subject to the conditions in Problem III., page 342. To the mean time at Greenwich, take out the corresponding equations in Table XLVI., Vol. II., page 597; which, being added to the mean sun's right ascension at the noon preceding the Greenwich time, the sum will be, the mean sun's correct right ascension. Now, this being added to the given mean time at ship, the sum (diminished by 24 hours if necessary) will be the right ascension of the meridian. Example. Required the right ascension of the meridian, May 6th, 1836, at 14:55 19: mean time; the longitude being 45:45:15 West ? Longitude 45:45:15" West, in time +3. 3. 1 Mean sun's R. A. at noon, May 6th, per Nautical Alm. 2:57:29:00 Equation answering to 17 hours in Table XLVI. = 2.47.56 9.03 0.05 3 025:64 14. 55. 19. 17:55:44:64 Remark. This problem solves the Example marked "vice versa," in page 502 of the Nautical Almanac for 1836; viz. : To convert 2:24:19:73 mean solar time, January 2nd, 1836, into sidereal time for the meridian of Greenwich : Mean sun's R. A. at noon, January 2nd = Equation answering to 2 hours in Table XLVI. Sidereal time = In the same manner may be solved the Example which is given in the first "Table of Equivalents," page 489 of the Ephemeris. Note. From the above it is clearly manifest that the right ascension of the meridian at any given place is the sidereal time at that place; and conversely.-See Explanatory Article 32, page 309, |