PROBLEM VII. Given the Right Ascension of the Meridian and the Longitude, to find the Mean Time at Ship, or Place ;-or, to Reduce the Sidereal Time at Ship to Mean Solar Time. RULE. Since the mean time at ship may be always estimated within two or three minutes of the truth; if, therefore, the longitude in time be applied thereto by addition when it is west, or by subtraction when east, the sum, or difference, will be the assumed time at Greenwich; subject to the conditions in Problem III., page 342. Take, from page II. of the Month in the Nautical Almanac, the mean sun's right ascension for the noon preceding the Greenwich time: which subtract from the given right ascension of the meridian (increased by 24 hours if necessary), and the remainder will be the approximate mean time at ship. Reduce this to the meridian of Greenwich, by Problem III.; and find the corresponding equation in Table XLV. (the first in page 597 of the second volume). Now, the equation thus found, being subtracted from the approximate mean time; the result will be the correct mean time at ship or place. Example. May 6th, 1836, in longitude 45:45:15" west, the right ascension of the meridian was 17:55 44:64, and the estimated mean time at ship 145356; required the correct mean time at ship. 14:53:56: Longitude 45:45:15" west, in time. +3. 3, 1 -The only use that is made of this time is to indicate the proper noon for taking out the mean sun's right ascension from the Nautical Almanac. Given right ascension of the meridian 17:55"44:64 Remark. This problem solves the Example in page 502 of the Nautical Almanac for the year 1836, viz. :— To convert 219 23:04 sidereal time, January 2nd, 1836, into mean solar time for the meridian of Greenwich. Approximate mean time at Greenwich 2:24:43:44 . . 2:24:43:44 And in the same manner may the example in the second "Table of Equivalents," page 491 of the Ephemeris be solved. From the above it appears evident that the sidereal time is the same as the right ascension of the meridian; and vice versa, that the right ascension at any place is the same as the sidereal time at such place. -See Explanatory Article 32, page 309. PROBLEM VIII. To find the Mean Time of a Star's Transit, or Passage over the Meridian of any known Place. Since the plane of the meridian of any given place may be conceived to be extended to the sphere of the fixed stars,-therefore, when the diurnal motion of the earth round its axis brings the plane of that meridian to any particular star, such star is then said to transit, or pass over the meridian of that place. This observation is applicable to all other celestial objects. The mean time of transit of a known fixed star is to be computed by the following RULE. Reduce the given mean time at ship or place to the meridian of Greenwich by Problem III., page 342 :-to the Greenwich time, thus found, let the mean sun's right ascension be corrected by Problem V., page 344. Reduce the right ascension of the star, as given in Table XLIV., to the given day; or, take it at once out of the Nautical Almanac': from which, increased by 24 hours, if necessary, subtract the mean sun's correct right ascension; and the remainder will be the mean time of the star's transit over the meridian of the given place. Example. At what time on the 2nd of January, 1836, will the star Rigel transit, or come to the meridian of a place 165:30: east of Greenwich, the mean time at such place being about 10:21 ? precediny day; viz., of the 1st of January. Mean sun's R. A. at noon Jan. 1st, per Nautical Almanac 18:40" 43:04 Equation answering to 23:19" in Table XLVI. = Right ascension of the star Rigel on the given day Mean time of the star's transit, as required. 3.49.82 + 18:44:32:86 5. 6.40.06 10:22 7:20 Remarks.-1. In the solution of the present problem, it will be quite sufficient, if the mean time at ship be known within 3 or 4 minutes of the truth; because an error to that amount in the estimated time will not affect the computed mean time of transit more than about half a second. 2. Should the mean time of the star's transit below the pole be required:-Let the constant quantity 11:58 1:72 (viz. 12 hours diminished by 1:58:28, or half the diurnal increase of the mean sun's right ascension) be added to the mean time of transit found as above; the sum, abating 24 hours, if necessary, will express the mean time of the star's transit below the pole.* PROBLEM IX.* To find what Stars will be on, or nearest to, the Meridian of a Ship or Place at any given Time. RULE. Since the right ascension of a fixed star is measured eastward, according to the order of the signs, from the first point of Aries to the point of the equinoctial, which is intersected by its circle of declination; and since the right ascension of the meridian signifies the point of the equinoctial which comes to the meridian of a place at any given time (Definitions 15 and 16); therefore, the right ascension of the meri * The author has invented a stellarium, which shows (without the trouble of calculation) the solution of Problems VIII. and IX. in all parts of the world. dian expresses the correct value of the sidereal right ascension of the heavens at any given time and place :-Hence, Let the right ascension of the meridian be found by Problem VI., page 345; then look for this among the right ascensions of the stars in the Nautical Almanac, or amongst the right ascensions in Table XLIV.; and it will show the stars that are either on or nearest to the meridian at the given mean time. The stars whose right ascensions are less than the computed right ascension of the meridian, will have passed, or be to the westward of the ship's meridian; but those whose right ascensions are greater, will be to the eastward of, and apparently approaching, the meridian of the ship or place. Example. What star will be nearest to the meridian, December 31st, 1836, at 10:12 41 mean time; the longitude being 70:45: West? Mean sun's right ascension, reduced by Problem V. 18:42 13:11 Given mean time at ship or place +10. 12. 41. Right ascension of the meridian. 4:54 54:11 Now, this being looked for among the right ascensions of the stars in Table XLIV., it will be seen that the star whose right ascension corresponds the nearest thereto is ß Eridani; which, therefore, is the star that is most contiguous to the meridian at the given time. Capella is 941 to the eastward of the meridian; and Aldebaran 28:22: to the westward of it—hence, the former is approaching, and the latter receding from, the meridian. Note. In general, the correction of the mean sun's right ascension may be dispensed with. The right ascension at noon of the given day, added to the mean time at ship, will show the value of the right ascension of the meridian to a sufficient degree of exactness for nautical purposes. The above problem will be found useful when the latitude is to be deduced from the meridional altitude of a fixed star. PROBLEM X. To compute the Mean Time of the Moon's Transit over the Meridian of Greenwich. Since the moon's transit over the Royal Observatory at Greenwich is only given to the nearest tenth of a minute in the Nautical Almanac ; and since it becomes absolutely necessary, on many astronomical occa sions, to have it more strictly determined; the following method is therefore given, by which the mean time of the moon's transit over the meridian of Greenwich may be obtained true to the decimal part of a second. RULE. From the moon's right ascension at noon of the given day (increased by 24 hours if necessary), subtract the mean sun's right ascension at that noon; the remainder will be the approximate time of transit. Find the difference between the moon's right ascension at noon, and at the hour which is next greater than the approximate time of transit : diminish this difference by the equation, in Table XLVI., corresponding to the hour which is next greater than the approximate time; the result will be the excess of the moon's motion over that of the mean sun in the interval betwixt noon and the approximate time of transit: then say, as the next greater hour diminished by this excess, is to the next greater hour, so is the approximate time of transit, to the correct mean time of transit over the meridian of Greenwich. Reduce all the terms to seconds; and then the proportion may be easily worked by logarithms, as in the following Example. Required the mean time of the moon's transit over the meridian of Greenwich on the 1st day of January, 1836? Approx. time of tran. 10 222:79 Difference = Do. reduced to seconds=36142.79 Next greater hour is 11 0 0 0 over mean sun's = 21.32.52 Equation to 11 hours Excess of D's mo- 10:3827:48=38307:48 Log. ar. comp.5. 416717 39600-Logarithm . 4.597695 36142.79 Logarithm 4.558021 Difference= Mean time of transit, in seconds 37362:24 Logarithm. 4.572433 Ditto, raised to hours, &c. . 10:22:42:24; which, therefore, is |