little distance, the position of the eye being fixed at some particular spot, such as at a small hole in a window-shutter nearly in the plane of the meridian; then if the observed times of disappearance correspond with the acceleration contained in the second column of the first compartment of the present Table, it will be an undoubted proof that the watch is well reguláted :-hence, if the watch be exactly true, the disappearance of the same star will be 356'. earlier every night ; that is, it will disappear 3:56: sooner the first night; 7* 52: sooner the second night; 11"48! sooner the third night, and so on, as in the Table.-Should the watch, or clock deviate from those times, it must be corrected accordingly; and since the disappearance of a star is instantaneous, we may thus determine the rate of a watch to at least half a second. The first compartment of this Table consists of two columns; the first of which contains the sidereal days, or the interval between two successive transits of a fixed star over the same meridian, and the second the acceleration of the stars expressed in mean solar time ; which is extended to 30 days, so as to afford ample opportunities for the due regulation of clocks or watches.—The five following compartments consist of two columns each, and are particularly adapted to the reduction of sidereal time into mean solar time:-the correction expressed in the column marked acceleration, &c. being subtracted from its corresponding sidereal time, will reduce it to mean solar time; as thus. Required the mean solar time corresponding to 14.40*55' sidereal time? Given sidereal time = 14:40:55: 2-24 .31 Do. 55 seconds 0.0.15 Mean solar time as required . 14.38.30'.69 Remark.-This Table was computed in the following manner ; viz., Since the earth performs its revolution round its orbit; that is, round the sun, in a solar year; therefore as 365!5.48"48: : 360 :::1! : 59'8".3; which, therefore, is the earth's daily advance in its orbit: but while the earth is going through this daily portion of its orbit, it turns once round on its axis, from west to east, and thereby describes an arc of 360:59'8".3 in a mean solar day, and an arc of 360: in a sidereal day. Hence, as 360:59'8".3: 24::360: : 23.56" 4'.09, the length of a sidereal day in mean solar time; and which, therefore, evidently anticipates 355:.91 upon the solar day as before-mentioned. Now, As one sidereal day, is to 3.55..91, so is any given portion of sidereal time to its corresponding portion of mean solar time :-and hence, the method by which the Table was computed. TABLE XLVI. To reduce Mean Solar Time into Sidereal Time, Since this Table is merely the converse of the preceding, it is presumed that it does not require any explanation farther than by observing, that the correction is to be applied by addition to the corresponding mean solar time, in order to reduce it into sidereal tine; as thus. . Required the sidereal time corresponding to 2015":33mean solar time? Given mean solar time = 20" 15"33; Corresponding to 20 hours is 35 17.. '.13) Do. 15 minutes 0. 2.465 Sum = + 39 19:. 68 Do. 33 seconds 0. 0.09 Sidereal time as required 20:18:52:.68 TABLE XLVII. Time from Noon when the Sun's Centre is in the Prime Vertical; being the instant at which the Altitude of that Object should be observed in order to ascertain the apparent Time with the greatest Accuracy. Since the change of altitude of a celestial object is quickest when that object is in the prime vertical, the most proper time for observing an altitude from which the apparent time is to be inferred, is therefore when the object is due east or west; because then the apparent time is not likely to be affected by the unavoidable errors of observation, nor by the inaccuracy of the assumed latitude.-This Table contains the apparent time when a celestial object is in the above position. The declination is marked at top and bottom, and the latitude in the left and right hand marginal columns : hence, if the latitude be 50 degrees, and the declination 10 degrees, both being of the same name, the object will be due east or west at 5:26" from its time of transit or meridional passage. Remark.This Table was computed by the following rule ; viz., To the log. co-tangent of the latitude, add the log. tangent of the declination; and the sum, abating 10 in the index, will be the log. co-sine of the hour angle, or the object's distance from the meridian when its true bearing is either east or west. Example. Let the latitude be 50 degrees, north or south, and the sun's declination 10 degrees, north or south; required the apparent time when that object will bear due east or west? Given latitude = 50. log. co-tangent = 9.923814 Declination of the sun = 10. log. tangent = 9.246319 Hour angle = 81:29:30"=log. co-sine = 9.170133 In time 5.25"58! ; which, therefore, is the apparent time when the sun bears due east or west. Note. During one half of the year, or while the sun is on the other side of the equator, with respect to the observer, that object is not due east or west while above the horizon ; in this case, therefore, the observations for determining the apparent time must be made while the sun is near to the horizon; the altitude, however, should not be under 3 or 4 degrees, on account of the uncertainty of the effects of the atmospheric refraction on low altitudes. TABLE XLVIII, Altitude of a Celestial Object (when its centre is in the Prime Vertical,) most proper for determining the apparent Tine with the greatest Accuracy. This Table is nearly similar to the preceding; the only difference being that that Table shows the apparent time when a celestial object bears due east or west, and this Table the true altitude of the object when in that position ; being the altitude most proper to be observed in order to ascertain the apparent time with the greatest accuracy :-thus, if the latitude be 50 degrees, and the declination 10 degrees, both being of the same name, the altitude of the object will be 13:6', when it bears due east or west from the observer ; which, therefore, is the altitude most proper to be observed, for the reasons assigned in the explanation to Table XLVII. Note.-- This Table was computed by the following rule ; viz., If the declination be less than the latitude ; from the log. sine of the former (the index being increased by 10), subtract the log. sine of the latter, and the remainder will be the log. sine of the altitude of the object when its centre is in the prime vertical :-But, if the latitude be less than the declination, a contrary operation is to be used ; viz., from the log. sine of the latitude, the index being increased by 10, subtract the log. sine of the declination, and the remainder will be the log. sine of the altitude of the object when its centre is in the prime vertical, or when it bears due east or west. Example 1. Let the latitude be 50., and the declination of a celestial object 10!, both being of the same name; required the altitude of that object when its centre is in the prime vertical. Declination of the object log. sine = 9. 239670 Latitude 50. log. sine = 9.884254 = 10: . Altitude required. 13:6:6 log. sine = 9.355416 Example 2. Let the latitude be 39, and the declination of a celestial object 14:, both being of the same name; required the altitude of that object when its in the prime vertical. Latitude, 3: Declination of the object = 14 log. sine = 8.718800 Altitude. required. 12:29:38 log. sine = 9.335125 Note.--Altitudes under 3 or 4 degrees should not be made use of in computing the apparent time, on account of the uncertainty of the atmospheric refraction near the horizon. And since the Table only shows the altitude of a celestial object most favourable for observation when the latitude and declination are of the same name; therefore during that half of the year in which the sun is on the other side of the equator, with respect to the observer, and in which he does not come to the prime vertical while above the horizon, the altitude is to be taken whenever it appears to have exceeded the limits ascribed to the uncertainty of the atmospheric refraction in page 120. TABLE XLIX. Amplitudes of a Celestial Object, reckoned from the true East, or West Point of the Horizon. The arguments of this Table are, the declination of a celestial object at top or bottom, and the latitude in the left, or right hand column; in the angle of meeting will be found the amplitude : proportion, however, is to be made for the excess of the minutes above the next less tabular arguments. Example 1. Let the latitude be 50:48: north, and the sun's declination 10:25 north h; required the sun's true amplitude at its setting ? True amplitude corresp. to lat. 50:, and dec. 10:, =W. 15. 40: N. 21! x48 Tab. diff. to 10 of lat. =21! ; now + 17, nearly; 60! 96' x 252 T.diff. to 1. of dec.=1:36:, or 96'; now 60 Sun's true amplitude as required = W. 16.37. N. + 40 Example 2. Let the latitude be 34:24! north, and the sun's declination 16:48: south ; required the sun's true amplitude at the time of its rising? True amplitude corresponding to latitude 34: N. and declination 16:30: S. = E. 20. 2: S. 15! x 24! Tab. diff. to 1: of lat. = 15! ; now t 6 60: Tab. diff. to 30 of dec. = 37!; now 37! *18! + 22, nearly. 30: Sun's true amplitude as required E. 20:30: S, Remark. This Table was computed agreeably to the following rule; viz., To the log. secant of the latitude, add the log. sine of the declination, and the sum, abating 10 in the index, will be the log, sine of the true amplitude. |