As one sidereal day, is to 355.91, 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 time; as thus. Required the sidereal time corresponding to 20:1533: mean solar time? Given mean solar time = 20:15:33: Sum = + 3:19.68 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? 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 Time 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. Let the latitude be 3o, 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. 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 argu ments. Example 1. Let the latitude be 50:48 north, and the sun's declination 10:25 north; required the sun's true amplitude at its setting? True amplitude corresp. to lat. 50, and dec. 10:, =W. 15: 40: N. Tab. diff. to 1 of lat. =21; now 21: × 48: = + 17, nearly; 60 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 349 N. and 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. Example. Let the latitude be 50:48, and the declination of a celestial object 10:25; required the true amplitude of that object? To find the Times of the Rising and Setting of a Celestial Object. This Table contains the semidiurnal arch, or the time of half the continuance of a celestial object above the horizon when its declination is of the same name with the latitude of the place of observation; or the time of half its continuance below the horizon when its declination and the latitude are of different denominations.-The semi-diurnal arch expresses the time that a celestial object takes in ascending from the eastern horizon to the meridian; or of its descending from the meridian to the western horizon. As the Table is only extended to 23 degrees of declination, being the greatest declination of the sun, and to no more than 60 degrees of latitude; therefore, when the declination of any other celestial object and the latitude of the place of observation exceed those limits, the semi-diurnal arch is to be computed by the following rule; viz., To the log. tangent of the latitude, add the log. tangent of the declination, and the sum, rejecting 10 in the index, will be the log. sine of an arch; which being converted into time, and added to 6 hours when the latitude and declination are of the same name; or subtracted from 6 hours when these elements are of contrary names; the sum, or difference, will be the semi-diurnal arch. Example 1. Let the latitude be 61 degrees, north, and the declination of a celestial object 25:10, north; required the corresponding semi-diurnal arch? |