from table XXVII., and with the arguments I., II., and III., take the corresponding equations from table XXVIII. The equation of the centre, and the three other equations, added to the mean longitude, give the true longitude, reckoned from the mean equinox. With the argument N, take the equation of the equinoxes, or, which is the same, the nutation in longitude, from table XXX., and apply it, according to its sign, to the true longitude already found, and the result will be the true longitude, from the apparent equinox. For the Hourly Motion and Semidiameter. With the sun's mean anomaly, take the hourly motion and semidiameter from tables XXV. and XXVI. For the apparent Obliquity of the Ecliptic. To the mean obliquity, taken from table XXIX., apply, according to its sign, the nutation in obliquity, taken from table XXX., with the argument N, and the result will be the Apparent Obliquity. For the Earth's Radius Vector. With the sun's mean anomaly and the arguments I., II., and III., take the corresponding quantities from table XXXI., and the small table on the same page, and the sum of these will be the Radius Vector. For the Right Ascension and Declination. To the log. cosine of the apparent obliquity of the ecliptic, add the log. tangent of the sun's true longitude, and reject 10 from the index of the sum; the result will be the log. tangent of the Right Ascension, which must always be taken in the same quadrant as the longitude. To the log. sine of the apparent obliquity, add the log. sine of the longitude, and reject 10 from the index of the sum; the result will be the log. sine of the Declination, which must be taken less than 90°; it will be north or south according as its sign is affirmative or negative. For the Equation of Time and the Apparent Time. To the sun's tabular mean longitude, increased by 2°, apply, according to its sign, the nutation in right ascension, taken from table XXX., with the argument N, and the result will be the sun's mean longitude from the true equinox. Take the difference between this longitude and the sun's right ascension, making it affirmative or negative according as the right ascension is less or greater than the longitude, and the result will be the Equation of Time in arc. This may be converted into time by Prob. III. The equation of time, applied, according to its sign, to the mean time, gives the apparent time. EXAM. 1. Required the sun's longitude, hourly motion, &c., on the 25th of October, 1836, at 10h. 37m. 10sec. A. M. mean time at Boston. * When one of these quantities is near to 0° and the other to 360°, the less must be increased by 360°, and the sum be regarded as its value. For Right Ascension and Declination. Obliq. Long. Obliq. 23° 27′ 45′′ 1. cos 9.96252 Sun's tab. M. Long. +2° For Equation of Time and Appar. Time. Sun's M. Long. from true equinox Sun's Longitude Dec. - 12° 16' 24" 1. sin 9.32751n 66 hourly mot. 66 semidiameter App. obliq. of ecliptic Declination Equat. of time Appar. Time Radius Vector 1. sin 9.60005 1. sin 9.72746n Equat. of Time, in arc Equat. of Time, in time 15m. 50.5sec. The equation of time, added to the given mean time, gives 10h. 53m. 0.5sec. A. M. for the apparent time. ++ EXAM. 2. Required the sun's longitude, hourly motion, &c., on the 15th of May, 1836, at 8h. 59m. 29sec. A. M., mean time at Philadelphia. Ans. PROBLEM VII. 214° 2′ 36′′ 214 2 26 210 4 48 3 57 38 54° 42′ 12′′ 225 15 50 23 27 44 52 20 23 18 57 45 N +3m. 56sec. 9h. 3m. 16sec. A. M. 1.01167 To find the Sidereal Time corresponding to a given Mean Time. To the sun's mean longitude from the true equinox, found as in the last problem, and converted into time by Prob. III., add the given mean time of the day, expressed astronomically, rejecting 24 hours from the sum, if it exceeds that quantity, and the result will be the sidereal time. When the sidereal time or right ascension of the zenith is required in arc, and not in time, it is most conveniently obtained by adding the mean time, expressed in arc, to the sun's mean longitude from the true equinox. Note. When the sidereal time has been found for a given mean time, it may be found for any other time, a few hours later or earlier, by simply adding or subtracting the sum of the quantities corresponding to the hours, minutes and seconds of the interval, taken from table X. EXAM. 1. What was the sidereal time at Philadelphia, on the 28th of May, 1840, at 3h. 19m. 20sec. P. M., mean time? M. Long. N Greenw. time 3h. gives 21m. 66 36.5sec. " 4 Sidereal time Tab. M.Long. 2 4 25 53 Sidereal time, When the sidereal time is required in arc, and not time. M. Long. from true equinox Given mean time, in arc 78 Sun's tab. M. Long. + 2° 66° 25′ 53′′ M. Long. from true equinox, 66 26 1 (C (C "in time Given mean time, 4h. 25m. 44.1sec. 3 19 20 9h. 45m. 4.1sec. Sidereal time, in arc 116 16 1 Let now the sidereal time be required at 6h. 40m. 56.5sec. P. M. Then the interval is 3h. 21m. 36.5sec. Table X., for, 66° 26' 1" To find the Sidereal Time corresponding to a given Mean Time, using the Nautical Almanac. In the Greenwich Nautical Almanac, the sidereal time at Greenwich mean noon is given for each day, and that for any other meridian may be found by adding if west, and subtracting if east, the sidereal acceleration. for the difference of longitude expressed in time. The acceleration may be found by taking from table X., the equivalents for the hours, minutes and seconds, rejecting the hours, minutes and seconds of the equivalents respectively, and adding the results. Or, by multiplying 9.8565sec. by the difference of longitude expressed in hours. Then, having found, as directed above, the sidereal time at mean noon, add to it the sidereal equivalents for the hours, minutes and seconds of the given mean time, taken from table X., and the sum will be the sidereal time required. EXAM. 1. What was the sidereal time at Philadelphia on the 17th of May, 1836, at 8h. 17m. 10.5sec. mean time? The acceleration for the diff. of Long. is by table X., for 5 hours 49$.282 "40 seconds 0$.5 sec. Sidereal time required 12h. 0m. 12.675sec. EXAM. 2. Required the sidereal time at St. Petersburg on the 20th of February, 1850, at 11h. 42m. 25sec. mean time, the Berlin sidereal time at mean noon being 22h. 0m. 6.36sec. 66 St. Petersburg east of Berlin 66 Sid. Equiv. for 11h. 66 66 42m. 25sec. St. Petersburg 49.392 h. m. sec. 3 40 51.11 3 41 40.502 8 1 18.852 17 2.793 10.027 0.501 h. m. sec. 2 1 16.0 53 35.5 Sidereal time required 9h. 44m. 15.633sec. EXAM. 3. Required the sidereal time at New Haven on the 10th of May, 1836, at 15h. 3m. 13.5sec., mean time. Ans. 18h. 19m. 45.04sec. |