4. Having found log. p cos ' and log. p sin o', for the place, by Prob. XV., find u, v, N, F, t and t' by the following formulæ. cos F (d+ v - q) sin N le cos (N + F) k Then will, T' ť" + t, be the corrected Greenwich mean time of the immersion. The difference between this and the observed time, will be the longitude in time; west, if the observed time is the earlier of the two, but east, if it is the later. In a similar manner we deduce the longitude from the observed emer sion, except, that, instead of t, we find t' = k cos (N-F) • When the immersion and emersion have both been observed, the longitude should be obtained from each, and the mean of the two results be taken. EXAMPLE. Suppose the observed immersion of i, Leonis, on Jan. 7th, 1836, at a place in latitude 52° 8′ 28′′ N., estimated longitude Oh. 1m. W., was 10h. 45m. 53.3 sec., mean time; required the longitude of the place. The observed time of immersion reduced to Greenwich time is, T = 10h. 46m. 53.3 sec. Taking T = 10.8h. = 10th., we easily find from the Nautical Almanac, A = 10 20 33.89 ; D = + 15° 49′ 31.2" A'= 10 23 26.39 ; D' = + 14 58 38.8 sec. A" = 122.905 = (in arc), 1843.6"; D" = log. 3.41288n D-D' c = .9079 .95804 a p' = d - d' = .5270 ; q = b + c = - .2101. Sidereal time at mean noon, Greenw. from N. Α., noon { to time T. (53.3.sec. “ 46 7.56 5 53 1.98 10 23 26.39 - 4 30 24.41 10. - 4 31 24.41 H-67°51′ 6′′ By Prob. XV., we have log. p cos φ' = 9.78888, log. p sin o' = 9.89538. p-u=+.1708 log. 9.23249 n F=118° 53′0′′ log. cos 9.68397 n N+F=230°37′ 10′′ log. cos 9.80241 n p - U log. 9.23249 n - p' Ar. Co. " 0.27819 t'-.3241 9.51068 n 9.51597 n To find the Heliocentric Longitude and Latitude, and the Radius Vector of Mercury, for a given time. Reduce the given time to mean time at Greenwich. Take from table LXIX., the mean longitude of Mercury, the longitudes of the aphelion and node, and the arguments II. and III., corresponding to the given year. Under the three former, place the motions for the months, days, hours, minutes, and seconds of the given time, taken from tables LXX. to LXXII.; and under each of the latter, place the number D, in table LXX., corresponding to the given month, and also the number expressing the day of the month, diminished by a unit. Add together the quantities in each column, rejecting 12 signs when either sum in the first three columns admits the rejection, but setting down the whole amount in each of the last two columns. Subtract the resulting longitude of the aphelion from the mean longitude of the planet, and the remainder will be the mean anomaly. With the mean anomaly as the argument, take the equation of the centre from table LXXIII., and applying it according to its signs to the mean longitude, add to the result the equations II. and III., taken from table LXXIV., with their respective arguments. The sum will be the orbit longitude of Mercury. From the orbit longitude subtract the longitude of the node, and the remainder will be the argument for the latitude; it will also be the argument for the reduction to the ecliptic. With this argument take the reduction from table LXXVI., and apply it according to its sign to the orbit longitude. The result will be the heliocentric ecliptic longitude, reckoned from the mean equinox. With the argument N, found from the solar tables for the given time, take the nutation in longitude from table XXX., and apply it according to its sign to the longitude from the mean equinox, and it will give the longitude from the true equinox. With the argument of latitude take, from table LXXVIII. the latitude, and also its secular variations. Multiply the secular variation by the number of years, the given time is subsequent to 1800,-and divide the product by 100. The result added to the latitude taken from the table, will give the correct heliocentric latitude. With the mean anomaly as the argument, take the radius vector from table LXXV. EXAM. 1. Required the heliocentric longitude and latitude, and the radius vector of Mercury, on the 13th of June, 1836, at 20h. 12m. 30sec. mean time at Greenwich. 2. Required the heliocentric longitude and latitude, and the radius vector of Mercury, on the 17th of November, 1837, at 11h. 29m. 20sec. mean time at Philadelphia. Ans. Long. 221° 47′ 1"; Lat. 0° 33′55′′N. Rad. Vect. 0.44780. 1. PROBLEM XXIV. The heliocentric longitude and latitude, and the radius vector of Mercury at a given time being given, to find its geocentric longitude and latitude and its horizontal parallax and semidiameter at that time. For the Geocentric Longitude. To the sun's longitude at the given time, found by Prob. VI., add 180° 0′ 20′′, and the sum will be the earth's longitude at the time.* Find the earth's radius vector, by adding to the radius vector taken from table XXXI., with the sun's mean anomaly as the argument, the perturbations taken from the small table on the same page, with the arguments I., II., * The sun's longitude found from the tables is the apparent longitude as affected by aberration, and is therefore 20" less than true longitude. and III. Subtract the longitude of the earth from the heliocentric longitude of the planet; the remainder, if less than 180°, will be the angle of commutation, to be marked west; but, if the remainder is greater than 180°, its supplement to 360° will be the angle of commutation, to be marked east. Take half the angle of commutation, and subtracting it from 90°, call the remainder A. Add together the log. cosine of the planet's heliocentric latitude, the logarithm of its radius vector, and the arithmetical complement of the logarithm of the earth's radius vector, rejecting the tens from the index of the sum, and the result will be the log. tangent of the arc B. To the log. tangent of the difference between the arc B. and 45° add the log. tangent of the arc A, rejecting ten from the index of the sum, and the result will be the log. tangent of an arc C. Subtracting C from A, the remainder will be the angle of elongation, of the same name as the angle of commutation. If the angle of elongation is east, add it to the sun's longitude increased by 20"; but if it is west, subtract it from the sun's longitude thus increased; and the sum or remainder will be the true geocentric longitude. Add together the arcs A and C, and the sum will be the annual parallax. With the elongation, annual parallax, and geocentric latitude as arguments, find the aberration in longitude from table LXXIX., and, applying it to the true longitude, the result will be the apparent longitude. For the Geocentric Latitude. Add together the log. tangent of the heliocentric latitude, the log. sine of the elongation, and the arithmetical complement of the log. sine of the commutation, rejecting ten from the index of the sum, and the result will be the log. tangent of the true geocentric latitude, which will be of the same name as the heliocentric latitude. With the angle of elongation increased by 270°, and the annual parallax and geocentric longitude, each increased by 90°, as arguments, take from table LXXIX., parts I., II., and III., respectively, of the aberration in longitude, and add them together, having regard to their signs. Multiply the sum by the multiplier, taken from a small table at the bottom of page 74, and the product will be the first three parts of the aberration in latitude. Take from the other small table on the same page, part IV. of the aberration in latitude, and add it to the former three, attending to the |