responding to the interval between noon and the time T, taken from table X. From the sum subtract A', the star's apparent right ascension, and, converting the remainder into degrees, it will be the value of H', at the time T. To this apply the longitude of the given place by adding when it is east, but subtracting when it is west, and the result will be the value of H, at the time T. The value of H at any other time T', may be found by applying to its value at the time T, the change in the right ascension of the zenith during the interval between T and T', taken from table XIV.; adding if T′ is later than T, but subtracting if it is earlier. 5. Having found the logarithms of p cos p' and p sin ', for the given place, by Prob. XV., find fand log. G, from the expressions log. ƒ = log. p sin log. cos D', and log. Glog. p cos + log. sin D'. 6. Taking the values of p, q, and H, at the time T, find the quantities u, v, &c., by the following formulä. log. at N log. (qv) + Ar. Co. log. (p'u'), log. log. cot N + log. (p—u); log. cos F=log. sin N+log. (d+v−q) + 0.5646 N and F both to be less than 180°. Jag, tlog, cos (NF)+ 9.4354 Ar. Co. log. (p'u'); ť log. t log. cos (NF) 9.4354+ Ar. Co. log. (p'-'); log. "log. (pu) Ar. Co. log. (p'u'). 7. Taking T' to stand for the approximate time of immersion, find p, 9, and H, for this time; and proceeding as in the last article, find tand t”, omitting t. Then T-tt, will be the time of immersion, very nearly. In like manner, finding, for the approximate time of emersion, t' and ', we have T' - tt for the time of emersion, very nearly. 8. With the values of u and v at the approximate time of immersion, find Q and V, as in Prob. XVI., art. 12. The arc V will designate the place of immersion, in reference to the moon's vertex, as seen through a telescope that inverts. A similar process will give the place of emersion EXAMPLE. Let it be required to calculate for Greenwich, latitude 51° 28′ 39′′ N., the occultation of i, Leonis, of Jan. 7th, 1836, having given the following data taken from the Nautical Almanac. Sidereal time at mean moon, 19h. 4m. 22.41sec. 10 25 3.92 14°58′ 38.8" 15 35 27.1 15 23 38.4 56 3.5 For Greenwich, log. p cos p' 9.79526, and log. p sin o' 9.89139 N+F=236° 56' log.cos 9.7369 n NF-9° 46′ log. cos 9.9937 log. 9.5623 T-t"+t=10.78=approx. time of immersion. puAr.Co." 0.3768 T-t"+t=11.77= emersion. PROBLEM XXII. To find the longitude of a place from an observed occultation of a fixed star by the moon. 1. Let A, A', &c., be as in the last problem, and .2725. Using the estimated longitude of the place, reduce the observed mean time of immersion to Greenwich time. Let T stand for this time, and T′ for the same time taken to the nearest tenth of an hour. From the Nautical Almanac, find, for the time T', by problem XVIII., the values of A, D, A", D", and, by proportion, the value of ; and also take out the values of A', D', and the sidereal time of mean noon. 2. With the values of A, D, &c., at the time T', find the values of p, q, p', and q', from the following formulæ.† 3. To the sidereal time at mean noon add the sidereal time corresponding to the interval that T is past noon, taken from table X., and from the sum subtract A'. To the remainder apply the longitude of the place in time, by adding if it is east, but subtracting if it is west, and, converting the result into degrees, it will be H, the star's hour angle at the observed time of immersion. * As the values of A and D are given in the almanac for every hour, the values at the time T' may be obtained, nearly, by proportion; and the values of A" and D" by taking the differences between the values of A and D, respectively, at the preceding and following hours. But it is more accurate and but little additional trouble to employ the problem for interpolation. † By differentiating the expressions for p and q (App. 75), we obtain the following expressions, very nearly, which are in accordance with the rule: |