E7°.32 H. at beg. centr. eclipse M' 21°26′ log. tang 9.5938n H, at end centr. eclipse M Then (Note), o' 51° 4' G = 17° .30 L': 50 39 +109° 50 +42 7 & +B+C=8° 12′ N., log. E 0.9735 L-S 0.1277 log. cos 9.7982 81° 13' log. sin 9.9949 log. G' 1.4450 1.2381 +42 G 27°.25 λH" EGL +97°.59 97° 35′ W. Assume L + 50°. Then (Note), B = + 21° 26′, C = + 29° 38′, p′ =51° 4′ N., ? : 51° 15'. & + 51° 4' log. sin 9.8909 8644 L 92° 46 7 50°. B 1. 31° 11' W. C G 0°.37 73°.50 29° 38′ log. sin 9.6941 8° 15' N. Assume L +130°. 51° 4' N., 51° 15' N., E 7°.32. = log. cos 9.9689 log. C′ 9.7252 log. cos 9.7982 S=1° 13' log. sin 8.3270 log. G' 1.4450 9.5702 31°.88 1.4354n 31° 53' E. By assuming for L various other values between its limits L' and L”, the latitudes and longitudes of a series of places at which the eclipse will be central, as given in the following table, may easily be found. The computation of a part of these may serve as an exercise for the student. L. 50° 39' 50 0 0 0 0 0 0 0 0 0 40 0 50 0 60 0 70 + 80 + 90 40 30 20 10 0 + 10 + 20 + 30 0 0 +100 0 +110 0 +120 0 +140 0 +150 0 +151 57 Lat. 8° 8 12 16 21 44 27 0 32 15 37 12 41 39 45 31 48 43 51 15 53 9 54 29 55 16 53 31 55 16 54 29 53 9 51 15 48 43 45 31 44 50 2' N. 15 9 42 PROBLEM XVIII. The value of a quantity at three consecutive whole hours, T — 1, T, and T+1, being given, to find its value at an intermediate time T", and its hourly variation at that time. Attending to the signs, subtract the value of the quantity at the time T 1, from its value at the time T; and its value at the time T, from its value at the time T + 1; and the remainders will be the first differences. Subtract the first of these from the second, and the remainder will be the second difference. Let a = the value of the quantity at the time T; 6 the half sum of the first differences; c= the second difference; and t the interval between T and T', expressed in the fraction of an hour, and marked negative when T' is earlier than T. Then the value of the quantity at the time T', will be And the hourly variation of the quantity at the time T, will be bt. c. с D EXAMPLE. Given the moon's declination, on a certain day, as follows: at 10h., +15° 58′ 50′′.1; at 11 h., D 15° 47′ 11′′.0; and at 12 h., 15° D: 35' 27".1. Required its value at 10g h. 10 11 12 b t. c a+t. b + +15° 47′ 11.0", 4 40.6 0.4 15 51 51.2, at time T. 11' 41.5" 1.9 11 39.6 D +15° 58' 50.1" 15 47 11.0 15 35 27.1 + to 2 IN C. 11' 41.5", c 2nd diff. 4.8" 4.8", t hourly variation at time T. PROBLEM XIX. To calculate an Eclipse of the Sun for a given place, obtaining the places, &c., of the moon and sun, from the Nautical Almanac. 1. Let A and A' represent the right ascensions of the moon and sun, respectively, D and D', their declinations, marked negative when south, я and ' their equatorial horizontal parallaxes, and 8 the sun's apparent semidiameter. Also let T be the Greenwich mean time of new moon, taken to the nearest whole hour; other contiguous whole hours, earlier and later being denoted by T1, T2, &c., and T+ 1, T2, &c. For the times T 2, T 1, T, T + 1, and T + 2, find the values of A, A', D, D', π, and S. And for the time T find the value of ', which may be regarded as constant during the eclipse. Also, for the times T2, T1, &c., find the log. tang ', log. sin D', log. cos D, log. A log. sin D' + log. tang s', and log. Blog, cos D' + log. tang s'. Arrange the quantities thus found in tables opposite the hours, as in the following example; and find a quantity g', from the formula. log. g' Ar. Co. log. sin (') log. sin 9.4353665. + л + For the times, T2, T1, &c. find by the following formulæ the quantities p, q, l, and l'. [log. sin (D — D'), log. p = Ar. Co. log. sin (π —— π') + log. sin (A — A') + log. cos D; log. a Ar. Co. log. sin (π — T') = log. b = log. p + log. sin (A — A') + log. sin D'; q = a + 1⁄2 b. [ + log. tang d'; log.dlog.p+ log. sin (A A') + log. cos D'; log. c Ar. Co. log. sin (π —π') + log cos. (D — D') l = c — } d + g' ; l' = c — } d — g'. find by the last The value of 7 or p From the values of p and q for the whole hours, problem, their values for the intermediate half hours. l', at each half hour may be taken equal to the half sum of its values at the preceding and following whole hours. Arrange the values thus found in a table, placing in columns adjacent to those containing the values of and q, the differences of their values for each half hour. The values of p p and L for any intermediate time may then be easily obtained by proportion. The values of p' and ', at each of the times T1, T, and T + 1, are respectively equal to the sums of the preceding and following differences. Their values at the times T 2 and T+2 become known from their hourly changes of value. Take from the Nautical Almanac, the sidereal time at mean noon at Greenwich, on the day of the eclipse. To this apply the sidereal time corresponding to the interval between Greenwich mean noon and the time T, by adding or subtracting, according as T is after or before noon. The result, converted into degrees, will be Z', the right ascension of the zenith of Greenwich, at the time T. All the preceding quantities are independent of the given place. 2. To the value of Z' or from it, add or subtract the longitude of the given place, according as it is east or west, and the sum or remainder will be Z, the right ascension of the zenith of the given place at the time T. From Z, subtract the value of A' at the time T, and the remainder will be H, the hour angle at the given place at the time T. Take the difference between the values of A' at the times T and T+1, and call it A”. Then the value of H at another time T' may be found by adding to its value at the time T, (T′ — T) × 15° + (T′ − T) × (2′ 28′′′ — A′). When A" does not differ more than two or three seconds from 2′ 28′′ the second product may be omitted without material error. Having found for the given place, the logarithms p cos p' and p sin ', by Prob. XV., the values of u, v, u', v', and h, may be found by the following formulæ : |