Reducing to Philadelphia, mean time, we have, h. m. Beginning of Eclipse at 7 3.0, A. M. 8 13.6 End 9 32.9 Digits eclipsed 7.9, on south limb. 66 66. Eclipse begins 85° 6', from vertex to the right To construct a figure representing the eclipse at the time of greatest obscuration, and showing the points of beginning and end. left. With a radius 6, taken from a scale of equal parts, describe a circle VBE, Fig. 67, to represent the sun's disc; then, taking a point V, at the top, to represent the vertex, draw the vertical diameter VV'. Make VB, or the angle VSB, equal to the value of the arc V, found for the time of beginning, and B will be the point at which the eclipse commences. Make the arc VBE, equal to the value of V at the end of the eclipse, or, which is the same, make the angle VSE, equal to its supplement to 360°, and E will be the point at which the eclipse ends. Make the arc VG equal to the value of the arc V, at the time of greatest obscuration, and drawing the diameter GG', make GD equal to the digits eclipsed, taken from the same scale. Then, as h 2732 : 2732 :: 6: a fourth term. Take DM equal to this fourth term, and with the centre M, and radius MD, describe the arc aDb. Then will aDbGa represent the quantity and position of the part eclipsed at the time of greatest obscuration. EXAM. 2. It is required to calculate for Philadelphia, an eclipse of the sun that occurred in May, 1854. h. m. 4 10.8 P. M. mean time Ans. Beginning at .. Ends 66 66 " 66 PROBLEM XVII. To find a series of places at which an eclipse of the sun will be central. 1. Take for p, q, p', and qʻ the 10000th parts of the values of these quantities, as found by the last problem, for the time T, and for D' its value at that time. To log. of q', add Ar. Co. log. of p', and the sum will be the log. cotangent of an affirmative arc N, less than 180°. To log. cot N add log. of p, and the sum will be the logarithm of a quantity d. To log. sin N add log. of (d q), and the sum* will be the log. cosine of an affirmative arc F, less than 180°. Find the intervals t, t, and t" from the formulæ, log. of t log. cos (NF) Ar. Co. log. of p'; log. of tlog. cos (NF)+ Ar. Co. log. of p'; and log. of t=log. of p+ Ar. Co. log. of p'. Then, Tť"' + 1⁄2 (t + t) will be T', the time of the middle of the central eclipse for the earth in general; T-t, will be the time of beginning; and T-t''t', will be the time of end. 2. To the value of H at the time T, found by the last problem, add (T′ T) .15°, and the sum will be the value of H' at the time T'. From this value of H', and to it, subtract and add, 1⁄2 (ť — t). 15°, and the remainder and sum will be the values of H' at the beginning and end of the central eclipse. 3. To log. sin (N+F+180°), add log. cos D', and the sum will be the log. sine of the geocentric latitude ', not exceeding 90°, and north or south, according as the sine is affirmative or negative. To q', add the reduction of latitude, taken from table XVI., with p' as the argument, and the sum will be, the latitude of the place at which the eclipse begins to be central. To log. cot (NF) add Ar. Co. log. sin D', and the sum will be the log. tangent of an hour angle H, less than 180°, and affirmative when (NF) is in the first or fourth quadrant, but negative when it is in the second or third. When H is to be negative, it must be taken greater than 90°, if its tangent is affirmative, but less than 90° if the tangent is negative. Subtract H from the value of H' at the beginning of the central eclipse, and the result will be 2, the longitude of the place at which the eclipse begins to be central. It will be west if affirmative, but east if negative. Using (NF) instead of (NF), and taking the value of H' at the * When this sum, after one 10 has been rejected from the index, is greater than 10, which is the greatest log. cosine, the eclipse cannot be central at any place. end of the central eclipse, we find, in like manner, the latitude and longitude of the place at which the eclipse ceases to be central. 4. To log. sin D', add log. tang N, and the sum will be the log. tangent of an arc M' not exceeding 180°, and with the same sign as D'. When M' is to be negative, it must be taken greater than 90°, if its tangent is affirmative, but less than 90°, if the tangent is negative. To log. sin D', add log. cot. N, and the sum will be the log. tangent of an arc N', less than 90°, and with the same sign as its tangent. Then take SM'+N'. Find log. tang D'Ar. Co. log. sin M' log. cos F+Ar. Co. log. cos D' + Ar. Co. log. sin N log. cos D'+ log. sin N+ log. cos N + log. of 15+ Ar. Co. log. of p'. log. G'log. sin N+ log. sin N + Ar. Co. log. cos N' log. of 15+ Ar. Co. log. of p'. log. B' log. C' log. E' 5. Add M' to the value of H' at the middle of the central eclipse, and call the sum H". Add M' to the value of H at the beginning of the central eclipse and call the sum L'; also add M' to the value of H at the end of the central eclipse and call the sum L". The quantities found in the last article and this, may all be regarded as constant. 6. Let L be the value of (H+ M') at any time during the central eclipse. Then, assuming for L any value at pleasure within its limits L and L", to log. sin L add log. B', and the sum will be the log. tangent of an arc B, not exceeding 90°, and with the same sign as its tangent. To log. cos B add log. C', and the sum will be the log. sine of an arc C less than 90° and with the same sign as its sine. Then will B + C be the value of ', the geocentric latitude of the place at which the eclipse is central when L has its assumed value. From ' we find ø, as in Art. 3. Add together log. sin p' and log. E', and the sum will be the logarithm of an arc E, in degrees and decimals of a degree. ', log. sin (Lanother arc G. Add together log. cos S), and log. G', and the sum will be the logarithm of Then will H" + E + G L, be λ, the longitude of the place. We may thus, by assuming a series of values for L within its limits L' and L", find a series of places at which the eclipse will be central. Note 1. When two assumed values of L differ only in the sign, the corresponding values of B will differ only in the sign, and the values of C will be precisely the same. When two assumed values of L are supplements of each other, the latitudes and values of E will be the same for each. By attention to these circumstances, the computation of a series of places may be considerably shortened. 2. If we subtract M' from any assumed value of L, the remainder will be the hour angle at the place at which the eclipse is then central. Required the times and places at which the eclipse of May, 1836, began and ceased to be central; and also a series of other places at which it was central. d. h. m. From the calculation of the last problem, we have T 15 2 0, D +18° 57′.7, and H' + 30° 59′; also dividing the values of p and at the time T, and the values of p' and q', by 10000, we have, p .1728, q.4403, p' + .4800, and q Չ + .1725. log. 9.2367 A. C." 0.3188 2.061 N+F188° 27' log.cos 9.9953 n 0.3141 n EXAMPLE. H', at time T, 15° X .027 N d h. = T = T — ¿'' + } (tt) = 2.027 = ť' + t 0.299 3.755 ť" + ť q F118° 13' ..3600 log. sin 9.9736 .5024 log. 9.7011 n log. cos 9.6747 n 1.395 A. C. " 0.3188 9.5563 n time of middle of central eclipse. 66 (6 (C beginning H', at middle of centr. eclipse H', at beginning of centr. eclipse + 5 28 +57 18 47° 59' log.cos 9.8257 A. C. log. 0.3188 0.1445 + 30° 59' 24 log. 9.2375 n +31 23 |