obscur. T―t' = 1.22 = approx. time of gr. For true time of greatest obscuration and approximate times of beginning p-u p'-u' t"-0.241 log. 2.9465 T'-t′′ + } (l+1') = 1.238=114.3=true time of gr. obs. A. C. 6.4349 T'—"+t= 0.005= approx. time of beg. V = 270°+Q-(N+F) 85° 6' V 270°+Q-(N-F) = 235° 45′ — = Eclipse begins 85° 6', from vertex to the right Greatest obscur. 157 49, Eclipse ends 66 66 124 15, 66 left. To construct a figure representing the eclipse at the time of greatest obscuration, and showing the points of beginning and end. 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 will occur in May, 1854. To find a series of places at which an eclipse of the sun will be central. 1. Take for p, q, p' and q', 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 cotangent of an affirmative arc N, less than 180°. To cot. N, add log of p, and the sum will be the logarithm of a quantitity d. To sin N, add log of (d-q), and the sum* will be the cosine of an affirmative arc F, less than 180°. Find the intervals t, t' and t" from the formula, log. of t = cos (N + F) Ar. Co. log. of p'; log. of t' = cos (N — F) + Ar. Co. log. of p'; and log. of t"=log. of p + Ar. Co. log. of p'. Then, T- t" + } (t + t') will be *When this sum, after one 10, has been rejected from the index, is greater than 10, which is the greatest cosine, the eclipse cannot be central at any place. |