t x = a + t (b + } c) + ' (t − 1 ) (c + d + į e) + (d + & e) + 2 Performing the multiplications indicated, and reducing, we have for the value x, at the time T + t, x = a + ib + c + 1 (1o − 1 ) d + t2 1) tb 2 6 24 (C) (D) 13 6 The interval t, may be taken of any value between 2 and +2. 63. The hourly variation of a quantity at any time, is the variation or change in its value which would take place in an hour, if the rate of change at that time were to be the same, throughout the hour. The average hourly variation of a quantity between two times T and T+t, is that hourly variation which, if continued during the interval t, would produce the change in the value of the quantity, that actually takes place during this interval. 64. Average hourly variation. It follows, from the definition, that the difference between the values of a quantity at the times T and T + t, divided by the interval t, must give the average hourly variation during the interval. Hence, if we put this average hourly variation, we have, from the formula (D), x' The average hourly variation or the values of x for the whole hours, are, 65. Hourly variation. If in the formula (D) we put tť instead From this value of x, subtracting its value at the time T + † (D), and dividing the remainder by t', we find the average hourly variation between the times Tt, and T + t + t', to be, Now, it is evident that the smaller the interval tis, the nearer will this average hourly variation approach to the hourly variation at the time T + t. Hence, if we now take x' to stand for the hourly variation at the time Tt, we have, by taking t = 0, in the above expression, ť The hourly variations, or the values of x at the whole hours, are, T2b2c+d-Ze (G) INVESTIGATION OF FORMULÆ FOR COMPUTING SOLAR ECLIPSES, OCCULTATIONS, AND TRANSITS. 66. Let O, Fig. 64, be the centre of the earth, A a place on its surface, S the centre of the sun, M that of the moon, and S', M', A', s, and m, the points in which OS, OM, OA, AS, and AM, produced, meet the celestial sphere. Then will S' and M' be the true places of the sun and moon, s and m, their apparent places, and A', the geocentric zenith of the place A. Let a be the zenith of the place A, OZ a straight line parallel to MS, meeting the celestial sphere in Z, EQ the equator, E the vernal equinox, P the north pole of the equator, and PB, PC, PF, and PK, declination circles through Z, S', M', and a and A'. Also, let BX and ZY be each a quadrant. Then, since BX is a quadrant, X is the pole of the declination circle YB; and, consequently, OX is perpendicular to OY and OZ. Also, since ZY is a quadrant, OY is perpendicular to OZ. Hence OX, OY, and OZ form a system of rectangular axes, having their origin at 0, the centre of the earth, and having the axis OZ parallel to MS, the line joining the centres of the moon and sun. distance of place A from earth's centre, 66 the moon from " sun's mean distance, distance between the centres of the moon and sun, right ascension of the moon, 66 66 68. To find the values of a, d, and g. Let EX' be a quadrant. Then OX', OP, and OE will evidently be another system of rectangular axes, having the same origin as the former system. On OZ, take OL MS, and let MG, SH, and LI be perpendicular to EOQ, the plane of the equator, meeting it in G, H, and I. Also let GU, HV, and IW be perpendicular to the axis OE, and let GN be parallel to it. As MS and OL are parallel and equal, their projections GH and OI are parallel and equal. Hence, as GN is parallel to OW, the right angle triangles GNH and OWI, are equal, and we have OW GN UV = OV OU. Consequently, that ordinate of the point L, which is parallel to the axis OE, is equal to the difference between the ordinates of S and M, parallel to the same axis. The same relation must evidently have place for the ordinates parallel to the axes OX' and OP. Hence, if we put a, a', and a" for the ordinates of M, S, and L, parallel to OX'; ß, p', and ß", for those parallel to OP; and y, y', and y", for those parallel to OE, we shall have, a" — a' a; ß" ß; and y" = y' Now, since OL MS= G, we have OI = OL cos BOZ G cos d; and, consequently, y′′ LI OL sin BOZ OW OI cos EOB — G cos d cos a. Also 3′′ Ꮐ cos d sin a. The co-ordinates of S and M will evidently have similar expressions. Hence, we have, R' cos D' cos A' R cos D cos A Consequently, G cos d cos a = G cos d sin a = R' cos D' sin A'R cos D sin A G sin d R' sin D' R sin D. Multiplying the first of these last three equations by cos A', and the second by sin A', and adding the products; and multiplying the first, by sin A', and the second by cos A', and subtracting the first product from the second, we obtain, These formulæ for determining the values of a, d, and g, are perfectly rigorous. But since r is always a small quantity, and since, at the time of an eclipse, A' is nearly equal to A, and D' to D, it is evident that a must be very nearly equal to A', and d to D'. Taking this into view, we may deduce from (C), the following more simple expressions, which will give the values of a and d, true to a small fraction of a second, and the value of g with corresponding precision. 69. To find the values of x, y, and z; and x", y", and z". Let Z and M' be joined by the arc of a great circle. Then, in the spherical triangle ZPM', we have (App. 34), cos ZM' or, cos ZOM' = cos M'P cos ZP + sin M'P sin ZP cos M'PZ, The right ascensions of X and Y are, a +90°, and a + 180°, and their declinations are, 0°, and 90° d. These right ascensions and declinations being substituted for a and d, in the expression for cos ZOM', will evidently give the expression for cos XOM' and cos YOM'. We shall thus have, COS XOM' cos D sin (A sin D cos d COS ZOM' a) cos D sin d cos (A —— a) sin D sin d + cos D cos d cos (A — a). If perpendiculars be let fall from the point M, on the axes OX, OY, and OZ, three right angled plane triangles will be formed, the common hypotenuse of which will be OM R 1 sin π and the bases will be |