Now, in the coefficient of f, if K be taken to denote (b −c) (c − a) (a−b), the coefficients of aw, Bw, yw, are, as may be easily ascertained, equal respectively to 27 γω and, in the coefficient of -g, the coefficients of aw,, Bw, yw as may be easily shewn, are respectively equal to KKK Thus the condition for an axis of kick is reduced to the equation az (af − g3) w, + bB (bƒ − g2) w ̧ + cy (cf — g3) w ̧ = 0. But the quantities w1, w, w, are connected together by the equation 2 2 a (aƒ — g2) w,2 + b (bƒ − g3) w," + c (cf — g2) w,” = 0. From these two results it follows that the intersections, when possible, of the axial cone a (af-g") x + b (bf — g3) y2 + c (cf — g3) z2 = 0, with the plane aa (af-g") x+bB (bf-g) y+cy (cf-g3) z = 0, are axes of kick, two in number. In order that the coexistence of these two equations may be possible, we must, as may be easily proved, have the condition (af – g*) (bf_g) (cf-g?) (co (af- g*)+b3 (bf-g*) +cy" (cf — g')} = or <0. From this condition it is evident that, when there is an axis of kick, the centre of gravity of the body does not lie within or without the axial cone, when the sign of (aƒ — g3) (bf — g3) (cf — g3) is respectively positive or negative. If af- g=0, then the equations to the axis of kick become b (b − a) y2 + c (c − a) z2 = 0, BB (b − a) y+c(c−a) ~ =0, and therefore y=0, z=0: thus the axis of kick coincides with the principal axis of greatest moment of inertia. 2 2 From the expressions for X, Y, Z, it may be easily seen that, w, w, being both zero, and w, equal to w2, when af=g, the equations to the reaction of the fixed point on the body are If cf-g=0, then it may be shewn in like manner that the axis of kick coincides with the principal axis of least moment of inertia, the equations to the direction of the reaction being If bf - g2 = 0, then the equations to the axis of kick are a (a - b) x = c(b− c) z2, aa (a−b) x = cy (bc) z. Thus the axis of kick lies anywhere in the plane denoted by the second of these equations, a condition for its existence being az2 (a - b) = cy" (b − c), which shews that the centre of gravity lies in this plane. In this case therefore the instantaneous axis is an axis of kick during its whole motion in its plane locus. July 20, 1870. ON THE ALGEBRA OF MAGIC SQUARES. No. II. By JOSEPH HORNER, M.A., Clare College. To prove that every odd zero arithmetical progression of po terms, where p =2r+1 and σ = 2s+1, may be separated into a zero columns of P terms each. This depends on the fact, that the equation R ̧x + R。 (s + x) + Ro (s - 2x + 1) = 0 is true for all integer values of x. Giving the σ values. -8, -8 +1,... 8-1, s, we obtain the following, a columns of zero triplets in different orders. There appears to be no other system of zero triplets, which can be formed by taking any odd zero series whatever three times in this manner. For some particular values of s however, there are other possible arrangements not comprised in the general formula above, and that seem to depend on complex laws to which I possess no clue. Thus when s=3, we have the symmetrical arrange ment 3210123 3 2 1 0 1 2 3 1 3 2 0 2 3 1 2 1 3 0 3 12. Also when 84, there are the following forms, one symmetrical, the other unsymmetrical, These two particular cases were suggested to me by William, only surviving child of the late W. G. Horner. Having formed the a zero triplets, we take any pair of the three rows, diminish each term of the one by σ, and increase each term of the other by that quantity. Thus we still have a zero triplets. For example, Above or below these we may write in rows the pairs of - 2rs-r-s, - 2rs-r-s+1, ... - 2rs - r+s the rows being in any order, and the terms of each arithmetical series in any order, provided the equal contrasignal terms of each pair be in the same column. Thus the problem is solved in various ways. Otherwise, we take any pair of the original three rows and diminish each term of the one by 20 or 30 or ..., and increase each term of the other by that same quantity; then write above or below these, with the licence just mentioned as to rows and terms, the arithmetical progressions or..., as the may be. Thus we obtain various other Ex. r=2, 8=2. The original rows are Reverting to the general arrangement of a zero columns of p terms each, and putting N-po, that is 2n+1=2+1.2s+1, and therefore n = 2rs+r+s, we ask attention to the fact that there is in the system one row consisting of all the terms that are greater than n − σ or 2rs + r− s −1; two rows consisting of all the terms greater than n-2σ or 2rs + r−3s-2; three rows of all that are greater than n-3σ; and so forth. e.g. In the example above, there is one row consisting of all the terms greater than 12-5 or 7 and two consisting of those which are greater than 12 - 10 or 2. - It follows, that every column contains only one number greater than n-o, two greater than n-20, three greater than n 30, and so on. Suppose any one of these zero columns arranged in descending order to be v1, v, V...V, and that we increase each of these terms by σ and then diminish v1o by N or po. The results still make a zero scries. The same is true if we increase each v by 2σ and then 17 |