THEOREM. From a relation of weights (m + n) R = mP+nQ we may infer one of twists (m + n) | R=m\P + n| Q. In fig. 2 we suppose OPQ to be the plane of the paper meeting the sphere in the circle C and P,Q in the lines LT, TM. The planes P, Q meet in a line P,Q perpendicular to the paper along which unit vector u is taken towards the reader. Let e denote the reciprocal of a point with respect to the circle. Then | PuP,│Q = u│cQ · But (m+n) c R = m│c P+n│cQ · Multiplying by u the theorem follows. The reciprocal of a couple P-Q may be shown to be a twist through O of intensity QP in a plane perpendicular to PQ and in the proper sense. SALMON's theorem again gives P│Q = Q│P. Coming to Regressive Multiplication I define the product of two twists as a force along their line of intersection, in the proper sense (See page 444 of the Book of the Heidelberg Congress), whose intensity is the product of the measures of the twists and the sine of the angle between their planes in the order of factors (thus αβ = βα). Here, P. Q = a forces along u reversed whose intensity is OP.OQ sinPOQ i. e. moment of PQ about O =PQ by above definition. It will be found by considering various positions of P, Q inside and outside sphere that the signs agree. (where | A has the meaning of this paper and is not equal to BCD). Now P, Qr=0 requires the coefficients of AB, AC etc. to vanish separately, PrQr because no forces along the edges of a tetrahedron can balance, therefore also Pr Qr=0. The theorem admits of independent proof. Let AB be one of a system of balancing forces and let it be replaced by an equal and parallel force OP together with a couple OAB. We have then in forces ΣΟΡ=0. Now let ONR be perpendicular to AB and AB, then Hence OP may be taken as the axis of this couple, and the first of the above. conditions then shows that 20|AB=0. Again | AB translated to O may be taken as the axis of tfie couple OAB; the second condition then shows that forces AB translated to O would balance. The two results show that ΣΑΒ = 0 . As a simple case it may be noticed that to every theorem concerning forces in one plane there is a reciprocal about forces meeting in a point P the pole of π; and if the first set be equivalent to a couple, the second set will reduce to a force perpendicular to л. Also to any theorem concerning forces acting along tangents to a sphere will correspond a theorem concerning forces respectively equal to them but turned through a right angle in the tangent planes containing them. The equation in space corresponding to (6) in plano is PQRS Q│PRS+R | PQSS | PQR = 0 or, if P', Q', R', S' be the poles, a, ẞ, y, d the tetrahedral co-ordinates of the centre O with respect to the tetrahedron PQRS. a PP' + ¿QQ'+y RR'+d SS' = 0. Mutatis mutandis, we have reciprocally a' P' P + ß' Q' Q + y' R' R + ƒ′ S′S = 0 or, if we understand by a the ratio of OQRS to PQRS, etc., then a a' etc. But we may derive another interpretation. By LEIBNITZ' theorem, O being the centroid of weights a,,y,d at P, Q, R, S. a PO+ẞQO+7RO+SSO = 0 Hence a P(P'- O) + ẞ Q(Q' — O)+ etc. = 0. P(PO) is a force through P parallel and equal to OP', the magnitude being equal to 1 the distance of O from the plane QRS, etc. We return thus to the old theorem that forces through P, Q, R, S perpendicular and proportional to the opposite faces are in equilibrium. A. MACFARLANE ON THE SQUARE OF HAMILTON'S DELTA In JoLy's Manual, the most recent treatise on the quaternion analysis, it is assumed, but nowhere proved, that 2 is a scalar operator. In this paper I propose to show that what has hitherto been taken for the square of is only the scalar part of the complete square. " This investigation was suggested by the more simple and logically prior one " » What is the complete square of a sum of vectors? Let, as is now usual, a, b, c denote three simple vectors; that is, vectors of the vector-analysts, not the right quaternions of HAMILTON (1). Their sum is denoted by a+b+c; their order in this sum may indicate a real order, in which case commutation is not indifferent. Let the square be denoted by (a + b + c)2. In works on vector-analysis it is assumed that this square is formed by writing one trinomial after the other, and then taking the several partial products, always preserving in the partial product the order in which the symbols appear in the trinomial factors. But That is Hence according to the above process of forming the square (a + b + c)2 = a2 + b2 + c2 + 2 } Sab + Sac + Sbc {. (1) To indicate clearly that a,b,c are vectors, some authors print them in fat Italic type. The three vector terms Vab, Vac, Vbc do not appear: they have been eliminated by the manner of effecting the multiplication. But the mode of multiplying is incorrect, for it destroys the natural order of the terms in the trinomial; for instance a is prior to b, but in one partial product a is placed after b. But the square can be formed so as to preserve the natural order, as follows: (a + b + c)2 = a2 + b2 + c2 + 2 (ab + ac+bc) a2 + b2 + c2 + 2 (Sab + Sac + Sbc) +21−1 (Vab +Vac + Vbc). That this is the correct square is proved by the fact that with it the exponential theorem remains true; whereas with the other form of square it does not. The exponentials eva, evib, eVic denote angles in space; and is true, provided the square and other powers are formed on the above principle, and not otherwise. The common way of forming the square of the sum of three vectors gives merely the scalar part of the complete square. Does a similar distinction hold in forming the square of HAMILTON'S operator V, which is well known to be a kind of symbolic vector? where () denotes the place for the function; and 2 is taken to be But the vector part has been cancelled by an artificial order of the factors, just as in the trinomial of vectors. When it is restored, the complete square has the additional term In the case of polar co-ordinates, the square of is considered to be |