It is clear that the rectilinear generators of any one system are all parallel to one plane, viz. the corresponding cyclic plane, and therefore the cyclic planes through the axis of the parabolic principal sections become asymptotic planes, to which the corresponding asymptotic cone in the hyperboloid degenerates. It is evident that any rectilinear generator does not cut any of its own system but that it does all of the other system. We may then generate the surface as follows. Take any two fixed lines not intersecting, and let a line move so as to always intersect them and to be parallel to a fixed plane; the locus of the line is an hyperbolic paraboloid, of which the two fixed lines are generators of the other system. A CONTINUED FRACTION FOR tannx. By J. W. L. Glaisher, B.A. I Do not remember to have seen anywhere the following continued fraction for the tangent of a multiple angle, viz. ON THE IMPACT OF ELASTIC RODS. By E. J. Nanson, B.A., Trinity College. IN a Paper on the "Imperfect Elasticity of perfectly Elastic Rods," published in a former number of the Messenger (Vol. I., p. 129), Dr. Hopkinson has considered the problem of the impact of two elastic rods. As some of the results arrived at are at variance with what is stated in Thomson and Tait's Natural Philosophy, it may be as well to give an elementary solution of the problem and to point out an error in the paper referred to. Let a be the velocity of a wave of longitudinal displacement in a uniform elastic rod; the elasticity, p the density, so that a p = ε. = If a longitudinal disturbance of any sort be made in an elastic rod, two waves will be generated, which will travel in opposite directions. In one of these ap ev, and in the other apv, when p is the pressure at any point, and v is the corresponding velocity; the first is the one which travels in the direction in which is estimated positive. Also at a free end a wave of compression is reflected as a wave of dilatation, and vice versa. These results have generally been obtained from the equation of motion, but they may also, I believe, be established in an elementary way. Let us now apply these results to the solution of the problem in question. Let m, m' be the masses of the rods, u, u' their velocities before impact, 7, 7' their lengths, 7 being the shorter of the two. Since we are only concerned with relative motion we may reduce m' to rest. Put then 2wu-u. Let A, B denote the extremities of the first rod, A', B' those of the second, and suppose that B impinges on A'. We may now consider AB' as one rod which has been disturbed so that p=0 throughout, and v=2w from A to B, v=0 from B to B'. Hence two waves will arise, each of length 7; one, in which ap=ew and v=w, will travel from B to B'; the other, in which ap=ew, and v=w, will begin to be reflected. at A at once, and in the reflected wave ap=ew, and v=w. We may therefore consider that we have one wave of length 21 27 whose front starts from B. After a time this wave will a have passed out of AB; also, its front will not have had time to reach A' after reflection at B, because A'B' is greater than AB. Hence at this instant the short rod is undisturbed; there is no pressure at B between the rods, and the impact will therefore cease. The short rod now remains undisturbed, but there is a wave of length 27 running backwards and forwards in the long rod. The impact really ceases at the end of a time but the rods will not separate till the front of the wave, after reflection at B', arrives at A'; that is, at the end of a time at this instant A' begins to move away from B with a velocity 2w. The velocity of the centre of inertia of the long rod, found by estimating its whole momentum at any time, is 2mw Restoring now the velocity u', which we took away, m' we find that if v, v' be the velocities of the centres of inertia of the two rods, Hence the motion of the centres of inertia will be given correctly by the ordinary formulæ for the impact of elastic spheres, provided we replace the coefficient of restitution by the ratio of the masses of the rods. The energy of the long rod is in general partly kinetic and partly potential. The sum of the two parts, together with the kinetic energy of the short rod, ought to be equal to the whole kinetic energy before impact. This may be verified thus: at the end of the time + the energy of the long rod is entirely kinetic, and a its amount = {} (m' — m) u′′ + {m (u' + 2w)”. Adding mu" to this, we get mu2 + {m*u'2. In the paper before referred to, the equation marked (4) only holds so long as the rods remain in contact. This will de be the case so long as the value of when x = l' does not dx vanish. Now, if we put x = l', at = 2 (1 − l'), it may be shewn GE dar that the series for vanishes. Hence, (4) does not neces sarily hold longer than the time 2 (1-1) 2a But even this is not sufficient to complete the solution; to do so we should have to show that the value of when xl could not 2 (1 + 1') de dx 2 (1-1') α = vanish before the time As a matter of fact, the equation (4), with the correction of a wrong sign, holds for a time : after that a new series must be found. Again, it is stated that both rods will rebound in a state 21 of vibration. This is deduced from (4) at the time We a α have already seen that (4) holds at this moment. Hence, a wrong result has here been deduced from a correct equation. The argument used only proves that if the rods rebound without vibration, then their lengths must be equal; the result stated, viz. that if the lengths are not equal, then there will be vibrations in both rods does not follow. In fact, if the series be examined, it will be found that there are vibrations in the long rod but not in the short one. CARTESIAN OVALS REGARDED AS PROJECTED INTERSECTIONS OF SURFACES OF THE SECOND DEGREE. By C. W. Merrifield, F.R.S. IF a sphere and right cone intersect, and the intersection be projected on a plane at right angles to the axis, the projection is a Cartesian oval. (x − m)2 + y2 + z2 = a”........ Let the cone be (z − 1)2 = k (x2 + y2)........... and the sphere Then, if ....(1), .........(2). x=r cose, y=r sin0, z= sine, z=1+rk. Substituting this in (2) we get 2 22 (1 + k) + 2r (l s/k — m cos 0) + l2 + m2 — a2 = 0... (3). Again, if we take the bipolar equation of a Cartesian oval, r+λη = μ...... then, if c be the distance between the foci, we have 2 (4); r2 (1 − x2) — 2r (μ — cλ2 cos() + μ2 — X2c2 = 0........(5), and (3) and (5) will represent the same curve if is = When 72+ m2 — a2 0, or when the vertex of the cone r = 2(a + B cos 0) and represents Pascal's limaçon. ....... (7), When the sphere touches the cone at its vertex, that is, when the sphere and cone have a common tangent plane, the curve is cusped and becomes the Cardioid, for which m2 = 12k. We may also derive the equations of the Cartesian ovals. from the intersection of a right cone and an equilateral hyperboloid of one sheet; or, indeed, from the intersection of a right cone with any quadric surface of revolution* about an axis parallel to that of the cone; for this merely involves the substitution of bz for z in one only of the equations * Singularities excepted; for then generality might be lost, as if the surface chosen were a right cylinder. |