material of the body, and is to be determined by calculating the twist velocity about one screw which would give the body one unit of kinetic energy. The same process is to be repeated for five other screws. Draw from the centre of the pitch hyperboloid six radii vectores parallel to these screws and equal to the corresponding twist velocities; thus six points on the ellipsoid and its centre are known, and it is, therefore, determined.

Draw the three common conjugate diameters of the pitch hyperboloid and the ellipsoid of equal kinetic energy. The three screws of S which are parallel to these diameters are the principal screws of inertia. This is proved as follows:

Let , x, y, be the three screws thus determined. If an impulsive wrench act about 4, the instantaneous screw about which the body commences to twist must be reciprocal to x and y, since o, x, y, are parallel to conjugate diameters of the ellipsoid of equal kinetic energy. But the only screw which belongs to S, and is reciprocal to x and, must be p, because they are parallel to conjugate diameters of the pitch hyperboloid. Hence an impulsive wrench about & must make the body commence to twist about. Similar reasoning applies to x and y.

§ 16. Construction for the instantaneous screw when the impulsive screw is given. All the screws in the screwcomplex S which are reciprocal to any screw (wherever situated) are parallel to a plane. This appears as follows: To introduce an additional constraint upon the body is equivalent to saying that all the screws about which the body can be twisted must be reciprocal to some screw n, in addition to all the screws of S'. In this case, the body

has only two degrees of freedom, and, therefore, all the screws about which it can be twisted are parallel to a plane (§ 4).

This plane, drawn through the centre of the pitch hyperboloid, may be called the reciprocal plane, with respect to n.

Let an impulsive wrench act about 7, then the body commences to twist about an instantaneous screw ŋ, which may be found as follows: Draw the reciprocal plane to n; find the diameter of the ellipsoid of equal kinetic energy which is conjugate to the reciprocal plane; then the screw belonging to S, which is parallel to this conjugate diameter, is the required instantaneous screw.

For let n,, be three conjugate screws of kinetic energy, and,,, be three corresponding impulsive screws. Since, are reciprocal to 7, they must be parallel to the reciprocal plane, and since n, §, y, are parallel to conjugate diameters of the ellipsoid of equal kinetic energy, the construction is manifest.

In the case of a body rotating around a fixed point, the construction reduces to Poinsot's well-known result.

§ 17. On the ellipsoid of equal potential energy. Let us suppose the rigid body to be in a position of stable equilibrium under the influence of a conservative system of forces. If the body receive a small twist about the displacement screw, it is no longer in a position of equilibrium, and the forces which act upon it constitute a wrench about a certain screw which we may call the restoration screw. Since the forces form a conservative system, the energy consumed in forcing the body from its initial position to any other position is quite independent of the route by which the displacement may have been effected. The potential energy due to any position can, therefore, be adequately expressed by a function of the coordinates of that position. It is easy to see that this function must be a homogenous expression containing λ, μ, v, in the second degree.

Proceeding as before we arrive at the conception of the ellipsoid of equal potential energy, which possesses the property that a twist about any screw of S, through a small angle proportional to the parallel semi-diameter of the ellipsoid, will do one unit of work against the external forces.

The three common conjugate diameters of the pitch hyperboloid and the ellipsoid of equal potential energy are parallel to the three screws of S, which possess the property that if the body be displaced by a twist about one of these screws, the forces of restoration constitute a wrench about the same screw.

The restoration screw corresponding to a given displacement screw can be found by the construction of § 16, if we replace the ellipsoid of equal kinetic energy by the ellipsoid of equal potential energy.

§ 18. Small oscillations. If released after the displacement, the body will commence to make small oscillations, the character of which we can completely determine. Draw the two ellipsoids of energy, find the three screws, u, v, w, belonging to S, which are parallel to their common conjugate diameters. If, then, the body be displaced originally by a twist about one of these screws, it will continue for ever to perform small twist oscillations about the same screw. In general, whatever be the initial displacement it may be decomposed into twists about u, v, w, and the small oscillations are compounded of three simple harmonic oscillations about the same screws. (u, v, w, are called harmonic screws at the suggestion of the Rev. Prof. Townsend, F.T.C.D.).

This is proved by the consideration that if u be the impulsive screw corresponding to u as an instantaneous screw, then, owing to the particular construction by which u is formed, precisely the same screw u corresponds as a restoration screw to u as a displacement screw.

R. S. BALL.

ON A GEOMETRICAL METHOD OF DEDUCING THE CENTRAL AND DIAMETRAL PROPERTIES OF CONICS FROM THOSE BELONGING TO THE FOCUS AND DIRECTRIX.

IN

N the following investigation a conic is defined as the locus of a point whose distance from a given point is in a fixed ratio to its distance from a given right line. The fixed point is, of course, the focus, and the fixed right line the directrix of the conic.

From the above definition the fundamental properties of the centre and diameters of conics can be easily deduced by means of the well-known geometrical problem. To find the locus of the vertex of a triangle whose base and the ratio of whose sides are given.

As the following investigation is founded altogether on the solution of this problem, it may be well to give that solution here.

Let AB be the given base, and APB any triangle satisfying the given conditions. Draw the bisectors of the internal and external angles at P, and they will be at right angles to each other and meet AB in points X, X', such that the ratios AX: XB and AX': BX' will be equal to the given ratio of the sides. Hence the points X and X' are found by dividing the base internally and externally in the ratio of the sides, and a circle described on XX' as diameter is the locus required.

It is easy by means of the above to construct the points in which any perpendicular to the directrix meets the conic. For, let (fig. 1) be the point in which the perpendicular meets the directrix, join FQ, and describe the circle which is the locus of the vertex of the triangle

whose base is FQ, and the ratio of whose sides is that of the distances from a point on the conic to the focus and directrix. The points PP' in which this circle meets QP are manifestly points on the conic, and it is also plain that they are the only points common to the conic and the line QP.

If DF (fig. 2), the perpendicular from F to the directrix, be divided internally in A and externally in A' in the given ratio, and if AД' be bisected in C, it is plain that parallels to the directrix through A and A' will divide any line FQ from the focus to the directrix internally in X and externally in X' in the given ratio; and, also, that a parallel to the directrix through C will bisect XX', and PP' since it is perpendicular to PP', and passes through the centre of the circle of which PP' is a chord. Hence, the locus of the middle points of chords of the conic which are perpendicular to the directrix is a parallel to the directrix through the point C.

It is obvious that the points in which a given parallel to the directrix meets the conic lie on a circle, having F for its centre; and, hence, that every such parallel can meet the conic in only two points, and that the locus of the middle points of chords parallel to the directrix is the line FD.

From hence it appears that the conic is perfectly symmetrical with respect to the line AF, and also perfectly symmetrical with respect to a perpendicular to AA', through C.

From this last consideration it appears that there must be a second point F' having properties similar to those of F, and situated on the line AA' at a distance CF' from C equal to CF. The point C is called the centre of the conic.

To determine the points in which any given right line meets the conic.

Let PQ (fig. 3) be the given line, Q the point in which it meets the directrix, and P a point on the conic.

M M