A Treatise on the Theory of Screws

317] THE GEOMETRICAL THEORY. 343 If a rigid body be free to move about screws of any system of the fourth order, we may determine its four principal screws of inertia as follows. We correlate the several screws of the system of the fourth order with the points in space. The points representing impulsive screws will be a system homographic with the points representing the corresponding instantaneous screws. If we have five pairs of correspondents (a, 77), (ß, £)> (% (8> #)> (6> ^)>in such a tomography we can at once determine the correspondent y/r to any other screw X. Draw a pencil of four planes through a, ß, and the four points 7, S, e, X, respectively. Draw also a pencil of four planes through 17, and the four points 0, cf>, y/r, respectively. These two pencils will be equianharmonic. Ihus we discover one plane which contains i/r. In like manner we can diaw a pencil of planes through a, 7, and ß, 8, e, X, respectively, and the equi- anharmonic pencil through 77, and 3. <fr, respectively. Ihus we obtain a second plane which passes through i/r. A third plane may be found by drawing the pencil of four planes through a, 8, and the four points ß, y, e, X, respectively, and then constructing the equianharmonic pencil through y, 3, and £, £ </>, respectively. From the intersection of these three planes, is known. In the case of two homographic systems in three dimensions, there are, of course, four double points. These may be determined as follows. Let 0 and O' be two corresponding rays. Then any plane through 0 will have, as its correspondent, a plane through O'. It is easily seen that these planes intersect on a ray which has for its locus a quadric surface 6, of which 0 and O' are also generators. This quadric must pass through the four double points. Let S' be the quadric surface which contains all the points in the second system corresponding to the points of S regarded as the first system. Then 0' will lie on S', and the rest of the intersection of S and S' will be a twisted cubic G, which passes through the four double points. Take any point P on G, and draw any plane through P. Then every ray of the first system of the pencil through P in this plane will have as its correspondent in the second system the ray in some other plane pencil L. One, at least, of the rays in the pencil L will cut the cubic G. Call this ray X', and draw its correspondent X in the first system passing through P. We thus have a pair of corresponding rays X and X', each of which intersects the twisted cubic C.