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), 

 (@&amp;gt; )&amp;gt; (% 0&amp;gt; (S, &), (e, &amp;lt;p), in such a homography we can at once determine 

 the correspondent ^ to any other screw X. 



Draw a pencil of four planes through a, /3, and the four points j, B, e, X, 

 respectively. 



Draw also a pencil of four planes through ij, , and the four points 

 , 0, (j), T/T, respectively. These two pencils will be equianharmonic. Thus 

 we discover one plane which contains i/r. In like manner we can draw 

 a pencil of planes through a, y, and /3, 8, e, X, respectively, and the equi 

 anharmonic pencil through 77, , and , 6, &amp;lt;f&amp;gt;, ^, respectively. Thus we 

 obtain a second plane which passes through ty. A third plane may be 

 found by drawing the pencil of four planes through a, S, and the four points 

 /3, y, e, X, respectively, and then constructing the equianharmonic pencil 

 through 77, 6, and , , 0, \^, respectively. From the intersection of these 

 three planes, ^r 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 and be two corresponding rays. Then any plane through will 

 have, as its correspondent, a plane through . It is easily seen that these 

 planes intersect on a ray which has for its locus a quadric surface S, of which 

 and are also generators. This quadric must pass through the four 

 double points. 



Let 8 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 will lie on 8 , and the rest of the intersection of 8 and S will be 

 a twisted cubic G, which passes through the four double points. 



Take any point P on C, 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 C. 

 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. 



