342 THE THEORY OF SCREWS. [316, 



The simplest method of solving this problem is that already given in 

 139, in which we regard the six points, suitably arranged, as the vertices 

 of a hexagon ; then the Pascal line of the hexagon intersects the circle in 

 two points which are the points corresponding to the principal screws of 

 inertia. 



317. Third and Higher Systems. 



We next investigate the principal screws of inertia of a body which has 

 three degrees of freedom. We have first, by the principles already ex 

 plained, to discover four pairs of correspondents. When four such pairs 

 are known, the principal screws of inertia can be constructed. Perhaps 

 the best method of doing so is to utilize the plane correspondence, as 

 explained in Chap. xv. The corresponding systems of impulsive screws 

 and instantaneous screws, in the system of the third order, are then repre 

 sented by the homographic systems of points in the plane. When four pairs 

 of such correspondents are known, we can construct as many additional pairs 

 as may be desired. 



Let a, 0, 7, 8 be four points in the plane, and let 77, , f, be the 

 points corresponding, so that ?/ represents the impulsive screw, and a the 

 instantaneous screw, and similarly for the other pairs. Let it be required 

 to find the impulsive screw &amp;lt;, which corresponds to any fifth instantaneous 

 screw e. Since anharmonic ratios are the same in two corresponding figures, 

 we have 



a (0, 7, 8, 6) = ,K ?, 0, (/&amp;gt;), 

 thus we get one ray r)&amp;lt;f&amp;gt;, which contains &amp;lt;. We have also 



(, 7, , &amp;lt;0 = f07, C, 0. &amp;lt;), 

 which gives a second ray &amp;lt;, containing &amp;lt;/&amp;gt;, and thus &amp;lt;f&amp;gt; is known. 



A construction for the double points of two homographic systems of 

 points in the same plane is as follows: 



Let and be a pair of corresponding points. Then each ray 

 through will have, as its correspondent, a ray through . The locus of 

 the intersection of these rays will be a conic S. This conic S must pass 

 through the three double points, and also through and . 



Draw the conic 8 , which is the locus of the points in the second system 

 corresponding to the points on S, regarded as in the first system. Then 

 since lies on S, we must have on S . But S must also pass through 

 the three double points. is one of the four intersections of S and S , and 

 the three others are the sought double points. Thus the double points are 

 constructed. And in this manner we obtain the three principal screws of 

 inertia in the case of the system of the third order. 



