311] THE GEOMETRICAL THEORY. 337 



Let us now think of 9 l as an instantaneous screw ; it lies on the cylindroid 

 A A lt and this cylindroid contains two principal screws of inertia. It follows 

 from 309 that the corresponding impulsive screw fa lies on the same cylin 

 droid. That screw fa can be determined by the construction there given. 

 In like manner we construct on the other four cylindroids the screws &amp;lt;j&amp;gt; 2 , fa, 

 fa, fa, which are the impulsive screws corresponding respectively to 2 , 3 , 

 # 4 , 6 , as instantaneous screws. 



Consider then the two pairs of corresponding impulsive screws and in 

 stantaneous J screws (77, a) and (fa, ^). We have arranged. the construction 

 so that 6 l is reciprocal to 77. Hence, by the fundamental principle so often 

 employed, a and d l are conjugate screws of inertia, so that a must be reciprocal 

 to fa. 



In like manner it can be proved that the instantaneous screw a for which 

 we are in search must be reciprocal to fa, fa, fa, fa. We have thus dis 

 covered five screws, fa, fa, fa, fa, fa, to each of which the required screw a 

 must be reciprocal. But it is a fundamental point in the theory that the 

 single screw reciprocal to five screws can be constructed geometrically ( 25). 

 Hence a is determined, and the geometrical solution of the problem is 

 complete. 



It remains to examine the failure in this construction which arises when 

 any one or more of the five screws fa ... fa becomes indeterminate. This 

 happens when 77 is reciprocal to two screws on the cylindroid in question. 

 In this case 77 is reciprocal to every screw on the cylindroid. Any one of such 

 screws might be taken as the corresponding fa and, of course, would have 

 been also indefinite, and a could not have been found. In this case 77 would 

 have been reciprocal to the two principal screws of inertia, suppose A , A l 

 which the cylindroid contained. Of course still more indeterminateness 

 would arise if 77 had been also reciprocal to other screws of the series A , A l} 

 A a , A 3 , A 4 , A 5 . No screw could, however, be reciprocal to all of them. If?; 

 had been reciprocal to five, namely, A l} A a , A s , A,, A 5 , then 77 could be no 

 screw other than A , because the six principal screws of inertia are co- 

 reciprocal ; 77 would then be its own instantaneous screw, and the problem 

 would be solved. 



We may therefore, under the most unfavourable conditions, take 77 to be 

 reciprocal to four of the principal screws of inertia A , A lt A a , A 3&amp;gt; but not to 

 A 4 or .-1 5 . We now draw the five cylindroids, A A t , A,A 4 , A 2 A 4 , A 3 A 4 , A A 5 . 

 We know that 77 is reciprocal to no more than a single screw on each 

 cylindroid. We therefore proceed to the construction as before, first finding 

 0, ... 5 , one on each cylindroid ; then deducing fa ... fa, and thus ultimately 

 obtaining a. 



Thus the general problem has been solved. 



B. 22 



