236] FREEDOM OF THE FIFTH ORDER. 



In like manner for the pectenoid relating to and /3, we have 



257 



where M = 0, N = also represent planes passing through 0, whence 

 eliminating p e we have 



(p a -pe) MM - kNM + k N M= 0. 

 The equation of the pectenoid can also be deduced directly as follows. 



Let five screws of the five-system be given. Take a point on one of 

 these screws (a) and through draw four screws /3, 7, 8, e which belong to 

 the four-system defined by the remaining four screws of the original five. 

 Let there be any three rectangular axes drawn through 0. Let ct l , 2 , a 3 be 

 the direction cosines of a and let /3 1( /3 2 , /3 3 be the direction cosines of /?, 

 and similarly for 7, 8, e. Let 6 be some other screw of the five-system 

 which passes through and let 1} 6 2 , 3 be its direction cosines, then if 

 twists of amplitudes a, ft , 7 , 5 , e , & neutralize we must have 



77i 



ee 2 



a 3 + + 773 + 3 + ee s 

 because the rotations neutralize, and also 



a.p a a. 2 



whence by elimination of a , /3 , . . . 6 , we have 

 a 2 @ 2 72 S 2 e 2 



83 &amp;gt;3 73 63 63 



i3 



2 = 0, 

 3 = 0, 



f p 9 1 = 0, 

 6 p e 6 z = 0, 

 p e & 3 = 0, 



= 0. 



This equation has the form 



Pe L e. + M e. + N e, 



Let p e = p /i and l = x-^- p, 6 z = y-^p, 3 = z H- p then by reduction and 

 transformation of axes we obtain 



where y and 2 are planes at right angles and a is constant. This is the 

 equation of the pectenoid. 



17 



a 



