228 THE THEORY OF SCREWS. [217- 



Of these the four screws with pitches a,b,+c,-c respectively are each 

 reciprocal to the cylindroid. Each of these four screws must thus belong to 

 the four-system. Further these four screws are co-reciprocal. 



If 0^ ... ({ be the six co-ordinates of a screw in the four-system referred 

 to these canonical co-reciprocals, then we have 



l = 0, 3 = 0. 

 For 0! = ie , but as the first screw of reference belongs to the reciprocal 



CL 



cylindroid we must have tzr ]0 = 0. In like manner vr 3g = 0, and therefore 

 0! = and #3 = are the two linear equations which specify this particular 

 four-system. 



The pitch of any screw on the four-system expressed in terms of its 

 co-ordinates is 



I^02^j&amp;gt;04 2 + c (0 5 2 - 6 2 ) 

 *f+0t + & + lff 



of which the four stationary values are a, b, + oo , GO . 



We may remark that if the four co-ordinates here employed be taken as 

 a system of quadriplanar co-ordinates of a point we have a representation 

 of the four- system by the points in space. P^ach point corresponds to one 

 screw of the system. The screws of given pitch p e are found on the quadric 



surfaces 



U+p e V=0, 



where ?7=0 is the quadric whose points correspond to the screws of zero 

 pitch and where V= is an imaginary cone whose points correspond to the 

 screws of infinite pitch. Conjugate points with respect to U = will cor 

 respond to reciprocal screws. A plane will correspond to a three-system 

 and a straight line to a two-system. 



The general theorem proved in 214 states that when is a screw of 

 stationary pitch in an ra-system to which any other screw &amp;lt; belongs, then 



^e* = Pe cos 0(/&amp;gt;. 



Let us now take a four-system referred to any four co -reciprocals and choose 



for $ in the above formula each one of the four co-reciprocals in succession, 



we then have 



Pi e, =p e {0, + 2 cos (12) + 6 3 cos (1.3) + 4 cos (14)}] 



p 2 2 = p e {6 l cos (1 2) + 2 +0 3 cos (23) + 4 cos (24)} 



p 3 3 = Pe {0i cos (1.3) + 2 cos (23) + a + 4 cos (34)} 



2&amp;gt;404 = p s {0i cos (14) + 2 cos (24) + 6, cos (34) + 4 ] 



Eliminating 1( 2 , 3) 4 we deduce a biquadratic for p e . But we have 



