148 THE THEORY OF SCREWS. [159 



surface ; it must also be reciprocal to all these ; and, as there are only two 

 screws of the given pitch, it will follow that 6 must cut at right angles every 

 generator of the species A. The same would have to be true for any other 

 reciprocal screw &amp;lt; similarly chosen ; but it is obvious that two lines 6 and &amp;lt; 

 cannot be found which will cut all the generators at right angles, unless, 

 indeed, in the extreme case when all these are coplanar and parallel. In the 

 general case it would require two common perpendiculars to two rays, which 

 is, of course, impossible. We hence see that S cannot be a surface of the 

 second degree. 



We have thus demonstrated that 8 must be at least of the third degree 

 in other words, that a line which pierces the surface in two points will pierce 

 it in at least one more. Let a and ft be two screws on 8 of equal pitch in, 

 and let 6 be a screw of pitch m which intersects a and ft. It follows that 



6 is reciprocal both to a and ft, and therefore it must be reciprocal to 

 every screw of 8. Let 6 cut 8 in a third point through which the screw 7 is 

 to be drawn, then 6 and 7 are reciprocal ; but they cannot have equal and 

 opposite pitches, because then the pitch of 7 should be equal to that of a 

 and ft. We should thus have three screws on the surface of the same pitch, 

 which is impossible. It is therefore necessary that 6 shall always intersect 7 

 at right angles. From this it will be easily seen that 8 must be of the 

 third degree ; for suppose that 6 intersected 8 in a fourth point, through 

 which a screw 8 passed, then 6 would have to be reciprocal to 8, because it is 

 reciprocal to all the screws of $; and it would thus be necessary for 9 to be 

 at right angles to 8. Take then the four rays a, ft, 7, 8, and draw across 

 them the two common transversals 6 and &amp;lt;. We can show, in like manner, 

 that (j) is at right angles to 7 and 8. We should thus have 6 and &amp;lt;/&amp;gt; as two 

 common perpendiculars to the two rays 7 and 8. This is impossible, unless 



7 and 8 were in the same plane, and were parallel. If, however, 7 and 8 be 

 so circumstanced, then twists about them can only produce a resultant twist 

 also parallel to 7 and 8, and in the same plane. The entire surface 8 would 

 thus degenerate into a plane. 



We are thus conducted to the result that 8 must be a ruled surface 

 of the third degree, and we can ascertain its complete character. Since any 

 transversal 6 across a, ft, and 7 must be a reciprocal screw, if its pitch be 

 equal and opposite to those of a and ft, it will follow that each such trans 

 versal must be at right angles to 7. This will restrict the situation of 7, 

 for unless it be specially placed with respect to a and ft, the transversal 6 

 will not always fulfil this condition. Imagine a plane perpendicular to 7, 

 then this plane contains a line / at infinity, and the ray 6 must intersect / as 

 the necessary condition that it cuts 7 at right angles. As 6 changes its 

 position, it traces out a quadric surface, and as I is one of the generators of 

 that quadric, it must be a hyperbolic paraboloid. The three rays a, ft, 7, 



