520 Proceedings of the Royal Irish Academy. 



second generator of zero pitch on S, and, tlierefore, in its original form 

 there must have been two screws of the given pitch m. 



Intersecting screws are reciprocal if they are rectangular, or if their 

 pitches be equal and opposite ; hence it follows that a screw 6 recipro- 

 cal to S must intersect S in certain points, the screws through which 

 are either at right angles to 6 or have an equal and opposite pitch 

 thereto. 



Prom this we can readily show that 8 must be of a higher degree 

 than the second; for suppose it were an hyperboloid, then a trans- 

 versal 6 which intersected two screws of equal pitch m must, when 

 it receives the pitch - m, be reciprocal to the entire system. We can 

 take for 6 one of the generators on the hyperboloid ; 6 will then inter- 

 sect every screw of the 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 one species. The 

 same would have to be true for any other reciprocal screw 6 lying on 

 the surface ; but it is obvious that two lines 6 and (jf) 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 ray which pierces the surface in two points will 

 at least pierce it in one more. Let a and /3 be two screws on 8 of equal 

 pitch m, and let ^ be a screw of pitch - m which intersects a and ^. 

 It follows that 6 is reciprocal both to a and to /?, and therefore it 

 must be reciprocal to every screw of 8. Let 6 cut -S' in a third point 

 through which the screw y is to be drawn, then 6 and y are reciprocal ; 

 but they cannot have equal and opposite pitches, because then the 

 pitch of y should be equal to that of a and /3. "We would thus have 

 three screws on the surface of the same pitch, which is impossible. 

 It is therefore necessary that shall always intersect y at right angles. 

 Prom this it will be easily seen that 8 must be of the third degree ; 

 for suppose that intersected >S' in a fourth point, through which a 

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

 reciprocal to all the screws of ^S* ; and it would thus be necessary for 

 6 to be at right angles to 8. Take then the four rays a, ft, y, S, and 

 draw across them the two common transversals 6 and ^. "We can show, 

 in like manner, that 4> is at right angles to y and 8. We would thus 

 have 9 and <^ as two common perpendiculars to the two rays y and 8. 

 This is impossible, unless y and 8 were in the same plane, and were 

 parallel. If, however, y and 8 be so circumstanced, then twists about 

 them can only produce a resultant twist also parallel to them, and 

 in the same plane. The entire surface 8 would thus degenerate to a 

 plane. 



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

 surface of the third degree, and it will now be easy to find out its com- 



