A Treatise on the Theory of Screws

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 9 must cut at right angles every generator of the species A. The same would have to be true for any other reciprocal screw </> similarly chosen; but it is obvious that two lines 9 and </> 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 $ 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 ß be two screws on S of equal pitch m, and let 9 be a screw of pitch — m which intersects a and ß. It follows that 9 is reciprocal both to a and ß, and therefore it must be reciprocal to every screw of >8. Let 9 cut $ in a third point through which the screw 7 is to be drawn, then 9 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 ß. We should thus have three screws on the surface of the same pitch, which is impossible. It is therefore necessary that 9 shall always intersect 7 at right angles. From this it will be easily seen that S must be of the third degree; for suppose that 9 intersected S in a fourth point, through which a screw 8 passed, then 9 would have to be reciprocal to 8, because it is reciprocal to all the screws of S; and it would thus be necessary for 9 to be at right angles to 8. Take then the four rays a, ß, y, 8, and draw across them the two common transversals 9 and </>. We can show, in like manner, that is at right angles to y and 8. We should thus have 9 and <(6 as two common perpendiculars to the two rays 7 and 8. This is impossible, unless 7 and 3 were in the same plane, and were parallel. If, however, y and 0 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 S would thus degenerate into a plane. We are thus conducted to the result that $ must be a ruled surface of the third degree, and we can ascertain its complete character. Since any transversal 9 across a, ß, and 7 must be a reciprocal screw, if its pitch be equal and opposite to those of a and ß, 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 ß, the transversal 9 will not always fulfil this condition. Imagine a plane perpendicular to 7, then this plane contains a line I at infinity, and the ray 9 must intersect I as the necessary condition that it cuts 7 at right angles. As 9 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, ß, 7,