24] RECIPROCAL SCREWS. 29 



generator through V, and that, therefore, the cone drawn from to the ellipse 

 TL VM is the cone required. 



We hence deduce the following construction for the cone of reciprocal 

 screws which can be drawn to a cylindroid from any point 0. 



Draw through a line parallel to the nodal line of the cylindroid, and 

 let T be the one real point in which this line cuts the surface. Find the 

 second screw LM on the cylindroid which has a pitch equal to the pitch of 

 the screw which passes through T. A plane drawn through the point T and 

 the straight line LM will cut the cylindroid in an ellipse, the various points 

 of which joined to give the cone required*. 



We may further remark that as the plane TLM passes through a gene 

 rator it must be a tangent plane to the cylindroid at one of the intersections, 

 suppose L, while at the point M the line LM must intersect another generator. 

 It follows (22) that L must be the foot of the perpendicular from T upon LM, 

 and that M must be a point upon the nodal line. 



24. Locus of a Screw Reciprocal to Four Screws. 



Since a screw is determined by five quantities, it is clear that when the 

 four conditions of reciprocity are fulfilled the screw must generally be confined 

 to one ruled surface. But this surface can be no other than a cylindroid. 

 For, suppose three screws X, //., v, which were reciprocal to the four given 

 screws did not lie on the same cylindroid, then any screw &amp;lt;f&amp;gt; on the cylindroid 

 (X, fi), and any screw -^r on the cylindroid (X, v) must also fulfil the conditions, 

 and so must also every screw on the cylindroid (&amp;lt;, ^) (22). We should thus 

 have the screws reciprocal to four given screws, limited not to one surface, 

 as above shown, but to any member of a family of surfaces. The construction 

 of the cylindroid which is the locus of all the screws reciprocal to four given 

 screws, may be effected in the following manner : 



Let a, /3, 7, 8 be the four screws, of which the pitches are in descending 

 order of magnitude. Draw the cylindroids (a, 7) and (13, 8). If &amp;lt;r be a linear 

 magnitude intermediate between pp and p y , it will be possible to choose two 

 screws of pitch or on (a, 7), and also two screws of pitch &amp;lt;r on (/3, 8). Draw 

 the two transversals which intersect the four screws thus selected ; attribute 

 to each of these transversals the pitch a, and denote the screws thus pro 

 duced by 6, &amp;lt;j&amp;gt;. Since intersecting screws are reciprocal when the sum of their 

 pitches is zero, it follows that 6 and &amp;lt;/&amp;gt; must be reciprocal to the cylindroids 

 (a, 7) and (13, 8). Hence all the screws on the cylindroid (6, &amp;lt;/&amp;gt;) must be re 

 ciprocal to a, /3, 7, 8, and thus the problem has been solved. 



* M. Appell has proved conversely that the cylindroid is the only conoidal surface for which 

 the feet of the perpendiculars from any point on the generators form a plane curve. Revue de 

 Mathematiques Speciales, v. 129 30 (1895). More generally we can prove that this property 

 cannot belong to any ruled surface whatever except a cylindroid and of course a cylinder. 



