Ball — Dynamical Problems in the Theory of a Rigid Body. 31 



of the two screws. From A and B set off distances AP and BQ 

 equal to the pitches of the two screws, and having thus found P 

 and Q, describe a circle through those points, so that the angle sub- 

 tended hj PQ at the circumference shall equal the angle between the 

 two screws. It thus appears that P and Q in this method cannot be 

 chosen arbitrarily, for PQ'^ must equal the square of the shortest per- 

 pendicular distance between the two screws, plus the square of the 

 difference between their pitches. 



B N 



Fiff. 1. 



It is now extremely easy to identify the various points on the 

 circle with their corresponding screws on the cylindroid. Any parallel 

 to iif will cut the circle in two points corresponding to two screws 

 of equal pitch; any perpendicular to iJf will cut the circle in points 

 corresponding to intersecting screws; a perpendicular to ZJ/ through 

 S, the centre, cuts the circle in points corresponding to the two prin- 

 cipal screws of the cylindroid ; a parallel to LM, through H, cuts the 

 circle in the points which correspond with the terminal screws of the 

 cylindroid, and the points in which LM cuts the circle give the screws 

 of zero pitch : the extremities of any diameter of the circle correspond 

 with a pair of screws at right angles ; and if a third screw Z be intro- 

 duced, then the twists or wrenches about P, Q, Z, which would neu- 

 tralize, are respectively proportional to the sides of the triangle PQZ. 



The interpretation of reciprocal screws on the cylindroid is some- 



