﻿Motion transmitted by Hooke's Joint, 



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2. In Hooke's Joint the point B moves in the great circle 

 CBN and the point A in the great circle CAN (rig. 1). The 

 arc AB is constant and equal to a quadrant of: the great 

 circle. The point A receives its motion from an arm OA 



set at right angles to the driving shaft OX, while B transmits 

 motion to the driven shaft OY. Let the angle between the 

 shafts be y • this w r ill be the angle between the planes of CAN 

 and CBN. In the spherical triangle ABC we have 



cos <• = cos a cos b + sin a sin b cos 7. 



Since c=7r/2, this becomes 



cos a cos Z> + sin a sin b cos y = 0. . . . (1) 



Plainly when B is at C, A will be at T; B will therefore 

 move through the angle a while A moves through the anple 



&-7T/2. 



Writing <£ for a and 6 for b — tt/2, equation (1) becomes 

 — cos (/> sin 6 + sin <f> cos cos y — 0. 



Put cos7 = (l — »)/(l + n), and we have 



(1 + n) cos </> sin 6 = (1 — n) sin cj> cos 6. 



.*. 11 (cos (/> sin 9 + sin (/> cos 6) = sin <£cos 6 — cos </> sin #. 

 sin (<£ — 0) = n sin (</> + 6). 



