Woodward — The Efficiency of Gearing under Friction. 99 



Ratio of Energy Lost to Energy Exerted. The whole 

 energy lost during the action of one pair of teeth is C7j + Z7 2 . 



The total energy exerted by the driver during the same 



time is M x multiplied by the arc (radius measure) described 



by the driver during the action. If we let the wheels have n Y 



2tt 

 and n 2 teeth, the arc required is — , and hence 



2tt 

 U=M.— (7). 



The ratio of work lost to energy exerted (which ratio I will 

 call R) is 



u - u 

 But the co-efficient 



r x \r x rj \rj V^ nj 



Substituting this, and (5), (6) and (7), we get 



, c ,, J~— log e (sin^— kcosO,)— k(\— d\ 

 B=(ljLl) 2 ^lf.( 7 j) l2 



+ 



1 +& 2 



log e (sin d 2 + h' cos 2 ) + h' g _ 2 ) 

 1 + Jc' 2 

 2r n \ m ,. / . 2r\ 



(8) 



in which k=(l — ~°)f, k' = (l + ^Af. 



The efficiency of the combination is found by subtracting R 

 from unity. 



Efficiency = 1 — R (9). 



Formula (8) is general and exact when but one pair of 

 teeth is in action at a time. 



7. In applying (9) to an ideal case the sum of X and d 

 must be calculated from the equation 



