Woodward — The Efficiency of Gearing under Friction. 



99 



Ratio of Energy Lost to Energy Exerted. The whole 

 enerfjy lost during the action of one pair of teeth is U^ + U^. 



The total energy exerted by the driver during the same 

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

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



27r 



and ?i, teeth, the arc required is — , and hence 



n^ 



U=M, 



(7). 



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

 call R) \Q 



R ^^ 



But the co-efl5cient 



^1 V*-! rj \rj \n, nj 



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



— loge(sin^^— ^cos^j)— A; ( „— ^i) 



ij=(L+l)?!^.(':p) 



\«j nJ TT \rj 



+ 



— log, (sin d, 4- h' cos 6^) + h' {^ — d^ 



1 + ^'2 



(8) 



in which h=(l-^-^y, ^•' = (l + ^o)/. 



The efficiency of the combination is found by subtracting R 

 from unity. 



Efficiency = 1 — i? (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 d^ and d 

 must be calculated from the equation 



