496 HENKY A. ROWLAND 



(b* - a^(y -y'T + P (x - x'J = f ( 2 - 8 sin 2 3) , 

 I 1 a* = Rr cos 2 a , 



R -\- r ir _ R 



simj = ^ sin a; cos -n = - cos a, 

 2a 2a 



= --, = -, 







T cos 7] sin r sin ft Rr . 



x=b - r ; v = a -. '- - - = r- sm in cos a , 



COS a Sin a COS a 



Vy (y -y'}+x (I* - a 2 ) (a; - aT) = (cos ,,. + cos 



26 2 (V + */ 2 + O = #r, 



- x')= (sin n + sin v) 



sin /jt + sin v cos a sin e 

 2a cos 5 = r cos /j. R cos y , 

 2a sin 5 = r sin /* R sin v . . 

 On substituting these values and reducing, we find 



2 2Rr cos a cos e 



~ r cos 2 y + R cos 2 n ' 



ds 



2 A more simple solution is the following: _ mnst be constant in the direction 



do 



in which the dividing engine rules. If the dividing engine rules in the direction of 

 the axis y, the differential of this with respect to y must be zero. But we can also 

 take the reciprocal of this quantity and so we can write for the equation of condi- 

 tion 



d d(R+ r) _ 



dy ds 



Taking a circle as our curve we can write 



(Z_X')2+ ( y yf)* = p* 



and (x x")* + (y y"V = -R 2 , 



(X - 2///)2 + (y - y'")1 = r 2 , 



+ r)_ i ( ,j*-x" x-x>\_ { ^_^ly-y" + y-v"'\) 



~~i\ (l/ y \2t - J \~~W~ ~r - j} 



(R + r) _ 1 r x x"x x'" , \~ x x")(y y"} 



dT~ ~yj~ R- ~T~ ~^~~ 



\ _<r 



Making x = 0, y = 0, y' = 0, x' p, 



we have x" x f " I x //2 x /// ~i\ 



~ ' ~ P ~ + ~ = ' 



_n cos p + cos v _ 2Rr cos a cose 



r cos" v + R cos 2 u r cos 2 v + R cos 2 u ' 



