PALMER: THE DESIGNING OF CONE PULLEYS. 49 
which is the middle term of each of the equations, except sof 
divided by z. 
For, AM-==d cos a, 
MQ=KN=PN sin a=da sin a, 
AQ=—d(a sin GI= (COS @ jc 
We have, then, all the values of this middle term (except 7 times 
too large) readily obtainable from this involute by varying a. We 
can now arrange to divide this expression by z for any value a may 
have, and thus secure the middle terms of equations (5) and (6) 
complete, as follows: 
Take the point F where the involute cuts BC, and drop a per- 
pendicular FH to AD. Assume the middle point O of FH and 
draw BO, producing it on to its intersection with AD. This line 
BO cuts RQ, already drawn, in a point G, and GR is equal the 
desired term 
Sarees Ol sin a—=- €OS' a): 
S y 
by 
“ : x d 
Bono = = d, by the property of the involute, and FO—— 
Z) 
construction. Then 
d 
BO I 
—.d 
2 
Then by similar triangles 
Ge. Oe ai ame 
hE Ee 2 RG mat BR 
But BR=AQ=d(a sin a+ cos a). 
RG——\(a sin a-++ cos a). 
AG 
We notice, now, that the line BO 1s inclined so that any ordinate 
; ee I ; 
drawn down from BC to it is the — part of the corresponding ab- 
7 
5 s I Letts : 
scissa, as GR=—\.BR. So, then, it is very easy to get the first 
T 
term of each of the equations (5) and (6) by laying off BL—l, the 
l 
half length of belt, when LS—~—_., and we have, at once 
7 
