104 Doolittle on the Steam Engine. 
62:66::125:22225— 133... 
" figure 3. the force being applied at a lever much short- 
er than that to which it is to be reduced, its effect at the 
extremity of the longer lever must be found by inverse be 
portion—thus— 
66:62:11: ag Oey 
Making similar constructions in the other points of divis- 
ion, and reducing the respective forces to AE same length © 
of lever, we have the following series.— | 
Forces at the points of application. | Forces reduced to an equal lever. 
= c = = 1 
97 - - = - 157 
124 - - = - 135 
172 - - - - 52 
G2 cela eae 11 
Ope eioueniie 0 
Dividing the sum by 8,the number of terms - 8 | 542 
We have, for the mean force utilized - ~- 68 for 200 applied. 
In this calculation, as in all which precede, to avoid 
fractions, where there were any, I have uniformly added an 
unit in their stead, in order to give the machine “a fair 
chance.” 
The mean force 68 is applied tangentially to the reduced 
circle, whose semi circumference is=207—the force that 
have supposed applied is 200, and the stroke of the piston 
is 131. Therefore force applied is to force utilized ::200 
X 131:68 x 207 or as 26:14, nearly; then say 
26:14::100:44°—54 
In the common crank the force applied is to the force 
utilized ::100:78, nearly. Therefore the effect of the new 
machine, is to the effect of the common crank, with the ap- 
plication of an equal force as 54:78 or ::9:13. 
We must observe also, that when the cylinder arrives in 
B or in B’ (fig. 1.) the piston has performed half its stroke. 
If, therefore, we consider the pressure of the steam as a 
weight, and multiply that weight by the distance gone 
through to find the quantity of force employed in giving 
