SECT, viii.] MANAGEMENT OF THE STEAM ENGINE. 259 



With these explanations we may now proceed to form rules from the equation. 

 The equation is 



32 P? 



20 



r n v 



but the dimensions of the rim will be more convenient than the weight ; and the 

 weight w = 2 x 3* 14 16 r A x 3'2 Ibs., for cast iron, = 20 r A, consequently, 



. 1'6 Pat, 



A = , the area ot the section of the nm in inches. 



r* n v 



If the fly wheel shaft makes N revolutions per minute, then in the time t it will 

 _N_ . . 6-2832 t 



60 '" 60 



t N 6-2832 t N r 



make - ' = - - and, 



15 Pa 



~r*Tir 



539. The next point to be considered is the degree of equalization a machine 

 requires. Its own parts have much effect, and the species of parts which act as 

 flies are most numerous in machines which require the equalizing power of the 

 fly the most. At a mean, perhaps a variation of one-tenth is nearly corresponding 

 with practice, and with this condition the rule is, 



p " 



A = 



540. CASE r. A double engine with a crank. In this case the variation is 

 from the full force of the steam to nothing at each quarter of the stroke ; hence, 

 the mean excess is one-fourth of the greatest force P on the piston, and the rule in 

 the nearest simple expression is, 



A -= 40Pg 



RULE. Multiply forty times the pressure on the piston in Ibs., by the radius 

 of the crank in feet, and divide this product by the cube of the radius of the fly 

 wheel in feet, and by the number of its revolutions per minute, the result is the 

 area of the rim of the fly in inches. 



The number of horse power, multiplied by 200, will be the greatest pressure 

 on the piston, nearly. 



Example. The pressure on the piston of an engine being 4000 Ibs., the 

 radius of the crank 2'5 feet, and the revolutions per minute 22, required the 

 section of the rim of a fly of 9 feet radius. In this case, 



40 



for the area of the section of the rim. 



