366 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



[NoVKMBSa, 



rim of the wheel, and the relative load per h. p. on the piston, (the 

 load of a 20 H. E. being 1,) having been given. 



Divide the constant number 37500 by the square of the mean velocity 

 of tne rim o/ the Jly-nheel in fett per second, and multiply the quotient 

 by the product of the length of stroke in feet, multiplied by the tabular 

 number P (see col. 7, table B.), (P representing the relative effective 

 load per h. p. on the piston, that of a 20 n. e. being ^,) the last product 

 rtill be the weight of the rim in lb. per h. p. 



Example 1. What will be the weight of the rim of the fly-wheel 

 of a 10 H. E., when the length of the stroke is 4 feet, the diameter of 

 the wheel in the middle of the rim 12 feet, and intended to make 48i 

 revolutions per minute ? 



when V= = 928-4 



P = 1-196 (see col. 7, table B.) 

 S = 4- 



therefore 



37500 



V- 

 37500 



928-4 

 Formula 5 is thus composed 



~^^^ \.r. 1000 

 450 = -^ 



Formula 5. 

 P S = weight of rim in lb. per h.p. 



4-784 = 193-2 = weight of rim in lb. per h.p. 



*^^ 144 



450 = 



1000 

 V2 



P 37-5 S = 



37500 



V- 



P S = weight of rim in lb. per h.p. 



Example 2. What will be the weight of the rim of the fly-wheel 

 of a 10 H. E., its diameter in the middle of the rim being 12-733 feet, 

 and making 26-7 turns per minute, the stroke being 3i feet. 



Observation ; this example corresponds with the proportion given 

 by Rule 1, as shown in the precedent table A. 

 then V= = 316-84 



P = 1-196 (see col. 7, table B) 

 S = 3i 



therefore -^^ P S = wt-tk-. 3-98G = 472 = weight of rim in lb. 



per H.P. 



For corn mills the constant number 37500 will be reduced to 



10=XZp19S 



700 „ ..„„„„ „^ . , , . 



P S = weight of rim 



Va 



450 = 



V= 



P 37-5 S = 



26250 



V- 



144 

 in lb. per h.p. 



Rule 5, requires a series of tabular numbers, representing the 

 relative effective load per H. p. on the piston, the load per h. p. on a 

 20 H. E. piston being considered as unity — this relative load varies 

 inversely as the speed of the piston, and will be obtained (considering 

 the speed of the piston of the 20 h. e., viz. 213 feet per minute as 

 the standard speed,) by dividing 213 by the effective speed of the 

 piston of the engine to be calculated ; the quotient will be the rela - 

 tive value P required : column 7 of table B has been calculated so 

 as to simplify the operation. 



Admitting 213 to be generally allowed for the speed of the piston 

 of a 20 H. e. in feet per minute, the calculation may be still farther 

 simplified, because a new constant number can be obtained by multi- 

 plying the constant number 37500 (given in formula 5) by 213, which 

 would create 



Formida 6. 



37500 



213 



"W 





37500 X 213 



s = 



7987500 

 V= W 



S = weight of the 



rim in lb. per h.p. 



when V2 = square of the velocity of the rim in feet per second. 



W = speed of the piston in feet per minute of the engine to be 

 calculated, ^i.'v-,.- ;;- 



S =' length ofthe'stroke in feet of the engine to be calculated, 



which gives us, '-y^^^fSi^i^t;:-. ' ~ 



Rule 6. To determine the weight in lb. per h. p. of the rim of a 



fly-wheel, its speed in feet per second, the speed of the piston in 

 feet per minute, and the length of stroke in feet being given. 



Divide the constant number 7,987,500 hy the product of tite sijuare of the 

 velocity in feet per second, of the rim of the fly-wheel, mulliplied iy the 

 speed of the piston in feet per minute, and the qnotient multiplied by the 

 length of the stroke in feet, will give the weight in lb, per h. r., of the rim 

 of the fly-wheel. 



Example. What will be the weight in lb. per h. p. of the rim of 

 a fly-wheel of a 10 h. e., its diameter in the middle of the rim being 

 12-733 feet, and making 2G-7 revolutions per minute, the length of 

 stroke being 3i feet? 

 then V= = 316-84 

 W = 178- 

 S = 3i 



■,987,500 3^ _ 4-2 = weight of rim in lb. 



therefore ^-S^S = 



V^ W " 56,397-52 



per H.P. 



For corn mills, &c. the formula would stand thus: 



5 591 250 

 ' „ ' S = weight of the rim in lb. per h.p. 



the constant number 7,987,500, being reduced to 5,591,250. 



To avoid misunderstanding it may be proper to repeat in this 

 place, that Rules 2, 3, and 4, are applicable only when the length of 

 the stroke and diameter of the rim of the fly-wheel, are made to cor* 

 respond with columns 2 and 5 of table A, according to Rule 1, 

 and when the fly-wheel makes the same number of revolutions as the 

 crank ; whereas the Rules 5 and 6 may be made use of on all occa- 

 sions, be the length of stroke, diameter of the wheel and its speed 

 what they may. 



It is occasionally desirable to make use of a fly-wheel, that has not 

 been expressly made for the engine to which it is intended to apply 

 it, and it is then requisite to determine the speed at which the rim 

 should travel to produce the usual effect on the motion of the engine; 

 the following rule will determine that speed ; the weight of the rim 

 in lb., and the circumference of the rim in feet being given. 



Rule 7. \st. Divide the total weight of the rim in lb. by tlie number of n.v. of 

 the engine, the quotient will be the weight of the rim in lb. per h. p., divide 

 this weight per h. p. by the length ofstroie of the engine in feet, and reserve 

 the quotient for a divisor. 



2nd. Divide the constant number 7,987,500, by the divisor found above, and 

 divide the quotient by the speed of the piston in feet per minute, which will 

 give the square of the velocity of the rim of the fly-wheel in feet per second, 

 and extract the square root thereof for the velocity of the rim in feet per 

 second. 



3rd. Divide the velocity of the rim in feet per second by the mean circum- 

 ference of the rim in feet, multiply the quotient by GO, and the product will 

 be the number of revolutions the fly-wheel should make per minute. 



This rule gives 



Formula 7. 



Vr, ,^r,M ,.,^n /total Weight of rim in lb. . , ., r . , . j... \ 

 7,987,500 -r- ( ^3— r length ofstrokeinft. 1 



speed of piston in feet per minute 

 V =: velocity of the rim of the fly-wheel in feet per second. 

 V 



circumference of rim in feet 

 wheel per minute, or 



X 60 =; number of revolutions of 



Formula 8. 



V 



- „«- „„ /^total weight of rim in lb. „ 1 r • . 



7,987,500 -~( „ „ ^ . ■ — r -■ , X speed of piston in 

 ' ' V H.P. X stroke in feet 



60 V 



I number of revo- 



feet per minute I ^ V, and -: — . . ,. , 



/ circumference in leet 



lutions of wheel per minute. 



Example. If a fly-wheel 15 feet in diameter, measured in the 

 middle of the rim, and weighing (the rim only) 5000 lb., is to be ap- 

 plied to a 20 H. E., what number of turns should this wheel make per 

 minute, the length of stroke of the engine being 5 feet, and the speed 

 of the piston 213 feet per minute? 



