1842.] 



THE CIVIL ENGINEER AND ARCHITECTS JOURNAL. 



36b 



section of the rim of a fly wheel, its diameter and velocity being 



determined by Rule 1. 



Divide the constant mimber 21300 by iJie cube of the relocily in feet 



per second, the quotient will be the area in square inches per h. p. 



Example. What will be the area in square inches per H. p. of the 



rim of the fly-wheel of a 20 H. E., the velocity of which taken in the 



middle of the rim is 21'3 feet per second ? 



When V =21-3 



V» = %(53-597 



21300 21300 „, . . , 



then J — =; , = 2-2 = area per h. p. ni square inches. 



For corn mills, saw mills, and other machinery not requiring great 

 regularity of motion, or which by its own peculiar character tends to 

 equalize the effect of the steam, 7 square inches of cross section may 

 be allowed instead of 10, and will be found to answer very well ; on 

 such occasions the constant mimber 21300, will be reduced to 



10-X7X21-3= 14910, and the formula No. 1 would be 



14910 



= A 



= Area of cross section of rim in square inches per h. p. 



The area of the cross section of the rim of the fly-wheel being 

 obtained as above, the total weight of the rim per H. P. will be found 

 by multiplying the area in square inches by the circumference of the 

 wheel in inches ; the product will represent the number of cubic 

 inches of metal in the rim per H. P. ; and this number divided by 

 1728, will give tlie number of cubic feet of metal, which multiplied by 

 450, (the weight in lb. of a cubic foot of cast iron,) will give the 

 weight of the rim in lb. per H. P. 



Therefore when V" = cube of velocity of rim of fly-wheel in feet 

 per second, t = circumference of ditto in feet, we shall find 



Formula 2. 



Formula 4. 



21300 



V3 



Xl2n 



1728 



450 = weight of rim in lb. per h. p., and by reducing 



the constant numbers to a simple denominator, we shall obtain, 



Formula 3. 



6651J2-5 



ir = weight of the rim per h. p. in lb. For corn mills, this 



constant number would be reduced to 53250, and formula 3 would be 



46593'8 • ui !• XI • • 7i_ 



— ir =: weight of the nm in lb. per h. p. 



Rule 3. To find the weight in lb. of the rim of a fly-wheel, its 

 velocity in feet per second, and circumference in feet, being given by 

 Rule 1st. 



Divide the constant number 66562'5 by the cube of the velocity of the 

 rim of the fy-mheel in feet per second, and multiply the quotient by the 

 circumference of the rim in feet, measured in the middle of its breadth, 

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



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

 of a fly-wheel of a 20 h. e., the circumference of the rim being 60 

 feet, and its speed 21-3 feet per second, (the stroke of the engine 

 being tJO inches) ? 



when V =: 21-3 

 V3 = 9663-597 

 IT = 60- 



then 



66o62'5 



66562-5 

 9663-597 



X 60 = 413-22 = lb. weight of 



per H.p. 



By the rule to find the area, the cross section of the rim of the fly- 

 wheel per H. p. would be 2-2x20 ^ 44 square inches, and 



By the rule to find the weight, it is shown that the weight of the 

 rim of the fly-wheel of a 20 h. e., would be 4 13-22 X 20 = 8264 lb. 



In formula No. 3, ir is the product of the length of stroke in feet, 

 multiplied by 3-82x3-1416 = 12, so that by multiplying the constant 

 number 66562-5 by 1-2, we shall obtain a new constant number 

 798750, which will admit of the following construction. 



798750 



S =r lb. vreight of the rim per h. p. 



In formula 4, S = the length of stroke in feet, this formula gives 

 Rule 4. To find the weight of the rim of the fly-wheel in lb. per H.P., 

 the length of stroke of the engine, and the velocity of the rim of the 

 fly-wheel, being determined by Rule 1. 



Divide the constant number 798750 by the cube of the velocity of {he 

 middle of the rim of the fy-whetl in feet per second, and multiply the 

 quotient by the length of the stroke in feet, the product Kill be the proper 

 freight of the rim in lb. per h. p. 



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

 the fly-wheel of a 20 h. e., the speed of the middle of the rim being 

 21-3 feet per second, and the length of the stroke 5 feet (see column 

 2 and 7 of table A) ? 



when V = 21-3 

 V^ = 0663-597 

 S = 5- 

 798750 

 " 966"-'97 ^ ~ 413-25 = lb. weight of rim per h.p- 



giving the same result as obtained by formula 3. 



For corn mills, &c. the constant number 79S750, will be reduced to 



559125, so that for such purposes the formula would be 



559125 



'3 S =: lb. weight of rim per h.p. 



The following table has been constructed according to the foregoing 

 rules, and column 7 has been added for formula 5. 



TABLE B. 



798750,. 

 then — 77— S = 



The precedent Rules 2, 3, and 4, will suflSce when the proportions 

 laid down by Rule 1 are adopted, but will not answer for general pur- 

 poses; I have given them because, having for several years adhered to 

 the length of stroke, and other dimensions shown in Table A, these 

 rules have answered perfectly so far as the regulation of the motion 

 of the engine itself is concerned, when exerting her full power. 



The following Rules 5, 6, and 7, may be employed in all cases for 

 the same purpose, let the stroke of the engine, the diameter and 

 speed of the wheel be what they may. 



Rule 5. To find the weight in lb. per h. p. of the rim of the fly- 

 wheel, the length of stroke, the mean diameter, and speed of the 



