236 OF SPECIFIC GRAVITY AND THE WEIGHING OF SOLID BODIES. 



_ sw* s'w 



w' w 



But the real or absolute weight of any body, is expressed by its 

 magnitude drawn into the specific gravity ; consequently, we have 



s w> s/ w 



_ 



W - W 



let this value of the real weight be compared with that above, and we 

 shall have 



.5236 d 3 (s w' s'w) _ sw' s' w 



w w s s' 



If the expression (sw' s'w) be suppressed on both sides of the 

 above equation, we shall obtain 



.5236rf_ 1 

 w' w s s' ' 

 and again, by suppressing the denominators, we get 



.5236(5 t')d* = u/ w; 

 therefore, dividing by . 5236 (s 5'), we have 



".5236(5 57 

 and finally, by extracting the cube root, we obtain 



.5236(5 5') (188). 



275. Now, the methods of reducing the equations (187) and (188), 

 or the practical rules derived from them, may be expressed as follows. 



1. When the absolute weight and specific gravity are given. 



RULE. Divide the absolute weight of the body, by .5236 

 times the specific gravity, and the cube root of the quotient 

 will be the diameter of the solid sought. 



2. When the weights indicated by the body in water and in air 



are given. 



RULE. Divide the difference between the weights, as obtained 

 from weighing the body in air and in water, by .5236 times 

 the difference between the specific gravities of water and air ; 

 then, the cube root of the quotient will be the diameter of the 

 solid sought. 



276. EXAMPLE 1. The absolute weight of a globular body is 14 Ibs.* 

 and its specific gravity 7 ; what is its diameter ? 



