148 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION 



EXAMPLE. The diameter of the bellows or cylindrical vessel, is 24 

 inches, the weight of the suspending fluid is 2 Ibs., and the weight 

 suspended on the bellows 8000 Ibs. ; what is the diameter of the 



tube? 



Performing according to the rule, we have 



and from this, by extracting the square root, we obtain 

 6?=V- 144 = - 38 of an i 



PROBLEM XXIV. 



157. In a hydrostatic bellows of a cylindrical form, the 

 diameter of the tube, the weight suspended, and the weight of 

 the suspending fluid, are given : 



It is required to determine the diameter of the bellows, so 

 that the whole may be in a state of equilibrium. 

 Let both sides of the general equation (113), be divided by w' the 

 weight of the suspending fluid, and we shall have 



__ 

 w' 



from which, by extracting the square root, we get 



w 1 ' (119). 



And from this equation, we obtain the following practical rule. 



RULE. Multiply the square of the diameter of the suspend- 

 ing tube, by the weight suspended, and divide the product by 

 the weight of the fluid which maintains the equilibrium; then, 

 the square root of the quotient will be the diameter of the 

 cylinder sought. 



EXAMPLE. The diameter of the suspending tube in a cylindrical 

 hydrostatic bellows, is half an inch, the weight of the suspending fluid 

 is 2 Ibs., and the weight suspended on the bellows board is 12000 Ibs. ; 

 what is the diameter of the bellows ? 



Here, by proceeding as directed in the foregoing rule, we get 



.5X. 5X12000 

 D*~ 1500, 



* This equation for the diameter of the bellows may be otherwise expressed ; thus : 



