OF THE HYDROSTATIC BELLOWS. 149 



and by extracting the square root, we have 



:rr 38.73 inches. 



158. The foregoing problems and rules, unfold every particular 

 respecting the calculation of the hydrostatic bellows, and from them 

 we may infer, that in the case of an equilibrium, if more fluid be 

 added : 



It will ascend equally in the suspending tube, and in the 

 cylindrical vessel composing the bellows, whatever may be 

 their relative magnitudes. 



The demonstration of this is very simple, for let A BCD be a vertical 

 section, passing along the axis of the cylindrical 

 vessel, and also along the axis of the suspending 

 tube KI ; and suppose that F and c are the points 

 to which the fluid rises in the vessel and the tube, 



when the bellows is in a state of equilibrium. 



Take ic equal to Da, and through the points a D C I 

 and c let a horizontal plane be drawn, intersecting 

 the vertical plane A BCD in the line ab ; then it is manifest, that the 

 weight w in the position EF, is equivalent to the weight of the fluid 

 column a&FE. Let more fluid be poured into the tube at K; the 

 equilibrium will then be destroyed, and the weight w will ascend, 

 until by discontinuing the supply, the equilibrium is restored, and the 

 fluid in the vessel and the tube becomes again quiescent at the points 

 n and K. 



Take IK equal to DA, and through the points A and K, let a hori- 

 zontal plane be drawn, cutting the vertical plane A BCD in the line 

 AB; then as before, the weight w in the position mn, is equivalent 

 to the weight of the fluid cylinder, of which A&nm is a vertical 

 section. 



Now, the weight w is not altered in consequence of the change of 

 position from EF to mn\ therefore, because EF is equal to mn, it 

 follows, that E is equal to mA; consequently, by taking away the 

 common space ma, the remainders Em and a A are equal to one 

 another; but by reason of the parallel lines ac and AK, the spaces 

 a A and CK are equal to one another; therefore CK is equal to 

 Em. 



From the principle here demonstrated, the resolution of the follow- 

 ing problem may readily be derived. 



