100 OF FLUID PRESSURE UPON TTTE INTERIOR OF A CYLINDER. 



and that of the upright surface, is 



a 3.1416DG?; 

 consequently, the pressure on the bottom becomes 



P = . r 7S54v*ds, (59). 



and for the pressure upon the upright surface, we have 



p=:3.Ul6vdSs; (60). 



therefore, by analogy, we obtain 

 P :p : : .7854 D*ds : 3.1416D^s; 

 from which, by omitting the common factors, we get 



P :p :: D : 43; 



now, by the construction of the figure, we have 3 z= \d ; therefore 

 4 zz 2d, and by substitution, we obtain 



P : p : : D : 2d : : |D : d. 



Hence it appears, that the pressure upon the bottom of a cylindrical 

 vessel, is to the pressure upon its upright surface, as the radius of the 

 base is to the perpendicular altitude. 



Since the entire pressure sustained by a cylindrical vessel, is equal 

 to the sum of the pressures on the bottom and the upright sides, it 

 follows, that 



P -f p = p'= .7854 (D + 43) vds, 

 or substituting \d for 3, we shall obtain 



p' = .7854 (D + 2d) vds. (61). 



It is demonstrated by the writers on mensuration, that the solid 

 content of a cylinder, is equal to the area of its base, drawn into its 

 perpendicular altitude ; therefore, we have 



where C denotes the solid content of the cylinder. 



Now, it is manifest, that the weight of an incompressible and non- 

 elastic fluid, is equal to its magnitude drawn into its specific gravity; 

 hence we have 



but this is precisely the expression which we have given in equation 

 (59), for the pressure perpendicular to the bottom of the vessel ; con- 

 sequently, the weight of the fluid, and the pressure on the bottom of 

 the vessel, are equal to one another ; hence, the following inference. 



110. When the sides of a vessel of any form whatever, are perpen- 

 dicular, and its base parallel to the horizon : 



