OF FLOATATION AND THE SPECIFIC GRAVITY OF BODIES. 217 



the fluid ascend when a sphere of 4 inches diameter is placed in it, 

 the specific gravities of the fluid and the sphere being equal to one 

 another ? 



lii this example there are given 5 ~ 6 inches and r zz 2 inches ; 

 therefore, by operating as directed in the rule, we shall have 



16r 3 =:16X2 3 128, the dividend, 

 and in like manner for the divisor, we get 



3S 2 = 3 X 12 2 432, the divisor ; 

 consequently, by division, we obtain 



x f Hi= 0.2962 of an inch. 



Hence it appears, that the height of the fluid in the vessel, is 

 increased by a quantity equal to 0.2962 of an inch, in consequence 

 of the immersion, and the whole height to which it rises, is 6.2962 

 inches. 



PROBLEM XXXII. 



239. A vessel in the form of a paraboloid, is placed with its 

 vertex downwards and its base parallel to the horizon; now, 

 supposing the vessel to be filled to the w th part of its capacity 

 with a fluid of known specific gravity, and let a spherical body 

 of a given size and substance be placed in it : 



It is required to ascertain the height to which the fluid 

 will rise, in consequence of the immersion of the spherical 

 segment. 



Let ABC represent a vertical section passing along the axis of the 

 vessel, whose form is that of a paraboloid, 

 generated by the revolution of the common 

 parabola ; and suppose the vessel to be filled 

 with an incompressible and non-elastic fluid 

 to the height sc, DE being its horizontal 

 surface when in a state of quiescence, before 

 the sphere whose diameter is mn is placed in 

 it; then will a b be the surface or the plane 

 of floatation after the immersion of the segment rnw, the fluid rising 

 to the height ts all around the spherical body. 



Now, it is obvious from the nature of the problem, that, the solidity 

 of the spherical segment rnw, together with the quantity of fluid in 

 the vessel, is equal to the magnitude of the paraboloid acb, whose 



