94 OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 



But the pressure perpendicular to any surface, whatever may be its 

 form, as we have already sufficiently demonstrated : 



Is equal to the area of the surface multiplied by the 

 perpendicular depth of the centre of gravity, and again by 

 the specific gravity of the fluid. 



Therefore, the pressure perpendicular to the convex surface of the 

 upper segment DAE, is 



P = 3.1416Da: X i* X * = 1.5708 DSX\ (51). 



and the pressure perpendicular to the convex surface of the lower 

 segment DBE, is 



(52). 



Now these two expressions, according to the conditions of the 

 problem, are equal to one another; consequently, by comparison, 

 we get 



1 .5708 D s x* 3.1416 D s {(D x) x -f g (D a;) 8 }, 

 and from this, by suppressing the common quantities, we have 



* a :=:2{(D-aO*+i(D-.af}; 

 therefore, by expanding the terms, we obtain 



2x* D S ; 

 consequently, by division and evolution, we get 



ar=zJDV2. (53). 



104. The ultimate form of this equation is extremely simple, and 

 the practical rule which it supplies, may be expressed as follows. 



RULE. Multiply the radius, or half the diameter of the 

 sphere by the square root of 2, and the product will give 

 the point in the vertical diameter through which the plane of 

 division passes, estimated downwards from the surface of the 

 fluid. 



105. EXAMPLE 22. A sphere or globe, whose diameter is 18 inches, 

 is immersed in a fluid, in such a manner, that the upper extremity is 

 just in contact with the surface ; through what point of the diameter 

 must a horizontal plane be made to pass, so to divide the sphere, that 

 the pressures on the curve surface of the upper and lower segments 

 may be equal to one another ? 



The square root of 2, is 1.4142, and half the given diameter is 9 

 inches ; consequently, by the rule we have 



x =1.4142 X 9 12. 7278 inches, 



