98 OF FLUID PRESSURE UPON THE INTERIOR OF A TETRAHEDRON. 



Consequently, the area of the base, and each of the containing 

 sides of the vessel, is expressed by 



therefore, the area of the three containing equilateral triangular 

 planes, becomes 



hence, for the pressure upon the base, we have 



P = 4JV3"X iV6 X s- iP 5V % (55). 



and the pressure upon the three containing planes, is 



j> = JJV3 X JV6 X s=il*s^2; (56). 



consequently, by analogy, we shall have 



P :p ::Psi/iTi JJt/2; 



and this, by suppressing the common factors, becomes 

 P : p : : 1 : 2. 



If the two equations marked (55) and (56) be added together, the 

 sum will express the aggregate pressure upon the vessel ; therefore 

 we have 



P + P=J>'=(i + 1) l*sV2 = $l*s^. (57). 



According to the principles of mensuration, the solid content of a 

 tetrahedronal vessel, is equal to the area of its base, multiplied by one 

 third of its perpendicular altitude ; therefore, we have 



IJViFx $1^/6 x i=ZTW2; 



now, the weight of the contained fluid, is manifestly equal to its 

 magnitude multiplied by the specific gravity ; consequently, we 

 obtain 



w = -frPsV2l (58). 



hence, by analogy, we get 



P : w : : %l s s j$ : ^s ^2, 

 and this, by suppressing the common factors, gives 



P : w : : 3 : 1. 



COROL. It therefore appears, that the pressure upon the base, is to 

 the pressure on the three sides, in the ratio of 1 to 2, and to the 

 weight of the contained fluid in the ratio of 3 to 1 ; consequently, the 

 weight of the fluid, the pressure on the base, and the pressure on the 

 sides, are to one another as the numbers 1, 3 and 6. 



