OF FLUID PRESSURE UPON THE INTERIOR OF A TETRAHEDRON. 97 



By the nature of the figure, the three containing planes A D E, ADC 

 and BDC, are equal to one another, and they are also equally inclined 

 to, or similarly situated with respect to the base ABC; consequently 

 DS, the perpendicular depth of the centre of gravity, is common to 

 them all. 



Now, by the property of the centre of gravity, we know, that or is 

 equal to two thirds of D n ; therefore, by reason of the parallel lines 

 np and rs, DS is also equal to two thirds of DP. 



Put a zn the area of the base and each of the other containing planes, 

 I ~ the length of the side of each triangular plane, or the edges 



of the figure, 

 d zz: DP, the perpendicular depth of the centre of gravity of the 



base ABC, 

 3 zr DS, the perpendicular depth of the centre of gravity of the 



side ADB ; 



P zz the pressure upon the b*ase, 

 i ' 



p zz the pressure upon 



wzzthe weight of the fluid contained in the vessel, and 

 s zz the specific gravity. 



Then, by the principles of Plane Trigonometry and the property of 

 the right angled triangle, we have 



DP d~ \/4 sec.* 30, 



but by the arithmetic of sines, we know, that 



secJ30zzli-; 

 consequently, by substitution, we have 



. i? d=-L V e. ' $ v; ::"::"; 



Now, according to the construction of the figure and the property 

 of the centre of gravity, it follows, that DS is equal to two thirds of 

 DP ; hence we get 



* 2J _ 

 DszzrSz:: ^6. 



By the nature of the figure, it is manifest, that the area of each of 

 the triangular sides is equal to the area of the base ; and by the prin- 

 ciples of mensuration : 



The area of an equilateral triangle, is equal to one fourth 

 the square of the side, multiplied by the square root. *f 

 three. 



X-oTST-^ 



/ or THE 



