OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 207 



Having thus assigned a particular value to the letter n, we shall 

 next proceed to illustrate the reduction of the equation ; for which 

 purpose, take the following example. 



226. EXAMPLE 3. The vertical transverse section of the wall which 

 supports the water in a reservoir, is 24 feet in perpendicular height ; 

 what is the thickness at the base of the wall, supposing the section to 

 be in the form of the frustum of an isosceles triangle, the slope or 

 inclination on each side, being equal to 2 feet, and the specific gravity 

 of the material If, that of water being expressed by unity ? 



Let the several numerical values here specified, be substituted 

 instead of the respective symbols in the equation (170), and we shall 

 obtain 



_24>O_ ___ nearly. 



The breadth of the section, or the thickness of the dyke at the 

 bottom, being thus determined, the breadth or thickness at the top 

 can easily be found, for we have 



6.571 4 = 2.571 feet. 



127. If the slope c should vanish; that is, if the side of the dyke 

 on which the water presses be perpendicular to the horizon ; then, the 

 equation (170), becomes 



(171). 



And if the opposite slope e becomes evanescent, while the slope c 

 remains ; then we have 



* = 2^7 +ic< (172). 



But if both slopes vanish, or the section of the wall becomes 

 rectangular; then, the equation (170) is 



(173). 



If therefore, the perpendicular altitude of the section, and the 

 specific gravity of the materials of which the dyke is composed, remain 

 as in the preceding example ; then we shall have 



b= 24X1 =4.571 feet. 

 2XfX 



228. When the section of the wall assumes the form of a right 

 angled triangle ; that is, when the slope c vanishes, and e becomes 

 equal to the whole breadth b ; then we have 



