OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 203 



219. There is still another case of very frequent occurrence that 

 remains to be considered, viz. that in which the section is in the form 

 of a right angled triangle, having its vertex on the same level with 

 the surface of the fluid. 



This case will also admit of two varieties, according as the perpen- 

 dicular side of the dyke is, or is not in contact with the fluid ; when 

 it is in contact with it c vanishes, and since the section is in the form 

 of a triangle, the breadth of the base b is equal to the remote slope e, 

 and the vertical pressure of the fluid on the dyke is evanescent; con- 

 sequently, the equation marked (149) becomes 



sd* = 3Ds'P DsV; (162). 



but by the nature of the problem e 2 is equal to 6 2 , and by the hypo- 

 thesis of equal altitudes c?=z= D ; therefore, in the case of water, whose 

 specific gravity is expressed by unity, we obtain 



Zs'tfi^d*; 

 and from this, by division, we get 



and finally, by extracting the square root, it is 



b d\/ _L 



V 2s'' (163). 



220. This is also a very simple expression for the base of the section, 

 and the rule for its reduction is simply as follows. 



RULE. Divide the specific gravity of the incumbent fluid, 

 by twice the specific gravity of the dyke or embankment, and 

 multiply the perpendicular depth of the fluid by the square 

 root of the quotient, for the required thickness of the dyke. 



Let the perpendicular altitude and the specific gravity of the wall, 

 be 20 feet and 1 j respectively, as in the foregoing cases, and we shall 

 have 



6 = 2 <V 2 -^ 5 = 10.68 feet. 



221. Lastly, if the fluid come in contact with, or press upon the 

 hypothenuse of the triangle ; then the slope e vanishes, and b and c 

 are equal ; consequently, equation (149) becomes 



and if the vertical pressure of the fluid be omitted, the first term on 

 the right hand side of the equation vanishes, and consequently, we get 



sd*=vs'b*; 



