204 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 



but C?:=D, and s=zl ; therefore, we shall have 

 and finally, by division and evolution, we obtain 



t=i ^ 



222. This equation is of a still simpler form than that which arises 

 when the perpendicular side is towards the pressure, and the rule for 

 its reduction is as follows. 



RULE. Divide the specific gravity of the fluid, by that of 

 the dyke or embankment, and multiply the perpendicular alti- 

 tude by the square root of the quotient, for the breadth of 

 the base. 



Therefore, by taking the altitude and specific gravities hitherto 

 employed, the rule will give 



6 = 20A/ =15.11 feet. 

 V 1.75 



223. This gives a thickness for the base of the section, exceeding 

 the thickness in the former case by 4.33 feet; which seems to be a 

 very great difference, when it is considered that both the form and 

 the perpendicular altitude of the wall are the same in both cases ; but 

 the reason of the difference will become manifest from the following 

 construction. 



Let ABC and a be be two right angled triangles, equal to one 

 another in every respect, but having 

 their perpendiculars opposed in such a 

 manner, that the water pressing in the 

 same horizontal direction, is resisted by 

 the perpendicular AB in the one case, 

 and by the hypothenuse ac in the other. 



Bisect the sides AB, AC in the point 

 M and N, and a b, ac in the points m 



and n respectively; draw the lines CM and BN intersecting in G, and 

 cm and bn intersecting in g; then are G and g, the centres of gravity 

 of the respective triangles ABC and a be. 



Demit the straight lines GH and gh, perpendicularly to BC and be; 

 then are HC and h b the levers, by which the weights of the sections, 

 supposed to be concentrated in their respective centres of gravity, 

 resist the horizontal pressure of the fluid which tends to turn them 

 round the points c and b. 



