196 OF THE PRESSURE OF FLUIDS ON DYKES AND EMBANKMENTS. 



Thus, let A B c D be a vertical section of 

 the dyke as before, and bisect the parallel /Hn\ 



sides AB and DC in the points m and w, 

 and join mn\ then, the straight line mn 

 will pass through the centre of gravity of 

 the figure ABCD. 



Take g a point such, that mg is to ng, as 2n c -j- A B is to BC -}- 2AD, 

 and g will be the centre of gravity sought ; through the points m and 

 g, draw the straight lines mr and gs, respectively perpendicular to 

 DC the base of the section, then is sc the length of the lever by which 

 the weight of the dyke or embankment opposes the horizontal pressure 

 of the fluid. 



From the points A and B, draw the straight lines AK and BL, re- 

 spectively perpendicular to D c ; then it is manifest from the principles 

 of geometry, that 



rw= (CL DK), 



and this, by restoring the symbols for CL and DK, becomes 

 rn~^(e c). 



But mrzrD; consequently, by the property of the right angled 

 triangle, we have 



m n* m m i* -f- n r 3 ; 

 or by restoring the analytical values, it is 



raw 2 =iD 9 4- \(e cf\ 

 therefore, by extracting the square root, we have 



m n J -Y/ 4o 2 -|- (e -= c)". 



By the property of the centre of gravity, and according to the 

 foregoing construction, the point g is determined in the following 

 manner. 



2DC-{- AB = 36 c e 

 36 2c 2e 



66 3c 3e : i^/4D 2 4-(e c) 8 : : 36 2c 2e : gn, 

 from which, by reducing the analogy, we get 



(36 2c 2e) v /4o 2 -f- (e c) s 

 9 n -~ 6 (26-c- e ) 



and by the property of similar triangles, it is 



2c 



