BY FINITE DIFFERENCES OF PHYSICAL PROBLEMS, ETC. 337 



2 = 0, is 1 3 '2 5 square metres. And this, at a density of 2'25, gives a mass of 

 29 '81 metric tons above the rock surface per slice 1 metre thick. The centre of this 

 mass was found by graphical construction to be at the point x = + 1'335,* z = 2'19. 

 The total horizontal thrust of the water on the same slice is 18'0 tons. In dividing 

 the forces acting on the structure into lake pressure and a force-at-a-point, after the 

 manner of 4'1'13, I have considered the lake to extend right up to the vertical face. 

 This leaves a force of 0125 ton acting upwards near the corner, due to the bevel 

 diminishing the depth of water just there to be included in the force at a point. 



In finding the force at a point, the total horizontal thrust of the water of 18 tons 

 acting through a point f of the height of the dam from its top, is compounded with 

 the weight of 29'82 tons acting through the centre of mass and with the 0'125 ton 

 acting upwards near the corner. The resultant of 34'74 tons strikes the base at an 

 angle, the tangent of which is 1'657 (= 58 54'), at a point the co-ordinates of which 

 are x = +2'55, z = 0. If there had been a pronounced curve at the bottom of the 

 front it would have been necessary to use a link polygon to find the load point of the 

 horizontal 2 = 0. 



4 '2 '2. Taking six co-ordinate differences in the height the surface conditions were 

 integrated by equations (13), (14), &c. A difficulty arose at top. For the dam there 

 being only one co-ordinate difference thick, that is, ^ being expressed by three 

 columns of numbers only, the four boundary conditions cannot necessarily be simul- 

 taneously satisfied. It was avoided by making the total change in S^/Bce over the 

 base equal to the total downward force, and by making the total change in f>-^/Sz over 

 the base equal to the horizontal thrust of the water, and then integrating the boundary 

 strips separately from in front and behind, and making ^ one- valued where they met 

 at the top. 



On the lower boundary ^ was made equal to the sum of the following four parts : 



(i.) For the lake pressure, the values of 6i// 3 , found directly from the co-ordinates 

 x and 2. They are given in Table IX. 



(ii.) For the force at a point 34'74i/; 2 , found from the factors r and r sin 6, measured 

 on a large sheet of scale paper. 34'74i// a is also given in Table IX. They may be of 

 use to anyone re-working the problem for a dam of a different shape. 



(iii.) A linear function added to the above two in order to make the sum agree 

 with the values of ^ on the air surface behind the dam. It would have been simpler 

 to have altered the surface values to make them agree with the base, but I did not 

 think of that in time. This linear function is most conveniently specified by its first 

 differences, which are 3^/3,r = +9'735, 8^/82= 15'35, and by the fact that it 

 vanishes where the point-force cuts 2 = 0. These first differences were calculated 

 from the first space-rates of the expression 34'747r~V0sin 6 along and normal to 

 6 = 58 54', and were used to calculate the linear function. 



(iv.) Finally, -g-2 3 was calculated and added. 



* By calculation more exactly 1 340, 

 VOL. OCX. A. 2 X 



