BACKWATEE FUNCTION 321 



of equation (1), p. 168. Assuming this to be done, we have, in the 

 preceding formulae 



Let the suffix (1) refer to a point just above the dam, and the suffix (2) 

 to the given point. Then since the positive direction of I is down-stream, 

 we have in equation (10), hi = 10 ; li = ; ^ = 2640, and from the 



tables we get <t> (?i) = </> ( J = '927, so that this equation becomes 

 10 - fa = -002 (2640) + '66 '927 - </> 



This equation can only be solved by trial. 

 Let /i2 - 



Then if fa = 5, y = 5 '6593 4'102 = + '2387 



hz = 4, y = 4 - '6927 - 4'102 = - '7947 



For a solution of the equation, y must equal 0, and since the value of 

 a continuous function such as y cannot change from + to without 

 passing through the value zero, it follows that for some value of fa 

 between 4 and 5, y = 0. 



Evidently, too, the correct value of fa is nearer 5 than 4. Try 

 fa = 4'75. 



If fa = 4-75, y = 4-75 - -6656 - 4-102 = - -0176. The value of fa 

 is then between 4 - 75 and 5*0. 



A close approximation to the correct result can then be obtained by 

 drawing a curve connecting those values of y and of h% already found. 

 Where this curve intersects the axis of fa, we shall have the value of fa 

 which makes y = 0, and therefore which satisfies the equation. In the 

 problem, h% = 4'78 provides a very close approximation to the correct 

 value. At a distance up-stream equal tc 4,000 feet the value of h, 

 determined in the same way, is 2'34 feet. Since the slope is *002 the 

 height of the bed at this latter point, above that at the dam, is '002 X 

 4,000 = 8'0 feet. 



The surface at this latter point is therefore '34 feet higher than at the 

 dam. With the dam removed and the flow per minute unchanged, the 

 flow being uniform and the depth of channel equal to T, the difference 

 of level instead of being '34 feet would be 8'0 feet. 



Figure 142 illustrates the form of backwater curve observed by 



H.A, y 



