214 IRRIGATION. 



ing actions of currents. These are either by being 

 overturned by the horizontal pressure of the water, or by 

 being forced from its position bodily by sliding upon its 

 base. The first alternative may be examined by con 

 sidering what power a certain structure a vertical wall 

 for instance exercises to resist the pressure of water, and 

 what the pressure amounts to for a certain hight. The 

 pressure of water upon any surface immersed in it, is 

 equal to the area of the surface multiplied by the depth 

 of its center of gravity below the level of the water, and 

 by the weight of a unit of water. The unit adopted in 

 these calculations is a foot, and a cubic foot of water 

 weighs 62 1 | a pounds. The resulting pressure is therefore 



. 101. GENERAL FORM OF DAM. 



readily found. Let it be supposed that a wall 10 feet 

 high is sustaining a body of water behind it, as shown in 

 fig. 102. One foot in length of the wall is taken as a 

 basis for the calculation. There is then 10 square feet 

 subject to pressure ; the depth of the center of gravity 

 is 5 feet ; and the weight of a foot of water is 62 1 | a pounds. 

 The product of these numbers is 3,125, which is the 

 number of pounds pressing upon one foot in length of 

 the wall. But this pressure, in this case, is not evenly 

 distributed over the whole wall, but in consequence of the 

 mobility of the water, the pressure is so distributed as to 

 be equal to, and to operate as, a single force acting at a 

 point one-third of the hight of the wall from the bot 

 tom. For this reason the product previously arrived at 

 should be multiplied by one-third of the hight, or 3 1 1 3 , 

 which will give as the total pressure exerted to overthrow 

 or push forward the wall, 10,406 pounds on every foot in 



