Let A B 6 F, represent the thickness of the wall. Its centre of gravity is at O, and is horizontally projected at m. The centre of gravity of the thrusting triangle of earth, B 4 6, is C, (found by the cutting of lines joining any two angles to the centre of the opposite sides,) is horizontally projected at Ca, and the horizontal component of the thrust is exerted at 2, tending to overthrow the wall with a leverage, 6 2.

Fig. 133.

The overthrowing power is, then, the area of the triangle B46 × the weight of the unit of area × the leverage 6 2. And the resisting power, the area AB × B6 × the weight of a unit of area by one half breadth, or m 6; or, calling w the weight of the wall, and w′ that of the triangle, B 4 6, and L and L′ the leverage respectively of the wall to resist and of the earth to overthrow; we must have at least

wL = w′L′,

and to insure stability,

wL > w′L′

or,

L = w′L′
w,

and as L = half base finally, the thickness, or