The weight of a cubic foot of sand is here taken as 0·056 ton. A. The weight of a lineal foot of the embankment when unsubmerged equals 201·60 tons, computed as follows:— Cubic ft. The central portion 30 ft. × 30 ft. × 1 ft. = 900 The two inclined 90 ft. × 30 ft. × 1 ft. = 2,700 portions ————— Ton. Tons. Cubic contents 3,600 × 0·056 = 201·60. B. At high water the weight of the embankment is reduced by the weight of the water displaced, which equals 84 tons, calculated as under. The submerged contents of the embankment are— Cubic ft. The central portion 30 ft. × 20 ft. × 1 ft. = 600 The two inclined portions 2/30 ft. × 20 ft. × 1 ft. = 1,200 60 ft. × 20 ft. × 1 ft. = 1,200 ————— Ton. Tons. Cubic contents 3,000 × 0·028 = 84. From this must be deducted the weight of the water resting upon the two slopes, which equals 33·60 tons— Cubic ft. Ton. Tons. 60 ft. x 20 ft. X 1 ft. = 1,200 X 0·028 = 33·60. C. Thus the insistent load at high water upon the whole area of the foundation is reduced by Tons. Tons. 84 – 33·60 50·40 —————————— = —————— = 0·25 = ¼. 201·60 201·60 D. At high water a vertical pressure is imposed upon the ground beyond the toe of the slope due to the 20 feet head of water— 20 ft. × 1 ft. × 1 ft. × 0·028 = 0·56 ton per square foot. This latter weight and element of stability tends to prevent movement of the ground, and also the toe of the slope, but is entirely removed at low water when the insistent pressure at the foot of the embankment is the greatest. For the purposes of illustrating the varying load upon the surface of the ground caused by a rise and fall of a tide, it will be sufficient to take one slope of the embankment. E. The weight of a lineal foot of one slope, if unsubmerged = Cubic ft. Ton. Tons. 90 ft. × 15 ft. × 1 ft. = 1,350 x 0·056 = 75·60. F. The weight of water resting upon the slope per lineal foot at high tide = Cubic ft. Ton. Tons. 60 ft. × 10 ft. × 1 ft. = 600 × 0·028 = 16·80. G. The weight of the water displaced by a lineal foot of the submerged portion of one slope of the embankment at high tide = Unsubmerged portion of one slope Cubic ft. Ton. Tons. 1,350 – (30 ft. × 5 ft. × 1 ft.) = 1,200 × 0·028 = 33·60. H. The area of the base of the slope per lineal foot =