90 ft. × 1 ft. = 90 square ft. I. Therefore the insistent pressure upon the surface of the ground at the base of one slope at low water, when the bed is presumed to be dry, is, E 75·60 ————— = 0·84 of a ton per square foot. 90 H J. At high water it is E F G (75·60 + 16·80) – 33·60 58·80 ——————————————————————— = ————— = 0·65 90 90 H of a ton per square foot, or 22·60 per cent. less. It has been shown that the vertical pressure of the water upon the ground beyond the slope is 0·56 ton per square foot at high tide, _vide_ D., therefore the weight upon the seat of the slope is only in excess of the normal weight of the water upon the ground beyond the slope, J D 0·65 – 0·56 = 0·09 of a ton. An inspection of the diagram shows that the flotation power of the whole of the shaded portion of the slope is balanced by the weight of water resting upon the whole of it, the areas of the triangles being similar; and that the portion W. of the slope is that which loses weight by immersion. Calculating the pressures at low and high water upon the base of the portion W., the relative vertical pressures would be— AT LOW WATER. Cubic ft. 30 ft. × 5 ft. × 1 ft. = 150 30 ft. × 20 ft. × 1 ft. = 600 ——— 750 × 0·056 = 42 tons. K. The area of the base = 30 ft. × 1 ft. = 30 square ft., consequently the pressure upon it = 42/30 = 1·40 ton per square foot. AND AT HIGH WATER. L. The unsubmerged portion = Cubic ft. Ton. Tons. 30 ft. × 5 ft. × 1 ft. = 150 × 0·056 = 8·40 The submerged portion = 30 ft. × 20 ft. × 1 ft. = 600 × (0·056 – 0·028) = 16·80/25·20 25·20 Consequently the pressure = ————— = 0·84 ton per square foot, 30 K a difference of 0·56 ton per square foot, or 40 per cent. less load. M. And upon the base of the central portion— AT LOW WATER. Cubic ft. Ton. Tons. 30 ft. × 30 ft. × 1 ft. = 900 × 0·056 = 50·40. 50·40 The pressure is, therefore, equal to ————— = 1·68 ton per square foot. 30 K AT HIGH WATER. N. The unsubmerged portion =