It is not my purpose here to discuss this law, or treat generally or specially of the theory of steam navigation. It will suffice that I point out clearly its existence and the prominent methods of its application only, as these are necessary to the general deduction which I propose making, that rapid steamships can not support themselves on their own receipts. The general reader can pass over these formulae to p. 69, and look at their results. I. TO FIND THE CONSUMPTION OF FUEL NECESSARY TO INCREASE THE SPEED OF A STEAMER. Suppose that a steamer running eight miles per hour consumes forty tons of coal per day: how much coal will she consume per day at nine miles per hour? The calculation is as follows: 8^3 : 9^3 :: 40 : required consumption, which is, 56.95 tons. Here the speed has increased 12-1/2 per cent., while the quantity of fuel consumed increased 42-1/2 per cent. Suppose, again, that we wish to increase the speed from 8 to 10, and from 8 to 16 miles per hour. The formula stands the same, thus: Miles. Miles. Tons Coal. Tons Coal. 8^3 : 10^3 :: 40 : _x_, = 78.1 8^3 : 16^3 :: 40 : _x_, = 320. II. TO FIND THE SPEED CORRESPONDING TO A DIMINISHED CONSUMPTION OF FUEL. Murray has given some convenient formulae, which I will here adopt. Suppose a vessel of 500 horse power run 12 knots per hour on 40 tons coal per day: what will be the speed if she burn only 30 tons per day? Thus: 40 : 30 :: 12^3 : V^3 (or cube of the required velocity,) Or, reduced, 4 : 3 :: 1728 : V^3, Equation, 3 x 1728 = 5184 = 4V^3, Or, 5184/4 = Cube root of 1296 = 10.902 knots = V, required velocity. Thus, we reduce the quantity of coal one fourth, but the speed is reduced but little above one twelfth. III. RELATION BETWEEN THE CONSUMPTION OF FUEL, AND THE LENGTH AND VELOCITY OF VOYAGE. The consumption of fuel on two or more given voyages will vary as the square of the velocity multiplied into the distance travelled. Thus, during a voyage of 1200 miles, average speed 10 knots, the consumption of coal is 150 tons: we wish to know the consumption for 1800 miles at 8 knots. Thus: 150 tons : C required Consumption :: 10^2 knots x 1200 miles : 8^2, Knots x 1800 miles. Then, C x 100 x 1200 = 150 x 64 x 1800,* Or, C x 120,000 = 17,280,000 Reduced to C = 1728/12 = 144 tons consumption. Suppose, again, that we wish to know the rate of speed for 1800 miles, if the coals used be the same as on another voyage of 1200 miles, with 150 tons coal, and ten knots speed: We substitute former consumption, 150 tons for C, as in the equation above, marked *, and V^2 (square of the required velocity) for 64, and have, 150 x 100 x 1200 = 150 x V^2 x 1800, Or, 120,000 = 1800V^2, Reduced, 1200/18 = V^2, And V = square root of 66.66 = 8.15 knots.