II. TO FIND THE SPEED CORRESPONDING TO A DIMINISHED CONSUMPTION OF FUEL.
Murray has given some convenient formulæ, 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 | :: | 123 | : | V3 (or cube of the required velocity,) | |
| Or, reduced, | 4 | : | 3 | :: | 1728 | : | V3, |
| Equation, | 3 × 1728 = 5184 | = | 4V3, | ||||
| Or, 5184/4 | = | ||||||
| 3√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 :: 102 knots × 1200 miles : 82 knots × 1800 miles.
| Then, | C × 100 × 1200 | = | 150 × 64 × 1800,* |
| Or, | C × 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 V2 (square of the required velocity) for 64, and have,