then we obtain by the following formula the amount of work done, the co-efficient of traction being multiplied by 2,240 (the number of pounds in a ton) to obtain the result in foot-tons.
(W + W1) × D ∕ (C × 2,240)
In ascending a height, a man raises his whole weight through the height ascended.
A regiment of soldiers marches ten miles, each carrying a weight of 60 lbs. What amount of work is performed by each soldier?
If we assume the average weight of each soldier to be 150 lbs., and that the march was at the rate of three miles an hour, then—
(150 + 60) × 10 × 5,280 ∕ (20 × 2,240) = 247·5 foot-tons.
In this example it is assumed that the march is on entirely level ground that all weights are carried in the most convenient manner, and that the rate of travel is three miles an hour. Velocity is gained at the expense of carrying power. It has been found that the amount of work is generally inversely as the square of the velocity. Haughton has determined from Weber’s calculations the co-efficient of resistance for three velocities.
| VELOCITY. | CO-EFFICIENT OF TRACTION OR RESISTANCE. |
|---|---|
| 1·818 miles per hour | 1 ∕ 28·27 |
| 4·353 miles per hour | 1 ∕ 13·70 |
| 10·577 miles per hour | 1 ∕ 7·51 |
Parkes has extended these calculations to show the distance in miles required to be travelled at various velocities to do work equal to 300 foot-tons, and the time required in each instance.
| VELOCITY IN MILES PER HOUR. | CO-EFFICIENT OF RESISTANCE. | DISTANCE FOR MEN OF 156 LBS. TO EQUAL 300 FOOT-TONS. | TIME REQUIRED IN HOURS AND MINUTES. | |
|---|---|---|---|---|
| 2 | 1 ∕ 26·74 | 12·2 | hrs. 10 | mins. 36 |
| 3 | 1 ∕ 20·59 | 16·3 | 5 | 24 |
| 4 | 1 ∕ 16·74 | 13·3 | 3 | 18 |
| 6 | 1 ∕ 12·18 | 9·6 | 1 | 36 |
| 8 | 1 ∕ 9·60 | 7·6 | 0 | 57 |
| 10 | 1 ∕ 7·89 | 6·3 | 0 | 38 |