┌─────────────────────┬────────────────────────┐ │ VELOCITY. │CO-EFFICIENT OF TRACTION│ │ │ OR RESISTANCE. │ ├─────────────────────┼────────────────────────┤ │ 1·818 miles per hour│ 1∕28·27 │ │ │ │ │ 4·353 „ „ │ 1∕13·70 │ │ │ │ │10·577 „ „ │ 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. ┌───────────────┬───────────────┬───────────────────┬────────────────┐ │ │ │DISTANCE FOR MEN OF│ TIME REQUIRED │ │ VELOCITY IN │CO-EFFICIENT OF│ 156 LBS. TO EQUAL │ IN HOURS │ │MILES PER HOUR.│ RESISTANCE. │ 300 FOOT-TONS. │ AND MINUTES. │ ├───────────────┼───────────────┼───────────────────┼────────────────┤ │ │ │ │ hrs. mins. │ │ 2 │ 1∕26·74 │ 12·2 │ 10 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 │ └───────────────┴───────────────┴───────────────────┴────────────────┘ The co-efficient 1∕20, corresponds very nearly to 3·1 miles per hour, and it appears that at this rate of travel the greatest amount of work can be done with the least expenditure of energy. _How much work is done by a man weighing 150 lbs. who walks 15 miles up an incline 1 in 200?_ The number of feet ascended in 15 miles = 5,280 × 15∕200 = 396. The amount of work done by the man in raising his own weight 396 feet high = 396 × 150∕2,240 = 26·5 foot-tons. The amount of work done in walking 15 horizontal miles at the rate of 3 miles an hour = 150 × 15 × 5,280/(20 × 2,240) = 265.2 foot-tons. Total amount of work done = 265.2 + 26.5 = 291.7 foot-tons. _Eight palanquin bearers carry an officer weighing 180 lbs. and a palanquin weighing 250 lbs., a distance of 25 miles. Assuming that each man weighs 150 lbs., what amount of work was done by each man? (Parkes.)_ 250 + 180 = 430