hat he would have to walk over eight miles at a rate of
three miles per hour in order to save money when he pays a 5-cent
carfare. (This, however, does not include the cost of shoe leather.)

The carrying of a load of 44 pounds is done at the same expenditure of
energy as the carrying of one's own body weight when the rate is three
miles an hour, so the soldier's equipment would call for the added
expenditure of 48 calories (44 x 1.1), making his total hourly
expenditure of energy nearly 300 calories (249 + 44) during a hike on a
level road. His daily requirement for energy might be:

Calories

Sleeping 8 hours at 70 calories per hour 560
Resting in camp 6 hours at 77 calories per hour 462
Hike of 30 miles, 10 hours at 300 calories per hour 3000
----
4022

This would be the heat production of a soldier on a day of a "forced
march." The ordinary day's march is only fifteen miles.

This assumes a level road. If, however, there are hills to climb and
the body weight and the pack are lifted 1000 feet during the hike, this
is done at the additional expense of approximately 0.96 calory of energy
per pound of weight lifted. If the man weighed 156 pounds and the pack
44 pounds, the additional fuel requirement would be 192 calories (200 x
0.96). The total energy requirement for this kind of a hike would have
been 4200 calories. Walking down hill is accomplished at an expenditure
of slightly less energy than walking on the level, but this factor need
not concern one.

Supposing, however, this individual were running, lightly clad, on a
level road in a race for a distance of 40 miles at the rate of 5.3 miles
per hour, he would complete the distance in seven hours and thirty-three
minutes, which is a reasonable record. His metabolism might thus be
calculated:

Calories

Sleeping 10 hours at 70 calories per hour 700
Resting 6 hours, 23 minutes, at 77 calories per hour 497
Running 7 hours, 33 minutes, at 561 calories per hour 4236
----
5433

It is a matter of record that a man has run between Milwaukee