The result is the horse-power needed, or
| 300 × 88 × 60 | = 48 H.P. |
| 33,000 |
Just as it takes more power to climb a hill than it does to run a car on the level, it takes more power to climb in the air with an airplane than it does to fly on the level. The more rapid the climb, the more power it will take. If the resistance remains 300 pounds and it is necessary to drive the plane at 90 miles per hour, we merely substitute proper values in the above formula and we have
| 300 pounds times 132 feet per second times 60 seconds in a minute | = 72 H.P. |
| 33,000 foot pounds per minute in one horse-power |
The same results can be obtained by dividing the product of the resistance in pounds times speed in feet per second by 550, which is the foot-pounds of work done in one second to equal one horse-power. Naturally, the amount of propeller thrust measured in pounds necessary to drive an airplane must be greater than the resistance by a substantial margin if the plane is to fly and climb as well. The following formulæ were given in “The Aeroplane” of London and can be used to advantage by those desiring to make computations to ascertain power requirements:
Fig. 1.—Diagrams Illustrating Computations for Horse-Power Required for Airplane Flight.
The thrust of the propeller depends on the power of the motor, and on the diameter and pitch of the propeller. If the required thrust to a certain machine is known, the calculation for the horse-power of the motor should be an easy matter.
The required thrust is the sum of three different “resistances.” The first is the “drift” (dynamical head resistance of the aerofoils), i.e., tan α × lift (L), lift being equal to the total weight of machine (W) for horizontal flight and α equal to the angle of incidence. Certainly we must take the tan α at the maximum Ky value for minimum speed, as then the drift is the greatest ([Fig. 1], A).