Factors Influencing Power Needed
Work is performed whenever an object is moved against a resistance, and the amount of work performed depends not only on the amount of resistance overcome but also upon the amount of time utilized in accomplishing a given task. Work is measured in horse-power for convenience. It will take one horse-power to move 33,000 pounds one foot in one minute or 550 pounds one foot in one second. The same work would be done if 330 pounds were moved 100 feet in one minute. It requires a definite amount of power to move a vehicle over the ground at a certain speed, so it must take power to overcome resistance of an airplane in the air. Disregarding the factor of air density, it will take more power as the speed increases if the weight or resistance remains constant, or more power if the speed remains constant and the resistance increases. The airplane is supported by air reaction under the planes or lifting surfaces and the value of this reaction depends upon the shape of the aerofoil, the amount it is tilted and the speed at which it is drawn through the air. The angle of incidence or degree of wing tilt regulates the power required to a certain degree as this affects the speed of horizontal flight as well as the resistance. Resistance may be of two kinds, one that is necessary and the other that it is desirable to reduce to the lowest point possible. There is the wing resistance and the sum of the resistances of the rest of the machine such as fuselage, struts, wires, landing gear, etc. If we assume that a certain airplane offered a total resistance of 300 pounds and we wished to drive it through the air at a speed of sixty miles per hour, we can find the horse-power needed by a very simple computation as follows:
The product of 300 pounds resistance times speed of 88 feet
per second times 60 seconds in a minute = H.P. needed.
divided by 33,000 foot pounds per minute
in one horse-power
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:
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).
Another method for finding the drift is D = K × AV2, when we take the drift again so as to be greatest.
The second “resistance” is the total head resistance of the machine, at its maximum velocity. And the third is the thrust for climbing. The horse-power for climbing can be found out in two different ways. I first propose to deal with the method, where we find out the actual horse-power wanted for a certain climbing speed to our machine, where
H.P. = climbing speed/sec. × W
550
In this case we know already the horse-power for climbing, and we can proceed with our calculation.
With the other method we shall find out the “thrust” in pounds or kilograms wanted for climbing and add it to drift and total head resistance, and we shall have the total “thrust” of our machine and we shall denote it with T, while thrust for climbing shall be Tc.
The following calculation is at our service to find out this thrust for climbing
Vc × W = H.P.,
550
thence
Vc = H.P. × 550 (1)
W
H.P. = Tc × V ,
550
then from (1)
Tc × V × 550
550
Tc × V
Vc =
,
W
W
thence,
Tc = Vc × W .
V
Whether T means drifts, head resistance and thrust for climbing, or drift and head resistance only, the following calculation is the same, only in the latter case, of course, we must add the horse-power required for climbing to the result to obtain the total horse-power.
Now, when we know the total thrust, we shall find the horse-power in the following manner:
We know that the
H.P. = P r 2π R
75 × 60
in kilograms, or in English measure,
H.P. = P r 2π R (Fig. 1, B)
33,000
where
| P | = | pressure in klgs. or lbs. |
| r | = | radius on which P is acting. |
| R | = | Revolution/min. |
When P × r = M, then
H.P. = M.R.2π ,
4,500
thence,
| M = | H.P. × 4,500 | = | 716.2 H.P. | in meter kilograms, |
| R2π | R |
or in English system
| M = | H.P. 33,000 | = | 5253.1 H.P. | in foot pounds. |
| R2π | R |
Now the power on the circumference of the propeller will be reduced by its radius, so it will be M/r = p. A part of p will be used for counteracting the air and bearing friction, so that the total power on the circumference of the propeller will be (M/r) × η = p where η is the mechanical efficiency of the propeller. Now η/tan α = T, where α is taken on the tip of the propeller.
I take α at the tip, but it can be taken, of course, at any point, but then in equation p = M/r, r must be taken only up to this point, and not the whole radius; but it is more comfortable to take it at the tip, as tan α = Pitch/r2π (Fig. 1, C).
Now we can write up the equation of the thrust:
| T = | 716.2 H.P. η | , or in English measure | 5253.1 H.P. η | , |
| R r tan α | R r tan α |
thence
| H.P. = | T × R × r tan α | , or in English measure | T × R × r tan α | . |
| 716.2 η | 5253.1 η |
The computations and formulæ given are of most value to the student engineer rather than matters of general interest, but are given so that a general idea may be secured of how airplane design influences power needed to secure sustained flight. It will be apparent that the resistance of an airplane depends upon numerous considerations of design which require considerable research in aerodynamics to determine accurately. It is obvious that the more resistance there is, the more power needed to fly at a given speed. Light monoplanes have been flown with as little as 15 horse-power for short distances, but most planes now built use engines of 100 horse-power or more. Giant airplanes have been constructed having 2,000 horse-power distributed in four power units. The amount of power provided for an airplane of given design varies widely as many conditions govern this, but it will range from approximately one horse-power to each 8 pounds weight in the case of very light, fast machines to one horse-power to 15 or 18 pounds of the total weight in the case of medium speed machines. The development in airplane and power plant design is so rapid, however, that the figures given can be considered only in the light of general averages rather than being typical of current practice.