Chapter XIII. Peculiarities of Airship Power.

As a general proposition it takes much more power to propel an airship a given number of miles in a certain time than it does an automobile carrying a far heavier load. Automobiles with a gross load of 4,000 pounds, and equipped with engines of 30 horsepower, have travelled considerable distances at the rate of 50 miles an hour. This is an equivalent of about 134 pounds per horsepower. For an average modern flying machine, with a total load, machine and passengers, of 1,200 pounds, and equipped with a 50-horsepower engine, 50 miles an hour is the maximum. Here we have the equivalent of exactly 24 pounds per horsepower. Why this great difference?

No less an authority than Mr. Octave Chanute answers the question in a plain, easily understood manner. He says:

"In the case of an automobile the ground furnishes a stable support; in the case of a flying machine the engine must furnish the support and also velocity by which the apparatus is sustained in the air."

Pressure of the Wind.

Air pressure is a big factor in the matter of aeroplane horsepower. Allowing that a dead calm exists, a body moving in the atmosphere creates more or less resistance. The faster it moves, the greater is this resistance. Moving at the rate of 60 miles an hour the resistance, or wind pressure, is approximately 50 pounds to the square foot of surface presented. If the moving object is advancing at a right angle to the wind the following table will give the horsepower effect of the resistance per square foot of surface at various speeds.

Horse Power Miles per Hour per sq. foot 10 0.013 15 0 044 20 0.105 25 0.205 30 0.354 40 0.84 50 1.64 60 2.83 80 6.72 100 13.12

While the pressure per square foot at 60 miles an hour, is only 1.64 horsepower, at 100 miles, less than double the speed, it has increased to 13.12 horsepower, or exactly eight times as much. In other words the pressure of the wind increases with the square of the velocity. Wind at 10 miles an hour has four times more pressure than wind at 5 miles an hour.

How to Determine Upon Power.

This element of air resistance must be taken into consideration in determining the engine horsepower required. When the machine is under headway sufficient to raise it from the ground (about 20 miles an hour), each square foot of surface resistance, will require nearly nine-tenths of a horsepower to overcome the wind pressure, and propel the machine through the air. As shown in the table the ratio of power required increases rapidly as the speed increases until at 60 miles an hour approximately 3 horsepower is needed.

In a machine like the Curtiss the area of wind-exposed surface is about 15 square feet. On the basis of this resistance moving the machine at 40 miles an hour would require 12 horsepower. This computation covers only the machine’s power to overcome resistance. It does not cover the power exerted in propelling the machine forward after the air pressure is overcome. To meet this important requirement Mr. Curtiss finds it necessary to use a 50-horsepower engine. Of this power, as has been already stated, 12 horsepower is consumed in meeting the wind pressure, leaving 38 horsepower for the purpose of making progress.

The flying machine must move faster than the air to which it is opposed. Unless it does this there can be no direct progress. If the two forces are equal there is no straight-ahead advancement. Take, for sake of illustration, a case in which an aeroplane, which has developed a speed of 30 miles an hour, meets a wind velocity of equal force moving in an opposite direction. What is the result? There can be no advance because it is a contest between two evenly matched forces. The aeroplane stands still. The only way to get out of the difficulty is for the operator to wait for more favorable conditions, or bring his machine to the ground in the usual manner by manipulation of the control system.

Take another case. An aeroplane, capable of making 50 miles an hour in a calm, is met by a head wind of 25 miles an hour. How much progress does the aeroplane make? Obviously it is 25 miles an hour over the ground.

Put the proposition in still another way. If the wind is blowing harder than it is possible for the engine power to overcome, the machine will be forced backward.

Wind Pressure a Necessity.

While all this is true, the fact remains that wind pressure, up to a certain stage, is an absolute necessity in aerial navigation. The atmosphere itself has very little real supporting power, especially if inactive. If a body heavier than air is to remain afloat it must move rapidly while in suspension.

One of the best illustrations of this is to be found in skating over thin ice. Every school boy knows that if he moves with speed he may skate or glide in safety across a thin sheet of ice that would not begin to bear his weight if he were standing still. Exactly the same proposition obtains in the case of the flying machine.

The non-technical reason why the support of the machine becomes easier as the speed increases is that the sustaining power of the atmosphere increases with the resistance, and the speed with which the object is moving increases this resistance. With a velocity of 12 miles an hour the weight of the machine is practically reduced by 230 pounds. Thus, if under a condition of absolute calm it were possible to sustain a weight of 770 pounds, the same atmosphere would sustain a weight of 1,000 pounds moving at a speed of 12 miles an hour. This sustaining power increases rapidly as the speed increases. While at 12 miles the sustaining power is figured at 230 pounds, at 24 miles it is four times as great, or 920 pounds.

Supporting Area of Birds.

One of the things which all producing aviators seek to copy is the motive power of birds, particularly in their relation to the area of support. Close investigation has established the fact that the larger the bird the less is the relative area of support required to secure a given result. This is shown in the following table:

Supporting
Weight Surface Horse area Bird in lbs. in sq. feet power per lb. Pigeon 1.00 0.7 0.012 0.7 Wild Goose 9.00 2.65 0.026 0.2833 Buzzard 5.00 5.03 0.015 1.06 Condor 17.00 9.85 0.043 0.57

So far as known the condor is the largest of modern birds. It has a wing stretch of 10 feet from tip to tip, a supporting area of about 10 square feet, and weighs 17 pounds. It. is capable of exerting perhaps 1-30 horsepower. (These figures are, of course, approximate.) Comparing the condor with the buzzard with a wing stretch of 6 feet, supporting area of 5 square feet, and a little over 1-100 horsepower, it may be seen that, broadly speaking, the larger the bird the less surface area (relatively) is needed for its support in the air.

Comparison With Aeroplanes.

If we compare the bird figures with those made possible by the development of the aeroplane it will be readily seen that man has made a wonderful advance in imitating the results produced by nature. Here are the figures:

Supporting
Weight Surface Horse area Machine in lbs. in sq. feet power per lb. Santos-Dumont . . 350 110.00 30 0.314 Bleriot . . . . . 700 150.00 25 0.214 Antoinette. . . . 1,200 538.00 50 0.448 Curtiss . . . . . 700 258.00 60 0.368 Wright. . . . .[4]1,100 538.00 25 0.489 Farman. . . . . . 1,200 430.00 50 0.358 Voisin. . . . . . 1,200 538.00 50 0.448

[4] The Wrights’ new machine weighs only 900 pounds.

While the average supporting surface is in favor of the aeroplane, this is more than overbalanced by the greater amount of horsepower required for the weight lifted. The average supporting surface in birds is about three-quarters of a square foot per pound. In the average aeroplane it is about one-half square foot per pound. On the other hand the average aeroplane has a lifting capacity of 24 pounds per horsepower, while the buzzard, for instance, lifts 5 pounds with 15-100 of a horsepower. If the Wright machine—which has a lifting power of 50 pounds per horsepower—should be alone considered the showing would be much more favorable to the aeroplane, but it would not be a fair comparison.

More Surface, Less Power.

Broadly speaking, the larger the supporting area the less will be the power required. Wright, by the use of 538 square feet of supporting surface, gets along with an engine of 25 horsepower. Curtiss, who uses only 258 square feet of surface, finds an engine of 50 horsepower is needed. Other things, such as frame, etc., being equal, it stands to reason that a reduction in the area of supporting surface will correspondingly reduce the weight of the machine. Thus we have the Curtiss machine with its 258 square feet of surface, weighing only 600 pounds (without operator), but requiring double the horsepower of the Wright machine with 538 square feet of surface and weighing 1,100 pounds. This demonstrates in a forceful way the proposition that the larger the surface the less power will be needed.

But there is a limit, on account of its bulk and awkwardness in handling, beyond which the surface area cannot be enlarged. Otherwise it might be possible to equip and operate aeroplanes satisfactorily with engines of 15 horsepower, or even less.

The Fuel Consumption Problem.

Fuel consumption is a prime factor in the production of engine power. The veriest mechanical tyro knows in a general way that the more power is secured the more fuel must be consumed, allowing that there is no difference in the power-producing qualities of the material used. But few of us understand just what the ratio of increase is, or how it is caused. This proposition is one of keen interest in connection with aviation.

Let us cite a problem which will illustrate the point quoted: Allowing that it takes a given amount of gasolene to propel a flying machine a given distance, half the way with the wind, and half against it, the wind blowing at one-half the speed of the machine, what will be the increase in fuel consumption?

Increase of Thirty Per Cent.

On the face of it there would seem to be no call for an increase as the resistance met when going against the wind is apparently offset by the propulsive force of the wind when the machine is travelling with it. This, however, is called faulty reasoning. The increase in fuel consumption, as figured by Mr. F. W. Lanchester, of the Royal Society of Arts, will be fully 30 per cent over the amount required for a similar operation of the machine in still air. If the journey should be made at right angles to the wind under the same conditions the increase would be 15 per cent.

In other words Mr. Lanchester maintains that the work done by the motor in making headway against the wind for a certain distance calls for more engine energy, and consequently more fuel by 30 per cent, than is saved by the helping force of the wind on the return journey.