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
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.