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PECULIARITIES OF AIRSHIP POWER.






From: Flying Machines Construction Operation

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.





Next: ABOUT WIND CURRENTS

Previous: HOW TO USE THE MACHINE.



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