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 moder
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


"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


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:


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



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.