Plane on a Treadmill

[NC Hokie]

Nope.....it illustrates the exact point of the test.  People think that a plane is a car.  That the interface between the ground and the wheels is as important to the plane as it is in a car.

The interface between the ground and the plane IS as important to the plane as it is to a car as long as the plane remains on the ground.

You and others have made the point that a plane is different from a car because the car transmits thrust directly to the wheels while the plane does not.  You're absolutely right, but HOW the thrust is transferred to the wheels is beside the point, because as long as the plane remains on the ground, those wheels are the only things transferring thrust to the immediate force of resistance (the ground).

Consider a plane doing a run-up before taking the active runway.  The pilot engages the wheel brakes and goes to full throttle.  The plane lurches forward a bit because the thrust is acting on the airframe, yet the plane remains stationary.  Why?  Because the wheels are locked, rendering the plane's effective speed zero.

You can apply the same principle to the perfect treadmill assumed by the original question.  If the treadmill is able to match the plane's ground speed, the plane's effective ground speed is zero, leaving the plane motionless and unable to rotate into it's natural environment.

[Eclipse]

Refer to the guidance hereinabove "The hypothetical does not state that the airplane cannot move forward, only that the speed (rotation) of the wheels is exactly matched by the treadmill. No one is arguing that the airplane will fly without forward motion into the wind."

Then this is a waste of time - if the plane can move forward, its not a test of the actual idea, just a longer runway.

[ßτε]

Can anyone explain exactly how the treadmill produces force on the airplane once the wheels start moving. It would need to create a force in the opposite direction of the thrust in order to prevent the airplane from accelerating forward.

[AirAux]

As noted previously in the thread, the treadmill moving at the same (but opposite) speed as the thrust of the A/C is the exact same as the A/C having the brakes locked and running up the engine.  The A/C will not fly as there is no relative wind travelling over the wings to induce lift..  Some people are trying to overthink this problem..  The A/C will not lift until the A/C is allowed to travel faster than the treadmill, thereby moving on the treadmill, thereby causing relative wind, hence lift..  Thrust and drag are two forces working on the A/C, but they are unimportant in this situation because we need relative wind to generate lift.  Thrust and drag will be used to determine how fast the A/C will fly once airborn..

[Major Lord]

Its a hard idea for people to wrap their minds around. You are quite right, there is no way in the real world that a treadmill could introduce enough force to make the airplane static, or to prevent it from accelerating. The arguments fall into two camps: The first camp believes that the airplane cannot accelerate over the treadmill, because in order for the wheels to be going at the same speed of the treadmill, the aircraft would be standing still. No one believes that the airplane would fly by itself absent forward velocity) ( Although an Aircraft with enough thrust to weight ratio could do exactly that, that rare exception is clearly not what the original questioner poses)

The "it can fly" camp generally takes the position that the airplane will accelerate to takeoff velocity over the treadmill, even if the treadmill exceeds the velocity of the wheels. ( which would be against the rules of the question) One area of ambiguity is that in real life, the treadmill and the airplane both require time to change velocities. As the airplane increase in speed going down the runway, the treadmill introduces an equal but opposite increase in velocity but because the aircraft wheels are in contact with the runway ( as long as their coefficient of friction, i,e, traction holds) they will be rotating the speed of the treadmill plus the forward velocity of the aircraft. Now the original question states that the treadmill will automatically change speed to match the velocity of the tires, and in some peoples minds this is a paradox.

Major Lord

[N Harmon]

Can anyone explain exactly how the treadmill produces force on the airplane once the wheels start moving. It would need to create a force in the opposite direction of the thrust in order to prevent the airplane from accelerating forward.

There a few different ways that it could. The resistance torque of the wheels would create a negative force on the aircraft while they are being accelerated by the treadmill. As would the rolling resistance of the tire against the conveyor belt. And as would the fluid drag of the oiled ball bearings in the wheel hub.

[AirAux]

According to the scenario given, although the wheels and the treadmill are moving and have overcome inertia, the fact that they are moving in opposite directions at the same speed means that the A/C has not overcome inertia in relationship to the earth and hence is not moving forward and will not create lift as there will be no relative wind between the earth and the A/C.. It seem so easy and I have explaned it many times in groundschool, but some people are trying to introduce quantum physics into a simple equation..  If the A/C is not moving in relationship to the earth, there will be no lifting, no flying, nada..  The wheels and threadmill are status quo and cancel each other out..

[N Harmon]

Nobody is introducing quantum physics into anything. This is simple high-school level newtonian mechanics. Quite simply, work equals force times distance...

W = F * d

The positive force in this equation is the force produced by the aircraft's engine. The negative force is the force produced by friction in the wheels. Remove the distance component, as if the aircraft was sitting still on a conventional runway, and it's just a battle of forces with the aircraft engine winning.

But the situation we have here is that the negative force produced by the friction in the wheels is multiplied over a MUCH greater distance than the distance of the engine. And at some point the Work/Distance function of the wheels intersects the Work/Distance function of the engines.

[AirAux]

And that has nothing to do with lift or flight.  You may have force in the engine, but if the A/C is standing still and not moving down the threadmill, then you have no distance and thus no work.. if the threadmill is going south at 60 miles and hour and the wheels are rolling north at 60 miles an hour, the A/C is not moving in relation to the earth and there is no relative wind and no lift..  The wheels and the treadmill are a distraction in this problem because without relative wind over the wing (and I don't mean propwash) there just ain't no lift..

[N Harmon]

if the A/C is standing still and not moving down the threadmill, then you have no distance and thus no work.

Not exactly... http://en.wikipedia.org/wiki/Thrust#Thrust_to_power_2

[SilverEagle2]

Imagine a plane sitting on a giant conveyor belt, as wide and long as a runway. The conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off?

No where does the question state the plane is stationary. We all agree that a stationary plane will not fly. What we do not agree is that the plane remains stationary.

It will not.

The treadmill and wheels would have to be spinning at a very high rate of speed in order to produce enough friction to keep a fully throttled up airplane static in relation to the earth/wind/air mass.

After all, what are wheels for? To remove/reduce friction of the plane as it moves forward over the ground.

Otherwise the treadmill would only have to be as long as the wheelbase of the plane. The question states it is a conveyor the size of the runway.

If there is no forward motion of the plane on the ground then there is no wheel speed for the conveyor to match. It must move forward to activate the question.

As the plane approaches Vr, the speed of the wheels is simply 2xVr. It really is that simple.

[N Harmon]

If there is no forward motion of the plane on the ground then there is no wheel speed for the conveyor to match. It must move forward to activate the question.

I think the way the problem presents it is that the treadmill starts at the exact time the aircraft's engine applies a force. But otherwise you would be correct.

[SilverEagle2]

So?

The plane still can accelerate down the belt, up to the point where the friction/drag equals thrust.

It will be long off the ground before that point.

[AirAux]

Consider this, if you had an A/C on an Aircraft carrier.  If the A/C had to reach 100 mph to take off and the Aircraft carrier was travelling in the oposite direction at 100 mph, would the A/C take off??  There would be no relative wind traveling over the airfoil to produce lift.  That is why an Aircraft carrier heads into the wind for take offs and landings..  I am not talking about acceleration, I am talking about constant speeds of both items.. They cancel each other out..

[SilverEagle2]

These are not the same. To many factors have been altered. The constants are no longer constants on a carrier.

Your logic is flawed as the start of the problem has the airplane sitting on the deck of a fast moving carrier.

Observer standing on the shore of the ocean looks out at a carrier and sees carrier and plane moving together. Catapult is then countering the carrier and the plane then looks static to the observer.

Observer turns around and sees plane static and the belt moving. Plane hits the throttle and then moves down the belt with high wheel speed.

They are not the same.

[N Harmon]

The plane still can accelerate down the belt, up to the point where the friction/drag equals thrust.

Somehow I think you're missing the premise of the question where the treadmill matches wheel speed exactly. Or are you arguing something else?

[SilverEagle2]

How...even if the plane is accelerating, is the distance that each revolution of the tire makes, not covering the same distance on the treadmill? They are equal.

[N Harmon]

How...even if the plane is accelerating, is the distance that each revolution of the tire makes, not covering the same distance on the treadmill? They are equal.

Because the speed of the wheels is the speed of the aircraft minus the speed of the treadmill. But since the premise of the problem says the speed of the treadmill is equal and opposite the speed of the wheels, we find that the speed of the wheels is the speed of the aircraft plus the speed of the wheels. This just doesn't work out for any nonzero aircraft speed.

[AirAux]

The point is the speed of the wheels is equal to the speed of the belt.  The A/C has no speed as it has no traction (in effect).  The premise was that the wheels and the belt travelled at the same speed.  So the faster teh wheels travel, if they do accelerate, the faster the belt would travel, causing no A/C speed and the A/C would be sitting in one location, not moving down the belt.  There is no need for the belt to be as long as a runway, it could be the lenght of the A/C in teh scenario.  The only way she will fly is "if" the wheels begin to move faster than the belt to allow air to flow over the wings and then you blow the premise of the wheels travelling the same speed as the belt.  If I read teh premise correctly (And I am now so dazed and confused, I am not sure of anything..)   

[SilverEagle2]

Before I concede your point...which is true...the question leaves out a vital point.

Since Speed or Velocity is the measurement of Distance/Time it is important to know in relation to where the observer is.

If the observer is outside the plane, then yes, the only time the conveyor speed and the wheel speed is equal is when they are at zero...no flight.

If you are the pilot in the cockpit and the equation is in relation to your position, you can get to Vr of 50 knots and see the ground under you going by at 100 knots with a wheel and conveyor speed of 100 knots. Since the pilot/observer is already at 50, his relative speed is to the plane is 0 but to the ground moving below him he is moving at 100 knots and his equation balances.

Since I am a pilot, I ran the equation from inside the plane. It appears that you are running it from outside the plane. Your answer is correct. So is mine.

It is all about where the observer is.

Tricky.