The airplane-treadmill problem is a physics thought experiment that has become a minor internet meme.  It goes something like this:

"There is an airplane at rest on a treadmill the size of a runway so that it is big enough for the airplane to take off. However, this is a special treadmill that is designed to move as fast as the airplane's wheels turn but in the opposite direction. So, as the airplane gets up to speed the wheels begin to turn as the airplane moves forward but the treadmill also begins to move just as fast as the wheels. Does the plane take off?"

The first known appearance of the meme was on a Russian forum on April 8, 2003.  Later it was re-posted in english to PhysOrg on July 19, 2005.  From there the question disseminated across the internet until it developed into a meme.

At first glance, the question seems to have an obvious answer: the plane will not move.  Unfortunately, that is the wrong answer and here's why.  For the purposes of simplicity I'm going to ignore the normal and gravitational forces as well as lift because they are perpendicular (and therefore irrelevant) to the range of motion which is important in the solution of this problem.  When driving in a car there are three main forces; the force of friction on the wheels, the air on the car, and the wheels on the road. If we drew a force diagram of this it would look something like this:

`                    __________                   /  |       \   Fac             _____/___|        \---------->     Fcr    /                   \  <--------|   __           __  |            \_/  \_________/  \_/   Frc              \__/         \__/------------->`

The only force giving thrust to the car is Fcr; the force of the car on the road.  So if you put a car on a treadmill like the airplane's, it would not move because Fcr and Frc cancel each other out. Since the force of the air on the car is zero when an object is at rest, there is no net force and therefore no motion.

However, an airplane operates differently.  Instead of pushing against the ground to produce its thrust, an airplane pushes against the air using a propeller or jet engine.  The wheels are present not to provide a means of propulsion but to reduce the friction between the airplane and the ground, increasing net thrust.  Again a force diagram:

`                           __                          /  |           __                         /   |          / |   Fpa       ___________/____|_________/  |<----------/_| oooooooooooooooooooooooo  |---------->          (____________\     |__________/     Fap                        \    |                                        \   |                          \__|`

As you can see from this diagram, the only way that net force can be zero is if the force of the thrust of the airplane is exactly equal to the air resistance.  Since we know airplanes can fly in the first place, it's clear that this point doesn't happen before airplanes reach their takeoff velocity.  So, simply put, the airplane will take off of the treadmill.

Now some of you may be saying "RedOmega you groovy dude, you just ignored the friction of the wheels!"  but the truth is that the presence or absence of friction does not change the results of this problem.  As the plane accelerates, the wheels begin to turn.  The treadmill then speeds up to match the speed of the wheels but in doing so the relative speed of the wheels increases causing another increase in the speed of the treadmill much in the fashion of Zeno's Paradox.  What we have here is an infinite acceleration loop in the presence of friction and unless the airplane has titanium-reinforced, Teflon-coated, out of this world space wheels, the tires and eventually the axles as well will melt into a pool of molten lava.  Since the airplane has no wheels now, the treadmill's speed drops to zero and the airplane takes off amidst a cloud of smoke with Samuel L. Jackson onboard.

Also all the frictional force is tangent to the point of rotation so it goes directly into making the wheels spin, not slowing the plane down.

Sources:

• http://waxy.org/2008/02/origins_of_the/
• Original Russian post: http://forum.ixbt.com/topic.cgi?id=64:417

Yes, I'm going there.

A plane is standing on a runway that can move (like a giant conveyor belt). This conveyor has a control system that tracks the plane's speed* and tunes the speed of the conveyor [relative to the Earth - clarification added] to be exactly the same (but in the opposite direction).

Will the plane be able to take off?

* NOTE: The reference frame for the measurement of the plane's speed is not stated, but unfortunately it makes a huge difference. This will be dealt with below. Read on.

### First principle

A plane takes off when it has sufficient lift. Lift comes from the wings, and is a function of the angle and geometry of the wing and, more importantly, the speed at which air passes over and under it.

The speed at which air passes over the wing depends on the speed of the plane relative to the air.

It does not depend on:

1. the speed of the plane relative to the runway, or
2. the speed at which the plane's wheels are turning.

The plane takes off when it reaches a specific (forward) takeoff speed, T, relative to the air.

In theory, a sufficiently strong gust of wind can provide the necessary lift to pull any plane off the ground. (And even hurl it backwards.) Since this a fairly pointless and unlikely boundary case, we shall discard it and assume that there is no wind.

Since there is no wind, the speed of the air relative to the Earth is zero.

Since the speed of the air relative to the Earth is zero, we can say without loss of generality that the plane takes off when it reaches a speed of T relative to the motionless Earth.

### Aeroplane engineering

Planes have no motors for their wheels. On the ground, they are sometimes towed around by dedicated towing vehicles - but this is not a situation we will consider since this particular plane is attempting to take off. A plane is more normally propelled by its engines, which act on the air itself. By pushing air backwards, the planes pushes itself forwards in accordance with Newton's Third Law of Motion.

Note that the engine generally pushes the air straight backwards. Aircraft don't generally take off by pushing the air downwards, with the exception of Vertical TakeOff and Landing craft and helicopters, which operate on entirely different principles. Remember: to take off, the plane needs lift, and to create lift, it needs to be moving at speed, and before takeoff, the only way to move at speed is horizontally, across the ground. So the logical direction for the engines to point is straight backwards.

Jet engines are extremely powerful, but they are not fans. They do not directly blow air across their own wings. They do not cause wind to appear. True, there are some negligible effects along these lines, depending on the configuration of the plane, but this is not the principle on which the engine works. The engine is not there to directly generate lift, but to make the plane move fast. For the sake of argument, then, it is safe to assume that all the engine does is provide a forward force for the plane. Since the engine acts on the air and not on the runway, we can say that, more specifically, the engine forces the plane forwards with respect to the air, not with respect to the runway.

### Newtonian mechanics

So we have an engine. And the engine provides a forward force. This force produces acceleration. Acceleration is an increase in speed. If the speed reaches T, the plane will take off. No argument there.

But!

What about other forces? There are lots of forces acting on the plane, but the most important ones are the ones which push the plane backwards, in the opposite direction to the push of the engine.

#### Air resistance

Firstly there is air resistance on the plane. This is unavoidable. The faster you go, the more drag the air creates. Eventually you reach a speed where all the forward force from the engines is completely cancelled out by air resistance pulling the plane to slow down. At this point the plane cannot go any faster and so it stops accelerating. This is its maximum speed.

Obviously, if the plane's maximum speed is lower than T, then the plane will never take off. Since planes do take off on a regular basis, however, it is reasonable to assume that our engines are powerful enough to overcome this threshold of air resistance by a wide margin.

There is one more force which slows the plane down, though.

#### Axle friction

During take-off, the plane is running on the ground, on wheels. The wheels are in contact with the tarmac of the runway. They are unpowered, but they are free to spin. Let's make an assumption: the wheels do not skid. They are "glued" to the runway. They have to turn as quickly as the runway passes below the plane. This is what happens in a real take-off, anyway; there is no circumstance in which the engines would pull the plane forward so suddenly that the wheels actually skidded.

```speed(wheels) = speed(plane relative to runway)
```

We know that, in order to reach take-off speed, the plane must be travelling at T relative to the Earth. If the runway is stationary relative to the Earth, as it usually is, this means that the plane must be travelling at T relative to the runway.

```speed(wheels at takeoff) = T
```

If the wheels were frictionless, then this would provide no slowing force. The wheels and runway treadmill could both be going infinitely fast and nothing (save air resistance) would stop the plane from accelerating forwards, and ultimately reaching T.

However, though they are free to spin, the wheels are not perfectly frictionless. There is resistance in the bearings/axle/whatever of the wheels. (Note that this is NOT friction between the wheel and the runway, which has negligible effect.) They don't "want" to spin. If they were spun up to a high speed and then left alone, they would slow down and stop because of this resistance. This frictional force is always there. It varies with the weight of the plane (the heavier the plane, the harder it is to spin those wheels). It also varies with the speed at which the wheels are already turning. The faster they turn, the more resistance there is. In the same way that eventually the plane cannot travel any faster due to air resistance, so there comes a point where the wheels physically cannot turn any faster.

Luckily, this speed, W, is extremely high. Since, again, we know that real planes are perfectly capable of taking off in reality, we know that W > T.

But!

What if the runway was actually a treadmill?

The runway and the air are completely separate beasts. They have no effect on each other. Let's assume that the air remains still even if the treadmill is running a million miles an hour. So the conditions for takeoff remain the same. The plane must travel at T speed relative to the air/Earth.

But, since the treadmill is moving backwards at high speed, the plane must move much faster relative to the runway. This means that, as long as they stay in contact with the ground and don't slip, the wheels must also turn much faster than normal in order for the plane to take off.

Here is where the poor wording of the question becomes critical. Exactly how fast is the treadmill going? There are two possibilities.

### Treadmill case 1: speed(treadmill relative to Earth) = -speed(plane relative to Earth)

This is an easy situation to imagine. Just imagine that there's a yellow line across the treadmill at the point where the plane starts. Then, for every metre the plane moves forward from its starting point, the yellow line moves a metre back from its starting point.

A simple calculation:

```speed(wheels) = speed(plane relative to treadmill)
= speed(plane relative to Earth) + speed(Earth relative to treadmill)
= speed(plane relative to Earth) - speed(treadmill relative to Earth)
= speed(plane relative to Earth) + speed(plane relative to Earth)
= 2 * speed(plane relative to Earth)
```

At takeoff, we know the plane is moving at T relative to the Earth, so:

```speed(wheels at takeoff) = 2T
```

So, at take-off speed, the treadmill simply means that the wheels must now be turning at twice the takeoff speed.

This is probably fine, since, probably, W > 2T. As long as the plane's engines can overcome the friction of the wheels turning twice as fast as normal, it will accelerate forwards relative to the air, achieve T and take off. No problem.

Another simple calculation:

```                                     speed(treadmill relative to Earth) = -speed(plane relative to treadmill)
speed(plane relative to Earth) = 0
```

This means the plane doesn't move, relative to the Earth. The treadmill is tuned so that the plane always stays in the same location on Earth.

This is a much more interesting situation. As you increase the speed of the treadmill, you will eventually reach W. In other words there comes a point where the wheels must spin faster than they are physically capable of spinning if they are to keep up with it. At this point, in order to gain speed, the pilot will increase the thrust from the engines. And two things can happen.

One is that the extra rotational force causes the wheels destroy themselves. If this happens then your answer is "No": the plane cannot take off. The treadmill is too powerful.

The other is that the wheels start skidding. The treadmill is going backwards faster than the wheels can turn, so the wheels lose contact with the treadmill and start skidding. They are still turning at maximum speed, W, but this is not fast enough.

When this happens there is now a third backward force:

#### Ground friction

There was air resistance (although actually there is none of this yet, because the plane is still stationary relative to the air!). There was friction in the bearings of the wheels. And now there is actual planar friction between the rubber of the tyres (turning at maximum speed) and the tarmac of the runway (turning even faster). In order to achieve take-off speed relative to the air, the plane must overcome all three:

1. air resistance from the air at take-off speed AND
2. maximal rotational friction in the wheels AND
3. planar friction between the skidding wheels and the runway.

Now here's the important bit. As we just derived, the treadmill always moves backwards quickly enough to prevent the plane from moving forwards, even by a centimetre. There is no stated upper limit to how fast it can go.

Planar friction between the skidding wheels and the runway, however, increases with the difference in speed between them. The faster the treadmill goes, the more resistance is provided. Since the treadmill can go arbitrarily fast, the amount of resistance provided can become arbitrarily large. (If the tyres get worn down, catch fire and explode, then again we have a failure condition. Let's assume, charitably, that they don't.)

Now.

If the combination of the three drag forces can become arbitrarily large, then the backward force can overcome any forward force provided by even the most powerful jet engines. However high the throttle goes, the treadmill operator can provide enough resistance -- just from the relatively small contact patches of the tyres -- to overcome those engines. And so the plane cannot accelerate forwards. It remains at 0 speed relative to the air. Therefore, in this situation, the plane cannot take off.

You may have seen this debunked somewhere with the opposite conclusion. Maybe even a real plane and treadmill. Trust me: reality is not a thought experiment. Any of a million things can go wrong with Case 2.

The most likely is if the treadmill isn't fast enough. To overcome a combined total of upwards of a million newtons using nothing but friction would require a treadmill capable of, ooh, let's say, a thousand kilometres per hour? The scenario with the plane's undercarriage self-destructing is actually far more likely. So if your treadmill isn't fast enough to destroy the plane's undercarriage, it's not fast enough for this experiment.

"No" is the solution to the problem as stated, but the problem as stated is one of mathematics and logic. Which means it's a problem to which there is a single, correct answer. "The Guide is definitive, reality is frequently inaccurate..."

### Conclusion

The long and the short of it is that, like many of the internet's favourite problems, it all depends on how the problem is interpreted. Imprecise wording leads to ambiguity and confusion and, here, in classic fashion, it leads to two diametrically opposed answers. No wonder there is no consensus on the answer! There isn't even any consensus on the question!

The airplane-treadmill problem is stated as follows:

A plane is standing on a runway that can move (some sort of band conveyor). The plane moves in one direction, while the conveyor moves in the opposite direction. This conveyor has a control system that tracks the plane speed and tunes the speed of the conveyor to be exactly the same (but in the opposite direction). Can the plane take off?1

This interesting gedankenexperiment has drawn quite a bit of discussion. This, in my opinion, is mainly because it is poorly worded. Hence, I'll first try to get unambiguous descriptions of the problem, before I'll try to answer the question.

The whole problem with the wording is in the word "speed". In physics, speed has no meaning if it isn't defined to a frame of reference, in other words, an observer. There are two sensible definitions of speed possible: speed with respect to the ground and speed with respect to the conveyor belt. The first case is easiest, so I'll tackle that first.

If the speed of the airplane is defined with respect to the ground, the conveyor belt will move in the opposite direction. Now, an airplane is powered by its engines, either jet engines or one or more propellers. These move the plane with respect to the air. A movement with respect to the ground is in itself not very relevant for the whole takeoff problem. The reason is that a plane lifts off when the speed of the air over its wings is high enough for the lift to be bigger than the weight of the airplane. The ground has little to do with that. Hence, the only thing that will happen is that the engines will have to work a bit harder as the tires rotate at twice the speed - for every meter the plane moves forward, the wheels must travel two meters, one with respect to the Earth and one because the conveyor belt moves in the opposite direction. In this scenario, the plane will simply take off, as the friction in the wheels is rather negligible compard to the drag induced by the air.

Now, another way of reading this is that we define speed as follows. The speed of the plane is the speed with respect to the conveyor belt. The speed of the conveyor belt is the speed with respect to us, an external viewer, sitting on a lawn chair with a parasol (this parasol will be important later),watching the whole spectacle. In this setup, the plane is stationary with respect to the observer - for every meter the plane moves forward on the belt, the belt moves back one meter, leaving at the exact same spot. This probably isn't the most obvious way of reading the question, but it is the most interesting one.

Before going to the cute theoretical answer to this paradox, it's perhaps more fun to consider what will happen in real life. Imagine our plane is passenger jet like a Boeing 767. This plane has a thrust of about 270 kN 2. All this thrust is being counteracted by the friction the wheels of the landing gear have with the conveyor belt (whether this friction is between the surface of the belt and the wheel or between the wheel and it axle is not too relevant at this point). There are, apart from gravity and the normal force that prevents the plane from sinking in the ground, no other forces acting on the airplane. This does imply two things:

1. The entire thrust has to be counteracted by the landing gear. I'm not an airplane engineer, so I can't tell whether the landing gear can hold 270 kN, or 27 tonnes, of force in that direction. Given the speed at which an aircraft can brake at the runway, I think it might.
2. The friction creates heat. Normally, the wheels of a landing gear only brake for a very short amount of time, and even then, occasionally, a tire pops. I wouldn't be surprised at all if the landing gear catches fire. Of course, such a burning landing gear will eventually ignite the plane and cause it to blow up. So, parts of it will take off - hence the need for a parasol.

From the discussion above, it is clear that such an experiment will be very difficult to conduct - the conveyor belt would need to run at a very high speed, potentially a lot higher than the normal takeoff speed of an airplane, to generate enough friction to counter the force of the engines. However, let's assume we live in an Ideal World where this is not a problem. Can we answer our paradox? It turns out that the answer depends on the conveyor belt used.

As discussed, an airplane takes off when the lift generated by its wings is larger than the weight of the airplane. Wings generate lift when air blows past them. We have just decided that the plane remains fixed with respect to the Earth. So, the only question remaining now is: Will it also remain fixed with respect to the air? In other words, does the air move?

There are two moving parts in our setup: the engines and the belt. The engines do move air. If it's a jet, the engines will move the air from in front of the wing to behind it, but the air will move through the engine, and it will not hit the front of the wing. In case of a propeller - driven aircraft, some air will hit the wings, but it will likely not be enough to cause liftoff. In other words, the engines can't help us. Now, let's look at our conveyor belt. Imagine I'm looking one millimeter above the conveyor belt. It stands to reason the air at that point has the same speed as the conveyor belt, as it is so close to it and the viscosity of the air is enough to keep it at (almost) the same speed.

Imagine the plane is on a set of very small conveyor belts just under the wheels, perhaps a few treadmills normally used for exercises, only reinforced and powered by a very powerful engine. The air very close to the treadmills will move at the speed of the treadmills; the rest of the air will just have the speed of the ground, which is zero with respect to the speed of the airplane. The plane won't take off.

Now, imagine the plane is on a gargantuan conveyor belt, one that has a surface of several square kilometers. Now, the speed of the air above this conveyor belt will be roughly equal to the speed of the conveyor belt - the larger the conveyor belt, the smaller the difference. It is possible to compute the minimum dimensions of the conveyor belt that is big enough for this the speed difference to be small, but I think this is not a very easy computation (the effect of the finite width of the conveyor belt are not difficult, that of the finite length seems harder) and I don't have a Computational Fluid Dynamics package handy here, so we'll assume that our conveyor belt is big enough - a few kilometers by a few kilometers seems more than big enough. In this case, the air will be blowing against the wings of the plane, and it will be just like a normal takeoff. As a matter of fact, if this is a really big conveyor belt, as big as the eye can see, the passengers wouldn't be able to tell they were taking off from a treadmill and not from a real runway (provided we paint nice runways on the rubber of the treadmill, that is).

In summary, we can state that most of the confusion of this paradox is in the poor wording, in particular the definition of speed. The outcome of the experiment - takeoff, no takeoff, or an accident - depend on the setup. If we say that the speed of the airplane is determined with respect to the ground, then the problem is not very difficult, and the plane will take off. If we define the speed with respect to the conveyor belt, it becomes more interesting, and we have three possibilities. If the wheels become too hot, an accident will happen. On a small conveyor belt - admittedly the most logical setup - the plane won't take off, while on a large enough conveyor belt, the plane will.

#### Sources:

1. http://mouser.org/log/archives/2006/02/001003.html
2. http://www.boeing.com/commercial/767family/pf/pf_200prod.html

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