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:

- the speed of the plane relative to the runway, or
- 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.

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*.

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.

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.

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) = 2`T`

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` > 2`T`. 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.

### Treadmill case 2: speed(treadmill relative to Earth) = -speed(plane relative to treadmill)

Another simple calculation:

speed(treadmill relative to Earth) = -speed(plane relative to treadmill)
speed(plane relative to treadmill) + speed(treadmill relative to Earth) = 0
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:

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:

- air resistance from the air at take-off speed AND
- maximal rotational friction in the wheels AND
- 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.

#### "But what about--"

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*!