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(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. (If there is, the plane may be able to move forward.)

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!

### Discussion (34)

#### 2008-10-28 01:04:57 by qntm:

I wrote this in about an hour. I had no idea what the answer was going to be going into the essay, I just started writing down the logic to see where it went. It turned out to be a more sophisticated problem than I thought!

#### 2008-10-28 01:30:50 by Ian:

Dammit, Sam, you ruin Mythbusters by adding so much thought into their experiments!

#### 2008-10-28 01:40:26 by Jake:

Considering all the other simplifying assumptions you've made, I'm surprised that you didn't idealize the friction in the wheels out (considering how the treadmill is moving so fast and IS NOT moving the air). You are right though. In reality, something will break. It may not be the materials, per se, but eventually something will become un-ideal, and invalidate the assumptions. Unstoppable force/immovable object anyone?

#### 2008-10-28 01:47:13 by qntm:

Actually, if you read the article, you see that I start with a highly idealised model and progressively add more complications. Up until the point where axle friction is introduced, there is indeed no reason why the plane can't take off, because the treadmill cannot slow the plane down at all even if it runs infinitely fast. Maybe I'll make that more explicit.

#### 2008-10-28 01:59:51 by Randall:

Sam, you're making a fairly simple mistake here. You state in your problem that the operator of the conveyor belt is making it go "as fast as the plane is moving forward." Not "as fast as he wants to try and keep the plane on the ground." So if the plane is moving forward at speed V, the most the conveyer belt can be going at is -V. But if the conveyer belt is moving backward at -V, the force it is applying on the plane is going to be less than the force from the engines, because the plane is on wheels and not stilts. Yes, there's friction in the wheels, so the conveyer belt is applying some backward force to the plane, but not as much force as the engines are applying forward. So the plane takes off. Now, yes, if the conveyer belt is allowed to go as fast as it wants, it can be increased to a speed such that the backward force it's applying actually does equal the forward force of the engines. But that doesn't seem to be in the spirit of the question.

#### 2008-10-28 04:40:16 by Daniel:

I thought the normal problem is if the treadmill is moving at the plane's takeoff velocity and the plane is stationary. Anyway, I just wanted to point out that in the normal friction model, force is independent of the relative speeds of what the friction is between (in this case the wheels and the treadmill). If you want to be realistic, I'd suspect that at a certain speed the treadmill would start burning the wheels, causing the plane to float on a cushion of smoke, and causing the friction to decrease.

#### 2008-10-28 10:27:33 by qntm:

Randall: > there's friction in the wheels, so the conveyer belt is applying some backward force to the plane, but not as much force as the engines are applying forward Did you not see the part where I proved that the friction from the conveyor belt can become arbitrarily large?

#### 2008-10-28 11:00:39 by Evan:

The main problem with this question is wording. If it is written, as in this article that the treadmill is adjusted such that the overall velocity of the plane relative to the earth is 0, then no, it cannot take off. If it is written such that the velocity of the runway is equal to the inverse velocity of the plane relative to the earth (i.e. V[plane-earth] = -V[treadmill-earth]), then yes it can take off.

#### 2008-10-28 11:39:03 by qntm:

Yup, I just realised this myself. I'm going to alter the essay to deal with both cases in turn, for clarity. Cheers.

#### 2008-10-28 15:16:10 by i:

sorry, but here you're wrong. see the straight dope.

#### 2008-10-28 17:11:02 by Dave:

Sam, I'm with Randall on this one. As you've posed the problem, the treadmill only moves at speed V, so the airplane's speed across the treadmill is just twice its airspeed. This reduces the airplane's effective acceleration but does not limit the maximum airspeed it can achieve while on the ground. "Up until the point where axle friction is introduced, there is indeed no reason why the plane can't take off, because the treadmill cannot slow the plane down at all even if it runs infinitely fast. Maybe I'll make that more explicit." This isn't correct. Part of the airplane engine's thrust must be used to counteract the backward force of friction on the tire, whether or not the tire is skidding, and even in the case of no treadmill and with a frictionless axle. This force is doing work to spin the tires up (as well as overcoming any frictional losses). If the tire is not skidding and the axle is frictionless, this force is proportional to the ground acceleration of the airplane, with a proportionality constant on the order of the wheel's rotating mass; if the airplane is moving at a constant speed relative to the runway, there is no backward force from the tires. However, if the treadmill speeds up, the tires must spin faster, and some of the energy to spin the tires up comes from the airplane's kinetic energy. This is probably easiest to see with a free-body diagram for the tire. The tire has a backward force f exerted on its contact patch by the runway (or treadmill), and a forward force t exerted at the axle by the strut (and so which must eventually be provided by the engine). For the airplane to accelerate, with acceleration a, requires t-f=m*a with m the mass of the wheel and tire. If the airplane is on a normal runway and the tire is not skidding, then the torque about the axle is f*r, where r is the tire's radius; and the wheel's angular acceleration is a/r. So f*r=I*(a/r), and f=(I/(r^2))*a, where I is the moment of inertia of the wheel-tire assembly. (I is typically on the order of 0.5*m*r^2, so this implies a frictional force on the order of 0.5*m*a; so even assuming a 1-g acceleration this acts as an extra drag force of half of the tire's weight, not all that big an effect.) If the treadmill is moving backward with the same speed, then this just doubles the tire's angular acceleration and so doubles this value. Since the tires are pretty light, relative to the airplane, this is not a major effect. If the treadmill is allowed to accelerate arbitrarily quickly, then this angular acceleration is similarly arbitrarily large, and so f gets very big and t must also get very big. However, f is limited by the maximum static frictional force supported at the tire-tarmac interface. This is typically modeled as proportional to the normal force between the surfaces (i.e., most of the weight of the airplane, considering all of the landing gear together), with a coefficient of static friction ~1 for rubber-on-concrete. This is of course very large, and most airplane engines do not have sufficient thrust to counteract this force. So in this case (which is not the case of the problem as posted), the airplane won't be able to take off. Also: "Did you not see the part where I proved that the friction from the conveyor belt can become arbitrarily large?" "Planar friction between the skidding wheels and the runway, however, is proportional to the difference in speed between them." Friction is a complicated issue, but these statements are a little oversimplified. Sliding friction is usually modeled in early physics courses as proportional only to the normal force between the two surfaces, with a proportionality constant depending on the materials (~0.5 for rubber on concrete). In this case, with this assumption, the frictional force is limited to something of the order of magnitude of the airplane's weight. Things are more complicated than this in reality, but most sliding friction is closer to a constant force than a velocity-proportional force.

#### 2008-10-28 18:19:05 by Randall:

Sam, yes, it can become arbitrarily large, *if the conveyer belt is moving arbitrarily fast.* But the original question doesn't state that the conveyer belt is moving arbitrarily fast, it says that it's moving as fast backward as the plane's forward velocity with respect to the earth. I'm going to repeat that, to be clear: I'm assuming that the conveyer belt has a maximum speed of however fast the plane is moving forward *with respect to the ground,* not *with respect to the conveyer belt.* If the conveyer belt is supposed to be moving as fast backward as the plane is moving forward *with respect to the conveyer belt,* it can be easily seen that as soon as the plane has any forward velocity whatsoever *with respect to the ground,* the belt will need to go backwards infinitely fast. Now, yes, if you're A. interpreting it this way, and B. allowing for the conveyer belt to move infinitely fast, then it will keep the plane on the ground. But the question says "as fast as the plane is going forward," not "as fast as the plane's wheels are spinning forward," so I think this interpretation is wrong.

#### 2008-10-28 22:45:32 by qntm:

The essay has been updated to take into account the two possible interpretations of that ambiguous wording. Thanks to everybody who pointed it out. Depending on how you interpret the question, the answer is "yes" or "no"! No wonder there is so much discussion.

#### 2008-10-29 12:56:44 by frymaster:

it also depends on the variation of the question. Personally I think the /intent/ of the question is "what happens if the treadmill is tuned such that the plane is stationary relative to whatever the treadmill is fixed to? (the earth, presumably)" in which case the obvious answer is "it won't take off, unless there's a mahoosive headwind", and the non-obvious answer probably involves things breaking ;) it's difficult for most people to get their heads round, as of course normally on a treadmill it's pushing against the treadmill surface that moves people relative to that surface, while a plane essentially "pushes against the air"

So, as far as I can tell we're all agreed that if the speed of the treadmill is set as equal and opposite to the speed of the plane, it takes off as normal, with the wheels spinning twice as fast as usual. So that leaves one or possibly two other interpretations. The first is the speed of the treadmill being matched to the speed of the wheels, which leads immediately to infinite speed for both the wheels and the treadmill and, presumably, mechanical failure. The final interpretation is usually phrased "the speed of the treadmill is set at such a value as to ensure the plane remains motionless relative to the earth", which is clearly begging the question. This interpretation doesn't set any conditions for detirmining what this speed might be, but assumes the outcome "the plane doesn't move" which, to all intents and purposes is the same as saying "the plane doesn't take off". Or am I missing something?

AARRGH! *Determining can has edit function?

#### 2008-10-29 21:44:45 by Mike:

I think the spirit of the question is for people to look at it and say, "The plane won't take off because there is no air over the wings to generate lift!" and think that they're smart because they used words like "generate" and "lift" and "the". Then the person with just a little common sense looks at it and says, "Just because the treadmill is moving doesn't mean the plane won't move forward, the wheels don't drive a plane!" and chastise the first person for being such a dullard. Of course, there's always that third person who makes things way more complicated than they need to be and somehow turns the question back against the person with common sense and say, "Actually, dullard here is right. When you bring in extra factors x, y, and z and consider various frames of reference blah blah blah..." I'm not saying you're wrong, Sam. I'm just saying that you don't need to assume that the treadmill can achieve infinite velocity just because the problem statement doesn't give an upper limit.

#### 2008-10-30 17:44:39 by Val:

You should check the Antonov An-2. It has such a low stall speed (it's not even in the manual, because it has no official stall speed), that you can fly with it backwards even in a not very strong wind.

#### 2008-10-30 23:46:12 by Daniel:

Are you sure that there's no upper limit on the friction of the conveyer belt? I'd expect that after the wheels start burning, increasing the speed of the conveyer belt would increase the speed at which the wheels burn, creating a thicker pocket of smoke, decreasing the friction.

#### 2008-10-31 02:07:45 by CJ:

I draw attention to the earlier comment: You will have a no-slip boundary condition for the air on the treadmill - running the treadmill backwards will cause an exponential profile in the air (of length scale... ugh, sqrt(t) of some kind) - eventually, if the plane is stationary by the earth's reference frame, there will be a sufficiently strong wind to lift it up.

#### 2008-10-31 10:44:33 by qntm:

If the treadmill has any kind of sensible finite upper limit to how fast it can spin then obviously the wheels can probably spin that fast too. One of the things about mathematics, though, is that you can't assume anything unless you're explicitly told that you can assume it. All we know is that the treadmill can go "as fast as necessary". If we don't have an upper limit to its speed, we can't assume that there is one. As for the treadmill *causing the air to move*, this is a rather interesting complication which I may add in to the problem. Thanks.

#### 2008-10-31 13:53:39 by skztr:

I enjoy the way this question has been turned into, basically, "what happens if you throw a jet plane going full-throttle onto a belt sander going at infinite velocity?" Visions of airplanes quickly and quietly melting into the ground replace the often boring associated images of planes simply sitting there (the "can it make a plane take off? of course not!" interpretation) or simply flying away (the "can it prevent a plane from taking off? of course not!" interpretation).

#### 2008-11-01 02:00:53 by Andrew:

This, when approached as an engineering problem, can have an easy solution, one way or the other. In theory, the wheels rotate freely and provide absolutely no resistive force, and thus can't ever hope to stop the plane. But as you point out, it's more interesting complicate it (why waste the question?). So, let's assume we can increase friction until it counters the plane. What, I wonder, is how fast this really is - what if the speeds are positively insane? A quick search shows that typically, the wheel bearings are designed to leap to 2k-3k rpm, but are only designed to withstand the already high rotational for short periods (5-10min), and takes over 5 times as long to cool. It would seem that in order to apply the back force on the plane through just the wheel axels, the bearings would almost certainly suffer immense mechanical failure first - most likely vaporizing the grease from the frictional forces. This would quickly lead to metal on metal friction, and unless the bearings are made of geometrium, they'll even more rapidly erode and destroy themselves into dust. (Centrifugal forces on the tire tube could also cause bursting from rpms hundreds of times too high - I'll need to work out what these speeds are). This is another case of the real world impinging it's boring pragmatism on theory. But there's still two places to go from here. One: assume the wheels are in fact made of geometrium: at what frictions levels does the increase in conveyor speed lead to relativistic wheel rotations? What is the friction level that is low enough that the wheels gain relativistic mass fast enough to prevent the plane from ever even HOPING to take off? Two: I think it's unfair to assume that the conveyor belt does not affect the air. Surely the no-slip boundary exists on it. At what speeds would the belt need to go to cause the plane to just hover in place? I'll need to reinstall OpenFOAM for this one I think...

#### 2008-11-10 11:46:37 by CJ:

It's pretty obvious by self-similarity that if z is the height above the moving conveyor belt, the typical air speed will be U*exp(-x sqrt(nu / t)). That might be wrong, but if it's wrong, then it's wrong and obvious. Write it as a dimensionless variable eta (f === f (eta)) and rejig the NS equations... WHY aren't I applying for PhD places? WHY?? I _need_ to get them done before the deadlines! Argh!

#### 2008-11-24 06:07:38 by AlbeyAmakiir:

I just sort of imagined that, as the engines are well above ground level, the plane would eventually tip over and it's nose would hit the ground. The resulting crash would be cool.

#### 2008-11-28 09:31:06 by Fractal:

If the plane wouldn't tip over with engines at full throttle but the brakes on, then the plane also won't tip over on the treadmill. It's only the force of the engines applying any torque to the plane.

#### 2008-12-24 03:45:12 by Karma:

I dont understand why so difficult. Assume plane needs lift. Lift needs velocity. Velocity is zeroed. a) wing vs air forward = normal gliders, wings b) wing vs air in immediate vicinity = helicopters or jet focuses force like harrier jet vto Only if we assume plane generates its own lift by engine then we have ascension. All quibbles aside, why make it more difficult? If this was to be university grade question there should be no room for bulldust.

#### 2008-12-24 03:49:48 by Karma:

Try it with an organic being like a duck. Duck on treadmill. Duck flaps wings, runs forward. Treadmill runs backward. Does duck fly? Forget all this friction and air resistance and so forth, get duck to fly, post video on youtube, MOT and QED or waddeva.

#### 2009-10-09 18:44:14 by Boter:

Or apply for a research grant. They hand out grant money for silly things in the States, so you might have to move here to do it.

#### 2009-10-28 17:24:43 by AJ:

" 2008-10-29 12:56:44 by frymaster: it also depends on the variation of the question. Personally I think the /intent/ of the question is "what happens if the treadmill is tuned such that the plane is stationary relative to whatever the treadmill is fixed to? (the earth, presumably)" in which case the obvious answer is "it won't take off, unless there's a mahoosive headwind", and the non-obvious answer probably involves things breaking ;) it's difficult for most people to get their heads round, as of course normally on a treadmill it's pushing against the treadmill surface that moves people relative to that surface, while a plane essentially "pushes against the air"" Um, why is it difficult? I don't understand all this complication. Wings need airflow over wings to generate lift. If the plane is still relative to the ground, and with no headwind, there is *no* airflow over the wings. So it can't take off. Oooh, how difficult.

#### 2012-10-29 23:04:44 by AndyMoore:

Great breakdown, Sam. You're totally right that it boils down to the interpretation of the treadmill speed. In this case, context comes into play to suss out the correct interpretation. From what I remember, this question was stated to engineering students that had been studying automotive mechanics as a "gotcha" question on an exam. The question is designed to prey on people's understanding of how an automobile works (with wheels and engine locked together to the ground) and trick them into giving the "no, it won't take off" answer. "Yes, it will take off" is the correct answer. I'm sure anyone that sussed out all the fine details and physics of *both* the yes and no answers (as you have done!) would get top marks. I wish someone would reformulate the question to be less ambiguous.

#### 2014-06-16 14:22:34 by Aegeus:

For what it's worth, when Mythbusters tested it they were very explicit about the speed. They stated the myth as "A treadmill set to the plane's takeoff speed," which is easy to test and limits it to the main "gotcha" of the question - that a plane's wheels spin freely and don't exert any force.

#### 2014-08-02 23:23:51 by simon:

I'm surprised no one seems to have taken issue with this assertion: "Planar friction between the skidding wheels and the runway, however, increases with the difference in speed between them." The usual approximation for sliding friction is that it is proportional to the normal force but independent of the speed. If the assumption is not true, then even an arbitrarily large conveyor belt speed can't prevent takeoff.

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