# What will happen if a plane trys to take off whilst on a treadmill?

So this has puzzled me for many a year... I still am no closer to coming to a conclusion, after many arguments that is. I don't think it can, others 100% think it will.

If you have a plane trying to take off whilst on a tread mill which will run at the same speed as whatever the planes tyres rotation speed is will it take off?

[edited to be more clear]

The question is simple. Will a plane take off if you put this plane onto a treadmill that will equity whatever speed the plane wheels are moving at. So the plane should not be able to move.

This is a hypothetical situation of course. But i am very interested.

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my thoughts, you need a certain amount of thrust on the wings, its the hole basis of flight. – Jamie Hutber Jul 17 '12 at 22:11

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– DJBunk Jul 17 '12 at 22:44

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What matters to an airplane is relative wind, period. (This is not rocket science ;-) – Mike Dunlavey Jul 18 '12 at 14:41

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– mwengler Jul 18 '12 at 15:11
mytherbusters version doesn't cut it for me though, as there is still friction between the floor and the tyres, this plus the fact we don't know how fast the treadmill is running at. – Jamie Hutber Jul 18 '12 at 16:07
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## 6 Answers

Idealizing the plane's wheels as frictionless, the thrust from the propeller accelerates the plane through the air regardless of the treadmill. The thrust comes from the prop, and the wheels, being frictionless, do not hold the plane back in any way.

If the treadmill is too short, the plane just runs of the end of it and then continues rolling towards take off.

If the treadmill is long enough for a normal takeoff roll, the plane accelerates through the air and rotates off of the treadmill.

UPDATE: Don't take Alfred's word for it. Mythbusters has actually done the experiment.

UPDATE 2: I've been thinking about how the problem is posed (for now as I'm typing this) and it occurred to me that the constraint "run at the same speed as whatever the planes tyres rotation speed" actually means run such that the plane doesn't move with respect to the ground.

Consider a wheel of radius $R$ on a treadmill. The treadmill surface has a linear speed $v_T$ to the right. The center of the wheel has a linear speed $v_P$ to the left. The CCW angular speed of the wheel is:

$\omega = \dfrac{v_T + v_P}{R}$

If "run at the same speed as whatever the planes tyres rotation speed" means:

$\omega = \dfrac{v_T}{R}$

then the constraint requires $v_P = 0$. That is, the question, as posed, is:

If the treadmill is run such that the plane doesn't move, will the plane take off?

Obviously, the answer is no. The plane must move to take off. Looking at mwengler's long answer, we see what is happening. The rotational speed of the tires and treadmill are not the key, it is the acceleration of the treadmill that imparts a force on the wheel axles (ignoring friction for simplicity here).

So, it is in fact the case that it is possible, in principle, ( don't think it is possible in practice though) to control the treadmill in such a way that it imparts a holding force on the plane, preventing it from moving. But, once again, this force is not proportional to the wheels rotational speed, but to the wheel's angular acceleration (note that in the idealized case of massless wheels, it isn't even possible in principle as the lower the moment of inertia of the wheels, the greater the required angular acceleration).

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You make the most important point, which is the plane is thrusted through the air. – mwengler Jul 17 '12 at 22:25

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@JamieHutber, the plane does move. The thrust from the propeller is far more significant that any nominal frictional force from the wheels. The plane moves forward. A plane isn't a car, the motor doesn't drive the wheels. – Alfred Centauri Jul 17 '12 at 22:44

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@Jamie: I think the idea is that you can't match the propeller, the friction on the wheels will never be that high. The plane needs just a little bit of thrust to stay stationary: The treadmill will move and the wheels will rotate, but the plane itself will stay in place. Any more thrust and it will move forward. – Javier Badia Jul 17 '12 at 23:15

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Another way to think about it is a plane taking off on ice - if the speed of the planes wheels mattered how would planes with skis manage? – Martin Beckett Jul 18 '12 at 4:17

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Alfred Centauri is 100% correct. If you are still confused you should re-read his answer, re-read it again, and if need be get a pilots license to fully understand that plane speed is not at all influenced by wheel speed. – Benjamin Horowitz Jul 18 '12 at 16:12
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Simplify. Suppose the air is still - no wind. Suppose the wheels are truly frictionless - like greased skids. (After all, that's why they have ball bearings.)

The aircraft starts from a standing position, and it accelerates to rotation airspeed, about 100 km/h. It does so by thrusting against the air, not against the surface it is standing on.

As it accelerates, the pickup truck pulls the fabric under the plane (simulating a treadmill) in the opposite direction, up to 100 km/h.

So, with respect to the fixed un-moving air, the aircraft is moving one way at 100, and the surface under the wheels is moving the opposite way at 100.

The plane takes off, because of its airspeed.

The wheels turn at 200 km/h, because somebody's dragging the runway backward. They don't care - they're frictionless.

All the "treadmill" has done is make the wheels turn faster.

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This is the correct answer.  If the plane is stationary, the plane will obviously experience zero aerodynamic lift. – QuantumDot Aug 23 '12 at 3:29

Taking this as a logical question rather than physics based it is clearly playing on the mis-assumption that motive force is can only be aplied through contact with the floor.

ie we walk forward by pushing on the floor, we drive by making car wheels push on the road.

However a solution is to realise that a commercial jet will gain the force from pushing on the air as explained elsewhere, the floor contact is irrelevant to the problem.

And so we say the plane takes off. Feel free to complicate the problem as you wish!

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Not just a commercial jet. All powered flying machines take there thrust from the air. – dmckee♦ Jul 18 '12 at 15:27
@dmckee: I was agreeing with you until I got a mental image of a leg-propelled hang-glider launching off a hillside :-) – Mike Dunlavey Jul 19 '12 at 18:04
@MikeDunlavey Uhm....er....I'm going to duck the issue by classifying that as unpowered. Yeah. That's it. – dmckee♦ Jul 20 '12 at 1:36

EDIT ADDED 7/18/12

Unfortunately, the original statement of the original question was totally different than the ACTUAL question the original poster intended to answer. That original question is simply asked and answerd by Mythbusters. If the original poster had simply referenced the source of his question, it would have been much clearer before I did my long answer below.

The actual question the poster wanted to ask, and the one asked and answered by Mythbusters is this: an airplane is on a conveyor belt runway that can run backwards. The forward speed of the airplane is monitored and the conveyor belt is run backwards at that forward speed as the airplane tries to take off. The wheels on the airplane are free rolling (no brakes, no motors). Can the airplane take off?

This is a WAY easier question than the one the poster originally asked in which the original question specified the conveyor belt would run at the speed of the WHEELS. So in the original question, the conveyor belt would run fast enough so that either the wheels were slipping on it (if the plane was moving forward) or the plane was forced to stand still (if the wheels were not slipping on it. That is the question I answered below.

The Mythbusters question is much easier. First, we know a plane doesn't even need wheels to take off, water planes and planes that land on snow or ice on skis do it all the time. The wheels are just a convenient way to have a connection to the ground which is low friction in the forward-backward direction. All the conveyor belt causes is the free-turning wheels to turn twice as fast as they normally would on take-off. Does this cause the engine to put a little more (OK, 4X as much) rotational energy into the rotation of the wheels? Yes it does. It is even vaguely questionable that a plane with a margin-of-error extra power enough to take off by pulling itself through the air can spin its (rather small, relative to the airplane mass) wheels twice as fast? No, the wheel mass is way too small to be a big part of the equation of motion of an airplane being pulled through the air by a propeller. Watch the youtube video and watch the plane take off from the conveyor belt no problem.

Below appears my answer to the original question, which was much more obscure, much more challenging to sort out from a physics perspective.

What a wild question!

The thing that determines take off is enough lift from the wings. The lift depends on the airspeed flowing over the wings. You might think on a windless day that the airspeed over the wings is zero if the airplane is not moving forward, but what if the airplane has a big propeller in front its wings? Then the propeller blows air over the wings. I don't know for sure, but maybe a very powerful acrobatic airplane can blow the wind across its wings with its propeller fast enough to create enough wing lift to take off, even when the airplane is not moving through the air itself. But certainly most front-propeller planes cannot do this, they need forward motion through the air to get enough airspeed across wings, and all jets and rear-propeller planes require forward motion to get airflow across wings.

So the next question is: does the airplane develop any forward motion as you define the problem? Suppose it is a jet. The jet engine is sending a lot of mass of air very fast backwards behind the plane. To conserve momentum, that reverse momentum must be going somewhere. On a normal runway (or a treadmill that can't keep up with the tires) much of that momentum would go in to the forward motion of the airplane.

Now we need to figure out something about what kind of force the treadmill can put on the airplane by running backwards. Suppose we had a tire (or a cylinder) on the treadmill, and the treadmill started running in such a direction as to start the tire spinning but not to translate the tire left or right. Would the tire move along the treadmill, or would the tire stay in place and simply turn as fast as the treadmill was moving? I feel as though I should stop here and let the students figure out their answer to this question. Instead I'll just continue.

Actually lets look at a SLIGHTLY simpler question first. We have a post holding that tire down against the treadmill. If the treadmill is stationary and the tire is stationary, we know there is no force on the post holding the tire. The tire sits still, the post is not being pulled forward backward or sideways.

Now what if the treadmill is running at a steady speed, then at steady state the tire is running at a steady rotational speed = to treadmill speed to stay in place as it will held in place by the post. But is there a forward or backward force on the post? If the bearing holding the wheel to its axle is frictionless, I am pretty sure there is no force. The tire is rotating at a contstant velocity, since the axle is frictionless, it doesn't need any force to keep it rotating at a constant speed. So in steady state, the tire rotating at a constant 100 kph on a treadmill running at a constant 100 kph puts no force one way or the other on the post holding it.

Now how the heck can we couple translation motion of the treadmill into any translational force on the airplane? Assuming frictionless axles on the wheels? In steady state we can't. But what about as we accelerate?

So we look at the problem where the wheel is on the treadmill stationary, and we speed the treadmill up to 100 kph. What happens

1. The wheel spins up slowly but does not forward or backward motion.
2. The wheel doesn't spin at all but does move in the direction of the treadmill
3. The wheel splits the difference, spinning up some as the treadmill accelerates, and picking up some forward motion as the treadmill accelerates.

Now those of us who have been around the block a few times KNOW the answer must be number 3, that is, unless it isn't. But how do we show that?

Consider a wheel in empty space, with its axle aligned with the x-axis, so it can spin freely through the y-z plane. At the lowest point (the most negative z point) we apply a force $+F\hat{y}$ for a time $t$, and then go back to applying zero force. $\hat{y}$ is a unit vector in the $y$ direction, that is the force we apply is along the surface of the wheel only. What does the wheel do?

Well we are imparting linear "impulse" into the wheel of $Ft$ so we change its linear momentum by $Ft$ so we change its linear velocity by $v=Ft/m$ where $m$ is the mass of the wheel.

But we are also putting torque around the axle of magnitude $Fr$ into the wheel where $r$ is the radius of the wheel. Thus we increase the angular momentum of the wheel by $Frt$. Which means we set the wheel to spinning with angular velocity $\omega = Frt/I$ where $I$ is the moment of inertia of the wheel about its axle.

Seeing the linear dependence of $v$ and $omega$ on $Ft$ we can see that no matter what force at what time we put in, the ratio is fixed: $$v/\omega = I/mr$$

The point being, a force applied along the surface of the wheel imparts some linear momentum into the wheel (and whatever it is attached to) and some angular momentum into the wheel (which spins the wheel).

So back to the airplane. We have this airplane with a powerful jet engine imparting a very large $-F\hat{y}$ into moving the airplane forward. If the treadmill is to keep the jet from accelerating forward, it will need to provide an equally large but opposite $F\hat{y}$ to the airplane. But as we saw above, whatever linear force the threadmill applies to the tire, it is applying a proportionally large torque to the wheel.

We note the mass of the airplane $M$ is much more than the mass of the tire, $m$, so $I/r = m\ll M$. So to counteract the force of the jet engine, the treadmill is going to have to accelerate a lot. That is, $\omega=Ct$ to counteract the linear force of the jet engine on the airplane. So the wheel is going to have to spin up really really REALLY quickly, and keep spinning up faster and faster as long as we have the jet engine going. My intuition suggests that long before the wheel reaches relativistic speeds, it will be flung apart by centrifugal forces overcoming the molecular forces that usually keep solid matter solid.

But until the wheel explodes (or the threadmill explodes), the jet is kept from having any linear acceleration, and so does not take off.

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This particular problem is somewhat ill defined but, as usually posed, the idea is that the treadmill speed matches the speed of the plane as if it were accelerating on a runway. So, for example, let there be two identical planes, one on a runway, and one on a treadmill the length of the runway. Both planes apply the same power and now, let the treadmill speed match the speed of the plane on the runway. This is, I think, what most have in mind when they pose this problem. – Alfred Centauri Jul 17 '12 at 23:44
I'm sorry if I'm not understanding something (entirely possible), but consider this: The wheel has two forces in the horizontal direction acting on it: the friction with the treadmill and the thrust from the plane. The whole system will move forward if the thrust is higher than the friction. But the friction has a maximum: a certain coefficiente times the weight of the plane. Therefore, all the plane needs to do to take off is to create a thrust larger than this maximum. Is this right? – Javier Badia Jul 18 '12 at 0:24
@JavierBadia Not completely. It could be moving, but slowed down enough by the friction to not be able to achieve take off speed. So while it might get to 200 mph being pushed by the jet when it is rolling, it might get to only 50 mph if there was some excess friction going on. The FACT is that the wheels can hold a commercial jet against its jet engines at maximum thrust. A commercial jet does not produce enough thrust to overcome tire friction, the brakes must be released for the jet to move. – mwengler Jul 18 '12 at 0:54
@AlfredCentauri your comment here is entirely different than the question you ask in your original post, which i answer in my answer. Your comment means the jet on the treadmill must spin its tires twice as fast as the jet on the regular runway to achieve the take off speed. It seems likely to me the tires might burst or fail in some other way since they are not designed for 2x the speed, and centrifugal force is 2x higher in this case. – mwengler Jul 18 '12 at 0:57

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It doesn't need to, unless the question assumes that the planes brakes are engaged, in which case this is a silly problem. Even with no treadmill, and a plane can't take off like that. – Colin K Jul 18 '12 at 1:51
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Quick answer: You better change your fitness center where you train or stop booze.

Serious answer:

The difference between the speed of air below and above the wing provides the lift. The velocity relative to the ground provides forward motion. The treadmill just zeros the latter.

Case 1: The plane is still relative to the treadmill. But somewhat the air keeps flowing around the wings. Then the plane takes off vertically (if the lift is greater than the weight).

Case 2: The plane is still relative to the treadmill. The air does not flow around the wings (the reactors are not good vaccum cleaners), or more likely, the lift is not greater than the weight, then you just burn fuel.

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It will if its a harrier :)

But if the treadmill is going fast enough so that the airplane has no velocity relative to the ground/air then it will not take off because there will be no lift from air flowing over the wings. This is of course not taking into the effects of propellers which shouldn't be able to lift the aircraft with 0 velocity relative to the ground/air

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The speed of the treadmill affects the rotational speed of the tires, i.e., it changes the speed from what it would be if the treadmill were off. For example, let's say that the ground speed is zero. With the treadmill stopped, the tires aren't rotating. With the treadmill on, the tires rotate and, assuming insignificant friction, the ground speed is still zero. Essentially, the only thing the treadmill speed does is to add or subtract from the rotational speed of the tires for a given ground speed. – Alfred Centauri Jul 18 '12 at 14:57

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My favorite (my only) airplane is a Cessna 172 Skyhawk. At takeoff power, it generates about 400lbs of thrust against the surrounding air. Since the plane weighs about 2000lbs, that's a pretty smart acceleration. It doesn't matter if somebody underneath is spinning up the wheels, forward or back. – Mike Dunlavey Jul 18 '12 at 18:13
@ Alfred, I don't know if the way I typed it was unclear but I totally agree with you and that was what I was trying to imply. – thebmags Jul 18 '12 at 19:23
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