Calling all mathmaticians

[Drakona]

Maybe. I was just assuming an ideal wheel with some nonzero coefficient of friction. Notionally, I could move the ground fast enough to get whatever force I wanted to counter the engine.

Practically, I have no idea what would happen. Snoopy likely knows much more about how wheels would fail in such circumstances than I do, seeing as he works on airplanes and all. Would it seize up all at once? Well, your plane still isn't taking off, then. At least, not in the intended direction . . .

[fliptw]

NB: mythbusters also did this conveyor airplane bit too.

[Duper]

NB: mythbusters also did this conveyor airplane bit too.

yeah, I saw that one. I was really amazed at how they did that.

[Krom]

No, Woodchip is right, as snoopy hinted. A sufficiently fast conveyor belt will prevent the airplane from taking off. The wheels might be low friction, but they aren't zero. You just have to spin them fast enough. I mean, really fast enough.

It's all in the posing of the problem.

If you say "the conveyor belt goes backwards as fast as the airplane goes forwards", then it takes off no problem.

If you say, "the conveyor belt goes backwards fast enough that the airplane holds still," then it doesn't take off. It's also an insanely fast conveyor belt.

Funny, every time I hear this problem posed, I'm always tempted to bring up the supersonic conveyor belt. I usually restrain myself, though, because it sows so much confusion, and apparently the mere distinction between ground speed and air speed sows enough confusion . . .

Strictly speaking the tires would explode and what remains of the wheels would turn into a spectacular shower of sparks, but that still wouldn't prevent the plane from taking off. Although it would jeopardize it making a safe landing again so no pilot would actually go through with it.

The problem is no matter how fast you ran the treadmill the wheels would never generate sufficient friction to stop the plane, by the time the wheels exploded they would be immediately converted to bare metal skids on the treadmill which still wouldn't have enough friction. Likely the ONLY time the wheels have enough friction to do the job is when the plane is already stopped. Once it is moving with the engines at full thrust the tires are just going to slip and burn even if you lock the brakes.

Here is an example realistically of how much friction the wheels generate:

[Thenior]

Now what if the plane was holding still, and allowed the conveyor belt to give it a backward momentum of 100 knots. The plane would then have to apply at least 100 knots of propulsion from the opposite direction to become 0 knots. But the instant it reaches 100 knots propulsion, it would then begin to gain forward momentum.

But what if it was going backward @ 100 knots and only applied 20 knots of forward propulsion?

[Lothar]

I used to know a friend who owned a Piper Cub. The plane had a very low takeoff speed. If the wind was blowing hard and constant enough, he could actually take off and fly backwards if he throttled back a little.

Pretty gutsy to take a 1000-pound airplane up in 33+ knot winds. They'd have to be remarkably constant...

[Duper]

A better question Thenoior, would be: how would the aerodynamics of the plane effect its ability to stay put while moving backwards at \"100 knots\" ( little more than 115 mph) that's a hurricane force tail wind. .. relative to a stationary object on the ground. That doesn't sound very promising for the plane.

But Like I said, the original question from which all this stemmed a couple of years ago is inherently flawed. it makes some radical assumption on certain forces while not taking into account others.

It would best be stated: If an forward vectored aircraft (to exclude helicopters and VETOLs <obviously>) is placed on a conveyor belt moving in the opposite direction of the aircrafts forward advancement; could the conveyor move fast enough to halt the craft's forward movement?

This isn't perfect, but it's better than the original.


LOL Krom. Great vid. This guy was at our air show this year pulling a full tanker fire truck.

[Spidey]

Now what if the plane was holding still, and allowed the conveyor belt to give it a backward momentum of 100 knots. The plane would then have to apply at least 100 knots of propulsion from the opposite direction to become 0 knots. But the instant it reaches 100 knots propulsion, it would then begin to gain forward momentum.

But what if it was going backward @ 100 knots and only applied 20 knots of forward propulsion?

Wouldn’t matter…as soon as you applied “any” forward thrust the aircraft would begin to slow its backward motion, and would just be a matter of time, before building forward air speed.

Assuming your 20 is enough to reach takeoff speed, it would take off.

Just a matter of time at that point.

[Floyd]

Now what if the plane was holding still, and allowed the conveyor belt to give it a backward momentum of 100 knots. The plane would then have to apply at least 100 knots of propulsion from the opposite direction to become 0 knots. But the instant it reaches 100 knots propulsion, it would then begin to gain forward momentum.

But what if it was going backward @ 100 knots and only applied 20 knots of forward propulsion?

first: there is no "backward momentum", only backward speed. momentum has to do with rotation ...
second: you don't apply "knots" propulsion, but force of propulsion. even if your scale was right, it assumes you drive through the wheels (like a car), not through the air. even if there were "20 knots" applied, it is a constant force which is necessary to be greater than the friction force between airplane and conveyor belt. a constant force means a constant acceleration. once the planes mass inertia is overcome (disregard wind=speed difference between air and ground), it starts to move forward, regardless of its position, until it reaches take off speed (air friction disregarded, otherwise this explanation would extend somewhat).
remember: the wheels are not connected to the planes engines.

[Tunnelcat]

Pretty gutsy to take a 1000-pound airplane up in 33+ knot winds. They'd have to be remarkably constant...

Lothar, it was funny to watch him do it. The runway ran in a northeasterly direction and the wind was almost always right down the runway. It was a private grass airstrip, so he didn't have to worry about other traffic most of the time. I saw him do it at least twice when the wind was the right speed. Strangest looking thing to watch.

You've all been sucked into the black vortex of endless debate over a nonsensical puzzle!

[Lothar]

first: there is no "backward momentum", only backward speed.

Wrong. Momentum is simply mass times velocity; it applies both to linear and rotational speed.

When you talk later about applying "force of propulsion", recall that force is mass times acceleration, and acceleration is a change in velocity -- in other words, force produces changes in momentum. When you talk about overcoming "mass inertia", that's equivalent to saying enough force has been applied to change the object's momentum from backwards to forwards (which is equivalent to saying enough acceleration has occurred to change the object's velocity from backwards to forwards.)

you don't apply "knots" propulsion, but force of propulsion

right -- but you can't neglect friction (drag) from the equation. In general, an aircraft (or any other object) will accelerate until thrust equals drag. If you don't apply adequate thrust, the final speed will be below (dirty or clean) stall speed, meaning the aircraft won't be able to take off.

In the case of an airplane on a backwards-moving conveyor belt with frictionless wheels, the airplane will accelerate and take off as normal (the conveyor belt produces zero drag.) In the case of an airplane on a backwards-moving conveyor belt that's moving fast enough to make the bearings seize up, the conveyor may produce sizeable drag. If the drag is high enough, the airplane won't be able to accelerate up to an adequate airspeed to take off -- but this will only happen in a case that produces a RIDICULOUS amount of drag from the conveyor belt.

[Floyd]

first: there is no "backward momentum", only backward speed.

Wrong. Momentum is simply mass times velocity; it applies both to linear and rotational speed.

oh. seems like a false friend. i should have looked up "momentum" before, which is "impuls" in german. what i have mistaken it for was "torque" (="moment" in german). thanks for pointing this out.
so the logic still stands anyway.
i thought i wasn't neglecting drag (other than that of air, as stated), however.

seems i have to work on my english then.

[Drakona]

No, Woodchip is right, as snoopy hinted. A sufficiently fast conveyor belt will prevent the airplane from taking off. The wheels might be low friction, but they aren't zero. You just have to spin them fast enough. I mean, really fast enough.

The problem is no matter how fast you ran the treadmill the wheels would never generate sufficient friction to stop the plane, by the time the wheels exploded they would be immediately converted to bare metal skids on the treadmill which still wouldn't have enough friction. Likely the ONLY time the wheels have enough friction to do the job is when the plane is already stopped. Once it is moving with the engines at full thrust the tires are just going to slip and burn even if you lock the brakes.

Oh, I've no doubt that's true in the practical case. Even if the static friction of tires on runway is enough to prevent takeoff--and I'd think they'd want to design airplanes so that was generally true--it doesn't surprise me to hear that the kinetic friction isn't. But that's with the runway holding still.

Perhaps my grasp of physics is incomplete. I had been thinking the coefficient of kinetic friction between the surfaces would be the same, regardless of how fast the plane is going relative to the runway/conveyor. If that's true, then the amount of drag the runway applies to the airplane increases linearly with the relative speed of the runway. At some speed, the conveyor really is fast enough to cancel out the force the engine applies and prevent the airplane from taking off.

(Yes, I know an impractically fast speed would be required to achieve this. It's just a notional criticism of the problem designed to make me arrogantly feel clever while confusing folks still struggling with the basic concept it's intended to illustrate. It would be most ironic if it actually revealed a physics deficiency on my part...)

[woodchip]

Krom, what do the wheels have to do with this? On a conveyor belt the plane is essentially parked. The rearward motion of the belt on a windless day is creating a negative lift force (much like the foil on a formula one race car that creates a downward force for better traction around curves). To balance the negative lift, does not the plane then have to be propelled forward to counter the negative lift?

[snoopy]

Perhaps my grasp of physics is incomplete. I had been thinking the coefficient of kinetic friction between the surfaces would be the same, regardless of how fast the plane is going relative to the runway/conveyor. If that's true, then the amount of drag the runway applies to the airplane increases linearly with the relative speed of the runway. At some speed, the conveyor really is fast enough to cancel out the force the engine applies and prevent the airplane from taking off.

I'll say that I think the drag of the wheels will increase as you increase in speed.

Here's why: The main thing that we're concerned about here is not so much coefficient of friction, but rolling resistance. At extremely low speeds rolling resistance is primarily driven by friction. As you increase in speed, two things become more and more of a factor: 1. The amount of energy lost in the action of squishing the tire as you roll along. 2. The amount of energy lost to vibration due to imperfections in the bearings and wheels.

Summary: as you increase in speed, rolling resistance is more of a factor of energy lost to deformation, rather than energy lost to slip.

I agree with Krom, I think that practically you will destroy your tire before you reach the point of the rolling resistance being able to keep you from taking off. Also, considering the imperfections inevitable to exist on the belt, you'd probably bounce around and lose control before you managed to make rolling resistance a factor. I don't think that makes make very stable cars, and they just hide it by getting stability from the airflow over their surfaces as they increase speed.

[Krom]

From what I was snooping around on the net when I wrote that up, friction (between solid objects, like wheels and pavement) is at is maximum at or near a complete stop and actually decreases beyond some speed threshold that is still a fairly slow. But after that it does not change very much but does slightly increase with speed. I don't know if it ever comes back to the same value that it was at a stop, but I imagine it doesn't.

In the impractical test of an airplane with indestructible wheels and a treadmill with no speed limit I don't know one way or the other actually. The treadmill could eventually build up enough rolling friction to keep the plane from moving forward, or the friction could reach some limit before it is enough to stop the plane. Although perhaps eventually the plane and the treadmill would simply move the atmosphere around the plane enough to generate sufficient lift to get it off the ground without actually moving forward anyway. All we need is air going over the wings, not actual forward motion.

But so far it stands that within practical means of measuring you can't build a treadmill to keep an airplane on the ground.

Now if you dropped the treadmill part, and instead built an enormous fan in front of the airplane, and ramped up the speed of the fan as the planes engines revved up. You could keep the plane from moving forward, but not necessarily from leaving the ground.