B Can a 747 Take Off on a Conveyor Belt?

[DrClaude]

PLANE CANNOT TAKE OFF: Taking off means leaving the ground and moving up into the air. Taking off = LIFTing off the ground. Lift is created by air rushing over the wings to create lift. If the conveyor belt is running the in the opposite direction at a speed equal to white wheels, the plane will not move forwards or backwards. This means that the air around the wing and the wing have about 0 speed. Remember, air needs to be moving at a RELATIVE velocity to the wings, meaning there need s to be a difference in the velocities/speeds. 0 relative velocity = 0 lift = No taking off.

Have you read the explanations given in this thread as to why the plane will fly?

 

[A.T.]

... speed equal to white wheels...

Whatever that means...

 

[Clausen]

The problem doesn't specify reference frames, which is part of the problem. But if the wheels are not sliding, there are three facts that exist no matter the interpretation:
1. The contact patches between the tires and conveyor are stationary with respect to each other.
2. The speed of the plane with respect to the ground is equal to and opposite the speed of the ground with respect to the plane.
3. The speed of the coneyor with respect to either the plane or ground is completely decoupled from the speed of the plane and ground with respect to each other.

This is a tortured mess, but the end result is that you've placed a hidden constraint on the plane's motion: *something* must tether it to the ground in order for it to not move in real life.
[see: right before a plane takes off from an aircraft carrier]
[edit] Actually, it's even worse: if you put numbers to this scenario, I think you will find that you are violating your own assumption.

Yes, it is ridiculous. Again, it takes a hidden mechanism somewhere to make the end result of this tortured mess happen. In this case, in order for the wheels to slide on the conveyor, the brakes must be applied. [see: right before most planes start their take-off roll]

I am too and I soooo disagree with that being how engineers minds should work.

What I see as the problem with the way many people are approaching this is that they are trying to make an interpretation of the problem happen without regard for reality. That's fine for a physics problem (assume you could crash Europa into Mars...), but this problem is supposed to be about real life and therefore the end conclusion must never be something that is impossible in real life.

You cannot use an arbitrary/nonsenscal choice of reference frame to prevent an airplane from taking off in real life.

Who says this is a problem about real life? When, in real life, have you ever seen a 747 on a conveyor belt?

This is all about reading and understanding the problem as it is presented. The fact is, if the belt moves backwards with respect to a point on the ground, the same amount that the circumference of the wheels turns through in trying to move forwards, the hub of the wheel remains stationary with respect to the ground and the air. The only way the plane can takeoff is by sliding while the wheels are spinning.

You may dispute the way an engineer is supposed to think in your view, but you cannot dispute that what I am saying is 100% correct.

 

[rootone]

The plane can only take off if there is airflow over the wings generating lift.
It makes no difference what kind of undercarriage arrangement it has, spinning the wheels will not generate any lift.

 

[russ_watters]

Who says this is a problem about real life?

The problem statement? Anyway, if it isn't about real life, the answer could be literally anything, including "yes it can take off if aliens teleport it into space."

In either case, just for the record, regardless of the problem as stated, do you agree that a *real* airplane on a *real* conveyor can take off?

When, in real life, have you ever seen a 747 on a conveyor belt?

Just because nobody's bothered to try it doesn't mean they couldn't. Mythbusters used a Cessna because that is easier/cheaper. Also, the scenario can be made functionally equivalent using wind: The plane is sitting stationary on the ground (speed of ground = speed of wheels = 0), with a 160kt headwind lifting it off the ground.

This is all about reading and understanding the problem as it is presented.

The problem, as specified, is mathematically/grammatically flawed/incomplete: 1=2 unless 0=0, in which case the problem is pointless (a plane sitting on the ground in no wind, with its engine off doesn't take off -- so what?).

The fact is, if the belt moves backwards with respect to a point on the ground, the same amount that the circumference of the wheels turns through in trying to move forwards, the hub of the wheel remains stationary with respect to the ground and the air.

Right. So with the engines at full throttle, what is keeping the plane stationary?*

The only way the plane can takeoff is by sliding while the wheels are spinning.

Or by breaking the problem because the problem is flawed. That's how it works in real life.

You may dispute the way an engineer is supposed to think in your view, but you cannot dispute that what I am saying is 100% correct.

Because the problem is ill-posed, "correct" is debatable in terms of how the problem statement works. But in terms of how real life works, the issue is not debatable: a plane on a conveyor can take off. So any answer that interprets the problem statement to yield an answer of no requires making up other assumptions and adding them to the problem*.

 

[russ_watters]

I'm going to try a different approach because up until the last exchange it actually hadn't occurred to me that someone would not be trying to make this problem real-world. That means there may not actually be any disagreement about any fully-defined scenario. So I'd like to run a handful of scenarios by people to see if there actually is any disagreement:

The basic, real life scenario:
1. A *real* plane on a *real* conveyor can take off regardless of the speed of the conveyor at least up to the speed where the wheels disintegrate.

Others that may or may not match real life and don't have to:
2. A plane sitting on a conveyor that rolls under it, with its engines on, tethered to the ground, with no wind will not take off.
3. A plane, sitting on the conveyor that rolls under it, with its engines off, tethered to the ground from the front, with sufficient wind will take off.
4. A plane sitting on a conveyor that rolls under it, with its engines on, while Luke Skywalker uses "The Force" to hold the plane in place will not take off.
5. A plane sitting on a conveyor that rolls under it, with its engines off, will take off if aliens use a tractor beam to yank it into space.

Does anyone disagree with any of these scenarios? If no, then the open issue is whether any particular person who asks the question is looking for a real life answer (even if they word the question badly) or not. And as 2 and 3 reflect, I've never seen anyone pose this problem in a complete way that enables the plane to remain stationary. It can be mathematically constrained, but if you apply physics to it, it requires another element, not specified, to provide the mathematical constraint.

Also, as can be seen, the motion of the conveyor(or not) is basically irrelevant to all of the scenarios! Whether the plane takes off or not depends entirely on the additional assumptions/constraints one adds to the problem!

 

[A.T.]

This is all about reading and understanding the problem as it is presented.

As it is presented? Let's see...

The fact is, if the belt moves backwards with respect to a point on the ground, the same amount that the circumference of the wheels turns through in trying to move forwards,

Where does the problem mention the circumference? Where does it define a reference frame? As it is presented, it could be any point on the wheel in any reference frame.

 

[jbriggs444]

2. A plane sitting on a conveyor that rolls under it, with its engines on, tethered to the ground, with no wind will not take off.

[...]
Just to explore the range of options similar to 2...

6. An automobile with its engine in gear drives forward at 30 miles per hour relative to a conveyor that is moving rearward at 30 miles per hour. The conveyor is light weight and [its mechanism is] nearly frictionless. As the driver presses on the accelerator, the car drives forward at 60 miles per hour and the conveyor naturally increases its rearward speed to 60 miles per hour. To compensate for frictional losses, the conveyor is slanted upward in front of the drive wheels. To ensure stability, it is also slanted upward to their rear.

This setup is also known as a dynamometer.

One then puts wings on the car, shifts into neutral, adds a propellor and takes off.

Edit: Added verbiage to clarify "frictionless" since @berkeman has correctly pointed out the ambiguity.

 

[berkeman]

The conveyor is light weight and nearly frictionless

That would mean that the bearings supporting the conveyor belt are nearly frictionless, not the surface of the conveyor belt that makes contact with the tires...

 

[OCR]

Whatever that means...

Can you show me where I made this quote... other than in this post ?

a speed equal to white wheels

I believe Tito did mention white wheels, though...

a speed equal to white wheels

How did you attribute that quote to me, anyway ? . Oh, and BTW I want to clarify something, when I said...
"think about this a little bit, just a little bit ..."
I really did mean a little bit... as in, don't "over think".

I thought the first video I posted would show everything... that was needed to know ?

My bad... apologies to boneh3ad, and all the rest.

Now...
Carry on.

 

[A.T.]

I believe Tito did mention white wheels, though... How did you attribute that quote to me, anyway ? .

Sorry, corrected now.

 

[Clausen]

The problem statement? Anyway, if it isn't about real life, the answer could be literally anything, including "yes it can take off if aliens teleport it into space."*.

Let's not get silly here. Lots of problems are posted as brain teasers that have little or nothing to do with real life. All I am trying to do is work within the constraints posed by the problem, regardless of how realistic those constraints are.

In either case, just for the record, regardless of the problem as stated, do you agree that a *real* airplane on a *real* conveyor can take off?

Yes, I have already stated that in my first post in this thread and I pointed out that in a real life situation there is no relation between the speed the wheels are turning and the ability of the plane to take off. Then, I followed that up by saying if we are to strictly be constrained by the condition that the belt moves backwards as fast as the wheels turn forwards, the only way the plane can advance is by sliding while the wheels are spinning. You still have not shown that to be wrong.

Just because nobody's bothered to try it doesn't mean they couldn't. Mythbusters used a Cessna because that is easier/cheaper. Also, the scenario can be made functionally equivalent using wind: The plane is sitting stationary on the ground (speed of ground = speed of wheels = 0), with a 160kt headwind lifting it off the ground.

You don't like the problem posed in the OP, so you want to change it into something you like. The OP specified a 747 on a conveyor belt. Nothing else.

The problem, as specified, is mathematically/grammatically flawed/incomplete: 1=2 unless 0=0, in which case the problem is pointless (a plane sitting on the ground in no wind, with its engine off doesn't take off -- so what?)..

OK. Then your answer is the question is flawed and unanswerable as stated. I am fine with that. But if you say, as others have, that the plane will definitely take off, under the constraints given, you need to explain how it does that. I say it can only do that if the wheels slide while spinning at the specified rate as dictated by the belt. That solution is in accord with all of the constraints of the problem.

Right. So with the engines at full throttle, what is keeping the plane stationary?.

I never said it will remain stationary. I said it will move forward with the wheels spinning and sliding and I gave some numbers but I do not know what the coefficient of resistance will be for a spinning and sliding wheel. Do you? I think it would be interesting to know that.

Or by breaking the problem because the problem is flawed. That's how it works in real life.

My life is not dictated to by your life. Instead of breaking the problem and solving something else, I try to come up with a solution to the problem as it is stated.

Because the problem is ill-posed, "correct" is debatable in terms of how the problem statement works. But in terms of how real life works, the issue is not debatable: a plane on a conveyor can take off. So any answer that interprets the problem statement to yield an answer of no requires making up other assumptions and adding them to the problem*.

So does any answer that yields a YES. You are all making up your own assumptions.

 

[A.T.]

Then, I followed that up by saying if we are to strictly be constrained by the condition that the belt moves backwards as fast as the wheels turn forwards,

Which makes no sense, because you are equating a linear velocity (belt moves) to an angular velocity (wheels turn). Also, specifying linear velocities requires a reference frame.

You are all making up your own assumptions.

But your assumptions are just as unclear as the original problem statement.

 

[sophiecentaur]

Which makes no sense, because you are equating a linear velocity (belt moves) to an angular velocity (wheels turn). Also, specifying linear velocities requires a reference frame.But your assumptions are just as unclear as the original problem statement.

Was the linear vs angular thing of any real importance? I think we all managed to cope with that.
Actually, the Original Problem was perfectly well specified. The only 'assumption' was that the conveyor belt is at least as long as the runway and that it's surface surface would be the same and would actually support the plane. What went wrong was that, later, people inserted some of their own constraints and also a few misconceptions about what factors were in fact of any importance.

 

[rootone]

I think the thing that has been learned here has nothing to do with planes.
It is however a good demonstration that if a question is imprecise then a precise answer cannot be expected.
In fact people may tend to make their own assumptions concerning imprecision in the question, and answer accordingly.
In my case had pictured the plane being somehow fixed in place with engine(s) off,
so the conveyor would have no effect other than turning the wheels.

 

[sophiecentaur]

imprecision in the question,

No imprecision in the question; you can assume the required takeoff airspeed of a specified aircraft. The only thing that was missing was the common sense of people who ignored the fact that is by far and away the major factor. The fact that there are 70+ posts goes to show how imprecisely many people tend to think about straightforward physical situations and how easily they can be misdirected. (That's how magicians earn their money.
If in doubt, draw a free body diagram and the answer will leap out of the page at you.

 

[jbriggs444]

Was the linear vs angular thing of any real importance? I think we all managed to cope with that.
Actually, the Original Problem was perfectly well specified. The only 'assumption' was that the conveyor belt is at least as long as the runway and that it's surface surface would be the same and would actually support the plane.

There is another assumption buried in the problem statement. That statement can be read in two parts.
1. There is a constraint that $v_{conveyor} = -v_{wheels}$
2. That constraint can be realized by appropriate engineering of a hypothetical conveyor.
For reference:

The conveyor best is designed to exactly match the speed of the wheels, but run in the opposite direction.

No such conveyor design is feasible when applied to a 747 on a takeoff run with the brakes off.

[Barring the unreasonable case where the wheels are made spin so rapidly that they disintegrate. That's another assumption buried in the problem statement: That a 747 on a takeoff run has wheels]

 

[sophiecentaur]

No such conveyor design is feasible when applied to a 747 on a takeoff run with the brakes off.

When you think of the serious consideration that the Space Enthusiasts give to some of their proposed schemes, I don't reckon it would be that unthinkable. But, if the OP had thought again about the question, he/she could have suggested a Learjet or Hawk and all the same theory would have applied. And all the same misconceptions would have been laid to rest.

[Barring the unreasonable case where the wheels are made spin so rapidly that they disintegrate.

I wouldn't imagine that any aircraft or car wheel assembly couldn't cope with twice normal operating speed.

 

[jbriggs444]

When you think of the serious consideration that the Space Enthusiasts give to some of their proposed schemes, I don't reckon it would be that unthinkable. But, if the OP had thought again about the question, he/she could have suggested a Learjet or Hawk and all the same theory would have applied. And all the same misconceptions would have been laid to rest.

I wouldn't imagine that any aircraft or car wheel assembly couldn't cope with twice normal operating speed.

There's another assumption you're bringing in -- that it's not wheel rotation speed that matters, but aircraft ground speed instead.

 

[sophiecentaur]

There's another assumption you're bringing in -- that it's not wheel rotation speed that matters, but aircraft ground speed instead.

That isn't an "assumption"; it's a consequence of the wording of the OP.
The wheel over conveyor speed will only be twice the wheel over ground speed (= takeoff airspeed). How could you suggest that it wouldn't be twice if the conveyor has equal and opposite velocity?

 

[jbriggs444]

That isn't an "assumption"; it's a consequence of the wording of the OP.
The wheel over conveyor speed will only be twice the wheel over ground speed (= takeoff airspeed). How could you suggest that it wouldn't be twice if the conveyor has equal and opposite velocity?

We are in violent agreement. The reason that it's a consequence of the wording of the OP is because of the absurdity of the alternative.

 

[sophiecentaur]

I am relieved about that. I mis-read your double-negative type structure.

 

[sophiecentaur]

83 now.

 

[Clausen]

83 now.

I don't like odd numbers.I would just like to have some of you people on a witness stand, under oath, and pose this question to you:

If during a given time interval, a wheel with a circumference of 1 meter rotates through one revolution in trying to roll to the left, on a conveyor belt that is moving to the right a distance of one meter, all with respect to a fixed point on the ground, does the center hub of that wheel move with respect to that fixed point?

A simple Yes or No will do please, no hemming and hawing.

 

[jbriggs444]

If during a given time interval, a wheel with a circumference of 1 meter rotates through one revolution in trying to roll to the left, on a conveyor belt that is moving to the right a distance of one meter, all with respect to a fixed point on the ground, does the center hub of that wheel move with respect to that fixed point?

Your honor, may I cross-examine the witness?

If one attempted to restrain the motion of the hubs on set of 747 wheels by means of a rearward acceleration of the conveyor belt upon which they ride, what acceleration would be required to match the static thrust of the craft's engines operating at takeoff power?

For how long could said acceleration be maintained without the wheels disintegrating?

Would you say that a 747 can operate its engines at such a power setting for longer than that?If it please the court, I now draw your attention to the question of statutory construction...

https://www.law.cornell.edu/wex/statutory_construction

"Courts generally steer clear of any interpretation that would create an absurd result which the Legislature did not intend."

 

[sophiecentaur]

If one attempted to restrain the motion of the hubs on set of 747 wheels by means of a rearward acceleration of the conveyor belt upon which they ride

To restrain the hubs, you would have to be moving the conveyor Forward and not Backwards. There would be no 'restraint' involved. the plane would just be moving forward, through the air and the conveyor would be keeping up with it. The wheels would have no tangential force on them.

 

[Nugatory]

does the center hub of that wheel move with respect to that fixed point?

A simple Yes or No will do please, no hemming and hawing.

No.

Of course the answer to this question has no bearing whatsoever on the question of whether the aircraft will in take off.

 

[jbriggs444]

To restrain the hubs, you would have to be moving the conveyor Forward and not Backwards. There would be no 'restraint' involved. the plane would just be moving forward, through the air and the conveyor would be keeping up with it. The wheels would have no tangential force on them.

If one were to attempt to restrain the hubs from moving forward, one would need the conveyor to be accelerating rearward.

This would be quite difficult, given the tendency of the wheels to freely roll with a forward angular acceleration, however a significant rearward force could be achievable with a sufficiently huge acceleration.

Accelerating the conveyor forward would instead result in a forward force on the contact patch at the bottom of the tires and an associated forward force on the hubs. It would not impede the forward motion of the craft.

Note that I am including the moment of inertia of the wheels in the model. The required tangential force on each wheel is equal to $\frac{static\_thrust}{number\_of\_wheels}$ and the resulting angular acceleration $\alpha = \frac{r_{wheel}\ static\_thrust}{number\_of\_wheels\ I_{wheel}}$ of each wheel would be rather large.

Edit: Back of the envelope...

747 static thrust is around 1,000,000 N, tire radius about 0.6 m, tire mass somewhat over 100 kg.
Estimated moment of inertia per wheel = $100\ 0.6^2$ ~= 36 kg m^2. 16 wheels on the main landing gear plus whatever is under the nose. Call it 1000 rad/s^2, give or take a factor of 2.

At their max rated speed (235 mph or around 100 m/s), a 0.6m radius tire will be rotating at about 30 rad/s. That means about 30 milliseconds into the takeoff attempt, the tires will be at their max rated speed -- well before the engines have even finished spooling up.

[Which calculation puts a similar upper bound on the duration of that chirp they dub into movies when the wheels touch down during landing of a jet aircraft -- a nice sanity check]

 

[sophiecentaur]

If one were to attempt to restrain the hubs from moving forward, one would need the conveyor to be accelerating rearward.

This needs clearing up. The plane is going forward so, to stop the wheels from rotating, the runway would need to go forward. That's true but not relevant to the OP which stipulates that the conveyor is going backwards at an equal speed to the forward speed of the wheels. At least, that's how I understood the OP. There is no wonder that we are shouting at each other if we interpreted the OP in directly opposite ways.
BUT, because the craft has wheels, which are very low drag (by design), it makes no difference one way or the other to the ability to take off. A forward moving conveyor would make things slightly easier but that's obvious (isn't it?).
Haha. 88 posts and we're only just getting down to what the OP meant. Good ole PF.

 

[jbriggs444]

This needs clearing up. The plane is going forward so, to stop the wheels from rotating,

Note the wording of the claim: "attempt to keep the hubs from moving forward". I am not speaking here of the rotational velocity, $\omega$ of the hubs but of their ground-relative linear velocity, $v$

It is, of course, the ground-relative linear velocity (and, hence, the air-relative linear velocity) that is of relevance to take-off. However, the constraint mentioned in the problem statement is more direct than that.

The constraint in the problem (if read as @Clausen would have it) is that $$v_{conveyor} = - \omega_{wheel} r_{wheel}$$.
The free-wheeling constraint I believe you adhere to (and with which I agree) is that $$v_{plane} = \omega_{wheel} r_{wheel} + v_{conveyor}$$
Putting those together, we have that $$v_{plane} = 0$$.
As I read the problem, a constraint that $v_{plane} = 0$ must be achieved entirely by design of the conveyor. This means that the plane must be restrained from forward motion by the rearward acceleration of the conveyor alone.

I am trying to point out that this chain of reasoning leads to an absurdity. So the starting interpretation that $v_{conveyor} = -\omega_{wheel} r_{wheel}$ is thrown into doubt.

[My suspicion is that pretty much everyone here agrees about the physics but disagrees about what point to make about it].