B Can a 747 Take Off on a Conveyor Belt?

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A 747 can take off from a conveyor belt designed to match the speed of its wheels in the opposite direction, as long as there is sufficient engine thrust to propel it forward relative to the air. The discussion highlights that the wording of the problem often leads to confusion, as the conveyor belt's motion does not significantly impede the plane's ability to generate lift. The thrust generated by the engines is the critical factor for takeoff, not the speed of the wheels. Comparisons to pontoons and water dynamics illustrate that the resistance faced by a plane on a treadmill is negligible compared to the thrust produced. Ultimately, the consensus is that the plane will take off regardless of the conveyor belt's presence, provided the engines generate enough thrust.
  • #51
This is a whole lot of hand-wringing and mental gymnastics over a concept that is relatively simple. There are basically 5 forces relevant to the plane as it tries to take off: thrust, drag, lift, weight, and some kind of friction or friction-like resistance to motion as a result of the plane's contact with the ground.

Thrust and weight are completely unaffected by whether the plane is on a treadmill or not. Lift and drag are based entirely on the motion of the plane relative to the air, so the question is really about whether the plane can achieve the required airspeed given a set of initial conditions and in opposite to whatever friction-like force exists between it and the ground. The friction force will depend on the motion of the plane relative to the ground. Let's ignore thrust and weight since they don't change. The problem boils down to a question of whether the aircraft can achieve and airspeed fast enough to generate lift greater than its weight. That means the lift requirement (and thus the airspeed and drag) do not change, and this is only a question of whether the friction force is great enough to prevent reaching that target airspeed.

Regarding the motion of the plane relative to the air, that depends on the motion relative to the ground in two ways: the initial airspeed (if the plane is initially moving with the moving ground) and the maximum achievable airspeed (since the friction force will be larger for larger motion relative to the ground, it essentially creates some terminal velocity). The initial velocity is essentially irrelevant to the final velocity and will only affect how long it takes to reach that terminal velocity, which must be greater than the airspeed required for takeoff.

That leaves us with the effect of friction with the ground. Typically, the thrust only has to overcome drag and rolling friction due to the wheels. If the ground is a treadmill opposing this motion, the only change is that the wheels are now turning faster, so there is likely to be more rolling friction. However, rolling friction is going to be incredibly tiny compared to the drag force and the thrust force as to be effectively irrelevant to the problem. This seems to be confirmed by that full-scale Mythbusters video I posted earlier. With pontoons on a seaplane, the situation is quite a bit more difficult because viscous drag on a pontoon from water is going to be much higher than rolling friction. It is conceivable that there could be a water velocity that is high enough that this drag is large enough, when combined with drag from the air, to prevent a plane from reaching takeoff velocity. I really can't think of a realistic scenario where this would be true of wheels.
 
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  • #52
Clausen said:
I'm not following you here. If you are using the tarmac as the reference it cannot be moving and matching the speed of the tires.

If the tires rotate through one meter going one way, the tarmac stays put and does not match the speed of the wheels.
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.
But that violates the stipulation that the tires must exactly match the belt speed! If that stipulation holds then as the tires rotate through one meter, the belt moves back one meter from some fixed reference (the ground) In that case, the plane cannot move forward unless it also slides as it rolls. At least, that is how I see it.
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.
Clausen said:
That's what I thought too, but the wheels will not be locked, they will be rolling and partially sliding at the same time. If they were locked up, the plane will not takeoff since the coefficient of sliding resistance, rubber on rubber is greater than one. There will not be enough thrust to move against the normal force of 3.5 million N. But sliding while rolling, it might be able to do it.

Yes, it is ridiculous.
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]
Sorry, that's just the way my mind works. I'm an engineer:smile:
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.
 
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  • #53
boneh3ad said:
I really can't think of a realistic scenario where this would be true of wheels.
The only realistic scenario I see doesn't really fit the spirit of the problem: spin the wheels/tires fast enough and they overheat and sieze-up or disintigrate.
 
  • #54
Bandersnatch said:
...the original wording is faulty..
This always seemed to be the whole issue in the discussions on this 10 years ago. Not much has changed it seems.
 
  • #55
russ_watters said:
The only realistic scenario I see doesn't really fit the spirit of the problem: spin the wheels/tires fast enough and they overheat and sieze-up or disintigrate.

Right. It seems like the only scenarios where it might be an issue would cause the wheels to fail anyway, which is not in the spirit of the original question.
 
  • #56
Airspeed, ground speed, tire speed (?... tire RPM...?)... c'mon guys, think about this a little bit, just a little bit ... :ok: ?

The only thing an airplane really cares about is... the relative wind .

Really, it's only the relative wind ... :wink:
 
  • #57
RandyD123 said:
There is a point of no return for any object in the water.
Yes there is, and in this case... I think it's Bud and his friend !
 
  • #58
OCR said:
Airspeed, ground speed, tire speed (?... tire RPM...?)... c'mon guys, think about this a little bit, just a little bit ... :ok: ?

The only thing an airplane really cares about is... the relative wind .

Really, it's only the relative wind ... :wink:

You... you do know the definition of airspeed, right?

Obviously lift only cares about airspeed, aka velocity relative to the wind, but sit down and think about the factors affecting that velocity when a plane is taxiing.
 
  • #59
boneh3ad said:
You... you do know the definition of airspeed, right?
I... I do know the definition(s) of airspeed.
I... I also know that...
The relative wind is of great importance to pilots because exceeding the critical angle of attack will result in a stall, regardless of airspeed.
You... you do the taxiing... I'll do the flying.. :ok:Carry on...
 
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  • #60
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.
 
  • #61
Tito said:
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?
 
  • #62
Tito said:
... speed equal to white wheels...
Whatever that means...
 
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  • #63
russ_watters said:
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.
 
  • #64
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.
 
  • #65
Clausen said:
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*.
 
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  • #66
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!
 
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  • #67
Clausen said:
This is all about reading and understanding the problem as it is presented.
As it is presented? Let's see...

Clausen said:
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.
 
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  • #68
russ_watters said:
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.
 
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  • #69
jbriggs444 said:
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... :wink:
 
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  • #70
A.T. said:
Whatever that means...
Can you show me where I made this quote... other than in this post ?
OCR said:
a speed equal to white wheels

I believe Tito did mention white wheels, though...
Tito said:
a speed equal to white wheels

How did you attribute that quote to me, anyway ? . :oldconfused: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.
 
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  • #71
OCR said:
I believe Tito did mention white wheels, though... How did you attribute that quote to me, anyway ? . :oldconfused:
Sorry, corrected now.
 
  • #72
russ_watters said:
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.

russ_watters said:
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.

russ_watters said:
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.

russ_watters said:
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.

russ_watters said:
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.
russ_watters said:
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.

russ_watters said:
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.
 
  • #73
Clausen said:
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.

Clausen said:
You are all making up your own assumptions.
But your assumptions are just as unclear as the original problem statement.
 
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  • #74
A.T. said:
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.
 
  • #75
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.
 
  • #76
rootone said:
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.
 
  • #77
sophiecentaur said:
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:
RandyD123 said:
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]
 
  • #78
jbriggs444 said:
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.
jbriggs444 said:
[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.
 
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  • #79
sophiecentaur said:
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.
 
  • #80
jbriggs444 said:
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?
 
  • #81
sophiecentaur said:
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.
 
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  • #82
I am relieved about that. I mis-read your double-negative type structure. :smile:
 
  • #83
83 now.
 
  • #84
sophiecentaur said:
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.
 
  • #85
Clausen said:
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."
 
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  • #86
jbriggs444 said:
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.
 
  • #87
Clausen said:
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.
 
  • #88
sophiecentaur said:
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]
 
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  • #89
jbriggs444 said:
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.
 
  • #90
sophiecentaur said:
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].
 
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  • #91
jbriggs444 said:
a constraint that vplane=0vplane=0v_{plane} = 0
How can it be suggested that plane can equal 0 if the wheels are free to rotate? This is getting out of hand.

jbriggs444 said:
disagrees about what point to make about it
What point is there to make except that, unless the wheels break up, the plane can take of. If the wheels are ideal then there is no difference between the stationary runway and a moving conveyor, (forward or reverse).
We seem to be indulging one or more contributors who seem to think that the wheels have a fundamental contribution to the situation. How many more times must it be pointed out that they don't?
 
  • #92
sophiecentaur said:
How can it be suggested that plane can equal 0 if the wheels are free to rotate?
Because that is the implication of the problem statement -- if read in a particular (and fairly natural) way. You take a seemingly reasonable premise, follow it to an absurd conclusion. That's the way "reductio ad absurdum" works.
sophiecentaur said:
What point is there to make except that
92 points, apparently.
 
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  • #93
Nugatory said:
No..
Thank you! You have restored my faith in humanity. I was beginning to think that there isn’t anyone here who would give a straight answer.
Nugatory said:
Of course the answer to this question has no bearing whatsoever on the question of whether the aircraft will in take off.
Taken by itself, this question alone does not say much about the ability of the plane to take off, but now that we have at least one small area of agreement, maybe we can expand that area a bit?

Can you please answer my follow up question in the same straight forward manner?

Remember, all of this follows from the first question, which I will repeat here for convenient reference:

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?

You answered "No", and I totally agree.

Now, the follow up:

If I now push on the hub of that wheel so that the hub Does move to the left (with respect to the same reference) and at the same time the rotation of the wheel and the speed of the belt is (somehow) maintained as in question 1, would the wheel need to be sliding on the belt as it moves?

I think it would have to be. What do you think?

This may not be clear how it applies to the 747 on the conveyor belt but once I have your answer to this I will explain how it all fits together. Please have a bit of patience.
 
  • #94
jbriggs444 said:
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."

I notice you didn't answer the question.

I will answer yours but can you please wait until I have Nugatory's answer to my follow up?
 
  • #95
Clausen said:
I notice you didn't answer the question.
Because the answer is not controversial. That is not our point of disagreement.
 
  • #96
It struck me that the misconception here is a bit like asking how long it would take to stop a bicycle by back-pedalling with an ideal freewheel.
 
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  • #97
If you have two different physical systems then one system one can only affect the other if there is a coupling mechanism between the two . The degree to which one system affects the other system depends on the characteristics and effectiveness of any such coupling mechanism .

In our present problem we have one physical system which is the aeroplane and another physical system which is the conveyor belt .

The only coupling mechanism between these two systems is the feeble one which comes from small interaction forces due to friction1 .

The two systems thus act and behave almost independently . No action of the conveyor belt can have any significant effect on the forces acting on or the motion of the aeroplane during the take off run .

The aeroplane takes off normally and the conveyor belt does whatever it likes .

Note 1 : The fact that the wheels may turn faster than normal has no significant effect on the aeroplane motion or the forces acting on the aeroplane .
 
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  • #98
This does not affect the actual problem answer but interesting to note that the conveyor cannot be running at constant speed . It must start at zero speed and accelerate at the same rate as the aeroplane . Not driven by the plane though - there would have to be an independent motor/engine , sensors to detect aeroplane speed and a control system .

Completely bonkers .
 
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  • #99
Bandersnatch has already touched on the key issue of this problem. If you assume ideal conditions, the instant the jet tries to move forward, the conveyor belt speed and rpm of the wheels would become infinite.
It suffers from the problem many scenarios do when you try to apply ideal conditions to them.
If you short an ideal battery with a perfect conductor, you get a similar problem. you should have zero voltage across a perfect short, but an ideal battery will always maintain a set voltage. So do you have zero voltage across the conductor and no current, or battery voltage across it and infinite current (Actually, using ohms law, I= V/R, which in this case is a division by 0, and is undefined. )
In the real world, the conveyor belt would not have an infinite power supply available to drive it, so it's top speed would be limited by that. The axle friction of the airplane's wheels would not be zero, so there would be a maximum speed at which they could spin before over heating and seizing up or failing for some other reason. In addition, the conveyor belt would not be able to instantly adjust to the tire rotation speed and would always lag behind by some amount. And even with a sufficient power supply, the conveyor belt would have limits on the strains it could endure.

It is only by knowing the real limitations of all the systems involved that you would be able to come up with an answer to this question under those conditions.
 
  • #100
There was another thread on this topic that I posted in a few years ago but I can't seem to find it. It surprises me that so many people get this wrong, even the pilot in the Myth Busters video. I think Nidum did a good job of describing the situation in his/her previous two posts. Barring obvious physical constraints, such as tire maximum ratings, the conveyor direction and speed has no bearing on whether the plane can take off or not.
 
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