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Yes. An airplane is not propelled forward by the wheels, but by the propeller(s) or jet engines. How fast the wheels are spinning is irrelevant. It may be a little trickier to steer but it'll take off.
If you think 'goto' is evil, try writing an Assembly program without JMP.
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wind speed is all that matters. Nothing else for take off. Not ground speed, not wheel speed. Only the speed with which the plane is moving in relation to the air. This is why planes prefer to take off into the wind. It reduces the amount of time on the ground before liftoff. But again ground speed doesn't matter. Air speed matters.
To err is human to really elephant it up you need a computer
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Yes. Unlike a car that uses it's engine to turn the wheels to propel it forward, on an airplane the wheels are not what propels the plane at all. Its the thrust being generated by moving air via propellers or jet engines that provides thrust for an airplane, and the medium they are moving in, and what that thrust is relative to, is a sea of air.
Put a boat on wheels, place it so it sits on a conveyor in water, and turn on the propeller. As long as the wheels on the boat that are in contact with the conveyor are free wheeling the boat will move forward.
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In general, no. The movement of the wheels has no effect on lift. To take flight, a craft needs either sufficient airflow across the wings or enough thrust to overcome gravity. A plane can take off from a stationary position, without a conveyor, if there is enough wind and the engine has enough thrust to counteract the drag; the plane would simply rise straight off the ground. Additionally, air craft with very powerful engines, like an F-16, can accelerate vertically. In this case, the wings do not generate lift; the engine itself provides all the lift. Theoretically, an F-16 could take off from a stationary but vertical position. Again, the wheels would not be used for this.
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Ignoring the possibility of a head wind strong enough to take of on the spot (or very close to it): No, it will not take off.
The engines will obviously apply a force to the aircraft. To match the speed of the wheels, the conveyer belt would have to keep the plane still - if the plane is moving forward, the wheels much move faster than the belt (ignoring maximum friction, so the wheels will not slip). This means the belt will have to apply enough reverse force on the tires that the increased tire rolling friction and bearing friction transferred to the landing gear is identical to the force applied by the engine. This would quickly require the wheels to spin so fast the centrifugal force will rip them apart - first the ties, then the wheel or bearings. Then anything remaining of the landing gear will be ripped off, and the aircraft will crash on its belly on top of a conveyer belt moving it rapidly backwards. Kind of hard to get in the air from that position.
Any limit to the available friction between belt and tires will allow the tires to slip over the belt - this means the aircraft could be moving forward while the belt is still matching the speed (but not position) of the wheels. But any friction available will be "used" to accelerate the wheels - so anything but the most minuscule friction would not allow the plane to reach takeoff speed before the wheels collapse. To take of, you should basically be able to do it with the breaks applied (ignoring the pesky nose or tail wheel without breaks).
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The conveyor stuff is just complication. The entire point of the conveyor is to prevent lateral movement. Mythbusters had no budget for a real conveyor and their facsimile wasn't a conveyor and didn't prevent lateral movement. The weight of the plane and the stretch of the material allowed for lateral movement. A dynamometer would have been better.
Sure, the wheels don't matter. So take the wheels off and jack the plane onto cinder blocks. Same concept, only without the fake-conveyor nonsense that lets people think you can have lift in no wind with no lateral movement.
Jet, plane, whatever... If you prevent it moving forward, it's not going to move upward just because you tilt a control surface. Tie a sea plane off by its rear to a dock. Throttle slightly to get all the slack/stretch of the line out, then push to full. It would remain more or less stationary.
"Prop wash" lift is a thing in R/C aircraft where thrust-to-weight blows pretty much all real planes out of the water. You can hover some R/C planes like a helicopter. You might be able to do that with some real sport planes, but I doubt it.
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jochance wrote: If you prevent it moving forward
But what would prevent it from moving forward? The engines, whether jet engines or propellers, thrust the plane in the forwards direction, and there is nothing from stopping it. The free running wheels will not stop it, even if they are spinning at a fairly high speed.
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That's the definition of an actual conveyor belt for purpose of this thought experiment.
If the conveyor moved backwards as fast as the plane would otherwise be moving forwards there would be 0 horizontal movement, and so, 0 air flow to generate lift.
Prop thrust is what gets the plane moving forward, it has nearly nothing to do with lift. So, simplify it further and take the plane bit out of the equation a moment. Make it a car instead.
If X RPM of a propeller will make it top out at 20 mph, use gearing to make that same RPM drive these car wheels at 20 mph. Put that car on conveyor moving the other way at 20 mph. You're telling me that car goes forward? No.
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jochance wrote: If the conveyor moved backwards as fast as the plane would otherwise be moving forwards there would be 0 horizontal movement, and so, 0 air flow to generate lift.
The only force from the conveyor belt on the plane to make it move backwards is the friction in the wheel bearings, which should be rather small. You have the same friction at a normal take of, so the engines are dimensioned to handle it. Compared to the force required to accelerate the plane to take off speed.
If the conveyor belt somehow has managed to get the plane up to take off speed in reverse, it must have taken a tremendously long time (and tremendously long conveyor belt) But once done, the turbines have no bigger problem pushing the plane up in speed on top of that conveyor belt. When they have done the same amount of work as is normally required to reach take off speed, the plane is standing still relative to the ground. Any further work done by the turbines will accelerate the plane relative to the ground, in the normal manner. The epsilon force from the conveyor belt, through the friction in the bearings, will not stop it.
So getting off the ground will take roughly twice as long (and twice as much fuel), but that doesn't prevent it from happening.
So, simplify it further and take the plane bit out of the equation a moment. Make it a car instead. That makes a completely different situation. The wheels of a car are not free running, but tightly connected to the car engine. The acceleration of the car is caused by the force of the rotating wheels on the ground / belt. Take the belt away, leaving the car floating in free air, and it can neither speed up nor brake down. Suspend a (motorized) model plane in a string, and start the propellers / turbines: It will pull ahead.
So if the original question was talking about a car, then you would be right. But it didn't. That is the essential 'trick question' part: Most of us will think of the way a car accelerates (completely dependent on a solid grip on the road) and overlook that planes are completely different (totally independent of any grip on the runway) with respect to propulsion.
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trønderen wrote: But once done
There is no "once done". It's a conveyor belt. It loops around and keeps happening. It also speeds up such that no, there is no "further work done by turbines" to move the plane relative to ground speed. The definition of conveyor for the thought experiment defines that.
It's no different than a car in neutral with wings slapped onto it and tied to a tree in front so it doesn't go backwards. I doubt you'd say that would lift off just because you spun the wheels fast.
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No.
The reason being Amarnath S wrote: match the speed of the wheels
the result of which is that, no matter how hard the plane pushes forward, the conveyor belt pushes the same wheels, with a nonzero mass, back exactly as fast as the wheels are moving forward relative to the conveyor belt, a process which is not limited by anything (say, the wheels will spin at an incredible rate, and they it were a real-world situation, the wheels would explode from centrifugal forces); so that the total speed of the plane relative to the ground is zero.
So the plane's thrust is used completely to drive the wheels' reaction force.
It cannot be anything different, or else the exactly matching speed in the opposite direction would no longer hold true.
So the wings do not catch any wind and the plane stays on the ground. All the force used by the plane for its attempt to take off is put into the wheels spinning.
And this conveyor belt, having to make those same crazy speeds in the other direction, would be a mighty impressive piece of work.
This is why it is important that you carefully read the question, be precise in what it states and what it does not state, not read carelessly, and not jump to conclusions.
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Martijn Smitshoek wrote: No.
unless
it is a plane designed to take of completely vertically.
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Martijn Smitshoek wrote: This is why it is important that you carefully read the question, be precise in what it states and what it does not state, not read carelessly, and not jump to conclusions.
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Martijn Smitshoek wrote: So the plane's thrust is used completely to drive the wheels' reaction force.
Do you happen to have a model (toy) airplane with free running wheels? If you also have an option to mount the the spare tire of you car so it can spin around, it can serve as a model conveyor belt.
Now start the tire (/conveyor belt) spinning, with the model plane on top of it, let your hand serve as model engines. You claim that there is no way that your hand can give the plane a thrust forward to throw it into the air. Even though the wheels are free running, in some magical way, they will convey a counter force against your hand making it impossible for the hand (/the engines of the plane) to push the plane forwards at a speed enough to give the plane a lift.
Obviously you can push the model plane forwards, even with the tire spinning ahead beneath it. You claim that if your hand is replaced with real engines, providing same thrusting power as your hand did, that thrust will be unable to move the plane ahead the way your hand did.
I do not see what makes the principal difference between the thrust from your hand on the plane body and the thrust from the engines on the plane body. You maintain that there is a principal difference, or alternately, that as long as those free running wheels touch the rotating tire conveyor belt, your hand can't possibly move the model plane forward.
I'd certainly like to know which one of these two alternatives you go for, along with a good justificaytion.
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Then you did not fully understand this part: designed to exactly match the speed of the wheels This is an extreme requirement and by no means I will claim that it is realistic. However, this is how it was was written down.
The moment that you try to push the model plane, the tire will accelerate with, possibly, hundreds, thousands of Gs, or even more until it satisfies the requirement that the tire's speed equates that of the plane's wheels. If your applied force creates a speed vector v to the wheels, the tire will make a speed of exactly -v to the wheels, which can only result in a standstill.
The result of the equation could be that while you are (maybe gently) pushing the plane, the tire might already be going at Mach 1 the other way, just so that the acceleration (and friction) on the tiny wheel becomes enough to stop that little plane from moving.
And, to get back to your question regarding the difference between hand force and engine thrust, there is no difference, both will not work.
Had the question been that the belt moved as fast as the plane's body, it would have been a different story and the plane would probably be able to take off the conventional way.
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I certainly understood the requirement, and also the lack of realism, but argued based on that premise.
Martijn Smitshoek wrote: he tire will accelerate with, possibly, hundreds, thousands of Gs, or even more until it satisfies the requirement that the tire's speed equates that of the plane's wheels
Why is that? "hundreds, thousands of Gs"?
I do not know the typical acceleration of a plane at take off, but doubt that it exceeds 1G. Whatever the value, it is the acceleration of the wheels during a normal take off. To follow up this, the moving conveyor belt would have to accelerate correspondingly. If you are capable of building a conveyor belt running at (backwards) take off speed under the plane, I am quite sure you can handle that quite moderate acceleration as well.
It also depends on how you read the initial premise: The conveyor belt should match he the speed of the wheels, but at what time? If the belt moves backwards at take off speed when the plane is standing still, and maintains this speed, the relative speed of the wheels to the belt would be twice the take off speed at actual take off. But if the belt's backwards take off speed relative to the wheels should be kept steady, the belt must be braked down when the turbines start working, down to standstill at the moment the plane lift, and is at takeoff speed.
As long as there is physical contact (friction) between the belt / runway and the wheels, you may assume that the wheel thread moves at speed zero relative to the belt / runway - any skidding is (or should be) at the microscopic level).
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trønderen wrote: Why is that? "hundreds, thousands of Gs"?
Because action equals reaction, so that if the plane attempts to accelerate with a typical force to move, say, 40 tonnes, at 100~200 km/h takeoff speed, the reaction force must be generated in its wheels because in these mechanics, the wheels will be free running so the only thing to stop the plane from moving is their rotational inertia. Now, considering that the wheels combined are much lighter than the rest of the plane, they will have
to spin much faster to counteract the engine's thrust and keep the speed equation intact.
trønderen wrote: match he the speed of the wheels, but at what time?
Doesn't say, therefore, at all times.
trønderen wrote: If the belt moves backwards at take off speed when the plane is standing still,
That is based on the way that you put an arbitrary restriction to the rules of the game, so you're trying to answer to a different question than what is being asked. I am not going there.
trønderen wrote: any skidding is (or should be) at the microscopic level).
I agree in that that effect is negligible.
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Martijn Smitshoek wrote: if the plane attempts to accelerate with a typical force to move, say, 40 tonnes, at 100~200 km/h takeoff speed, the reaction force must be generated in its wheels because in these mechanics, the wheels will be free running so the only thing to stop the plane from moving is their rotational inertia.
The engines transfer no energy at all to the wheels to accelerate the plane. The wheels just roll along, with as little friction as possible. The turbines / propellers are what pushes the plane ahead, and work directly on the plane body, with no mechanical coupling to the ground / belt. Action / reaction considerations relate to the plane with its turbines or propellers and the surrounding air, not at all to the wheels / belt.
Sure, if the conveyor belt has managed to accelerate the plane backwards to take off speed, when the turbines are fired up, they must first stop that movement. It takes as much energy to decelerate 40 tons from 100 km/h to zero as it takes to accelerate 40 tons from zero to 100 km/h. But when that is achieved, the turbines continue working, pushing the plane up to (positive) take off speed, relative to the ground and air - but twice the take off speed relative to the belt. That doesn't matter, as the only connection between the belt and the plane is through the free running wheels that just roll along at twice the speed they are used to. Maybe the will be getting hotter than normal, but they will not keep the plane from rising into the air.
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I've been trying to explain this like 3, 4, 5 times and I am not willing to do it again. I have been polite enough and made it perfectly clear that you are frivolously skipping parts of the question just so that you can make up your own story, and I am done with that. I'm sorry.
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Wrong 
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Could be my view is too simplistic but I imagine a plane sort of like a submarine, needing to be buoyant in much lighter fluids (air). For this to happen you need quite a volume of air flowing under the wings as opposed to over it, giving it 'lift'.
Having no significant flow of air while basically staying in one place, I think, is not gonna help. Maybe you'd need a vertical vortex?
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JP Reyes wrote: Having no significant flow of air while basically staying in one place,
Are you saying that with the engines under the wings running at full power, those free-running will have the power to hold back the plane with the same force (but in the backwards direction) as the thrust from the engines in the forwards direction, to make the plane stand still?
I guess that would cause so much stress on those wheels that the would break apart. With no wheels on that conveyer belt, the plane would be free to fly away 
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Well if I understood correctly, the conveyer belt is meant to match the speed of the wheels even at full engine thrust. I Don't know if the wheels have a speed threshold with all that weight, one would imagine the rubber does have it's limits (heck I even bet the conveyer belt would buckle long before the jet engines go to full power)
Realistically I can only imagine the most catastrophic take off (I think the wheels would be useless for landing and the huge conveyer belt tarmac, broken and in tatters). Nonetheless I would have to agree with you.
But referring the original (very hypothetical) question:
Quote: If an airplane is positioned on a conveyor belt as wide as a runway, and this conveyor belt is designed to exactly match the speed of the wheels, but moving in the opposite direction, ...
Can the airplane take off?
Barring the unlikely existence of such a conveyer belt and matching powerful set of wheels, I would still say it doesn't take off. Unless convinced otherwise, it's the volume of the air flowing under the wings that matter, not the volume of air flowing through the turbines.
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JP Reyes wrote: Well if I understood correctly, the conveyer belt is meant to match the speed of the wheels even at full engine thrust.
Sure, but that doesn't null the thrust.
If the conveyor belt is running at takeoff speed before the engines are started, then you fire up the engines and zip down the runway (/conveyor belt), when the plane lifts off the ground the wheels is spinning at twice the takeoff speed (unless the conveyor belt has been slowed down as the plane accelerates, to maintain the 'wheels spinning at takeoff speed).
I am not into construction of air planes, but I wouldn't be surprised if twice the normal takeoff speed is well within the safety margins for the wheels. In any case, it doesn't affect the principal question of whether the plane could take off.
Unless convinced otherwise, it's the volume of the air flowing under the wings that matter, not the volume of air flowing through the turbines. The air flowing through the turbines would give the plane a forward speed that would cause an airflow over and under the wings.
Or are you suggesting that the thrust from the turbines are nulled out because the free running wheels are spinning around? What are the mechanism behind this canceling? Assume that the wheels for some reason started spinning mid-air, would this cancel the thrust from the turbines as well, so the plane crashes? Or does it require the wheels to be in contact with the ground for the thrust to be nulled out?
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