# So, you have a plane on a conveyor belt...

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6 hours ago, sevenperforce said:

What I'm taking issue with is the feedback loop in the conveyor belt's programming. You have to define what the conveyor belt is attempting to do in order to figure out what is going to happen.

A PID loop on position of the plane OR wheel speed will result in constraints I've given within reasonable error.

You need to keep in mind that what you really have direct control over is torque applied to the belt's driving axles, not the speed. So this is inherently a second order differential, which is exactly what PID is designed to deal with. And any errors that build up in PID loop decay exponentially, so long as parameters are correctly tuned.

1 hour ago, Mad Rocket Scientist said:

It seems like part of it is confusion over acceleration and speed, and how the belt exerts backwards force on the plane both as a constant, tiny friction due to speed, and a greater backwards impulse that only occurs when the belt accelerates (I think). However, the thrust from the engines is always a constant acceleration, meaning that the belt must continuously and rapidly accelerate if it wants to overcome the acceleration of the engines.

Correct on all points. The caveat I'm trying to point out is that all of this applies to a car on the belt as well. The "thrust" output of car's engine does drop off with wheel speed, and there is an upper limit in theory, but it's way beyond the red line. Since a car is designed to overcome significant air resistance at cruising speeds, which isn't going to be present in the belt setup, it will still take an accelerating belt to keep a car in place if the driver is gunning it, and either the RPM limiter will kick in or something will mechanically fail long before this system reaches any sort of a steady state.

So for both a gas engine car and an airplane, if full throttle is applied, unless the belt constantly and very rapidly accelerates, the vehicle will be able to move forward. And if the belt does accelerate, within seconds mechanical limits of either vehicle will be exceeded.

Electric cars are a more interesting case, as their motor speed is inherently limited by battery voltage, and their top no-load RPM tends to be at much more reasonable speeds. You probably could keep something like a Nisan Leaf on a belt/dynamo with the driver pressing accelerator into the floor without anything breaking.

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11 hours ago, K^2 said:

A PID loop on position of the plane OR wheel speed will result in constraints I've given within reasonable error.

You need to keep in mind that what you really have direct control over is torque applied to the belt's driving axles, not the speed. So this is inherently a second order differential, which is exactly what PID is designed to deal with. And any errors that build up in PID loop decay exponentially, so long as parameters are correctly tuned.

So what is the programming?

1 Get(VehiclePosition);
2 IF (VehiclePosition > 0) BeltTorque++;
3 ELSE IF (VehiclePosition < 0) BeltTorque--;
4 GOTO 1

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25 minutes ago, sevenperforce said:

So what is the programming?

1 Get(VehiclePosition);
2 IF (VehiclePosition > 0) BeltTorque++;
3 ELSE IF (VehiclePosition < 0) BeltTorque--;
4 GOTO 1

Nah, that would lead to loads of over- and undershooting, making the thing incredible unstable until it will just stop working eventually.

You'd use a PID as @K^2 described to adjust the speed of the belt. It would work like this: The P (proportional) measures the distance of the plane to its "target" position, like if it moves half a meter behind it's target, p will equal -0.5. D, derivative, measures the speed at which the plane is moving away from its target, and finally, I, integral, measures the error over time of the plane to its target position. For the last step, you'd multiply P, I and D with unique constants to balance them and then add P, I and D together to get an output value for the belt.

Edited by Kartoffelkuchen

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20 minutes ago, Kartoffelkuchen said:

Nah, that would lead to loads of over- and undershooting, making the thing incredible unstable until it will just stop working eventually.

You'd use a PID as @K^2 described to adjust the speed of the belt. It would work like this: The P (proportional) measures the distance of the plane to its "target" position, like if it moves half a meter behind it's target, p will equal -0.5. D, derivative, measures the speed at which the plane is moving away from its target, and finally, I, integral, measures the error over time of the plane to its target position. For the last step, you'd multiply P, I and D with unique constants to balance them and then add P, I and D together to get an output value for the belt.

Sure.

With a car on a treadmill, it is a battle between the torque in the belt gearing and the torque in the vehicle transmission. The wheels are geared to the vehicle transmission; if the vehicle transmission cannot keep up with the rotation rate of the belt. With a plane on a treadmill, the wheels are decoupled and so it is a battle between wheel inertia and the belt gearing.

If the materials and components in the belt are anywhere close to the materials and components in the wheels, then the belt will fail before the wheels do.

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3 hours ago, sevenperforce said:

If the materials and components in the belt are anywhere close to the materials and components in the wheels, then the belt will fail before the wheels do.

Who says that it needs to be the same materials? The wheels usually fail due to overheating and centrifugal stress on the tires. In this setup, nothing is stopping us from using a kevlar belt running over titanium drums. That stuff can get to mach 1 before there is a problem. And I'd probably use linac drive to just pull on the surface of the belt, since this whole thing just needs a few seconds of operation before the plane is completely destroyed.

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18 minutes ago, K^2 said:

Who says that it needs to be the same materials? The wheels usually fail due to overheating and centrifugal stress on the tires. In this setup, nothing is stopping us from using a kevlar belt running over titanium drums. That stuff can get to mach 1 before there is a problem. And I'd probably use linac drive to just pull on the surface of the belt, since this whole thing just needs a few seconds of operation before the plane is completely destroyed.

In that case we can also optimise the aircraft parts to avoid destruction and make them more like frictionless skids.

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Come to think of it, once the wheels spin up high enough, they're going to have trouble maintaining traction. I bet traction fails before the wheels do, and at a high spin rate they'll have a different coefficient of friction than if they were braked and skidding.

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6 minutes ago, sevenperforce said:

Come to think of it, once the wheels spin up high enough, they're going to have trouble maintaining traction. I bet traction fails before the wheels do, and at a high spin rate they'll have a different coefficient of friction than if they were braked and skidding.

That's called making stuff up. There is absolutely nothing in the physics of a tire that allows for a significant reduction of traction at high speeds, until you start to experience equally significant deformations, which leads to failure under load pretty much instantly. If anything, heating up the tires will slightly improve traction.

The reason aircraft have lower traction at high speeds is due to air speed, which reduces loading factor on the landing gear, and traction is always proportional to load. Stationary plane on a conveyor belt will experience no such reduction in traction.

@Reactordrone Yes, skids are a different story. They experience an almost constant friction regardless of speed, which a plane at full thrust should be able to overcome. They have a whole lot of other problems when used as landing gear for an airplane, but none of them are relevant to this particular question. So if we are talking a plane on skids on a conveyor belt, it's definitely impossible to keep it fixed in place by varying speed of the belt. Still possible to destroy it, though, as heat generated by friction is still proportional to the relative speed of the surfaces.

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53 minutes ago, K^2 said:
1 hour ago, sevenperforce said:

Come to think of it, once the wheels spin up high enough, they're going to have trouble maintaining traction. I bet traction fails before the wheels do, and at a high spin rate they'll have a different coefficient of friction than if they were braked and skidding.

That's called making stuff up. There is absolutely nothing in the physics of a tire that allows for a significant reduction of traction at high speeds, until you start to experience equally significant deformations, which leads to failure under load pretty much instantly. If anything, heating up the tires will slightly improve traction.

If you're driving a car on a treadmill, rolling resistance losses sap from the engine and require a greater power output. Rolling resistance losses for free-rolling wheels, as with an airplane, are going to contribute to slippage, which robs the belt of its ability to slow the airplane.

53 minutes ago, K^2 said:

The reason aircraft have lower traction at high speeds is due to air speed, which reduces loading factor on the landing gear...

Obviously.

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1 hour ago, sevenperforce said:

If you're driving a car on a treadmill, rolling resistance losses sap from the engine and require a greater power output. Rolling resistance losses for free-rolling wheels, as with an airplane, are going to contribute to slippage, which robs the belt of its ability to slow the airplane.

That's not how it works. That's not how any of it works. Rolling resistance in landing gear is low enough that a human being can pull an airliner. Compare that with traction, which is roughly equal to airplane's weight. More importantly, rolling resistance doesn't really increase with speed by any significant amount. It just isn't a factor. We can take wheels to be frictionless, and it will be a fair approximation.

The only reason the belt can apply a significant force on the plane, is because wheels are not massless, and making them turn requires energy. That energy has to come from work done by plane, conveyor belt, or some linear combination (as usual, depending on your choice of coordinate system). Either way, it means that there is a force applied by the belt on the aircraft, and that force is directly proportional to acceleration, as previously derived in this thread.

For a car, you have added rotational inertia of the transmission and the engine. That can be rolled into the moment of inertia I, so that doesn't really change the physics, but does dramatically reduce acceleration of the belt required to keep the car put. In addition, there are viscous forces in the engine associated with air movement through manifolds. This is why a car is capable of engine-braking. As well as reduction in engine efficiency at high RPMs. This further makes it easier for the belt to keep the car in place at higher speeds, but the car on the belt easily exceeds maximum RPM for both the engine and the tires.

So we are still back to the situation where there is no qualitative difference between vehicles placed on the belt. The critical factor remains the amount of mass that has to be turning for the vehicle to roll forward. It's pretty small, but not insignificant, for an airplane, and much higher for a car. And that's that.

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Gut feeling, based on my experiences with actual aircraft ground handling: You'll be able to take off, albeit it'll take a bit longer to break ground due to the need to overcome the slightly increased drag on the wheels and the fact that the aircraft will probably have to stop its rearward motion first. But, if you've ever been around any sort of airplane, you'll know that it doesn't take much throttle to keep you rolling once you actually get going, if any. Breakaway thrust, though, is another matter altogether, but that's brief.

Next time you go near an airport, watch and listen closely to taxiing aircraft. You'll hear them throttle up some to get rolling, but then almost immediately throttle back down to near idle to taxi.

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On 5/21/2018 at 9:58 AM, mrfox said:

The effect would have been even better had they had some sort of speed indication for the turntable, plane and car wheels, and forward ground/airspeeds for the vehicles, but nevertheless a fun museum exhibit (photo lifted from newspaper article)

Hey, I remember seeing this exhibit at a museum before.

Yeah, it basically proves you can take off from a treadmill.   When the engine is off, the treadmill is going pretty fast, and the plane is going backwards.  Then you start the engine, the planes gathers forward speed and eventually takes off.

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11 hours ago, K^2 said:

Rolling resistance in landing gear is low enough that a human being can pull an airliner.

At low speeds, yes.

11 hours ago, K^2 said:

Compare that with traction, which is roughly equal to airplane's weight. More importantly, rolling resistance doesn't really increase with speed by any significant amount. It just isn't a factor.

But that simply isn't true. Rolling resistance includes hysteresis, wheel bearing resistance, vibration/oscillation losses, and wheel slippage. Of these, only wheel bearing resistance is not a function of RPM. Wheel slippage increases as a second-order effect of velocity due to the increase in hysteresis and vibration/oscillation.

The faster the wheels (and the belt) are spinning, the more total rolling resistance you will have.

11 hours ago, K^2 said:

For a car, you have added rotational inertia of the transmission and the engine. That can be rolled into the moment of inertia I, so that doesn't really change the physics, but does dramatically reduce acceleration of the belt required to keep the car put. In addition, there are viscous forces in the engine associated with air movement through manifolds. This is why a car is capable of engine-braking. As well as reduction in engine efficiency at high RPMs. This further makes it easier for the belt to keep the car in place at higher speeds, but the car on the belt easily exceeds maximum RPM for both the engine and the tires.

So we are still back to the situation where there is no qualitative difference between vehicles placed on the belt. The critical factor remains the amount of mass that has to be turning for the vehicle to roll forward. It's pretty small, but not insignificant, for an airplane, and much higher for a car. And that's that.

But there is a qualitative difference between the vehicles placed on the belt.

If you have a car on a belt, then the car's engine is using the transmission and wheels to exert a force on the belt in order to push the car forward against the rearward travel of the belt. Traction losses and rolling resistance reduce the engine's ability to fight the belt.

If you have a plane on a belt, then the belt is using the plane's wheel inertia to exert a force on the plane in order to push the plane backward against the forward travel produced by its engines. Traction losses and rolling resistance reduce the belt's ability to fight the engines.

That is the qualitative difference.

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I hate this question so much. Not the OP, nor the people having intelligent discussion. But I hate the question and questions like it.

The conveyor is merely a source of resistance to movement, and since its not a physical object, it can have properties such as arbitrary speed without any limitations.

The fact that the question usually says something like it "matching" the "speed" of the wheels is moot, effectively its a source of friction that can be scaled to any amount.

If the conveyor is moving a shade under the speed of light backwards, no, no plane will be able to successfully take off.

If the conveyor is moving backwards at only a few cm/s than off course the plane can take off.

Obviously there is a turning point at which the speed of the conveyor completely precludes its use as a runway.

Whether or not the conveyor running at "the speed of the planes wheels" is before or after this limit is just a case of knowing the values of the various physical characteristics of the system, like friction coefficients of all the various wheel-and-axle components, the surface of the conveyor etc.

It is possible to construct cases which match the question, and match both versions of the answer "yes it will" and "no it cant". But it sounds like there should be a definite and single answer, hence its possible for the discussion to go on to infinity.

So the next time someone smugly asks this, just challenge them "You KNOW this question is unanswerable without knowing the coefficient of friction in the planes main wheel and nosewheel bearings,...and you havnt even told us what the conveyor is made of and what its power limitations are! Stop asking silly questions and get your facts straight! What is the static and dynamic thrust curve of the engines? What is the breakaway resistance?"

Edited by p1t1o

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There are much cheaper ways to destroy airplanes.

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6 hours ago, sevenperforce said:

If you have a plane on a belt, then the belt is using the plane's wheel inertia to exert a force on the plane in order to push the plane backward against the forward travel produced by its engines. Traction losses and rolling resistance reduce the belt's ability to fight the engines.

All of these factors make it easier to pull the plane back. Think what happens if you apply brakes on a plane that's already sitting on top of a moving treadmill. Does the plane go forward or backward as a result?

You keep making up arguments, but it sounds like you have not taken the time to draw force diagrams and consider refernce frames. You have to be able to do this to analyze dynamics of a system. This problem is good practice.

Accelerating belt can keep in place a plane at full throttle EVEN if wheels are frictionless. Friction makes it a LOT EASIER. If full brakes are applied, treadmill doesn't even need to move.

Under your analysis, applying brakes should be part of takeoff procedure, as it reduces ground's ability to slow down the plane.

If your analysis fails trivially under a change of coordinate system, you made a misstake. Always check.

A car has more sources of drag and higher moment of inertia attached to wheels. That makes it easier for the belt, but that is the definition of quantitative difference.

Finally, while there are always non-constant sources of friction, for landing gear, they just aren't significant enough to consider at speeds it can possibly reach before failing. Otherwise, planes would have a lot of trouble taking off. If rolling resistance increased significantly, as load decreases, wheels would fail to roll during large chunk of takeoff, that means dragging the gear on every takeoff. The wear would make next landing hazardous. So landing gear is designed to have minimal folling resistance in wide range of speeds.

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21 minutes ago, K^2 said:
7 hours ago, sevenperforce said:

If you have a plane on a belt, then the belt is using the plane's wheel inertia to exert a force on the plane in order to push the plane backward against the forward travel produced by its engines. Traction losses and rolling resistance reduce the belt's ability to fight the engines.

All of these factors make it easier to pull the plane back. Think what happens if you apply brakes on a plane that's already sitting on top of a moving treadmill. Does the plane go forward or backward as a result?

You're conflating static friction and kinetic friction. A wheel which is slipping while rolling has less purchase than a wheel which is not slipping while rolling.

Naively, one might imagine that an increase in rolling resistance in the wheel would make it easier for the belt to exert a force on the body of the airplane. However, the sort of rolling resistance which increases with respect to velocity (hysteresis and oscillation/vibration) will produce slippage, which will reduce traction, which will reduce the belt's ability to transfer energy to the wheels via rotation.

21 minutes ago, K^2 said:

You keep making up arguments, but it sounds like you have not taken the time to draw force diagrams and consider refernce frames.

You're the one advancing a reference frame under constant acceleration.

21 minutes ago, K^2 said:

Accelerating belt can keep in place a plane at full throttle EVEN if wheels are frictionless. Friction makes it a LOT EASIER. If full brakes are applied, treadmill doesn't even need to move.

If the wheels are frictionless and both the wheels and the treadmill components have infinite structural integrity, then yes. But if the wheels are frictionless and the wheels and treadmill components have finite structural integrity, then the treadmill components will fail before the wheel components and the plane will take off. Remember that your model requires the belt to be under constant acceleration, so the airplane need only hold its engines at a constant thrust in order to force the belt to accelerate straight up to relativistic speeds.

If the wheels have ordinary friction, then yes, the acceleration rate of the treadmill can initially be lower than with frictionless wheels. However, if the wheels have friction, then second-order effects like vibration will rapidly reduce traction to the point that the treadmill will be unable to act on the wheels.

21 minutes ago, K^2 said:

...while there are always non-constant sources of friction, for landing gear, they just aren't significant enough to consider at speeds it can possibly reach before failing.

But this begs the question.

Your construction requires the belt to accelerate continually while the engines need only maintain static thrust. The engines can maintain full thrust indefinitely. Thus, one of two things will happen:

A) The belt will accelerate until the speeds involved exceed the structural limitations of its components. Assuming comparable structural integrity between the belt and the wheels, the belt will fail first. The plane will then proceed to a normal rolling takeoff.

B) The belt will accelerate until second-order effects induce slippage and reduce traction. The belt will no longer be able to continue accelerating the wheels, and the plane will take off.

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On 5/23/2018 at 4:22 AM, Gargamel said:

The effect would have been even better had they had some sort of speed indication for the turntable, plane and car wheels, and forward ground/airspeeds for the vehicles, but nevertheless a fun museum exhibit (photo lifted from newspaper article)

I'll just bump and quote this again, in case some of you missed the significance of it.

This exhibit puts a plane.... on a treadmill..... and it flies......

It's not a simulation or a video.  It's an actual tethered airplane, powered by a prop engine, on a circular treadmill.

It flies.....

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I guess whether or not the aeroplane will fly depends almost entirely* on the amount of air passing by the wing, resulting (or not) in lift either via the Bernoulli effect (pressure differential) or Newtonian deflection (conservation of momentum)

To achieve the outcome of air moving past the wing you could either:

• move the wing through the stationary air
• move the air around the stationary wing

The first option is precluded by the conditions of the experiment (i.e. a fictional magically-powered conveyor that can instantaneously reach any speed such that friction is provided to prevent the craft from moving forward.)

However the second option is still viable, and can be achieved via several means such as a gust of wind, prop wash, or even an extreme boundary layer of air propelled by the moving surface of the belt as it approaches warp speed!

*Thrust vectoring or a high AOA of might also cause the aeroplane to break free of the conveyor's clutches through sheer brute force!

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1 hour ago, Gargamel said:

The treadmill isn't an ideal treadmill, nor does it match the speed of the wheels.

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15 minutes ago, DAL59 said:

The treadmill isn't an ideal treadmill, nor does it match the speed of the wheels.

Be it a circle or a belt, a treadmill is a treadmill.   The parts not in contact with the plane have no bearing on the plane or the question.

While in contact with the treadmill, the wheels will by default be matching the speed of the treadmill.   At startup in this case, the treadmill is going much faster than the plane is, so it has more of an 'effect' on the plane than the hypothetical one.

This exhibit was part of the mythbusters traveling exhibit that toured around 3-4 years ago, seen at many science museums.

I got to see it, and I flew the plane off the treadmill.

EDIT:  I googled the image trying to find the article it came from, and I found it: http://www.nwitimes.com/niche/shore/shore-newsletter-article/mythbusters-exhibit-comes-to-chicago/article_98ae270c-917f-5d21-999f-4a6d1cae7630.html

But the funny part is, google image search returns it's best guess as air hockey.  When you look at the photo, you can see how it made that guess.

EDIT2:

Found a video of it:

You can see that there is also a slot car on the treadmill, and it is affected by the treadmill.

Edited by Gargamel

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14 hours ago, sevenperforce said:

You're conflating static friction and kinetic friction. A wheel which is slipping while rolling has less purchase than a wheel which is not slipping while rolling.

Naively, one might imagine that an increase in rolling resistance in the wheel would make it easier for the belt to exert a force on the body of the airplane. However, the sort of rolling resistance which increases with respect to velocity (hysteresis and oscillation/vibration) will produce slippage, which will reduce traction, which will reduce the belt's ability to transfer energy to the wheels via rotation.

Au contraire. Landing gear is going to remain firmly in the static friction regime. It's what it's designed to do. A slipping wheel cannot provide lateral forces allowing any sort of control. And while rudder does assist at all in early stages of landing or late stages of takeoff, you rely on wheels for much of steering during runup. If wheels began to even just noticeably slip during takeoff before you are ready to rotate, you would not be able to keep the plane on the runway. This I can tell you from first-hand experience.

The second piece of relevant information is that I'm aware of no plane that has landing gear that will survive more than twice the takeoff speed. In other words, the wheels will be rolling at most twice the speed at which they are designed to operate routinely, before they fail. At these speeds you are not suddenly going to develop slipping, excessive friction, etc. The wheels will continue to operate as wheels, providing static friction with the ground, with minimal resistance to rolling, and almost no slipping, save for that necessary for deformation at contact point.

We have race cars that operate at higher speeds than landing gear failure point. Other than additional reinforcement, their physical properties are not that different. If what you were suggesting was remotely true, these cars would not be able to keep themselves on the race track.

14 hours ago, sevenperforce said:

If the wheels are frictionless and both the wheels and the treadmill components have infinite structural integrity, then yes. But if the wheels are frictionless and the wheels and treadmill components have finite structural integrity, then the treadmill components will fail before the wheel components and the plane will take off. Remember that your model requires the belt to be under constant acceleration, so the airplane need only hold its engines at a constant thrust in order to force the belt to accelerate straight up to relativistic speeds.

The landing gear will fail within seconds. We don't need to be talking about relativistic speeds. We can deal with real materials and real vehicles here. And real airplanes have very real limitations. Their landing gear is not designed for very high speeds, because at these speeds the aircraft is not normally in contact with the ground.

Your statement that belt has to fail first has zero ground. There are just two places where stress is necessary. Centrifugal stress on rollers themselves, which are likely to be solid metal and can far, far outlast landing gear tires, and the belt itself as it bends around at end-points. Belt has numerous advantages. It can be made from better materials than rubber. It can have more reinforcement. It does not experience the maximum stress in the same place as maximum curvature. And it does not experience heating. If that's not enough, I can always increase the diameter of the rollers, reducing curvature at end points, minimizing stress. None of these mitigating factors apply to the landing gear. I can easily design a treadmill that can survive transsonic speeds, but real landing gear fails at much lower speeds. And if I were to try and design better landing gear, I'd still be unable to make one that's going to both perform its primary function sufficiently well and survive at higher speeds as the treadmill.

Heck, even if we use the same metal for landing gear hubs as treadmill rollers, same rubber for main surface on both, and same cables for reinforcements on both, the treadmill is still going to be under far less stress and survive higher speeds. And at this point, we've morphed landing gear into something monstrous, and the treadmill is still just a treadmill.

14 hours ago, sevenperforce said:

Your construction requires the belt to accelerate continually while the engines need only maintain static thrust. The engines can maintain full thrust indefinitely. Thus, one of two things will happen:

A) The belt will accelerate until the speeds involved exceed the structural limitations of its components. Assuming comparable structural integrity between the belt and the wheels, the belt will fail first. The plane will then proceed to a normal rolling takeoff.

Again, completely unfounded. Real landing gear will fail way earlier, and one that's specifically designed to last as long as possible, made from same materials, merely slightly earlier. In both cases, in matter of seconds or less, I might add.

14 hours ago, sevenperforce said:

B) The belt will accelerate until second-order effects induce slippage and reduce traction. The belt will no longer be able to continue accelerating the wheels, and the plane will take off.

And this one's a plain impossibility. We are talking about thrust of the plane times hundreds of meters per second for this scenario, which is enough power delivered to wheels to melt not just the rubber, but the metal hubs of the landing gear. Meanwhile, the belt remains cool, since that's not where energy is released, and there is quite a bit of it, allowing cooling along the entire surface. If somehow you came up with magical materials that allowed landing gear not to fail in scenario A), this is where it fails. You simply cannot have sufficient slippage for takeoff without generating all that heat.

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Quick google search shows max rated belt speeds today to be around 100m/s - around 200 knots - so it seems for high speed airplanes (most large airliners today have rated max tire speeds around 230 knots) both the belt and the wheels would fail around the same time.

The belt would heat up just as much as the tires would due to the mechanical deformation of the material as it bends around the rollers.

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18 minutes ago, mrfox said:

Quick google search shows max rated belt speeds today to be around 100m/s - around 200 knots - so it seems for high speed airplanes (most large airliners today have rated max tire speeds around 230 knots) both the belt and the wheels would fail around the same time.

The belt would heat up just as much as the tires would due to the mechanical deformation of the material as it bends around the rollers.

What on Earth does one use a 200kt belt for?!

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4 hours ago, Gargamel said:

I'll just bump and quote this again, in case some of you missed the significance of it.

This exhibit puts﻿ a plane.... on a treadmill..... and it flies......

It's not a simulation or a video.  It's an actual tethered airplane, powered by a prop engine, on a circular treadmill.

It flies..... ﻿

Sir, I think you miss the point.

This is the internet.  We will not have you derailing a perfectly good science discussion with such nonsense as facts and reality!  Take that silliness elsewhere!

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