Randazzo

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

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1 hour ago, razark said:

What if we replace the wheels with treadmills?

1200px-B-36_tracked_gear_edit.jpg

In all seriousness, the higher rolling friction of treads and velocity-dependent drag might make this more like the "upstream seaplane" than the "jet on a treadmill".

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So, how many more pages will it take before we can agree on how wheels, propellers/jet engines, and treadmills work? :)

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7 hours ago, razark said:

What if we replace the wheels with treadmills?

I believe the plane would perform, what we call in the business, a 'face plant'.

 

Edited by Cunjo Carl
Got wrong quote! :D Fixed.

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5 minutes ago, 55delta said:

So, how many more pages will it take before we can agree on how wheels, propellers/jet engines, and treadmills work? :)

2,113.3

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What if the conveyor belt is massive enough and moving quickly enough that frame dragging becomes significant? :sticktongue:

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4 hours ago, 55delta said:

So, how many more pages will it take before we can agree on how wheels, propellers/jet engines, and treadmills work? :)

A number greater than 2+5i.

(N.B. I did not say "greater magnitude".)

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On 5/17/2018 at 4:05 PM, YNM said:

Bloody mate, how can this thread gets to page 13 without resolving ? Is this truly greater than any conspiracy ?

Because everybody thinks they understand what's going on, while not being able to write down the equations of constraints or equations of motion relevant to the problem. Once you define these in mathematical terms, this is a trivial problem.

But very few people can do either, so we end up with people who still think that there is a difference between a car and a plane in this problem. What's worse, most think that their fallacy is backed up by physics.

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35 minutes ago, K^2 said:

so we end up with people who still think that there is a difference between a car and a plane in this problem.

Well, isn't it ?

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The only thing that determines if a plane will take off is AIRSPEED, something that is not determined by whether it is a regular surface or a treadmill, or whether the wheels are powered or not.

 

Lift is the only determining factor that dictates whether an aircraft will take off or not.

 

I really fail to see how this thread has gone on for so long as i felt the need to intervene.

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3 hours ago, Hyperant said:

I really fail to see how this thread has gone on for so long...

Because many of us see the pointlessness of the question and are just having fun, others like to answer questions, and even others just like to be right and can't resist poking their noses into an argument, which the internet makes so very easy.

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14 hours ago, YNM said:

Well, isn't it ?

Of course not. The question, as originally stated, is defined fully by constraints. It doesn't have anything to do with means of propulsion.

There are two constraints. First, is the rolling constraint of whatever vehicle.

vv = rω + vb

Where vv is forward velocity of the vehicle with respect to ground, vb is the same for surface of the belt, and ω is angular velocity of the wheels, r being their radius. For simplicity, lets just deal with wheels of equal size.

Second constraint comes from the problem. "Belt matches speed of the vehicle." There are several ways to interpret that.

1) Surface of the belt moves with respect to ground in reverse at the same speed as the vehicle moves forward:

vb = -vv

In this case, the vehicle is free to move forward. We can solve for ω = (vv - vb) / r = 2vv / r. In other words, if the car is moving forward at 20mph, the belt is moving in reverse at the same speed, and the car's speedometer shows 40mph. The car has no trouble just driving off the belt in this case. But in the original thread, where this whole story starts, it is said that "in a way that would keep car stationary." Which leads to a second popular interpretation of this constraint.

2) Surface of the belt matches speed of rotation of the wheel. That is, for a car, matching speedometer speed.

vb = -rω

Again, it is easy to solve for the missing variable. vv = rω - rω = 0. The vehicle remains stationary. And it doesn't matter at all what kind of propulsion we're dealing with. The very statement of the problem demands that the vehicle is stationary. If the vehicle moves, we have not satisfied the conditions of the problem. Is it possible to move the belt in such a way? Well, I'll get back to that in a moment. But first, there is one final interpretation, which I wouldn't even have thought of, but it's what Mythbusters have gone with, so it bears mentioning.

3) Surface of the belt matches some constant reference speed.

vb = v0

Mythbusters considered two cases. Electric toy car and airplane. For electric car, v0 = rωmax of electric motor, which lead to the condition vv = rωmax - vb = 0 preventing the electric car from going up the belt. This does not generally work the same with internal combustion cars, by the way, but they never went full scale with this. And for an airplane, they chose v0 = vtakeoff and so simply had airplane wheels spinning at twice the rate they were meant for. That can cause wheels to fail, but not on a tiny plane they used in the show, so naturally, they had no problems seeing the plane fly. What they were trying to demonstrate with all of this, I have no idea.

 

Ok, so the first interpretation is straight forward enough. Neither a car nor a plane have any trouble making it off the treadmill if it simply matches the vehicle's forward speed with respect to ground. So is the third. No matter how fast the belt moves, that either exceeds some mechanical limitation of the vehicle, or it doesn't. Plane has advantage, as its propulsion system is decoupled, so the only way to prevent the plane from taking off is by moving the belt fast enough to shred the wheels. But what about second interpretation? Can the belt be moving in such a way as to match airplane's wheel speed always, thereby, preventing it from taking off? This is where we take our equations of constraints and add them to equations of motion. I'm not going through conventional derivation from Lagrangian, as that is needlessly mathy, but go with Newton's laws instead. First, we need to reframe constraints in terms of acceleration. (By taking derivatives on both sides.)

av = rα + ab

ab = -rα

Moreover, from Newton's first.

mvav = ΣF= T - Fb

Here, T is thrust and Fb is the force with which the belt pulls on the wheel. The wheel is free to spin, and if it was massless, that force would be zero. But the wheel does have some rotational inertia. So we have the following.

Iα = rFb 

Here, I is combined moment of inertia of the wheels. And putting it all together, we can now solve this mess.

Fb = T - mvav = T - (rα + ab) = T - (rα - rα) = T

ab = -rα = Fbr2 / I = Tr2 / I

So since T, r, and I are finite numbers, there is definitely the right amount of acceleration with which the surface of the belt can be moving, such that due to "drag" generated by the pull of the belt as it tries to spin up the wheels of the plane, the thrust completely cancels out, keeping the airplane put. For a real airplane, that works out to something in the broad neighborhood of 100g. There are two things I can say about that number. First, it's doable. You could construct a treadmill that will provide you with ~100g of acceleration at a sustained force matching thrust of the airplane. Second, this situation can't last for very long. Putting aside difficulties of maintaining these conditions on the treadmill itself, within about a second, almost any airplane's landing gear will disintegrate due to combination of heat and centrifugal forces.

On the net, we have a problem that's fully resolved in constraints before we even have to talk about physics. If constraints are matched, the behavior does not depend at all on what kind of vehicle is placed on the belt. The only place where the choice of vehicle matters is in how spectacular the failure will be when the constraints can no longer be physically maintained. For a car, if the driver's gunning it, the RPM limiter's going to kick in pretty quickly, and probably prevent anything bad happening. For an airplane, most likely a big fireball.

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1 hour ago, K^2 said:

Of course not. The question, as originally stated, is defined fully by constraints. It doesn't have anything to do with means of propulsion.

Even with this ?

Bloodhound_SSC_(19).jpg

The bare aluminum wheels might well be skis for all it's matter.

My puzzlement is "why would changing the skis to be a wheel affect anything".

 

EDIT : Also, AFAIK threadmills work by centering the object. It can't detect the rolling speed of anything other than the belt itself.

Edited by YNM

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5 hours ago, YNM said:

My puzzlement is "why would changing the skis to be a wheel affect anything".

EDIT : Also, AFAIK threadmills work by centering the object. It can't detect the rolling speed of anything other than the belt itself.

The answer to both of these is that there is a rolling constraint between the wheel and the surface on which it is rolling. Whether the axle to which wheel is attached moved forward a meter, or the surface was dragged the opposite direction underneath, the wheel will have to turn a certain amount. And because the wheel has mass, making it rotate faster requires an input of energy. That energy has to come from somewhere, and that means there is a sort of "drag" between surface and vehicle that depends on relative acceleration between the two. And that's what letting us apply a force to the vehicle via the belt. That force simply has to counter whatever propulsion system is in place, whether it's propeller or torque delivered directly to wheels from a car engine.

I've outlined the relevant equations in the post above. The formal way of deriving this is via Lagrangian.

L = mvvv2 / 2 + I ω2 / 2 - T xv - Fb xb + λ1(rθ + xb - xv) + λ2(xb + rθ)

Euler-Lagrange equation (∂L/∂xi - d/dt(∂L/∂vi)) and differentiated constraints give equations of motion.

mvav = T - λ1 - λ2

α = (λ1 + λ2)r

Fb = λ1 + λ2

av = rα + ab

ab = -rα

The lambdas here are the undetermined multipliers that take care of the constraint physics. And once you used the 3rd equation to substitute for sum of lambdas in first and second, you'll see that this becomes the same equations I've marked above.

 

In terms of the control system for the conveyor belt / treadmill, it doesn't really matter. If you hook it up to a sensor that's trying to keep a vehicle centered, if any force is applied to the vehicle, this will result in the belt accelerating. This is equivalent to keeping track of the wheel velocity, because the rolling constraint guarantees us that any relative velocity between vehicle and belt is the rolling velocity of the wheel.

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6 hours ago, YNM said:

The bare aluminum wheels might well be skis for all it's matter.

They are designed to work more like ice skates than skis.  You can barely make it out, but you can see the middle of the wheel is raised above the edges.

They did an episode of startalk a couple years ago where the pilot/driver of this thing talks through the whole procedure of driving it.   It's a 5 minute monologue that is absolutely delightful to listen to.  I think the episode is behind a paywall now though  :(.  

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51 minutes ago, K^2 said:

The answer to both of these is that there is a rolling constraint between the wheel and the surface on which it is rolling.

Interesting.

Does this mean that perhaps it's sometimes better just to use skis rather than wheels ?

 

EDIT : Also, is there a difference between powered wheels and unpowered wheels ? True that by inertia there's some force that needs to be transferred, but unpowered wheels only transfers direct force through friction between the races, while a direct transmission (powered wheels) goes to connect with the prime mover itself.

Edited by YNM

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1 hour ago, YNM said:

Interesting.

Does this mean that perhaps it's sometimes better just to use skis rather than wheels ?

 

EDIT : Also, is there a difference between powered wheels and unpowered wheels ? True that by inertia there's some force that needs to be transferred, but unpowered wheels only transfers direct force through friction between the races, while a direct transmission (powered wheels) goes to connect with the prime mover itself.

There is an fundamental difference between powered wheels on an car and unpowered on an plane. On an car you have to accelerate the drive train not only the wheels and you will reach maximum speed as your rpm even on highest gear get redlined, you will have friction in engine and drive train in addition to the rolling friction. 
This will happen much faster than an normal acceleration as you don't have to accelerate the heavy car just the wheels and drive train. 
The treadmill has to speed up its not insignificant mass of rollers and the belt as fast as the car can accelerate it wheels and drive train up to speeds above 300 km/h. 
Doable but hard you need to reinforce the belt who increases mass and you have to keep inverting its direction on the end rollers.  

On an plane you have an constant trust, the only thing restricting you is the rolling resistance of wheels and friction in bearings, in short the treadmill has to fight jet engines 
To be real evil put an high performance fighter jet on the treadmill, its capable of an 1g acceleration, have fun trying to combat this rotating the wheels, yes all of the 1g acceleration of an 20 ton plane now goes to accelerating the wheels. 
You now have an engineering challenge :)  

 

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4f6251871c224.image.jpg

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)

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19 hours ago, Hyperant said:

The only thing that determines if a plane will take off is AIRSPEED, something that is not determined by whether it is a regular surface or a treadmill, or whether the wheels are powered or not.

Whether the wheels are powered or not is important. If the wheels are geared to something that is RPM-limited, then the plane will not be able to take off.

Of course, no plane actually has wheels geared to anything...or, if they do, they are in neutral during the takeoff roll.

10 hours ago, K^2 said:

In the original thread, where this whole story starts, it is said that "in a way that would keep car stationary." Which leads to a second popular interpretation of this constraint.

2) Surface of the belt matches speed of rotation of the wheel. That is, for a car, matching speedometer speed.

vb = -rω

Again, it is easy to solve for the missing variable. vv = rω - rω = 0. The vehicle remains stationary. And it doesn't matter at all what kind of propulsion we're dealing with. The very statement of the problem demands that the vehicle is stationary. If the vehicle moves, we have not satisfied the conditions of the problem. Is it possible to move the belt in such a way? Well, I'll get back to that in a moment.

In the earliest formulation I've found, the phrasing is, "designed to match the speed". So even if you're interpreting the speed as the rotation of the wheels, as above, you are left with a treadmill which will attempt to keep the vehicle (plane or car) stationary, but its ability to do so is not a given.

10 hours ago, K^2 said:

Since T, r, and I are finite numbers, there is definitely the right amount of acceleration with which the surface of the belt can be moving, such that due to "drag" generated by the pull of the belt as it tries to spin up the wheels of the plane, the thrust completely cancels out, keeping the airplane put. For a real airplane, that works out to something in the broad neighborhood of 100g. There are two things I can say about that number. First, it's doable. You could construct a treadmill that will provide you with ~100g of acceleration at a sustained force matching thrust of the airplane. 

I would question this.

In order to match the interpretation above, the belt must be programmed thus: 

FOR GetSpeed(Wheels);
IF (GetSpeed(Wheels) > GetSpeed(Belt)), IncreaseSpeed(Belt);
ELSE IF (GetSpeed(Wheels) < GetSpeed(Belt)), DecreaseSpeed(Belt);
RETURN;

We can suppose that the belt runs this subroutine an arbitrary number of times each second.

Now, with the above programming, the belt cannot turn at all until it receives a signal that the plane has begun moving forward. Once the plane has begun moving forward, the belt will immediately detect GetSpeed(Wheels) > GetSpeed(Belt) and will run IncreaseSpeed(Belt); as a result. However, this will cause the result of GetSpeed(Wheels) to increase in response, because the wheels are held to an axle which is not force-coupled to the treadmill; increasing belt speed directly increases wheel speed. So the only way to satisfy the programming is for the speed of the belt to increase to infinity, instantly.

It's like running IF (GetSpeed(Belt) = GetSpeed(Belt)), IncreaseSpeed(Belt).

 

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Aircraft tires are not generally designed for perfect friction at high speeds, so even if you had a magical treadmill that could somehow increase rolling-friction by increasing speed, the aircraft tires would still have some slip, especially at such high speeds as the treadmill would need to be spinning at(with the air being drug along by the surface of the treadmill adding an additional cushion, and if the treadmill is smooth enough not to drag air, then it is smooth enough for the tire to drag more easily).

Any dragging that the tire does on the treadmill equals forward movement of the aircraft, so even if the rotational speed of the surface of the tires matches the linear speed of the tread-mill, it still could not prevent the aircraft from moving forward due to friction between the wheel and the treadmill not being perfect.

Without the coefficients of friction and the actual speed of the wheel/treadmill being provided, we cannot say if this slippage is sufficient for the aircraft to take off, but with enough friction between the tires and the treadmill, the tires will eventually melt/burn, causing the treadmill to stop(when the entire tire is stuck to the treadmill, the treadmill must come to a stop to prevent the melted tire material sticking to the treadmill from moving), at which point the aircraft can take off.

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4 hours ago, sevenperforce said:

So the only way to satisfy the programming is for the speed of the belt to increase to infinity, instantly.

Demagoguery. I've given equations of motion for this setup. They clearly show finite acceleration. If you are going to argue with that, you need to point out an error in equations or do your own derivation.

Your reasoning is on par with Xeno's paradoxes. I hope, I don't have to explain why these don't work.

P.S. Earliest mention of the question is on a Russian forum. Some of the translations have been iffy. But that's not really relevant, so long as we agree on which constraints we are considering. If belt is matching wheel speed, or alternatively, tries to keep vehicle in place, acceleration of the belt is finite, albiet, very high.

Edited by K^2

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14 hours ago, YNM said:

"why would changing the skis to be a wheel affect anything".

It wouldn't.  Either way, it doesn't take off.

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2 hours ago, Terwin said:

Any dragging that the tire does on the treadmill equals forward movement of the aircraft

To build up takeoff speed in this case, would be equivalent to taking off with brakes set. If, somehow, you manage to gain speed, your tires will burst long before you can manage it. More realistically, since traction far exceeds thrust, and any slipping is due to deformation,  the forward motion you are talking about will not even reach taxi speeds.

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33 minutes ago, K^2 said:

Demagoguery. I've given equations of motion for this setup. They clearly show finite acceleration. If you are going to argue with that, you need to point out an error in equations or do your own derivation.

Your reasoning is on par with Xeno's paradoxes. I hope, I don't have to explain why these don't work.

There's no problem with your maths; the maths are fine. I'm saying that the case you've set up doesn't match any as-yet-defined formulation of the problem.

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.

think that you could arrive at the formulation you're describing by using the following programming:

IF (AirplanePosition > 0) BeltSpeed++;
ELSE IF (AirplanePosition < 0) BeltSpeed--;
ELSE Return;

...and repeat.

 

 

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What would happen to the plane if P=NP?

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On 5/18/2018 at 2:27 PM, 55delta said:

So, how many more pages will it take before we can agree on how wheels, propellers/jet engines, and treadmills work? :)

This thread was actually dead in 2015 on page 6. I remember posting on it:

On 3/24/2015 at 9:47 PM, Mad Rocket Scientist said:

Before I started playing KSP I would have been stumped by this problem; however now it's obvious: you need a F-35 to take off from a conveyer belt!

 

On 5/14/2018 at 6:23 PM, Snark said:

No.  The plane is stationary with respect to the ground.

  • Case A:  It's sitting on a conveyor belt, stationary with respect to the ground and the air (until the plane's engines move it forward).
  • Case B:  It's sitting on the ground, stationary with respect to the ground and the air (until the plane's engines move it forward).

At no point does the plane move backwards.  No tailwind.

[...]

This brings up an interesting analogy:

If a plane is sitting on a infinite conveyor belt, which is already moving backwards at a fixed, constant speed, and the plane is moving along with it, backwards relative to the ground and the air, it is indistinguishable from a tailwind. I believe this can help to narrow down the exact source of the confusion, as everyone can agree that most planes can take off in reasonable tailwinds.

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.

Personally I think it depends on how plausible the conveyor, wheels, and plane are.

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