B Can a 747 Take Off on a Conveyor Belt?

[RandyD123]

Imagine a 747 sitting on a large conveyor belt, as long and as wide as the runway.
The conveyor best is designed to exactly match the speed of the wheels, but run in the opposite direction.

CAN THE PLANE TAKE OFF?

 

[DrClaude]

If the airplane was on a frictionless surface, do you think it would be able to take off?

 

[phinds]

Imagine a 747 sitting on a large conveyor belt, as long and as wide as the runway.
The conveyor best is designed to exactly match the speed of the wheels, but run in the opposite direction.

CAN THE PLANE TAKE OFF?

This exact question was answered fairly recently. I suggest a forum search.

 

[RandyD123]

This exact question was answered fairly recently. I suggest a forum search.

I tried that first, but unfortunately I didn't find that thread. Sorry

 

[phinds]

I tried that first, but unfortunately I didn't find that thread. Sorry

Hm ... I don't have a link. Maybe someone else will remember it or know what search term to use.

 

[Bandersnatch]

Or... the OP could try and work it out himself instead. Why not start by answering DrClaude's question?(By the way there's an issue with the wording of this version of the problem - the conveyor belt should be matching plane speed w/r to the ground, otherwise the proposition is faulty and results in infinities)

 

[phinds]

Or... the OP could try and work it out himself instead. Why not start by answering DrClaude's question?

Good point

(By the way there's an issue with the wording of this version of the problem - the conveyor belt should be matching plane speed w/r to the ground, otherwise the proposition is faulty and results in infinities)

Another good point. I missed that by dismissing the question too quickly.

 

[russ_watters]

Imagine a 747 sitting on a large conveyor belt, as long and as wide as the runway.
The conveyor best is designed to exactly match the speed of the wheels, but run in the opposite direction.

CAN THE PLANE TAKE OFF?

This "riddle" went around the internet a few years ago and we have a few old threads on it. Mythbusters actually tested it with a Cessna. As you worded it, there is no reason the plane wouldn't be able to take off, but "exactly match the speed of the wheels, but run in the opposite direction" doesn't actually make any sense because the wheels are already rolling when the runway is stationary and the contact patch is always stationary - so as worded, either the conveyor isn't moving or the plane has to be tethered to prevent it from moving while the conveyor rolls at any random speed under it. And that's the only thing difficult about the various incarnations of the "riddle": they are badly or impossibly worded, which triggers arguments about trying to reconcile the wording with reality.

 

[Clausen]

The plane will take off. Wheel speed is not even relevant as long as there is sufficient engine thrust to push the plane forward with respect to the air. The plane doesn't even need wheels, pontoons will do just as well.

 

[RandyD123]

The plane will take off. Wheel speed is not even relevant as long as there is sufficient engine thrust to push the plane forward with respect to the air. The plane doesn't even need wheels, pontoons will do just as well.

Wheels have nothing to do with solving this question...more of less. Think of this, let's use a SEA PLANE...NO WHEELS and let's put that plane in water, near the edge of a waterfall. There is a point of no return for any object in the water. Now let's put power to the plane weather it's jet power or propeller power. The water going over the edge of the waterfall is the "conveyor" belt. The plane, in that water has a point of no return. That point will change based on the THRUST of the engines. But if the THRUST MATCHES the force of the water going over the falls, the PLANE DOES NOT MOVE. Therefore it creates NO LIFT. NO LIFT... NO FLY.

 

[A.T.]

match the speed

MATCHES the force

Even given the already mentioned ambiguity of the original formulation, this is a completely different one.

 

[Clausen]

Wheels have nothing to do with solving this question...more of less. .

Yes, that is what I said. Pontoons will do just as well

Think of this, let's use a SEA PLANE...NO WHEELS and let's put that plane in water, near the edge of a waterfall. There is a point of no return for any object in the water. Now let's put power to the plane weather it's jet power or propeller power. The water going over the edge of the waterfall is the "conveyor" belt. The plane, in that water has a point of no return. That point will change based on the THRUST of the engines.

Yes, and that thrust must be able to push the plane with respect to the air, in order to create lift on the wings.

But if the THRUST MATCHES the force of the water going over the falls, the PLANE DOES NOT MOVE. Therefore it creates NO LIFT. NO LIFT... NO FLY.

The force of the water going over the falls has nothing to do with the plane taking off. The engine thrust only needs to be greater than the force of the water on the pontoons. The engine thrust will most certainly be sufficient to overcome that frictional resistance as that resistance is approximately constant over a large range of velocities.

 

[RandyD123]

Not sure if the "wheels" make the difference. If it's a waterfall or black hole, in both cases there is a point of no return, no matter what the thrust is. If the plane is on the "edge" of the point of no return then the plane can't fly. Why would a conveyor belt mean any less than the other 2?

 

[A.T.]

Not sure if the "wheels" make the difference.

Nothing is sure if the assumptions are not clear. What horizontal force are the wheels assumed to produce? How does this force depend on the relative speed between plane and surface? How is the belt speed defined exactly? That determines if the plane can take off, or not.

 

[RandyD123]

I say the plane can't fly, if it could then conveyors would be on all aircraft carriers.

 

[russ_watters]

Not sure if the "wheels" make the difference. If it's a waterfall or black hole, in both cases there is a point of no return, no matter what the thrust is. If the plane is on the "edge" of the point of no return then the plane can't fly. Why would a conveyor belt mean any less than the other 2?

The conveyor belt applies virtually no force to the plane because the plane is on wheels. If you think otherwise, that is your error.

I say the plane can't fly, if it could then conveyors would be on all aircraft carriers.

That makes no sense at all, however aircraft carriers do have catapaults...and also conduct flight operations while in motion...

It is starting to feel like you are going to keep altering the scenario until you figure out a way to keep the plane from flying. Certainly you can, but why do that except to be argumentative?

 

[boneh3ad]

I would like to contend that there is a difference (in practice) between wheels and pontoons, though it's a difference in magnitude rather than in kind. That difference has to do with the type and amount of "friction" opposing motion.

I think the fundamental issue that @RandyD123 is missing here is that the plane's motion has everything to do with the thrust, which is completely decoupled from the motion of the treadmill underneath. It has little to do with that friction force from wheels since that will be many, many orders of magnitude less than the thrust. The treadmill could tend to drag the plane along with it a little bit, but it won't be much, and, depending on the situation, it's possible the conveyor could simply spin the wheels without moving the plane. So relative to the moving treadmill, the plane could have some motion, but this motion is irrelevant when it comes to lift. The motion relative to the air is what matters, and that is going to depends pretty much entirely on the thrust in this case. In other words, once that engine starts, that plane is going to move forward nearly identically whether it is on a runway or a treadmill. The only difference will be the rotation rate of the wheels, not the speed of the plane relative to the air.

With pontoons it is slightly different because the drag of the water moving over pontoons will be much larger (probably several orders of magnitude) than rolling friction on wheels. In that case, there may be a noticeable difference between a boat with pontoons taking off, say, upstream on a river, compared to a plane with wheels on a treadmill. In that case, for a given engine thrust, there could be some water velocity that would result in the drag exactly matching the thrust and the plane not going anywhere, but the water would have to be moving pretty quickly to do that.

At any rate, here's a video of a plane taking off on a treadmill:

 

[Nugatory]

Why would a conveyor belt mean any less than the other 2?

I say the plane can't fly, if it could then conveyors would be on all aircraft carriers.

What is the net force on the plane? If it is positive the plane will accelerate until it reaches a speed sufficient to take off. If it is negative, the plane will be dragged backwards against the best efforts of its howling engines. If it is zero, the plane will sit there until runs out of fuel.

As this problem is most often presented (it is all over the Internet, although I'd trust the explanations you're getting here more than some of what's out there) the conveyor belt and wheels are frictionless so apply no backwards force; the net force is the engine thrust as on a normal takeoff. A pontoon system applies a significant backwards force acting against the engine thrust so will delay (normal float plane takeoff) or prevent (absurd hypothetical in which the plane takes off against an absurd current) the takeoff. The catapult on an aircraft carrier applies a significant force in the same direction as the engine and assists the takeoff - that's why aircraft carriers use catapults instead of conveyor belts.

So bottom line: yes, the mechanism matters, if different mechanisms apply different forces to the plane. And yes, the plane will take off.

 

[RandyD123]

I would like to contend that there is a difference (in practice) between wheels and pontoons, though it's a difference in magnitude rather than in kind. That difference has to do with the type and amount of "friction" opposing motion.

I think the fundamental issue that @RandyD123 is missing here is that the plane's motion has everything to do with the thrust, which is completely decoupled from the motion of the treadmill underneath. It has little to do with that friction force from wheels since that will be many, many orders of magnitude less than the thrust. The treadmill could tend to drag the plane along with it a little bit, but it won't be much, and, depending on the situation, it's possible the conveyor could simply spin the wheels without moving the plane. So relative to the moving treadmill, the plane could have some motion, but this motion is irrelevant when it comes to lift. The motion relative to the air is what matters, and that is going to depends pretty much entirely on the thrust in this case. In other words, once that engine starts, that plane is going to move forward nearly identically whether it is on a runway or a treadmill. The only difference will be the rotation rate of the wheels, not the speed of the plane relative to the air.

With pontoons it is slightly different because the drag of the water moving over pontoons will be much larger (probably several orders of magnitude) than rolling friction on wheels. In that case, there may be a noticeable difference between a boat with pontoons taking off, say, upstream on a river, compared to a plane with wheels on a treadmill. In that case, for a given engine thrust, there could be some water velocity that would result in the drag exactly matching the thrust and the plane not going anywhere, but the water would have to be moving pretty quickly to do that.

At any rate, here's a video of a plane taking off on a treadmill:

That whole video is bogus. Does not even come close to the physics of the real question.

 

[RandyD123]

The conveyor belt applies virtually no force to the plane because the plane is on wheels. If you think otherwise, that is your error.

That makes no sense at all, however aircraft carriers do have catapaults...and also conduct flight operations while in motion...

It is starting to feel like you are going to keep altering the scenario until you figure out a way to keep the plane from flying. Certainly you can, but why do that except to be argumentative?

Probably only until someone can provide real world physics to this question. Maybe someone already has and I just don't understand it. Can we all at least agree that in a waterfall or black hole scenario, the plane won't EVER fly?

 

[sophiecentaur]

I say the plane can't fly, if it could then conveyors would be on all aircraft carriers.

They use a 'steam catapult' for takeoff and an arrestor wire for landing. What would the conveyor achieve in the limited takeoff length that's available.
But the OP seems so blindingly obvious to me. What have I missed? The only difference with and without the conveyor is that the wheels would rotate at twice the speed. Assuming the bearings are ok and the rolling friction is small enough, what difference would that make to the plane taking off?

 

[boneh3ad]

That whole video is bogus. Does not even come close to the physics of the real question.

Ah yes, putting a real, full-size plane on an real, full-size moving surface going in the opposite direction of its takeoff direction clearly doesn't match the real world. I can't believe I ever thought that it would.

"It doesn't matter how beautiful your guess is; it doesn't matter how smart you are, who made the guess, or what his name is. If it disagrees with experiment, it's wrong." -Richard Feynman

 

[russ_watters]

Probably only until someone can provide real world physics to this question. Maybe someone already has and I just don't understand it.

It is difficult when you keep changing the scenario and don't explain how you think the new scenario works!

Can we all at least agree that in a waterfall or black hole scenario, the plane won't EVER fly?

I don't know about "ever", but given a strong enough current, a pontoon plane might be kept from taking off. I don't know how much is needed.

The black hole thing is too silly to discuss though.

 

[russ_watters]

That whole video is bogus. Does not even come close to the physics of the real question.

What do you think is missing or not "real" about this real demonstration?

 

[A.T.]

I say the plane can't fly...

Doesn't matter what you say, until you clarify the details, because it's not clear what you talk about. Here the questions again:

What horizontal force are the wheels assumed to produce? How does this force depend on the relative speed between plane and surface? How is the belt speed defined exactly?

 

[A.T.]

That whole video is bogus. Does not even come close to the physics of the real question.

Or the question is bogus, since everyone interprets it differently.

 

[lightandmatter]

The plane has to take off doesn't it, the jet provides a thrust of air backwards, hence the plane moves forward relative to the air until it reaches a balance or speed at which the thrust backwards matches the air friction whilst in forward flight, it doesn't matter what speed the wheels are doing relative to the runway on takeoff if we are assuming there is no friction due to a faster or twice than normal rotating wheel. The things that would affect the plane taking off is a head or tailwind as this affects the speed of air over the wing and hence its lift, and even then the planes takes off, it just needs less or more runway. Remember a plane can only fly relative to the air flowing over its wings and not relative to the ground or moving ground/conveyor. For those that think the plane doesn't takeoff then where does the energy go from the take off thrust??

 

[Clausen]

Probably only until someone can provide real world physics to this question.

As long as the engine thrust is able to overcome the rolling resistance of the tires and other drag forces, the plane will move forward and eventually reach takeoff speed.

The question is: does forcing the tires to spin at the same speed as the conveyor belt cause more drag than engine thrust? If so, the plane won’t take off. Taking some typical real-world numbers, a 747 with a mass of 350,000 kg will have a normal force of 3.5 Million N. The four engines may produce a thrust of 1 Million N. If the coefficient of rolling resistance is less than 1/3.5, (and ignoring air resistance) the plane will takeoff. Typically, coefficients of rr are much less than this value but this is not a typical situation. I think the actual coefficient would need to be determined experimentally in this situation because the wheels would be both rolling and slipping.

I still think it will take off but I can see a possibility it will not be able to reach takeoff speed.

Nothing is sure if the assumptions are not clear. What horizontal force are the wheels assumed to produce? How does this force depend on the relative speed between plane and surface? How is the belt speed defined exactly? That determines if the plane can take off, or not.

Right. Nothing is sure here if the assumptions are wrong or unclear.(The bigger question here : is takeoff one word or two?)

 

[sophiecentaur]

I still think it will take off but I can see a possibility it will not be able to reach takeoff speed.

Doesn't it have to be true that the aerodynamic drag forces at takeoff speed must be much higher than any resistive forces from the wheels. If resistive forces were present, the plane would surge forward in an alarming way when the wheels leave the tarmac (even when the rotation of the wheels is only half of what they would be in this scenario). I have never been aware of this and it is not something that's ever mentioned. Indeed, if it were significant, undercarriages would be designed differently.
Also, at touchdown, there would be a similar backwards jerk. Have you ever noticed that?
(PS After takeoff, I take off my dark glasses. )

 

[Clausen]

Doesn't it have to be true that the aerodynamic drag forces at takeoff speed must be much higher than any resistive forces from the wheels. If resistive forces were present, the plane would surge forward in an alarming way when the wheels leave the tarmac (even when the rotation of the wheels is only half of what they would be in this scenario). I have never been aware of this and it is not something that's ever mentioned. Indeed, if it were significant, undercarriages would be designed differently.
Also, at touchdown, there would be a similar backwards jerk. Have you ever noticed that?)

I haven't noticed it but then again I have never taken off or landed on a conveyor belt.

How do you figure the wheels will be turning at twice the normal takeoff rate? Won't they need to slide as well as spin if the plane moves?

 

[russ_watters]

Doesn't it have to be true that the aerodynamic drag forces at takeoff speed must be much higher than any resistive forces from the wheels. If resistive forces were present, the plane would surge forward in an alarming way when the wheels leave the tarmac (even when the rotation of the wheels is only half of what they would be in this scenario).

I've never ridden a pontoon plane, so maybe someone else can comment on that, but with aerodynamic drag increasing on rotation, positive g's being applied, reduced rolling resistance due to reduced weight on the tires and the sudden smoothness of no longer rolling, I don't think such an effect would be noticeable even if moderately significant. Big airliners don't accelerate very fast as it is.

 

[russ_watters]

...forcing the tires to spin at the same speed as the conveyor belt...
[Snip]
How do you figure the wheels will be turning at twice the normal takeoff rate? Won't they need to slide as well as spin if the plane moves?

Cleaning-up the wording sloppiness of the original question, the usual formulation is that the plane rolls forward with respect to the ground and the conveyor slides backwards with respect to the ground at the same speed (this is what Mythbusters attempted to duplicate). So the speed of the wheels on the conveyor is X-(-X)=2X

[Edit] There is a silliness about this scenario in that it implies an inherent relationship between the speed of the plane and conveyor where none need exist. In practice, it would require adjusting the speed of the conveyor to match the speed/acceleration of the plane. Mythbusters just had a car pull the conveyor, without any real regard to speed matching. But the conveyor speed could be made to be anything.

 

[sophiecentaur]

I haven't noticed it but then again I have never taken off or landed on a conveyor belt.

How do you figure the wheels will be turning at twice the normal takeoff rate? Won't they need to slide as well as spin if the plane moves?

Apart from the actual force involved, which would be higher, what would suddenly make it noticeable if you were on a conveyor belt?
The only difference between normal takeoff and the conveyor belt thing is that the wheels would be turning at twice the rate by the time takeoff speed has been reached. If you want them to slip (on an icy conveyor belt, perhaps), then so much the better. What difference could that make except to reduce the resistive force?
Why twice the speed? They would have to be turning at the same rate in order to keep up with the conveyor (remain stationary) and then the same again for taking off speed. The engine power required to stay stationary would be very low (only to overcome the wheel friction.
I think you are looking at this problem in the wrong way. What is to stop the wheels from rotating easily on the conveyor belt?

 

[rexregisanimi]

Guys, this isn't a difficult question but it does test our ability to communicate precise concepts.

Here's the Physics:

We make a free body diagram of the forces on the plane: before the engines or conveyer belt turn on, there is a force pulling it straight down onto the ground (gravity). Another force, the normal force, is the ground pushing up on the plane as a reaction to the plane pushing down in the ground. No other forces exist on the plane so it sit, stationary, until something changes.

So we start the engines. Now there is a third force pushing forward on the plane (the force created by the engines). No force balances it so the plane starts to move forward. However, at that moment, the conveyer belt starts to move in the opposite direction. The only question we need to worry about is this:

Does that conveyer belt create enough force on the plane to balance out the force created by the plane's engines?

To answer this, we need to think of a free body diagram of the wheel.

There is one force, at the center of the wheel, that results from the engines. This force is transferred through the structure if the plane to the hub of the wheel. It pushes the wheel forward from the center. Another force, created by the conveyer belt, pushes on the bottom edge of the wheel and pushes backwards. This causes the wheel to begin to spin. However, and this is the key to understanding this question, that force only causes the wheel to spin.

If you're holding a bike off the ground and you have a friend make one of the wheels spin, the bike doesn't move forward. The wheel spins, however. In fact, you could hold a bike wheel by the middle (holding onto an axle or something) and an outside force could cause it to spin quite rapidly and you would have no problem holding into it because forces that make a wheel spin do not make the wheel move from one location to another.

So, back to the plane. The wheel is being made to spin by the conveyer belt but that force only makes the wheel spin. It does not cause any force to be transferred into the hub of the wheel and, subsequently, into the structure if the plane.

The force of the engine pushes on the plane as a whole but the conveyer belt does not affect the movement of the plane because the wheels isolate the rest of the plane from the conveyer belt.

Since there is no balancing force created against the force from the engines, the plane accelerates forward. Once that acceleration creates a speed sufficient to produce enough lift, the plane takes off.

Sent from my SM-G935T using Physics Forums mobile app

 

[Joseph Riley]

Imagine a 747 sitting on a large conveyor belt, as long and as wide as the runway.
The conveyor best is designed to exactly match the speed of the wheels, but run in the opposite direction.

CAN THE PLANE TAKE OFF?

Mythbusters answered this a long time ago. Short answer yes.

 

[ddjj77]

Let's assume that the plane is sitting on the conveyor with its brakes on, and the conveyor and plane are moving at 60mph opposed to the plane's heading (the plane is going in reverse). Now the pilot decides to take off. The only difference between a normal takeoff and the one just described is the "unusual" takeoff starts at speed -60mph, and will take a bit longer than usual to reach takeoff speed, and then up, up, and away! The higher the conveyor speed, the longer it would take to reach takeoff speed.

 

[sophiecentaur]

I've never ridden a pontoon plane, so maybe someone else can comment on that

That's a very different matter. I have never been on one, either but I do know that the 'stickiness' of the water is a severe problem.
Just pushing a regular seaplane through the water at twice takeoff speed would just not be easy,. The pontoons need to have enough volume and to be long enough for a reasonable design speed (max displacement speed for a reasonable power input). That in itself would require a pontoon length of about √2 times the original. The hulls then need to plane over the water in order to get to takeoff airspeed. I have read (Francis Chichester 'Lonely Sea and Sky' - which dates me) that it can be so marginal that you may need chop on the water to break away and start planing. A 2:1 water speed would require some very special pontoons or a very powerful engine. Plus some pretty clever air control surfaces to stop you diving in , head first. (Even vector thrust !?)
I was complaining, earlier about the over analysis of the "buckets" thread and look what I've just done!

 

[EspressoDan]

Yes, wheel speed is not important. But, a lot of replies here state that the important factor is the the forward speed of the aircraft, or the thrust of the engines.

Fundamentally, the problem has to do with neither, the only important factor is the speed of the airflow over the wings. Engines and runways are only there to allow the aircraft to move fast enough relative to the air mass in which it flies to make velocity of air over the wings increase to the point where it allows the airfoil at a given angle of attack to the free steam flow to generate enough lift to overcome the weight of the aircraft.

 

[The Wizard]

Interesting theories etc being put forward.. the 4 Elements relative to flight of an aeroplane are simple; Thrust and lift have to exceed weight and drag
Thrust being power available from the engine(s).
Lift is the Lifting force available produced by the speed of the air over and below the Wing Surface
Weight is the total weight of the Aeroplane acting through Gravity
Drag is the resistance of the mass on the Surface, in this case the wheels on the belt.

So, in respect of the original question the belt Speed is directly linked to the speed of the wheels. So, assuming the belt and the wheel are motionless at the beginning, as the aircraft generates thrust it will start to move forward and the belt will run in the opposite direction so the aircraft is accelerating due to the increased thrust and consequently the belt is also increasing in speed, relative to the speed of the tyres.

The end result will be that the aircraft (747) will reach a speed of about 160 knots at which point it will lift off., the Point where thrust and lift exceed weight and drag. The speed of the wheels at this point will depend on the size of the tyre and that will then determine the speed of the belt. The tyre / belt speed is different to the actual speed of the aircraft.

By the Airspeed I did not take into account different weights or amount of Flap used, it's just an average figure, used to indicate aircraft speed is different to speed of wheels/belt.

 

[jbriggs444]

So, assuming the belt and the wheel are motionless at the beginning, as the aircraft generates thrust it will start to move forward and the belt will run in the opposite direction so the aircraft is accelerating due to the increased thrust and consequently the belt is also increasing in speed, relative to the speed of the tyres.

The end result will be that the aircraft (747) will reach a speed of about 160 knots at which point it will lift off., the Point where thrust and lift exceed weight and drag. The speed of the wheels at this point will depend on the size of the tyre and that will then determine the speed of the belt. The tyre / belt speed is different to the actual speed of the aircraft.

How do you imagine the problem description requires the conveyor and tires to move?

As I understand the problem, the requirement is that the conveyor speed is equal and opposite to the plane's ground speed (speed relative to the control tower, for instance). So if winds are calm and the plane is moving at 160 knots the conveyor will be moving at -160 knots and the tires will be turning at the equivalent of 320 knots.

Of course as has been repeatedly affirmed, the tire speed and the plane's ground speed are irrelevant to the plane's airspeed, thrust and lift. [With exceptions for float planes, planes with tires that have not been lubed in the last ten years and Air France 4590]

 

[The Wizard]

@ jbriggs444,
Where does the question state the conveyor belt has to match the planes ground speed? The question states

"The conveyor best is designed to exactly match the speed of the wheels.

The speed of the wheels is not the ground speed of the aircraft. The speed of the wheels will be measured at a fixed point on the circumference of the tyre.

The tyre circumference will determine how many revolutions it will make within a certain distance so if the aircraft is traveling at 160 kilometers per hour, that is
160000 meters an hour.
As the average runway distance needed for a loaded 747 to take off is about 3300 meters @ 160 kmh it would take about 75 seconds from stillstand to take off.

 

[jbriggs444]

The conveyor best is designed to exactly match the speed of the wheels.

But that would be a meaningless requirement. The speed of the conveyor always matches the speed of the wheels [as long as we are rolling and not skidding]. Even if the "conveyer" is tarmac firmly in place on the ground. The only meaningful requirement is "in the opposite direction".

Edit: Let me try to amplify this a bit.

Suppose the plane is moving at 1 meter per second forward and the tires would ordinarily be rotating at 1 meter per second over the pavement. You put the conveyor under the tires. The conveyor must now, per your reading of the specification, be moving at 1 meter per second rearward. But that means that the tires will now be moving at 2 meters per second. But that means that the conveyor must be moving at 2 meters per second rearward. But that means that the tires are now moving at 3 meters per second...

 

[Clausen]

@ jbriggs444,
Where does the question state the conveyor belt has to match the planes ground speed? The question states

"The conveyor best is designed to exactly match the speed of the wheels.

The speed of the wheels is not the ground speed of the aircraft. The speed of the wheels will be measured at a fixed point on the circumference of the tyre.

That is my interpretation also.

But that would be a meaningless requirement. The speed of the conveyor always matches the speed of the wheels [as long as we are rolling and not skidding]. Even if the "conveyer" is tarmac firmly in place on the ground. ...

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.

Edit: Let me try to amplify this a bit.

Suppose the plane is moving at 1 meter per second forward and the tires would ordinarily be rotating at 1 meter per second over the pavement. You put the conveyor under the tires. The conveyor must now, per your reading of the specification, be moving at 1 meter per second rearward. But that means that the tires will now be moving at 2 meters per second. But that means that the conveyor must be moving at 2 meters per second rearward. But that means that the tires are now moving at 3 meters per second...

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.
I still think the plane will take off, but the wheels will need to use some combination of sliding while rolling to meet the stipulation about matching the belt speed at all times, and I have no idea how to calculate the rolling resistance in that situation.

 

[jbriggs444]

but the wheels will need to use some combination of sliding while rolling to meet the stipulation about matching the belt speed at all times, and I have no idea how to calculate the rolling resistance in that situation.

But that is completely ridiculous. The problem then reduces to: "Can the plane take off with its wheels locked?"

 

[Clausen]

But that is completely ridiculous. The problem then reduces to: "Can the plane take off with its wheels locked?"

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.

 

[PeterO]

Imagine a 747 sitting on a large conveyor belt, as long and as wide as the runway.
The conveyor best is designed to exactly match the speed of the wheels, but run in the opposite direction.

CAN THE PLANE TAKE OFF?

Of course it can.
The wheels on the plane will just rotate at a higher speed that normal when the conveyor belt is used - it is not as if the wheels are being driven by the engines, they just spin at what ever speed the surface is passing at.

Just to fill in for some of your earlier apprehension.

lets suppose a 747 usually takes of at 200 knots. That means all of the plane is traveling forward at 200 knots, the wings, the body, the wheels, the passengers, the luggage .. everything.
When it comes to the wheels, you might choose to look at the top and bottom of the tyres separately, in which case you will find that the bottom of the tyre is actually traveling at zero speed, and the top of the tyre is traveling forward at 400 knots.
With your conveyor belt set to travel at the same speed, but opposite direction, as the wheels - that means the conveyor belt is traveling at 200 knots in the opposite direction of the 747, as it is about to take off.
In that case, the wheels are traveling forward at 200knots (with the rest of the 747), the bottom of the tyre is traveling backwards at 200 knots (with the conveyor belt it is touching) while the top of the tyre is traveling forward at 600 knots.

 

[Bandersnatch]

Where does the question state the conveyor belt has to match the planes ground speed? The question states

"The conveyor best is designed to exactly match the speed of the wheels.

As was pointed out a few times already, the original wording is faulty. It only makes sense if it's the ground speed that is matched.

Let's take the original wording at face value.

It means either that the conveyor matches the speed of the wheel hub w/r to the conveyor surface, or that it matches the linear speed of tire contact point with the conveyor surface. The difference is in the direction the conveyor is moving, which will be against plane velocity in the former case, and in the same direction as the plane in the latter. Both result in the same problem. Although it's worth noting that this ambiguity is yet another issue with how the question is formulated.

Since I think it's generally supposed for the conveyor to move against the direction of takeoff, let's assume the wheel velocity is measured between their hub and the conveyor surface.

$V_w$ - speed of the wheels
$V_c$ - speed of conveyor (measured at top surface w/r to the ground)
$V_p$ - speed of the plane w/r to the ground

When the plane is stationary, all velocities = 0 and there's no problem.
But as soon as the plane engages its engines and start moving w/r to the ground with any non-zero velocity $V_p$, it momentarily causes the wheels to have velocity w/r to the conveyor $V_{w0}=V_p$.

In accordance with the wording of the question, this in turn causes the conveyor to match the speed of the wheels in the opposite direction:
$V_{c0}=-V_{w0}=-V_p$.
Only now, the motion of the conveyor changes the speed of wheels:
$V_{w1}=V_{w0}-V_{c0}=2V_{w0}$
This makes the conveyor speed up to match the new velocity:
$V_{w2}=V_{w1}-V_{c1}=2V_{w1}=4V_{w0}$
which again makes the wheels roll faster, and so on, without limit. All of this happens regardless of how fast the plane is moving, as long as it's not 0.

Changing the meaning of the speed of wheels to the linear velocity of tire contact point changes only the sign of:
$V_{c0}=-V_{w0}=V_p$.

That's why the original wording of the question is faulty - it makes the conveyor and the wheels roll at infinite velocities as soon as the plane starts moving. The only sensible wording of the question is to make the conveyor match the ground speed of the plane.

(I've just noticed @jbriggs444 wrote pretty much the same thing in his edited post. Redundant redundancy is redundant.)

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.

You're giving the question too much credit. There's no indication that slipping was to be taken into account. If the question included e.g. something along the lines of 'assume thrust is X, rolling resistance is Y while slipping resistance is Z - can the plane take off?' then we could ponder how to make it work. Heck, why not assume there's no slipping, but the belt can only accelerate realistically, so that the question requires figuring out whether the plane can take off before the belt accelerates enough for the wheel bearings to blow off and the undercarriage to collapse or the rubber catches fire. We'd only need to know another dozen variables to solve it.

Remember that this is not a well-though out textbook problem. It's a question that's been circulating around Facebook and the internet at large. Somebody must have at some point copied it without understanding, and changed the wording to what they thought was equivalent, but was in fact physically faulty.

 

[Clausen]

You're giving the question too much credit. There's no indication that slipping was to be taken into account. If the question included e.g. something along the lines of 'assume thrust is X, rolling resistance is Y while slipping resistance is Z - can the plane take off?' then we could ponder how to make it work. Heck, why not assume there's no slipping, but the belt can only accelerate realistically, so that the question requires figuring out whether the plane can take off before the belt accelerates enough for the wheel bearings to blow off and the undercarriage to collapse or the rubber catches fire. We'd only need to know another dozen variables to solve it.

Remember that this is not a well-though out textbook problem. It's a question that's been circulating around Facebook and the internet at large. Somebody must have at some point copied it without understanding, and changed the wording to what they thought was equivalent, but was in fact physically faulty.

Sorry, that's just the way my mind works. I'm an engineer

I considered all of those possibilities

 

[russ_watters]

Yes, wheel speed is not important. But, a lot of replies here state that the important factor is the the forward speed of the aircraft, or the thrust of the engines.

Fundamentally, the problem has to do with neither, the only important factor is the speed of the airflow over the wings. Engines and runways are only there to allow the aircraft to move fast enough relative to the air mass in which it flies to make velocity of air over the wings increase to the point where it allows the airfoil at a given angle of attack to the free steam flow to generate enough lift to overcome the weight of the aircraft.

One starting assumption though is that the air is stationary with respect to the ground.

 

[jbriggs444]

One starting assumption though is that the air is stationary with respect to the ground.

... because otherwise the plane could conceivably take off while rolling backward.