B Can a 747 Take Off on a Conveyor Belt?

[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.

 

[boneh3ad]

This is a whole lot of hand-wringing and mental gymnastics over a concept that is relatively simple. There are basically 5 forces relevant to the plane as it tries to take off: thrust, drag, lift, weight, and some kind of friction or friction-like resistance to motion as a result of the plane's contact with the ground.

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

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

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

 

[russ_watters]

I'm not following you here. If you are using the tarmac as the reference it cannot be moving and matching the speed of the tires.

If the tires rotate through one meter going one way, the tarmac stays put and does not match the speed of the wheels.

The problem doesn't specify reference frames, which is part of the problem. But if the wheels are not sliding, there are three facts that exist no matter the interpretation:
1. The contact patches between the tires and conveyor are stationary with respect to each other.
2. The speed of the plane with respect to the ground is equal to and opposite the speed of the ground with respect to the plane.
3. The speed of the coneyor with respect to either the plane or ground is completely decoupled from the speed of the plane and ground with respect to each other.

But that violates the stipulation that the tires must exactly match the belt speed! If that stipulation holds then as the tires rotate through one meter, the belt moves back one meter from some fixed reference (the ground) In that case, the plane cannot move forward unless it also slides as it rolls. At least, that is how I see it.

This is a tortured mess, but the end result is that you've placed a hidden constraint on the plane's motion: *something* must tether it to the ground in order for it to not move in real life.
[see: right before a plane takes off from an aircraft carrier]
[edit] Actually, it's even worse: if you put numbers to this scenario, I think you will find that you are violating your own assumption.

That's what I thought too, but the wheels will not be locked, they will be rolling and partially sliding at the same time. If they were locked up, the plane will not takeoff since the coefficient of sliding resistance, rubber on rubber is greater than one. There will not be enough thrust to move against the normal force of 3.5 million N. But sliding while rolling, it might be able to do it.

Yes, it is ridiculous.

Yes, it is ridiculous. Again, it takes a hidden mechanism somewhere to make the end result of this tortured mess happen. In this case, in order for the wheels to slide on the conveyor, the brakes must be applied. [see: right before most planes start their take-off roll]

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

I am too and I soooo disagree with that being how engineers minds should work.

What I see as the problem with the way many people are approaching this is that they are trying to make an interpretation of the problem happen without regard for reality. That's fine for a physics problem (assume you could crash Europa into Mars...), but this problem is supposed to be about real life and therefore the end conclusion must never be something that is impossible in real life.

You cannot use an arbitrary/nonsenscal choice of reference frame to prevent an airplane from taking off in real life.

 

[russ_watters]

I really can't think of a realistic scenario where this would be true of wheels.

The only realistic scenario I see doesn't really fit the spirit of the problem: spin the wheels/tires fast enough and they overheat and sieze-up or disintigrate.

 

[A.T.]

...the original wording is faulty..

This always seemed to be the whole issue in the discussions on this 10 years ago. Not much has changed it seems.

 

[boneh3ad]

The only realistic scenario I see doesn't really fit the spirit of the problem: spin the wheels/tires fast enough and they overheat and sieze-up or disintigrate.

Right. It seems like the only scenarios where it might be an issue would cause the wheels to fail anyway, which is not in the spirit of the original question.

 

[OCR]

Airspeed, ground speed, tire speed (?... tire RPM...?)... c'mon guys, think about this a little bit, just a little bit ... ?

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

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

 

[OCR]

There is a point of no return for any object in the water.

Yes there is, and in this case... I think it's Bud and his friend !

 

[boneh3ad]

Airspeed, ground speed, tire speed (?... tire RPM...?)... c'mon guys, think about this a little bit, just a little bit ... ?

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

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

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

Obviously lift only cares about airspeed, aka velocity relative to the wind, but sit down and think about the factors affecting that velocity when a plane is taxiing.

 

[OCR]

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

I... I do know the definition(s) of airspeed.
I... I also know that...

The relative wind is of great importance to pilots because exceeding the critical angle of attack will result in a stall, regardless of airspeed.

You... you do the taxiing... I'll do the flying.. Carry on...

 

[Tito]

PLANE CANNOT TAKE OFF: Taking off means leaving the ground and moving up into the air. Taking off = LIFTing off the ground. Lift is created by air rushing over the wings to create lift. If the conveyor belt is running the in the opposite direction at a speed equal to white wheels, the plane will not move forwards or backwards. This means that the air around the wing and the wing have about 0 speed. Remember, air needs to be moving at a RELATIVE velocity to the wings, meaning there need s to be a difference in the velocities/speeds. 0 relative velocity = 0 lift = No taking off.