Can a 747 Take Off on a Conveyor Belt?

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In summary: 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.
  • #36
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.
 
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  • #37
russ_watters said:
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!
 
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  • #38
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.
 
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  • #39
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.
 
  • #40
The Wizard said:
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]
 
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  • #41
@ 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.
 
  • #42
The Wizard said:
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...
 
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  • #43
The Wizard said:
@ 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.

jbriggs444 said:
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.
jbriggs444 said:
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.
 
  • #44
Clausen said:
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?"
 
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  • #45
jbriggs444 said:
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.
 
  • #46
RandyD123 said:
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.
 
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  • #47
The Wizard said:
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.)

Clausen said:
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.
 
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  • #48
Bandersnatch said:
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:smile:

I considered all of those possibilitieso0)
 
  • #49
EspressoDan said:
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.
 
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  • #50
russ_watters said:
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.
 
  • #51
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.
 
  • #52
Clausen said:
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.
Clausen said:
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:smile:
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.
 
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  • #53
boneh3ad said:
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.
 
  • #54
Bandersnatch said:
...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.
 
  • #55
russ_watters said:
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.
 
  • #56
Airspeed, ground speed, tire speed (?... tire RPM...?)... c'mon guys, think about this a little bit, just a little bit ... :ok: ?

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

Really, it's only the relative wind ... :wink:
 
  • #57
RandyD123 said:
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 !
 
  • #58
OCR said:
Airspeed, ground speed, tire speed (?... tire RPM...?)... c'mon guys, think about this a little bit, just a little bit ... :ok: ?

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

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

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.
 
  • #59
boneh3ad said:
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.[COLOR=#black].[/COLOR] :ok:Carry on...
 
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  • #60
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.
 
  • #61
Tito said:
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.
Have you read the explanations given in this thread as to why the plane will fly?
 
  • #62
Tito said:
... speed equal to white wheels...
Whatever that means...
 
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  • #63
russ_watters said:
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.

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.

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]

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.

Who says this is a problem about real life? When, in real life, have you ever seen a 747 on a conveyor belt?

This is all about reading and understanding the problem as it is presented. The fact is, if the belt moves backwards with respect to a point on the ground, the same amount that the circumference of the wheels turns through in trying to move forwards, the hub of the wheel remains stationary with respect to the ground and the air. The only way the plane can takeoff is by sliding while the wheels are spinning.

You may dispute the way an engineer is supposed to think in your view, but you cannot dispute that what I am saying is 100% correct.
 
  • #64
The plane can only take off if there is airflow over the wings generating lift.
It makes no difference what kind of undercarriage arrangement it has, spinning the wheels will not generate any lift.
 
  • #65
Clausen said:
Who says this is a problem about real life?
The problem statement? Anyway, if it isn't about real life, the answer could be literally anything, including "yes it can take off if aliens teleport it into space."

In either case, just for the record, regardless of the problem as stated, do you agree that a *real* airplane on a *real* conveyor can take off?
When, in real life, have you ever seen a 747 on a conveyor belt?
Just because nobody's bothered to try it doesn't mean they couldn't. Mythbusters used a Cessna because that is easier/cheaper. Also, the scenario can be made functionally equivalent using wind: The plane is sitting stationary on the ground (speed of ground = speed of wheels = 0), with a 160kt headwind lifting it off the ground.
This is all about reading and understanding the problem as it is presented.
The problem, as specified, is mathematically/grammatically flawed/incomplete: 1=2 unless 0=0, in which case the problem is pointless (a plane sitting on the ground in no wind, with its engine off doesn't take off -- so what?).
The fact is, if the belt moves backwards with respect to a point on the ground, the same amount that the circumference of the wheels turns through in trying to move forwards, the hub of the wheel remains stationary with respect to the ground and the air.
Right. So with the engines at full throttle, what is keeping the plane stationary?*
The only way the plane can takeoff is by sliding while the wheels are spinning.
Or by breaking the problem because the problem is flawed. That's how it works in real life.
You may dispute the way an engineer is supposed to think in your view, but you cannot dispute that what I am saying is 100% correct.
Because the problem is ill-posed, "correct" is debatable in terms of how the problem statement works. But in terms of how real life works, the issue is not debatable: a plane on a conveyor can take off. So any answer that interprets the problem statement to yield an answer of no requires making up other assumptions and adding them to the problem*.
 
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  • #66
I'm going to try a different approach because up until the last exchange it actually hadn't occurred to me that someone would not be trying to make this problem real-world. That means there may not actually be any disagreement about any fully-defined scenario. So I'd like to run a handful of scenarios by people to see if there actually is any disagreement:

The basic, real life scenario:
1. A *real* plane on a *real* conveyor can take off regardless of the speed of the conveyor at least up to the speed where the wheels disintegrate.

Others that may or may not match real life and don't have to:
2. A plane sitting on a conveyor that rolls under it, with its engines on, tethered to the ground, with no wind will not take off.
3. A plane, sitting on the conveyor that rolls under it, with its engines off, tethered to the ground from the front, with sufficient wind will take off.
4. A plane sitting on a conveyor that rolls under it, with its engines on, while Luke Skywalker uses "The Force" to hold the plane in place will not take off.
5. A plane sitting on a conveyor that rolls under it, with its engines off, will take off if aliens use a tractor beam to yank it into space.

Does anyone disagree with any of these scenarios? If no, then the open issue is whether any particular person who asks the question is looking for a real life answer (even if they word the question badly) or not. And as 2 and 3 reflect, I've never seen anyone pose this problem in a complete way that enables the plane to remain stationary. It can be mathematically constrained, but if you apply physics to it, it requires another element, not specified, to provide the mathematical constraint.

Also, as can be seen, the motion of the conveyor(or not) is basically irrelevant to all of the scenarios! Whether the plane takes off or not depends entirely on the additional assumptions/constraints one adds to the problem!
 
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  • #67
Clausen said:
This is all about reading and understanding the problem as it is presented.
As it is presented? Let's see...

Clausen said:
The fact is, if the belt moves backwards with respect to a point on the ground, the same amount that the circumference of the wheels turns through in trying to move forwards,
Where does the problem mention the circumference? Where does it define a reference frame? As it is presented, it could be any point on the wheel in any reference frame.
 
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  • #68
russ_watters said:
2. A plane sitting on a conveyor that rolls under it, with its engines on, tethered to the ground, with no wind will not take off.
[...]
Just to explore the range of options similar to 2...

6. An automobile with its engine in gear drives forward at 30 miles per hour relative to a conveyor that is moving rearward at 30 miles per hour. The conveyor is light weight and [its mechanism is] nearly frictionless. As the driver presses on the accelerator, the car drives forward at 60 miles per hour and the conveyor naturally increases its rearward speed to 60 miles per hour. To compensate for frictional losses, the conveyor is slanted upward in front of the drive wheels. To ensure stability, it is also slanted upward to their rear.

This setup is also known as a dynamometer.

One then puts wings on the car, shifts into neutral, adds a propellor and takes off.

Edit: Added verbiage to clarify "frictionless" since @berkeman has correctly pointed out the ambiguity.
 
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  • #69
jbriggs444 said:
The conveyor is light weight and nearly frictionless
That would mean that the bearings supporting the conveyor belt are nearly frictionless, not the surface of the conveyor belt that makes contact with the tires... :wink:
 
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  • #70
A.T. said:
Whatever that means...
Can you show me where I made this quote... other than in this post ?
OCR said:
a speed equal to white wheels

I believe Tito did mention white wheels, though...
Tito said:
a speed equal to white wheels

How did you attribute that quote to me, anyway ?[COLOR=#black] .[/COLOR] :oldconfused:Oh, and BTW I want to clarify something, when I said...
"think about this a little bit, just a little bit ..."
I really did mean a little bit... as in, don't "over think".

I thought the first video I posted would show everything... that was needed to know ?

My bad... apologies to boneh3ad, and all the rest.

Now...
Carry on.
 
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