Will it take off?

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posted on Feb, 14 2006 @ 09:07 PM
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OK! everybody sitdown and SHUDUP!:MAD:...jeez

Forget about the bloody conveyer belt or whatever the wheels are doing to counter the motion of the belt. in terms of how a plane creates lift in order to become airbourne, THE BELT AND THE WHEELS ARE USELESS!

Any pilot or aviator with minimal hours will explain that unless there is airflow over the surface area of the aerofoil (the wing), there is no lift/upwards force vector i.e. NO FLIGHT! Suggested reading includes Bernoulie's theorum and the Venturi Effect. I suggest that anybody who even remotely thinks that flight is possible in this infernal arrangment of conveyer belts should seriously re-research their response/theories. I am completely astounded that there are so many people that would just blatently post a response to this sort of thing without actually looking into why aircraft fly in the first place. If flight was as simple as conveyer belts, we'd all be flying well before 1903. To the original author of the hypothetical which has atracted WAY too much attention for what it is (should have been debunked a bit earlier), it sounds like it was coming from the mind of a 12y/o to be quite frank, if so, i commend his ingenuity and initiative into aviation.

Ok so, no airflow, no lift, no flight. Got it!? good. PLEASE CLOSE THIS TOPIC

p.s. anyone who wants proof, take a hike to your nearest flight school. I haven't been flying for the past 5 years for someone to tell me that relative motion and NO AIRFLOW can make an aircraft fly.




posted on Feb, 14 2006 @ 09:16 PM
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Let me see if I can explain it to you.

It is a trick question.

The way the question is worded, you think that the plane is motionless on the conveyor belt.

It is not. The belt speed is related to the forward speed of the craft. If the craft were motionless, the belt would be motionless also.


Think of it this way. Imagine that instead of a plane, you have a rocket on wheels on a conveyor on the moon.

No air.

Will the rocket move forward?

Answer: yes, since the motion of a rocket in space is entirely dependent of Newton's third law. The opposite reaction is the forward motion of the rocket, not the retrograde motion of the belt.

The plane will take off.



posted on Feb, 14 2006 @ 09:19 PM
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Originally posted by av8or
OK! everybody sitdown and SHUDUP!:MAD:...jeez

Forget about the bloody conveyer belt or whatever the wheels are doing to counter the motion of the belt. in terms of how a plane creates lift in order to become airbourne, THE BELT AND THE WHEELS ARE USELESS!

Any pilot or aviator with minimal hours will explain that unless there is airflow over the surface area of the aerofoil (the wing), there is no lift/upwards force vector i.e. NO FLIGHT! Suggested reading includes Bernoulie's theorum and the Venturi Effect. I suggest that anybody who even remotely thinks that flight is possible in this infernal arrangment of conveyer belts should seriously re-research their response/theories. I am completely astounded that there are so many people that would just blatently post a response to this sort of thing without actually looking into why aircraft fly in the first place. If flight was as simple as conveyer belts, we'd all be flying well before 1903. To the original author of the hypothetical which has atracted WAY too much attention for what it is (should have been debunked a bit earlier), it sounds like it was coming from the mind of a 12y/o to be quite frank, if so, i commend his ingenuity and initiative into aviation.

Ok so, no airflow, no lift, no flight. Got it!? good. PLEASE CLOSE THIS TOPIC




I'm a pilot. The relative motion of a plane is not controlled by the force that the gear has on the runway, it's by the propellor/engine moving air in the opposite direction. The gear are essentially free-moving and although with their increase in speed would come a slight increase in friction, it would in no way impede the aircraft from taking off...

Think of it this way...there's an aircraft in the air and some sort of device causes the wheels to spin up in the opposite direction of flight, it would have absolutely no affect on the aircraft's performance since the aircraft is moving due to the force of the prop/engine against the AIR MASS, NOT THE GROUND.


Originally posted by av8orp.s. anyone who wants proof, take a hike to your nearest flight school. I haven't been flying for the past 5 years for someone to tell me that relative motion and NO AIRFLOW can make an aircraft fly.


5 years of flight experience ehh? that kinda scares me.



posted on Feb, 14 2006 @ 09:22 PM
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Ok, I understand what people are saying, there is no way for the conveyor belt to counteract the forward thrust of the engines, the tires will move at twice the speed, but the plane will still take off, I understand now! The plane will need the exact same amount of runway to lift off though.



posted on Feb, 14 2006 @ 09:23 PM
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You all missing the same thing I was.

Even with the conveyor belt moving against the plane it still moves forward, exactly as it would without a conveyor.

It does not stand still...

Take a long peice of paper lay it on the ground put a toy car (matchbox car) which wheels are free. (like a plane) You push the car forward have someone pull the paper backwards slowly but as equally as possible.

The car moves forward! omgs! (Thus the airplane would and it would create lift)

You are the propellors/jet your buddy is the conveyor.

edit..

mx got it!



[edit on 14-2-2006 by Xerrog]



posted on Feb, 14 2006 @ 09:39 PM
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Whatever,

>>
The wheels on a plane are there simply to support the aircraft while it is on the ground. They do not provide the force that moves the plane forward such as what a car would do.
>>

Which is exactly why I said there is NO TORQUE beyond what the wheels are receiving as a function of THRUST from the engine, they are not 'pulling' the conveyor back. They are simply freewheeling sufficient to balance the thrust with the rate of their own rotation.

If the engine (at max) THRUST is only sufficient to make the jet stay in one place, relative to it's position on the conveyor, it will never achieve sufficient speed of airflow over the wings to move down the runway.

People are stupid. They assume that an aircraft which can fly must have more thrust than weight. It does, but not at zero airspeed because the thrust is actually an upwards pressure differential across the airfoil rather than a kinetic force vector out the tailpipe or behind the prop.

Which is why a B-2 with a thrust to weight ratio of less than .2 can still take off.

What makes the foolishness obvous is that any relatively strong person can tug and shift a 500lb refrigerator (which he or she could /never/ lift over their heads) out from a wall.

BUT NOT if they were fighting against an opposed force provided by a conveyor belt 'sucking it back'.

With the high weights and low installed thrust of many modern airplanes it doesn't take much frictive counter force applied /thru the wheels/ to indeed _keep them right where they are_. As surely as if they were chained down.

>>
The force to move forward is provided solely by the propellor and/or jet engine against the air mass around the aircraft. Thus, the force of the engine against the air mass is not affected by wheels that are essentially free spinning and the aircraft would take off normally with just the speed of the wheels themselves being affected.
>>

But if, as implied, the friction with the conveyor is sufficient to counteract initial forward motion so that the relative position of the airframe to a static observer on the ground remains unchanged (5mph ->:5mph>
How else would planes take off on ice???
>>

Why do airplanes on _snow_ need skis to displace more of their footprint across a wider area than the tires can provide? Because the friction with the surface would 'plant' the tired and lock the plane in place or rip the struts off otherwise.

On ice where there is no counterforce applied as the surface is not moving backward towards the aircraft, the aircraft will takeoff. Even if it were, it /might/ be possible to cause the tires to skid forward faster than the ice retreated, due to a lack of positive friction.

For much the same reason it is not a case of someone 'pulling themselves up' a treadmill on rollerblades using a rope because the implication is that for /every acceleration/ applied (as tractor thrust) dragging you forward, the machine is matching with an equal and opposed relative FORCE negative acceleration through the increased rotation rate on the wheels.

Sooner or later you would reach a position whereby you could not PULL HARDER than the treadmill was pushing back and you would either have equilibrium of balanced forces. Or be pulled down on your face. The BEST you could do would be to pull one foot up and place it forward in an environment of ZERO coeffcient of friction, changing the spatial location on the treadmill while maintaining similar or lessened pull forces (more weight on one set of tires being the operative problem here) until you were sufficiently well forward to step off it.

Airframes cannot hop from leg to leg while striding forward however.

The original post and Straight Dope solution is, IMO, a case of someone with poor language skills trying to prove his how smart his math is in a way that deliberately does not match the contextual scenario he is describing.

You can call it a 'trick question' all you like, but it if he put it correctly, there would be no confusion between whether the aircraft is remaining stationary or _skidding_ forward faster than the tires can rotate, relative to a fixed observer.

Such is not science because it's fails to underline the base problem _or it's evolution_ in a way which leaves no doubt as to the proof being tested.

And so you can write the ending anyway you want, based upon the initial incomplete observation of the phenomena at-start.


KPl.



posted on Feb, 14 2006 @ 09:43 PM
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Originally posted by G8tsoHellrOpn4me


I'm a pilot. The relative motion of a plane is not controlled by the force that the gear has on the runway, it's by the propellor/engine moving air in the opposite direction. The gear are essentially free-moving and although with their increase in speed would come a slight increase in friction, it would in no way impede the aircraft from taking off...

Think of it this way...there's an aircraft in the air and some sort of device causes the wheels to spin up in the opposite direction of flight, it would have absolutely no affect on the aircraft's performance since the aircraft is moving due to the force of the prop/engine against the AIR MASS, NOT THE GROUND.



ok.....shall i draw in crayon this time.....

you missed my point COMPLETELY. In saying the gear was usless, i ment in regards to what the question was asking. don't you think i know that the wheels do SFA for the aircraft's motion!? they're just there for a smooth landing :p If the question is really saying the a/c isn't moving in relation to the conveyer belt then yes, all we're talking about there is a simple catapult, like an aircraft carrier. The way it was interpreted and i'm pretty sure it keeps getting interpretted is that the plane is stationary relative to the viewer/air and that the wheels were moving freely over the belt like something running on top of a rolling log, in that case then, no becasue there would obviously be no airflow ergo no lift, no flight. Thats all i'm trying to say



posted on Feb, 14 2006 @ 09:46 PM
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The conveyor does not PULL back on the plane.

It turns the wheels.

Which may slightly pull back on the plane but hardly.

The wheels are FREE TURNING. Pulling on them does not directly pull on the plane!

I'm done arguing.. stop and think or do the experiment I suggested or those many others have. Do not assume you are right.

My first conclusion was that it wouldnt take off either. It didnt make sense. Until you realize the conveyor turning the wheels has little to no bearing on the plane moving forward.



posted on Feb, 14 2006 @ 09:48 PM
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I haven't read the answer yet, but theoretically it would be possible for an aircraft to lift off under these circumstances. However, when you actually start crunching the numbers, you realize that the scale of the required conditions is quite extreme.

As stated by previous posters, the only thing we need to consider is motion of air over the aircraft's wings, regardless of reference frame. Due to the fact that there is a "no-slip" boundary condition at the runway surface, the moving runway will pull air along with it at its surface and over a short vertical distance above the runway until the velocity of the air is once again zero. This can be seen as some kind of reverse boundary layer. The thickness of this "boundary layer" will be determined by the speed of the runway movement, the length of the runway, the place on the runway at which the aircraft is placed and the roughness of the runway.

The requirement for lift-off will be this:

- the properties and motion of the runway will need to be such that the air velocity at the wing height is equal to or greater than the take-off velocity of the aircraft.

Once this threshold is met, the aircraft will lift off the runway but will immediately begin to be pushed backwards by the air from the runway (due to drag) now that the wheels can no longer force the aircraft forwards. This would quickly reduce the relative velocity of the air across the wing, dropping it below the required take off velocity.

The craft would then touch down on the runway again, and we start over as seen in this image:




Up to this point though, this is completely theoretical. Let's try putting in some numbers to test this out. The two equations that are of primary interest are:

and


These equations come from the One-Seventh Power Law for estimating the velocity profile of a turbulent boundary layer over a smooth flat plate, where delta is the BL thickness, x is the horiz. distance from the right of the runway in this case and y is the vertical distance from the runway. If we say that the vertical height of the wings from the ground is 3 m, and assume a lift-off speed of 50 m/s, then it turns out that we would need a runway that is 2 km long and moving at 1000 m/s in order to generate enough lift to get the aircraft to lift off... not very practical. There is also the problem that the wind velocity would be far greater closer to the runway so that when the craft lifts off, it is possible that the much stronger winds acting on the landing gear would cause a large pitching moment which would cause the nose to come slamming down. An interesting question to think about.



posted on Feb, 14 2006 @ 10:09 PM
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Technically, the plane could take off, but maybe not stay up. If it was moving, it still has momentum, so it if you raise it up to take off, it might just fly off since the jets will be pushing against the floor, but it might not be enough to keep it up, actually it might only do like a "jump" but fall back down.



posted on Feb, 14 2006 @ 10:11 PM
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I just read the "answer", the interpretation of the question that I was using was that the movement of the treadmill and the rotation of the landing gear exactly cancel one another such that the aircraft remains stationary relative to the ground. No jet engines or props needed. I think that this is the more interesting interpretation of the question, otherwise the conveyor belt is just a trick and has no effect on the result of lift-off other than the wheels having to rotate at twice their normal takeoff speed (as most posters have stated).

Feynman's underwater sprinkler question is actually a lot more difficult than this.

[edit on 14-2-2006 by Maser]



posted on Feb, 14 2006 @ 10:14 PM
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Bhadhidar is absolutely correct. lift is the key to flight, and no lift is generated in this situation. no wind flowing over the wings, no lift. period. it has absolutely nothing to do with the movement of the wheels.



posted on Feb, 14 2006 @ 11:18 PM
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HR,

>>>
As postulated, if the aircraft generates no relative motion along the length of the runway, the velocity of the airfoil _moving through the air_ will never be sufficient to take off.
>>>

>>
Actually that is not what is postulated. What is postulated is that the belt moves backward at the same speed that the plane is moving forward. Therefore, if the plane was truly stationary, then the belt would be also.
>>

The initial thrust by the impulse of the engine is against the mass of the airframe. If the airframe is sitting statically, weight on wheels, and the conveyor moves backwards with equal acceleration to equal velocity _as a force_. The airframe won't move at all.

If the airframe moves forwards and the conveyor responds, the velocities )(as forces) of each are not in balanced equilibrium.

If the aircraft is moving with equal speed as it attempts to 'board' the active, it will bounce or tuck under for much the same reason as up-the-down-escalator is 'tricky'.

If friction drag (on the tires) is overcome via slip, the question becomes the overspeed burst values on the tires vs. the required takeoff distance effect on rearwards progression of an airframe still 'attached sufficiently' to a runway suffer a negative airflow effect (a tailwind) due to rearwards progression of the runway vs. limited traction on the tires.

My bet is on the tires going before the plane runs out of room or gas to takeoff.

>>
It is a trick question.
>>

It is a poorly worded physics statement designed to be interpreted the way the author wishes it to be without adequate reference to starting conditions or balance/progression of the dynamics in the experiment.

>>
Let’s look at the problem another way. Let’s say that instead of an airplane you have a rocket car. That is a car that is powered by rocket exhaust. Would the car move forward to the end of the belt?
>>

Only if the instantaneous impetus from behind (in the forward plane of motion) was greater than that applied from the front in overcoming a static inertia of weight-on-wheels vs. tire slip condition.

Forces out of balance.

>>
Let’s say that instead of a rocket car you took a sidewinder missile, removed the warhead and mounted the missile on a four wheel cart. Would the missile-cart reach the end of the belt?
>>

See above. I would guess that the imbalance of vectors (slip vs. direct impulse locations above and under the cart) would cause the cart to fall over, which direction, would depend on how great the acceleration disparity was vs. tire traction.

>>
If you fired just the missile itself, above the belt it wouldn’t make any difference at all how fast the belt was moving.
>>

Yes it would. For much the same reason that a guns muzzle velocity is additive to the firing airframes own forward motion. Many forward fire missiles have minimum launch speeds to adequately clear the rail and energize the aerodynamic controls (Sidewinder needs about 105 knots minimum off an AH-1W).

That said, if the missile could overcome a given 'zero state' of opposed forces a start, it would fly independently through the air thereafter.

>>
What if you had the cart, but put wings on the missile? What then?
>>

Given that the missile will fly without the wings and the missile alone is flying, the question becomes one of stability with the added attachments. The cart will fly only if the rearwards acceleration to maximum velocity was out of phase with the forward impulse.

This-

>
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?"
>

Does not specify that.



posted on Feb, 14 2006 @ 11:25 PM
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Originally posted by av8or

Originally posted by G8tsoHellrOpn4me


I'm a pilot. The relative motion of a plane is not controlled by the force that the gear has on the runway, it's by the propellor/engine moving air in the opposite direction. The gear are essentially free-moving and although with their increase in speed would come a slight increase in friction, it would in no way impede the aircraft from taking off...

Think of it this way...there's an aircraft in the air and some sort of device causes the wheels to spin up in the opposite direction of flight, it would have absolutely no affect on the aircraft's performance since the aircraft is moving due to the force of the prop/engine against the AIR MASS, NOT THE GROUND.



ok.....shall i draw in crayon this time.....

you missed my point COMPLETELY. In saying the gear was usless, i ment in regards to what the question was asking. don't you think i know that the wheels do SFA for the aircraft's motion!? they're just there for a smooth landing :p If the question is really saying the a/c isn't moving in relation to the conveyer belt then yes, all we're talking about there is a simple catapult, like an aircraft carrier. The way it was interpreted and i'm pretty sure it keeps getting interpretted is that the plane is stationary relative to the viewer/air and that the wheels were moving freely over the belt like something running on top of a rolling log, in that case then, no becasue there would obviously be no airflow ergo no lift, no flight. Thats all i'm trying to say


=o, Whoops, well all apologies sirs, after reading the last 2 pages of posts about how planes on the ground behave just like cars; I somewhat skimmed through yours without realizing the point you were making was completely valid...no relative wind = no lift; however, this doesn't mean that a giant conveyor belt would somehow keep a plane from lifting off as many here still believe.



posted on Feb, 15 2006 @ 01:50 AM
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Here's a challenge..

Since we cant possibly hope to replicate this for real...

Any programmers here want to take a shot at building or altering a flight sim to show what would happen. Using various existing aircraft.



posted on Feb, 15 2006 @ 03:53 AM
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Sorry but i haven't read all the posts, this may be linked already
but
forum.physorg.com...

Has the most extensive debates on this topic from extremely brainy people, whom go into full physics details about it.
I was quite surprised how it stumped alot of people and was even debated by them.

For....... TWO HUNDRED+ PAGES

People even made mini scale models to test the hypothesis

I think the end result was that it actually would take off.



posted on Feb, 15 2006 @ 04:24 AM
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sob.. Cant help it I gotta try to prove the fly point..

Lets look at the forces at work and specifically what they are working on.

First factor is the conveyor. Lets say it is moving.

It is acting on the wheels of the plane.

The wheels in turn act on the plane via friction only. They turn freely and have no gearing, driveshafts or any other "hard" connection to the plane.

If propellors they act on the air near the plane pulling/pushing the plane through the air, and forcing more air over the wings at the same time.

If a jet it acts solely on the plane pushing it forward.

The plane itself acts on the wheels through gravity causing some friction. (Otherwise the wheels would not act against the plane at all)

There also may be some airflow caused by the conveyor moving. This would be towards the front of the plane causing some lift, but also causing some drag. Both would be negligible.

..................

So lets say its the jet..

We fire the engine and the plane begins to move forward at say 20mph. This in turn causes the conveyor to move backwards at 20mph. The wheels will for a short time turn at 40mph. If we stay at the same amount of thrust the friction from the wheels to the plane will cause it to slow its forward movement slightly, so the conveyor slows and the plane speeds back up. Eventually it would even out somewhere just below 20mph each. The whole time the plane would continue moving forward however.

Now if we continue to give more thrust the plane continues to gain speed. The jet is pushing DIRECTLY against the plane. The conveyor is acting on the plane only through friction from the wheels to the plane. The plane continues to move forward but gaining speed slightly slower then normal. The wheels are turning at almost double the normal speed. If the wheels and their assembly can withstand the forces then the plane will take off. Albeit needing a slightly longer runway.

......................

With a propellor plane its the same scenario but since the forward thrust isnt directly working on the plane it will experience more "lag" from the conveyor.

................

Now lets look at a car.

The conveyor again acts against the wheels.

The wheels this time however work agaisnt the drive shafts, the transmission, engine, and eventually the car itself.

The Engine works oppositley agains the drivetrain, then the wheels.

Then the wheels work against the conveyor.

The car itself works against the wheels through gravity and friction.

................

As the car tries to move forward the conveyor belt acting against the wheels, drivetrain, car push it back. The car goes nowhere. But this is soley because its source of propulsion is connected directly to the wheels.

..............

If we add a jet, or fan or propellors to the car instead of a normal engine/drivetrain and rip all of that old stuff out leaving the wheels "free moving" (or put it in nuetral!) The car will move down the conveyor and eventually off of it.







[edit on 15-2-2006 by Xerrog]



posted on Feb, 15 2006 @ 05:06 AM
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I've kinda already explained why it will take off, but didn't make a good job of it.

There are two forces upon the aircraft.

The thrust from the engine, and the friction from the wheels.


To simplify things we will make the assumption that the engine thrust does not vary with airspeed.

The friction from the wheels is dependant on wheel rotational speed, as wheels at rest produce no friction, so it would be something like F(wh) = mu*Vg.

Now, the rotational speed of the wheels will be twice the airspeed of the aircraft.


Since the engine thrust will be greater than the friction from the wheels initially, the aircraft will accelerate (and assuming it stays on the ground) to a point where the ground speed (wheel rotational speed) Vg*mu = engine thrust.


The key is whether this wheel rotational speed is greater or less than twice the take-off speed (V rotation [Vr]) of the aircraft.

If its greater (i.e. the aircraft reaches Vr before the wheel friction balances the engine) the aircraft takes off.

If its less, then the aircraft won't take off.



My money is on the aircraft taking off quite easily. There is also the boundary layer effect helping as mentioned earlier too.

[edit on 15-2-2006 by kilcoo316]



posted on Feb, 15 2006 @ 05:25 AM
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Sorry folks, but (Asumeing the speed of the treadmill increases as the engine is throttled up) the plane is not going anywhere!

As several other have said, the Backwards speed of the treadmill is Equal to the forward speed of the treadmill.

Yes, it is true that the wheels are not part of the propultion system, but under the circumstances, that is irrevent! Untill the Plane achieves lift, it's wheels support ALL of its weight! As was stated earilier, the Backwards speed of the treadmill is Equal to the forward speed of the treadmill. This makes the net speed of the plane in relation to the stationary ground 0. With a net speed of zero, you will not get any airflow over the wings! Without airflow, there can be NO lift!

No lift means the plane will Not fly! It's that simple. You would get the same resualt from tieing the plane down with chains like they do in a ground run!

As for Newton's Third Law, Remember, it's being applied Twice: Once between the engine and the plane, and again between the plane and the ground. The two action reaction sets have a cancelling effect!

Tim

[edit on 15-2-2006 by ghost]



posted on Feb, 15 2006 @ 06:00 AM
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Originally posted by ghostAs was stated earilier, the Backwards speed of the treadmill is Equal to the forward speed of the treadmill. This makes the net speed of the plane in relation to the stationary ground 0. With a net speed of zero, you will not get any airflow over the wings! Without airflow, there can be NO lift![edit on 15-2-2006 by ghost]

The net speed of the aircraft in relation to stationary ground is not zero, and is in fact not even affected by the treadmill. Which means the plane is actually moving, will generate airflow over the wings, will generate lift, and will fly.

[edit on 15-2-2006 by Travellar]






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