Mythbusters punk'd whole internet

[LTC8K6]

How is the conveyor applying force to the plane?

It must apply a force to the plane to keep the plane from moving forward.

It must counter the engine's thrust, or the plane will move.

The only contact the conveyor has with the with the plane is at the wheels.

The wheels are free to spin.

The conveyor cannot exert much force on the plane at all.

Almost 100% of the conveyor's force will be lost to wheel rotation.

If we had magic perfect frictionless bearings, 100% of the force would be lost.

In reality, there is a little force, because we don't have perfect bearings, but that's it.

The plane's engine can easily overcome wheel bearing friction.

We know this because wheel bearing friction is basically all the plane has to overcome to take off on a runway.

If you want to pick nits, there is a little friction we are leaving out, and that is the tire's rolling resistance, and the tire's grip on the surface.

These are all small though, and will not change the outcome. They are the same things the engine has to overcome on a normal takeoff.

 

[BoberFett]

Originally posted by: mrSHEiK124

Originally posted by: waggy

Originally posted by: smack Down

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

thats the problem. THIS IS NOTHING LIKE A CAR.

really its not. a car gets its thrust by the wheels. a plane by the thrust of the engines.

For every action, there is an equal and opposite reaction. the thrust of the engine has to have its opposite action. the wheels/treadmill ARE NOT IT.

Thank you, best explanation ever.

THE AIRPLANE TAKES OFF. The wheels moving backwards, forwards, sideways; fuck, take off the wheels and make the damn plane slide across the treadmill, IT WILL TAKE OFF. None of the forces I just mentioned would do ANYTHING to counter the thrust of the jet engines. NOTHING.

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

 

[BucsMAN3K]

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

It's speed is based on the acceleration due to the sum of the forces on the plane. The sum of the forces is essentially the thrust minus and friction due to the wheel bearings. So there IS a net force on the plane, which causes acceleration, which means the plane is moving....

...so there's lift.

http://youtube.com/watch?v=-EopVDgSPAk

 

[LTC8K6]

"Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift."

What is preventing the plane from moving forwards and aquiring lift?

It can't be the conveyor belt, because all it can do is rotate the plane's wheels, which will not stop the engine from moving the plane forward.

So, what is countering the engine's thrust?

 

[Mrvile]

Why is this thread still up? It's getting a bit toasty in here.

 

[zanejohnson]

holy crap, i thought the people who didnt think it would take off were just joking...

who could be that stupid lol...

of course the plane will take off.

the force that makes a plane take off is the lift, from air rushing under the wings, "lifting" it up....

no matter how fast the conveyer belt moves the thrust from the jets/propellors will STILL push the plane forward, because a plane is not pushed with a driveshaft to the wheels like a car, the wheels have nothing to do with the plane taking off.

it would take very little thrust to keep the plane stationary on the treadmill, and just a little more to push it forward on the treadmill.... and as soon as it starts moving forward air will flow beneath the wings, applying lift.

 

[jmmtn4aj]

Originally posted by: smack Down

Originally posted by: Skeeedunt
I think this finally makes sense. If the wheels are frictionless, the plane takes of. Otherwise, the conveyor can move the wheels so fast that the downward force of the wheels' rotational inertia is so great that it counteracts the thrust of the engine against the air. Am I right am I right?? Do I get a cookie??

Almost but friction is the force that matters as long as the wheels have mass the plane would be unable to take off. If the wheels do not have mass then that would mean it violates the conditions of the problem.

Friction isn't only dependent on mass, roughness also matters, which is why in ideal-case mechanics you can have all sorts of calculation involving objects with mass, gravity, but no friction.

The treadmill, by the way, can match the speed of the plane. In fact, it can even exceed it, yet the plane will still take off.

The question is tricky because it involves a moving ground, and since people can't see air they tend to relate a plane taking off by it's relative speed to the ground. Instead of a moving ground scenario, imagine it another way: both the ground AND the plane are horizontally motionless, zero motion relative to each other on the horizontal plane (mathematical plane, not aircraft plane) ;

Imagine a scenario where the plane is stationary with a strong headwind. Theoretically if the wind is laminar and blowing slightly upwards relative to the ground, and at a certain speed, and the control surfaces are manipulated to keep the plane level, the plane would be able to lift off without actually moving relative to the ground (too strong and the plane is blown backwards, too weak and not enough airflow passes over and under the wing). The wheels wouldn't spin, but the plane would lift off anyway.

Flight does not depend on the ground, it depends on pressure differences between air flow on the top surface of the wing and air flow on the bottom surface. Lower pressure on the top surface 'sucks' it upwards, and higher pressure on the bottom surface pushes. This phenomenon is called lift. An illustration flight's dependence on relative motion through the air and not to the ground.. roof panels getting blown away in a hurricane, wind tunnels, wind blowing sheets of paper off your desk.. just to name a few.

The lack of relationship between ground speed and airspeed is why aircraft carriers accelerate into headwind when launching aircraft, why you feel a lurch downwards in a microburst or when entering tailwind even though ground speed remains the same, and why jet airliners have increased ground stall speeds when entering tailwind.

 

[jagec]

Originally posted by: Skeeedunt
I never actually understood why the force applied by linear momentum was backwards any more than it was, say, forward or upward. Is it simply a fact that the linear momentum would be in the opposite direction of a force applied at a wheel's edge (i.e. the treadmill)?

Well, momentum is kind of an imaginary force...it only exists when we're drawing a dynamic, not a static, system. If forces balance and there is no acceleration, there is no momentum term. Momentum is basically the "reaction" force to any "action" force we place on a system...e.g. the whole "you push the cart, the cart pushes back on you" principle.

Originally posted by: mobobuff
the belt and plane will only be moving at 150-200mph (if we assume there is no wind) by the time the plane achieves the required airspeed and takes off. Arguing anything beyond that point is arguing a different question altogether.

I agree. In the original question, the plane moves forward at 200mph, the belt moves backward at 200mph, and the plane takes off without a hitch.

The question that I've been arguing is no longer "Does the treadmill in the OP prevent the plane from taking off?", but "CAN a treadmill prevent a plane from taking off?"

The answer is yes, due to an interesting coupling between linear and angular movement.

OK, so a force being applied on an object away from its center of mass can be written equivalently as the sum of a torque around the center of mass, and a proportionally smaller force through the center of mass. The two situations are equivalent, from a dynamics perspective.

Now, consider three satellites in space (crappy MS paint warning). The first one has a thruster on each side, thrusting in opposite directions. This is equivalent to two linear forces, both displaced from the center of mass. If the forces are equal and are equally distant from the center of mass, this simplifies to a torque around the center of mass, and a net linear force of zero through the center of mass.

Eliminate one of these thrusters. The torque around the center of mass remains, but it is smaller, and now there is a net linear force through the center of mass as well.

Now attach a string to a hub in the center of the satellite, and pull. You can apply a force sufficient to cancel out the linear portion of the force through the center of mass, keeping the satellite in one place, but the torque term is still there and therefore there will be angular acceleration.

This last scenario is exactly how I propose to keep the airplane on the treadmill.

Remember, the wheels are NOT fully decoupled from the plane. They are perfectly able to transfer linear force in all three axes, and torque in two axes. The ONLY thing that a wheel does is eliminate the coupling along one ROTATIONAL axis. To prove this, try pushing, pulling, lifting, and twisting a bicycle by holding on to the front wheel. All these things are possible. The only thing that you cannot make a bicycle do through the wheels is flip it over the handlebars...and you can do this, too, if you take advantage of momentum.

 

[jmmtn4aj]

Oh, yeah, to those who think friction caused by the wheels might stop the plane, it won't. A jet airliner only has to reach an airspeed of about 250-300km/h to take off. If a treadmill were to match that, relative speed would be 500-600km/h, someone can convert that to rotational motion for me. The bottom line is that might be enough to blow the tires or seriously destroy the landing gear struts and braces, but it won't generate enough frictional force to overcome engine thrust. Acceleration might be slower, but the average airliner turbofan is pretty freaking powerful.

 

[jmmtn4aj]

Originally posted by: jagec
The question that I've been arguing is no longer "Does the treadmill in the OP prevent the plane from taking off?", but "CAN a treadmill prevent a plane from taking off?"

The answer is yes, due to an interesting coupling between linear and angular movement.

OK, so a force being applied on an object away from its center of mass can be written equivalently as the sum of a torque around the center of mass, and a proportionally smaller force through the center of mass. The two situations are equivalent, from a dynamics perspective.

Now, consider three satellites in space (crappy MS paint warning). The first one has a thruster on each side, thrusting in opposite directions. This is equivalent to two linear forces, both displaced from the center of mass. If the forces are equal and are equally distant from the center of mass, this simplifies to a torque around the center of mass, and a net linear force of zero through the center of mass.

Eliminate one of these thrusters. The torque around the center of mass remains, but it is smaller, and now there is a net linear force through the center of mass as well.

Now attach a string to a hub in the center of the satellite, and pull. You can apply a force sufficient to cancel out the linear portion of the force through the center of mass, keeping the satellite in one place, but the torque term is still there and therefore there will be angular acceleration.

This last scenario is exactly how I propose to keep the airplane on the treadmill.

Remember, the wheels are NOT fully decoupled from the plane. They are perfectly able to transfer linear force in all three axes, and torque in two axes. The ONLY thing that a wheel does is eliminate the coupling along one ROTATIONAL axis. To prove this, try pushing, pulling, lifting, and twisting a bicycle by holding on to the front wheel. All these things are possible. The only thing that you cannot make a bicycle do through the wheels is flip it over the handlebars...and you can do this, too, if you take advantage of momentum.

I'm an engineering (aeronautics) student and I have no idea what you've just suggested o.o

So what exactly, does your treadmill do to keep the plane on the ground?

 

[Suspicious-Teach8788]

I believe we shouldn't even talk about friction. It's not a fixed number and it's really just a small force there. It's like assuming when the airplane will roll to a stop after it lands without reverse thrust/brakes.

When we talk about physics and acceleration and everything, we usually assume ideality do we not? Unless of course the problem states so, we usually just ignore friction, and I believe it should be ignored here. Even if it were accounted for, it's just a small force that's negligible in the end here.

Originally posted by: BoberFett

Originally posted by: mrSHEiK124

Originally posted by: waggy

Originally posted by: smack Down

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

thats the problem. THIS IS NOTHING LIKE A CAR.

really its not. a car gets its thrust by the wheels. a plane by the thrust of the engines.

For every action, there is an equal and opposite reaction. the thrust of the engine has to have its opposite action. the wheels/treadmill ARE NOT IT.

Thank you, best explanation ever.

THE AIRPLANE TAKES OFF. The wheels moving backwards, forwards, sideways; fuck, take off the wheels and make the damn plane slide across the treadmill, IT WILL TAKE OFF. None of the forces I just mentioned would do ANYTHING to counter the thrust of the jet engines. NOTHING.

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

Think skateboard my friend. Tie yourself to the wall and you're on a skateboard on a treadmill. Pull on the rope. You move forward. Free moving wheels FTW. Get out of the car analogy because it's not an analogy but a misinterpretation that results in mistakes. Ground speed is NOT affected by the treadmill.


For those of you who don't get it yet, please draw a Free Body Diagram. Where does the backwards force go into? It goes into pure rotation, not translational motion. Thus there is NOTHING that pushes the plane backwards while the engines push forward.

 

[jmmtn4aj]

Originally posted by: BoberFett

Originally posted by: mrSHEiK124

Originally posted by: waggy

Originally posted by: smack Down

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

thats the problem. THIS IS NOTHING LIKE A CAR.

really its not. a car gets its thrust by the wheels. a plane by the thrust of the engines.

For every action, there is an equal and opposite reaction. the thrust of the engine has to have its opposite action. the wheels/treadmill ARE NOT IT.

Thank you, best explanation ever.

THE AIRPLANE TAKES OFF. The wheels moving backwards, forwards, sideways; fuck, take off the wheels and make the damn plane slide across the treadmill, IT WILL TAKE OFF. None of the forces I just mentioned would do ANYTHING to counter the thrust of the jet engines. NOTHING.

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

Ground speed and airspeed are independent of where the plane is. The only difference between ground speed and airspeed is ground speed is measured with radio beacons and GPS, while airspeed is measured with a pitot static tube. The only reason ground speed values are available to pilots is because it's required for navigation, knowing how much ground they've covered in a certain amount of time based on ground speed. Take off, landing, cruising, descent and ascent speeds, are all calculated using true airspeed. Because TAS is the through speed of the air actually passing over the lift surfaces. Who cares if the plane is shooting a thousand kilometres per hour over land, if the wind blowing a thousand kilometres per hour in the same direction along with it is going to cause it to crash because zero lift is generated? In that situation, ground speed is WAY above the stall speed of any plane in existence (we're not talking about bullets or missiles), yet airspeed is zero.

Similarly, who cares if a plane is shooting at the same speed on land, with wind blowing the same way? Zero lift = the plane ain't going to leave the ground because airspeed is zero, even though ground speed is enough to make you shit your pants.

 

[Ricemarine]

I think this whole page already covered that the plane does take off...

So let it die .

 

[jagec]

Originally posted by: jmmtn4aj

I'm an engineering (aeronautics) student and I have no idea what you just suggested o.o

So what exactly, does your treadmill do to keep the plane on the ground?

I'm an engineering graduate, but my Chem E degree has as little do with planes on treadmills as your does, aside from the physics classes that we have both taken.

Did you follow and agree with the FBDs I did on the satellite scenarios? If you disagree, with which part do you disagree? If you're having trouble understanding what I'm getting at, what needs to be made clearer? I fear that sometimes I have a tendency not to set things out in as straightforward a manner as is necessary for accurate communication.

The third satellite scenario is practically equivalent to the plane on a treadmill.

Originally posted by: DLeRium

For those of you who don't get it yet, please draw a Free Body Diagram. Where does the backwards force go into? It goes into pure rotation, not translational motion. Thus there is NOTHING that pushes the plane backwards while the engines push forward.

Take a wheel out into space, where there are no other forces interacting with it. Tie a string to the outer edge of the wheel. Pull the string.

The wheel starts spinning, but it also starts drifting towards you. There is a translational AND a rotational component to the resulting momentum you just imparted to the wheel.

 

[jmmtn4aj]

Originally posted by: jagec

Originally posted by: jmmtn4aj

I'm an engineering (aeronautics) student and I have no idea what you just suggested o.o

So what exactly, does your treadmill do to keep the plane on the ground?

I'm an engineering graduate, but my Chem E degree has as little do with planes on treadmills as your does, aside from the physics classes that we have both taken.

Did you follow and agree with the FBDs I did on the satellite scenarios? If you disagree, with which part do you disagree? If you're having trouble understanding what I'm getting at, what needs to be made clearer? I fear that sometimes I have a tendency not to set things out in as straightforward a manner as is necessary for accurate communication.

The third satellite scenario is practically equivalent to the plane on a treadmill.

I agree with the FBDs, I just don't see how it relates to a plane. Where is the torque? A plane has it's centre of mass on the same plane as it's engines, and on take off both engines output almost equal amounts of thrust, so there isn't torque around any axis due to it's engines. It's wheels provide zero thrust, and only serve to provide an equal reaction to gravitational force (i.e holding the plane up) whilst allowing it to move freely. After it lifts off, the lift surfaces provide all the vertical forces to keep the plane level.

..Ok, back to the ground scenario. Since frictional force in the wheels are negligible, and there are wheels both forwards and back, and sideways of a plane's centre of mass (otherwise it'd tip over whilst on the ground), I don't see torque generated by those around the aircraft's centre of mass either..

???

 

[jagec]

Originally posted by: jmmtn4aj
I agree with the FBDs, I just don't see how it relates to a plane. Where is the torque? A plane has it's centre of mass on the same plane as it's engines, and on take off both engines output almost equal amounts of thrust, so there isn't torque around any axis due to it's engines. It's wheels provide zero thrust, and only serve to provide an equal reaction to gravitational force (i.e holding the plane up) whilst allowing it to freely. After it lifts off, the lift surfaces provides all the vertical forces to keep the plane level.

..Ok, back to the ground scenario. Since frictional force in the wheels are negligible, and there are wheels both forwards and back, and sideways of a plane's centre of mass (otherwise it'd tip over whilst on the ground), I don't see torque generated by those around the aircraft's centre of mass either..

???

Sorry, those FBDs are supposed to be analogs to the wheels of the plane, not the whole thing. The torque is around the CoM of the wheel, being applied to the wheel itself. The force being applied to the outer edge of the wheel is the treadmill pushing the wheel backwards (no-slip condition), and the force exerted by the string represents the force of the engines, which is applied to the wheel at and via the hub (which is, coincidentally, the wheel's CoM). Final crappy MS paint illustration. The third satellite scenario is included at the top for ease of reference. Yes, I am possibly the world's worst artist.

Force exerted by engines has been moved to wheel hubs to clarify. Yes, I shouldn't move forces around without adding the equivalent torque to the system, but that will just cause the front tires to be loaded more than the rear tires, and we both agree that it is unimportant to the discussion.

OK, I'm going to bed now.

 

[Suspicious-Teach8788]

Originally posted by: jagec

Originally posted by: jmmtn4aj

I'm an engineering (aeronautics) student and I have no idea what you just suggested o.o

So what exactly, does your treadmill do to keep the plane on the ground?

I'm an engineering graduate, but my Chem E degree has as little do with planes on treadmills as your does, aside from the physics classes that we have both taken.

Did you follow and agree with the FBDs I did on the satellite scenarios? If you disagree, with which part do you disagree? If you're having trouble understanding what I'm getting at, what needs to be made clearer? I fear that sometimes I have a tendency not to set things out in as straightforward a manner as is necessary for accurate communication.

The third satellite scenario is practically equivalent to the plane on a treadmill.

Originally posted by: DLeRium

For those of you who don't get it yet, please draw a Free Body Diagram. Where does the backwards force go into? It goes into pure rotation, not translational motion. Thus there is NOTHING that pushes the plane backwards while the engines push forward.

Take a wheel out into space, where there are no other forces interacting with it. Tie a string to the outer edge of the wheel. Pull the string.

The wheel starts spinning, but it also starts drifting towards you. There is a translational AND a rotational component to the resulting momentum you just imparted to the wheel.

Your wheel analogy... Doesn't that assume that you're pulling on it for some length of time? If it's a sudden force, shouldn't it translate purely into rotation? Maybe not. I'm studying for my stats final, so I can't really spend too much time dwelling on it, but doesn't tugging it at the edge cause rotation about the axis of rotation (where the axle would be)? Your assumption that it starts driting towards you is based on either:

a) Your string is not perpendicular with the radius of the wheel

or

b) It's a real human pull where you can't let go in time, so part of your time pulling is where the wheel has already begun spinning.

I believe that if it's the ideal pull we learned in Physics class it should be purely rotational.

About your bicycle question, holding onto the front wheel you can definitely push the bicycle, but what about the edge. The very edge that makes contact with the ground so that your pushing is perpendicular with the radius. Obviously we cannot simulate the very edge, and like I said this treadmill question should be considered using ideality. The amount of frictional force that pulls the plane back is negligible. You need to assume a very well oiled machine where the wheels are free spinning.

Can a treadmill slow down a plane in REAL LIFE? Yes. Without a doubt. But once agian, that friction value is so dependent on your plane's wheels, what kind of plane, etc etc. It might slow down a 747 better than a Cessna, or whatever but who knows? That's not really what we're concerned with and it shouldn't even be discussed. To me, all it causes is trouble.

I think based on the amount of force your engine can produce, all that really matters is whether it takes off or not.

 

[Anubis]

fing A lock this allready they fing plane takes off get over it

plane takes off people are correct
everyone else is a moron

also .9999999...=1

 

[Fritzo]

Maybe I'm missing part of the question (because I never got in on this the first time around). The plane is on a conveyor, and I'm assuming that the conveyor is moving in the opposite direction of the plane at the exact same speed that the plane is moving forward.

Is this correct? Why do I even care about this?

 

[911paramedic]

? = 0³

 

[MasonLuke]

Originally posted by: BoberFett

Originally posted by: mrSHEiK124

Originally posted by: waggy

Originally posted by: smack Down

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

thats the problem. THIS IS NOTHING LIKE A CAR.

really its not. a car gets its thrust by the wheels. a plane by the thrust of the engines.

For every action, there is an equal and opposite reaction. the thrust of the engine has to have its opposite action. the wheels/treadmill ARE NOT IT.

Thank you, best explanation ever.

THE AIRPLANE TAKES OFF. The wheels moving backwards, forwards, sideways; fuck, take off the wheels and make the damn plane slide across the treadmill, IT WILL TAKE OFF. None of the forces I just mentioned would do ANYTHING to counter the thrust of the jet engines. NOTHING.

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

Thank you. This is the logical answer and the only answer!

 

[hdeck]

you kids are CRAZY. this is turning into what, the 5th thread on this subject?

 

[Fritzo]

Originally posted by: MasonLuke

Originally posted by: BoberFett

Originally posted by: mrSHEiK124

Originally posted by: waggy

Originally posted by: smack Down

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

thats the problem. THIS IS NOTHING LIKE A CAR.

really its not. a car gets its thrust by the wheels. a plane by the thrust of the engines.

For every action, there is an equal and opposite reaction. the thrust of the engine has to have its opposite action. the wheels/treadmill ARE NOT IT.

Thank you, best explanation ever.

THE AIRPLANE TAKES OFF. The wheels moving backwards, forwards, sideways; fuck, take off the wheels and make the damn plane slide across the treadmill, IT WILL TAKE OFF. None of the forces I just mentioned would do ANYTHING to counter the thrust of the jet engines. NOTHING.

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

Thank you. This is the logical answer and the only answer!

That's what I was saying, and three people called me an idiot. It hurt my feelings.

 

[randay]

Originally posted by: Fritzo

Originally posted by: MasonLuke

Originally posted by: BoberFett

Originally posted by: mrSHEiK124

Originally posted by: waggy

Originally posted by: smack Down

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

thats the problem. THIS IS NOTHING LIKE A CAR.

really its not. a car gets its thrust by the wheels. a plane by the thrust of the engines.

For every action, there is an equal and opposite reaction. the thrust of the engine has to have its opposite action. the wheels/treadmill ARE NOT IT.

Thank you, best explanation ever.

THE AIRPLANE TAKES OFF. The wheels moving backwards, forwards, sideways; fuck, take off the wheels and make the damn plane slide across the treadmill, IT WILL TAKE OFF. None of the forces I just mentioned would do ANYTHING to counter the thrust of the jet engines. NOTHING.

Wrong. Until there is sufficient lift to get the plane off it's wheels, it's speed IS it's ground speed and IS dependent on the wheels. Unfortunately when there's a conveyor belt that is matching it's ground speed in the opposite direction, the ground speed relative to the conveyor will rise but it's ground speed relative to the actual ground would be zero. And since the actual ground speed is zero the air speed (the speed of the air running under/over the wings) will never be enough to cause lift.

Thank you. This is the logical answer and the only answer!

That's what I was saying, and three people called me an idiot. It hurt my feelings.

Its because this is a very old topic and everyone is tired of explaining it. Its easier just to call the "not-take-offers" idiots.

The ground speed is not the air speed. They are independant of each other. For instance, there can be a head wind or tail wind and the airplane will take off at different ground speeds but always the same air speed. Say an airplanes air speed must be 100mph to take off. Now add a 100mph head wind from some cold front or hurricane. The airplanes ground speed is 0. Does it have sufficient lift to take off? Yes, and its ground speed is still 0.

 

[bignateyk]

Originally posted by: zanejohnson
holy crap, i thought the people who didnt think it would take off were just joking...

who could be that stupid lol...

of course the plane will take off.

the force that makes a plane take off is the lift, from air rushing under the wings, "lifting" it up....

no matter how fast the conveyer belt moves the thrust from the jets/propellors will STILL push the plane forward, because a plane is not pushed with a driveshaft to the wheels like a car, the wheels have nothing to do with the plane taking off.

it would take very little thrust to keep the plane stationary on the treadmill, and just a little more to push it forward on the treadmill.... and as soon as it starts moving forward air will flow beneath the wings, applying lift.

The whole point of the problem is that the plane is stationary on the treadmill with respect to the ground. No shit it will take off if it starts moving foward on the treadmill. If it is just cruising along at the same speed as the treadmill and not moving foward, there will be no air moving over the wings, and it will have no lift. It wont take off. There will be thrust, but just because there is thrust doesnt mean there is lift.