A Plane Problem

[Ste]

Some of you have probably seen this before.

"A plane equipped with fixed horizontal engines and wheel landing gear is placed on a huge treadmill runway. The treadmill has a clever design and always matches the speed of the plane, but runs in the opposite direction. Will the plane take off and fly or not?"

A crappy diagram



It's confused me for a while but I think I have my answer...

[Knight Theamion]

first response: it won't take off, because he has no speed, there is no lift


which is wrong, it will take off, which is explained later.

[MrL_JaKiri]

Of course it will. It'd take off in exactly the same way, assuming that the wheels are ideal (ie have zero friction).

The only reason it wouldn't take off (as theamion claims) was if the impetus to get moving was provided by the wheels, which is not the case.

[Tomkat]

There's no forces from the air running against the plane. The plane needs the oppostie resistance from the air to be able to take off.

My understanding of the technical words to describe how it works isn't that great, but I know roughly how planes take off. There's no air circulation over the wings (as it's exactly stationary, due to the treadmill moving at EXACTLY the same speed as the wheels), so it would just keep rolling on and on and on and on.

[MrL_JaKiri]

Originally Posted by Tomkat

There's no forces from the air running against the plane. The plane needs the oppostie resistance from the air to be able to take off.

Why would it be stationary relative to the air?

[Tomkat]

Originally Posted by MrL_JaKiri

Why would it be stationary relative to the air?

Because Ste said the treadmill works in a way where the plane never moves forward - whatever speed it goes out, it is counteracted by the treadmill's speed.

[MrL_JaKiri]

Originally Posted by Tomkat

Because Ste said the treadmill works in a way where the plane never moves forward - whatever speed it goes out, it is counteracted by the treadmill's speed.

Why would it be counteracted by the treadmill's speed? What you said would be true for a car, not for a plane.

[Tomkat]

Originally Posted by MrL_JaKiri

Why would it be counteracted by the treadmill's speed? What you said would be true for a car, not for a plane.

How can the plane push air downwards beneath its wings if it's standing still?

[Ste]

I was hoping Jakiri would take longer to post

I agree with him btw...

[MrL_JaKiri]

Originally Posted by Ste

I was hoping Jakiri would take longer to post

I agree with him btw...

I can delete my posts if you prefer!

[Ste]

Originally Posted by MrL_JaKiri

I can delete my posts if you prefer!

Nah it's ok - So long as people have a go at the problem before reading the posts below I'm sure it'll be ok

(Or just don't trust that we're correct)

[lokken]

I get the feeling it would take off but not very safely as there will be force against the plane to generate lift.

[wu_trax]

tomkat is right.

[deerbarn]

Someone explain this please. I can't see how tomkat could possibly be wrong, but i have a sneaky feeling he is. Im confused.

[deerbarn]

Oh its ok I get you both now.

[Ste]

The wheels are simply moving twice as fast.

The plane itself isn't affected if you assume the friction between the wheels and the plane is zero.

The plane will move forward and will take off when it reaches the required air speed.

[Tomkat]

Originally Posted by Ste

The wheels are simply moving twice as fast.

The plane itself isn't affected if you assume the friction between the wheels and the plane is zero.

The plane will move forward and will take off when it reaches the required air speed.

Why would you assume the friction between the wheels and the plane is zero?

There would be no air speed. The treadmill stops the plane from moving forward.

You have to read the initial question again. This isn't a regular treadmill, that the plane can eventually accelerate against and take off. It is being held stationary by the magical super forces of exact speed.

[MrL_JaKiri]

Originally Posted by Tomkat

Why would you assume the friction between the wheels and the plane is zero?

Because it's the simplest way to model it. You don't have to, which is why I demonstrated that under any consideration of real materials, ie not a frictionless world, it can still take off.

Originally Posted by Tomkat

There would be no air speed. The treadmill stops the plane from moving forward.

Stop saying this. Please.

How are you modelling friction?

Originally Posted by Tomkat

You have to read the initial question again. This isn't a regular treadmill, that the plane can eventually accelerate against and take off. It is being held stationary by the magical super forces of exact speed.

Oh gods.

[MrL_JaKiri]

Lets go over this in detail.

Let us first assume that there is a first order relationship between friction and velocity (that's y=mx + c, where y is friction, x is velocity and m and c are the coefficients of kinetic and static friction respectively, for all you GCSE maths types out there)..

This is a pretty safe assumption, because it's true.

Next: if you have a static force on an object, then it will take the same amount of time and distance to accelerate to a velocity as the object would to come to a stop from that velocity when the force opposes the movement.

For the next bit, I'll explain it as if the forces were constant (which is fine, the net impulses aren't changed by assuming this, and we're only effectively dealing with the plane at two points, when it's stopped and when it's taking off, so we're not losing any particularly useful information).

Lets call the force from the engines E.

Let us call the (constant in this example) friction force F.

The net force when taking off is E - F.

The net force when landing, from the wheels alone (no breaks, or flaps or whatnot) is F.

Now, let us assume that the plane doesn't need breaks or flaps when landing. It just stops of its own accord.

In this case E - F = F

E = 2F.

Would it take off on a rolling runway made of the same materials? Lets investigate!

The friction between the wheels and the track works as if the velocity was twice what it usually is, so y(2) = 2mx + c, not just mx + c. However, if c is greater than 0, which it is, y(2) is less than y, because 2mx + c is less than 2mx + c + c, obviously.

So, the friction in that case would be less than 2F, so there would still be some resultant force on the plane, so it would still accelerate. It would still take off, albeit over a very long time if c is small.

Remember that this is assuming that a plane will just roll to a halt once you've got it down on the ground.

Do I really have to go into the real world, where you have to use breaks and whatnot, in which case E - F = F + K, where K is the retardation force provided by the breaks (etc), and is in all likelihood larger than F?

Please say I don't.

[Tomkat]

Originally Posted by MrL_JaKiri

Lets call the force from the engines E.

Let us call the (constant in this example) friction force F.

The net force when taking off is E - F.

The net force when landing, from the wheels alone (no breaks, or flaps or whatnot) is F.

Now, let us assume that the plane doesn't need breaks or flaps when landing. It just stops of its own accord.

In this case E - F = F

E = 2F.

As the treadmill is moving at the exact speed, then it is exerting exactly the same force as the engines are thrusting, so surely E = F.

[MrL_JaKiri]

Originally Posted by Tomkat

As the treadmill is moving at the exact speed, then it is exerting exactly the same force as the engines are thrusting, so surely E = F.

You clearly haven't understood what I posted, as the F is the friction between the plane and a stationary runway.

[Tomkat]

Originally Posted by MrL_JaKiri

You clearly haven't understood what I posted, as the F is the friction between the plane and a stationary runway.

Ok, so then E = F + T

Where T is the force exerted by the treadmill on top of friction (F) to counteract whatever forces the engines can exert (E).

[MrL_JaKiri]

Originally Posted by Tomkat

Ok, so then E = F + T

Where T is the force exerted by the treadmill on top of friction (F) to counteract whatever forces the engines can exert (E).

Originally Posted by MEEE AGAIN

The friction between the wheels and the track works as if the velocity was twice what it usually is, so y(2) = 2mx + c, not just mx + c. However, if c is greater than 0, which it is, y(2) is less than y, because 2mx + c is less than 2mx + c + c, obviously.

So, the friction in that case would be less than 2F, so there would still be some resultant force on the plane, so it would still accelerate. It would still take off, albeit over a very long time if c is small.

[Arachnidman]

There was no need for that.

The frictional force between the wheel and the track that you go on about is irrelevant - that is not what halts a rolling plane.

It is the frictional force between the wheel and the axel that is of importance, and that you neglect to talk about.

Originally Posted by MrL_JaKiri

Lets go over this in detail.

Let us first assume that there is a first order relationship between friction and velocity (that's y=mx + c, where y is friction, x is velocity and m and c are the coefficients of kinetic and static friction respectively, for all you GCSE maths types out there)..

This is a pretty safe assumption, because it's true.

Next: if you have a static force on an object, then it will take the same amount of time and distance to accelerate to a velocity as the object would to come to a stop from that velocity when the force opposes the movement.

For the next bit, I'll explain it as if the forces were constant (which is fine, the net impulses aren't changed by assuming this, and we're only effectively dealing with the plane at two points, when it's stopped and when it's taking off, so we're not losing any particularly useful information).

Lets call the force from the engines E.

Let us call the (constant in this example) friction force F.

The net force when taking off is E - F.

The net force when landing, from the wheels alone (no breaks, or flaps or whatnot) is F.

Now, let us assume that the plane doesn't need breaks or flaps when landing. It just stops of its own accord.

In this case E - F = F

E = 2F.

Would it take off on a rolling runway made of the same materials? Lets investigate!

The friction between the wheels and the track works as if the velocity was twice what it usually is, so y(2) = 2mx + c, not just mx + c. However, if c is greater than 0, which it is, y(2) is less than y, because 2mx + c is less than 2mx + c + c, obviously.

So, the friction in that case would be less than 2F, so there would still be some resultant force on the plane, so it would still accelerate. It would still take off, albeit over a very long time if c is small.

Remember that this is assuming that a plane will just roll to a halt once you've got it down on the ground.

Do I really have to go into the real world, where you have to use breaks and whatnot, in which case E - F = F + K, where K is the retardation force provided by the breaks (etc), and is in all likelihood larger than F?

Please say I don't.

[All Systems Go]

Originally Posted by Arachnidman

There was no need for that.

there was (and still is) no need for this thread to be four pages long and counting.

[MrL_JaKiri]

Originally Posted by Arachnidman

There was no need for that.

The frictional force between the wheel and the track that you go on about is irrelevant - that is not what halts a rolling plane.

It is the frictional force between the wheel and the axel that is of importance, and that you neglect to talk about.

I didn't neglect it. I included it with the other sources of (speed of track dependent) friction.

In modelling terms, there's no difference between the friction between the road and the wheel, and the friction twixt wheel and axel, if they increase with respect to velocity in the same way, which they do.

Like I said, more elegent.

[Arachnidman]

You do realize that rolling resistance (a static frictional force) is independant of velocity don't you?

You are confusing this with the viscal frictional force caused by air resistance, which indeed is a function of velocity.

Originally Posted by MrL_JaKiri

I didn't neglect it. I included it with the other sources of (speed of track dependent) friction.

In modelling terms, there's no difference between the friction between the road and the wheel, and the friction twixt wheel and axel, if they increase with respect to velocity in the same way, which they do.

Like I said, more elegent.

[MrL_JaKiri]

You do realise it makes more sense to put what you're quoting first?

Originally Posted by Arachnidman

You do realize that rolling resistance (a static frictional force) is independant of velocity don't you?

You are confusing this with the viscal frictional force caused by air resistance, which indeed is a function of velocity.

My argument doesn't work without considering static frictional forces, and the fact that I took them into account without making a comparison of scale of static and kinetic frictional forces once again demonstrates to me that my argument was more elegent.

[Tomkat]

I just thought of an example that might help people see how I'm thinking (I admit I might be wrong, but then the entire thing is hypothetical anyway).

If you're running on a road, then you feel the air blow against your face. Because you're pushing through it.

If you're running on a treadmill, you feel no air blow against your face. Because you aren't pushing against it (ignoring the air being moved round by your arms and legs moving).

With no significant air movement, the plane wouldn't be able to take off.


I dunno if that helps you see how I'm seeing the problem, but hopefully it might.

[MrL_JaKiri]

Originally Posted by Tomkat

I just thought of an example that might help people see how I'm thinking (I admit I might be wrong, but then the entire thing is hypothetical anyway).

If you're running on a road, then you feel the air blow against your face. Because you're pushing through it.

If you're running on a treadmill, you feel no air blow against your face. Because you aren't pushing against it (ignoring the air being moved round by your arms and legs moving).

With no significant air movement, the plane wouldn't be able to take off.


I dunno if that helps you see how I'm seeing the problem, but hopefully it might.

Whether or not there is resultant airflow is the answer of the question, not what you have to base your argument upon.

[deerbarn]

Originally Posted by MrL_JaKiri

Whether or not there is resultant airflow is the answer of the question, not what you have to base your argument upon.

No. How easy was that!

[MrL_JaKiri]

Originally Posted by horn

how many posts should it take a good teacher to get this across to the children tomkat? you're rubbish!

I hope this is just a teaching exercise to make me think about the problem in more detail!

[Mek]

must admit i agree with tomkat here. the treadmill is controlling the speed of the ground the plane traveling on is moving at, it is not controlling the air around it.

as tomkat says the airplane takes off by manipulating the air around it to create lift. the air around the plane whilst on the treadmill is essentially stationary...right?

[MrL_JaKiri]

I give up. I prove it with actual proper maths and you still don't believe me.

Did you know that 0.9 recurring is equivilent to 1?

[Phil^]

im in agreement with tomkat, there would be no forward velocity relative to the air around the plane, and thus no lift.
Think of it this way. If you are running on a treadmill which matches your speed, do you still feel the same air resistance against your face as you would if you were running on ground?

[Tomkat]

I think the main problem we have here is that Mark is thinking about this in a realistic sense, in which case the treadmill could never possibly exert the exact same force back. In that case, yeah, the plane would probably eventually build up enough speed and power to pull off the treadmill, and have accumulated enough to be able to take off.

But in this hypothetical situation, the treadmill can move at an infinite speed, thus preventing the plane from moving anywhere, no matter how much force it exerts from its engines.

[MrL_JaKiri]

Originally Posted by Tomkat

I think the main problem we have here is that Mark is thinking about this in a realistic sense, in which case the treadmill could never possibly exert the exact same force back.

No.

Originally Posted by MEEEEE

Strictly speaking, the problem is meaningless if you don't assume it can take off - because if you believe it makes it so the plane can't take off because the air is stationary around it, the speed of the plane is zero and thus the ground isn't moving. Of course, instantaneously, the plane has a speed relative to the air again, and so you get an infinitely fast progression.

The only sensible way of looking at it, if you want to consider what people think it means, is that if the ground has the same speed as the speed of the outer surface of the wheels.

Originally Posted by Tomkat

But in this hypothetical situation, the treadmill can move at an infinite speed, thus preventing the plane from moving anywhere, no matter how much force it exerts from its engines.

You still don't appear to know how planes work, or how friction works, or both.

[Phil^]

Actually, my analogy has given me food for thought.
In the running on treadmill example, the force is applied against the ground ( running ) , with a plane that force is applied to the air.

Im back to the "hmm, not sure" stage. It would need to be shown through experiment for me

perhaps the root of the confusion is in the wheels, so lets remove those from the equation.
Assume the plane's wheels and the threadmill surface are replaced with magnets of equal charge ( therefore the plane hovers ).
Would it take off now? I suspect it would

[Tomkat]

I can't really emphasise my argument without repeating myself again and again so :shrug: .

I found this though:

Originally Posted by http://www.meignorant.com

The aircraft will not fly. Wheels don't make an aircraft fly, nor does thrust (look at a glider). The force that makes an aircraft fly is the air that passes over the wings creating lift (air moves at more velocity over the top of the wing than the lower surface, creating a low pressure area - therefore, lift).

The treadmill always moves at the same speed as the aircraft - so the aircraft can never gain forward momentum in relation to the static (unmoving) neighboring air, therefore the wings can not create lift. In effect, the treadmill, the wheels and the engines don't even matter in the 'problem' - if the aircraft can not move in relation to the air (and create lift) they might not as well even be moving. Even an aircraft using very heavily 'blown' wings and flaps (using engine bleed air) require at least a small amount of forward motion to create the extra lift which is used for very low speed takeoffs. (Search the NASA site for more details...)

Simply put it's not an aircraft's engines that make it fly, it's the result of the movement of the air over the wings (the engines simply move the aircraft forward, creating that movement.) No movement no lift. No lift, no fly.

Mike Kopack
FAA Airframe and Powerplant Technician
EX USAF F-16A/B/C/D Crew Chief
http://www.lucky-devils.net

[MrL_JaKiri]

Originally Posted by Tomkat

I can't really emphasise my argument without repeating myself again and again so :shrug: .

You don't have an argument. You just state it won't happen.

I have an argument. You can't have read it, because it really was very simple.

Originally Posted by Tomkat

I found this though:

He's wrong, as I proved on the previous page.

Come on guys, MATHS IS NOT OPINION BASED.

[edit]

If you want to make an argument for your position, explain why the plane would be stationary. Don't just say that it is.

[Deepflow]

The plane will not remain stationary, it will take off.

\o/

[Tomkat]

After a lengthy discussion with Mark being very patient and understanding ( ) I now see his point.

The thing is that most of the people agreeing with him (Ste and a few others) aren't actually saying the same thing as him...

The way I imagined it (for those stupid enough to agree with me!) is that the runway isn't necessarily tarmac. Imagine it as a big block of ice that the plane is moving against. Eventually the plane will speed up and take off (assuming that the treadmill is long enough).

When I said it didn't take off I assumed there was enough friction exerted constantly against the plane to keep it stationary. The question doesn't mention friction so you can't assume that really.

[MrL_JaKiri]

A plane will take off faster than usual from a big block of ice. The wheels are only there to reduce the friction against the tarmac, and ice has less friction than tarmac.

[Tomkat]

Originally Posted by MrL_JaKiri

A plane will take off faster than usual from a big block of ice. The wheels are only there to reduce the friction against the tarmac, and ice has less friction than tarmac.

Look bucko I only just agreed with you about it taking off AT ALL. Don't push your luck about it taking off fast or slow or whatever.

[Dace]

The plane doesnt move (in relation to a fixed point to the side of the treadmill) so there's no lift so it doesn't take off.

IMO anyways.

[Dace]

Tom you started off correct and now it's all gone terribly wrong for you ... ;'(

[Tomkat]

Originally Posted by Dace

Tom you started off correct and now it's all gone terribly wrong for you ... ;'(

I did, but only if you assume the treadmill is made of something that provides enough friction to keep the plane stationary on it (tarmac would do). If you think about it though, if the treadmill is made of something like ice or glass or something, then the plane would eventually gather enough momentum to start accelerating.

Obviously you are meant to assume that it DOES have enough friction (which is what my entire argument was based on). As it doesn't say that though, we can't assume it.

I wasn't wrong, but I also wasn't right, as I assumed things that shouldn't be assumed.

[MrL_JaKiri]

Originally Posted by Tomkat

I wasn't wrong

Yes you were, the maths post on the last page showed that.

[Tomkat]

Originally Posted by MrL_JaKiri

Yes you were, the maths post on the last page showed that.

I think that for the purpose of the poorly worded question, we aren't supposed to take friction into account. Just the idea of a plane moving forward and being pulled back at the same time.

I was right if we assume the friction is high enough to prevent the plane from moving forward. What Ste and others thought is that the plane could still take off from the thrust in its engines, even though it wouldn't be moving anywhere. Which it couldn't :crymeariver:

[Structural Integrity]

This thread sucks.
It started with a silly trick question, and now we're at a point where people start throwing in formula's and Tomkat is totally losing it.
World wars have been fought over less!