Will it take off?

[oldtimer]

What you are missing is as I have said before (and others) is that once the thrust (force) is enough to counteract the gravity (drag) of the plane, then more thrust will propel it forward, regardless of how fast things underneath are spinning.

[sputnik]

You were onto something about wheel speed versus plane speed. Now you are going backwards. Could it be that you are confusing an accelerating force (plane's engine acting against air) with a non-accelerating force (spinning conveyor belt)? What part of freely spinning wheels don't you get? Dr.Who had it on the very first page by saying that the wheels would simply spin twice as fast.

Dr. Who did not get it right. If you run on the normal ground at 15 miles an hour, you move forward at 15 miles an hour. If you suddenly jump onto the treadmill (which is going 15 miles an hour contraflow) and you keep running at 15 miles an hour, you go nowhere. That does not mean your feet are now moving 30 mph.

Ah! But you'll say wheels are different than feet. Not at all. Here's why. Put the wheel on the conveyor and run it at 100 mph. The wheel spins at 100 mph. But does it? No, it will not - not unless you apply some force to the axle to keep the wheel stationary. So, let's say you run the conveyor belt with the plane on it. The plane WILL go backwards with the belt UNLESS there is a counter-force. This counter-force can be a rope attached to an imaginary ceiling. In this case, this new force will allow the wheel to spin and have no movement of the plane - forward or backward - IF AND ONLY IF the force equals the opposing force. So, when the rope tied to the ceiling stops the plane from going backward at 100 mph, it is actually pulling the plane forward at 100 mph. Thus, the wheel spins and the plane does not move. Substitute the rope for an equal force of jet propulsion, and you get THE EXACT SAME RESULT.

No, you are missing the point. Force is force is force. There are different causes of force, but it all works together in the same way. Equal opposing forces cancel each other out. It does not matter that you pit a jet engine against a treadmill vs. a car against a treadmill. In both cases, there is a force required to move the thing forward.

Where everyone is getting it wrong is they forget about this little thing called gravity. They assume that if you put a plane on a treadmill and the engines are TURNED OFF, the plane wheels will just sin and spin, and the plane will go nowhere. That is NOT the case. Then, they add little examples about how, with the wheels in place, all it takes is a nominal force to keep the plane stationary on the treadmill. Again, this is wrong. If that plane is on the treadmill - wheel and all - it will move 100 mph backwards if the engine is off. Period. We should ALL agree with that, and if you don't, well....... (it really is that basic).

Now, you say, "Oh, but it only takes a little bitty force to start the wheel spinning and that little, tiny force will allow the plane to sit motionless while the wheel spins and spins. Not true. It takes whatever force is necessary to offset the force of the treadmill. Anybody heard of Newton's law? (I forget which one it is; he had a few).

Jeff,

You're just embarrassing yourself. Instead trying so hard to prove how you're right, consider the possibility that you may be wrong.

[Jeff Matthews]

You all are just plain missing the question. I don't car if it takles a little, tiny force to start the wheel spinning and leave the jet motionless on the treadmill. No matter what force you add to the back end (being a jet engine, a rocket engine or whatever), you have EXACTLY THAT MUCH MORE FORCE THAT WILL BE EMITTED BY THE TREADMILL. Read the question. It says NO MATTER HOW FAST THE PLANE GOES, THE TREADMILL GOES JUST AS FAST THE OTHER WAY.

So, you can open up the burners and really get that plane going, but when you do that, that treadmill is going to use its nuclear engines and go faster than Dallas. It will offset BECAUSE THERE WILL ALWAYS - ALWAYS - ALWAYS BE COMMON POINT OF CONTACT. That point is the wheel and the belt. Until that wheel can outspeed the belt enough - which by definition it cannot - the plane will not move forward and cannot fly.

[Jeff Matthews]

What you are missing is as I have said before (and others) is that once the thrust (force) is enough to counteract the gravity (drag) of the plane, then more thrust will propel it forward, regardless of how fast things underneath are spinning.

No. You are NOT ALLOWED to have enough force to counteract the gravity. That is the way the problem was presented. Read it carefully and think about it some more. That force (which is supposed to counteract the gravity) is constantly offset. Read the question.

[Jeff Matthews]

You were onto something about wheel speed versus plane speed. Now you are going backwards. Could it be that you are confusing an accelerating force (plane's engine acting against air) with a non-accelerating force (spinning conveyor belt)? What part of freely spinning wheels don't you get? Dr.Who had it on the very first page by saying that the wheels would simply spin twice as fast.

Dr. Who did not get it right. If you run on the normal ground at 15 miles an hour, you move forward at 15 miles an hour. If you suddenly jump onto the treadmill (which is going 15 miles an hour contraflow) and you keep running at 15 miles an hour, you go nowhere. That does not mean your feet are now moving 30 mph.

Ah! But you'll say wheels are different than feet. Not at all. Here's why. Put the wheel on the conveyor and run it at 100 mph. The wheel spins at 100 mph. But does it? No, it will not - not unless you apply some force to the axle to keep the wheel stationary. So, let's say you run the conveyor belt with the plane on it. The plane WILL go backwards with the belt UNLESS there is a counter-force. This counter-force can be a rope attached to an imaginary ceiling. In this case, this new force will allow the wheel to spin and have no movement of the plane - forward or backward - IF AND ONLY IF the force equals the opposing force. So, when the rope tied to the ceiling stops the plane from going backward at 100 mph, it is actually pulling the plane forward at 100 mph. Thus, the wheel spins and the plane does not move. Substitute the rope for an equal force of jet propulsion, and you get THE EXACT SAME RESULT.

No, you are missing the point. Force is force is force. There are different causes of force, but it all works together in the same way. Equal opposing forces cancel each other out. It does not matter that you pit a jet engine against a treadmill vs. a car against a treadmill. In both cases, there is a force required to move the thing forward.

Where everyone is getting it wrong is they forget about this little thing called gravity. They assume that if you put a plane on a treadmill and the engines are TURNED OFF, the plane wheels will just sin and spin, and the plane will go nowhere. That is NOT the case. Then, they add little examples about how, with the wheels in place, all it takes is a nominal force to keep the plane stationary on the treadmill. Again, this is wrong. If that plane is on the treadmill - wheel and all - it will move 100 mph backwards if the engine is off. Period. We should ALL agree with that, and if you don't, well....... (it really is that basic).

Now, you say, "Oh, but it only takes a little bitty force to start the wheel spinning and that little, tiny force will allow the plane to sit motionless while the wheel spins and spins. Not true. It takes whatever force is necessary to offset the force of the treadmill. Anybody heard of Newton's law? (I forget which one it is; he had a few).

Jeff,

You're just embarrassing yourself. Instead trying so hard to prove how you're right, consider the possibility that you may be wrong.

You, too, Sput. You are wrong. Consider that possibility. It's Newton's law.

[oldtimer]

"I think I've finally come to understand the problem. There is a finite amount of intelligence in the world, and the population keeps increasing." Scorch points to Ray.

Jeff, you are mistaken that the belt DOES counteract the force. It does not.

[sputnik]

You all are just plain missing the question. I don't car if it takles a little, tiny force to start the wheel spinning and leave the jet motionless on the treadmill. No matter what force you add to the back end (being a jet engine, a rocket engine or whatever), you have EXACTLY THAT MUCH MORE FORCE THAT WILL BE EMITTED BY THE TREADMILL. Read the question. It says NO MATTER HOW FAST THE PLANE GOES, THE TREADMILL GOES JUST AS FAST THE OTHER WAY.

So, you can open up the burners and really get that plane going, but when you do that, that treadmill is going to use its nuclear engines and go faster than Dallas. It will offset BECAUSE THERE WILL ALWAYS - ALWAYS - ALWAYS BE COMMON POINT OF CONTACT. That point is the wheel and the belt. Until that wheel can outspeed the belt enough - which by definition it cannot - the plane will not move forward and cannot fly.

Jeff,

My question again. Simply put, how does any horizontal force from the conveyor acting on the contact patch of the tire get beyond the wheel bearing of the landing gear to oppose thrust? It can't happen - the bearing will not transfer a torque. It just spins the wheel about the hub. Once again the conveyor speed is completely irrelevant. It's a red herring inteded to throw you off.

[Ray Garrison]

The question doesn't say "The belt moves fast enough to hold the plane stationary relative to its surroundings." It says the belt moves as fast as the plane, in the other direction. So if the plane is moving right to left at 50 mph, relative to everything else in the picture, the belt is moving left to right at 50 mph, relative to everything else. Relative to each other, the plane and the belt are moving past each other at 100 mph. If the plane could take off at 50 mph (a Piper Cub?) it would fly away.

[oldtimer]

One last time (I doubt it). How does the belt present increasing gravitational force to the plane? It does not. Just like your friend Newton would explain. A spinning conveyor belt touching the wheels is not the same force as the thrust of the engine on the air, once the threshold of drag has been reached.

[Champagne taste beer budget]

Force is force is force, that much you have right. But there are so many different types of force. And they can act on so many different things, in so many different ways. Would you agree that if a runner were on a treadmill, with a constantly adjustable speed treadmill, and a large hammer, say 30 feet long, with the head being the size of a 55 gallon drum, were pivoted over his head much like a swing on a swing set, if said hammer were let loose from horizontal, so that gravity caused it to arc down, gaining force as it sped up, if said hammer were to strike the runner in the back, would the runner no doubt be knocked off the treadmill? It's a force, one to be reckoned with, and it could care less how fast any stinking treadmill is going. Much like a jet engine.

That happened because a body in motion tends to stay in motion until acted upon by some other force. A body at rest tends to stay at rest. Relative to the hammer, the runner is at rest. Relative to the body of the treadmill, the runner is at rest. The only thing that is in motion in relation to anything is the treadmill itself. The hammer, on the other hand, is in motion. It has force. Or energy. I believe knetic, but that's been a long time ago. If all the forces in this question are acting upon the treadmill only, it would be a different question. Once you throw in the jet engine, which acts upon the air, all bets are off.

[oldtimer]

Here's a suggestion. We pose the question in the form of a myth and send it in to Mythbusters of Discovery Channel fame. They will first make a small scale experiment, then a larger one, hopefully with a way to blow it up at the end.

[sputnik]

I need to take myself out of the game for a while and go Christmas shopping. Jeff, I admire your tenacity in the face of futility. If were guilty of a capital crime, I'd hire you defend my worthless case.

[pauln]

There is a certain kind of physical intuion that some of us have developed through experience, study, and aptitude. This tends to be a fundamentally mathematical sense and allow us to analyse this kind of problem instantly and get the correct answer - which is "yes, it flys".

There is another kind of mis-applied approach to this kind of problem which is just irritating to us who get it - that is the verbal approach well exampled by Jeff (no offense, don't sue me). Here is the difference between the two approaches:

The first approach is a logical approach which means something very specific - that is, absolutely each and every step of the argument must be 100% correct - like the steps in a mathematical proof. One wrong step or error and the entire structure falls. If each connecting idea and principle is correct and logically related the result is confirmed.

In the verbal argument, the goal is to succeed in making a persuasive argument ending in convincing someone of the idea. This appraoch can succeed even if the arguments are wrong or subject to logical fallicies. It is often the case that some or even most of the individual arguments fail, but one or a couple that sound right will succeed in gaining a complete agreement. This approach has it's place in the courtroom where decisions may be based on past rulings and points of existing law, but this appraoch is not sufficienty precise or rigorous to get the right answer to a physical, chemical, mechanical, electrical.etc. type of problem.

The result of this kind of mismatch is State Legislatures attempting to pass laws decreeing the value of pi. http://www.straightdope.com/classics/a3_341.html

[Jeff Matthews]

Hey, Paul. I'm back. Here's the deal.... The plane will not move forward because the treadmill will not let it. How many steps do you need?

I'm still looking via Google for a definitive answer. You'd be surprised at how there is not a quick find on this topic. All I see are arguments both way, with both sides accusing each other of being idiots. And these opposing arguments span many, many websites/forums/blogs, etc.

[Jeff Matthews]

I followed some posts until I bumped into this guy's reasoning. The question, according to him, calls for an imprecise interpretation, so he posits a few:

Different people are getting different answers to this question because the basic assumptions can be interpreted in different ways.

Suppose we actually built a treadmill like that described in the OP and put a 747 on it. Would the 747 take off? If "exactly matching the speed of the wheels" means that the treadmill matches the hub speed of the wheels, then yes. The plane takes off, and the treadmill's going twice the speed in the opposite direction.

OK, as Manduck say, that problem is trivial. Let's assume that "exactly matching the speed of the wheels" means "matching the outer diameter surface velocity" Would the 747 take off? Almost certainly it would, but only because the treadmill wouldn't be able to keep up with the thrust transmitted to the plane by the engines--in other words, we violate the spirit of the question, because the treadmill isn't matching the wheel velocity.

OK, that's stupid. It's a thought experiment. Posit a magic treadmill that can accelerate as fast as desired. All right. Multiple things could happen. To begin with, the plane would stay stationary as the thrust power was dissipated in the wheel bearings (as friction), tires (hysteresis), and in accelerating the wheel to ever-increasing speeds. Since all the power is dissipated in the wheels, eventually either the bearings would overheat, the tires would blow, or the wheel would rip itself apart due to inertial forces, and the plane crashes and burns. Then you've destroyed a rather expensive magic treadmill.

Thought experiment, I said! Let's posit ultra-strong and heat resistant tires. All right. Again, multiple things. If the treadmill is a long, runway-sized treadmill, it will eventually, running thousands of miles an hour, pull in air at high enough velocity that the plane will lift off at zero ground speed (but substantial air speed). However, now you're running into trans-sonic compressibility effects...

No speed of sound effects! And assume magic air that doesn't become entrained with the treadmill motion. In that case, the treadmill speeds up until it and the wheels are running near light speed, and relativistic effects takes over. The wheels get smaller, I suppose...

None of that! No relativity-- Hey, wait a minute. Back up. Suppose we have zero friction bearings and tires. And while we're at it, make them massless, so there's no inertial effects. Well, zero friction tires would mean they just skid on the runway, since nothing turns them. So the plane will take off, tires motionless, and the treadmill won't move.

OK, then. There's friction between the tires and treadmill, but not in the bearings or sidewall. No energy is lost in the wheels and tires, in other words. Now you've got an unstable runaway system. There's no resistance to treadmill motion, and a positive feedback circuit. Imagine, now, the poor mechanic who bumps a wheel, setting it in motion. A very slight roll by the tire is sensed, and the treadmill luches forward. The tire goes faster, the treadmill goes faster, the tire goes faster.... Since we've posited an instantly-accelerating treadmill and no relativity and no air resistance and no wheel inertia, the treadmill goes from zero to infinity in no time flat. Try to keep your balance on that. The same thing happens, of course, when the engines light off. There's nothing coupling the plane to the treadmill--no bearing friction, no inertial effects, no air resistance, and no way for the treadmill . So the plane takes off, leaving the infinite-speed treadmill behind.

[sputnik]

There is a certain kind of physical intuion that some of us have developed through experience, study, and aptitude. This tends to be a fundamentally mathematical sense and allow us to analyse this kind of problem instantly and get the correct answer - which is "yes, it flys".

There is another kind of mis-applied approach to this kind of problem which is just irritating to us who get it - that is the verbal approach well exampled by Jeff (no offense, don't sue me). Here is the difference between the two approaches:

The first approach is a logical approach which means something very specific - that is, absolutely each and every step of the argument must be 100% correct - like the steps in a mathematical proof. One wrong step or error and the entire structure falls. If each connecting idea and principle is correct and logically related the result is confirmed.

In the verbal argument, the goal is to succeed in making a persuasive argument ending in convincing someone of the idea. This appraoch can succeed even if the arguments are wrong or subject to logical fallicies. It is often the case that some or even most of the individual arguments fail, but one or a couple that sound right will succeed in gaining a complete agreement. This approach has it's place in the courtroom where decisions may be based on past rulings and points of existing law, but this appraoch is not sufficienty precise or rigorous to get the right answer to a physical, chemical, mechanical, electrical.etc. type of problem.

The result of this kind of mismatch is State Legislatures attempting to pass laws decreeing the value of pi. http://www.straightdope.com/classics/a3_341.html

Paul,

That was very well put. I think this thread is very illuminating - it reveals a lot about how we analyze problems, communicate ideas, and face disagreement. In this case there is a clear correct answer. Those of us with engineering and science backgrounds can sift through the problem easily. Those that have other backgrounds must devise their own means of understanding the physics involved. I admire those people without a technical background that stuck with the problem and figured it out in their own way - Phil Mays and dwilawyer come to mind. I'm sure that Jeff will get it eventually.

What I find interesting is that this is an objective problem with a definite correct answer yet it is debated as hotly as a subjective argument over religion or politics. How many have just jumped in without reading through previous posts and confidently defended the incorrect answer or simply disregarded clear explanation in favor of defending the intuitive conclusion that they felt comfortable with. If we have this much trouble communicating objective concepts with clear answers, how can we ever hope to have a real exchange of ideas in areas where the answers (if there are any) aren't so clear.

[Jeff Matthews]

Here's some more:

Quote:

Originally Posted by scr4

If the tire and treadmill surfaces are not frictionless (i.e. tire is not slipping across the treadmill), then there will be horizontal force, even if the bearing is frictionless. Because a bearing does not need friction to transmit horizontal force.

Quote:

Originally Posted by Eyer8

Not true.

Yes, true.

How else would automobiles move? The traction force from the pavement on the tire is transmitted to the axle through the wheel bearing. Or the other way around, if you prefer.

Same principle.

[Champagne taste beer budget]

The folks on this forum achieved consensus a while ago, I believe. It would fly.

Then some pesky Bob guy brought the ugly head back to life again, and started another 4 pages of "stuff". (No offense to BEC, I loved the pic and the link I got from it.)

Once again, in B.A.S.I.C.

I am standing on a treadmill with rollerskates on my feet.

As fast as I skate, the treadmill keeps up.

I strap a rocket to my a$$.

I ignite the rocket.

I fly off the front of the treadmill, hitting the wall, likely breaking every bone I have as I go through the wall, and if I do NOT go through the wall, I get myself burnt to oblivion.

The treadmill didn't know I had a rocket; it was just trying to keep up to my speed of skating.

REMEMBER!! The speed of the conveyor matches that of the wheels, not the other way around.

[Jeff Matthews]

You all are just plain missing the question. I don't car if it takles a little, tiny force to start the wheel spinning and leave the jet motionless on the treadmill. No matter what force you add to the back end (being a jet engine, a rocket engine or whatever), you have EXACTLY THAT MUCH MORE FORCE THAT WILL BE EMITTED BY THE TREADMILL. Read the question. It says NO MATTER HOW FAST THE PLANE GOES, THE TREADMILL GOES JUST AS FAST THE OTHER WAY.

So, you can open up the burners and really get that plane going, but when you do that, that treadmill is going to use its nuclear engines and go faster than Dallas. It will offset BECAUSE THERE WILL ALWAYS - ALWAYS - ALWAYS BE COMMON POINT OF CONTACT. That point is the wheel and the belt. Until that wheel can outspeed the belt enough - which by definition it cannot - the plane will not move forward and cannot fly.

Jeff,

My question again. Simply put, how does any horizontal force from the conveyor acting on the contact patch of the tire get beyond the wheel bearing of the landing gear to oppose thrust? It can't happen - the bearing will not transfer a torque. It just spins the wheel about the hub. Once again the conveyor speed is completely irrelevant. It's a red herring inteded to throw you off.

Sput, here's your answer. The guy lays out all the physics formulas. The force to spin the wheel is transmitted to the plane.

All right, then. Apropos of a number of posts in this thread (no *single* post, but the cumulation of multiple posts), let's talk about forces and force transfer. Let's talk about forces in the context of, say, a wheel:

Code:

*******

** **

** **

* *

* *

* *

* O *

* *

* *

* *

** **

** **

*******

The wheel is round, and it has some mass, m, with a center of mass in the center of the wheel at "O". It's also got rotational inertia, I. One way to think of mass is as the resistance of the wheel to motion when a force is applied. In the same way, rotational inertia can be thought of as the resistance of the wheel to rotation when a torque is applied. With me so far?

In addition, let's assume that the wheel has a bearing in the center, at the "O". The other side of the bearing is connected to Something. We haven't defined what Something is, yet, but we will. Be patient. Let's make this simple and assume the bearing is frictionless. We can add friction in later if we want, but let's keep it simple for now. Now, the bearing is a pin joint--it allows both horizontal and vertical forces between the wheel and the Something that it's connected to, but does not allow torques to be transferred.

One last thing. Let's make this wheel and the associated forces as simple as possible. So let's ignore gravity and everything else. No gravity, no treadmill, no Earth, no nothing. This is, right now, just a wheel in space connected to Something through the bearing. OK. Got it?

Now let's change the story.

Code:

*******

** **

** **

* *

* *

* *

* O *<----F

* *

* *

* *

** **

** **

*******

I've added a force F on the wheel; this force F passes through the center of mass. This force is, as drawn, unbalanced, and it isn't yet the whole story. From here on out, a few different things could happen.

Suppose the wheel isn't connected to anything at the bearing (the Something is Nothing, in other words). In this case, there's no way the force F can be resisted, and the entire wheel will accelerate to the left. The acceleration, in fact, will obey Newton's law, F = ma.

OK, but what if the wheel is actually connected to Something? What if that Something was the ground, and the wheel is prevented from actually moving to the left? In that case, there will be a reaction force at the bearing. The wheel will push on the ground, through the bearing, and the ground will push back:

Code:

*******

** **

** **

* *

* *

* *

* R---->O *<----F

* *

* *

* *

** **

** **

*******

The ground will push back, in fact, so that R exactly equals F. This is the key: because the wheel does not move, the forces must be add to zero. If they didn't add to zero, the wheel would accelerate to the left or right. But it doesn't. So the forces must be balanced, and the force F is transmitted to the ground.

Another possible thing that could happen would be that the Something that the wheel is connected to is something very massive, but moveable, which has a mass M. In that case, as soon as the force F is applied, everything accelerates--both the wheel and the Something--at an acceleration of F = (M+m)a. In this case, there will still be a reaction R on the bearing, but the reaction R will be less than F. How much less is left as an exercise to the reader.

All right, got all that? Now let's do something a little different. Let's move the force:

Code:

*******

** **

** **

* *

* *

* *

* O *

* *

* *

* *

** **

** **

*******<----F

Now the force F is no longer a the center of mass. How does that change things?

First or all, let's go back to the assumption that the wheel is just free in space--nothing is connected to the bearing. In this case, the wheel will still accelerate to the left in accordance with F = Ma. The force F is still unbalanced, and the unbalanced force must result in an acceleration. However, the wheel will also rotationally accelerate. The force F is not through the center of mass, and that creates a torque T = Fr. This torque rotationally accelerates the wheel through the rotational equivalent of Newton's law: T = Ia (a being rotational acceleration). So the wheel will begin to move to the left, and spin clockwise.

Now let's assume, instead, that the wheel is connected to the ground at the bearing--the Something is the ground again. This gives a case like so:

Code:

*******

** **

** **

* *

* *

* *

* R---->O *

* *

* *

* *

** **

** **

*******<----F

Here's the key question: does the center of mass of the wheel accelerate to the left? No! it's connected to the ground! It can't move to the left. And because it can't move to the left, the forces must add to zero, and the force R equals the force F.

Of course, the difference between this case and the one above where the force F was pointed toward the center of mass is that now the wheel will spin. And we can calculate exactly how much it will spin: the two forces set up a torque T = Fr, which rotationally accelerates the wheel through the rotational equivalent of Newton's law: T = Ia.

So all that leads to the real question: how does this apply to an airplane? The point is, to keep an airplane from moving, you need to apply a force F to the plane that is equal and opposite the engine thrust. Assuming the plane won't flip end-for-end or anything silly like that (adding back gravity and so forth), it doesn't matter if that force is applied to the nose of the airplane, or the tail, or the landing gear.

It doesn't even matter if the force is applied at the bottom of the tires. That force will still be transmitted to the plane and oppose the engine force. Given that this is a thought experiment, it is possible to apply an opposing force to keep the plane in place via the tires. However, this will come at the cost of rotationally accelerating the wheels. And that acceleration is apt to be very large.

[Jeff Matthews]

Okay, Sput and CTBB, I gave you some very detailed posts (not my own math for sure, but I think they set out the problem fairly well and under some scenarios debunk your argument). At best, maybe we are thinking of different scenarios. Did you read these posts? What do you think?