billoday.net » Blog Archive » Airplane-Treadmill Problem Solution

Airplane-Treadmill Problem Solution - Nihilist

This is not what all the talking heads are saying, but I’m almost positive this would be the end result of this mental exercise. For the uninitiated, look at BoingBoing:

David Pogue at the NYT has presented this classic airplane on a giant treadmill problem, and people are arguing about whether or not the plane would take off or not. Here’s the problem:

“Imagine a plane is sitting on a massive conveyor belt, as wide and as long as a runway. The conveyer belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off? “I say no, because the plane will not move relative the the ground and air, and thus, very little air will flow over the wings. However, other people are convinced that since the wheels of a plane are free spinning, and not powered by the engines, and the engines provide thrust against the air, that somehow that makes a difference and air will flow over the wing.”

I say yes. Let’s assume the friction in the wheel bearings is negligible. Putting a plane on a treadmill is like putting it on an icy lake. When you fire up the jets, the plane is going to shoot down the lake and take off just like it would on a runway.

The most important issue is that even if the the friction is negligiblein most situations, it does exist. Therefore it is important in this case. The detail being missed by most people is the way the question is set up: the treadmill is set to match the speed of the wheels, not the land speed (actual or normal) of the aircraft. So, in order to answer this question you have to get technical. Let us assume that there is enough tread on the wheels so that the friction of the wheels does matter (meaning that when the plane goes forward, the wheels move with it). This is to defeat the “icy lake” hypothesis which is based on the fact that one an icy lake, the wheels don’t turn at the land speed, mainly because the plane is in effect in a long slide. Here’s the solution:

The engines start up creating a forward force in the upper half of the plane body, because the land speed is not yet big enough to create lift and the engines are parallel to the ground, this force pushes down onto the wheels with the help of our friend gravity.
The wheels move as the plane’s body begins to be acted upon that force.
The treadmill matched the speed of the wheels movement, creating a backwards force that initially matches the forward force.
2 and 3 get into a bit of a Cold War era armament situation: because the forward force is not generated by the wheels, the wheels spin free, meaning their speed is increased beyond what it would take in a standard situation to take off. The treadmill speed matches the wheel speed. Now, remember that in order for the plane to move forward and the wheels to meet the friction needed in the axiom (because the treadmill is a surface that would give traction, there is friction) the wheels would need to spin faster than the treadmill for forward momentum. Therefore the backward force generated by the treadmill would get significantly higher. So rinse and repeat until 5.
The treadmill is designed (by nature of the problem) to be able to handle infinitely high speeds. The wheels, being true physical objects have a failure speed. The wheels fail, the plane makes physical contact with the treadmill, and is thrown back into the airport. Everyone dies.

The end result before death would be the forward force remaining relatively constant and the backward force being several orders of magnitude stronger. Which of course would beg the question, why don’t we use this magical treadmill to launch these planes into the air. If it generates so much force, short flights wouldn’t even need engines…

EDIT: I suppose that I should directly address Mark’s point. The wheels in a normal situation act as the icy lake, they remove the friction from the ground for the plane as a whole. What happens with the treadmill, is that by moving at would be around an infinite speed (hence the implausibility of this problem) it would in fact create a whole lot of friction for the plane, making it the exact opposite of the frozen lake. Oh and before anyone attacks my idea as overly simplistic, I forgot to give a few assumptions:

1-5 occur almost instantly.
The wheels won’t slip. If they do, the plane will take off or a wheel will blow, killing everyone. If it takes off, I wouldn’t want to be on the plane. We still need that wheel when we land.
If you ignore the practical limits of airplanes, this question becomes totally meaningless, as both the wheel and the treadmill will have broken the speed of light and saved Lois Lane from an earthquake. Therefore, they both heroes and should be left alone for their retirement.
Either the tire will blow or the bearings will break: the plane will never take off, only crash and be flung back into the airport.

This entry was posted on Tuesday, December 12th, 2006 at 7:37 am and is filed under Writings. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

40 Responses to “Airplane-Treadmill Problem Solution - Nihilist”

Ken Says:
December 12th, 2006 at 10:07 am
I’d say no, the plane would not take off, even with no equipment failures. The air sucked in by the turbines is not enough to create a localized flow over the wing. I’m also guessing that the air disturbed by the treadmill runway is also not enough, and that the plane’s wings are well above the boundary condition.

So what do we have left? A plane traveling at zero miles an hour. No lift. The jets would reach their maximum output (equipment failures notwithstanding) and remain there unless you removed ALL friction - wheel to ground, wheel to axle, engine to localized air flow, turbine blades to air, etc. Unless all friction sources are discounted, infinity speed could not be reached.
Brian Says:
December 12th, 2006 at 10:15 am
Lift is a result of air moving over the wings. This is usually created by the forward motion of the plane. If there is no forard motion in this treadmill scenario, I don’t see where the lift is going to come from.
Brian Says:
December 12th, 2006 at 10:24 am
I suppose if it’s a front mounted, prop driven airplane, that would create enough airflow over the wings for lift. However, a rear jet propelled aircraft wouldn’t have the same effect.
poopdog Says:
December 12th, 2006 at 10:25 am
This is very simple.

When you run on a treadmill, do you feel the wind on your face?

No? No flying then.
Gareth Says:
December 12th, 2006 at 10:39 am
You people are retarded. The question itself is retarded.

When the engines produce thrust, the body of the plane begins to move forward due to the force acting against it. The wheels only move forward because ground provides friction against which they turn. If this super-retarded treadmill matches the movement of the wheels perfectly, they will not turn but instead stay still - however the plane itself will move forward because the thrust is still there. So the treadmill will provide the means by which the plane advances down the runway, and it will take-off as normal.

You are all approaching this problem as if it was a car and the forward momentum was coming from the wheels as they turn. The plane produces its own momentum and doesn’t need the wheels to turn. Countering their turning motion won’t stop the plane. The only way to visualize stopping the plane they way the problem intends it, is to have the plane try to produce thrust in a vacuum. No air = no thrust = no movement = no flying.
brian Says:
December 12th, 2006 at 10:41 am
the plane would take of if it were not for the stipulation that the treadmill speed matched the wheel speed. because the treadmill speed matches the wheel speed the plane cannot move, meaning the wheels cannot move, and the treadmill will also not move. it is impossible to accelerate under these conditions. i’m not familiar with airplane mechanics, but firing the engines at full speed with the wheels impossibly locked to the treadmill sounds like it would result in engine failure or the wheels being ripped off the body of the airplane. this would result in the airplane hitting the treadmill, which by the stipulation of this problem will never move [not launching the plane backwards], resulting in what i believe would be a failure to launch.

however, if the stipulation that the treadmill moves at the speed of the wheels is removed the plane would take of successfully as it would be a normal takeoff where the wheels just spin faster as per the speed of the airplane moving forward plus the speed of the treadmill moving in reverse.

regardless of simple physics, the solution is always death.
brian Says:
December 12th, 2006 at 10:53 am
i resign my post gareth has a good point, for some reason i didn’t think about that and was looking at this problem from the wrong angle. the plane would launch as gareth points out

for some reason i was associated the wheels not moving with the plane not moving, this is incorrect. the plane would move and the wheels wouldn’t.
John Says:
December 12th, 2006 at 10:55 am
It won’t take off. It’s all about airflow over a wing. An airplane is not a rocket. It needs the negative pressure over the wing to allow lift.

Here’s another way to think about it. If you’re a pilot and have flown on very windy days, you have enjoyed some interesting phenomena that underscores this viewpoint. You can point the nose of your aircraft into the wind and experience a substantially reduced takeoff roll. If you are just walking around the pad, you can watch some aircraft (those that face into the wind) actually attempt liftoff whilst they are unmanned and tied down! This is somewhat the opposite of the conveyor belt conundrum.

Quite simply, wind (natural or manufactured) over the wings is what makes an airplane take off.
Ray setzer Says:
December 12th, 2006 at 10:57 am
Why is everyone so focused on the wheels.

Airplanes fly because air is flowing over the wings and the shape of wings create lift. No air moving over the wings, no lift. Or, in really simple terms. No liftee - no flyee.
Bill O'Day Says:
December 12th, 2006 at 10:58 am
@Gareth:

I almost admitted defeat, but then I look at the problem one more time. My edit addresses the fact that the thrust comes from the engines. Think about the problem visually. Try to imagine a go cart moving down a hill without the wheels turning. It’s a sled, it won’t move or at least not much. The thrust doesn’t come from the wheels, but from gravity.

Now, let’s use the same mental exercise again. If the go cart is on the treadmill and the treadmill is matching wheel speed, the go cart’s wheels and the treadmill (in the theoretical world that ignores friction in bearings and the like) both reach infinite speed. Adding in friction of bearings and the like, but not accepting equipment failure, friction will keep the go cart stationary even though the speed of the wheels and the treadmill will be far, far beyond the terminal velocity of the go cart and unhappy occupant.

Now, change back to the plane. The situation is just the same. If the treadmill was exerting a force equal to forward force, yes the plane would take off, but it isn’t. It’s matching the wheel speed.

The solution is quite simple as brian says: death.
David Says:
December 12th, 2006 at 11:04 am
Let’s look at the scenario. Assume that the rolling coefficient of force is that of a car - 0.03, that is, that’s the percentage of the normal force needed to roll the tires. Normal force is (mass times gravity)/number of wheels.

First, roll the treadmill but don’t fire up the engines. The tires will rotate at .03 x the force generated by the treadmill, once inertia is overcome, due entirely to the friction of the tires against the treadmill and the internal bearings in the wheels. The plane will roll backwards with 97% of the force applied to the tires. It will NOT roll backwards with the exact force of the treadmill. That’s the static case with the plane engines off.

Now stop the treadmill and return to start. Fire up the engines. The wheels will not move until the engines overcome intertia and the .03 factor of resistance. As they start to roll, the treadmill starts to turn at the same rate. But of the force on the plane, only .03 of it is used to roll the wheels! That force is being compensated by the treadmill, so the wheels rotate that much faster than usual over the same distance. The rest of the energy moves the plane forward over the treadmill (but now with a .06 coefficient to overcome, not .03).

The only way the wheels will rotate is if the plane moves forward, since the wheels are not direct-drive like a car. So the movement of the treadmill is dependent on the plane actually covering ground. If the overall friction of the tires and wheels were so great that it could not be overcome by the engines, it would stay still and never take off, but because it’s less than the total force available, the plane simply has more force to overcome - but only twice as much, since the treadmill is only compensating for the resistance of the tires, not exceeding it.

The tires will rotate faster than normal but the plane will take off. That’s my take.
Bill O'Day Says:
December 12th, 2006 at 11:18 am
@David:

The trick is that there can be no movement of the plane itself without the wheels slipping, because each time the wheels would go faster, the treadmill matches the speed. So, you are right about the coefficient being increased, only the increase will still increase. What would happen is that because the thrust is independent of the wheel motion, the plane wants to move anyway. Therefore, because the wheels provide little friction, the effect of treadmill movement is many, many orders of magnitude greater than effect of engine output. With no failures, and any friction in the bearings, a point will be hit where there is equilibrium and the plane itself will be stationary even though the wheels would be spinning at a speed Superman wouldn’t be able to reach. Failure is far more likely, however. The problem is interesting because it is the intuitive answer for nonintuitive reasons.
Gareth Says:
December 12th, 2006 at 11:22 am
Look, I’m sorry if I sounded insulting by calling people “retarded”, but I am frustrated by this problem because from a purely logical standpoint it seems clear as day to me that the plane would take-off without a hitch. David brought up a good point with the rolling coefficient of friction, but I still think it is even simpler than that.

Imagine, if you will, that the plane does not have wheels but is instead being supported by simple legs that end in flat plates (like the lunar lander). The problem did not state that the treadmill would _only_ move if the wheels moved, just that it would move to match them in reverse. So if our plane sans wheels fires up its engines wouldn’t the treadmill itself act as the wheels and allow the plane to move forward? This would provide the “liftee” that Ray Setzer mentioned, and allow for take-off. Now put the wheels back on the plane and imagine the same scenario - the wheels of the plane would not be compelled to rotate because their rotation would be countered by the treadmill and they would essentially stay still. The plane, however, would still advance down the runway and take-off.

I don’t know how I could explain it more simply, but try to visualize a small-scale experiment to recreate this situation - say an R/C plane sitting on a regular treadmill that has been disengaged from the drive motor and allowed to move freely and ideally without friction. Now fire up that R/C plane and see what happens. Where does the thrust go if not towards advancing the position of the plane? If the wheels were truly “locked” as such and could not rotate, wouldn’t the thrust act on the movement of the treadmill surface itself? Think about it. The plane would still move forward.
David Says:
December 12th, 2006 at 11:26 am
Well, again, I’m wondering how the wheels turn without the plane actually moving.

Picture the reverse treadmill situation - the treadmill rotates *forwards* against wheel rotation as the plane accelerates. This will keep the tires from rotating as quickly, but it will not nullify their motion entirely, since the rolling coefficient is not equal to the normal force. It will simply reduce the rotation of the tires. The only question is, how much will it reduce that? I think it’s related to the rolling coefficient, not an assumed 100% of the force involved.

The treadmill matches the speed of rotation, but not the force of acceleration. Right?
bill Says:
December 12th, 2006 at 11:32 am
Gareth, I see what you are saying. The thing is that according to the problem, the treadmill would match the speed of the wheels. So, while the wheels remained stationary, the treadmill would be forced to remain stationary. Having the treadmill move with the plane while the wheels are not rolling is outside of the bounds.

As for the lunar lander, it’s a non starter. That is a vertical takeoff system. Therefore, you are dealing gravity being the limiting force exclusively (aside from wind resistance, which is really not an issue on a lunar surface).

For a plane, we don’t challenge gravity directly, we allow for forward motion to create lift. The wheels act in a way to minimize friction on the ground. By adding the treadmill, what occurs is a situation that whenever inertia is about to be overcome to allow for forward movement what’s needed is increased.

Don’t worry about the barb. To be honest your first comment was the most difficult challenge I’ve had to my view yet.
Gareth Says:
December 12th, 2006 at 11:32 am
“Well, again, I’m wondering how the wheels turn without the plane actually moving.”

The plane does not require the wheels to move in order to move, and the wheels do not require the plane to move in order to move.
bill Says:
December 12th, 2006 at 11:34 am
David: yes
David Says:
December 12th, 2006 at 11:36 am
I have an error in the above - I confused tire speed and force. Let me think about it again.
Gareth Says:
December 12th, 2006 at 11:40 am
“The thing is that according to the problem, the treadmill would match the speed of the wheels. So, while the wheels remained stationary, the treadmill would be forced to remain stationary. Having the treadmill move with the plane while the wheels are not rolling is outside of the bounds.”

If this is true, then the entire problem is a non-issue. If the wheels are essentially locked, and the treadmill is essentially locked, then the plane would produce thousands of pounds of thrust that would be acting against, essentially, whatever mechanism is holding the treadmill still (some sort of brake, I assume?). So either the brake would fail and the plane would move, or the plane would just sit there making lots of noise and upsetting the captain. The problem, I think, is the problem. It’s not clear because the way I read it there is no such restriction.
Jason S. B. Says:
December 12th, 2006 at 11:48 am
The plane will not take off. Gareth is retarded because he doesn’t understand the point of the experiment: If you have a radar gun pointing at the plane, the speed will ALWAYS say ZERO MPH because even though the plane would like to move forward because of the thrust from the engines, the treadmill is moving it backwards at the same speed.

If somehow the radar gun shows that the plane isn’t going ZERO MPH, then the plane has a chance of taking off. But the point of this is that no amount of thrust is going to allow a jet to take off at ZERO MPH.
David Says:
December 12th, 2006 at 11:54 am
“The plane does not require the wheels to move in order to move, and the wheels do not require the plane to move in order to move.”

Right. But neither does the entire force of the engines go to the rotation of the wheels.

You are postulating that because there is *some* friction, that the rotation of the wheels will increase to the point where the friction is equal to the normal force. The problem I’m trying to get across is that while that is occuring, the plane is able to move forward, and I believe that the rate of the plane’s movement will govern the increase in rotation of the tires. I think your scenario implicitly assumes that the treadmill will actually speed the tires up.

I’m arguing that if the rotation speed of the tires is 1000 rpm at liftoff, then the effect of the treadmill will be to rotate the tires at an additional 1000rpm at that point, thus doubling the force the plane has to push against, rather than increasing it infinitely. This would not actually stop the plane

The rotation of the tires and the tradmill cannot increase by more than the rolling coefficient plus the same amount from the treadmill, right? So the increase in resistance to be overcome is finite, not infinite.
Jason S. B. Says:
December 12th, 2006 at 12:02 pm
Think of it this way, Gareth: You turn on this giant treadmill and the plane starts to slowly move backwards… so you fire up the engines to counteract that backwards movement. Now you’ve got a slowly moving treadmill and engines on with enough power to move the plane “forward”, but the end result is the plane is moving ZERO MPH. Now you speed up the treadmill a little bit and the plane starts to drift backwards again… so you throttle up to counteract that backwards movement and the plane is still moving ZERO MPH (according to the cop sitting off to the side with a radar gun). When you eventually get to the point where the engines are at full throttle, any forward movement that full throttle would have given you is counteracted by the treadmill trying to move the plane backwards.

End result: the plane is throttled up like crazy, the treadmill is going backwards like crazy, the actual airspeed of the plane is ZERO MPH, and the plane goes nowhere (especially not up).

Comprende? I think the key phrase above is “the actual airspeed of the plane is ZERO MPH”.
David Says:
December 12th, 2006 at 12:02 pm
“The plane does not require the wheels to move in order to move, and the wheels do not require the plane to move in order to move.”

First part, I agree with. The second does not follow since the scenario states that the plane’s force drives the treadmill. That force does not move the wheels without moving the plane as well - the wheels will not rotate unless the plane is moving, since the treadmill is motionless until the wheels rotate. Only a part of the plane’s engine power is used to rotate the wheels.

Put another way - the rotation of the wheels depends on the distance traveled. Increasing the rotation of the wheels by increasing the movement of the runway merely lengthens the takeoff roll. You’d have to speed the treadmill up to apply a force equal to the engine force to keep the plane from moving, but that force would far exceed the force the plane is using to turn the wheels.
Stephen Ingrum Says:
December 12th, 2006 at 12:06 pm
I agree that people are think of this like a rocket ship instead of an airplane. Wind must pass over the wing to provide lift. When you are on a treadmill- no matter how fast you run, you will never feel any wind. Therefore, no lift off.

You could make the plane take off even without the wheels moving if it was in a wind tunnel, but without actually moving against the wind- there will be no takeoff.
Jason S. B. Says:
December 12th, 2006 at 12:16 pm
The question should be distinguised between: Does a plane (not in a wind tunnel) have a chance of taking off if its airspeed is ZERO MPH?

vs.

Would a plane on a treadmill move forward when you crank up the treadmill, force the wheels to go backwards at an incredible rate (that rate not matching the actual reverse speed of the plane if the wheels were locked–I understand that), and then power up the plane enough that it attains actual airspeed above ZERO MPH?

These are two different questions.
David Says:
December 12th, 2006 at 12:23 pm
Suppose the plane is landing at (say) 200mph instead of taking off. If the treadmill can infinitely accelerate to match the rotation of the wheels as you argue, would not the plane simply stop dead upon contacting the ground? I think in fact what would happen is that the wheels would simply spin faster, but the plane would still move over the runway, because the force transmitted through the rotation of the wheels is not equal to the force possessed by the plane as it moves. The treadmill would slow with the plane, but the distance traveled by the wheels would be twice normal.
Jason S. B. Says:
December 12th, 2006 at 12:30 pm
Fine, I’m retarded.

The plane will take off. Why? Because we have to stop assuming that it won’t move forward. Yes, a plane at ZERO MPH airspeed will never take off, but in this case the plane will achieve airspeed via forward movement because the engine thrust easily overcomes the canceling-out below of the free-spinning airplane wheels and the fast-moving treadmill.

The plane takes off because it moves forward. That’s where people (including myself) get confused.
gabe M. Says:
December 12th, 2006 at 1:15 pm
The confusion between gareth and others is due to opposite assumptions in terms of the direction the treadmill is moving. Gareth is assuming a forward direction relative to the plane, everyone else is assuming a backward direction relative to the plane.
bill Says:
December 12th, 2006 at 1:30 pm
Jason: what I’m saying is that it won’t move forward. Or if it does, it does due to the tires slipping. While it is true that mph for mph isn’t equal between engine force and treadmill force, the treadmill (if done the way I’ve stated it) will provide the same amount of force backwards. Because otherwise, the tires will turn more(or slip, thus negating the entire comments’ line of reasoning). If the tires turn more, the the treadmill turns more, thus equilibrium is maintained.

Gareth:
“The conveyer belt is designed to exactly match the speed of the wheels, moving in the opposite direction.” While true that it says designed, not “the conveyer belt exactly matches the speed of the wheels, moving in the opposite direction,” the spirit of the experiment (IMHO) is the latter.

The “lockdown” theory you have put forward is the thing I’m not following. As soon as thrust is applied, the wheels and the treadmill will begin moving at equal speed, and within fractions of a second either break (likely) or reach a frictional equilibrium where the force of the friction of the bearings equals the force of the thrust.

So to simplify, lets say that it takes 500 mph of force on the wheels to have friction slow the speed by 1 mph. And let’s say that the engines are putting out 500 mph of thrust.

In a normal situation: a force graph would look like this:
500mph–>Plane<--1mph (wheels turning at 500mph)
Friction would cause the plane to move forward at 499 mph.

In this situation:
500mph-->Plane<–500mph (wheels turning at 250,000 mph)
Friction would balance the forward force.

Before that happened, obviously something would break, but there is the theoretical limit of the situation.
Russ Says:
December 12th, 2006 at 1:53 pm
the wheels don’t turn unless the plane moves forward. the engines push the plane forward. I’m still not sure if the wheels turn at twice their normal speed (makes the most sense to me), or at infinite speed, but the plane will take off because the only way the wheels can turn is if the plane moves forward.
Jason S. B. Says:
December 12th, 2006 at 2:01 pm
The other thing I think people aren’t grasping (or maybe it’s just a matter of the definition of the terms of the experiment) is that the “conveyor belt” is more of a treadmill in that it is powered and moving the opposite direction of the plane.

Just think of it this way: You stand on a treadmill while wearing a pair of roller blades. You’re holding onto a rope tied to the front of the treadmill. You turn the treadmill on and crank it up to super high speed. Eventually when you’re balanced and everything has evened out, the force on the rope you’re holding onto isn’t all that great. If we picture an airplane in place of the person on rollerblades, the forward force that the airplane engines can produce will be far greater than the force on the rope. Therefore the plane will move forward with ease. And then do you know what it will do?

Take off, hoser.
Gareth Says:
December 12th, 2006 at 2:22 pm
But if the plane is on a big, frictionless treadmill, then it can move forward without the wheels turning at all - that’s my point.
bill Says:
December 12th, 2006 at 2:23 pm
Russ: Ok, you bring an interesting point to the table. Let’s say the plane does move forward but at an infinitely small amount. It’s not going to take off because the treadmill effect kicks well before it has enough velocity to lift off.
bill Says:
December 12th, 2006 at 2:35 pm
Gareth: If the treadmill was frictionless, it wouldn’t be a treadmill. It would be a slippery surface.

Jason: In reality, the plane would take off if some one actually tried to implement this. However, that is only because it would be practically impossible to get a treadmill to go as fast as it needed to get. By the limits stated in the original problem, it would reach balance and not fly away.
Gareth Says:
December 12th, 2006 at 2:38 pm
Bill: Sorry, I meant the inner-workings of the treadmill were frictionless, not the surface
Gareth Says:
December 12th, 2006 at 2:40 pm
Bill: I meant the inner-mechanics of the treadmill were frictionless, not the surface itself.
Jason S. B. Says:
December 12th, 2006 at 2:54 pm
Bill, please state the experiment as you understand it. Is it a free turning conveyor belt, or is it a powered treadmill that runs in the reverse direction of the direction the plane would travel to take off under normal circumstances?
bill Says:
December 12th, 2006 at 2:56 pm
Gareth, now I’m confused. Let’s just say that friction exists everywhere, because it would have to. Therefore the whole thing is kind of dumb, as has been said by a few (you included) before. What’s going to happen is that either the treadmill will be able to get to a speed where the backwards force will be equal to the thrust provided by the engines (the plane remains Earth-bound) or the friction of the bearings in the treadmill will cause the treadmill to be unable to keep up with the plane’s forward momentum (the plane takes off).

Either way, we are (at least in my view) debating the most minute of minutia… Thanks for keeping me on my toes.

Jason: I saw it as a powered treadmill, otherwise it really isn’t capable of even really slowing down the plane significantly as it would adjust physical plane displacement more than wheel speed.
curious bystander Says:
December 12th, 2006 at 10:34 pm
Reduction in wheel friction and frictional losses, sure. These are all important and potential opportunities in modern day flight. How do they apply to a conveyor belt or etc.? Forward air speed over and under the wings of an aircraft provide lift. Not a moving surface below. Am I missing something? If there is no friction between the ground and an aircraft, we still need air moving over and under a wing surface to provide lift. Helicopter for example!

Someone please enlighten me here…
David Says:
December 13th, 2006 at 2:42 pm
Hopefully everyone got a look at Cecil Adams take on this at the Straight Dope website. Just for closure.

You must be logged in to post a comment.

Airplane-Treadmill Problem Solution - Nihilist

40 Responses to “Airplane-Treadmill Problem Solution - Nihilist”

Leave a Reply