Airplane on a converter belt question.

[sushijerk]

I recently heard this and gave it a lot of thought:

Suppose a normal commercial airplane is ready to begin takeoff. The thrusters are turned on and the wheels of the airplane begin to move forward. However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels. Image that there is a electronic device gauging the speed of the wheels and then immediately relays the information to the belt to accelerate/deccelerate accordingly. Will the plane ever be able to achieve takeoff?

[MidGe]

use the search facilty... been discussed to death already

[Lestat]

My answer would be no.

So I see I'm wrong. This was a cool problem! Glad you posted it. Thanks.

[sushijerk]

[ QUOTE ]
use the search facilty... been discussed to death already

[/ QUOTE ]

Really? Doh. A mechanical engineering friend of mine said the answer is yes, but I'm still not sure why.

[Bork]

conveyor belt

If the plane moves forward fast relative to the air then it will take off, if not no it won't take off.

Air needs to flow rapidly over the wings to generate lift.

[MidGe]

plane + conveyor belt (http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=4011389&page=0&fpart=1&v c=1)

BTW, whether an aeroplane takes off or not has nothing to do with runway speed (A/c carrier), aeroplane ground speed, or torque on the wheels. If it's airspeed is greater than take-off speed it will take off. I have seen planes flying backwards relative to the ground and they were not falling. lol.

[madnak]

The plane will take off. The wheels aren't moving the plane forward, they're just spinning idly.

[purnell]

[ QUOTE ]
I recently heard this and gave it a lot of thought:

Suppose a normal commercial airplane is ready to begin takeoff. The thrusters are turned on and the wheels of the airplane begin to move forward. However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels. Image that there is a electronic device gauging the speed of the wheels and then immediately relays the information to the belt to accelerate/deccelerate accordingly. Will the plane ever be able to achieve takeoff?

[/ QUOTE ]


Clearly the answer is yes. The planes engine(s) are not connected to its wheels. The same low pressure ahead of the intakes would pull the plane forward, and it would move along the runway and take off as usual, regardless of how fast the conveyor belt runs in any direction.

edit: err, "...belt runs in either direction" (parallel to the engines' force). /images/graemlins/crazy.gif

[purnell]

[ QUOTE ]
plane + conveyor belt (http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=4011389&page=0&fpart=1&v c=1)

BTW, whether an aeroplane takes off or not has nothing to do with runway speed (A/c carrier), aeroplane ground speed, or torque on the wheels. If it's airspeed is greater than take-off speed it will take off. I have seen planes flying backwards relative to the ground and they were not falling. lol.

[/ QUOTE ]

While this is correct, aircraft carriers use a steam-powered catapult to accelerate the aircraft to the required speed in the very short distance available. If the catapult fails, the aircraft nearly always falls into the sea.

[MidGe]

[ QUOTE ]
While this is correct, aircraft carriers use a steam-powered catapult to accelerate the aircraft to the required speed in the very short distance available. If the catapult fails, the aircraft nearly always falls into the sea.

[/ QUOTE ]

This true of most high performance aircrafts, it is not so, of some other aircrafts (and I am talking about conventional a/c not STOL or VTOL). The reason is that whatever the plane, if it can reach its minimum take-off speed minus headwind by the end of the deck, there are no issues except the margin of safety. There are many examples of such unassisted take-offs from a/c carriers for demonstration, exhibition or routine actions. I don't even think that all a/c carriers have catapults... (see russia).

[Go Blue]

I've thought about this problem for who knows how long and I recently did a presentation which involved explaining the answer to this question. Let me attempt to give a clear answer and put this issue to rest.

The answer is: Yes.

Why? Because a plane generates speed (thrust) from its engines; it does not step on the gas so that the wheels turn. As a matter of fact, the wheels surve no purpose other than to reduce friction. First, imagine this scenario: The speed of the wheels is zero. In other words, the runway doens't move backwards (it moved at speed zero) and the wheels don't move either. The plane turns its engines on. If they can provide enough force, the plane will just begin to slide on the runway until it slides fast enough to take off.

From this, you can see where I'm going. If the wheel speed was, say 100 mph, the runway would move back at 100 mph. But this doens't mean that the plane itself couldn't move faster. If the plane travels at 250 mph, the wheels basically compensate for 100 mph of that and the rest is just from sliding.

So, basically, yes the plane can take off if its engines are powerful enough to overcome the friction of the "still" wheels.

[Nottom]

Note that the way this is phrased is what causes a lot of the confusion. The conveyer belts can't move at the same speed as the wheels in this case. As stated above, the wheels on a plane don't have any power applied to them to move when taking off, they just spin.

A simple comparison would be you holding a toy car on a treadmill. Assuming friction is negligable, the treadmill can go as fast as it wants and the wheels of the car will simply spin in place. Now move the car foward, the wheels of the car go faster than the treadmill and it is impossible for the treadmill to go as fast (in theory the treadmill would be going infinitly fast since it would keep trying and never catch up). This is basically the same deal with a plane, except instead of your hand the engines on the plane provide the foward momentum so the plane will still move foward and take off.

[Ray Zee]

in this case the wind isnt traveling over the wings the wind is staionary so the plane will not rise. if the plane was held stationary and the wind was driven over the wings at liftoff speed the plane would rise verically.

[DougShrapnel]

[ QUOTE ]
in this case the wind isnt traveling over the wings

[/ QUOTE ] Pretty sure the jet engines would still create wind traveling over and under the wings. The motion of the converter belt would also create some mininal lift.

[Sharkey]

As long as the entire weight of the plane is on the wheels, and the tires don’t skid, the plane cannot move forward.

In fact, the wings don’t even have to be there until some significant forward motion is achieved.

[purnell]

[ QUOTE ]
As long as the entire weight of the plane is on the wheels, and the tires don’t skid, the plane cannot move forward.

In fact, the wings don’t even have to be there until some significant forward motion is achieved.

[/ QUOTE ]

This would only be true if the pilot were applying the brake while attempting takeoff (which would define him as an idiot, I suppose).

[Sharkey]

The tires will skid if the wheels are turning at a different speed than the plane is moving.

[purnell]

The wheels spin on an axle with very low friction (much lower than the tires against the belt). Thus, the tires spin with [are driven by] the belt, in either direction, at any speed, but the plane is still pulled forward by the engines.

[Sharkey]

Since the plane is attached to the wheels, for ever little increment the plane tries to move forward, the conveyor belt will speed up and pull it back, assuming the tires don’t skid.

[purnell]

[ QUOTE ]
Since the plane is attached to the wheels, for ever little increment the plane tries to move forward, the conveyor belt will speed up and pull it back, assuming the tires don’t skid.

[/ QUOTE ]

Are you just [censored] with me?

The tires spin against the belt, having no effect on the movement of the rest of the plane. This is so elementary I'm having a hard time believing that you don't get it.

[BCPVP]

[ QUOTE ]
[ QUOTE ]
Since the plane is attached to the wheels, for ever little increment the plane tries to move forward, the conveyor belt will speed up and pull it back, assuming the tires don’t skid.

[/ QUOTE ]

Are you just [censored] with me?

The tires spin against the belt, having no effect on the movement of the rest of the plane. This is so elementary I'm having a hard time believing that you don't get it.

[/ QUOTE ]
There was a monster thread on this (I think in OOT) and there were many people who couldn't get over the fact that the wheels aren't driving the plane forward.

[Sharkey]

[ QUOTE ]
The tires spin against the belt, having no effect on the movement of the rest of the plane.

[/ QUOTE ]

In the ideal case, the inertia of the plane is independent of what the wheels are doing, but if there’s the least friction in the whole ensemble, the conveyor belt will simply speed up however much is necessary to counteract the thrust of the engines.

[CORed]

[ QUOTE ]
The wheels spin on an axle with very low friction (much lower than the tires against the belt). Thus, the tires spin with [are driven by] the belt, in either direction, at any speed, but the plane is still pulled forward by the engines.

[/ QUOTE ]

Of course, if the conveyor belt moves sufficiently fast, either the tires will overheat and blow out, the wheel bearings will burn out or the wheels thenselves will fly apart from centrifugal force, or sufficient frictional forces will be developed to slow the plane down. Of course, similar engineering consideratios will limit the maximum speed atainable by the conveyor belt.

[MidGe]

Once more,

For take-off only the relative airspeed is relevant. The undercarriage plays no role. Given a sufficiently strong head wind a small plane can take off with its wheel brakes locked and thus zero ground speed.

[Sharkey]

[ QUOTE ]
Of course, if the conveyor belt moves sufficiently fast, either the tires will overheat and blow out, the wheel bearings will burn out or the wheels thenselves will fly apart from centrifugal force, or sufficient frictional forces will be developed to slow the plane down. Of course, similar engineering consideratios will limit the maximum speed atainable by the conveyor belt.

[/ QUOTE ]

Since none of the component specs are in evidence, that is an irrelevant argument.

[Sharkey]

[ QUOTE ]
Once more,

For take-off only the relative airspeed is relevant. The undercarriage plays no role. Given a sufficiently strong head wind a small plane can take off with its wheel brakes locked and thus zero ground speed.

[/ QUOTE ]

The plane will never attain sufficient air speed because its acceleration is being reduced by friction with the conveyer belt.

[purnell]

From the OP:
[ QUOTE ]
However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.

[/ QUOTE ]

BTW, WTF is a converter belt? It sounds suspiciously like something that might have been used during one of the inquisitions, in which case the whole discussion is moot. /images/graemlins/crazy.gif

[CORed]

[ QUOTE ]
Once more,

For take-off only the relative airspeed is relevant. The undercarriage plays no role. Given a sufficiently strong head wind a small plane can take off with its wheel brakes locked and thus zero ground speed.

[/ QUOTE ]

I'm not disputing that. I'm only pointing out that a backward moving conveyor belt does generate some frictional force on the wheels of the plane, which will increase with the speed of the conveyor belt, and the wheels of the plane cannot turn infinitely fast. Now, I'll admit that the conveyer belt would have to move bacwards at an undgodly speed to overcome the thrust of the jet engines. Actually, the whole excercise is pretty silly, because the notion of a conveyor belt that can increase it's speed instantaneously to match the forward speed of the plane is physically impossible. The conveyor belt has mass and therefore inertia, and it takes a force acting over time to accelerate it. The conveyor belt also presumably has a tnsile strength and that limits the rate of acceleration, because if the force accelerating it excedds its tensile strength the belt will break. As a practical matter, I don't think you could build a conveyor belt that could accelerate backwards fast enough to prevent a plane from taking off, or even come close to doing so, assuming anybody was foolish enough and had enough money to waste to try.

[Sharkey]

[ QUOTE ]
From the OP:
[ QUOTE ]
However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.

[/ QUOTE ]

BTW, WTF is a converter belt? It sounds suspiciously like something that might have been used during one of the inquisitions, in which case the whole discussion is moot. /images/graemlins/crazy.gif

[/ QUOTE ]

Of course. The wheels and the conveyor belt have to move at the same speed under all circumstances, given that there’s no skidding. What about it?

A converter belt is like a girdle, I think.

[purnell]

[ QUOTE ]
[ QUOTE ]
From the OP:
[ QUOTE ]
However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.

[/ QUOTE ]

BTW, WTF is a converter belt? It sounds suspiciously like something that might have been used during one of the inquisitions, in which case the whole discussion is moot. /images/graemlins/crazy.gif

[/ QUOTE ]

Of course. The wheels and the conveyor belt have to move at the same speed under all circumstances, given that there’s no skidding. What about it?

A converter belt is like a girdle, I think.

[/ QUOTE ]

The "What of it" is that if the belt is rigged to run at the same speed as the wheels [linear speed of a point on the belt equal to the instantaneous linear speed of a point on the perimeter of the tire], the belt never reaches a speed at which it will have a significant effect on the speed of the plane. /images/graemlins/tongue.gif

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
From the OP:
[ QUOTE ]
However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.

[/ QUOTE ]

BTW, WTF is a converter belt? It sounds suspiciously like something that might have been used during one of the inquisitions, in which case the whole discussion is moot. /images/graemlins/crazy.gif

[/ QUOTE ]

Of course. The wheels and the conveyor belt have to move at the same speed under all circumstances, given that there’s no skidding. What about it?

A converter belt is like a girdle, I think.

[/ QUOTE ]

The "What of it" is that if the belt is rigged to run at the same speed as the wheels [linear speed of a point on the belt equal to the instantaneous linear speed of a point on the perimeter of the tire], the belt never reaches a speed at which it will have a significant effect on the speed of the plane. /images/graemlins/tongue.gif

[/ QUOTE ]

Not so, because the wheels and the belt are coupled (by way of the non-skidding tires) the speeds of the two will always be equal, no matter how fast either one goes.

[purnell]

[ QUOTE ]
The "What of it" is that if the belt is rigged to run at the same speed as the wheels [linear speed of a point on the belt equal to the instantaneous linear speed of a point on the perimeter of the tire], the belt never reaches a speed at which it will have a significant effect on the speed of the plane.



Not so, because the wheels and the belt are coupled (by way of the non-skidding tires) the speeds of the two will always be equal, no matter how fast either one goes.


[/ QUOTE ]

But the belt's speed is dependent upon the speed of the tire. The wheels are not driven by anything, they only come along for the ride. It is the belt that must be driven (by its own motor) to match the speed of the tire, and during takeoff a point on the perimeter of the tire never reaches a speed which will cause any significant losses to friction at the axle. You do understand that the plane's wheels are not driven, that they transmit no force to the ground other than weight during takeoff, right? At landing time, when the brake is applied, they transmit horizontal force to the rest of the plane, but not during takeoff.

[Borodog]

[ QUOTE ]
[ QUOTE ]
The "What of it" is that if the belt is rigged to run at the same speed as the wheels [linear speed of a point on the belt equal to the instantaneous linear speed of a point on the perimeter of the tire], the belt never reaches a speed at which it will have a significant effect on the speed of the plane.



Not so, because the wheels and the belt are coupled (by way of the non-skidding tires) the speeds of the two will always be equal, no matter how fast either one goes.


[/ QUOTE ]

But the belt's speed is dependent upon the speed of the tire. The wheels are not driven by anything, they only come along for the ride. It is the belt that must be driven (by its own motor) to match the speed of the tire, and during takeoff a point on the perimeter of the tire never reaches a speed which will cause any significant losses to friction at the axle. You do understand that the plane's wheels are not driven, that they transmit no force to the ground other than weight during takeoff, right? At landing time, when the brake is applied, they transmit horizontal force to the rest of the plane, but not during takeoff.

[/ QUOTE ]

<pedantic nit>Well, that's of course not strictly true, or else the tires wouldn't rotate during takeoff.</pedantic nit>

[purnell]

Pont taken. Nit. /images/graemlins/laugh.gif

[Borodog]

/images/graemlins/wink.gif

[CORed]

[ QUOTE ]
at the exact speed of the wheels.

[/ QUOTE ]

When I first read this, I interpreted it to meand that the conveyor belt was moving backward at the same speed the plane is moving forward. I suppose I should qulify that with "relative to the ground". I think that might be what the OP meant, but it's not what he said. Of course, he also said "converter belt". We all assumed that he meant "conveyor belt" I'm not sure what "speed of the wheels" means. The speed of the contact point of the wheel with the belt is zero relative to the belt, given that the wheel is spinning freely, not skidding. This, of course is true whether the belt is stationary or moving relative to the groung. The speed of the hub of the wheel is, of course, the speed of the plane, again, relative to the belt. The speed of the top of the tire is twice the speed of the hub, again, relative to the belt. I would maintain that if we place no engineering constraints on the backward acceleration of the belt, it would be possible to accelerate the belt backwards fast enough to stop the plane, if we assume a normal commercial airliner that does not have magical fricionless wheels to go along with our magical conveyor belt capable of any finite acceleration. At some speed, something in the wheel tire assembly will fail and the plane will stop rolling. If we assume that the conveyor belt is made of commerically available materials and power souce, I would say that it is probably impossible to make a conveyor belt that would accomplish this. It would damn sure be prohibitively expensive.

Edit:
I'm thinking about this way too much. On further reflection, given the way this problem is stated, if we assume that the conveyor belt moving backward at the "speed of the wheels" to mean that the backward ground speed of the belt is the same as the forward ground speed of the plane, this would only require the wheels to rotate twice as fast as they normally would, which is probably within the safety margin of the landing gear, and would not prevent the plane from taking off. I was thinking of accelerating the belt backward ever faster in an effort to stop the plane, which, while theoretically possible, if we place no engingeering constraints on the speed and acceleration of the belt, would probably be unachievalbe in the real world. Besides which, it's stupid, so nobody is going to try.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
The "What of it" is that if the belt is rigged to run at the same speed as the wheels [linear speed of a point on the belt equal to the instantaneous linear speed of a point on the perimeter of the tire], the belt never reaches a speed at which it will have a significant effect on the speed of the plane.



Not so, because the wheels and the belt are coupled (by way of the non-skidding tires) the speeds of the two will always be equal, no matter how fast either one goes.


[/ QUOTE ]

But the belt's speed is dependent upon the speed of the tire. The wheels are not driven by anything, they only come along for the ride. It is the belt that must be driven (by its own motor) to match the speed of the tire, and during takeoff a point on the perimeter of the tire never reaches a speed which will cause any significant losses to friction at the axle. You do understand that the plane's wheels are not driven, that they transmit no force to the ground other than weight during takeoff, right? At landing time, when the brake is applied, they transmit horizontal force to the rest of the plane, but not during takeoff.

[/ QUOTE ]

The OP does not say that the tires turn the belt, but rather that the belt “is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels”, which is logically implied by “the tires don’t skid”.

It is not the case that the wheels are not driven by anything. They are made to turn at a speed equal to the difference in speed between the plane and the belt.

There is a corresponding friction from the tires-and-belt, the wheel bearings etc for every speed of the wheels, which can be made arbitrarily high.

[Sharkey]

By “speed of the wheels” is meant the linear speed of the wheels at their circumference.

The frictional backward force on the plane from the conveyor belt etc is the same that would be encountered by the plane with its engines off, coasting to a stop after landing (on a very long runway), excluding air friction.

[madnak]

Can we not do this again?

[CORed]

[ QUOTE ]
By “speed of the wheels” is meant the linear speed of the wheels at their circumference.

The frictional backward force on the plane from the conveyor belt etc is the same that would be encountered by the plane with its engines off, coasting to a stop after landing (on a very long runway), excluding air friction.

[/ QUOTE ]

Unless the wheel is skidding, the speed of the belt is allways going to matnch the "linear speed of the wheels at their circumference."

[Sharkey]

[ QUOTE ]
[ QUOTE ]
By “speed of the wheels” is meant the linear speed of the wheels at their circumference.

The frictional backward force on the plane from the conveyor belt etc is the same that would be encountered by the plane with its engines off, coasting to a stop after landing (on a very long runway), excluding air friction.

[/ QUOTE ]

Unless the wheel is skidding, the speed of the belt is allways going to matnch the "linear speed of the wheels at their circumference."

[/ QUOTE ]

Let’s say the belt is moving very slowly and the plane is standing still under low thrust.

At the moment of the slightest throttle up to start the plane moving forward, the belt sees the wheels begin to move along its length, which implies a change in wheel speed.

To compensate, the belt then accelerates in the other direction. If the wheels continue changing their position, the belt will continue increasing speed, and so on until the limiting value is reached where the friction of the entire plane-belt ensemble stops the plane’s advance.

[Sharkey]

On the other hand, maybe any constant speed on the part of the plane corresponds to a constant wheel speed against the belt.

Accordingly, the plane would reach take-off speed, leaving the belt running at the same speed backwards.

[CORed]

I can't think of a stupider, less eficient way to try to prevent an airplane from taking off than putting it on a conveyor belt and trying to move the belt backwards fast enough to prevent takeoff. Given a conveyor belt with near magical ability to reach any speed with nealy infinite acceleration, I suppose it could be done, but I suspect you would have to move the belt backwards so fast as to cause a catastrophic failure -- tire blowout, bearing siezure, or wheel breakup, to accomplish it. Now of course, if you're dealing with a car, truck, bus, railroad locomotive, horse and buggy, person on foot or anything else driven by the wheels, or walking or running animal or person, it's easy. The wheels or walking person or animal are already pushing the belt backwards. You just have to apply enough power to the belt to overcome its inertial and frictional resistance. But for plane, dirven by a jet engine or propeller, the only backward force you can generate is frictional force through the tires and wheels, which are, of course, designed to minimize that frictional force.

[Sharkey]

It’s called a “thought experiment” because the challenge is to find a solution given a few parameters without worrying about all the practical considerations.

[mindflayer]

The answer is generally no.
Wind speed over the wings creates lift.

(modern) Air plane engines do Not pull air over the wings, they suck air in (at whatever speed the plane is moving at) and throw it out at high velocity, creating thrust, pushing the plane. It is the air speed over the wings that create lift.
Wheel speed is not relavent to takeoff speed.

A plane with a normal (no wind) takeoff speed of say 100mph in a 60mph headwind would lift off with a wheelspeed of 40mph.
The same plane in a 60mph tailwind would need a wheelspeed of 160 to lift off.

The conveyor has the effect of creating a tailwind of equal speed to the wheel speed.
So at a wheel speed of 100mph the conveyor essentially creates a tailwind of 100mph. (zero wind speed over the wing) At 200mph wheel speed .. tail wind of 200mph (zero wind speed over the wing etc) The plane cannot take off.

Ok that said .. if you take a say Fokker Triplane (the Red Baron's plane) with short wings and a propeller engine, you can make a case that the enging actually increases the wind speed over the wings (to some degree and has a chance of taking off.)

[Sharkey]

There are two forces acting on the plane: thrust and friction.

The engines provide the (forward) thrust.

The (backward) frictional force (from the tires on the conveyor belt, the wheel assembly, etc) is the same as would be encountered by the plane coasting to a stop after landing, excluding air friction.

Any forward motion by the plane indicates a wheel speed faster than the belt speed, so the conveyor belt compensates by accelerating and continues doing so as long as the plane advances, until backward friction equals forward thrust, and the plane stops.

[CORed]

[ QUOTE ]
BTW, WTF is a converter belt? It sounds suspiciously like something that might have been used during one of the inquisitions, in which case the whole discussion is moot.

[/ QUOTE ]]

You mean like, if you don't convert, I'll whack you with this belt?

[purnell]

[ QUOTE ]
[ QUOTE ]
BTW, WTF is a converter belt? It sounds suspiciously like something that might have been used during one of the inquisitions, in which case the whole discussion is moot.

[/ QUOTE ]]

You mean like, if you don't convert, I'll whack you with this belt?

[/ QUOTE ]

Yarr. Repent, heathen, or its into the pit with ye. /images/graemlins/tongue.gif

[mindflayer]

you need at least two more forces.. gravity and lift.
even though a plan may "stop" in the horizontal direction.. it can still take off vertically if the lift is greater than the force of gravity....
think helicopter.

[Sharkey]

You’re correct, of course. I didn’t include them because weight and lift are constant and perpendicular to the motion required in this particular experiment.

[Sharkey]

Also...

No offense to other puzzlers, but excluding friction is just “plane” ridiculous.

Friction is what causes the relationship between conveyer belt and wheels to have any significance. The condition that belt speed equals wheel speed would be pointless in the absence of friction, which would allow the wheels and the belt to be turning any which way, independently of each other, with no effect on or by the motion of the plane.

The presumed relevance of all included conditions trumps any strained attempts at ad hoc, fudge-factor tomfoolery by the other side of this debate.

[HigherAce]

How could a plane not moving forward or backward take off? If the wheels could spin fast enough and the plane never moved where would it get lift from. I dont feel wind when im on a treadmill wether I sprint or walk. I say its grounded.

[Jeremy517]

[ QUOTE ]
How could a plane not moving forward or backward take off?

[/ QUOTE ]

The plane is moving forward. There is no force exerted on the plane by the wheels. The angular velocity of the wheels has nothing to do with the velocity of the plane.

[Sharkey]

[ QUOTE ]
There is no force exerted on the plane by the wheels.

[/ QUOTE ]

No plane has yet been invented that would not coast to a stop on a long enough runway, air resistance discounted.

[Jeremy517]

You're nitpicking. The friction between the wheel and the plane is irrelevant. The wheels are designed to minimize friction. Also, there is a huge difference between coasting and having jet engines going strong. OK I'll rephrase though: no significant force.

The point remains the same. The angular velocity of the wheels has nothing to do with the velocity of the plane. The plane isn't staying still relative to an observer (the atmosphere). It is moving and will take off.

[Sharkey]

The main friction is between the tires and the runway and in the wheel bearings. Both sources increase with wheel speed. Neither can be entirely eliminated.

The feedback mechanism controlling the speed of the conveyor belt will indirectly increase the backward frictional force at the wheels as much as necessary to stop any forward motion by the plane.

[purnell]

[ QUOTE ]
The main friction is between the tires and the runway and in the wheel bearings. Both sources increase with wheel speed. Neither can be entirely eliminated.

The feedback mechanism controlling the speed of the conveyor belt will indirectly increase the backward frictional force at the wheels as much as necessary to stop any forward motion by the plane.

[/ QUOTE ]

I think I can see where you're coming from now.

At the instant the plane starts to be pulled forward by the low pressure created by the jet or prop, the tire begins to roll, and the belt begins to move, thus increasing the tire's angular velocity (rpm if you like), thus increasing the speed of the belt, ad nauseam. The premise is a practical impossibility, but if such a machine could be built, it would instantaneously accelerate the belt and the tire to infinite speed. Why do you assume that the conveyor can operate without friction (impossible- the belt won't move without friction between it and the drum) while the wheel cannot?

[Sharkey]

It’s not the wheels rolling per se, but the wheels rolling faster than the belt that increases the belt speed. What does a wheel speed greater than the belt speed look like? The plane moves forward. If the wheel speed and the belt speed are the same, the plane doesn’t go.

The following feedback loop is implied by “belt speed will be kept equal to wheel speed”:

1. Check wheel speed.

2. Is wheel speed greater than belt speed? [Or, “Is the plane advancing?”]

3. If “yes” then increase belt speed to make up the difference. [This will also increase the backward friction acting against the forward thrust by increasing the relative speed between belt and plane.]

4. Go to step 1.

Accordingly, whenever the plane tries to advance, backward friction will be increased by however much is necessary to stop the plane by bringing wheel speed and belt speed back to equality.

[purnell]

In step three: When the belt speed is increased "to make up the difference", the wheel speed is increased simultaneously. Thus the belt can never make up the difference, thus the instantaneous infinite feedback loop (at least until something breaks).

[Sharkey]

Not quite.

Wheel speed does increase when belt speed increases, but at a lesser rate. This is so because the deceleration between belt and plane is subtracted from the wheels’ acceleration.

This continues until belt speed catches up to wheel speed and equilibrium is restored.

[purnell]

[ QUOTE ]
Not quite.

Wheel speed does increase when belt speed increases, but at a lesser rate. This is so because the deceleration between belt and plane is subtracted from the wheels’ acceleration.

This continues until belt speed catches up to wheel speed and equilibrium is restored.

[/ QUOTE ]

Regardless of the relative speeds of the belt and the plane, for each unit of distance the belt travels, a point on the perimeter of the tire MUST travel the same unit of distance.

The only equilibrium possible in this system happens when the frictional resistance of the wheel system is equal to the thrust of the engines. This is not going to happen, because the wheel system will fail due to overheating long before equilibrium is reached- unless the axle is assumed to be frictionless, and the only friction is between the tire and the belt, in which case the plane takes off as normal- unless the tire and/or the belt fail due to overheating.

I'm done with this. God bless, and good night.

[Sharkey]

It works like this:

wheel speed = plane speed + belt speed,

which means

plane speed = wheel speed - belt speed.

It’s apparent that if belt speed is increasing faster than wheel speed, the plane is decelerating.

And, once more, part specs not included in the OP are irrelevant.

God speed, and come out to play again soon, won’t you?

[purnell]

Arrgh. I guess I'm not done. Scratch what I said about equilibrium. If the belt is running toward the rear of the plane, the frictional force at the axle is translated by the wheel system into horizontal force in the same direction as the engine's force on the body of the plane. There is no equilibrium. In the given scenario, the friction at the axle actually propels the plane forward faster than normal, and thus it will reach takeoff speed in less time.

If the belt is somehow capable of instantaneous acceleration, we still have the problem of the infinite feedback loop, and if there is any friction anywhere, the system catches fire and/or explodes. If we are talking about real-world equipment, the plane simply accelerates faster.

Forgive me, I'm slow.

edit: now I think this is wrong too! /images/graemlins/laugh.gif

[Lestat]

Here's how dumb I am with all this stuff...

Suppose I'm holding a hot wheel car in my right hand. There is a conveyor belt in front of me running from my right to left. I place the hot wheel car on the conveyor belt with the front of the car pointing to to my left.

I now start rolling the car along the conveyor belt with my right hand in a left to right direction. The conveyor belt like the one in this example, starts matching the rpm's of the wheels in the reverse direction.

I CAN'T EVEN FIGURE OUT IF THE CAR WILL MOVE!! Is that sick?!?!?!?!?!?!?

I'm thinking the car HAS to move because the conveyor belt certainly wouldn't stop my hand/arm from moving left to right. And the hot wheels car is in my hand, so it would also have to move left to right. In other words, it would have to achieve some forward (left to right), velocity. Yet, I can't comprehend it. Or is this the same reason the plane moves? If I release the car, and the conveyor belt continued, the car would immediately reverse direction.

Sometimes I can't believe I'm this stupid. I'm really not, but there are some things which my brain just can't grasp.

[Sharkey]

As far as the plane is concerned, the friction caused by the wheels on the belt is the same as would be experienced on a conventional runway.

The belt doesn’t have to accelerate instantaneously, just fast enough to restore equilibrium before the plane reaches takeoff speed.

[MidGe]

Hi Sharkey,

[ QUOTE ]
The belt doesn’t have to accelerate instantaneously, just fast enough to restore equilibrium before the plane reaches takeoff speed.

[/ QUOTE ]

The takeoff speed has nothing to do with ground speed, all to do with airspeed. The question really is what is the wind strength and direction. The rest is a red herring.

[purnell]

[ QUOTE ]
Hi Sharkey,

[ QUOTE ]
The belt doesn’t have to accelerate instantaneously, just fast enough to restore equilibrium before the plane reaches takeoff speed.

[/ QUOTE ]

The takeoff speed has nothing to do with ground speed, all to do with airspeed. The question really is what is the wind strength and direction. The rest is a red herring.

[/ QUOTE ]

I've been assuming the airspeed is zero relative to the stationary plane.

[Sharkey]

Friction between the car’s wheels and the conveyor belt would only become a factor at very high rpm.

A belt sander might be more like it.

[Sharkey]

[ QUOTE ]
wheel speed = plane speed + belt speed,

which means

plane speed = wheel speed - belt speed.

It’s apparent that if belt speed is increasing faster than wheel speed, the plane is decelerating.

[/ QUOTE ]

How can belt speed increase faster than wheel speed?

Because:

wheel speed = plane speed + belt speed,

and plane speed will increase at a slower rate than belt speed due to the offsetting effects of greater friction with higher wheel speed.

[purnell]

[ QUOTE ]
It works like this:

wheel speed = plane speed + belt speed,

which means

plane speed = wheel speed - belt speed.

It’s apparent that if belt speed is increasing faster than wheel speed, the plane is decelerating.

And, once more, part specs not included in the OP are irrelevant.

God speed, and come out to play again soon, won’t you?

[/ QUOTE ]

Plane speed does not equal wheel speed minus belt speed, unless the brake is applied to lock the wheel.

Plane speed = the sum of the forces acting on the plane * time elapsed

The non-negligible forces acting on the plane are thrust, friction, and gravity.

The engines produce force in the forward direction.

The wheel system produces a much smaller force, when in contact with the ground (or the belt) in the rearward direction.

The belt applies force to the bottom of the wheel system in the rearward direction, causing the wheel system to spin counter-clockwise, which results in a clockwise frictional force. To the extent that this force works in the opposite direction of the belt, it produces a force in the forward direction.

Gravity is constant and works vertically, with no horizontal component.

That's all there is.

Clearly the sum of the forces is in the forward direction, and will produce forward acceleration over time.

edit: Thank you for making me clarify my thinking on this- I'm ashamed that it took me so long to produce this fundamental result. /images/graemlins/blush.gif

[Marko Schmarko]

[ QUOTE ]
Suppose I'm holding a hot wheel car in my right hand. There is a conveyor belt in front of me running from my right to left. I place the hot wheel car on the conveyor belt with the front of the car pointing to to my left.

[/ QUOTE ]

Sharkey, can you address this? Are you asserting that your arm would be immobile? This seems like a perfect analogy, and it seems obvious that you could move the car.
- mark

[Sharkey]

[ QUOTE ]
Plane speed does not equal wheel speed minus belt speed, unless the brake is applied to lock the wheel.

The wheel system produces a much smaller force, when in contact with the ground (or the belt) in the rearward direction.

The belt applies force to the bottom of the wheel system in the rearward direction, causing the wheel system to spin counter-clockwise, which results in a clockwise frictional force. To the extent that this force works in the opposite direction of the belt, it produces a force in the forward direction.

[/ QUOTE ]

The wheels move the plane forward, and the belt moves it back. So: plane speed = wheel speed - belt speed. What’s not to get?

The wheel system produces an arbitrarily great frictional force. It’s up to the conveyer belt to determine that.

The belt contributes to a friction force in its own direction: backward. Just like the road under a car.

[purnell]

[ QUOTE ]
[ QUOTE ]
Suppose I'm holding a hot wheel car in my right hand. There is a conveyor belt in front of me running from my right to left. I place the hot wheel car on the conveyor belt with the front of the car pointing to to my left.

[/ QUOTE ]

Sharkey, can you address this? Are you asserting that your arm would be immobile? This seems like a perfect analogy, and it seems obvious that you could move the car.
- mark

[/ QUOTE ]

Physics I.

change in velocity = sum of forces * elapsed time.

[purnell]

[ QUOTE ]
The wheels move the plane forward, and the belt moves it back.

[/ QUOTE ]

The wheels do not move the plane forward. I thought you got that already.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Suppose I'm holding a hot wheel car in my right hand. There is a conveyor belt in front of me running from my right to left. I place the hot wheel car on the conveyor belt with the front of the car pointing to to my left.

[/ QUOTE ]

Sharkey, can you address this? Are you asserting that your arm would be immobile? This seems like a perfect analogy, and it seems obvious that you could move the car.
- mark

[/ QUOTE ]

If the belt was going around fast enough, you couldn’t drag the car over the surface of the belt against the friction being generated. It’s like swimming against the current.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
The wheels move the plane forward, and the belt moves it back.

[/ QUOTE ]

The wheels do not move the plane forward. I thought you got that already.

[/ QUOTE ]

That’s right, the engines do. A semantical issue.

The wheels move forward at the difference in speed between the plane and the belt,

or:

wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[Nottom]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Suppose I'm holding a hot wheel car in my right hand. There is a conveyor belt in front of me running from my right to left. I place the hot wheel car on the conveyor belt with the front of the car pointing to to my left.

[/ QUOTE ]

Sharkey, can you address this? Are you asserting that your arm would be immobile? This seems like a perfect analogy, and it seems obvious that you could move the car.
- mark

[/ QUOTE ]

If the belt was going around fast enough, you couldn’t drag the car over the surface of the belt against the friction being generated. It’s like swimming against the current.

[/ QUOTE ]

OK you win. In your scenario where the wheels have to deal with friction and the belt doesn't, the belt will win because the wheels will eventually disintigrate and the plane will crash to the ground and fly backwards at some rediculous speed killing everyone onboard in a fiery explosion.

If your intention was to ask an impossible question where you want to win an argument via a nitpick then you succeeded. The classic example of the problem usually assumes frictionless wheels and in that case the plane will take off for all the reasons mentioned above.

[Nottom]

[ QUOTE ]


wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[/ QUOTE ]

This is the crux of the problem. If "plane speed" > 0 then this is an impossible equation.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Suppose I'm holding a hot wheel car in my right hand. There is a conveyor belt in front of me running from my right to left. I place the hot wheel car on the conveyor belt with the front of the car pointing to to my left.

[/ QUOTE ]

Sharkey, can you address this? Are you asserting that your arm would be immobile? This seems like a perfect analogy, and it seems obvious that you could move the car.
- mark

[/ QUOTE ]

If the belt was going around fast enough, you couldn’t drag the car over the surface of the belt against the friction being generated. It’s like swimming against the current.

[/ QUOTE ]

OK you win. In your scenario where the wheels have to deal with friction and the belt doesn't, the belt will win because the wheels will eventually disintigrate and the plane will crash to the ground and fly backwards at some rediculous speed killing everyone onboard in a fiery explosion.

If your intention was to ask an impossible question where you want to win an argument via a nitpick then you succeeded. The classic example of the problem usually assumes frictionless wheels and in that case the plane will take off for all the reasons mentioned above.

[/ QUOTE ]

Yet again: part specs are not in evidence. This is a thought experiment (maybe using some futuristic materials).

[Lestat]

[ QUOTE ]
Friction between the car’s wheels and the conveyor belt would only become a factor at very high rpm.

A belt sander might be more like it.

[/ QUOTE ]

Even so. Picture running a hot wheel car over a belt sander. No matter how fast the sander goes in the opposite direction of the car's wheels, you will be able to move the car forward, because the car's wheels are not what's generating the forward torque. Why wouldn't thrust work the same way?

[Sharkey]

[ QUOTE ]
[ QUOTE ]


wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[/ QUOTE ]

This is the crux of the problem. If "plane speed" > 0 then this is an impossible equation.

[/ QUOTE ]

You’re riding a bicycle down the road:

wheel speed = bike speed + road speed. Obvious.

You’re riding a bicycle 5 mph at the airport and get on a 4 mph conveyor belt gong the wrong way (so that you’re going forward at 1 mph):

wheel speed = bike speed + belt speed,

or

5 mph = 1 mph + 4 mph.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Friction between the car’s wheels and the conveyor belt would only become a factor at very high rpm.

A belt sander might be more like it.

[/ QUOTE ]

Even so. Picture running a hot wheel car over a belt sander. No matter how fast the sander goes in the opposite direction of the car's wheels, you will be able to move the car forward, because the car's wheels are not what's generating the forward torque. Why wouldn't thrust work the same way?

[/ QUOTE ]

The belt sander is 5 feet wide and you’re pushing a refrigerator.

[Jeremy517]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The wheels move the plane forward, and the belt moves it back.

[/ QUOTE ]

The wheels do not move the plane forward. I thought you got that already.

[/ QUOTE ]

That’s right, the engines do. A semantical issue.


[/ QUOTE ]

It is hardly a semantical issue.

[Nottom]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[/ QUOTE ]

This is the crux of the problem. If "plane speed" > 0 then this is an impossible equation.

[/ QUOTE ]

You’re riding a bicycle down the road:

wheel speed = bike speed + road speed. Obvious.

You’re riding a bicycle 5 mph at the airport and get on a 4 mph conveyor belt gong the wrong way (so that you’re going forward at 1 mph):

wheel speed = bike speed + belt speed,

or

5 mph = 1 mph + 4 mph.

[/ QUOTE ]

Yes but in your scenario, conveyer speed = wheel speed.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The wheels move the plane forward, and the belt moves it back.

[/ QUOTE ]

The wheels do not move the plane forward. I thought you got that already.

[/ QUOTE ]

That’s right, the engines do. A semantical issue.


[/ QUOTE ]

It is hardly a semantical issue.

[/ QUOTE ]

Okay, it’s an earthquake.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[/ QUOTE ]

This is the crux of the problem. If "plane speed" > 0 then this is an impossible equation.

[/ QUOTE ]

You’re riding a bicycle down the road:

wheel speed = bike speed + road speed. Obvious.

You’re riding a bicycle 5 mph at the airport and get on a 4 mph conveyor belt gong the wrong way (so that you’re going forward at 1 mph):

wheel speed = bike speed + belt speed,

or

5 mph = 1 mph + 4 mph.

[/ QUOTE ]

Yes but in your scenario, conveyer speed = wheel speed.

[/ QUOTE ]

Which scenario and how do you mean? Please be specific.

[Lestat]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[/ QUOTE ]

This is the crux of the problem. If "plane speed" > 0 then this is an impossible equation.

[/ QUOTE ]

You’re riding a bicycle down the road:

wheel speed = bike speed + road speed. Obvious.

You’re riding a bicycle 5 mph at the airport and get on a 4 mph conveyor belt gong the wrong way (so that you’re going forward at 1 mph):

wheel speed = bike speed + belt speed,

or

5 mph = 1 mph + 4 mph.

[/ QUOTE ]

First, I'll easily admit that you're much more proficient at this stuff than I am. However, this is where I think you're going wrong.

The reason your 5 mph = 1 mph + 4 mph works, is because the 5 mph is emanating from the wheels of the bike. In other words, the torque is attached to the bike's wheels which is resting on top of the conveyor belt.

But instead of someone pedaling the bike at 5 mph, suppose they were moving at 5 mph while suspended from the ceiling above the conveyor belt holding on to the bike's handle bars. Do you have any doubt that the bike would be "pulled" forward in the opposite direction of the conveyor belt at more than 1 mph.?

The wheels of a plane only facillitate forward motion, they do not cause it. The thrust is a completely seperate force acting on the plane generating forward momentum. The wheels just go along for the ride.

[Nottom]

[ QUOTE ]

Which scenario and how do you mean? Please be specific.

[/ QUOTE ]

The one you have been arguing about for the last 20 posts. If the conveyer doesn't "match" the speed of the wheels than what is keeping the plane from moving foward?

If the wheels are moveing faster than the conveyer than obviously the plane moves and eventually takes off.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]


wheel speed = plane speed + belt speed

(belt speed being measured in the negative direction).

[/ QUOTE ]

This is the crux of the problem. If "plane speed" > 0 then this is an impossible equation.

[/ QUOTE ]

You’re riding a bicycle down the road:

wheel speed = bike speed + road speed. Obvious.

You’re riding a bicycle 5 mph at the airport and get on a 4 mph conveyor belt gong the wrong way (so that you’re going forward at 1 mph):

wheel speed = bike speed + belt speed,

or

5 mph = 1 mph + 4 mph.

[/ QUOTE ]

First, I'll easily admit that you're much more proficient at this stuff than I am. However, this is where I think you're going wrong.

The reason your 5 mph = 1 mph + 4 mph works, is because the 5 mph is emanating from the wheels of the bike. In other words, the torque is attached to the bike's wheels which is resting on top of the conveyor belt.

But instead of someone pedaling the bike at 5 mph, suppose they were moving at 5 mph while suspended from the ceiling above the conveyor belt holding on to the bike's handle bars. Do you have any doubt that the bike would be "pulled" forward in the opposite direction of the conveyor belt at more than 1 mph.?

The wheels of a plane only facillitate forward motion, they do not cause it. The thrust is a completely seperate force acting on the plane generating forward momentum. The wheels just go along for the ride.

[/ QUOTE ]

Same thing, but instead off pedaling, the power comes from a powerful fan attached to the back of the bicycle.

You’re riding a bicycle down the road:

wheel speed = bike speed + road speed. Obvious.

You’re riding a bicycle 5 mph at the airport and get on a 4 mph conveyor belt gong the wrong way (so that you’re going forward at 1 mph):

wheel speed = bike speed + belt speed,

or

5 mph = 1 mph + 4 mph.

[Nottom]

[ QUOTE ]
First, I'll easily admit that you're much more proficient at this stuff than I am. However, this is where I think you're going wrong.

The reason your 5 mph = 1 mph + 4 mph works, is because the 5 mph is emanating from the wheels of the bike. In other words, the torque is attached to the bike's wheels which is resting on top of the conveyor belt.

[/ QUOTE ]

Actualy "wheel speed = belt speed + plane/bike speed" is about the only correct thing he has said. Since the wheels don't have any power of their own, their speed is simply how fast the belt is moving plus how fast the object is moving down the belt.

In your example, with a wind powered bike on a 4 mph belt. The bike moves foward at whatever rate the fan can propel it and the wheels move at that rate + 4mph.

[Sharkey]

[ QUOTE ]
[ QUOTE ]

Which scenario and how do you mean? Please be specific.

[/ QUOTE ]

The one you have been arguing about for the last 20 posts. If the conveyer doesn't "match" the speed of the wheels than what is keeping the plane from moving foward?

If the wheels are moveing faster than the conveyer than obviously the plane moves and eventually takes off.

[/ QUOTE ]

As soon as the wheels go faster than the belt, the belt accelerates to compensate.

Read the OP.

[Nottom]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

Which scenario and how do you mean? Please be specific.

[/ QUOTE ]

The one you have been arguing about for the last 20 posts. If the conveyer doesn't "match" the speed of the wheels than what is keeping the plane from moving foward?

If the wheels are moveing faster than the conveyer than obviously the plane moves and eventually takes off.

[/ QUOTE ]

As soon as the wheels go faster than the belt, the belt accelerates to compensate.

Read the OP.

[/ QUOTE ]

I have no clue why I've bother to respond to you at all. I'm done with this thread becasue its pointless to argue with trolls.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

Which scenario and how do you mean? Please be specific.

[/ QUOTE ]

The one you have been arguing about for the last 20 posts. If the conveyer doesn't "match" the speed of the wheels than what is keeping the plane from moving foward?

If the wheels are moveing faster than the conveyer than obviously the plane moves and eventually takes off.

[/ QUOTE ]

As soon as the wheels go faster than the belt, the belt accelerates to compensate.

Read the OP.

[/ QUOTE ]

I have no clue why I've bother to respond to you at all. I'm done with this thread becasue its pointless to argue with trolls.

[/ QUOTE ]

I thought you wanted to learn. That’s what I’m doing here.

[bunny]

What's a troll?

Just a generic term of abuse? Or some kind of internet jargon to label people on an internet forum?

I know I'm showing my age, but I've seen the term several times now and cant make all the different contexts fit the name. /images/graemlins/frown.gif

[BCPVP]

[ QUOTE ]
What's a troll?

Just a generic term of abuse? Or some kind of internet jargon to label people on an internet forum?

I know I'm showing my age, but I've seen the term several times now and cant make all the different contexts fit the name. /images/graemlins/frown.gif

[/ QUOTE ]
Whenever you're confused by some internet jargon, check Urban Dictionary (http://www.urbandictionary.com/define.php?term=troll)

[bunny]

[ QUOTE ]
Whenever you're confused by some internet jargon, check Urban Dictionary (http://www.urbandictionary.com/define.php?term=troll)

[/ QUOTE ]
Thanks!

[Kevin J]

I've been following this thread with interest.

Why are we even concerned with wheel speed? Wheel speed should not even factor into the equation. The missing element so far, is gravity (g). It is gravitational force which acts upon the object and keeps it in contact with the conveyor belt.

I'm not a physicist and can't provide an equation, but it should look something like...

Forward momentum = thrust propulsion - (gravitional force upon object's weight + speed of conveyor belt).

I'm sure other things may need to be factored in to be exact, but that's the bulk of it. Wheel speed is not a variable when calculating the bike's forward momentum (speed), because the object is not being propelled by it's wheels.

[DougShrapnel]

Shouldn't this question be phrased. Can a helicopter take off?

[Kevin J]

[ QUOTE ]
That’s right, the engines do. A semantical issue.

[/ QUOTE ]

This is much more than a semantical issue Sharkey.

[ QUOTE ]
The wheels move forward at the difference in speed between the plane and the belt,

[/ QUOTE ]

Even though the wheels are spinning, they have no forward or backward momentum in their own right. They are attached to the plane and make forward or backward progress only as the plane does.

[ QUOTE ]
wheel speed = plane speed + belt speed

[/ QUOTE ]

This is like calculating a room's temperature to get it's area. It is an irrelevant equation to the problem at hand.

Gravity is the acting force which keeps the plane (and it's wheels), in contact with the conveyor belt. So the main thing thrust needs to overcome is the gravitational force of the plane's weight upon the conveyor belt, Not... Wheel speed. Although, wheel spin will cause the belt to move in an opposite direction, thrust force will overcome this and the plane will eventually take off.

[Kevin J]

Possibly. I don't know, because I'm not a physicist. I would think a helicopter is operating under a different set of laws. A plane takes off due to air flow over it's wings. This necessitates forward velocity. A helicopter doesn't have wings and takes off for different reasons, so no forward velocity is required.

[Kevin J]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Friction between the car’s wheels and the conveyor belt would only become a factor at very high rpm.

A belt sander might be more like it.

[/ QUOTE ]

Even so. Picture running a hot wheel car over a belt sander. No matter how fast the sander goes in the opposite direction of the car's wheels, you will be able to move the car forward, because the car's wheels are not what's generating the forward torque. Why wouldn't thrust work the same way?

[/ QUOTE ]

The belt sander is 5 feet wide and you’re pushing a refrigerator.

[/ QUOTE ]

I think you might have countered yourself here. Why would it be more difficult for the refrigerator to overcome the belt sander? Not because of it's wheel's speed, but because of it's weight! IOW- Gravity, not wheel speed.

[Nottom]

[ QUOTE ]
Shouldn't this question be phrased. Can a helicopter take off?

[/ QUOTE ]

Helicopters don't have wheels so the belt wouldn't move. Even if it did a helicopter creates its own lift by spinning the rotor. There is really no arguement for it not to take off.

An somewhat equivalent question for a helicopter would be that it was sitting on a giant spinning circle that spun at the same speed as the rotor. Would the helicopter take off?

(I would think no)

[DougShrapnel]

[ QUOTE ]
Helicopters don't have wheels so the belt wouldn't move. Even if it did a helicopter creates its own lift by spinning the rotor. There is really no arguement for it not to take off.

[/ QUOTE ] Right, essentialy you are turning an airplane into a helicopter. If the engine is powerfull enough to create enough lift, the plane will take off. I have no idea if a standard commercial engine is powerful enough, but my guess is that it is. The fluid speed, in this case airspeed, is what matters. Forward motion in realtion to a marble gound, a grass gound, or a converyor belt is not the only cause of lift. Forward motion in relation to the air is the cause of lift.

[ QUOTE ]
An somewhat equivalent question for a helicopter would be that it was sitting on a giant spinning circle that spun at the same speed as the rotor. Would the helicopter take off?


[/ QUOTE ] I think this one is no. The fluid speed would be dictated by the helicopter it self. I don't think a helicopters engine is srong enough to create enough lift from the body of the helicopter instead of the rotor.

[TomTom]

Rather then considering belt speed locked to wheel speed, I first back off and think HOW they would be locked. Being an engineer, I would do this by measuring the position of the wheels. Say the plane points to the left, and the belt revolves to the right. If my control system saw the wheel move to the left, the belt would speed up. If the wheel moves to the right, the belt slows down. A classic single loop control feedback loop.

Thus a plane just beginning to move (to the left) would cause the belt to speed up (to the right) to return to the equilibrium state of wheel at the starting point.

Now wheels are by design low friction bearings, not the best (meaning the worst) way to impart a force is by spinning the wheels. If we let this friction go to zero we quickly reach the point where the slightest movement of the plane will require an infinite belt speed to overcome.

This infinite speed is the given of the thought experiment (the “converter belt which is rigged to move/spin in the opposite direction” part). So by the very definition of this problem, the plane sits exactly where it is on the ground, it has no air speed, it does not take off.

And of course, this assumes the belt itself doesn’t generate a back wind to raise the plane. But since the plane never moves, the belt can be as small as the contact area of each individual wheel, so this isn’t a limiting assumption. Either use multiple small belts, or one large one with wheel cut outs. No movement, no wind, no rise.

If we let this friction go above zero as we transition from pure thought to some kind of physical experiment, the wheel bearings become the critical item. The belt now need to turn fast enough (but now finitely so) to keep the plane in place by using this friction as a brake. The wheel bearings would need to absorb whatever force the engines are imparting to the entire plane.

I don’t see why this would be a problem at all. After all, I seem to be able to drive my car to and from work every day, and those bearings seem to handle the load acceptably well.

Here’s another way to look at the whole thing: Don’t think of belt/plane relative motion, be the guy standing next to the belt on the runway doing this experiment. Since he knows he wouldn’t see the plane move relative to him, instead of buying an expensive high speed conveyer belt, he buys a couple of bottle of crazy glue, and sticks the plane to the runway via the wheels.

De plane don’t move. De plane don’t take off.

[Sharkey]

[ QUOTE ]
Shouldn't this question be phrased. Can a helicopter take off?

[/ QUOTE ]

The question really boils down to the following:

Can the plane acquire the minimum speed necessary for takeoff when, with every increase in plane speed, there follows an increase in belt speed by however much is necessary to increase the backward frictional force on the plane by however much is necessary to stop the plane?

Given that the answer is “no”, the plane will not move forward. Lift is not a factor.

[Kevin J]

Hopefully, you are an engineer who is not employed in the aerodynamic or jet propulsion fields. /images/graemlins/laugh.gif

Everything you say is absolutely correct IF.... The wheels were what propelled the plane forward. However, thrust is what propells a plane, not it's wheels. The wheels only facillitate forward movement, they are not the cause of forward movement.

The plane's forward progress is hampered by the loss of surface to wheel friction, but it is not arrested altogether. Thrust force, which operates independent of wheel rotation (or any part of the plane for that matter), will eventually supercede this loss of friction and the plane WILL take off I assure you.

[Sharkey]

Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[madnak]

[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

There is no backward force. The bearings prevent any force from being exerted.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

There is no backward force. The bearings prevent any force from being exerted.

[/ QUOTE ]

No plane has ever been invented that would not coast to a stop on a long enough runway, air drag discounted.

[Sharkey]

[ QUOTE ]
The wheel bearings would need to absorb whatever force the engines are imparting to the entire plane.

[/ QUOTE ]

Some good points in your post, except be advised that the tires on the runway are also a significant source of friction.

[Kevin J]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

There is no backward force. The bearings prevent any force from being exerted.

[/ QUOTE ]

No plane has ever been invented that would not coast to a stop on a long enough runway, air drag discounted.

[/ QUOTE ]

You cannot discount air drag, because it is this, and gravity which cause the plane to eventually stop. Remember, "An object in motion tends to stay in motion". How long of a runway would you need for the same plane to coast to a stop in a complete vaccuum?

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

There is no backward force. The bearings prevent any force from being exerted.

[/ QUOTE ]

No plane has ever been invented that would not coast to a stop on a long enough runway, air drag discounted.

[/ QUOTE ]

You cannot discount air drag, because it is this, and gravity which cause the plane to eventually stop. Remember, "An object in motion tends to stay in motion". How long of a runway would you need for the same plane to coast to a stop in a complete vaccuum?

[/ QUOTE ]

No plane has ever been invented that would not coast to a stop in a vacuum on a long enough runway. How long of a runway I would need is irrelevant, since no speed limitation was put on the belt being used to create the same friction.

Gravity has nothing to do with stopping the plane once it’s on the runway, except that it leads to the very friction I’m talking about.

[Kevin J]

[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

The wheels are not the "source" of the backward force either. The conveyor belt is the "source" of backward force. The thrust is the "source" of forward force. Gravity is the "source" that keeps the wheels in contact with the conveyor belt.

You need to let go of the notion that the wheels are some primary causal force. This would be the case with a drive train in a car, but not with a plane. A plane moves by thrust. Not by the drive from it's wheels.

Unfortunately, I am not smart enough/educated enough to explain this in further detail.

[Kevin J]

[ QUOTE ]
Gravity has nothing to do with stopping the plane once it’s on the runway, except that it leads to the very friction I’m talking about.

[/ QUOTE ]

Exactly. And that (gravity), is a very important factor. It's what is holding the plane's wheels to the conveyory belt.

Like I said, I don't know of any other way to explain this. You can prove this to yourself, but performing your own experiment..

Attach powerful fan to a skate board. Set the fan on high and measure the MPH of the skate on a stationary surface. Now get a treadmill and set it to the same MPH that the skateboard achieved on the stationary surface. Place the skateboard on the treadmill with the fan at the same setting.

I'm willing to bet that the skateboard is not dumped backwards off the treadmill. You might have to slowly increase the treadmill speed to that of the skateboard (to give the skateboard time to catch up). Regardless, you should be able to clearly see that the speed of the skateboards wheels are not the only factor in what happens to the skateboard.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

The wheels are not the "source" of the backward force either. The conveyor belt is the "source" of backward force. The thrust is the "source" of forward force. Gravity is the "source" that keeps the wheels in contact with the conveyor belt.

You need to let go of the notion that the wheels are some primary causal force. This would be the case with a drive train in a car, but not with a plane. A plane moves by thrust. Not by the drive from it's wheels.

Unfortunately, I am not smart enough/educated enough to explain this in further detail.

[/ QUOTE ]

The word “source” is being used somewhat loosely in this thread, so don’t get hung up on it.

There are two relevant forces acting on the plane: thrust and friction.

The engines provide the (forward) thrust.

The (backward) frictional force (from the tires on the conveyor belt, the wheel assembly, etc) is the same as would be encountered by the plane coasting to a stop in a vacuum.

[CORed]

[ QUOTE ]
[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

There is no backward force. The bearings prevent any force from being exerted.

[/ QUOTE ]

Provided you have magical zero friction bearings to go along with your magical conveyor belt that can accelerate instantaneously to whatever speed necessary to counteract the thrust of the jet engines.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Gravity has nothing to do with stopping the plane once it’s on the runway, except that it leads to the very friction I’m talking about.

[/ QUOTE ]Attach powerful fan to a skate board. Set the fan on high and measure the MPH of the skate on a stationary surface. Now get a treadmill and set it to the same MPH that the skateboard achieved on the stationary surface. Place the skateboard on the treadmill with the fan at the same setting.

[/ QUOTE ]

In your example, the skateboard would be stationary with wheel speed and belt speed both equal to that of the skateboard on the pavement previously.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Incorrect.

The wheels may not be causing the forward motion, but they are the source of an arbitrarily great backward force which will be raised by way of an increase in belt speed whenever the plane begins to move foreward.

[/ QUOTE ]

There is no backward force. The bearings prevent any force from being exerted.

[/ QUOTE ]

Provided you have magical zero friction bearings to go along with your magical conveyor belt that can accelerate instantaneously to whatever speed necessary to counteract the thrust of the jet engines.

[/ QUOTE ]

The tires on the runway are a very significant source of friction.

The belt doesn’t have to accelerate instantaneously, just fast enough to restore equilibrium before the plane reaches takeoff speed.

[DougShrapnel]

If anyone is going to say that the plane doesn't take off.
Please tell me the the ammount of thrust provided by a commercial jet engine at full capacity. And the wind speed needed to lift a commercial jet airplane.

[CORed]

[ QUOTE ]
The belt doesn’t have to accelerate instantaneously, just fast enough to restore equilibrium before the plane reaches takeoff speed.

[/ QUOTE ]
Okay, instantaneous is not correct, but it would require enormous speed to generate enough frictional force to counter the thrust of the jet engine at full throttle and enormous acceleration to achieve that speed quickly enough to prevent the plane from moving forward. Crazy glue (or a wheel chock) would be a lot more practical.

[Benman]

[ QUOTE ]
The plane will take off. The wheels aren't moving the plane forward, they're just spinning idly.

[/ QUOTE ]

Yes and no. This is entirely true if the conveyor belt's velocity adjustment happens quickly/simultaneously with the increasing thrust of the engines. The best analogy is this: imagine you're standing on an area rug wearing well-oiled rollerblades. If a very strong person "pulled the rug out from under you", you wouldn't accellerate backwards at the same speed of the rug. You'd essentially stay in place relative to the room as the rug rolled out from under you, a sort of skidding effect. As the coveyor belt accelerates simultaneously with the increase in thrust, the same thing would likely occur with the plane, and most of the increased backwards velocity of the conveyor belt would be converted to increased wheel speed rather than increased speed of the plane itself. So far so good.

But, imagine if the conveyor belt began to move backwards very slowly, before you turned the engines on. If the initial acceleration of the conveyor belt was slow enough, there wouldn't be any skidding/spinning effect, and the weight of the plane would keep the whole plane "glued" to the conveyor belt, which is accellerating backwards at a very small rate. Even if there was a small amount of skid/spinning at first, if the backwards movement of the belt was gentle enough that skidding/spinning would eventually end and the plane would eventually be moving backwards at the same rate as the conveyor belt (just as you would be moving backwards relative to the room if the guy didn't yank the carpet out from under you but rather ever so slowly started to drag it backwards with you on it). Once the plane reaches -10 mph relative to its surroundings, then that is a true negative velocity that would require a certain amount of thrust just to get the plane back to relative motionless (vis a vis the surrounding ground/air). At that point, a given level of thrust from the engines that would normally move the plane forward at 10mph would only suffice to keep you stationary relative to the surrounding air, thus creating no lift.

As the speed of the conveyor belt increases backwards (again slowly enough that the weight of the plane cancels out the skidding/spinning effect) to a sufficient degree, other forces will come into play that may aid the plane. One, the backwards moving plane will encounter negative wind drag, and this will begin to create skidding/spinning that will slow the planes backwards motion.

The real answer to this overall paradox depends on how suddenly the conveyor belt adjusts to increases in thrust. If the conveyor belt doesn't start until the pilot guns the engines, he could probably take off by applying maximum thrust as quickly as possible. Because jets have so much thrust, the conveyor belt would have to begin its backwards motion so rapidly that most of the backwards motion would be converted to wheel spin, which does not move the plane backwards (even in the absence of forward thrust). If the pilot applied thrust too slowly, he might never make it however, as the conveyor belt could be accelerated slowly enough to avoid/minimize spin.

[Sharkey]

[ QUOTE ]
If anyone is going to say that the plane doesn't take off.
Please tell me the the ammount of thrust provided by a commercial jet engine at full capacity. And the wind speed needed to lift a commercial jet airplane.

[/ QUOTE ]

Engine maximum is irrelevant, because there is nothing in the setup conditions preventing belt speed being increased to match any thrust with friction.

Minimum airspeed is also irrelevant, since the plane won’t move forward.

[CORed]

You know, I don't know how much thrust a jet engine puts out, and I don't have access to the ground speed vs. rolling resistance graph for the landing gear, either. But my intuitive take is, that, using commercially available materials and power sources, and a budget smaller than the GNP of a large state, it would be impossible to construct a conveyor belt that, placed under the wheels of a jet airliner (or a piper cub, for that matter), would prevent it from taking off. However given a budget the size of the national debt, hoover damn and a couple spare nuclear power plants to power it, the flow of the Mississippi river to cool it, and a conveyor belt of pure unobtanium, with 1,000,000 times the tensile strength of carbon nanotube (leftover after building our geosynchronous elevator, perhaps), it could be done, most likely by blowing out the tires or overheating and seizing the wheel bearings. A hell of a way to run an airline, if you ask me.

[Sharkey]

The condition put forward in the OP says the conveyor belt “is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.”

Why start side bets on details not in evidence, like performance data?

[CORed]

[ QUOTE ]
The condition put forward in the OP says the conveyor belt “is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.”

Why start side bets on details not in evidence, like performance data?

[/ QUOTE ]

The use of the phrase, "rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels” leads me to believe that the OP was working under the erroneous assumption that the wheels are what drives the plane. I think it is a valid point that, while a backward moving conveyor belt, at a sufficient speed, could in theory produce enough frictional force to prevent the plane from moving forward, that as a practical matter, it would be nearly impossible to accomplish.

[DougShrapnel]

[ QUOTE ]
Minimum airspeed is also irrelevant

[/ QUOTE ] It is the only thing relevent. The engine is still pushing wind over the wings. If the airspeed created by the engine is greater than the minimum airspeed for a plane to lift off, the plane will take off.

[CORed]

[ QUOTE ]
[ QUOTE ]
Minimum airspeed is also irrelevant

[/ QUOTE ] It is the only thing relevent. The engine is still pushing wind over the wings. If the airspeed created by the engine is greater than the minimum airspeed for a plane to lift off, the plane will take off.

[/ QUOTE ]

The enginge doesn't "push air over the wings" (a propellor does to some extent, but that's not the primary means of generating lift). The engine (jet or propellor) generates thrust to move the plane forward. Any fixed-wing aircraft requires a minimum air speed to take off.

[DougShrapnel]

The engine (jet or propellor) generates thrust to move the plane forward thru the air. Any fixed-wing aircraft requires a minimum air speed to take off. So what is the airspeed of the plane? Not the ground speed.

[Sharkey]

[ QUOTE ]
I think it is a valid point that, while a backward moving conveyor belt, at a sufficient speed, could in theory produce enough frictional force to prevent the plane from moving forward, that as a practical matter, it would be nearly impossible to accomplish.

[/ QUOTE ]


The belt speeds required could be brought down significantly by increasing surface traction.

[Benman]

[ QUOTE ]
The condition put forward in the OP says the conveyor belt “is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.”

Why start side bets on details not in evidence, like performance data?

[/ QUOTE ]

Think of it this way. Let's say a plane can go from 0 to 100 knots in 10 seconds (I'm making this figure up). If the OP's conditions are to be applied, the conveyor belt would have to go from 0 to 100 knots in reverse in 10 seconds as well. The problem is your assumption that such a rapid reverse acceleration by the conveyor belt would translate into -100 knots of airplane velocity relative to teh surroundings (thus keeping it in place). It wouldn't. Read my analogy again about the rollerskates on a rug. Most of reverse accelleration would be translated into mere spinning of the tires. Meanwhile, the forward thrust would move the plane forward. The tires would thus always be moving forward at a greater rate than the conveyor belt was capable of moving backwards. This is not through any fault in the power or capability of the conveyor belt, but rather the fact that most of the rapid backwards motion is converted into meaniningless wheel spin. Thus, the forward thrust adds a speed component to the wheels that's always one step ahead of the conveyor belt. In a sense, the OP's conditions are a theoretical impossiblity.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
The condition put forward in the OP says the conveyor belt “is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels.”

Why start side bets on details not in evidence, like performance data?

[/ QUOTE ]

Think of it this way. Let's say a plane can go from 0 to 100 knots in 10 seconds (I'm making this figure up). If the OP's conditions are to be applied, the conveyor belt would have to go from 0 to 100 knots in reverse in 10 seconds as well. The problem is your assumption that such a rapid reverse acceleration by the conveyor belt would translate into -100 knots of airplane velocity relative to teh surroundings (thus keeping it in place). It wouldn't. Read my analogy again about the rollerskates on a rug. Most of reverse accelleration would be translated into mere spinning of the tires. Meanwhile, the forward thrust would move the plane forward. The tires would thus always be moving forward at a greater rate than the conveyor belt was capable of moving backwards. This is not through any fault in the power or capability of the conveyor belt, but rather the fact that most of the rapid backwards motion is converted into meaniningless wheel spin. Thus, the forward thrust adds a speed component to the wheels that's always one step ahead of the conveyor belt. In a sense, the OP's conditions are a theoretical impossiblity.

[/ QUOTE ]

If the plane and the belt are capable of the same acceleration, then: belt speed = wheel speed at (almost) all times, as in the OP.

What you seem to think of as mere “spinning of the tires” actually corresponds to a backward force on the plane which can be raised arbitrarily high to stop any forward motion. It is not the case that the wheels can “always be moving forward at a greater rate than the conveyor belt was capable of moving backwards”, because at some very high belt speed, the engines would be incapable of pushing the plane forward.

The reverse speed of the belt wouldn’t pull the plane suddenly backward any more that the (relative) reverse speed of a conventional runway instantly stops a coasting plane after landing.

[CORed]

In order to prevent the plane from moving forward, the belt wouold have to be able to accelerate much faster than the plane. It would take an extremely high speed of the belt to generate enough frictional force through the landing gear to counteract the thrust of the engines.

[CORed]

[ QUOTE ]
[ QUOTE ]
I think it is a valid point that, while a backward moving conveyor belt, at a sufficient speed, could in theory produce enough frictional force to prevent the plane from moving forward, that as a practical matter, it would be nearly impossible to accomplish.

[/ QUOTE ]


The belt speeds required could be brought down significantly by increasing surface traction.

[/ QUOTE ]
Huh? I suppose a rough an/or sticky surface on the belt would increase the rolling resistance of the tires somewhat (and generate more heat, so the tires would blow sooner), but I still don't think this is doable from a practical engineering standpoint.

[Sharkey]

[ QUOTE ]
In order to prevent the plane from moving forward, the belt wouold have to be able to accelerate much faster than the plane. It would take an extremely high speed of the belt to generate enough frictional force through the landing gear to counteract the thrust of the engines.

[/ QUOTE ]

Your “extremely high speed” is merely a function of the coefficient of friction between the tires and the belt, which can be increased by as much as necessary.

[CORed]

[ QUOTE ]
[ QUOTE ]
In order to prevent the plane from moving forward, the belt wouold have to be able to accelerate much faster than the plane. It would take an extremely high speed of the belt to generate enough frictional force through the landing gear to counteract the thrust of the engines.

[/ QUOTE ]

Your “extremely high speed” is merely a function of the coefficient of friction between the tires and the belt, which can be increased by as much as necessary.

[/ QUOTE ]

Well sure, or as someone else suggested, we can just glue the tires to the runway and forget about the silly conveyor belt.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
In order to prevent the plane from moving forward, the belt wouold have to be able to accelerate much faster than the plane. It would take an extremely high speed of the belt to generate enough frictional force through the landing gear to counteract the thrust of the engines.

[/ QUOTE ]

Your “extremely high speed” is merely a function of the coefficient of friction between the tires and the belt, which can be increased by as much as necessary.

[/ QUOTE ]

Well sure, or as someone else suggested, we can just glue the tires to the runway and forget about the silly conveyor belt.

[/ QUOTE ]

Modern chemistry has no shortage of glue-like substances that would allow the tires to roll. The point is that the objection that too-high belt speeds would be necessary to generate sufficient friction is wrong because it ignores this variable.

[Marko Schmarko]

I think the problem with Sharkey's argument is that he's rejecting perfect bearings, but espousing a perfect conveyer belt. Admittedly, a perfect conveyer belt was assumed in the problem, but that is just as ridiculous as frictionless bearings. Choose hypothetical or practical—don't mix and match.

Also, as has been pointed out previously, there is a limit to the conveyer belt's ability to apply a backwards force on the plane (which, assuming there is friction in the wheel bearings, it certainly can do): the friction between the belt and the tires.

There would be a point at which the wheels would hit their maximum speed, and the tires would lose traction. At this point, the plane's thrusters would prevail and the plane would slide to takeoff (assuming the tires didn't explode).

[purnell]

Wow, you guys are still at it? You (were supposed to have) learned this in the intro to classical physics course you took when you were a freshman.

If the plane starts with velocity = 0,

The velocity of the plane is equal to the product of the sum of the forces acting on it and the elapsed time.

v = (Sigma F) (t1 - t0)

Nothing else is relevant to this problem.

[bunny]

I love this thread - I have no idea who's arguing what anymore, but it has this hypnotic appeal and I doggedly read everything posted each time I log, still with no clue about what the answer is. /images/graemlins/tongue.gif

[Marko Schmarko]

Purnell: I, too, thought this problem was simpler than I now think it is.

The fact that the speed sensor is attached to the wheels, rather than the plane itself, means that the belt will spin arbitrarily quickly until it exerts some force on the plane. Assuming there is any friction in the bearings at all, this force won't be negligible given the belt's apparently infinite capability.

(Sigma F) = Thrust - friction from wheels until the thrust was = [coefficient of kinetic friction of the tires to the belt][normal]
.. I think.

[purnell]

/images/graemlins/laugh.gif This thread has gotten too big. I made the same point you just did last night- it's in here somewhere.

[Sharkey]

[ QUOTE ]
v = (Sigma F) (t1 - t0)[/b]

[/ QUOTE ]

That’s more or less the point of comparing a fast enough belt to a long enough runway considering the friction they both cause to stop the plane.

[Marko Schmarko]

Sharkey: do you agree or disagree that if the wheels are truly frictionless, the plane will take off as normal?

You may have posted this before, but I don't remember seeing it.

- mark

[Sharkey]

[ QUOTE ]
Assuming there is any friction in the bearings at all, this force won't be negligible given the belt's apparently infinite capability.

[/ QUOTE ]

There’s no need for “infinite capability”. The belt is only fighting the engines.

[purnell]

[ QUOTE ]
[ QUOTE ]
v = (Sigma F) (t1 - t0)[/b]

[/ QUOTE ]

That’s more or less the point of comparing a fast enough belt to a long enough runway considering the friction they both cause to stop the plane.

[/ QUOTE ]

Agreed. But if you get a magic conveyor belt, I want my magic wheel /images/graemlins/laugh.gif.

[Sharkey]

[ QUOTE ]
Sharkey: do you agree or disagree that if the wheels are truly frictionless, the plane will take off as normal?

You may have posted this before, but I don't remember seeing it.

- mark

[/ QUOTE ]

Without friction, the plane would take off.

What follows is pasted from a previous post of mine:

Friction is what causes the relationship between conveyer belt and wheels to have any significance. The condition that belt speed equals wheel speed would be pointless in the absence of friction, which would allow the wheels and the belt to be turning any which way, independently of each other, with no effect on or by the motion of the plane.

The presumed relevance of all included conditions trumps any strained attempts at ad hoc, fudge-factor tomfoolery by the other side of this debate.

[purnell]

[ QUOTE ]
[ QUOTE ]
Sharkey: do you agree or disagree that if the wheels are truly frictionless, the plane will take off as normal?

You may have posted this before, but I don't remember seeing it.

- mark

[/ QUOTE ]

What follows is pasted from a previous post of mine:

Friction is what causes the relationship between conveyer belt and wheels to have any significance. The condition that belt speed equals wheel speed would be pointless in the absence of friction, which would allow the wheels and the belt to be turning any which way, independently of each other, with no effect on or by the motion of the plane.

The presumed relevance of all included conditions trumps any strained attempts at ad hoc, fudge-factor tomfoolery by the other side of this debate.

[/ QUOTE ]

The frictional force is transmitted by the wheel bearing. Given a frictionless bearing, no force is transmitted to the plane's body.

[Sharkey]

[ QUOTE ]
The frictional force is transmitted by the wheel bearing. Given a frictionless bearing, no force is transmitted to the plane's body.

[/ QUOTE ]

So you’re going to eliminate tire friction but not wheel bearing friction?

On what basis do you make that distinction?

[purnell]

[ QUOTE ]
[ QUOTE ]
The frictional force is transmitted by the wheel bearing. Given a frictionless bearing, no force is transmitted to the plane's body.

[/ QUOTE ]

So you’re going to eliminate tire friction but not wheel bearing friction?

On what basis do you make that distinction?

[/ QUOTE ]

I'm eliminating bearing friction, not tire friction.

As i said before, if you get a magic conveyor belt, I want my magic wheel.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The frictional force is transmitted by the wheel bearing. Given a frictionless bearing, no force is transmitted to the plane's body.

[/ QUOTE ]

So you’re going to eliminate tire friction but not wheel bearing friction?

On what basis do you make that distinction?

[/ QUOTE ]

I'm eliminating bearing friction, not tire friction.

As i said before, if you get a magic conveyor belt, I want my magic wheel.

[/ QUOTE ]

No magical devices allowed.

There’s nothing impossible in a conveyor belt matching the power output of a jet plane’s engines, especially since there are no size limitations on the belt’s drive mechanism.

[Marko Schmarko]

[ QUOTE ]
There’s no need for “infinite capability”. The belt is only fighting the engines.

[/ QUOTE ]

Sort of. It fights against the engines as a function of the friction of the wheel bearings. Given that this friction is incredibly minimal, the speed of the belt would have to be incredibly fast. While by no means infinite, that is one incredible [censored] belt. So incredible, in fact, that I believe it non-existent.

Also, I believe you have yet to address this problem:
[ QUOTE ]
[T]here is a limit to the conveyer belt's ability to apply a backwards force on the plane (which, assuming there is friction in the wheel bearings, it certainly can do): the friction between the belt and the tires.

[/ QUOTE ]

[Sharkey]

The friction originates at both the tires and the bearings.

Implying that the tires would start skidding at some point introduces an arbitrary parameter into the experiment.

[Marko Schmarko]

The only reason that the friction from the tires has any effect on the plane is through the friction of the bearings. It's not really arbitrary any more than you saying that the friction of the bearings slows the plane. Neither will happen until some unspecified point. That doesn't make it irrelevant.

[mostsmooth]

[ QUOTE ]
An somewhat equivalent question for a helicopter would be that it was sitting on a giant spinning circle that spun at the same speed as the rotor. Would the helicopter take off?
(I would think no)

[/ QUOTE ]
the helicopter would take off

[Sharkey]

[ QUOTE ]
The only reason that the friction from the tires has any effect on the plane is through the friction of the bearings. It's not really arbitrary any more than you saying that the friction of the bearings slows the plane. Neither will happen until some unspecified point. That doesn't make it irrelevant.

[/ QUOTE ]

Okay. I think we’ve been talking past each other to some extent. I’ll try again:

The condition given in the OP that belt speed equals wheel speed means belt speed will be increased by however much is necessary to transmit whatever force is necessary to the plane to stop its forward motion, because as long as the plane has such motion, there is a positive difference between wheel speed and belt speed.

The capacity of the tires and the wheel bearings to transmit up to the maximum output of the engines is not addressed in the OP, and therefore, since possible, cannot be discounted.

[madnak]

If this is a "perfect hypothetical" and the conveyor belt is magical and can move at infinite speed, then we can assume the wheels are also magical and there is no friction except the perfect friction at the contact point between the wheels and conveyor. If this is the case, then the wheels are irrelevant.

The main problem is that under such a scenario, the conveyor belt by definition cannot match the speed of the wheels. As the wheels move forward, their speed increases relative to the speed of the conveyor belt. This is the only problem, and the question should have been posed in such a way as to account for it.

If the conveyor belt isn't magical, then the friction at the bearings will be far insufficient to counter the engine thrust. The friction at the bearings is the only backward pressure applied to the stationary plane, and the bearings are specifically designed to minimize friction. That friction is also minimal in the first place due to the small mass of the wheels.

Any plane will coast to a stop on a sufficiently long runway provided there is no engine thrust. However, with even minimal engine thrust most planes will continue indefinitely on a flat runway.

Even if the backward thrust at the bearings completely counteracts the forward thrust from the engine, in this "imperfect" setup we can assume some significant skid. You can't have it both ways. If the friction between the wheels and the conveyor is perfectly efficient, then the bearings are also perfectly frictionless. You can't just toss in certain physical impossibilities and refuse to allow others.

If the conveyor belt is magical and the wheels are not, then the friction will generate great heat and not only will the wheels lock with the bearings, they will eventually melt allowing the plane to smoothly slide forward.

There is no scenario in which the plane will not take off provided sufficient engine thrust. There is only one nit to pick here, and you seem to have missed it.

Did you read through this thread (http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=4011389&page=0&fpart=1&v c=1)? Honestly, if you're going to debate this you should read through the whole thing.

[Sharkey]

Magical: anything “frictionless”.

Not magical: The conveyor belt doesn’t need to move at an “infinite” speed, only fast enough to generate the friction required to offset the thrust from the engines.

Two questions:

1. How do you figure belt speed can’t keep up with wheel speed?

2. Where does the energy dissipated by the friction between the tires and the belt come from?

And I did read the earlier thread.

[purnell]

Durr. Velocity can't equal Force * time. This product would be in units of mass * distance / time. I guess you have to divide by the mass of the plane in this case. I'm going to go sit in the corner now. /images/graemlins/blush.gif

[Sharkey]

Actually, it’s this funny little thing called “impulse”, but who’s keeping score?

[madnak]

That speed would be incredibly fast, to the point of screwing up the mechanics (and specifically generating a lot of heat).

There is always going to be skid. It is magical thinking to assume otherwise. There is always going to be energy lost as heat. It is magical thinking to assume otherwise. Unless you don't agree with basic thermodynamics?

1. It can, if there is skid.

If there's no skid (impossible in the first place), then the wheel speed will be determined by the belt speed, so it's not a matter of whether the belt can keep up with the wheel speed. It's a matter of whether it's physically possible for the belt to accelerate fast enough to create enough pressure at the wheel bearings to counter the increased engine thrust. I believe that kind of acceleration to be impossible (and with sufficient engine thrust the conveyor would have to move faster than the speed of light regardless, which is equally impossible). I might be mistaken because I don't know the specific numbers, but I know that a plane's engine can generate a tremendous amount of thrust in a very short time, and the conveyor would have to move orders of magnitude faster in order for the pressure on the bearings to compensate.

2. I'm not sure I understand the question. Are you trying to say friction doesn't generate heat? Every transaction of energy results in some energy lost to heat.

I don't think the bearings would be able to withstand such forces regardless. The wheels would probably break right off the plane under that kind of pressure.

[kbfc]

God, what a bunch of retards still talking about this....

Well, since I'm a self-aware retard, here we go.

Anyone who says 'yes it will take off' gets 0 points.
Anyone who says 'no it will not take off' has a chance at some points if they give a good explanation.
Anyone who says 'this is a retarded nonsensical question' gets full credit.

Now I run away for a few hours, and if my life is as lame as it could very possibly be, I'll come back and prove it with equations and pictures and everything.

[madnak]

The answer to the question as posed is "it depends on variables not accounted for in the OP." The answer to the question as it's supposed to be posed is "Yes, it will take off." The answer to the question as posed in the other thread is "the problem as stated is inherently paradoxical."

[Sharkey]

[ QUOTE ]
That speed would be incredibly fast, to the point of screwing up the mechanics (and specifically generating a lot of heat).

There is always going to be skid. It is magical thinking to assume otherwise. There is always going to be energy lost as heat. It is magical thinking to assume otherwise. Unless you don't agree with basic thermodynamics?

1. It can, if there is skid.

If there's no skid (impossible in the first place), then the wheel speed will be determined by the belt speed, so it's not a matter of whether the belt can keep up with the wheel speed. It's a matter of whether it's physically possible for the belt to accelerate fast enough to create enough pressure at the wheel bearings to counter the increased engine thrust. I believe that kind of acceleration to be impossible (and with sufficient engine thrust the conveyor would have to move faster than the speed of light regardless, which is equally impossible). I might be mistaken because I don't know the specific numbers, but I know that a plane's engine can generate a tremendous amount of thrust in a very short time, and the conveyor would have to move orders of magnitude faster in order for the pressure on the bearings to compensate.

2. I'm not sure I understand the question. Are you trying to say friction doesn't generate heat? Every transaction of energy results in some energy lost to heat.

I don't think the bearings would be able to withstand such forces regardless. The wheels would probably break right off the plane under that kind of pressure.

[/ QUOTE ]

The work done by the tire-belt friction is translated into heat. Where does that energy come from?

Look at it this way: Is it conceivable that wheel bearings could be made such that the plane could roll down a runway in a vacuum with (the equivalent of) its engines at full power? I don’t think that’s too magical.

Now, for the conventional runway above, substitute a conveyor belt with a (backward) speed equal to that of the plane. What have you got?

Well, it’s exactly the situation I’m describing: a stationary plane under full power.

[Marko Schmarko]

[ QUOTE ]
Now, for the conventional runway above, substitute a conveyor belt with a (backward) speed equal to that of the plane. What have you got?

Well, it’s exactly the situation I’m describing: a stationary plane under full power.

[/ QUOTE ]
WRONG.

Your entire argument thus far has relied on the belt going the speed of the wheels. That's very different from the speed of the plane.

[madnak]

The conveyor belt needs to move much faster than the speed of the plane. The backward thrust applied at the bearings is going to be only a fraction of the thrust provided by the engines if the conveyor is moving at the speed of the plane. The nature of the bearings and wheels is that the wheels can spin at a much faster rate than the plane is moving. This would be true regardless, due to skid. But even without skid the wheels can spin very quickly and exert extremely minimal pressure on the plane.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
Now, for the conventional runway above, substitute a conveyor belt with a (backward) speed equal to that of the plane. What have you got?

Well, it’s exactly the situation I’m describing: a stationary plane under full power.

[/ QUOTE ]
WRONG.

Your entire argument thus far has relied on the belt going the speed of the wheels. That's very different from the speed of the plane.

[/ QUOTE ]

I’m not wrong.

For a stationary plane, wheel speed = belt speed.

[madnak]

[ QUOTE ]
WRONG.

Your entire argument thus far has relied on the belt going the speed of the wheels. That's very different from the speed of the plane.

[/ QUOTE ]

Do you disagree that such high speeds would cause skid and heat in such quantities that a reasonably powerful engine would still be able to move the plane forward?

[Aitcho]

Consider this:

A car is stationary on an ice lake, assume there is no friction between the ice and the tyres.

The car will try to move but the wheels will just spin and it wont go anywhere.

A plane on the ice lake will not have the same problem as it does not require friction with the ground, it has jet engines.

Can you know see why it is relevant that the wheels do not drive the plane? The plane will move forward on the conveyor belt and will eventually take off.

[Marko Schmarko]

[ QUOTE ]
Do you disagree that such high speeds would cause skid and heat in such quantities that a reasonably powerful engine would still be able to move the plane forward?

[/ QUOTE ]
Have you read anything I've said?

[madnak]

Are you responding to me? I agree, that's the same argument I'm trying to make.

[Marko Schmarko]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Now, for the conventional runway above, substitute a conveyor belt with a (backward) speed equal to that of the plane. What have you got?

Well, it’s exactly the situation I’m describing: a stationary plane under full power.

[/ QUOTE ]
WRONG.

Your entire argument thus far has relied on the belt going the speed of the wheels. That's very different from the speed of the plane.

[/ QUOTE ]

I’m not wrong.

For a stationary plane, wheel speed = belt speed.

[/ QUOTE ]
Right, but the plane wouldn't be stationary if the belt only matched the plane's speed.

I thought this was the crux of your argument all along.

[madnak]

[ QUOTE ]
I’m not wrong.

For a stationary plane, wheel speed = belt speed.

[/ QUOTE ]

But belt speed != plane speed. Unless the speed is 0 for everything.

[Sharkey]

plane speed = wheel speed - belt speed

under any circumstances. I hope we can all agree on something that basic.

[madnak]

[ QUOTE ]
plane speed = wheel speed - belt speed

[/ QUOTE ]

No. Skidding and slipping will prevent this from being true.

The wheel speed and belt speed will cancel each other out, however. That much is true.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
plane speed = wheel speed - belt speed

[/ QUOTE ]

No. Skidding and slipping will prevent this from being true.

[/ QUOTE ]

Show me where slipping or skidding is inevitable from the conditions set up in the OP.

[madnak]

The OP never stated there was perfect friction between the wheel and the belt. I don't even think that's physically possible. Like I said before, if there is perfect friction then we may as well assume frictionless bearings as well. The friction between the wheels and the bearing would probably cause some skid regardless.

[kbfc]

The other thread made it very clear that no-slip/no-deformation rolling and frictionless bearings were to be assumed. This post didn't state those explicitly, but the question is pretty stupid without those assumptions (it's stupid with them as well, but marginally less-so).

If you're not going to assume those conditions, (not to mention the magical "instant feedback loop"), then the answer is pretty clearly "it depends on those variables." But then it would be less of a stupid thought problem and more of a normal physics problem.

With these assumptions, it is clearly a non-sensical question:

Under non-slip conditions, the lateral velocity of the plane relative to the belt (Vpb) is equal in magnitude to the tangential velocity of the plane's wheels (Vw). The entire belt reference frame moves relative to the air (Vba) at that same magnitude, but in the opposite direction.

Given this, the lateral velocity of the plane relative to the air (Vpa) = Vpb + Vba = Vpb - Vpb = 0.

Clearly Vpa is the important factor in determining whether there is lift off.

Now plenty of retards on here with all sorts of self-promoted credentials have spewed out [censored] like "don't even look at the wheels. they're not important. this isn't a car....blah blah blah. draw a FBD....blah blah." Ignoring our little belt for a second, they are correct. Unfortunately the belt is LISTED IN THE GODDAMNED PROBLEM AS A CONSTRAINT.

So what you do if you draw your pretty little FBD is you realize that the plane will of course move forward as usual, but the belt WILL NOT match its speed. (Vw != Vba) That part is impossible. Hence the question is retarded and meaningless.

oh, and someone owes his friend his 400 dollars back from the other thread. nlsoldier i think.


This all being said, it is very possible that there is confusion with the direction of the conveyor belt. Problems using velocity without specifying direction are often subject to this, especially when using something like the wheel's tangential velocity which is different at every point along the wheel's circumfrence.

Because the actual wheel rim is moving opposite the plane at the bottom of the wheel where the wheel contacts the ground, one could take it to mean the opposite of what I listed above. In this case, the equation becomes Vpa = Vpb + Vba = Vpb + Vpb = 2Vpb. Solving for Vpb or Vba gives you 0.5 Vpa. Here the plane takes off as usual, and the conveyor belt and plane wheels are travelling exactly half as fast as the plane relative to the air.

So in summary:
- for the case in this horribly worded problem where the belt moves in the opposite direction of the plane, the problem is meaningless and retarded.

- for the case in this horribly worded problem where the belt moves in the same direction as the plane, the problem is trivial (yes it lifts off), and likewise retarded.

I find it hard to believe that the 2nd case is what is intended by the framer of the question, though. The question only has legs if you interpret it the first way and get a bunch of retards to yell at each other for 8 million posts.

QED.

edit to add: i went back and looked at the first post in this thread and it clearly states opposite the plane. i don't believe the other thread was that clear....so my first case holds for this thread, and the caveat part only applies to the other thread.

[Sharkey]

What do you suggest the belt will do if and when the plane advances?

[kbfc]

who the hell knows? it certainly won't instantaneously match the speed, though. you could probably solve for the cut-off point where the lag between the belt and the wheels is at its smallest possible value before other concerns arise (such as relativity.....).

[madnak]

See my earlier post:

[ QUOTE ]
The answer to the question as posed is "it depends on variables not accounted for in the OP." The answer to the question as it's supposed to be posed is "Yes, it will take off." The answer to the question as posed in the other thread is "the problem as stated is inherently paradoxical."

[/ QUOTE ]

Sounds like you're agreeing with me.

The way the question is classically posed is simple. Imagine a loose conveyor belt that will roll with the wheels. Can a plane take off from such a belt?

This question is valuable because it highlights the difference between the plane and a car. A car on such a belt would just push the belt backward and remain stationary. A plane will move forward. This is because a plane's wheels are based on the plane's movement, while on the other hand the car's movement is based on the car's wheels.

That is the question that Slow Play Ray and sushijerk presumably meant to pose.

You can also substitute the conveyor belt with a frictionless surface.

[madnak]

[ QUOTE ]
What do you suggest the belt will do if and when the plane advances?

[/ QUOTE ]

That depends on many factors. If we're assuming "magical" conditions, the problem is paradoxical and meaningless.

Otherwise, my first guess is that the conveyor belt would suffer from a power failure or mechanical failure. If that didn't happen the wheels would fuse to the bearings from the heat and the conveyor belt would stop moving. Or the sensors might be destroyed and everything could go haywire. Or the wheels could break off.

It's not a very stable setup.

[kbfc]

whether or not that's what they meant to pose, it's not what they did. not only that, but in the original post, there were further clarifications that specifically stated no-slip, frictionless bearings, etc.....

i've also seen the question in other places where these same stipulations were made.

the question is just a simple classical physics problem if you don't make those assumptions....and the answer is not necessarily 'yes.' (imagine the case where the wheel friction is HUGE)

i think it's pretty clear that the problem is posed as some sort of thought exercise or riddle. anyway, gotta go....

[Sharkey]

I’ll put up a few items, and you tell where you disagree.

1. plane speed = wheel speed - belt speed, at all times.

2. Therefore, the plane moving forward means wheel speed > belt speed.

3. According to the OP, under the circumstances in (2), the belt will increase its speed to that of the wheels.

4. According to (1), since the acceleration of the belt is greater than that of the wheels (3), the plane is decelerating.

5. The plane will be made to decelerate to v = 0 whenever it begins to advance.

[madnak]

[ QUOTE ]
I’ll put up a few items, and you tell where you disagree.

1. plane speed = wheel speed - belt speed, at all times.

2. Therefore, the plane moving forward means wheel speed > belt speed.

3. According to the OP, under the above circumstance, the belt will increase its speed to that of the wheels.

4. According to (1), since the acceleration of the belt is greater than that of the wheels (3), then the plane is decelerating.

5. The plane will be made to decelerate to v=0 whenever it begins to advance.

[/ QUOTE ]

1. Untrue unless we assume perfect friction.

2. Untrue unless we assume perfect friction and bearings.

3. Assumes belt is magical.

4. The plane doesn't decelerate at all. Based on this reasoning the plane can't decelerate, any more than it can accelerate. It must remain stationary.

5. Untrue. If the plane begins to advance under the magical conditions specified, we have a paradox.

Wheel speed = belt speed (OP)
Plane speed = wheel speed - belt speed (1)

Since belt speed is wheel speed, plane speed = 0. If plane speed > 0, then wheel speed - belt speed > 0. If wheel speed - belt speed > 0, wheel speed != belt speed.

If the plane engine generates thrust, then the plane moves forward. The only thing preventing such is the friction between the bearings and the wheels, but any such friction would result in heat and/or skid. But the plane speed must be 0. Therefore it is impossible to have thrust in such a scenario. There are two ways to make these "magical" assumptions work.

First, the plane may not be generating thrust. The engines are turned off.

Second, there is a gale force wind or some other external pressure equal to the thrust generated by the plane but pressing in the opposite direction.

These assumptions seem to violate the common-sense assumptions of the OP. I would rather violate the "magical" assumptions and accept skidding/energy loss than violate the common-sense assumptions. However the "engine off" solution may be the simplest.

At any rate, so long as the plane is generating thrust the plane "won't not" take off. Either the plane takes off, or there is a mechanical failure, or the situation is a paradox, or the plane is not generating thrust.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
I’ll put up a few items, and you tell where you disagree.

1. plane speed = wheel speed - belt speed, at all times.

2. Therefore, the plane moving forward means wheel speed > belt speed.

3. According to the OP, under the above circumstance, the belt will increase its speed to that of the wheels.

4. According to (1), since the acceleration of the belt is greater than that of the wheels (3), then the plane is decelerating.

5. The plane will be made to decelerate to v=0 whenever it begins to advance.

[/ QUOTE ]

1. Untrue unless we assume perfect friction.

2. Untrue unless we assume perfect friction and bearings.

3. Assumes belt is magical.

4. The plane doesn't decelerate at all. Based on this reasoning the plane can't decelerate, any more than it can accelerate. It must remain stationary.

5. Untrue. If the plane begins to advance under the magical conditions specified, we have a paradox.

Wheel speed = belt speed (OP)
Plane speed = wheel speed - belt speed (1)

Since belt speed is wheel speed, plane speed = 0. If plane speed > 0, then wheel speed - belt speed > 0. If wheel speed - belt speed > 0, wheel speed != belt speed.

If the plane engine generates thrust, then the plane moves forward. The only thing preventing such is the friction between the bearings and the wheels, but any such friction would result in heat and/or skid. But the plane speed must be 0. Therefore it is impossible to have thrust in such a scenario. There are two ways to make these "magical" assumptions work.

First, the plane may not be generating thrust. The engines are turned off.

Second, there is a gale force wind or some other external pressure equal to the thrust generated by the plane but pressing in the opposite direction.

These assumptions seem to violate the common-sense assumptions of the OP. I would rather violate the "magical" assumptions and accept skidding/energy loss than violate the common-sense assumptions. However the "engine off" solution may be the simplest.

At any rate, so long as the plane is generating thrust the plane "won't not" take off. Either the plane takes off, or there is a mechanical failure, or the situation is a paradox, or the plane is not generating thrust.

[/ QUOTE ]

1. If you’re riding a bicycle powered by a fan (no pedaling), are you saying it takes “perfect” friction for bike speed = wheel speed - road speed?

2. Is there such a thing as frictionless anything?

3. What’s stopping the belt from accelerating to the speed of the wheels?

4. The plane is decelerated back to equilibrium by the feedback mechanism controlling belt speed. It is not, nor does it have to be, instantaneous.

5. See (4).

Wheel speed is obviously not always equal to belt speed. That is the equilibrium point the non-instantaneous feedback mechanism is trying to achieve.

How about some non-magical responses, i.e. no “frictionless” anything.

[madnak]

[ QUOTE ]
1. If you’re riding a bicycle powered by a fan (no pedaling), are you saying it takes “perfect” friction for bike speed = wheel speed - road speed?

2. Is there such a thing as frictionless anything?

3. What’s stopping the belt from accelerating to the speed of the wheels?

4. The plane is decelerated back to equilibrium by the feedback mechanism controlling belt speed. It is not, nor does it have to be, instantaneous.

5. See (4).

Wheel speed is obviously not always equal to belt speed. That is the equilibrium point the non-instantaneous feedback mechanism is trying to achieve.

How about some non-magical responses, i.e. no “frictionless” anything.

[/ QUOTE ]

1. Yes, of course. Unless friction is perfect there is always skid as the tires move along the surface of the road. Skid doesn't have to be noticeable. The tires might need to wear down before they start skidding significantly, but after enough friction is applied the tires will be worn to such a state that they will skid noticeably on a regular basis. Of course it usually takes months of wear for the treads to reach such a point, but the speeds and forces involved in holding back a jet engine would strip the tires down in seconds, if not sooner.

2. I'm not sure. Ask a surface chemist. I'm pretty sure there's no such thing as frictionless bearings however.

3. The laws of physics. It's much easier if it doesn't have to be instantaneous, but that doesn't change much. The belt has to accelerate to such a degree that the force it applies on the wheels transfers enough pressure to the bearings to counteract the full force of an engine capable of quickly accelerating many tons of weight to extremely high speeds. Since virtually none of the conveyor's movement is actually going to translate into backward pressure on the bearings, the speed at which the belt must move is astronomical. It is probably below the speed of light but fast enough to require relativistic mechanics and warping of time-space (about which I know nothing).

Well before then, however, the extreme speed would melt the wheels into puddles, destroy the bearings, ruin any mechanism causing the belt to run (barring futuristic technology), and possibly even tear the plane apart.

[Sharkey]

The belt only has to achieve the speed that the plane would rolling down a runway in a vacuum with (the equivalent of) its engines at full power. This is nowhere near the speed of light. Nor does the belt have to accelerate particularly fast, only somewhat faster than the wheels do.

Here’s the difference between our positions: I need a very high tolerance wheel assembly, while you need one that dissipates no heat. My position requires improbable engineering, yours requires impossible physics.

On second thought, even if the wheels do fail, the plane still doesn’t take off.

Since you have chosen to argue with my previous reasoning, I’ll resort to the obvious.

1. Give me a situation with a fan-powered bicycle going the wrong way on an airport conveyor belt where it is not the case that bike speed = wheel speed - belt speed.

2. Give me a situation with a fan-powered bicycle going the wrong way on an airport conveyor belt where it is not the case that the bike moving forward means wheel speed > belt speed.

3. Give me a situation with a fan-powered bicycle going the wrong way on an airport conveyor belt where, under the circumstances in (2), it cannot happen that the belt will increase its speed to that of the wheels.

[madnak]

The wheels on an airplane are specifically designed not to transfer force. The conveyor would need to move orders of magnitude faster than the plane would move in order to give the wheels (which have a relatively small mass) enough power to move the plane. Of course, since the wheels are designed to transfer as little energy as possible to the bearings, most of that power wouldn't reach the plane anyhow, so we need even more speed. Finally, quite a lot of energy will be lost as heat and much of the pressure on the bearings will not be backward pressure, so we need even more speed.

A bike is much smaller than a jet and a fan is much less powerful than a jet engine. If a jet engine were attached to the bike I'm not sure the bike could handle that kind of force. If so it would definitely skip, jump, and squeal on the conveyor.

Finally, you have already admitted that the plane can move forward. So all we need to do is increase the engine power in order to make that forward movement sufficient to achieve take-off.

[BCPVP]

The plane will take off. The conveyor belt is irrelevant.

An article (The plane will take off. The conveyor belt is irrelevant.) from some folks who might know a thing or two about airplanes explains it.

[Sharkey]

All three items concerning the bicycle on the airport conveyor belt are directly relatable to the jet plane situation. It’s not enough to say margins of error will increase, some slippage will occur, etc. The principles are the same, and belt speed can be raised as high as necessary to take advantage of the least amount of the slightest hint of a trace of friction, phrases like “designed to transfer as little energy as possible” notwithstanding.

Besides, the stresses at the bearings will not exceed those encountered if the plane was rolling down the runway in a vacuum with (the equivalent of) its engines at full power: not an impossible criterion.

[purnell]

[ QUOTE ]
The plane will take off. The conveyor belt is irrelevant.

An <a href="The plane will take off. The conveyor belt is irrelevant." target="_blank">article</a> from some folks who might know a thing or two about airplanes explains it.

[/ QUOTE ]

This link is broken- Could you please re-post?

[BCPVP]

[ QUOTE ]
[ QUOTE ]
The plane will take off. The conveyor belt is irrelevant.

An <a href="The plane will take off. The conveyor belt is irrelevant." target="_blank">article</a> from some folks who might know a thing or two about airplanes explains it.

[/ QUOTE ]

This link is broken- Could you please re-post?

[/ QUOTE ]
Arg, wtf?
Here it is: http://www.avweb.com/news/columns/191034-1.html

[madnak]

From the article:

"The only time it could be a problem is if the wheel speed got so high that the tires blew out."

[Kevin J]

Sharkey-

Ok.. I did some research on the net and it turns out there is a paradox which is preventing us from reaching a resolution... It all depends on how the question is worded.

If the conveyor belt is set to match the speed of the plane... Then the plane takes off normally as I have stated. However...

If the belt is set to match the speed of the wheels... Well, now there is an unreconcilable paradox. In short, this is an impossible scenario. Can't believe I didn't catch this myself. The reason it's impossible is because if the wheels are traveling 10 MPH, then the belt goes 10 MPH in the opposite direction, however... If the belt goes 10 MPH in the opposite direction, then this causes the wheels to rotate at 20 MPH. Etc. etc. Ad-nausium. Of course, this is ann impossibility.

So in the latter scenario I suppose you are correct. The plane wouldn't take off, mainly because it's an impossible situation. In the former scenario, I am correct. If the conveyor belt simply matches the speed of the plane, the plane will will take off.

Do you agree with this?

[kbfc]

Given the problem stated in that article, of course the plane takes off. Unfortunately, as someone pointed out, that is a DIFFERENT problem.

Both posts on here used the wheel's tangential velocity as the feedback control for the conveyor, NOT the plane's lateral velocity. For that case, the question is most certainly meaningless.....and it has nothing to do with wheels blowing up due to overrotation or anything....it's simply nonsense. There were a lot of people blowing a lot of steam in the last thread saying it would take off AND the conveyor would match the wheel speed. They are simply idiots.

[kbfc]

[ QUOTE ]
I’ll put up a few items, and you tell where you disagree.

1. plane speed = wheel speed - belt speed, at all times.

2. Therefore, the plane moving forward means wheel speed > belt speed.

3. According to the OP, under the circumstances in (2), the belt will increase its speed to that of the wheels.

4. According to (1), since the acceleration of the belt is greater than that of the wheels (3), the plane is decelerating.

5. The plane will be made to decelerate to v = 0 whenever it begins to advance.

[/ QUOTE ]

Blah blah....ignore all that crap about nonperfect friction. The problem is clearly assuming NON-SLIP rolling with frictionless bearings. It was at least made explicit in the other thread. If you don't make these assumptions then the answer is "it depends." Not a very good question.

As for your points. #1 is true. #2 is true. After that you go wrong. One of the few things that the people getting the problem wrong have been able to manage is that the conveyor is unable to transfer lateral force to the plane. The conveyor cannot MAKE the plane decelerate. It can't make the plane do much of anything except not fall on its belly. Your #3 is inaccurate. The circumstances in #2 break the constraints of the problem, so it's irrelevant what might happen in that case.

[kbfc]

For everyone convinced that the plane can take off and there is no issue with any paradox in the problem, what would you say if I asked the following question:

Can a normal airplane take off while the pilot shoves a square peg through a round hole? (cheeky answers will be punished by eternity in the christian fantasy of hell)

[BCPVP]

[ QUOTE ]
For everyone convinced that the plane can take off and there is no issue with any paradox in the problem, what would you say if I asked the following question:

Can a normal airplane take off while the pilot shoves a square peg through a round hole? (cheeky answers will be punished by eternity in the christian fantasy of hell)

[/ QUOTE ]
http://img100.imageshack.us/img100/5964/squareround9oa.png

[Sharkey]

[ QUOTE ]
Sharkey-

Ok.. I did some research on the net and it turns out there is a paradox which is preventing us from reaching a resolution... It all depends on how the question is worded.

If the conveyor belt is set to match the speed of the plane... Then the plane takes off normally as I have stated. However...

If the belt is set to match the speed of the wheels... Well, now there is an unreconcilable paradox. In short, this is an impossible scenario. Can't believe I didn't catch this myself. The reason it's impossible is because if the wheels are traveling 10 MPH, then the belt goes 10 MPH in the opposite direction, however... If the belt goes 10 MPH in the opposite direction, then this causes the wheels to rotate at 20 MPH. Etc. etc. Ad-nausium. Of course, this is ann impossibility.

So in the latter scenario I suppose you are correct. The plane wouldn't take off, mainly because it's an impossible situation. In the former scenario, I am correct. If the conveyor belt simply matches the speed of the plane, the plane will will take off.

Do you agree with this?

[/ QUOTE ]

Actually, there is no paradox. All that is necessary is for the belt to accelerate at a greater rate than the wheels do, and the speed of the belt will catch up to the speed of the wheels.

What this looks like is the belt accelerating at a certain rate to bring its speed up to that of the wheels, which acceleration is added to the wheels as you say, but then the increased backward frictional force on the plane decelerates the plane, which adds a negative acceleration to the wheels. The result is an acceleration for the wheels that is less than the acceleration for the belt. The speeds will equalize.

plane speed = wheel speed - belt speed.

As far as the conveyor belt being set to match the speed of the plane instead, yes, as I have said in a earlier post, the situation would be different.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
I’ll put up a few items, and you tell where you disagree.

1. plane speed = wheel speed - belt speed, at all times.

2. Therefore, the plane moving forward means wheel speed > belt speed.

3. According to the OP, under the circumstances in (2), the belt will increase its speed to that of the wheels.

4. According to (1), since the acceleration of the belt is greater than that of the wheels (3), the plane is decelerating.

5. The plane will be made to decelerate to v = 0 whenever it begins to advance.

[/ QUOTE ]

Blah blah....ignore all that crap about nonperfect friction. The problem is clearly assuming NON-SLIP rolling with frictionless bearings. It was at least made explicit in the other thread. If you don't make these assumptions then the answer is "it depends." Not a very good question.

As for your points. #1 is true. #2 is true. After that you go wrong. One of the few things that the people getting the problem wrong have been able to manage is that the conveyor is unable to transfer lateral force to the plane. The conveyor cannot MAKE the plane decelerate. It can't make the plane do much of anything except not fall on its belly. Your #3 is inaccurate. The circumstances in #2 break the constraints of the problem, so it's irrelevant what might happen in that case.

[/ QUOTE ]

The conveyor belt can make the plane decelerate by increasing the backward frictional force accompanying wheel speed on the belt.

The circumstances in (2) do not break the constraints of the problem. Every time the plane begins to move forward, wheel speed > belt speed, a situation the feedback mechanism controlling belt speed then proceeds to correct.

[Kevin J]

[ QUOTE ]
What this looks like is the belt accelerating at a certain rate to bring its speed up to that of the wheels, which acceleration is added to the wheels as you say, but then the increased backward frictional force on the plane decelerates the plane, which adds a negative acceleration to the wheels.

[/ QUOTE ]

This in a nutshell is where you're thinking goes errant (or at least one our thinking does). The thrust is not attached to the plane or it's wheels, therefore there is no backward frictional force!! What happens to the wheels and/or the conveyor belt is irrevelant to the force of thrust.

[ QUOTE ]
The result is an acceleration for the wheels that is less than the acceleration for the belt.

[/ QUOTE ]

How can this make sense? The problem specifically states that the belt exactly matches the speed of the wheels. If the wheels at any point deccelerate, then so does the belt.

At least you acknowledge that if the belt matches the speed of the plane, it will take off. That tells me you understand this more than I originally thought, so it concerns me that you understand the wheel issue as well and maybe I'm wrong. But I'm pretty sure that the belt matching the wheel speed is an impossible scenario, since both would have to reach infinite speed.

Anyway, I enjoyed thinking about this.

[Sharkey]

Whenever the plane rolls forward over any surface, be it the belt or a conventional runway, there is a frictional force opposing that motion. In addition, this backward frictional force increases with wheel speed.

The OP states: “Image that there is a electronic device gauging the speed of the wheels and then immediately relays the information to the belt to accelerate/deccelerate accordingly.”

I take it the mechanism described is not instantaneous, so wheel speed can, at least momentarily, exceed belt speed. This is the only condition under which the plane can move forward, anyway.

Once the feedback mechanism controlling belt speed detects that wheel speed > belt speed, it will make belt acceleration > wheel acceleration. How is this possible? Total wheel acceleration is being caused by belt acceleration and the plane deceleration being caused by the increasing wheel speed.

[madnak]

The frictional force opposing the motion is absorbed by the wheels. That is the point of the wheels. It doesn't translate to the plane. A small amount of that force might make it to the bearings, but as the article stated the tires would blow out well before that force became significant. The amount transferred to the bearings would also be so small as to require extremely high conveyor speeds which would destroy the entire scenario.

The only way the conveyor belt affects the plane is through the amount of energy the wheels fail to absorb and instead transfer to the bearings. It is only where the wheels are failing at their intended task that the plane gets any of the energy. Much of the energy transferred to the bearings would also represent heat and pressure applied in other directions. There is no realistic way to transfer enough force to the wheel bearings to affect the plane's movement.

[Sharkey]

[ QUOTE ]
The frictional force opposing the motion is absorbed by the wheels. That is the point of the wheels. It doesn't translate to the plane.

[/ QUOTE ]

Before we continue, are you telling me that a plane coasting on a runway in a vacuum would keep rolling on forever?

[madnak]

If the engine was on, yes.

[Sharkey]

[ QUOTE ]
If the engine was on, yes.

[/ QUOTE ]

Why must the engine be on?

[madnak]

Primarily because of the loss of energy due to friction. Also due to skid. And possibly to a much smaller degree because of pressure on the bearings from the wheels. The loss due to friction would be much greater.

Regardless, a plane in a vaccuum would coast for a very long time, probably many miles. Even with the ground moving backwards relative to the plane at very high speeds. And the loss of momentum due to friction is relatively invariable. As the initial momentum of the plane increases, the length of coasting increases almost proportionally.

Even with the ground moving against the wheels at faster than the speed of sound, it will still take the plane hours to come to a complete stop without skidding.

[madnak]

To clarify, when I say loss of energy due to friction is relatively invariable, I mean relative loss of energy due to friction. That is to say, the decreasin efficiency will not be enough to counter the increased momentum of the plane itself. At reasonable speeds, anyhow.

I don't know what would happen at relativistic speeds, but I doubt it would be pretty.

[Sharkey]

The rate of decrease of the plane’s kinetic energy (into the heat of friction, which you acknowledge happens) is a function of wheel speed, i.e. more or less proportional to it. The rate of increase of the plane’s kinetic energy is a function of engine power.

With every increase in engine output to advance the plane, the feedback mechanism controlling belt speed will (indirectly) increase wheel speed to counteract it, until the engines max out and the system is in equilibrium.

[madnak]

[ QUOTE ]
The rate of decrease of the plane’s kinetic energy (into the heat of friction, which you acknowledge happens) is a function of wheel speed, i.e. more or less proportional to it. The rate of increase of the plane’s kinetic energy is a function of engine power.

[/ QUOTE ]

No. Not remotely. The wheels are absorbing some of the plane's momentum in the coasting scenario, so their speed does tend to reflect the energy lost from the plane in that situation. But even without thrust, with nothing more than momentum, the wheels would need to "be spinning" at unrealistic speeds in order to stop the plane in a short amount of time. It would take literally hours for the plane to stop under these conditions. So even if the rate of the wheels were proportionate with the deceleration of the plane it would still require great speed. (it isn't, gravity and skid play a major role in the frictional loss)

If it takes 4 hours to coast to a stop with the wheels going the same speed as the plane, then it would take 2 hours if the wheels were going twice as fast. If the wheels are going 4 times as fast, it takes 1 hour. In order to stop the plane in 3.5 minutes, the wheels must move 64 times the speed of the plane. (Again, assuming that it's directly proportionate which it isn't. The reality is the wheels would need to be spinning hundreds of times as fast as the plane is moving). This is just to handle the existing momentum, no force is added by the engine.

But most importantly, the energy being lost comes from the plane's momentum. The energy lost from the plane is limited by the speed of the plane relative to the ground. The plane's energy is spinning the wheels, thus the plane's energy is lost to the friction. In a conveyor belt that energy comes from the conveyor, and it is the conveyor belt that is going to slow down. Not the plane. The conveyor actually adds energy to the plane, rather than taking it away. It's the opposite effect. Skid likewise helps the plane in this situation. If the conveyor is somehow offering more resistance than the skid, the plane will just skid forward with its engine thrust. The wheels will squeal along in protest.

[ QUOTE ]
With every increase in engine output to advance the plane, the feedback mechanism controlling belt speed will (indirectly) increase wheel speed to counteract it, until the engines max out and the system is in equilibrium.

[/ QUOTE ]

In the first place, we have established that the wheels need to move orders of magnitude faster than the plane. And the conveyor must move as fast as the wheels. The heat generated by this friction will be very intense, as will the pressures on the wheel bearings and the tires. Moreover the conveyor belt is losing the energy to the friction, not the plane. The plane is actually gaining energy. If we assume that energy being transferred from the belt to the plane applies a backward pressure, then it might theoretically counteract the thrust. However, now it is only the energy being specifically transferred backward, minus the energy transferred forward, minus the energy lost to friction, that is providing a backward pressure on the bearings (which probably can't handle the full opposing forces anyhow). The result is that any speed that could hypothetically match the force of the engine would be orders of magnitude greater than the original speed, which itself is orders of magnitude greater than the speed the plane would achieve on a normal runway. No real conveyor belt can run at anywhere near that kind of speed. And if it could, friction + insane forces on the belt and wheel would destroy the plane. And if the plane managed to get all the pressure it might just shear apart anyhow.

Every transaction loses energy. Here are the transactions in the separate scenarios.

Coasting: plane -> wheels -> ground
Conveyor: belt -> wheels -> plane

Or put another way.

Coasting: plane has energy, ground gets energy
Conveyor: belt has energy, plane gets energy

The setup with the conveyor is the opposite of the setup with the coasting.

[Sharkey]

Where does this force resisting the tendency of the engines to move the plane against the belt reside? At the wheel bearings. What form does that resistance take? Friction. What is the mechanism of that resistance? The conversion of the kinetic energy of the plane with respect to the belt into heat energy at the wheel bearings.

Agreed that the wheels would have to spin extremely fast: the same speed they would go if the plane was traveling without air drag under full power down the runway. This is a thought experiment, where extreme engineering is permitted (but not physical impossibility).

As far as the plane is concerned, the conveyor belt is indistinguishable from a conventional runway without looking out the windows. Rather than the plane going forward on a stationary surface, the surface goes backward under a stationary plane. It’s the famous principle of relative motion. Either way, the plane will lose speed relative to the surface it’s running on by converting kinetic energy into frictional heat at the wheel bearings.

[madnak]

That mechanic is constant and based on the plane's speed relative to the ground. The conveyor speed has nothing to do with the energy from the engines.

They would have to go much faster than that.

The surface doesn't go backward because there is a static frame of reference. The belt surface doesn't even affect the plane, the plane isn't remotely concered with the surface underneath it. The plane is only concerned with the air surrounding it.

[Kevin J]

[ QUOTE ]
The surface doesn't go backward because there is a static frame of reference. The belt surface doesn't even affect the plane, the plane isn't remotely concered with the surface underneath it. The plane is only concerned with the air surrounding it.

[/ QUOTE ]

This is exactly how I understand it. What we really need is someone who is adept at physics to spell it out mathematically. There must be a way to put this into the form of an equation, but I sure don't know how to do that.

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
The surface doesn't go backward because there is a static frame of reference. The belt surface doesn't even affect the plane, the plane isn't remotely concered with the surface underneath it. The plane is only concerned with the air surrounding it.

[/ QUOTE ]

This is exactly how I understand it. What we really need is someone who is adept at physics to spell it out mathematically. There must be a way to put this into the form of an equation, but I sure don't know how to do that.

[/ QUOTE ]
we all agree the problem is worded wrong. it should say the belt moves opposite the plane speed. plane takes off.
one could look at the problem and claim "wheel speed" is the speed of the wheel assembly as a whole, which would equal plane speed. plane takes off.
the other option of "wheel speed" would be the one you are all using now, the angular velocity (i think thats the right term?). what sharkey is saying is true as far as my little brain can tell. the belt is capable of stopping the plane, if the belt spins fast enough, and the force imposed can overcome the jets. suppose the plane was just sitting on the belt, no engines on, not moving, and the belt starts moving backward, the plane will eventually start moving backwards as well. we should all be able to agree on that. this proves that the belt does impose a force on the plane. now, our problem is that we have a lot of unknowns preventing us from clearing up our problem.
in the OP, the plane starts moving and then the belt starts moving. we dont know how fast the belt speeds up, thus we cant tell if the plane would get up to take off speed before the belt could prevent it.
we dont know how powerful the jets are, and we dont know how much backward force would be generated by the belt at high speed. the force may not be enough to overcome the jets.
if the belt can go from 0 to the speed required to start slowing the plane (if there is such a belt speed) before the plane can take off, then the plane wont take off.
hopefully that all made sense.
as stated in the OP, i think the answer is "not enough information"

[Lestat]

My impression was that the belt immediately attained the necessary speed. You're right, the question is, necessary speed for what?

I don't see how belt speed could ever equal the angular velocity of the "tires". It's a paradox. If the tires go 10 MPH, the belt goes 10 MPH in the opposite direction, which mean the tires go 20, which mean the belt goes 20, which mean the tires go 40, and so on, until reaching infinite speed. This is an impossible scenario.

Under any other scenario the plane takes off. Thrust doesn't care what's happening with the wheels or belt. The wheels of a plane only "facillitate" forward motion, they don't "cause" it.

This thread is far too long. We all know what each other is talking about and what's worse, I think we all agree (still not sure about Sharkey) with each other! We just keep finding ways to say the same things. We need to let this thread die. Nothing new can be said. Most of us know the correct answer. Those who don't, never will.

[Matt R.]

The forces acting on the plane are the thrust and the friction due to the conveyor belt. Note that if you try to ignore friction (which some people seem to be arguing) the wheels won't even spin. Then the plane will just slide along the conveyor belt and it will take off.

So the net force on the plane is F = T - f, T is thrust and f is friction. The wheels are spinning without slippage, so there is a torque being applied to the wheels torque = f*R, R is the radius of the wheel. Substitution gives F = T - torque/R. R is clearly constant. So if torque/R equals T the plane will not accelerate, if T > torque/R the plane will accelerate.

From the conditions set forth in the problem, the conveyor belt's velocity increases as the tangential velocity of the wheel increases, to a point where tang. vel. of the wheel = belt velocity. So as thrust is applied to the plane, the belt must accelerate. Thus the accelerating belt is clearly applying a torque to the wheels. Since the belt can accelerate indefinitely, the torque can increase indefinitely. The problem is very clear that the velocity of the wheel is never greater than the velocity of the belt. Therefore torque/R must equal T and the plane does not accelerate. The plane will not take off.

Note that it is impossible to ignore friction if you take the wheels as spinning. This is where the confusion comes in I think.

[Matt R.]

Without reading the thread (it may have been pointed out), I just realized a huge problem with the original post as it was stated. It says that the conveyor belt matches the wheel speed. I assumed this meant the tangential velocity of the wheel. If it simply means the translational velocity of the wheel, that is completely different. The translational velocity is of course equal to the speed of the plane.

If the translational velocity of the plane is say, 100 mph relative to the air, then the conveyor belt is -100mph. But then, of course, the tangential velocity of the wheel is 200 mph. This is a completely different scenario. t/R would be less than thrust since the translational velocity is > 0. The plane could conceivably take off. This is *completely* different than saying the belt speed matches the tangential velocity of the wheels.

The question isn't precise enough to answer as-is. Maybe it was clarified at some point in the thread that I'm unaware of.

[Matt R.]

One last point,

The distinction between tangential and translational velocity is clearly where *all* of the argument is coming from in this thread. The problem is actually very simple once you distinguish between the two. The plane can clearly take off if the OP means translational velocity. If the original question refers to tangential velocity of the wheel, then the problem is slightly more difficult as you have to consider torque, but the plane will not take off.

[Sharkey]

It’s tangential velocity. If it was translational velocity (i.e. plane speed), the plane would take off (with belt speed = 2 * plane speed, if I’m not mistaken).

The condition in the OP that belt speed = wheel speed doesn’t have to be realized instantaneously. I just assumed a very fast feedback mechanism. All that is necessary is that belt acceleration be significantly greater than wheel acceleration, according to the following:

plane speed = wheel speed - belt speed.

[Lestat]

If OP meant tangential velocity (I'll assume this term is correct), it's a moot point, because such a scenario can't exist. It's paradox.

If OP meant translational velocity (again I'm assuming this term is correct), not only does the plane take off, it takes off quite normally.

[Sharkey]

If the OP meant plane speed instead of (tangential) wheel speed, it would have been easy enough to talk about the plane directly and not mention the wheels.

The paradox you’re probably referring to has be cleared up already. As I just mentioned in my previous post, all that is necessary is that belt acceleration be significantly greater than wheel acceleration, according to the following:

plane speed = wheel speed - belt speed,

and belt speed can catch up to wheel speed as plane speed decreases due to increasing friction.

[Matt R.]

[ QUOTE ]
If OP meant tangential velocity (I'll assume this term is correct), it's a moot point, because such a scenario can't exist. It's paradox.

[/ QUOTE ]

It can exist. It simply directly implies that the plane is NOT moving -- i.e. the translational velocity of the wheel is zero. There's actually no need to invoke torque or any other force, so I was kind of incorrect in stating that. It's simply a relative velocity problem. But to *understand* what force is opposing thrust, you need to realize that it is the torque being applied to the wheels via the conveyor belt.

It's not really a paradox. It is just logically inconsistent to say that the tangential velocity is equal to the belt velocity, yet the translational velocity is > 0. It's sort of like making the statement, "The plane moves but the plane does not move." The equation plane speed = (tangential) wheel velocity - belt velocity is just the mathematical way to describe this. The ONLY way to satisfy the equation for when tang. wheel velocity = belt velocity is plane speed is zero.

[Matt R.]

And as far as what would happen in a "real world" scenario of the problem... this may help clear up any misunderstandings.

We've already determined that the plane can take off if the OP is referring to translational velocity of the wheel. What if it is referring to tangential?

The original equation I typed out was F = T - f. f is the frictional force, and the wheels do not slip so it is f = (mu)*n. mu is the coefficient of static friction, and n is the normal force. n is just the weight of the plane in this instance. Obviously f has to be HUGE to overcome the thrust of a commercial airline, so mu has to be huge as well. I'm somewhat sure that no materials exist for mu to be large enough, and still allow normal turning of the wheel in contact with the belt (the wheel would have to be physically stuck to the belt I think). So in a *real world* scenario, the assumption of zero slippage would not be met. The belt would turn at the tangential velocity of the wheel, but the plane/wheel would skid along the belt. So it would take off.

Of course since this is a thought experiment sharkey and I (in my original post) assumed no slippage. The coefficient of static friction has to be unfeasibly huge for this to be true, however. So the assumption of zero slippage given the scenario would realistically be invalid.

[Sharkey]

About the “paradox”:

The supposed paradox that has come up is that once the plane starts moving forward, and thus wheel speed > belt speed (momentarily), the belt can never again match the speed (tangential velocity) of the wheels, because every time the belt accelerates, the wheels accelerate with it, ad infinitum.

In fact, since the plane is being decelerated by increasing wheel speed, referring to the equation wheel speed = plane speed + belt speed, the direct increase in wheel speed caused by an increase is belt speed is to some extent offset by a decrease in plane speed.

Belt speed can catch up to wheel speed.

About the traction issue:

Yes, the concept of tire friction etc supporting the full thrust of jet engines is a bit of a stretch, but maybe someday the right materials will be available. Around the time of the space elevator.

[Lestat]

Set me straight on this...

If tangential velocity refers to the RPM of the tires (which translates into MPH), then how is this not impossible? For every MPH the conveyor belt increases to match tire speed, it results in tire speed increasing (by 2X's). Hence, the paradox. They both reach infinite speed. This is impossible.

[mostsmooth]

[ QUOTE ]
Set me straight on this...

If tangential velocity refers to the RPM of the tires (which translates into MPH), then how is this not impossible? For every MPH the conveyor belt increases to match tire speed, it results in tire speed increasing (by 2X's). Hence, the paradox. They both reach infinite speed. This is impossible.

[/ QUOTE ]
eventually the backward force from the belt will stop the plane.
what we dont know is if it can stop the plane before it takes off.

[Matt R.]

[ QUOTE ]
Set me straight on this...

If tangential velocity refers to the RPM of the tires (which translates into MPH), then how is this not impossible? For every MPH the conveyor belt increases to match tire speed, it results in tire speed increasing (by 2X's). Hence, the paradox. They both reach infinite speed. This is impossible.

[/ QUOTE ]

Angular velocity refers to the RPM of the tire. Tangential velocity refers to the velocity of the outside edge of the tire. They are related by v = r(omega), omega is the angular velocity (which you seem to be referring to) and r is the radius of the wheel. The outside edge of the tire clearly has to rotate at the same speed as the conveyor belt if there is no slippage (I assume this much is obvious?) Thus the tangential velocity of the wheel equals the conveyor belt velocity. Edit -- this is assuming the plane's velocity is zero... this is really important. The tangential velocity of the wheel more generally equals the plane velocity minus the conveyor belt velocity. Re-edit for clarity -- again assuming no slippage.

So... assuming we're talking tire speed = tangential velocity, I don't know where the 2x factor came in.

I am also assuming instantaneous correction by the belt to an increase in wheel speed -- the conveyor belt increases its speed at the same time that the wheel speed increases. It seems some other posts are saying there's a delay in this correction? That would complicate matters somewhat... but I think the fundamental idea remains the same.

ONE MORE EDIT!! -- I think precisely where you're confused is that RPM translates directly to MPH. This isn't true. RPM refers to angular velocity, but only after you multiply times the radius of the wheel (and do other unit conversions from minutes to hours) can you get MPH out of this. Maybe somewhere in here is where you got a factor of 2x?

[kbfc]

Everything you've written in this thread is misguided.

Torque is irrelevant. Yes people have been mixing up the term angular velocity (change in angle over time) with tangential velocity (instantaneous velocity of the wheel edge), but that hasn't caused any errors because the tangential velocity is EQUAL to the translational velocity under no-slip conditions. You don't need to do any more equations involving the radius of the wheel, etc...

There are a few conditions that were stated explicitly in the other thread, and must be assumed in this one for the problem to have any relevance:

- frictionless bearings,
- no-slip rolling.

Note that these conditions, which are fairly standard for physics thought problems, do NOT imply a complete lack of friction. Given a complete lack of friction, the plane clearly takes off as usual, with the wheels simply sliding along the ground not rotating at all.

If you don't assume either of these, you end up with a bunch of retards debating how the tires are going to explode, when the answer to the problem is pretty clearly, "it depends what those values actually are."

[kbfc]

Lestat,

Please stop listening to him. You are close.

It's not that the belt eventually speeds up to infinity; if that were the case somewhere along the way the plane might take off. It is that there is no possible (and I mean this in the a priori sense) mechanism by which the belt can instantaneously match wheel speed for any plane velocity > 0.

[kbfc]

Come on. It is specifically stated in the problem: "INSTANTANEOUSLY".

That doesn't mean, "very fast." That means "instantaneously."

As I wrote somewhere earlier, if you don't want to assume this condition, why don't you go solve for the smallest possible value of belt lagtime where classical physics is still somewhat appropriate? With lag, you will get a feedback loop between the wheel and the belt, but it is possible, depending on the values, that the plane will take off successfully before this feedback gets out of control.

[mostsmooth]

[ QUOTE ]
Come on. It is specifically stated in the problem: "INSTANTANEOUSLY".


[/ QUOTE ]
instantaneously is not stated in the OP

[Lestat]

Sorry. I meant angular velocity was impossible, not tangential velocity.

The 2X's came in by assuming that the belt automatically matched the "spin" of the tires. I admit I'm confused by the terms angular and tangential. All I know is that if the tires spin at 100 RPM and the belt matches this speed in the opposite direction, then this would cause the tires to spin at 200 RPM, which creates a belt speed of 200 RPM, which cause the tires to spin at 400 RPM, and so on to infinity.

I believe we (everyone?) are getting caught up in semantics. If the belt speed is set up to match the plane's forward velocity in the opposite direction, the plane takes off. It's as simple as that. Example: When the plane's forward velocity is 40 MPH, the belt is traveling at 40 MPH in the opposite direction. Under this scenario there is no question that the plane takes off. Right or wrong?

The whole point of this thought problem is to understand the concept that thrust works independently from the plane's wheels. It doesn't matter what's happening to the wheels underneath. Under normal circumstances, round tires only facillitate the plane's ability to gain enough forward momentum for take off. This is certainly preferrable to having the plane rest on it's fuselage. But the fact is, the conveyor belt could be traveling 1000 MPH in the opposite direction. As soon as engine thrust became sufficient to overcome the "weight" of the plane on it's wheels, it would eventually gain the required forward momentum to take off. Right or wrong?

[Sharkey]

The OP states: “...there is a electronic device gauging the speed of the wheels and then immediately relays the information to the belt to accelerate/deccelerate accordingly.” Nothing about “instantaneously”.

All that is required is a reasonably short reaction time on the part of the above device and the capacity for belt acceleration > wheel acceleration, in which case, whenever plane speed > 0 (momentarily), the state of belt speed = wheel speed can be restored to stop the plane.

[Lestat]

This thread is so nested I'm having a hard time telling who's who's responding to who (problem with my browswer?).

Are you referring to me kbfc?

[Lestat]

Got it. I just saw your reply now. The nesting is really getting screwed up on my screen. Thanks kbfc.

[kbfc]

No. I believe that was addressed to Matt R. I'm reading in flat-mode, so it's a bit confusing as well, but I've tried to reply to the specific posts I'm addressing.

[kbfc]

For you and mostsmooth, since you both had the same objection:

Apparently it doesn't say "instantaneously" in this post. It must have in the other thread. That being said, "instantaneously" is clearly implied. You cut part of this OP out, but ignored the previous line: "However, the plane is on a converter belt which is rigged to move/spin in the opposite direction of the plane at the exact speed of the wheels."

Given that you assume some sort of lag in the speed guage, perhaps you could interpret it as non-instantenously overall, even if the relay of information is "immediate". The sentence you posted is only meant to be an example of the condition imposed in the preceeding sentence. That sentence makes it pretty clear that no matter what the mechanism is, the belt moves at the exact same speed as the wheels. If you want to interpret this to mean "the exact speed the wheels used to be going a few ticks ago," then you're insane. I suggest you direct some of that insanity into solving for that minimum lag value.

[Sharkey]

Including the “instantaneously” makes the whole problem a lot less interesting, because the plane wouldn’t budge an inch. Also, “instantaneously” is a theoretical impossibility.

A certain lag in the feedback mechanism is not a problem, as I described in my previous post.

[kbfc]

No. It still makes the problem interesting enough, because you get people to act like retards for 500 posts or however long these threads are combined. Also you said 2 things here, 1 of which was misguided and 1 of which was completely accurate:

Misguided: "because the plane wouldn't budge an inch." It would budge....the conveyor just wouldn't keep up.

Accurate: "'instantaneously' is a theoretical impossibility."

Adding lag to the feedback absolutely breaks the problem. It makes it a completely different problem. Much more of a straight calculation affair, and not a stupid thought riddle.

[Sharkey]

[ QUOTE ]
Misguided: "because the plane wouldn't budge an inch."

[/ QUOTE ]

Describe a situation in which the plane is rolling forward, yet belt speed = wheel speed.

[kbfc]

Please stop butchering quotes:

[ QUOTE ]
Misguided: "because the plane wouldn't budge an inch." It would budge....the conveyor just wouldn't keep up.


[/ QUOTE ]

[Sharkey]

Still, describe how the belt not keeping up is consistent with the OP.

[Matt R.]

[ QUOTE ]
Torque is irrelevant.

[/ QUOTE ]

Torque is absolutely relevant. If you take the wheels as spinning, what do you think is causing the wheels to spin? Further, since the torque is being caused by a frictional force, it must be in the opposite direction of the planes acceleration. Thus it is the ONE single force that is opposing the thrust in this problem. This is pretty clear if you imagine a free body diagram of the forces.

I see that you also wrote that translational v equals tangential v under no slip conditions. This is absolutely correct, and is what I stated before. But if you want to understand what is opposing the acceleration of the plane, you need to understand that it is the frictional force being applied to the wheels. If you ignore this force, and thus the torque on the wheels, the only force you will see is thrust. And it will look like the plane must accelerate.

[Matt R.]

[ QUOTE ]
The 2X's came in by assuming that the belt automatically matched the "spin" of the tires. I admit I'm confused by the terms angular and tangential. All I know is that if the tires spin at 100 RPM and the belt matches this speed in the opposite direction, then this would cause the tires to spin at 200 RPM, which creates a belt speed of 200 RPM, which cause the tires to spin at 400 RPM, and so on to infinity.

[/ QUOTE ]

Lestat,
Don't think of the belt speed matching the 'rpm' of the wheel. The conveyor belt isn't circular like the wheel, and trying to think of this solely in terms of angular velocity will get really confusing. Convert everything to tangential velocities. Just imagine a point at the very edge of the wheel. The tangential velocity is the speed at which this is rotating. If you think about it, you will see that if the velocity of the 'rim' of the wheel is the same as the edge of the conveyor belt, then the plane cannot be moving forward.

Think of it this way. For a circle, the circumference is related to the radius by C = 2pi*r. 2 pi is just the angle of rotation. So if you want the distance along the circumference, s, it is related by the angle through which you are rotating and the radius of the circle... s = r*(theta). Thus the velocity of this edge (such as on a wheel) is related by v = r*omega where omega is the angular velocity. Divide both sides by time to see this.

It is incorrect to think of the wheel matching the belt speed in terms of rpm of the wheel. You need to use tangential velocity of the wheel to determine what speed the belt is 'matching'. The factor that you came up with, 2x, will depend on the radius of the wheel. This factor isn't a constant (okay, it IS a constant in the mathematical sense since the radius isn't changing. What I really mean is that it will depend on the radius... i.e. it's not always 2x).

[Lestat]

Matt-

I appreciate you spelling this all out for me and I'm definitely learning something. However, this is all really beyond my scope and education. I hate for you to waste your time typing to someone who's not *fully* grasping what you're saying.

My main point is that I don't think whoever first posed this question ever intended it to become so convoluted with semantical details. I'm quite sure the gist of it is that the wheels do not drive the plane. Sloppy or lazy thinkers will not readily reaize this. They will incorrectly assume the plane will be unable to gain forward momentum and therefore can't take off. Which would be wrong according to the puzzle's original intention.

I am not an engineer or a physicist, so I do not understand all the angular, tangential terms and equations. But I am quite sure that given the question's intended context, the correct answer is that the plane takes off.

[Matt R.]

Yes, based on a link that BCVP posted earlier in the thread that "answered" the question, I think the original intent of the question is much much simpler than the various interpretations that we've had in this thread. The question is poorly worded, but I think you understand the original intent without going into the tangential vs. angular velocity jargon.

[madnak]

Sharkey, I'll certainly concede that this might be possible under certain hypothetical conditions (reinforced uber-materials).

As far as I can tell, we're all agreed about everything except the specific assumptions implied by the OP. And I don't think that can be resolved. "It depends" seems like the clear answer to this particular thread on the subject.

[bunny]

Heh - I've been doggedly lurking through thirteen pages of arguments and the answer turns out to be "it depends" /images/graemlins/tongue.gif

[kbfc]

If torque is so important, then what is the moment of inertia of the wheels?

The answer is ZERO...who cares? It's clearly assumed in the problem. Frictionless bearings. Zero-mass wheels. No-slip rolling. Obviously these things don't exist in real life, but they're standard assumptions in physics problems like this.

[kbfc]

He's just confusing you with irrelevant crap. The following is the correct "answer":

The problem describes an impossible situation. However, if you change it slightly to say, "some guy CLAIMS the belt will match the wheel speed exactly," then you could argue that the answer is, "yes, the plane takes off normally, and that guy's claim turns out to be false."