Can the airplane take off?

[dkozloski]

The basic premise of the problem is that the belt doesn't move until the aircraft moves forward. The aircraft is always ahead of the belt. When the aircraft reaches flying speed the belt is moving the opposite way at flying speed. The wheels are turning twice flying speed as they separate from the belt. You can't change the conditions of the problem in mid stream. The only additional impediment of the belt to the aircraft besides the frictionless bearings is the spinup loads on the wheels which is roughly half the mass of the wheels and tires which are a very small percentage of the total mass of the flying machine. This is not a very clever or tricky problem. It's actually pretty dumb. It was probably put forward by an individual just about as dumb.

On to more worthy discussions. I'm tired of casting my pearls before swine.

 

[CSMR]

The question didn't specify whether the wheels are free to turn or not. That is the main cause of confusion, though there are other ambiguities too.

 

[Born2bwire]

Originally posted by: smack Down

Originally posted by: Born2bwire

Originally posted by: smack Down

Originally posted by: dkozloski
It makes no difference what the conveyor belt does. It is irrelevant. The belt can go forward, it can go backward, it can start and stop, it can alternate back and forth, and it makes no difference other than being an annoyance.

That is simpley wrong. By definition if the treadmill is going backwards at the same speed the wheels are turning then plane can't go anywhere.

Why? The plane's only contact with it's wheels are free rolling bearings.

I stole this from the other thread because he says it better and even includes the FBD.

Originally posted by: jagec

Originally posted by: dxkj
Correct, a small amount of energy is used to rotate the mass of the wheels. but wouldn't that energy be generated by the super fast super powered tread mill, and not by the plane? As long as the friction is zero on the axles the wheels could move infinitely fast with the tread mill, and the plane wouldnt move at all even under 0 power.


edit: IE, the wheels have a frictioned connection to the conveyer belt with the rubber touching rubber, but the ball bearing connection to the plane is frictionless, thus the energy and angular momentum provided to the wheels comes completely from the surface the wheels are touching.

If you draw the FBD, you find that the conveyor exerts a linear force on the base of the wheel, and the engines exert a linear force in the opposite direction through the axis of rotation. This creates a torque. If the force of the engines is not balanced perfectly by the force of the conveyor, you'll get linear as well as angular acceleration. This would be the case in any real-world scenario, since it's nearly impossible to exert the same force on the wheels with the conveyor, as you can on the plane with the engines. F=ma; when you're working against the large mass of the plane, that's fine. When you're working against the small rotational inertia of the wheels, the acceleration term becomes exceedingly large.

But in magical physics land, where we can accelerate the conveyor as fast as we bloody well please, we're able to equalize the forces. Thus, there is no net lateral force, just a huge torque that's spinning the wheels faster and faster, until the plane runs out of fuel.

Even with frictionless bearings, the wheels are able to exert a lateral force on the plane through the angular momentum term. That's the secret. It's easier to think of a massively weighty wheel on a treadmill...if we exert no additional forces on it, the wheel will begin to slowly turn and slowly move backwards at the same time when we turn the treadmill on. If we add a force through the axis of rotation, we're able to keep the wheel in one place, but it builds angular momentum that much faster.

You're still missing the point. Nobody is going to argue that the wheels will exert a force back onto the plane. But this force is going to be negligible for our situation. The whole point of the rollerskates on a treadmill thought experiment is that it's no real feat of strength to pull yourself forward even if the treadmill is moving at a very high speed. The fact here is realize that when the plane takes off with no net effect from the wind, the wheels are already going to be turning at x mph. The use of the conveyor belt is only going to increase this speed by a simple factor, which in our statement of the problem is 2. If a plane is able to take off with ease despite the fact that the wheels are providing a counter force Y when they are moving at a lateral velocity equal to the take off speed, then there still is not going to be any significant factor when we double or triple this lateral velocity. You would have to probably assume that the conveyor belt would have to move at orders of magnitude faster than the required take off velocity. But this is obviously in violation of the spirit of our Gedankenexperiment.

The only time that I have seen this situation to be interpreted where the plane does not take off is when the problem is stated incorrectly. I think it was previously stated as such in the OT forum where the conveyor belt moves such that the plane doesn't move. Obviously these conditions negate the experiment. Another incorrect way to state the problem is to say that the belt matches the speed of the wheels. But the wheel's velocity is the sum of the plane's plus the belt. So this is another impossibility because your are saying that V_belt = V_wheel + V_plane = V_wheel. The only valid setup that would ensure the plane would not take off would be to require that the conveyor belt accelerates in such a way that the force by the wheels on the plane overtakes the force of the engines before it can reach take off velocity. But it does not take much of a stretch of the imagination to see that this would require the belt to ramp up to an incredible speed very quickly, certainly not something as small as the ground speed of the aircraft.

 

[Born2bwire]

Originally posted by: CSMR
The question didn't specify whether the wheels are free to turn or not. That is the main cause of confusion, though there are other ambiguities too.

It's an airplane. I can only imagine the spectacle that would happen if the wheels were not free motion but coupled to an engine of some kind and the pilot attempts to land with the wheels engaged to the engine.

 

[beansbaxter]

I asked my friend who is a pilot and he said:

He said that as long as there is nothing to restrain the airplane, ie. a tie down that will prevent forward momentum, the airplane will take off and fly. This is due to the fact that unlike our bikes and cages that develop drive from power being transmitted to the ground through the driven wheel(s) an aircraft develops drive through thrust of its jet engine or propeller. The wheels are not connected to the drive mechanism so do not affect the speed of the airplane. There are various airplanes out there that have a very slow take-off speed and are able to fly with a headwind of around 20kts. So if a natural wind is blowing at 20 kts and the plane is stationary, all the pilot has to do is just crack the throttle and the plane will fly sort of like a helocoptor, until forward momentum is attained.

 

[Zorba]

Originally posted by: enwar3
I can't believe we're still discussing this.

A plane takes off because of the lift of air passing under its wings. If it's taxiing on a conveyor belt like you said, it's motionless - air does not pass under the wings and there is no lift.

Anyone that claims that a plane takes off because of air passing UNDER its wings, does not have the education to respond to this thread.

A wing generates lift from air moving around the wing. The air moving over the top of the wing is moving faster than the air on the bottom, due to the shape of the wing. Due to conservation of momentum, the faster moving air has a lower static pressure than the slower moving air on the bottom of the wing. This difference of pressure on the top and bottom of the wing is what generates lift. Basically the wing is being sucked upward.

BTW: The plane will take off, no questions asked. The reason why Mythbusters doesn't do a show on it, is because it is completely stupid. It is like asking if a car would move forward if the wind was going the same speed in the opposite direction.

 

[Toastedlightly]

I hereby apologize for posting a link to this in OT. I offer my life to the people of Highly Technical.

/bows

 

[smack Down]

Originally posted by: Born2bwire

Originally posted by: smack Down

Originally posted by: Born2bwire

Originally posted by: smack Down

Originally posted by: dkozloski
It makes no difference what the conveyor belt does. It is irrelevant. The belt can go forward, it can go backward, it can start and stop, it can alternate back and forth, and it makes no difference other than being an annoyance.

That is simpley wrong. By definition if the treadmill is going backwards at the same speed the wheels are turning then plane can't go anywhere.

Why? The plane's only contact with it's wheels are free rolling bearings.

I stole this from the other thread because he says it better and even includes the FBD.

Originally posted by: jagec

Originally posted by: dxkj
Correct, a small amount of energy is used to rotate the mass of the wheels. but wouldn't that energy be generated by the super fast super powered tread mill, and not by the plane? As long as the friction is zero on the axles the wheels could move infinitely fast with the tread mill, and the plane wouldnt move at all even under 0 power.


edit: IE, the wheels have a frictioned connection to the conveyer belt with the rubber touching rubber, but the ball bearing connection to the plane is frictionless, thus the energy and angular momentum provided to the wheels comes completely from the surface the wheels are touching.

If you draw the FBD, you find that the conveyor exerts a linear force on the base of the wheel, and the engines exert a linear force in the opposite direction through the axis of rotation. This creates a torque. If the force of the engines is not balanced perfectly by the force of the conveyor, you'll get linear as well as angular acceleration. This would be the case in any real-world scenario, since it's nearly impossible to exert the same force on the wheels with the conveyor, as you can on the plane with the engines. F=ma; when you're working against the large mass of the plane, that's fine. When you're working against the small rotational inertia of the wheels, the acceleration term becomes exceedingly large.

But in magical physics land, where we can accelerate the conveyor as fast as we bloody well please, we're able to equalize the forces. Thus, there is no net lateral force, just a huge torque that's spinning the wheels faster and faster, until the plane runs out of fuel.

Even with frictionless bearings, the wheels are able to exert a lateral force on the plane through the angular momentum term. That's the secret. It's easier to think of a massively weighty wheel on a treadmill...if we exert no additional forces on it, the wheel will begin to slowly turn and slowly move backwards at the same time when we turn the treadmill on. If we add a force through the axis of rotation, we're able to keep the wheel in one place, but it builds angular momentum that much faster.

You're still missing the point. Nobody is going to argue that the wheels will exert a force back onto the plane. But this force is going to be negligible for our situation. The whole point of the rollerskates on a treadmill thought experiment is that it's no real feat of strength to pull yourself forward even if the treadmill is moving at a very high speed. The fact here is realize that when the plane takes off with no net effect from the wind, the wheels are already going to be turning at x mph. The use of the conveyor belt is only going to increase this speed by a simple factor, which in our statement of the problem is 2. If a plane is able to take off with ease despite the fact that the wheels are providing a counter force Y when they are moving at a lateral velocity equal to the take off speed, then there still is not going to be any significant factor when we double or triple this lateral velocity. You would have to probably assume that the conveyor belt would have to move at orders of magnitude faster than the required take off velocity. But this is obviously in violation of the spirit of our Gedankenexperiment.

The only time that I have seen this situation to be interpreted where the plane does not take off is when the problem is stated incorrectly. I think it was previously stated as such in the OT forum where the conveyor belt moves such that the plane doesn't move. Obviously these conditions negate the experiment. Another incorrect way to state the problem is to say that the belt matches the speed of the wheels. But the wheel's velocity is the sum of the plane's plus the belt. So this is another impossibility because your are saying that V_belt = V_wheel + V_plane = V_wheel. The only valid setup that would ensure the plane would not take off would be to require that the conveyor belt accelerates in such a way that the force by the wheels on the plane overtakes the force of the engines before it can reach take off velocity. But it does not take much of a stretch of the imagination to see that this would require the belt to ramp up to an incredible speed very quickly, certainly not something as small as the ground speed of the aircraft.

If V_belt = V_wheel + V_plane = V_wheel then V_plane = 0 and the plane doesn't take off. The problem assumes that the treadmill can go infinitly fast and can track the speed of the plane 100%. The fact that no real treadmill can do this doesn't matter.

If you assume that the treadmill tracks the velocity of the plane relitive to ground yes the plane will take off and a car can drive. On the other hand if you assume that the treadmill tracks the velocity of the plane relitive to the treadmill the plane or a car will not move.

 

[CSMR]

Originally posted by: Born2bwire

Originally posted by: CSMR
The question didn't specify whether the wheels are free to turn or not. That is the main cause of confusion, though there are other ambiguities too.

It's an airplane. I can only imagine the spectacle that would happen if the wheels were not free motion but coupled to an engine of some kind and the pilot attempts to land with the wheels engaged to the engine.

Who is talking about landing?

 

[Gibsons]

Originally posted by: smack Down


If V_belt = V_wheel + V_plane = V_wheel then V_plane = 0 and the plane doesn't take off. The problem assumes that the treadmill can go infinitly fast and can track the speed of the plane 100%. The fact that no real treadmill can do this doesn't matter.

If you assume that the treadmill tracks the velocity of the plane relitive to ground yes the plane will take off and a car can drive. On the other hand if you assume that the treadmill tracks the velocity of the plane relitive to the treadmill the plane or a car will not move.

How are you measuring V_wheel and V_plane?

The plane's engines creates thrust. this will propel the plane forward in a newtonian fashion (m1v1 = m2v2, plane is m1, jet exhaust is m2) neglecting friction (from the air and the wheels spinning on the ground). All the treadmill does is increase the friction from the tires by causing them to spin faster than usual. If takeoff speed is 100 mph, the wheels will spin at an apparent rate (assuming you attempt to measure groundspeed by rpm of the tires) of 200 mph, but the plane will be moving forward with an airspeed of 100 mph - and it will take off.

Or, re-read drkozloski's explanation. I thought it was perfectly clear, but I guess not.

 

[smack Down]

Originally posted by: Gibsons

Originally posted by: smack Down


If V_belt = V_wheel + V_plane = V_wheel then V_plane = 0 and the plane doesn't take off. The problem assumes that the treadmill can go infinitly fast and can track the speed of the plane 100%. The fact that no real treadmill can do this doesn't matter.

If you assume that the treadmill tracks the velocity of the plane relitive to ground yes the plane will take off and a car can drive. On the other hand if you assume that the treadmill tracks the velocity of the plane relitive to the treadmill the plane or a car will not move.

How are you measuring V_wheel and V_plane?

The plane's engines creates thrust. this will propel the plane forward in a newtonian fashion (m1v1 = m2v2, plane is m1, jet exhaust is m2) neglecting friction (from the air and the wheels spinning on the ground). All the treadmill does is increase the friction from the tires by causing them to spin faster than usual. If takeoff speed is 100 mph, the wheels will spin at an apparent rate (assuming you attempt to measure groundspeed by rpm of the tires) of 200 mph, but the plane will be moving forward with an airspeed of 100 mph - and it will take off.

Or, re-read drkozloski's explanation. I thought it was perfectly clear, but I guess not.

There are simplely two ways to read the problem.

If the treadmill tracks the speed of the plane relitive to the treadmill it will speed up to such a speed that the plane is stationary due to the force of spinning the wheels backwards. Think a car on a treadmill where the speed of the treadmill is the same as read by the speedopeter.

The other way to read the problem is the treadmill tracks the speed of the plane relitive to ground. This is the case where you get the wheels are spinning at 2x the speed relivitave to ground. This case can be thought about as a treadmill going at a fixed speed at where x is the take off speed. If you place a car on that treadmill which with a top speed greater then 2x it will also reach speed x.

 

[Born2bwire]

Originally posted by: CSMR

Originally posted by: Born2bwire

Originally posted by: CSMR
The question didn't specify whether the wheels are free to turn or not. That is the main cause of confusion, though there are other ambiguities too.

It's an airplane. I can only imagine the spectacle that would happen if the wheels were not free motion but coupled to an engine of some kind and the pilot attempts to land with the wheels engaged to the engine.

Who is talking about landing?

If the wheels are not free motion it implies that the wheels must be engaged to some sort of drive system. And if this is the case, which would be totally unnecessary for an airplane, I could only imagine the unique set of accidents that could result from this.

 

[Born2bwire]

Originally posted by: smack Down

Originally posted by: Gibsons

Originally posted by: smack Down


If V_belt = V_wheel + V_plane = V_wheel then V_plane = 0 and the plane doesn't take off. The problem assumes that the treadmill can go infinitly fast and can track the speed of the plane 100%. The fact that no real treadmill can do this doesn't matter.

If you assume that the treadmill tracks the velocity of the plane relitive to ground yes the plane will take off and a car can drive. On the other hand if you assume that the treadmill tracks the velocity of the plane relitive to the treadmill the plane or a car will not move.

How are you measuring V_wheel and V_plane?

The plane's engines creates thrust. this will propel the plane forward in a newtonian fashion (m1v1 = m2v2, plane is m1, jet exhaust is m2) neglecting friction (from the air and the wheels spinning on the ground). All the treadmill does is increase the friction from the tires by causing them to spin faster than usual. If takeoff speed is 100 mph, the wheels will spin at an apparent rate (assuming you attempt to measure groundspeed by rpm of the tires) of 200 mph, but the plane will be moving forward with an airspeed of 100 mph - and it will take off.

Or, re-read drkozloski's explanation. I thought it was perfectly clear, but I guess not.

There are simplely two ways to read the problem.

If the treadmill tracks the speed of the plane relitive to the treadmill it will speed up to such a speed that the plane is stationary due to the force of spinning the wheels backwards. Think a car on a treadmill where the speed of the treadmill is the same as read by the speedopeter.

The other way to read the problem is the treadmill tracks the speed of the plane relitive to ground. This is the case where you get the wheels are spinning at 2x the speed relivitave to ground. This case can be thought about as a treadmill going at a fixed speed at where x is the take off speed. If you place a car on that treadmill which with a top speed greater then 2x it will also reach speed x.

The problem that the belt moves at the same speed as the plane relative to the belt is the same impossibility that I had stated before. It defeats the entire purpose of the question because the only set of conditions under which these can hold is that the plane remains stationary. But this has no real physics related to it because it does not imply anything about how the treadmill affects the plane. It does not mean that the belt is going to move faster than the plane can counteract. It only means that the only set of conditions that satisifies the problem statement is that the plane speed is zero. Further more, we also have to have our initial conditions be satisfactory as well. So if the plane starts at rest, it must remain at rest because we cannot allow it to have a forward motion because we would run into something similar to a Zeno's paradox.

But all of this is immaterial, the original problem statement is fine here. It says that the treadmill will match the plane's speed, when the plane moves forward. This implies forward motion of the plane. This absolves us from having to consider the other possible nonsensical interpretations because if the treadmill was to match the plane's speed in reference to the treadmill, then we could never allow the plane to have any motion relative to the ground or treadmill.

 

[smack Down]

Originally posted by: Born2bwire

Originally posted by: smack Down

Originally posted by: Gibsons

Originally posted by: smack Down


If V_belt = V_wheel + V_plane = V_wheel then V_plane = 0 and the plane doesn't take off. The problem assumes that the treadmill can go infinitly fast and can track the speed of the plane 100%. The fact that no real treadmill can do this doesn't matter.

If you assume that the treadmill tracks the velocity of the plane relitive to ground yes the plane will take off and a car can drive. On the other hand if you assume that the treadmill tracks the velocity of the plane relitive to the treadmill the plane or a car will not move.

How are you measuring V_wheel and V_plane?

The plane's engines creates thrust. this will propel the plane forward in a newtonian fashion (m1v1 = m2v2, plane is m1, jet exhaust is m2) neglecting friction (from the air and the wheels spinning on the ground). All the treadmill does is increase the friction from the tires by causing them to spin faster than usual. If takeoff speed is 100 mph, the wheels will spin at an apparent rate (assuming you attempt to measure groundspeed by rpm of the tires) of 200 mph, but the plane will be moving forward with an airspeed of 100 mph - and it will take off.

Or, re-read drkozloski's explanation. I thought it was perfectly clear, but I guess not.

There are simplely two ways to read the problem.

If the treadmill tracks the speed of the plane relitive to the treadmill it will speed up to such a speed that the plane is stationary due to the force of spinning the wheels backwards. Think a car on a treadmill where the speed of the treadmill is the same as read by the speedopeter.

The other way to read the problem is the treadmill tracks the speed of the plane relitive to ground. This is the case where you get the wheels are spinning at 2x the speed relivitave to ground. This case can be thought about as a treadmill going at a fixed speed at where x is the take off speed. If you place a car on that treadmill which with a top speed greater then 2x it will also reach speed x.

The problem that the belt moves at the same speed as the plane relative to the belt is the same impossibility that I had stated before. It defeats the entire purpose of the question because the only set of conditions under which these can hold is that the plane remains stationary. But this has no real physics related to it because it does not imply anything about how the treadmill affects the plane. It does not mean that the belt is going to move faster than the plane can counteract. It only means that the only set of conditions that satisifies the problem statement is that the plane speed is zero. Further more, we also have to have our initial conditions be satisfactory as well. So if the plane starts at rest, it must remain at rest because we cannot allow it to have a forward motion because we would run into something similar to a Zeno's paradox.

But all of this is immaterial, the original problem statement is fine here. It says that the treadmill will match the plane's speed, when the plane moves forward. This implies forward motion of the plane. This absolves us from having to consider the other possible nonsensical interpretations because if the treadmill was to match the plane's speed in reference to the treadmill, then we could never allow the plane to have any motion relative to the ground or treadmill.

No the bolded really does not imply forward motion it only implies that the treadmill has a control system which matches the speed of the plane. At best you could say that there is some lag in the control system. Which speed it matchs is total up to the user to determine.

So your say that because the plane plane is stationary it is wrong to read it that way. Have you ever used a treadmill before? Guess what you stay pretty much in the same spot. Your just trying to read the question in such a way to say it takes off when the question doesn't give important detials.

 

[pm]

My answer is, it depends. I know one specific case, where it does seem to work.

I have an electric model airplane with a thrust to weight ratio of about 2.5:1 (plane weighs ~12oz, prop/motor generates ~30oz of thrust), if I remove the rear wheels so that the plane is supported merely by 2 wires in the back, and one wheel in the front, plunk it in the middle of a grassy field and fire up the motor to full, it will take off. As I throttle up, it will sit, and sit and then it will start to wobble and then at some point on the very high end of the throttle it will take off (or it will flip forward on it's nose... but if I get lucky, with full elevator, it will go up). The only trick is keeping it from flipping forward - which is why I need a wheel on the front. I have done this a couple of times (a wheel falls off, can't fix it, but still want to fly) and it works... landings are tough though. If I fly normally, with the wheels on, the plane takes off in about 2 feet... so I can put in on one side of a sidewalk and take off before I hit the other side. If you guys want a video, I can see what I can do.

My explanation for this is that the plane has a very large prop compared to the wing size. The prop that I fly with is a 10" prop and the total wing from tip to tip is 31.5". The prop/motor at full throttle generates a ridiculous amount of thrust for a wing this size which corresponds to a ridiculous amount of airflow over the wing. When the speed of the prop wash is faster than the stall speed (which is very low... 32" wing and 12oz...) and there is sufficient air movement over the wing, and the elevator is full back (and it's a huge elevator relative to the size on a "real" plane), then the plane just lifts itself up. The motor that I'm using is a hand-made (by myself), custom-wound, 3-phase brushless outrunner made with neodymium magnets. For what it's worth, the plane that I am using is: Mountain Models Switchback 3D

At one point, I put floats on it (for water take-offs and landings) and I had enough thrust to go straight vertical with it immediately after leaving the water surface. I was flying it inverted (plane upside down, floats pointed skyward)... it was pretty funny to watch.

 

[CSMR]

Originally posted by: Born2bwireIf the wheels are not free motion it implies that the wheels must be engaged to some sort of drive system.

No a braking system is not free motion and is not a drive system either.

And if this is the case, which would be totally unnecessary for an airplane, I could only imagine the unique set of accidents that could result from this.

http://en.wikipedia.org/wiki/Clutch

 

[OCedHrt]

Originally posted by: smack Down

There are simplely two ways to read the problem.

If the treadmill tracks the speed of the plane relitive to the treadmill it will speed up to such a speed that the plane is stationary due to the force of spinning the wheels backwards. Think a car on a treadmill where the speed of the treadmill is the same as read by the speedopeter.

The other way to read the problem is the treadmill tracks the speed of the plane relitive to ground. This is the case where you get the wheels are spinning at 2x the speed relivitave to ground. This case can be thought about as a treadmill going at a fixed speed at where x is the take off speed. If you place a car on that treadmill which with a top speed greater then 2x it will also reach speed x.

I really hate people who can't admit that they are wrong. These kind of people could possibly be the source of all the problems in the world.

There is only one way to read this problem. The speed of the plane relative to the treadmill or the ground are the same. The treadmill is not moving anywhere.

On the otherhand, the speed of the plane relative to the treadmill surface will be 2x the velocity of the plane relative to the treadmill or ground, simply because the speed of the treadmill surface will be going at -V, where V is the velocity of the plane.

However, the plane does not care about its relative velocity to the treadmill, but rather only the lift generated by the pressure difference between the air below and above the wings. You can shoot a plane out of a canon at liftoff speed and it will fly (assuming you don't damage the plan with the cannon).

The wheels on the otherhand, due to friction, will spin at 2x the plane's velocity relative to the ground or treadmill. This is further supporting evidence that the difference in speed in the treadmill's/ground's frame of reference is 2V, where once again, V is the relative velocity of the plane in this frame.

 

[Calin]

Can a stationary plane lift off (stationary - zero ground speed, no wind)? Yes, with some conditions:
1. a Harrier is designed for that - they can hover in place just like a helicopter can
2. There are other planes that have more thrust than weight (can accelerate in a vertical climb). If you put them in a good position (straight up) and can stabilize them, they will start to increase their altitude
3. A commercial jet will never do that - they lift themselves by the wings (and body for some), and the wing lift comes only from airspeed.
4. Hypothetically, a plane powered by big propellers could create (if the propellers are in front of the wings) enough airspeed so the lift of the wings will launch it (at zero ground speed and no wind). If the propellers are behind the wings, they still pull air in front of the wings, so there is some "airspeed" over parts of the wing.
5. commercial jets keep the air/hot gasses pulled/pushed by engines out of the wing's trajectory (they are below the wing, and in front of it)

 

[smack Down]

Originally posted by: OCedHrt

Originally posted by: smack Down

There are simplely two ways to read the problem.

If the treadmill tracks the speed of the plane relitive to the treadmill it will speed up to such a speed that the plane is stationary due to the force of spinning the wheels backwards. Think a car on a treadmill where the speed of the treadmill is the same as read by the speedopeter.

The other way to read the problem is the treadmill tracks the speed of the plane relitive to ground. This is the case where you get the wheels are spinning at 2x the speed relivitave to ground. This case can be thought about as a treadmill going at a fixed speed at where x is the take off speed. If you place a car on that treadmill which with a top speed greater then 2x it will also reach speed x.

I really hate people who can't admit that they are wrong. These kind of people could possibly be the source of all the problems in the world.

There is only one way to read this problem. The speed of the plane relative to the treadmill or the ground are the same. The treadmill is not moving anywhere.

On the otherhand, the speed of the plane relative to the treadmill surface will be 2x the velocity of the plane relative to the treadmill or ground, simply because the speed of the treadmill surface will be going at -V, where V is the velocity of the plane.

However, the plane does not care about its relative velocity to the treadmill, but rather only the lift generated by the pressure difference between the air below and above the wings. You can shoot a plane out of a canon at liftoff speed and it will fly (assuming you don't damage the plan with the cannon).

The wheels on the otherhand, due to friction, will spin at 2x the plane's velocity relative to the ground or treadmill. This is further supporting evidence that the difference in speed in the treadmill's/ground's frame of reference is 2V, where once again, V is the relative velocity of the plane in this frame.

Well smart guy if the treadmill and ground are the same maybe just maybe I was talking about the part of a treadmill that moves. You know like you did. So fine the other way to read it is to consider that the treadmill tracks the speed of the plane relitive to the top moving surface of the treadmill you happy. Now when are you going to admit you are wrong?

 

[AeroEngy]

Let's say the maximum speed of the plane is 300mph. You place the aircraft on the belt, and set it in reverse for 300mph. With the engines off, the plane will fall off the back of the belt rather quickly.

Turn on the engines, put them to full power. The speed of the plane FORWARD is 300mph, but that's being counteracted by the speed of the belt BACKWARDS of 300mph. Include the friction of the wheels, and you see that there is NO WAY FOR LIFT TO FORM UNDER THE WINGS! Therefore it cannot take off! Simple physics.

This is completely wrong.

If you draw a free body Force diagram around the plane. The only things acting on the plane in the horizontal direction would be rolling resistance acting on the wheels opposite the force generated from the engine. So the only way to stop the plane form moving would be if the rolling resistance was equal to the thrust of the plane. Since rolling resistance in most models is NOT A FUNCTION of angular velocity (Rolling Friction = C_roll * N ; C_roll ~ 0.01 to 0.015; N = Weight of the plane). So a loose calc for rolling friction for an F-16 at max takeoff weight is approximately 0.015*42,300lbs = 634 lbs., again independent of wheel speed. The thrust from an afterburning F110 GE turbofan engine on an F-16 is about 28,500 lbs. So you have a net force of 27866 lbs acting to accelerate the plane. Since F=m*A the plane would accelerate forward relative to the ground and eventually take off.

For your above scenario the rolling resistance would have to equal 28,500lbf to negate the thrust. That would mean your friction coefficient would have to be 28,500/42,300 = .67. This is around the range of rubber sliding not rolling on asphalt. So basically even if the wheels were locked and would not rotate the plane would still accelerate relative to the ground and take off( after leaving stinking burning rubber all over the conveyer belt).

Side Note:
There are some very complicated models for friction coefficient in bearings that do factor in angular velocity at extremely high speeds. Where the viscosities of the lubricant inside the bearing are considered, sliding friction from the deformation of the contact points inside the bearings as well as many other factors. However, the overall friction coefficient in these models is still a weak function of velocity. But even if you accelerated the belt to a high enough speed to lock up the bearing the F110 turbine would overcome the friction between the belt and the tires.

PS This topic/argument is ridiculous anyway and hopefully someone will close this thread.

 

[sao123]

Originally posted by: BriGy86
I don't see how it could take off, no one has mentioned the force of gravity that holds the plane to the ground, if your on a convayor belt with roller skates (never mind the rope) and you turn the belt on you will start to go backwards, because there is friction in bearings and gravity is pulling you down.

the rope idea I see as being flawed, you're pulling on a stationary object.

I also looked over at the mythbusters forums to see if this is covered, they have a 53 page thread, why the hell hasn't it been aired yet?
one good way for them to test this would be with a small model airplane (that can actually take off and fly) on some sort of long tread mill

the engines pull on a stationary object also... its called the atmosphere.

 

[smack Down]

Originally posted by: AeroEngy

Let's say the maximum speed of the plane is 300mph. You place the aircraft on the belt, and set it in reverse for 300mph. With the engines off, the plane will fall off the back of the belt rather quickly.

Turn on the engines, put them to full power. The speed of the plane FORWARD is 300mph, but that's being counteracted by the speed of the belt BACKWARDS of 300mph. Include the friction of the wheels, and you see that there is NO WAY FOR LIFT TO FORM UNDER THE WINGS! Therefore it cannot take off! Simple physics.

This is completely wrong.

If you draw a free body Force diagram around the plane. The only things acting on the plane in the horizontal direction would be rolling resistance acting on the wheels opposite the force generated from the engine. So the only way to stop the plane form moving would be if the rolling resistance was equal to the thrust of the plane. Since rolling resistance in most models is NOT A FUNCTION of angular velocity (Rolling Friction = C_roll * N ; C_roll ~ 0.01 to 0.015; N = Weight of the plane). So a loose calc for rolling friction for an F-16 at max takeoff weight is approximately 0.015*42,300lbs = 634 lbs., again independent of wheel speed. The thrust from an afterburning F110 GE turbofan engine on an F-16 is about 28,500 lbs. So you have a net force of 27866 lbs acting to accelerate the plane. Since F=m*A the plane would accelerate forward relative to the ground and eventually take off.

For your above scenario the rolling resistance would have to equal 28,500lbf to negate the thrust. That would mean your friction coefficient would have to be 28,500/42,300 = .67. This is around the range of rubber sliding not rolling on asphalt. So basically even if the wheels were locked and would not rotate the plane would still accelerate relative to the ground and take off( after leaving stinking burning rubber all over the conveyer belt).

Side Note:
There are some very complicated models for friction coefficient in bearings that do factor in angular velocity at extremely high speeds. Where the viscosities of the lubricant inside the bearing are considered, sliding friction from the deformation of the contact points inside the bearings as well as many other factors. However, the overall friction coefficient in these models is still a weak function of velocity. But even if you accelerated the belt to a high enough speed to lock up the bearing the F110 turbine would overcome the friction between the belt and the tires.

PS This topic/argument is ridiculous anyway and hopefully someone will close this thread.

The treadmil doesn't act on the plane by friction (well it does but it is ignored in this problem) The treadmill applies a torque to the wheels which will then move the plane if the torque is large enough to overcome friction.

 

[videogames101]

....Look. No matter how fast any jet engine goes the engine itself does not move air around the wing. The plane stays where it is, air is not being forced around it. It's not that hard to understand... The plane will never take off. To those who say it will, what exactly, is pushing air around the wings?

Lets get mythbusters in here? lol, Someone should submit this to discovery. I will admit I may be wrong in my theory.

 

[brikis98]

it'll take off. here's a non-technical (and not 100% accurate, but good enough) way to picture why:

you have the same plane and treadmill, except now the plane is suspended from the ceiling by some cables such that its wheels are just making contact with the treadmill. so, there is basically no weight resting on the wheels. moreover, imagine the cables are connected to slide rails on the ceiling of this magic building so that they can slide (friction free) along the length of the building (in the direction of the treadmill). now, remember that in an airplane, the wheels are NOT connected by any kind of drive train to the actual plane - they are just free to spin. so, assume the plane's engines are off and the treadmill starts rolling backwards... what happens to the plane?

essentially, nothing. it's wheels start rolling, but since they are free to spin (ignoring, for now, friction and torque from the wheels), they don't move the plane backwards at all. now, what happens when you start the jet engines of the airplane?

the engines create thrust and will start pushing it forward on the slide rails. the wheels spinning backwards would have NO effect on the plane (again, we're still ignoring friction/torque), the plane would build up velocity as usual and take off (straight into the ceiling )...

now, lets throw friction and wheel torque back into the equation... the plane is still suspended by cables, so at most, this will slightly push the plane backwards. nothing major, really, it still takes off, right? ok, so now lets remove the whole cable assembly. what has changed?

the only real change is... more friction because of more mass pushing down! however, this won't be nearly enough to overcome the engine thrust and the plane will take off. the key is just to realize that the engines, not the wheels, propel the plane and that the wheels are just a way of reducing friction (as opposed to having the belly of the plane skidding on the runway). they just spin freely and won't play any role in preventing the plane's takeoff.

 

[randay]

To all of those who have lost faith in HT due to dumbness of some of the replies here, take note, the really smart guys are quite absent here.