Can the airplane take off?

[smack Down]

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

 

[brikis98]

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

 

[randay]

Im bored so I guess I'll chime in with my answer as well. The plane takes off "normally" with the exception that the wheels are spinning at a rate equal to as fast as the plane is moving.

It all depends on reading comprehension and interpretation of the question.

"An airplane is on a conveyur belt, the belt matches the airplanes speed in reverse. Can the airplane take off?"

You have to think of the airplane as a car in neutral with a rocket attached to it, not as a car in gear using its tires to move forward. The job of the wheels here would be to reduce friction between the car and the ground/belt. As long as the wheels are decently efficient are reducing friction, there should be absolutely no problems with overcoming the friction created by the wheels and the ground/belt.

If the airplane moves forward at 1 mph, then the belt moves backwards at 1 mph. The wheels are then spinning at a rate of 2 mph. This does not affect the airplanes acceleration except for the small amount of friction that the belt exerts on the wheels. Many have said to ignore this friction, for the sake of figuring out this rediculous riddle. Either way, I don't think that the amount of friction can stop an airplanes forward motion, unless the wheels break.

So say the airplane needs to be moving at 100 mph to take off. The above situation reoccurs, mph for mph until the airplane is moving 100 mph forward, the belt is moving 100mph backward, and the wheels are spinning at a rate of 200mph. Assuming that the wheels are performing thier job correctly, the airplane will take off normally after compensating for the small amount of extra friction on the wheels.

All those that think the belt would stop the airplane from moving are incorrect, because if the belt matches the airplanes speed, and the airplane is not moving, then the belt is not moving, and if the belt is not moving, then its not going to stop the airplane from moving. /headexplode

 

[AeroEngy]

Originally posted by: 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.

You have to be joking right. How else does the treadmill act on the plane if not through the rolling friciton in the wheels of the plane. What exactly do you think I just explained in my earlier post. The torque that you mentioned would just be the Force of Friction I calculated times the distance from the exterior of the tire to the hub. However, in a free body diagram who cares what the torque on the tire is, just sum the forces acting on the plane and that will equal the Mass * Acceleration.

Originally posted by: 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.

I agree that if there is no air moving over the wing then the plane will not lift off . However, the plane will acclerate and generate wind speed. There is not enough friction in the wheel bearings no matter how fast the treadmill goes to overcome the thrust of an airplane. Rolling resistance, friction, torque or whatever you want to call it is not a function of wheel speed (get a physics book and look up the equation for rolling friction F = C*N) so even if the tread mill is spinning at 5,000mph the plane will still accelerate relative to the ground/wind and take off when it had enough wind speed. (Assuming the wheels don't explode first)

This will be my last post on the subject unless someone directly asks me a legitimate question .... It would be kind of cool to see on Mythbusters though.

 

[smack Down]

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

 

[brikis98]

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

i'm not sure i see your point... the car's wheels are connected to the drive train... so if it's going 100mph relative to ground and the treadmill is going 100mph relative to ground in the other direction, the car will not actually move forward. this is just like you running up a down escalator...

on the other hand, an airplane's wheels are NOT connected to any drive train and so in the same scenario, it WILL move forward. yes, the wheels will be going 200 mph, but they are just there to reduce friction...

 

[smack Down]

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

i'm not sure i see your point... the car's wheels are connected to the drive train... so if it's going 100mph relative to ground and the treadmill is going 100mph relative to ground in the other direction, the car will not actually move forward. this is just like you running up a down escalator...

on the other hand, an airplane's wheels are NOT connected to any drive train and so in the same scenario, it WILL move forward. yes, the wheels will be going 200 mph, but they are just there to reduce friction...

If a car is going 100 mph relitive to ground then by definition it is going forward. The car's spedometer will read 200 mph just like the planes.

 

[dkozloski]

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

i'm not sure i see your point... the car's wheels are connected to the drive train... so if it's going 100mph relative to ground and the treadmill is going 100mph relative to ground in the other direction, the car will not actually move forward. this is just like you running up a down escalator...

on the other hand, an airplane's wheels are NOT connected to any drive train and so in the same scenario, it WILL move forward. yes, the wheels will be going 200 mph, but they are just there to reduce friction...

If a car is going 100 mph relitive to ground then by definition it is going forward. The car's spedometer will read 200 mph just like the planes.

An airplane doesn't have a speedometer. It has an airspeed indicator. The only parameter the aircraft responds to is it's relative speed through the air.

 

[smack Down]

Originally posted by: dkozloski

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

i'm not sure i see your point... the car's wheels are connected to the drive train... so if it's going 100mph relative to ground and the treadmill is going 100mph relative to ground in the other direction, the car will not actually move forward. this is just like you running up a down escalator...

on the other hand, an airplane's wheels are NOT connected to any drive train and so in the same scenario, it WILL move forward. yes, the wheels will be going 200 mph, but they are just there to reduce friction...

If a car is going 100 mph relitive to ground then by definition it is going forward. The car's spedometer will read 200 mph just like the planes.

An airplane doesn't have a speedometer. It has an airspeed indicator. The only parameter the aircraft responds to is it's relative speed through the air.

You could put one in the wheels if you want and then it would read the same are you happy? Was there a point to your post?

 

[Toastedlightly]

OMG I AM YELLING!

The plane takes off.

 

[Toastedlightly]

Originally posted by: smack Down

Originally posted by: dkozloski

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

i'm not sure i see your point... the car's wheels are connected to the drive train... so if it's going 100mph relative to ground and the treadmill is going 100mph relative to ground in the other direction, the car will not actually move forward. this is just like you running up a down escalator...

on the other hand, an airplane's wheels are NOT connected to any drive train and so in the same scenario, it WILL move forward. yes, the wheels will be going 200 mph, but they are just there to reduce friction...

If a car is going 100 mph relitive to ground then by definition it is going forward. The car's spedometer will read 200 mph just like the planes.

An airplane doesn't have a speedometer. It has an airspeed indicator. The only parameter the aircraft responds to is it's relative speed through the air.

You could put one in the wheels if you want and then it would read the same are you happy? Was there a point to your post?

Do you think the plane takes off?

 

[The Boston Dangler]

If the plane is motionless relative to the air or that motion is less than take-off speed = no Bernoulli = no flight. Lateral motion relative to the ground is irrelevant.

 

[smack Down]

Originally posted by: Toastedlightly

Originally posted by: smack Down

Originally posted by: dkozloski

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: brikis98

Originally posted by: smack Down

Originally posted by: 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.

That doesn't matter at all no matter how you define the treadmills controls the plane will act just like a tank, car, or cat.

care to explain that?

If you assume that the belt matches the speed of the plane relitive to ground then when the car is going 100 mph relitive to the ground the tread mill will be going backwards 100 miles per hour. Note that the car is still moving 100 mph forward relitive to ground so the wheels will be driven such that they spin at 200 mph.

i'm not sure i see your point... the car's wheels are connected to the drive train... so if it's going 100mph relative to ground and the treadmill is going 100mph relative to [/b]ground in the other direction, the car will not actually move forward. this is just like you running up a down escalator...

on the other hand, an airplane's wheels are NOT connected to any drive train and so in the same scenario, it WILL move forward. yes, the wheels will be going 200 mph, but they are just there to reduce friction...

If a car is going 100 mph relitive to ground then by definition it is going forward. The car's spedometer will read 200 mph just like the planes.

An airplane doesn't have a speedometer. It has an airspeed indicator. The only parameter the aircraft responds to is it's relative speed through the air.

You could put one in the wheels if you want and then it would read the same are you happy? Was there a point to your post?

Do you think the plane takes off?

Depends if the treadmill tracks the speed of the plane relitive to ground then it takes off or if the treadmill tracks the speed of the plane relitive to treadmill's surface then it does not.

 

[Gibsons]

Originally posted by: smack Down


Depends if the treadmill tracks the speed of the plane relitive to ground then it takes off or if the treadmill tracks the speed of the plane relitive to treadmill's surface then it does not.

The speed of the treadmill is practically irrelevant.

I think AeroEngy's post on the previous page explains it well:

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.

 

[smack Down]

Originally posted by: Gibsons

Originally posted by: smack Down


Depends if the treadmill tracks the speed of the plane relitive to ground then it takes off or if the treadmill tracks the speed of the plane relitive to treadmill's surface then it does not.

The speed of the treadmill is practically irrelevant.

The size of the force doesn't matter or where it comes from doesn't matter if you read the question such that the speed of the treadmill tracks the speed of the plane relitive to the treadmill surface you get this equation. This is the case most people think about when talking about treadmills

VPlane to treadmill = Vground + -Vtreadmill
Now by definition of the problem Vtreadmill = -VplanetoTreadmill
We get Vtreadmill = -Vground + Vtreadmill
Therefor Vground = 0 and the plane doesn't take off.



Where Vtread

 

[sao123]

The plane will take off.

Jet engines provide thrust relative to the atmosphere... not relative to the ground.
The treadmill is meaningless.

 

[BrownTown]

I think I should start a parody thread where I try to argue that 2+2=5 and ignore all evidence to the contrary. I mean seriously, if you are arguing that the plane doesn't take off then you are either stupid, stubborn, or both. Just because the fact that you are wrong is not as immediately obvious as the fact that "2+2=5" is wrong does not mean that you are any less wrong.

 

[AeroEngy]

Originally posted by: smack Down
The size of the force doesn't matter or where it comes from doesn't matter if you read the question such that the speed of the treadmill tracks the speed of the plane relitive to the treadmill surface you get this equation. This is the case most people think about when talking about treadmills

VPlane to treadmill = Vground + -Vtreadmill
Now by definition of the problem Vtreadmill = -VplanetoTreadmill
We get Vtreadmill = -Vground + Vtreadmill
Therefor Vground = 0 and the plane doesn't take off.

I know I said that my previous post was going to be my last. But LMAO ... are you realling trying to argue that size and direction of the forces don't matter. When they are the only thing that matters.

Newton's Second Law of motion- The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

 

[OCedHrt]

Originally posted by: 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?

The speed of the plane would never be equal to the speed of the treadmill surface relative to the plane unless the plane was stationary. The speed of the treadmill surface relative to the plane, where T is the speed of the treadmill surface, would be T = -2V. Only when V is 0 would T = V. However, once again, V is not 0 in this case. Therefore, the treadmill CANNOT track the speed of the plane relative to the treadmill moving surface unless the treadmill was off and this surface was moving at a speed of 0. Because if the plane where to go at V = 100, then the treadmill relative to the plane would be at T = -200 (speed of treadmill surface relative speed of plane = speed of treadmill surface - speed of plane = -100 - 100 = -200).

 

[smack Down]

Originally posted by: AeroEngy

Originally posted by: smack Down
The size of the force doesn't matter or where it comes from doesn't matter if you read the question such that the speed of the treadmill tracks the speed of the plane relitive to the treadmill surface you get this equation. This is the case most people think about when talking about treadmills

VPlane to treadmill = Vground + -Vtreadmill
Now by definition of the problem Vtreadmill = -VplanetoTreadmill
We get Vtreadmill = -Vground + Vtreadmill
Therefor Vground = 0 and the plane doesn't take off.

I know I said that my previous post was going to be my last. But LMAO ... are you realling trying to argue that size and direction of the forces don't matter. When they are the only thing that matters.

Newton's Second Law of motion- The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

Just show how my equations are wroung. It is simple algebra how or why it happens doesn't matter because the question defines it as happening.

 

[smack Down]

Originally posted by: OCedHrt

Originally posted by: 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?

The speed of the plane would never be equal to the speed of the treadmill surface relative to the plane unless the plane was stationary. The speed of the treadmill surface relative to the plane, where T is the speed of the treadmill surface, would be T = -2V. Only when V is 0 would T = V. However, once again, V is not 0 in this case. Therefore, the treadmill CANNOT track the speed of the plane relative to the treadmill moving surface unless the treadmill was off and this surface was moving at a speed of 0. Because if the plane where to go at V = 100, then the treadmill relative to the plane would be at T = -200 (speed of treadmill surface relative speed of plane = speed of treadmill surface - speed of plane = -100 - 100 = -200).

Where did you ever get the idea that T = -2V?

Follow your little work out another step if V = 100 then as you showed T = -200, but now the speed relitive to the treadmill surface relitive to the plane is 300 so T = 300. The speed of the treadmill will increase until the plane stops.

 

[gsellis]

Can we just permaban folks who think the answer is No? It sure would improve the intelligence at AT.

 

[BrownTown]

Originally posted by: smack Down

Originally posted by: AeroEngy

Originally posted by: smack Down
The size of the force doesn't matter or where it comes from doesn't matter if you read the question such that the speed of the treadmill tracks the speed of the plane relitive to the treadmill surface you get this equation. This is the case most people think about when talking about treadmills

VPlane to treadmill = Vground + -Vtreadmill
Now by definition of the problem Vtreadmill = -VplanetoTreadmill
We get Vtreadmill = -Vground + Vtreadmill
Therefor Vground = 0 and the plane doesn't take off.

I know I said that my previous post was going to be my last. But LMAO ... are you realling trying to argue that size and direction of the forces don't matter. When they are the only thing that matters.

Newton's Second Law of motion- The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

Just show how my equations are wroung. It is simple algebra how or why it happens doesn't matter because the question defines it as happening.

Yes, your equations are wrong because you say "Now by definition of the problem Vtreadmill = -VplanetoTreadmill", this is not correct, Vtreadmill = -Vground". This is obtained through a little reading comprehension.

"when the airplane moves forward, the conveyor matches its speed in reverse"

"Vground = -Vtreadmill"

 

[AeroEngy]

Originally posted by: smack Down

Originally posted by: AeroEngy

Originally posted by: smack Down
The size of the force doesn't matter or where it comes from doesn't matter if you read the question such that the speed of the treadmill tracks the speed of the plane relitive to the treadmill surface you get this equation. This is the case most people think about when talking about treadmills

VPlane to treadmill = Vground + -Vtreadmill
Now by definition of the problem Vtreadmill = -VplanetoTreadmill
We get Vtreadmill = -Vground + Vtreadmill
Therefor Vground = 0 and the plane doesn't take off.

I know I said that my previous post was going to be my last. But LMAO ... are you realling trying to argue that size and direction of the forces don't matter. When they are the only thing that matters.

Newton's Second Law of motion- The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

Just show how my equations are wroung. It is simple algebra how or why it happens doesn't matter because the question defines it as happening.

Ok ..
p_t = plane with respect to surface of treadmill, t = tread mill, p_g = plane wrt ground

Vp_t = Vp_g + Vt ...... I think in your equation you had a sign error here.
(Ex. to prove validity of above equation only - if you are running on a treadmill set at 5 MPh then your Vp_t is 5 and vp_g is zero and you are stationary. If you are running 1 mph faster than the treadmill then Vp_t=1+5 =6. You are moving faster than the treadmill by 1 mph and you are going to run off the end of it so Vp_g = 1)

define Vt = Vp_g through the problem statement "when the airplane moves forward, the conveyor matches its speed" NOT Vt = -Vp_t as you stated ........it does not say when the airplane moves forward with respect to treadmill, it just says moves forward.

Vp_t = Vp_g +Vp_g substituting given into original equation
Vp_t = 2*Vp_g So if the plane is moving at 5mph WRT to ground then the Vp_t is 10mph

Nowhere does this prove that Vp_g is 0

Now, I challenge you to try to find fault in the logic of summing the forces acting on a real world plane on a real world treamill. I bet that you will not do so, becuase you can not.

Originally posted by: BrownTown
I think I should start a parody thread where I try to argue that 2+2=5 and ignore all evidence to the contrary. I mean seriously, if you are arguing that the plane doesn't take off then you are either stupid, stubborn, or both. Just because the fact that you are wrong is not as immediately obvious as the fact that "2+2=5" is wrong does not mean that you are any less wrong.

You should start the old post that 2=1
a=b Define
a^2 = b*a Multiplication Property of Equality
a^2-b^2 = b*a - b^2 Subtraction Property of Equality
(a-b)*(a+b)=b*(a-b) Distributive Property
a+b = b Division Property of Equality
b+b = b Substitution Property
2b=b Combine like terms
2=1 Division Property of Equality

That should confuse some of the fools.

 

[sao123]

Originally posted by: smack Down

Originally posted by: AeroEngy

Originally posted by: smack Down
The size of the force doesn't matter or where it comes from doesn't matter if you read the question such that the speed of the treadmill tracks the speed of the plane relitive to the treadmill surface you get this equation. This is the case most people think about when talking about treadmills

VPlane to treadmill = Vground + -Vtreadmill
Now by definition of the problem Vtreadmill = -VplanetoTreadmill
We get Vtreadmill = -Vground + Vtreadmill
Therefor Vground = 0 and the plane doesn't take off.

I know I said that my previous post was going to be my last. But LMAO ... are you realling trying to argue that size and direction of the forces don't matter. When they are the only thing that matters.

Newton's Second Law of motion- The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

Just show how my equations are wroung. It is simple algebra how or why it happens doesn't matter because the question defines it as happening.

Did you read my last post with diagram?

The plane will take off.

Jet engines provide thrust relative to the atmosphere... not relative to the ground.

The question does not and can not define things happening as you have described them. Its impossible.
Your equations are wrong because they dont have a common frame of reference:
You must have the common frame of reference the "ATMOSPHERE."

Velocity of the plane to the ground = (VplanetoATM - VGroundtoATM)
= (+300 + 0) = 300
Velocity of the Treadmill to the ground = (VGroundtoATM - VTreadtoATM)
= (0 - 300) = -300
Velocity of the plane to the Treadmills = (VPlanetoATM - VTreadtoATM)
= (300 - -300) = 600

There is no physical way for the engines to be on and the plane not be moving forward because the engines push against the atmosphere, not the ground. Yuor equations dont even have the atmosphere in them as a reference point.