That plane, and it taking off.

[FeuerFrei]

My vote was a conditional "No."
Theoretically possible. Hard to imagine IRL.

 

[NanoStuff]

Originally posted by: jagec
The only part of that phrase that's unclear is the "belt compensates for the rotation...", but that is clarified by the second half of the sentence, which eliminates the angular velocity interpretation.

It does indeed, great for you to point that out. Thus the plane takes off. It would NOT have taken off in the rotational interpretation, which demonstrated the influence of the belt on the plane.

 

[JujuFish]

Originally posted by: NanoStuff

Originally posted by: jagec
The only part of that phrase that's unclear is the "belt compensates for the rotation...", but that is clarified by the second half of the sentence, which eliminates the angular velocity interpretation.

It does indeed, great for you to point that out. Thus the plane takes off. It would NOT have taken off in the rotational interpretation, which demonstrated the influence of the belt on the plane.

In the rotational interpretation, you would have an impossibility, so you cannot say wether it would or wouldn't, since it obviously isn't based on the laws of physics.

 

[NanoStuff]

Originally posted by: JujuFish

Originally posted by: NanoStuff

Originally posted by: jagec
The only part of that phrase that's unclear is the "belt compensates for the rotation...", but that is clarified by the second half of the sentence, which eliminates the angular velocity interpretation.

It does indeed, great for you to point that out. Thus the plane takes off. It would NOT have taken off in the rotational interpretation, which demonstrated the influence of the belt on the plane.

In the rotational interpretation, you would have an impossibility, so you cannot say wether it would or wouldn't, since it obviously isn't based on the laws of physics.

Again, the cogs. It's very possible and it's very straight forward. Like you said yourself if I remember right, on a regular runway the belt moves backwards in relationship to the wheels at the same speed the wheels are moving forwards in relationship to the runway, yet I'm sure you'd agree that's very possible.

 

[JujuFish]

Originally posted by: NanoStuff

Originally posted by: JujuFish

Originally posted by: NanoStuff

Originally posted by: jagec
The only part of that phrase that's unclear is the "belt compensates for the rotation...", but that is clarified by the second half of the sentence, which eliminates the angular velocity interpretation.

It does indeed, great for you to point that out. Thus the plane takes off. It would NOT have taken off in the rotational interpretation, which demonstrated the influence of the belt on the plane.

In the rotational interpretation, you would have an impossibility, so you cannot say wether it would or wouldn't, since it obviously isn't based on the laws of physics.

Again, the cogs. It's very possible and it's very straight forward. Like you said yourself if I remember right, on a regular runway the belt moves backwards in relationship to the wheels at the same speed the wheels are moving forwards in relationship to the runway, yet I'm sure you'd agree that's very possible.

Precisely. The cogs. Two different sized, connected cogs will never move with the same rotational speed.

 

[NanoStuff]

Originally posted by: JujuFish

Originally posted by: NanoStuff

Originally posted by: JujuFish

Originally posted by: NanoStuff

Originally posted by: jagec
The only part of that phrase that's unclear is the "belt compensates for the rotation...", but that is clarified by the second half of the sentence, which eliminates the angular velocity interpretation.

It does indeed, great for you to point that out. Thus the plane takes off. It would NOT have taken off in the rotational interpretation, which demonstrated the influence of the belt on the plane.

In the rotational interpretation, you would have an impossibility, so you cannot say wether it would or wouldn't, since it obviously isn't based on the laws of physics.

Again, the cogs. It's very possible and it's very straight forward. Like you said yourself if I remember right, on a regular runway the belt moves backwards in relationship to the wheels at the same speed the wheels are moving forwards in relationship to the runway, yet I'm sure you'd agree that's very possible.

Precisely. The cogs. Two different sized, connected cogs will never move with the same rotational speed.

If that doesn't satisfy you, use cogs of same sizes. And before you argue the belt is much bigger, replace the belt with wheels that are the same size as the plane's. Same end result.

 

[SparkyJJO]

Yes it will take off

propulsion is not from the wheels in contact with the ground but from the prop or jet, so no matter how fast backwards the conveyor runs it'll still move forward. Now you might burn up a wheel bearing if it gets too extreme but...

 

[JujuFish]

Originally posted by: NanoStuff
If that doesn't satisfy you, use cogs of same sizes. And before you argue the belt is much bigger, replace the belt with wheels that are the same size as the plane's. Same end result.

*sigh*

You just don't get it. The world you're creating is physically impossible because the conveyor cannot hold back the plane unless the plane was limited to a finite speed and the conveyor was not.

 

[NanoStuff]

Originally posted by: JujuFish

Originally posted by: NanoStuff
If that doesn't satisfy you, use cogs of same sizes. And before you argue the belt is much bigger, replace the belt with wheels that are the same size as the plane's. Same end result.

*sigh*

You just don't get it. The world you're creating is physically impossible because the conveyor cannot hold back the plane unless the plane was limited to a finite speed and the conveyor was not.

I could tell you the same thing and provide no reason why. We both believe we're equally right, thus our argument; much like the plane, remains stationary.

 

[JujuFish]

Originally posted by: NanoStuff

Originally posted by: JujuFish

Originally posted by: NanoStuff
If that doesn't satisfy you, use cogs of same sizes. And before you argue the belt is much bigger, replace the belt with wheels that are the same size as the plane's. Same end result.

*sigh*

You just don't get it. The world you're creating is physically impossible because the conveyor cannot hold back the plane unless the plane was limited to a finite speed and the conveyor was not.

I could tell you the same thing and provide no reason why. We both believe we're equally right, thus our argument; much like the plane, remains stationary.

The difference is I have physics on my side.

 

[mrSHEiK124]

Originally posted by: NanoStuff

Originally posted by: blahblah99

Originally posted by: 91TTZ

Originally posted by: deathkoba


The plane will take off only if it's engine can generate enough thrust to push the plane even harder than it already is doing, (which is what's keeping it visually stationary in the first place) to the point where there is enough airspeed for take off. So the real answer is YES or NO depending on the performance of the engine and how fast that conveyor belt is moving and whether or not the conveyor belt is adjusting it's speed in realtime as the plane's engine makes adjustments. That plus other environmental factors that affect flight performance directly or indirectly. It's common sense folks.

For the most part, it doesn't matter what the belt is doing, since the plane is just rolling.

My previous example was taken to the extreme, and isn't realistic.

If a jet's takeoff speed is 200 mph, and the belt is moving backwards at 200 mph, the plane will still take off just as easily. The only difference will be that the wheels will be turning at 400 mph.

Bingo! Why is that so hard to comprehend for some people?

Because some people are aware of the 200mph clocked on the belt while other choose to ignore it's impact on the plane. The plane would take off, but only if it exerts twice the thrust it otherwise would have.

Which a 747 is quite capable of...
THE DAMN THING TAKES OFF!

Even if you did have a conveyor belt that could keep up with the thrust of two fvcking JET ENGINES, the thrust from the engines themselves would be what makes the plane fly, not the movement of the wheels.

 

[NanoStuff]

Originally posted by: mrSHEiK124

Originally posted by: NanoStuff

Originally posted by: blahblah99

Originally posted by: 91TTZ

Originally posted by: deathkoba


The plane will take off only if it's engine can generate enough thrust to push the plane even harder than it already is doing, (which is what's keeping it visually stationary in the first place) to the point where there is enough airspeed for take off. So the real answer is YES or NO depending on the performance of the engine and how fast that conveyor belt is moving and whether or not the conveyor belt is adjusting it's speed in realtime as the plane's engine makes adjustments. That plus other environmental factors that affect flight performance directly or indirectly. It's common sense folks.

For the most part, it doesn't matter what the belt is doing, since the plane is just rolling.

My previous example was taken to the extreme, and isn't realistic.

If a jet's takeoff speed is 200 mph, and the belt is moving backwards at 200 mph, the plane will still take off just as easily. The only difference will be that the wheels will be turning at 400 mph.

Bingo! Why is that so hard to comprehend for some people?

Because some people are aware of the 200mph clocked on the belt while other choose to ignore it's impact on the plane. The plane would take off, but only if it exerts twice the thrust it otherwise would have.

Which a 747 is quite capable of...
THE DAMN THING TAKES OFF!

Even if you did have a conveyor belt that could keep up with the thrust of two fvcking JET ENGINES, the thrust from the engines themselves would be what makes the plane fly, not the movement of the wheels.

You just opposed your own argument. If the belt could keep up with the thrust from the jet engines, the plane would not fly because the belt would be 'keeping up'

 

[JujuFish]

Originally posted by: NanoStuff
You just opposed your own argument. If the belt could keep up with the thrust from the jet engines, the plane would not fly because the belt would be 'keeping up'

No, because a plane doesn't drive forward like a car.

 

[ultimatebob]

So... Who's working on building one of these giant 200 MPH treadmills? It sounds like they could revolutionize the whole airport industry... letting planes take off without hardly moving and all

 

[waggy]

Originally posted by: ultimatebob
So... Who's working on building one of these giant 200 MPH treadmills? It sounds like they could revolutionize the whole airport industry... letting planes take off without hardly moving and all

who needs planes? 200mph belt you can keep stuff moving at a constant rate!

 

[smack Down]

Originally posted by: jagec

Originally posted by: Kenazo

Whatever the case, using the second stipulation, there is for sure no way that this could happen. If the belt is just moving the same speed as the wheels, but in the opposite direction the plane could not be moving forward, unless the plane was skidding across the conveyer.

Angular velocity. Linear velocity. They're not the same thing. "Forward" is linear velocity" "Turning", or "around", is angular velocity.

Originally posted by: NanoStuff
The plane increases it's thrust and the wheels begin to rotate. The belt compensates for the rotation of the wheels in reverse, as in the belt moves in reverse exactly as fast as the wheels move forward.

Y HLO THAR, it's our friend Mr. LINEAR velocity!!!

The only part of that phrase that's unclear is the "belt compensates for the rotation...", but that is clarified by the second half of the sentence, which eliminates the angular velocity interpretation.

Unless you consider the forward to be forward on the belt then the angular velocity is the same as the linear velocity and the plane doesn't move.

 

[skyking]

Originally posted by: BrokenVisage
I love how everyone on here saying yes is suddenly an avionics expert in addition to being a computer geek.

Too bad we can't simulate this in Flight Simulator to prove you guys wrong.

I don't think anybody claimed to be an expert about aircraft electronics and radio navigation equipment

 

[smack Down]

Originally posted by: blahblah99

Originally posted by: 91TTZ

Originally posted by: deathkoba


The plane will take off only if it's engine can generate enough thrust to push the plane even harder than it already is doing, (which is what's keeping it visually stationary in the first place) to the point where there is enough airspeed for take off. So the real answer is YES or NO depending on the performance of the engine and how fast that conveyor belt is moving and whether or not the conveyor belt is adjusting it's speed in realtime as the plane's engine makes adjustments. That plus other environmental factors that affect flight performance directly or indirectly. It's common sense folks.

For the most part, it doesn't matter what the belt is doing, since the plane is just rolling.

My previous example was taken to the extreme, and isn't realistic.

If a jet's takeoff speed is 200 mph, and the belt is moving backwards at 200 mph, the plane will still take off just as easily. The only difference will be that the wheels will be turning at 400 mph.

Bingo! Why is that so hard to comprehend for some people?

Because the question states that the treadmil is going in the same speed but opisite direction as the wheels so the treadmill can't turn at 200 mph and 400 mph at the same time.

 

[thirdlegstump]

Originally posted by: NanoStuff

Originally posted by: blahblah99

Originally posted by: 91TTZ

Originally posted by: deathkoba


The plane will take off only if it's engine can generate enough thrust to push the plane even harder than it already is doing, (which is what's keeping it visually stationary in the first place) to the point where there is enough airspeed for take off. So the real answer is YES or NO depending on the performance of the engine and how fast that conveyor belt is moving and whether or not the conveyor belt is adjusting it's speed in realtime as the plane's engine makes adjustments. That plus other environmental factors that affect flight performance directly or indirectly. It's common sense folks.

For the most part, it doesn't matter what the belt is doing, since the plane is just rolling.

My previous example was taken to the extreme, and isn't realistic.

If a jet's takeoff speed is 200 mph, and the belt is moving backwards at 200 mph, the plane will still take off just as easily. The only difference will be that the wheels will be turning at 400 mph.

Bingo! Why is that so hard to comprehend for some people?

Because some people are aware of the 200mph clocked on the belt while other choose to ignore it's impact on the plane. The plane would take off, but only if it exerts twice the thrust it otherwise would have.

Man I'm full. Burp.

Anyway Nano, you've got the right idea although your calculation would probably be off by a bit as thrust isn't always linear but you're still on the right track.

The plane cannot 'roll' without using some level of engine power. That's the equation almost EVERYONE in this thread is missing. How the hell can the plane even begin to roll on it's own? The wheels would not rotate unless 1 of 2 things happen: 1. Someone is pushing the plane from the back. 2. Engines do some work.

Let's put it in real world perspective:

You're sitting on the tarmac with engines at idle. The ground starts to move towards you. Your wheels will first not rotate but your aircraft's position in terms of latitude and longitude would change....backwards.....creating TAILWIND........UNLESS you apply some power. TAILWIND is basically negative airspeed which has a direct negative impact on an aircraft's flight performance. In order for the plane to take off, it would have to generate enough thrust to counter that tailwind (backwards longitudinal movement) PLUS meet the airspeed requirement for normal take off.

Here are some artificial numbers assuming aircraft takes off at 0 deg. heading or North for simplicity's sake:

Speed of conveyor belt: 20 knots moving South.

Nominal take off airspeed of hypothetical aircraft: 120 knots.

Power required for take off: 100%

Power required to push/taxi aircraft ON THE GROUND at 20 knots: 20% (this would only negate the 20 knots induced by conveyor belt giving the aircraft an airspeed of zero knots)

Total power required to reach an airspeed of 120 knots: 120%

Total required groundspeed relative to conveyor belt to reach an airspeed of 120 knots: 140 knots

Wow so difficult to understand eh? Just stop it already.

 

[JujuFish]

Originally posted by: deathkoba
You're sitting on the tarmac with engines at idle. The ground starts to move towards you. Your wheels will first not rotate but your aircraft's position in terms of latitude and longitude would change....backwards.....creating TAILWIND........UNLESS you apply some power.

This may or may not be true. It depends on how quickly the ground starts to move. However, assuming a slow enough acceleration, I'm in agreement with you so far.

Power required to push/taxi aircraft ON THE GROUND at 20 knots: 20% (this would only negate the 20 knots induced by conveyor belt giving the aircraft an airspeed of zero knots)

This, however, is where you're very wrong. This would be the case if the plane were a car and you were trying to reach the appriopriate speed. However, a plane is not a car and does not require to push off the ground to move.

 

[NanoStuff]

Originally posted by: deathkoba
Man I'm full. Burp.

Anyway Nano, you've got the right idea although your calculation would probably be off by a bit as thrust isn't always linear but you're still on the right track.

I damn well know the calculations would be off, but they'd be a whole lot closer to the right number than the inexistant calculations for the belt of so many other people

Originally posted by: deathkoba
This, however, is where you're very wrong. This would be the case if the plane were a car and you were trying to reach the appriopriate speed. However, a plane is not a car and does not require to push off the ground to move.

No he's not, he explained it perfectly and his simple while effective calculation makes perfect sense. 20 + (-20) = 0. Math is not necessary to understand how a belt works, but it sure explains it nicely.

 

[cKGunslinger]

Originally posted by: NanoStuff

Originally posted by: JujuFish
This, however, is where you're very wrong. This would be the case if the plane were a car and you were trying to reach the appriopriate speed. However, a plane is not a car and does not require to push off the ground to move.

No he's not, he explained it perfectly and his simple while effective calculation makes perfect sense. 20 + (-20) = 0. Math is not necessary to understand how a belt works, but it sure explains it nicely.

No, no, no!

Like was mentioned in the last thread: Take a Hot Wheels car on your home treadmill. If you turn teh belt on and run it at 20mph, do you honestly think you have to push the Hot Wheels car at 20mph in the opposite direction to keep the car in one place? That's rediculous.

Now, suppose the treadmill is going 200mph. Do you think you'd have to push the car at 201mph to make the car inch forward in relation to the earth? Of course not, the only thing you need to do is gently push your hand and finger forward at a ~1mph rate in order for the car to also travel at 1mph. Almost the exact same amount of pressure you'd use if the treadmill was off and completely stationary.

Spinning wheels are not related to foward momemtum. Your hand (analogous to the plane's jets) is not affected (by any noticeable amount) by the treadmill's rotation, and neither is the momentum and speed of the car (plane.)

It's not rocket science, people You can make up bullsh1t equations and conditions, but the fact remains is that the external force of the engine is NOT noticeably affected by the rotation of the treadmill. And certainly not in a 1-to-1 amount. :roll:

 

[smack Down]

Originally posted by: cKGunslinger

Originally posted by: NanoStuff

Originally posted by: JujuFish
This, however, is where you're very wrong. This would be the case if the plane were a car and you were trying to reach the appriopriate speed. However, a plane is not a car and does not require to push off the ground to move.

No he's not, he explained it perfectly and his simple while effective calculation makes perfect sense. 20 + (-20) = 0. Math is not necessary to understand how a belt works, but it sure explains it nicely.

No, no, no!

Like was mentioned in the last thread: Take a Hot Wheels car on your home treadmill. If you turn teh belt on and run it at 20mph, do you honestly think you have to push the Hot Wheels car at 20mph in the opposite direction to keep the car in one place? That's rediculous.

Now, suppose the treadmill is going 200mph. Do you think you'd have to push the car at 201mph to make the car inch forward in relation to the earth? Of course not, the only thing you need to do is gently push your hand and finger forward at a ~1mph rate in order for the car to also travel at 1mph. Almost the exact same amount of pressure you'd use if the treadmill was off and completely stationary.

Spinning wheels are not related to foward momemtum. Your hand (analogous to the plane's jets) is not affected (by any noticeable amount) by the treadmill's rotation, and neither is the momentum and speed of the car (plane.)

It's not rocket science, people You can make up bullsh1t equations and conditions, but the fact remains is that the external force of the engine is NOT noticeably affected by the rotation of the treadmill. And certainly not in a 1-to-1 amount. :roll:

The treadmil isn't at a fixed speed. It increases speed as the speed of the wheels increases, so no forward movement is pausable with out the wheels going faster then the treadmil.

 

[NanoStuff]

Originally posted by: cKGunslinger

Originally posted by: NanoStuff

Originally posted by: JujuFish
This, however, is where you're very wrong. This would be the case if the plane were a car and you were trying to reach the appriopriate speed. However, a plane is not a car and does not require to push off the ground to move.

No he's not, he explained it perfectly and his simple while effective calculation makes perfect sense. 20 + (-20) = 0. Math is not necessary to understand how a belt works, but it sure explains it nicely.

No, no, no!

Like was mentioned in the last thread: Take a Hot Wheels car on your home treadmill. If you turn teh belt on and run it at 20mph, do you honestly think you have to push the Hot Wheels car at 20mph in the opposite direction to keep the car in one place? That's rediculous.

It's not ridiculous, simple conservation of energy. You have to apply a force on the car that would be equivalent to pushing it at 20mph to keep it still.

I can imagine why this is so difficult to understand, but it just takes a little thought.

 

[Rubycon]

That plane will never take off!



























The tower will NEVER give clearance to take off from a conveyor. :laugh: