That plane, and it taking off.

[KillerCharlie]

First of all, I will begin this by saying I'm an aerospace engineer. More specifically, I'm an aerodynamicist for Boeing. I analyze and help design the aerodynamic configuration for commercial planes. Take my word for what it is, but it's obvious that lots of people here are spewing crap out of their butt and don't have a background in science and engineering. I work in designing the shape of the aircraft and am not a performance engineer. I know some people who work in performance, but I don't personally know anyone who works with takeoff performance. There are a couple points I must make.

First of all, the question is poorly worded. Whoever wrote it is an idiot and needs to learn how to write questions that a well-posed. There are 2 interpretations I have seen:

1) The treadmill keeps the plane from moving in the reference frame of an observing sitting on the ground (not standing on the treadmill)

2) The treadmill just goes the opposite speed of the airplane

For condition 1, the answer is obvious - no lift because of no airflow over the wing. For condition 2, if the airplane goes forward, then you can takeoff. The problem here is determining if the aircraft can go forward.

Here is how aircraft are typically modeled for takeoff:
Forces acting on aircraft:
Lift upwards (proportional to V^2)
Weight downward, thrust forward, drag backwards, and resistance from the wheels pulling you backward as well.

The resistance force from the wheels, as a first approximation, is proportional only to the normal force on the aircraft (weight-lift) and nothing more. Therefore, the treadmill does not do anything.


Of course, a higher-fidelity model for wheel resistance will include things like tire rotational speed, tire air pressure, and so on. However, the resulting changes are small. I'd love to ask a performance engineer how they calculate these things, but I have the feeling they just do it with a flight test and don't worry about the physics of it.


Oh, and AlienCraft, pitot tubes do not measure groundspeed. It will not read 30mph - your airspeed indicator will read 130 mph. Your relative airspeed it what matters, not ground speed. Some advanced aircraft can get their ground speed using things such as GPS. Also, determining airspeed from a pitot tube does NOT require any differential equations.

 

[KillerCharlie]

Originally posted by: AlienCraft

Originally posted by: PurdueRy

Originally posted by: WHAMPOM
Forward motion is counteracted, what other forces are there? A stationary plane cannot take-off.

counteracted by what?

The FBD will show you the force of the treadmill does not oppose the body of the plane/engines...it only spins the wheels. Therefore there is no opposing force.

Yes, Gravity. The wheels are spinning, but without forward motion, there is no air movement across the wings.
Remember, the engine ( especially a rear mounted jet as in the original example) is not providing lift.


What is this acronym you keep using "FBD"?

There IS forward motion - the engines are providing thrust. The treadmill doesn't do anything to pull the plane back.

FBD = free body diagram - a picture showing the forces on an object. Anyone who has taken a physics course should know this.

 

[NanoStuff]

Oh look, it's another 'aerospace engineer'... damn, you guys are a dime a dozen in conveyor belt threads.

 

[KillerCharlie]

Originally posted by: NanoStuff
Oh look, it's another 'aerospace engineer'... damn, you guys are a dime a dozen in conveyor belt threads.

Who else in this thread claims to be one? I would certainly say someone who designs airplanes would be a good source of information. What do you do for a living?

 

[NanoStuff]

At least one other person somewhere in this thread, but they pop up all over the internet really. Ok, so two dimes a dozen

But seriously, this is fundamentally a question of wheels and belts, not aerodynamics. We don't need to know what the plane's takeoff speed is or whether it has delta-wings, or whether the pilot has a blood alcohol level unsuitable for flight.

What do you do for a living?

I eat cheetos in my underpants, how about you?

... oh wait I just remembered, you're an 'aerospace engineer'

 

[KillerCharlie]

Every resource I have puts the wheel resistance directly proportional to the normal force. Even though it's not 100% true, nothing in engineering ever is. Though the force from the wheel friction is subject to other factors, they are mostly negligible.

"Aircraft Performance and Design," John D. Anerson
Navy Flight Testing Link

 

[WHAMPOM]

In this hypothetical situation, I am now standing(at a safe distance from the plane and conveyer)watching the hypothetical plane, it's engine running at full throttle. The hypothetical conveyer moving north at the exact same speed the plane's wheels are spinning south. The plane is stationary to my position. Gravity is King, it will stay there til the fuel runs out. I forgot inertia, but can not see how it applies except as a stabilizing force.

 

[PurdueRy]

Originally posted by: KillerCharlie
First of all, I will begin this by saying I'm an aerospace engineer. More specifically, I'm an aerodynamicist for Boeing. I analyze and help design the aerodynamic configuration for commercial planes. Take my word for what it is, but it's obvious that lots of people here are spewing crap out of their butt and don't have a background in science and engineering. I work in designing the shape of the aircraft and am not a performance engineer. I know some people who work in performance, but I don't personally know anyone who works with takeoff performance. There are a couple points I must make.

First of all, the question is poorly worded. Whoever wrote it is an idiot and needs to learn how to write questions that a well-posed. There are 2 interpretations I have seen:

1) The treadmill keeps the plane from moving in the reference frame of an observing sitting on the ground (not standing on the treadmill)

2) The treadmill just goes the opposite speed of the airplane

For condition 1, the answer is obvious - no lift because of no airflow over the wing. For condition 2, if the airplane goes forward, then you can takeoff. The problem here is determining if the aircraft can go forward.

Here is how aircraft are typically modeled for takeoff:
Forces acting on aircraft:
Lift upwards (proportional to V^2)
Weight downward, thrust forward, drag backwards, and resistance from the wheels pulling you backward as well.

The resistance force from the wheels, as a first approximation, is proportional only to the normal force on the aircraft (weight-lift) and nothing more. Therefore, the treadmill does not do anything.


Of course, a higher-fidelity model for wheel resistance will include things like tire rotational speed, tire air pressure, and so on. However, the resulting changes are small. I'd love to ask a performance engineer how they calculate these things, but I have the feeling they just do it with a flight test and don't worry about the physics of it.


Oh, and AlienCraft, pitot tubes do not measure groundspeed. It will not read 30mph - your airspeed indicator will read 130 mph. Your relative airspeed it what matters, not ground speed. Some advanced aircraft can get their ground speed using things such as GPS. Also, determining airspeed from a pitot tube does NOT require any differential equations.

I see you said the conveyor does practically nothing...but then how did you get interpretation #1?

 

[PurdueRy]

Originally posted by: WHAMPOM
In this hypothetical situation, I am now standing(at a safe distance from the plane and conveyer)watching the hypothetical plane, it's engine running at full throttle. The hypothetical conveyer moving north at the exact same speed the plane's wheels are spinning south. The plane is stationary to my position. Gravity is King, it will stay there til the fuel runs out. I forgot inertia, but can not see how it applies except as a stabilizing force.

Draw the FBD people! PLEASE!

That is all I ask...

 

[AlienCraft]

Originally posted by: KillerCharlie
First of all, I will begin this by saying I'm an aerospace engineer. More specifically, I'm an aerodynamicist for Boeing. I analyze and help design the aerodynamic configuration for commercial planes. Take my word for what it is, but it's obvious that lots of people here are spewing crap out of their butt and don't have a background in science and engineering. I work in designing the shape of the aircraft and am not a performance engineer. I know some people who work in performance, but I don't personally know anyone who works with takeoff performance. There are a couple points I must make.

First of all, the question is poorly worded. Whoever wrote it is an idiot and needs to learn how to write questions that a well-posed.

I couldn't agree with this more.

There are 2 interpretations I have seen:

1) The treadmill keeps the plane from moving in the reference frame of an observing sitting on the ground (not standing on the treadmill)

2) The treadmill just goes the opposite speed of the airplane

For condition 1, the answer is obvious - no lift because of no airflow over the wing. For condition 2, if the airplane goes forward, then you can takeoff. The problem here is determining if the aircraft can go forward.

That condition 1 is the way I interpreted the question.

Here is how aircraft are typically modeled for takeoff:
Forces acting on aircraft:
Lift upwards (proportional to V^2)
Weight downward, thrust forward, drag backwards, and resistance from the wheels pulling you backward as well.

The resistance force from the wheels, as a first approximation, is proportional only to the normal force on the aircraft (weight-lift) and nothing more. Therefore, the treadmill does not do anything.


Of course, a higher-fidelity model for wheel resistance will include things like tire rotational speed, tire air pressure, and so on. However, the resulting changes are small. I'd love to ask a performance engineer how they calculate these things, but I have the feeling they just do it with a flight test and don't worry about the physics of it.


Oh, and AlienCraft, pitot tubes do not measure groundspeed. It will not read 30mph - your airspeed indicator will read 130 mph. Your relative airspeed it what matters, not ground speed. Some advanced aircraft can get their ground speed using things such as GPS. Also, determining airspeed from a pitot tube does NOT require any differential equations.

Well, I knew my ass was doing some talking there as it's been some 30 years since I took my flight lessons, so, I concede the point to you. And I thank you for correcting in a gentlemanly fashion.
My ass heard all the others flappin and decided to butt in...... BWAHAHAHAHAHAHAA sorry couldn't resist that pun.:laugh:

 

[WHAMPOM]

Originally posted by: PurdueRy

Originally posted by: WHAMPOM
In this hypothetical situation, I am now standing(at a safe distance from the plane and conveyer)watching the hypothetical plane, it's engine running at full throttle. The hypothetical conveyer moving north at the exact same speed the plane's wheels are spinning south. The plane is stationary to my position. Gravity is King, it will stay there til the fuel runs out. I forgot inertia, but can not see how it applies except as a stabilizing force.

Draw the FBD people! PLEASE!

That is all I ask...

All forces are cancelled out , the sum is zero.

 

[PurdueRy]

Originally posted by: WHAMPOM

Originally posted by: PurdueRy

Originally posted by: WHAMPOM
In this hypothetical situation, I am now standing(at a safe distance from the plane and conveyer)watching the hypothetical plane, it's engine running at full throttle. The hypothetical conveyer moving north at the exact same speed the plane's wheels are spinning south. The plane is stationary to my position. Gravity is King, it will stay there til the fuel runs out. I forgot inertia, but can not see how it applies except as a stabilizing force.

Draw the FBD people! PLEASE!

That is all I ask...

All forces are cancelled out , the sum is zero.

Nope. There are two systems here. One is the plane one is the wheels. The force of the treadmill acts only on the wheels. Therefore all it can do is cause the wheels to spin and has no affect on the plane body itself. The engines propel the plane body forward with an unopposed force causing acceleration. This is the main concept people have to understand to see why it takes off

 

[PurdueRy]

Originally posted by: AlienCraft

Originally posted by: KillerCharlie
First of all, I will begin this by saying I'm an aerospace engineer. More specifically, I'm an aerodynamicist for Boeing. I analyze and help design the aerodynamic configuration for commercial planes. Take my word for what it is, but it's obvious that lots of people here are spewing crap out of their butt and don't have a background in science and engineering. I work in designing the shape of the aircraft and am not a performance engineer. I know some people who work in performance, but I don't personally know anyone who works with takeoff performance. There are a couple points I must make.

First of all, the question is poorly worded. Whoever wrote it is an idiot and needs to learn how to write questions that a well-posed.

I couldn't agree with this more.

There are 2 interpretations I have seen:

1) The treadmill keeps the plane from moving in the reference frame of an observing sitting on the ground (not standing on the treadmill)

2) The treadmill just goes the opposite speed of the airplane

For condition 1, the answer is obvious - no lift because of no airflow over the wing. For condition 2, if the airplane goes forward, then you can takeoff. The problem here is determining if the aircraft can go forward.

That condition 1 is the way I interpreted the question.

Here is how aircraft are typically modeled for takeoff:
Forces acting on aircraft:
Lift upwards (proportional to V^2)
Weight downward, thrust forward, drag backwards, and resistance from the wheels pulling you backward as well.

The resistance force from the wheels, as a first approximation, is proportional only to the normal force on the aircraft (weight-lift) and nothing more. Therefore, the treadmill does not do anything.


Of course, a higher-fidelity model for wheel resistance will include things like tire rotational speed, tire air pressure, and so on. However, the resulting changes are small. I'd love to ask a performance engineer how they calculate these things, but I have the feeling they just do it with a flight test and don't worry about the physics of it.


Oh, and AlienCraft, pitot tubes do not measure groundspeed. It will not read 30mph - your airspeed indicator will read 130 mph. Your relative airspeed it what matters, not ground speed. Some advanced aircraft can get their ground speed using things such as GPS. Also, determining airspeed from a pitot tube does NOT require any differential equations.

Well, I knew my ass was doing some talking there as it's been some 30 years since I took my flight lessons, so, I concede the point to you. And I thank you for correcting in a gentlemanly fashion.
My ass heard all the others flappin and decided to butt in...... BWAHAHAHAHAHAHAA sorry couldn't resist that pun.:laugh:

except...interpretation 1 can't happen without some SERIOUS speed from the treadmill...and the plane would have already taken off by then.... Notice he said the "treadmill does nothing"

 

[Suspicious-Teach8788]

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?


That said.... The problem with this question is that some people interpret it as.... a plane turning on its engines and a conveyor belt starting up simultaneously. In that case, the plane SHOULD and WILL take off. I think that cant be disputed.

Then there are people who assume the situation is already in play. The plane's velocity without a conveyor belt with respect to the ground is v. And then it is placed on a conveyor belt that moves at -v. Then the net velocity is 0?

This was "Condition 1" as stated by someone. However, whichever way you interpret it, there is still 0 force upon the entire aircraft due to the conveyor belt. THus even if you placed a plane that is already pushing with velocity v on a conveyor belt, the belt still does nothing.

Forget the relative velocity argument.

That argument only stands if we assume friction in the bearings of the gears.

 

[PurdueRy]

Originally posted by: DLeRium
Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?

Damnit why can't we discuss in one damn thread only? Think of how long of a thread we could make if we condensed it all into 1?


That said.... The problem with this question is that some people interpret it as.... a plane turning on its engines and a conveyor belt starting up simultaneously. In that case, the plane SHOULD and WILL take off. I think that cant be disputed.

Then there are people who assume the situation is already in play. The plane's velocity without a conveyor belt with respect to the ground is v. And then it is placed on a conveyor belt that moves at -v. Then the net velocity is 0?

This was "Condition 1" as stated by someone. However, whichever way you interpret it, there is still 0 force upon the entire aircraft due to the conveyor belt. THus even if you placed a plane that is already pushing with velocity v on a conveyor belt, the belt still does nothing.

Forget the relative velocity argument.

That argument only stands if we assume friction in the bearings of the gears.

The net velocity would still be v. You can't add velocities that do not act on the same object. The v is acting on the body of the plane. The -v is acting on the wheels...spinning them..but can't pull the body of the plane with it.

Therefore the net velocity is still v and the plane takes off...which looks like you agree with

 

[Witchfire]

I just cannot believe how many people voted that it wouldn't take off... This just reaffirms my belief that some people should never, ever breed...

 

[WHAMPOM]

Opposite and equal reactions cancel each other out. The speed of the planes wheels turning is the speed of the plane and the opposite matching speed of the conveyer cancels each other out. You have a basic mistake in figuring out this problem, adding too many factors to a simple problem. No matter how fast the plane goes(wheel speed) the conveyer cancells it out.

 

[OvErHeAtInG]

Originally posted by: UlricT
WTF guys... the planes wheels are free-rolling. It does NOT drive the plane. The wheels will just spin twice as fast as the plane is moving forward (plane speed + conveyor belt speed)!

OK this is a great example of why seemingly intelligent people actually think the plane would take off. They misread the question and think the plane is actually moving relative to the air.

* If the plane is moving (fast enough) relative to the atmosphere, it will generate lift and take off.
* If the plane is not moving (fast enough) relative to atmosphere, it will not generate lift and take off.

See? Not so complicated. No differential equations, no general relativity, no damn Lorentz transformation. :roll:

The question, as I originally read it on ATOT months ago (maybe year+) stated that the speed of the conveyor would always match the speed of the plane relative to the conveyor. Meaning that the plane won't go anywhere relative to the atmosphere. However, if you wish to read the (very poorly worded) question as allowing for the increased velocity of the plane without a corresponding increase relative to the conveyor belt, you would believe that the plane could take off.

 

[PurdueRy]

Originally posted by: WHAMPOM
Opposite and equal reactions cancel each other out. The speed of the planes wheels turning is the speed of the plane and the opposite matching speed of the conveyer cancels each other out. You have a basic mistake in figuring out this problem, adding too many factors to a simple problem. No matter how fast the plane goes(wheel speed) the conveyer cancells it out.

LOL...you're right....physics is wrong...

What about if I pointed out to you that if the plane is moving at 4 MPH the treadmill is moving at 4 MPH but the wheels are moving at 8 MPH(net velocity)?

Ah don't worry about it....Free body diagrams are never right....physics is wrong...

 

[PurdueRy]

Originally posted by: OvErHeAtInG

Originally posted by: UlricT
WTF guys... the planes wheels are free-rolling. It does NOT drive the plane. The wheels will just spin twice as fast as the plane is moving forward (plane speed + conveyor belt speed)!

OK this is a great example of why seemingly intelligent people actually think the plane would take off. They misread the question and think the plane is actually moving relative to the air.

* If the plane is moving (fast enough) relative to the atmosphere, it will generate lift and take off.
* If the plane is not moving (fast enough) relative to atmosphere, it will not generate lift and take off.

See? Not so complicated. No differential equations, no general relativity, no damn Lorentz transformation. :roll:

The question, as I originally read it on ATOT months ago (maybe year+) stated that the speed of the conveyor would always match the speed of the plane relative to the conveyor. Meaning that the plane won't go anywhere relative to the atmosphere. However, if you wish to read the (very poorly worded) question as allowing for the increased velocity of the plane without a corresponding increase relative to the conveyor belt, you would believe that the plane could take off.

Put a toy car on a treadmill...push it forward and try to make it stop by increasing treadmill speed...see what happens...

I must be crazy for actually showing the physics behind this problem...

 

[Witchfire]

Look people... The force applied from the engines is in the form of thrust into the air. There is no driving force whatsoever supplied to the wheels/landing gear of any aircraft. You guys that say it won't take off and 'all forces are cancelled out' (LMFAO on that one) can spin your hampster wheels on those treadmills as fast as you want, and it makes absolutely no difference. The force those engines are applying are no different than if you replaced those engines with a mechanical winch. You could again spin your treadmill as fast as you wanted, and the winch would overcome it.

 

[smack Down]

Originally posted by: Witchfire
Look people... The force applied from the engines is in the form of thrust into the air. There is no driving force whatsoever supplied to the wheels/landing gear of any aircraft. You guys that say it won't take off and 'all forces are cancelled out' (LMFAO on that one) can spin your hampster wheels on those treadmills as fast as you want, and it makes absolutely no difference. The force those engines are applying are no different than if you replaced those engines with a mechanical winch. You could again spin your treadmill as fast as you wanted, and the winch would overcome it.

The question as stated has the treadmil speed increasing as the rotation of planes wheels increase such that there is no forward movement of the plane. So a plane can't take off unless it moves forward.

 

[PurdueRy]

Originally posted by: smack Down

Originally posted by: Witchfire
Look people... The force applied from the engines is in the form of thrust into the air. There is no driving force whatsoever supplied to the wheels/landing gear of any aircraft. You guys that say it won't take off and 'all forces are cancelled out' (LMFAO on that one) can spin your hampster wheels on those treadmills as fast as you want, and it makes absolutely no difference. The force those engines are applying are no different than if you replaced those engines with a mechanical winch. You could again spin your treadmill as fast as you wanted, and the winch would overcome it.

The question as stated has the treadmil speed increasing as the rotation of planes wheels increase such that there is no forward movement of the plane. So a plane can't take off unless it moves forward.

Show me where the original question says there is "No forward movement"

You make that assumption and it is incorrect.

The treadmill speed matches that of the aircraft...that does NOT mean the airplane is not moving forward.

 

[AlienCraft]

Originally posted by: PurdueRy

except...interpretation 1 can't happen without some SERIOUS speed from the treadmill...and the plane would have already taken off by then.... Notice he said the "treadmill does nothing"

Notice I kept refering to "perfect conveyor".
Of course the damns thing will take off if you're putting all aspects into a "reality" based POV. I thought it was meaning to be perfect conditions in all aspects.

Oh...



Never mind..... </ Rose Ann Roseanadana>

 

[OvErHeAtInG]

Originally posted by: PurdueRy

Originally posted by: OvErHeAtInG

Originally posted by: UlricT
WTF guys... the planes wheels are free-rolling. It does NOT drive the plane. The wheels will just spin twice as fast as the plane is moving forward (plane speed + conveyor belt speed)!

OK this is a great example of why seemingly intelligent people actually think the plane would take off. They misread the question and think the plane is actually moving relative to the air.

* If the plane is moving (fast enough) relative to the atmosphere, it will generate lift and take off.
* If the plane is not moving (fast enough) relative to atmosphere, it will not generate lift and take off.

See? Not so complicated. No differential equations, no general relativity, no damn Lorentz transformation. :roll:

The question, as I originally read it on ATOT months ago (maybe year+) stated that the speed of the conveyor would always match the speed of the plane relative to the conveyor. Meaning that the plane won't go anywhere relative to the atmosphere. However, if you wish to read the (very poorly worded) question as allowing for the increased velocity of the plane without a corresponding increase relative to the conveyor belt, you would believe that the plane could take off.

Put a toy car on a treadmill...push it forward and try to make it stop by increasing treadmill speed...see what happens...

I must be crazy for actually showing the physics behind this problem...

Oh fvck. Nevermind. I finally figured it out in my head and then realizeed you posted the answer above:

The force of the treadmill acts only on the wheels.

Why the fvck didn't I get that. I guess it didn't help that most people here are posting nonsense... sorry