Aircraft On A Conveyor Belt

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Member for

16 years 10 months

Posts: 3,214

Guys - im sure this has been done to death

A plane is standing on a runway that can move (some sort of conveyer belt). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction).

Can the plane take off?

Someone please tell me im right in saying the following

No - It wouldnt - IF the circumstances are explained as above, then the aircraft is stationary :nod:

The wheels may be rotating however fast but the aircraft is not moving through the air in any sense of the word - friction of the bearings has fcukall to do with it

If it is stationary, then there is NO airflow over the wings, if there is no air going over the wings, then no lift is produced and as a result the aircraft will NOT takeoff

The forumla for lift - L = (1/2).d.v^2.s.CL

where

  • L = Lift, which must equal the airplane's weight in pounds
  • d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
  • v = velocity of an aircraft expressed in feet per second
  • s = the wing area of an aircraft in square feet
  • CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

Density of air, coefficent of lift, wing area all mean nothing when "V" = 0 since multiplying any active part of a forumla by 0 = 0 then you can also say that L = 0

V will be = 0 as if the wheels are turning but the aircraft is not, no forward motion is made

If the aircraft was on a treadmill and you were blowing the required speed of air over the wings for it to lift - then it would lift with no forward motion, it would lift vertically.

An example of something similar is when aircraft takeoff from aircraft carriers, their takeoff speed is reduced by two factors - firstly the speed of the carrier through the air will reduce the required takeoff speed relative to the carrier. secondly any headwind coming accross the deck will further reduce the speed needed for the aircraft to lift

carrier travelling at = 20mph
headwing = 30 mph

takeoff speed = 150 mph - 30mph -20mph = 100mph

Original post

Member for

16 years 10 months

Posts: 3,214

I figure that if you strip away all the irrelevant parts to that question

the main question - Is there any airflow over the wing to create lift - no

therefore aircraft wont lift at all

Member for

16 years 4 months

Posts: 144

Interesting conversation Nash!

Member for

16 years 10 months

Posts: 3,214

not a problem lol

ON ANOTHER NOTE CAN SOMEONE PLEASE ANSWER MY QUESTION?!?!?!

Member for

16 years 10 months

Posts: 3,214

Saw that experiment - Imo its a rubbish example of this experiment, varying rates of acceleration between the conveyor belt and aircraft etc.

So far I came up with this (rather sarcastic) theory

Right, I’ve gone away and thought about this some more and I still come up with the same conclusions as I did earlier on although slightly edited

In answer to the original question “Will the plane take off or not” My answer is now

Theoretically – No the aircraft will not lift
However it would be impossible to conduct the experiment in a 100% controlled environment

I’ve taken the time to think about this and to type it up so that my opinions/points of view and theories could (hopefully be) best understood, all I ask it that you read through it carefully before making any comments :thumbs:

A plane is standing on a runway that can move (some sort of conveyer belt). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction).

Can the plane take off?

Difficult question

Before I go any further I want to make sure that we all understand the difference when I say “Closed” and “Open” systems

A system is commonly defined as a group of interacting units or elements that have a common purpose. The units or elements of a system can be cogs, wires, people, computers, and so on. Systems are generally classified as open systems and closed systems and they can take the form of mechanical, biological, or social systems. Open systems refer to systems that interact with other systems or the outside environment, whereas closed systems refer to systems having relatively little interaction with other systems or the outside environment.

In this case – Closed system = No variables etc we are imagining the experiment at its simplest.

Again, please take some time to read this before saying “pfft load of crap” :thumbs: Im taking some time and effort to explain what I percieve this experiment to be.

Firstly I can say that the reason the mythbusters experiment worked is because of the multitude of variables that are not taken into account in conducting the experiment.

The exact wording from the experiment imo illustrates that – conveyor belt and aircraft, not only have the exact same acceleration but the exact same velocity. The mythbusters experiment has already gone wrong here, as there is no way for the rate of acceleration of the car to match that of the aircraft. As a result the aircraft was able to gain momentum and airflow over its wings and lift. They also chose an aircraft in its design that would utilise the airflow over its foil from the prop to its maximum effect.

So to completely simplify the experiment down to the key point – is there any airflow going over the wings.

Prior to the introdcution of the aircraft, lets first imagine when you go running on the treadmill at the gym – this is EXACTLY the same principle as the question that has been asked above – you’re running at whatever speed relative to the treadmill but you arent moving forwards through the air. You are effectively running on the spot

Lets look at this from the point of view of a closed system/experiment.

Closed System
In this thought experiment, let us imagine the following. There is no wind, the air is perfectly still, the manner of propulsion of the aircraft, be it turbine or prop does not cause any air to flow over the wings at all (in real life, this is negligable anyway) for this experiment “friction = 0”

So, lets put our aircraft on the conveyor belt, the aircraft accelerates to 1mph and stays there, at the exact same moment, the conveyor belt accelerates in the opposite direction to 1mph and stays there. Rate of acceleration for aircraft and conveyor belt are both equal. Has the aircraft moved from the point upon which it was placed on the conveyor belt?

The aircraft is technically travelling at 1mph in relation to the conveyor belt. However it speed relative to the ground/air about it is 0mph.

When you run on a treadmill at 10mph – you are running at a speed of 10mph on the treadmill, but at a speed of 0mph relative to the ground, do you feel air moving past you when you run on a treadmill no because you arent moving through the air you are running on the spot.

So let have the aircraft and conveyor belt now accelerate to the takeoff speed lets say 150mph :nod: the rate of acceleration is the exact same until both arrive at 150mph, at this point the aircrafts speed relative to the stationary conveyor belt is 150mph, however it speed relative to the ground and the air surroungind it is still 0mph – taking from where I said above that we arent assuming that the engines are providing a flow of air over the wings – then as a result there has been no airflow over the wings at all.

As I have already ountlined using the formula for lift

L = (1/2) d v2 s CL
• L = Lift, which must equal the airplane's weight in pounds
• d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
• v = velocity of an aircraft expressed in feet per second
• s = the wing area of an aircraft in square feet
• CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

(1/2)d.v^2.s.CL = L
IF v=0
Then
L=0

As if you multiply anything by 0 or have any active component of a formula such as this at 0 – your outcome will be 0
Definition of lift

Lift is created by a difference in pressure between two surfaces – in an aerofoil (wing/propellor) it works by forcing the air on the top surface to travel faster (further to travel) than the lower surface. The result is a considerable difference in pressures between the upper and lower surfaces resulting in the wing being sucked up into the air due to lower pressure on the upper surface

From this we can deduce that to create lift in any form from an aircraft wing, we need an airflow.

So in this closed system – would the aircraft takeoff? No as above. Is this experiment realistic? No.

Open System

Concerning the open system, there are a whole shedload of variables to take into account.

To name a few
• Unnacountable airflow from – wind/engines etc
• Multiple types of friction – friction of air against the aircraft as a structure, friction of the tyres with the conveyor, friction of the wheelbearings with the spindles
• Density of the air

+ a great shedload of other variables that are uncontrollable

If we take the mythbusters example as supplied by Rubix :thumbs: this is an open system – the rates of acceleration were different between the aircraft and the conveyor, no account was taken for windspeed and the design of the aircraft was such that would allow the prop wash to assist the wing in producing lift :nod: the same would apply should you use turbines to power your aircraft, however obviously the amount of air sucked over the wing cause by the intake of the engines would be dependant on their placement. Wing mounted, tail mounted etc.

To summarise, in a perfectlty controlled experimental world, the aircraft would not lift, in the real world however it is a different matter. You could improve upon the experiment greatly to make it as close to controlled as possible and then you would be closer towards proving the theory of the experiment that I outlined above :nod:

But… No one is going to spend millions of pound to prove someone else in the internet wrong! :wack:

Member for

14 years 11 months

Posts: 519

Oh dear, oh dear, oh dear. Not this one again.

Nashio, yes the plane WILL TAKE OFF. Forget all the myth-busters crap, the treadmill crap and what ever else is out there crap. Just put your brain into gear and figure it out. Let's take it in easy steps (sorry if I'm sounding patronizing)

We'll stick with a prop powered plane. Firstly ask yourself, what makes the plane move on the ground? It sure isn't anything to do with driving the wheels as in a car is it? If not, then it's got to be something to do with the prop working the air, right? OK. Now the same reasoning goes for when the plane is flying through the air. The prop verses air struggle propels the aircraft forward. Therefore we can rule out any mechanical/direct link to the runway in any senerio. Bare with me. Sure you say, but the runway is moving in the opposite direction to the pulling power of the prop. Yes, but as there is no mechanical link between aircraft movement and the runway, what's going to drag the aircraft backwards, or at least, hold it back to counter-act the pulling power of the prop? The prop will still cause the plane to move forward. THE AIR ISN'T MOVING BACKWARDS AT THE SAME RATE AS THE RUNWAY. IT'S MOTIONLESS AND THE PROP/PLANE WORKS AGAINST IT, NOT THE GROUND.The wheels, being in touch with both the r/w and plane, will rotate at twice their normal speed for sure, but the plane will gain forward motion and eventually take off.

Another way to look at it is, imagine a plane coming into land. Just at the moment of touch-down it's traveling at say 55mph and therefore, the r/w is traveling at 55mph in the opposite direction with 'respect to the plane'. Does the plane care? No, because the wheels just spin up to the required speed once the tires hit the dirt. If the r/w was going backwards at 110mph, would it make any difference? No, the wheels would just spin twice as fast. The plane wouldn't and couldn't come to a sudden stop.

I hope that is clear enough for you and anyone else that is in doubt. Once again, sorry if I come over as patronizing.

Member for

20 years 5 months

Posts: 4,674

Of course the plane will take off.

A plane aint a wheel driven vehicle. Only a car would stand still.

The only thing that might happen is that the tires blow before lift off.

More tricky version of that stupid question: A flying boat on a river with a *very* fast current. What happens if the floatplane gets on the step before aerodynamically stable - will it fall over? Hehehe.

Member for

16 years 10 months

Posts: 3,214

yeah cheers guys lol I worked it out just as I was falling alseep last night :)

DP - Thanks for explaining it :) I dont appreciate being told to put my brain in gear, all ive done is completely overthink the experiment.

Member for

14 years

Posts: 1,234

Didn't this topic hold a record on another Aeroplane forum? 20,000 posts or similar?

Member for

16 years 10 months

Posts: 3,214

not a clue chap, I got all worked up about it having read it just as i got in from work at about 7am - ranted for a while, then woke up at about 5pm and was like

Oh ffs - major facepalm lol

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18 years 1 month

Posts: 894

I must confess, I dont get it? Maybe because it's 1am, and I should be sleeping, but lift is caused by airflow over the wings.... Its nothing to do with the engine. The engine simply provides forward movement through the air to allow the air to pass over the wing.

If you assume that there is no wind, then what causes the airflow over the wing? The 'treadmill' cancels out any forward movement, the aircrafts position relative to the runway remains the same and thus there is no increase in airflow.

If you liken it to a person running on a treadmill. If you run on the open road you'll feel air passing over your face.... You can run at any speed on a treadmill, but you will never feel any movement of air passing you.

How does the aircraft take off? Unless someone else know another way to take off without air passing over the surface of the wing, then I dont see how it's possible.

Member for

16 years 10 months

Posts: 3,214

I had this going over and over and over in my mind, for hours the other night :(

The U/C Wheels are not driven. they're castors - as a result the loss in forward momentun will be minimal and be dependant on the friction of the aircraft against the belt :)

The forward motion of the aircraft unlike that of a car doesnt have to act against the full opposite force of the conveyor.

So the wheels will spin REALLY fast but the aircraft will move forward eventually etc

With me? lol

Member for

14 years 11 months

Posts: 519

yeah cheers guys lol I worked it out just as I was falling alseep last night :)

DP - Thanks for explaining it :) I dont appreciate being told to put my brain in gear, all ive done is completely overthink the experiment.

No offense intended mate, just our way of saying 'clear your mind', which you eventually did. Good onya.

Member for

14 years 11 months

Posts: 519

I must confess, I dont get it? Maybe because it's 1am, and I should be sleeping, but lift is caused by airflow over the wings.... Its nothing to do with the engine. The engine simply provides forward movement through the air to allow the air to pass over the wing.
Correct, and that produces lift.
If you assume that there is no wind, then what causes the airflow over the wing? Wind has nothing to do with it. Only the forward motion of the wing through the air is necessary.The 'treadmill' cancels out any forward movement, the aircrafts position relative to the runway remains the same and thus there is no increase in airflow.Only if the wheel brakes are on.

If you liken it to a person running on a treadmill. If you run on the open road you'll feel air passing over your face.... You can run at any speed on a treadmill, but you will never feel any movement of air passing you. Forget treadmills, that's a totally different set of circumstances.

How does the aircraft take off? Unless someone else know another way to take off without air passing over the surface of the wing, then I dont see how it's possible.

Look at it this way. Forget the wheels and put the aircraft on skis, on the most slippery ice ever formed so that there is no friction between the skis and the ice. The slightest touch and the aircraft moves. OK, start the engine and the a/c moves forward (No wind situation) Does it matter if the ice moves in the opposite direction, can the ice drag the a/c backwards? Remember, no friction equals no connection of any sort. The engine is still pulling against the air and the a/c is still moving forward. It doesn't matter how fast the two surfaces are passing each other (ski and ice) the a/c keeps accelerating until it takes off.

Now, imagine the ice is covered with a layer of small ball bearings then look very closely at the interface of ski and ice with the bb's in between. As the a/c moves over them, they'll rotate, very fast, without affecting the a/c nor the ice, right?
The faster the a/c goes, the faster the ice goes in the opposite direction, the faster the balls rotate when the a/c skis reaches and passes over them. That is exactly what the wheels do in the standard description, they rotate at twice the speed of the normal rotational speed for any given airspeed. They are just another very slippery interface between the a/c and the runway. They cannot slow anything down unless the brakes are applied and friction is added into the scenario. The a/c keeps moving forward at an every increasing speed, the runway matches that speed in the opposite direction and the wheels spin like ****. THE PLANE TAKES OFF.:diablo::diablo:

Member for

24 years 2 months

Posts: 16,832

Guys - im sure this has been done to death

It has elsewhere, but not here

Someone please tell me im right in saying the following

You are wrong. It can take off unless it is a very odd aircraft that uses powered wheels to reach flying speed.

Moggy

Member for

19 years 9 months

Posts: 690

This is a good one to drop into a drunken conversation then go and chat up the women while all the blokes shout at each other. I agree with the consensus by the way, the u/c just keeps the thing off the tarmac.

Another one is "does a jet engine suck the aircraft along or blow it along?" I spent days in an argument about this with someone who claimed to be an expert in fluid dynamics. I gave up in the end - okay air is sucked in but the expanding gases from combustion provide the thrust surely?

If am being thick please forgive me :)

Member for

16 years 10 months

Posts: 3,214

It has elsewhere, but not here

You are wrong. It can take off unless it is a very odd aircraft that uses powered wheels to reach flying speed.

Moggy

Moggy old chap, if you look at the date when I posted this and my reply to VX above you'll find I realised I was wrong last week :)

Member for

18 years 1 month

Posts: 894

With any experiment we have to make some assumptions, or have a control... I read this with the assumption that any forward movement by the aircraft on its wheels is equalled by the reverse movement of the treadmill on which the aircraft is standing. I also assume zero wind. Final assumption is that we're looking for the aircraft to fly through conventional means... By that I mean we're looking for the wing to produce the lift, not the engine.

The engine is simply a means of propelling the aircraft through the air so that the wing can generate lift.

The wheels are simply a mechanism to allow the aircraft the travel along the ground in order that air can flow over the wings, thus producing a change is pressure, and thus producing lift. If any forward movement of the aircraft is equally countered by opposite movement of the ground (the treadmill), then how does the aircraft get the necessary airflow over the wing to produce the lift?

None of the answers I've read actually answer that question? Lift is produced by airflow over the wing. The aircraft doesn't move the air because it's forward movement is equalled by a reverse movement of the ground. Therefore the aircraft cant fly. The only way I can see the aircraft flying is if the airflow produced by the prop pulling air over the wing were sufficient to create that change in pressure, but in most cases that would not happen.

Another scenario... If my aircraft gets airborne at 35Kts, and I travel down the runway with a 35Kt tail wind, will the aircraft fly?... NO. because there isn't the necessary fly of air over the wing.

Assuming that you guys are right, then what part of my logic breaks down?

Member for

16 years 10 months

Posts: 3,214

It took me a LONG time to get it right in my head

If the wheels were driven and moving at 150mph in the opposite direction then it would stay still.

However the wheels are castors, so if you were to move the conveyor belt backwards, the aircraft would move, yes but not at exactly 150mph? the wheels would spin etc.

The thrust from the engine, is not a direct force acting against the conveyor belt - it will still pull the aircraft through the air regardless of the speed at which the belt rotates/speed at which the wheels spin.

In short, the bearings might give out/tyres pop - she will fly :)

Member for

24 years 2 months

Posts: 16,832

Moggy old chap, if you look at the date when I posted this and my reply to VX above you'll find I realised I was wrong last week :)

Apologies - didn't read the whole thread. After all, you chaps are trustworthy :)

Moggy