Can an aeroplane on a conveyor belt get airborne?

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Offline burning

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I'm still hoping that someone can prove to me, once and for all, if that plane on the treadmill can lift or not?

I say it will not..

But..?

Awh, come on now, myth-busters has defined it as lifting, and the majority of that physics community I was on then seemed to think it should too?

==the exact quote;



A plane is standing on runway that can move (some sort of band conveyer). 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 opposite direction).

The question is:

Will the plane take off or not? Will it be able to run up and take off?

==end.

Jump right in, but prove your point sientiflicly please, I do have some standards?
I think?


I hope people will forgive me for moving this to another thread.  I want to answer this one thoroughly, and I don't want to threadjack the other discussion since it may still continue.

Right.  *cracks knuckles*  Get ready for a wall o' text.

One problem that bedevils most of the discussions I have seen about this topic is that the problem as stated is ambiguous.  Depending on how you interpret the problem gives you a very different answer.  This leads to a lot of discussion at cross-purposes.  I'm going to look at the two simplest interpretations of the problem and work them out in detail so that you see what is involved.  There will be a number of simplifying assumptions, but I will point them all out as I make them.

The ambiguity is in the phrase "tracks the plane's speed."  We are measuring the speed with respect to what?  Are we measuring it with respect to the ground or with respect to the conveyor belt?  We'll take them both in turn.

(Two comments: 1. Obviously the same ambiguity applies to the speed of the conveyor belt, but I've never seen anyone interpret the conveyor's speed as being anything other than with respect to the ground, and I'll be sticking with that.  2. The two interpretations of the plane's speed I gave are not the only ones, and indeed neither is the one the Mythbusters tested.  The Mythbusters tested using the interpretation the plane's speed with respect to the ground if it were on a stationary runway.  This is more difficult to analyze, but I think the ones I show you will be sufficiently illustrative.)

Before we get into the conveyor belt cases let's be sure we understand the simple case of a plane taking off from a stationary runway.  Now the important thing for an airplane to take off is that its wings need to be moving quickly with respect to the air, and for most practical purposes that means moving quickly with respect to the ground.  We'll assume that we're dealing with dead calm so that the plane's velocities relative to the two are identical.

The force acting on the plane in the direction that it wants to go is the thrust from the engines; specifically the propellers or the jets push air backwards and the air pushes the plane forwards by Newton's third law.  All other forces on the jet are either working against the thrust or are perpendicular to the thrust.

I'm making the following simplifying assumptions:
1) There is no atmospheric drag
2) The plane's wheels roll without slipping
3) There is no friction in the wheel bearings
4) There is no "rolling friction" between the wheels and the runway (rolling friction is a dissipative force due to deformation of the rolling object and the surface it rolls over)

These assumptions add up to two results
1) The only force working opposite the direction of the thrust is the force of static friction between the wheel and the runway.
2) The work done by the friction all goes into making the wheels spin.

In equation form we have

F - f = Ma
and
τ = Iα

Where F is the thrust, f is the force of friction, M is the mass of the entire plane, a is the plane's acceleration, τ is the torque on the wheels from the friction, I is the net moment of inertia of the wheels, and α is the angular acceleration of the wheels.

As another simplifying assumption, we will assume all the wheels are identical.  This means each has the same formula for its moment of inertia, and we can treat the problem as if the plane had only one wheel.  (We would actually reach this point of simplification anyway if we treated all the wheels separately, but the final formulas are ugly if the different wheels have mass distributed differently).  Finally, I'm going to treat the wheels as uniform disks, so the moment of inertia is I = ½mR², where m is the mass of the wheels and R is their radius.  (This is not the worst case scenario for the plane being able to take off, by the way.  Worst case scenario is treating the wheels as hoops (I = mR²).  I suspect the disk approximation comes closer to the truth, but you should be able to work out the worst case scenario for yourself if I succeed in explaining everything clearly.  *touch wood*)

Having a single radius also lets us rewrite the torque as τ = fR and we can rewrite
τ = Iα
as
fR = ½mR²α
or
f = ½mRα

Right now we have three unknowns: f, a, and α.  However, we assumed above that the wheels are rolling without slipping.  Which gives the relationship
a = Rα
so
f = ½ma

We can now solve for either f or a in terms of the knowns:

f = F / [1 + 2M/m]
or
a = F / [M + ½m]

Both equations imply that good airplane design is going to involve having the wheels account for as small a fraction of the total mass as possible.  This would be true even if we put back in all the dissipative forces we decided to neglect above.  I don't have any figures on realistic airplane wheel masses.  Below I'm going to crunch some numbers using an assumption of m = .1 M, which sounds like an overestimate to me, but as I said I have no data.

We'll get into the conveyor cases in the next post.


[MOD EDIT - TITLE RE-FORMATTED AS A QUESTION - PLEASE TRY TO DO THIS IN FUTURE, IN LINE WITH THE FORUM POLICY]
« Last Edit: 29/03/2011 17:01:17 by Meera »

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Offline burning

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Airplane on a conveyor belt
« Reply #1 on: 29/03/2011 17:10:13 »
In this post we move on to ---

Case 1: The speed of the conveyor belt with respect to the ground matches the speed of the plane with respect to the conveyor belt.

This case is actually trivial, but by the same token it's a cheat.  If we grant that it is possible for the conveyor belt to move at this speed, then the motion of the conveyor belt cancels the motion of the plane and the speed of the plane with respect to the ground is zero.  That's just basic from the meaning of relative velocity.

But if we use this interpretation, that means that we are begging the question.  The problem boils down to "If we move the conveyor belt fast enough to hold the plane still, does the plane hold still?"  Not very interesting, eh?

The second problem with this interpretation is the assumption that it is possible to run a conveyor belt fast enough.  This assumption we will see is almost certainly wrong.  First let's note that while the problem talks about speed, what is really important is that we cancel out the acceleration.  If the conveyor belt is moving at a constant speed the problem reduces to that of the stationary runway we looked at above.  The plane may have to travel farther, but it will eventually reach take off speed with respect to the ground (think of the constant speed conveyor belt as equivalent to a stationary runway and a tail wind).

So we have acceleration of plane with respect to the conveyor is equal an opposite of the acceleration of the conveyor with respect to the ground, resulting in acceleration of the plane with respect to the ground being zero.  Well if a = 0, that implies that the force of static friction has increased so that f = F.  (Side note: Remember static friction can take any value up to a maximum limit; as long as that limit is not exceeded, it is whatever it needs to be to prevent the two surfaces from slipping over each other.  I looked up masses of jets, thrusts of jets, and typical coefficients of friction.  It might actually be possible for the static friction to hold the jet still.)  If we substitute this into the torque equation above, we can find the angular acceleration of the wheel in terms of the thrust:

α = 2F/mR

Now rolling without slipping is still assumed to be in play, so the acceleration of the plane with respect to the conveyor must be

a = 2F/m

If a0 is the plane's acceleration in the static runway case, and a1 is the acceleration relative to the conveyor in this case, we find that the ratio of the two accelerations is

a1/a0 = 1 + 2M/m

If we plug in my guess for the ratio of the masses of M/m = 10 (recall that I suspect this to be unrealistically high) we get

a1/a0 = 21

And this is how much the conveyor belt has to accelerate.  It's got to keep that up constantly, so it's going to get faster and faster.  And the wheels on the plane will have to spin faster and faster, much faster than they would be forced to spin on take off if we keep this up for any length of time.  Whether the conveyor or the plane's wheels give out first, I think it's going to be messy.  And yeah, the plane won't take off, but eventually that's going to be because everything is in flames rather than because of precision force balancing.

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Offline burning

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Can an aeroplane on a conveyor belt get airborne?
« Reply #2 on: 29/03/2011 17:45:24 »
Case 1 should have made you suspicious about the likelihood of ---

Case 2: The speed of the conveyor belt with respect to the ground matches the speed of the plane with respect to the ground.

First note that this must mean that the plane succeeds in moving.  If the plane remained stationary, that would mean the conveyor remained stationary.  But we know that a plane can take off on a stationary runway, so that's absurd.

And since we are really matching accelerations, not just velocities, the plane keeps accelerating.  As long as the wheels and the conveyor belt can take the acceleration (see above) the plane will eventually take off.  That is assuming it doesn't run out of runway, so let's look at how much further the plane needs to travel to reach take off speed.

If the accelerations of the plane and the conveyor with respect to the ground are equal and opposite, their acceleration with respect to each other has twice the magnitude.  Once again invoking rolling without slipping for the wheels, if a is the plane's acceleration with respect to the ground, in this case

α = 2a/R

which yields a force of friction

f = ma

or twice that in the case of the stationary runway.

Solving for the acceleration in terms of the thrust and masses yields

a = F/(M + m)

If a0 is the ground acceleration with the stationary runway, and a2 is the ground acceleration in this case, we find

a2/a0 = (M + ½m)/(M + m)

Once again looking at the special case of m = .1M we get

a2/a0 = 1.05/1.10 ≈ .955

Not, I think you will agree, a big reduction in ground acceleration.  Since we are interested in how far the plane has to travel to reach a given speed from rest under a given acceleration, we use the elementary kinematics equation

a Δx = v2 - v02

The velocities are the same in the two cases, so we eliminate them and get

Δx2/Δx0 = a0/a2 ≈ 1/.955 ≈ 1.05

We only need a runway that is 5% longer than the stationary one.

In the interests of full disclosure, this case is a lot more vulnerable than case 1 if my assumptions about reasonable wheel masses are false.  In case 1, the wheel mass assumptions lead to an impressive acceleration ratio, but even if the conveyor belt could accelerate much slower, it would eventually reach a speed that was too much for either the conveyor or the wheels.  It is conceivable here however, that we might be able to (with massive enough wheels) have a situation where the runway becomes too long without the anything having to spin too fast.  Also, in this case if we reduce the net force enough, the assumption of no atmospheric drag might really cause problems.

So, I hope I have been clear, informative, and maybe even slightly entertaining.  Bottom line, it really looks to me that if we have a set up that can really prevent take off, it is likely to be one that would be impossible to build and maintain in real life.

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Offline yor_on

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Can an aeroplane on a conveyor belt get airborne?
« Reply #3 on: 30/03/2011 00:05:33 »
I loved it Burning, but I will have to reread your latest post :)
But it's the best approach to it I've seen yet :)

Quite impressive..

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Offline yor_on

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Can an aeroplane on a conveyor belt get airborne?
« Reply #4 on: 30/03/2011 00:23:10 »
Case 1: The speed of the conveyor belt with respect to the ground matches the speed of the plane with respect to the conveyor belt.

Here you define the ground as the 'static' observer of both speeds, and assume that both, more or less instantly will adapt to the same speed, relative that 'static' observer?

You mean that the conveyor belts speed relative the ground is equal to the planes speed relative the conveyor belt? That must be it, right?

You know, I thought I could read, but, I'm having to reread it again :)
Why can't people learn Swedish?


=
Case 2: The speed of the conveyor belt with respect to the ground matches the speed of the plane with respect to the ground.

I take that to mean that if the plane relative a unmoving surface (e.g the tarmac) move at a 100 km/h, then the conveyor belt will have the same speed in the opposite direction? But you are no longer considering it as 'instant'? You expect there to be a slight difference in time, in the conveyor belt adapting its speed to the planes acceleration?

Ah : "First note that this must mean that the plane succeeds in moving.  If the plane remained stationary, that would mean the conveyor remained stationary.  But we know that a plane can take off on a stationary runway, so that's absurd."

That got me wondering a little though :)
« Last Edit: 30/03/2011 00:28:22 by yor_on »
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Can an aeroplane on a conveyor belt get airborne?
« Reply #5 on: 30/03/2011 00:48:19 »
But it is thought provoking :)

It seems to boil down to a description in where we can't expect any real life scenario to get the conveyor belt to adapt instantly and so always be a 'step' behind in its acceleration relative the plane?

But; "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 opposite direction)." Here is another ambiguity I think, there is no definition of what he mean by 'the same'.

If I by 'the same' expect him to have meant 'instantaneous', or at least so near to it we can come, then that conveyor belt will have to be infinitely long it seems to me, as there will be a infinite amounts of accelerations (jerks) relative the 'unmoving' ground (for this purpose I will define the ground as unmoving). That should mean that we would have a skeleton sitting in that pilot seat, right?

If I on the other hand expect him to have meant a real life sceniario, then we need to define the mass of the plane, it's lift ratio relative that mass (how easy it can 'take to wing' sort of), the wheels resistance relative the ground and its own bearings, relative wind and real wind, and also what difference it might bring if we assume it to be nose-propeller driven plane that thrust air over the wings relative a jet propelled aircraft with its engines mounted under the wings. We also need to consider the amount of thrust that plane can bring on. Like if it is a rocket laid sideways on that conveyor belt, or if it is a Airbus. And I'm sure Ive missed a lot of other stuff too.

But in the case of it being a near 'instantaneous' adaption I do not expect it to take of, unless of course, it is a rocket :)
Or a Helicopter/VTOL craft.
=

In a real scenario you either have to assume that the acceleration is 'constant' unceasing so to speak, or, you might assume that there comes instants where the plane and the conveyor band matches each other speed, which seem a reasonable approach as all engines have a power curve as far as I know? If that is so then it will have to restart from zero at each such 'instant'. The other scenario where the acceleration is constant reminds me more of a rocket?

==

Maybe one can say it boils down to this?

In the case of a rocket or similar, the wheels are of no real importance, as the thrust generated is acting directly against the rocket itself. Then it certainly will take off. Ah, assuming that the hull holds up to constant grinding the conveyor belt will have on its downward side resting on the conveyor belt? At some time, depending on thrust it should reach a velocity enabling it to leave the conveyor belt though, a little unbalanced by that time perhaps, due to that constant grinding, but I'm sure it bravely would continue, its gyros working overtime :)

Or in the case of a 'normal' aircraft, needing to get up to a certain speed relative its 'ground' for getting that downward downwash. The downwash is relative the wind over its wings which means that if the aircraft is defined to need 250 km/h for a takeoff, then that speed has to be reached relative the conveyor belts opposite direction/speed too. And there I wonder how much of a advantage a (nose) propeller driven plane would have over one with the engines situated under the wings? But here it also becomes a matter of the engine power, the stronger the engine relative the aircraft, the better its acceleration relative the conveyor belt, and the greater chance of that final take off, at some time in the future :) depending of course on the durability of the constructions. I mean, after some years going on, out there? As a passenger I'm quite sure I would find it quite unbearable to be served the same meals, for months and months to come? Not to mention the toilets?

And this should go for 'normal' Jets too, although some military versions seems more like rockets :)

Or?

==
ah Burning, love this, a subject worthy of a deeper study.
I'm pleased that we started it :)
« Last Edit: 30/03/2011 01:50:29 by yor_on »
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Can an aeroplane on a conveyor belt get airborne?
« Reply #6 on: 30/03/2011 04:38:05 »
Perhaps it would be simpler to see if the question was.

Can you get a conveyor band to neutralize the acceleration of a car upon that conveyor band, using the ground the combined contraption stands on as the referent for zero motion?

Assume that the conveyor band will answer any motion of the wheels with the exact opposite motion of the conveyors band, inside 1 thousand of a second from the motion gets detected (the detector-speed included).

How long would it take for that car to move 1000 meters on that band, depending on its acceleration. And will it make a difference if it 'accelerates faster', using more engine power?
==

As long as we're not assuming that using a propeller, or a jet, distributes the force in some other manner, than to its wheels, that is :)
==

And the final Q, referring to the aircraft takeoff velocity of 250 km/h.

What acceleration would it need to overcome the bands constant adaption to its motion, and what time would it take, for it to reach a velocity of 250 km at the end of those 1000 meters. And what would that acceleration compare to, if using the same acceleration/energy expenditure on the ground? That is, how many meters would the car need to reach 250 km/h, and what time would it take?

(Phieww, had to think to get this right. But I hope I covered it now?)
 
And, as an extra. How many horse powers?
And lastly, will you sell it to me :)
« Last Edit: 30/03/2011 05:13:11 by yor_on »
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Offline Bored chemist

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Can an aeroplane on a conveyor belt get airborne?
« Reply #7 on: 30/03/2011 06:57:56 »
"Perhaps it would be simpler to see if the question was.

Can you get a conveyor band to neutralize the acceleration of a car upon that conveyor band, using the ground the combined contraption stands on as the referent for zero motion?" It might be simpler, but it's a totally different problem.

Anyway, this has already been discussed
http://www.thenakedscientists.com/forum/index.php?topic=12548.0
« Last Edit: 30/03/2011 07:01:57 by Bored chemist »
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Offline Geezer

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Can an aeroplane on a conveyor belt get airborne?
« Reply #8 on: 30/03/2011 08:20:55 »
Yup. Basically, it does not make a damn bit of difference.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ĉther.

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Offline Jolly- Joliver

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Can an aeroplane on a conveyor belt get airborne?
« Reply #9 on: 30/03/2011 14:15:15 »
Depends on the plane, a Harrier would. [:)]
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Can an aeroplane on a conveyor belt get airborne?
« Reply #10 on: 30/03/2011 14:24:59 »
Yep, but if you want to argue that it takes of, it might be enlightening to see what happens when we put in some figures in our discussion, and then also compare it to a comparative effect done on the ground instead of the treadmill.

That's why I doubt.

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Offline Geezer

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Can an aeroplane on a conveyor belt get airborne?
« Reply #11 on: 30/03/2011 20:58:12 »
There are only two things that could make any difference.

1) If the tires and bearings produced significant amounts of friction, it would reduce the acceleration of the plane.

2) If the belt was really wide, it would tend to create a bit more headwind, which would actually help the plane take off!
There ain'ta no sanity clause, and there ain'ta no centrifugal force ĉther.

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Offline yor_on

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Can an aeroplane on a conveyor belt get airborne?
« Reply #12 on: 31/03/2011 20:26:17 »
Pretty nice thread BC. Still I withhold that what makes lift is the speed relative the surface. Without that speed you won't get any downward pressure, aka 'lift' of the aircraft. The aircraft is assumed to depend on its wheels to get this speed relative the surface it traverse.

The possible acceleration will reduce the aircrafts weight, relative the surface of course, making the weight resting on a conveyor tread smaller. But assuming a car we look at a simpler example I think.

And to create a speed relative the air, you will need a motion relative that same atmosphere. You can ignore the belt and the ground if you like, instead defining it as the atmospheres motion relative the aircraft (wings primarily), just as we do in a air-tunnel. And there I still think that, even though that you with a sufficiently long runway (tread belt), possibly, smelling of burning metal, would see it take off, I don't think you will see it for a normal jet, with a reasonably spaced tread belt, under a reasonable amount of time :)
« Last Edit: 31/03/2011 20:28:29 by yor_on »
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Offline Bored chemist

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Can an aeroplane on a conveyor belt get airborne?
« Reply #13 on: 31/03/2011 20:49:23 »
"Pretty nice thread BC. Still I withhold that what makes lift is the speed relative the surface. "

How does the plane know that the ground is there?
I know about the ground effect.
I means, once you are at 20,000 feet, how does it know?
In particular, it's possible that, at that height the wind is (for a ground based observer) travelling West while the air below it, nearer the ground, is travelling South.

How can the 'plane "see through" all that air to the ground?

" But assuming a car we look at a simpler example I think. "
Simpler, but unhelpful.
A plane isn't driven by its wheels, but a car is.
So "The aircraft is assumed to depend on its wheels to get this speed relative the surface it traverse." isn't true.
The two systems are not comparable.


For a normal take-off the wheels offer very little friction- they are, after all, wheels.
So, in the case of a plane on a belt, the fact that the "ground" is moving under the wheels doesn't exert much (drag) force on the aircraft. With "ideal" wheels it would be zero. (they would exert a force upwards to counter the weight of the plane.)
Since there's little or no force acting on the plane due to the action of the belt, how would the plane "know" not to take off?
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Offline Geezer

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Can an aeroplane on a conveyor belt get airborne?
« Reply #14 on: 31/03/2011 21:31:04 »
Anyway, as I already pointed out, if the belt has any influence on airspeed relative to the plane (which I doubt) ,it can only increase the airspeed and therefore increase the lift.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ĉther.

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Can an aeroplane on a conveyor belt get airborne?
« Reply #15 on: 31/03/2011 21:34:05 »
As I said before, somewhere, the thread should be declared 'immortal' :)

As long as it's rolling on the ground it has to know it, if we assume that this is the way a plane defines itself relative its surroundings :) Otherwise it won't know the difference between lift and 'no lift'.

I think?
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Offline Geezer

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Can an aeroplane on a conveyor belt get airborne?
« Reply #16 on: 31/03/2011 21:56:22 »
Yoron, as you recently pointed out, a plane takes off because it accelerates air down to produce a reaction force up.

I think F = m.a might have something to do with it  [::)]
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« Reply #17 on: 31/03/2011 22:40:27 »
"The lift of a wing is equal to the change in momentum of the air it is diverting down. Momentum is the product of mass and velocity. The lift of a wing is proportional to the amount of air diverted down times the downward velocity of that air. Its that simple. (Here we have used an alternate form of Newton’s second law that relates the acceleration of an object to its mass and to the force on it; F=ma) For more lift the wing can either divert more air (mass) or increase its downward velocity. This downward velocity behind the wing is called "downwash"."

Quite so, and to get it from a 'standing position' you need ???
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Offline Geezer

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Can an aeroplane on a conveyor belt get airborne?
« Reply #18 on: 31/03/2011 22:44:04 »

Quite so, and to get it from a 'standing position' you need ???


It's not standing when it takes off. It has an airspeed relative the air.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ĉther.

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Offline Bored chemist

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« Reply #19 on: 01/04/2011 06:56:41 »
"Quite so, and to get it from a 'standing position' you need ???"
A F***ing great engine pushing it, and nothing but air resistance holding it back.


(whatever the wheels and the ground are doing)
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« Reply #20 on: 01/04/2011 07:02:27 »
"Quite so, and to get it from a 'standing position' you need ???"
A F***ing great engine pushing it, and nothing but air resistance holding it back.


(whatever the wheels and the ground are doing)

LOL!
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« Reply #21 on: 01/04/2011 17:39:01 »
Well, that's where we diverge BC.
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Offline Madidus_Scientia

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Can an aeroplane on a conveyor belt get airborne?
« Reply #22 on: 02/04/2011 10:03:31 »
Still I withhold that what makes lift is the speed relative the surface. Without that speed you won't get any downward pressure, aka 'lift' of the aircraft.

No, the lift is created by airflow over the wings. The ground has nothing to do with it.

http://www.youtube.com/watch?v=JslkmigogKY

The aerofoil in this video has no 'speed relative to the surface'. Well, except for when it goes up.

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Offline Bored chemist

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« Reply #23 on: 02/04/2011 15:06:09 »
Well, that's where we diverge BC.

I asked a couple of questions; could you please answer them?
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« Reply #24 on: 02/04/2011 16:09:00 »
Still I withhold that what makes lift is the speed relative the surface. Without that speed you won't get any downward pressure, aka 'lift' of the aircraft.

No, the lift is created by airflow over the wings. The ground has nothing to do with it.

http://www.youtube.com/watch?v=JslkmigogKY

The aerofoil in this video has no 'speed relative to the surface'. Well, except for when it goes up.

Sweet one MS :) And a good point.
It doesn't matter where the direction of the lift comes from for a lift I agree.

From a hurricane or the wheels of the aircraft. That's, if you excuse me, a relative thing. I already pointed out that you can get the same lift in a air-tunnel, using a 'stationary aircraft'. That's how we build them.

But excluding that hurricane you still need to get a speed for the aircraft to get its lift. So the question, as I sees it, becomes if the plane really get that speed relative the atmosphere. And there I severely doubt it. And that's also why I suggested that we could use the cars instead to see at what 'real distance' relative the ground they would get up to a certain speed, still relative the ground, as defined from where the cars, and tread mill, starts to move opposite each other.

Take the car, draw a vertical line, or use a straight ruler, down to the ground. We define that ground as 'unmoving', then use the point where the ruler intersect the ground as our definition of 'zero motion'. Now start the contraption (car/tread mill) and see how far the car reach, under what time, and at what speed, relative that 'zero mark'. If we define a planes takeoff as 250 km/h relative the tarmac, then the cars speed would need to reach the same speed relative the 'starting point(zero)'. How far does it have to move relative the zero mark at the ground, on the tread mill for that? And how long time will we measure it too take?



« Last Edit: 03/04/2011 00:03:05 by yor_on »
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« Reply #25 on: 02/04/2011 21:23:32 »
Do you realise there's a difference between cars which have powered wheels and aircraft which use props or jets?

(and I'd still like an answer to the questions I asked earlier too)
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« Reply #26 on: 02/04/2011 21:48:13 »
Airspeed is what produces lift. If the air moves over the wing fast enough, it will lift the plane. If the plane is stationary relative to the ground, it may get some lift due to the fact that the conveyor belt's friction generates wind. If lift-off speed is 80 knots, the belt will have to go considerably faster than that, and the wind it produces will only go up a short distance. An airplane hovering above the belt would be no different from hovering above the ground in an 80 knot wind, except it would have to hover very low where the wind is strong.
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« Reply #27 on: 02/04/2011 23:47:55 »
Are you saying that a aircraft don't rest on its wheels on ground BC?
Or do you mean the fact that as they get up to speed (take off) the weight resting on the wheels get progressively smaller? If it is the later I agree, my car idea is in no way a perfect fit but it will give a general idea of the condition for takeoff. As for the question if I see the difference between cars and props or jets. Yes, I believe I do? I'm not saying that a car is as a aircraft?

I'm interested in putting some few figures to it, seeing the time needed for that possible, as judged by some, inevitable takeoff. To do that we can use a 'takeoff'-speed by a prop or jet, and just look at the time it takes to reach that speed. Or a car, doesn't really matter for this question. when it comes to how much a lift will matter before takeoff I don't really know? But that's not the point either, I'm just interested in how long it will take to reach those 250 km/h on that treadmill.
as for the discussion how the atmosphere might move at different levels? What has that to do with the aircraft standing on the treadmill? If you don't want to assume a hurricane coming?
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« Reply #28 on: 02/04/2011 23:53:43 »
Maybe it could get a 'lift' on air pushing it up, from the treadmills motion, but my guess is that it would be rather unstable. I think Rosy wrote something about that? (The thread BC linked to.)
==

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« Reply #29 on: 03/04/2011 05:25:03 »
OK Yoron, try this.

The conveyor belt has a layer of ice on it (yes, I know that's a bit difficult to achieve) and the aeroplane has ice skates instead of wheels.

So, what happens when the conveyor belt starts moving and the plane's engines are off? I'll tell you what will happen. Not a dang thing! Because of its inertia, the plane will stay in the same position while the belt moves under it.

The only difference between wheels and skates is that wheels produce a little more friction, so the engines have to generate a little bit of thrust to compensate for that. Other than that, whether there is a moving belt or not makes almost no difference.

 

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« Reply #30 on: 03/04/2011 05:33:23 »
Well, if it was ice and skates that plane would find it a lot easier to get into a 'real motion' relative the earths 'starting-mark' I think. But when it comes to the tread belt, and the invariant mass of that aircraft resting on its wheels I'm not as sure of that motion.

« Last Edit: 03/04/2011 05:35:28 by yor_on »
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« Reply #31 on: 03/04/2011 05:42:23 »

Well, if it was ice and skates that plane would find it a lot easier to get into a 'real motion' relative the earths 'starting-mark'


No, it would not!! You seem to be making a very bad assumption about how much friction wheels produce. They are hardly any different from ice skates.

The amount of thrust that the engines have to produce to overcome the rolling resistance of the wheels is negligible compared with the thrust required to accelerate the mass of the plane and the thrust required to overcome wind resistance.
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« Reply #32 on: 03/04/2011 05:52:21 »
I'm not speaking about friction?
You are Geezer.

I'm speaking of the mass, the weight resting on the wheels before it gets its lift. Unless you mean friction to be the sole cause? It can mean many things friction? If the texture of the treadmills band is sandpapered you will find a greater friction too.

But assuming no oil leaks, just something resembling a normal surface where the wheels can get a grip then it will be the props, jets, that makes those wheels start to turn, and in my scenario make the tread mill move the exact same amount in the opposite direction one thousand of a second slower/after. And that's what I'm curious about. Assuming a magical lift and no weight leaves me sort of wondering?
« Last Edit: 03/04/2011 05:53:54 by yor_on »
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« Reply #33 on: 03/04/2011 05:54:06 »
No, it would not!! You seem to be making a very bad assumption about how much friction wheels produce. They are hardly any different from ice skates.

The amount of thrust that the engines have to produce to overcome the rolling resistance of the wheels is negligible compared with the thrust required to accelerate the mass of the plane and the thrust required to overcome wind resistance.

Indeed. You could stand at the back of the conveyor belt and hold the plane on it

yor_on, if you rolled a toy car down an escalator do you really think it's going to go much slower than if the escalator was turned off?
« Last Edit: 03/04/2011 05:57:41 by Madidus_Scientia »

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« Reply #34 on: 03/04/2011 06:00:24 »
This is not a good comparison :)
The wheels will turn, skates don't.

Sh* nobody but me that sees it?
As for the rest, let's take that as a new question as it's not the same.

Because with skates it would be different, as you say.
==

If we really want to prove a point assume a frictionless material.
Then the bigger the mass the more inertia. and so it never will move.

Probably :)
« Last Edit: 03/04/2011 06:05:24 by yor_on »
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« Reply #35 on: 03/04/2011 06:02:52 »
I'm not speaking about friction?
You are Geezer.

I'm speaking of the mass, the weight resting on the wheels before it gets its lift.


Then you ARE speaking of friction. If the wheels have no rolling resistance, the weight resting on the wheels is completely irrelevant, so if you are concerned about that weight, you must be making a false assumption about the significance of friction.
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« Reply #36 on: 03/04/2011 06:06:15 »
No rolling resistance?
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« Reply #37 on: 03/04/2011 06:08:29 »
Am I to understand this to mean that, according to you, there is no difference between 10 kg resting on three wheels, or 10 tons?
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« Reply #38 on: 03/04/2011 06:10:13 »
This is not a good comparison :)
The wheels will turn, skates don't.

Sh* nobody but me that sees it?
As for the rest, let's take that as a new question as it's not the same.

Because with skates it would be different, as you say.

No, it wouldn't be different. Wheels and skates do the same thing. They support a mass and minimize friction at the interface between the mass and some other surface.

I suggest you go back and study some basic Newtonian Mechanics.
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« Reply #39 on: 03/04/2011 06:13:47 »
:)

We'll see. Now tell me what you meant by 'no rolling resistance'?
You deem it as if the wheels 'friction' is what counts right? And somehow the mass resting on them has no importance? So I could mount Titanic on those wheels then? And give it a flying start? Well, sort of?
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« Reply #40 on: 03/04/2011 06:14:27 »
Ignoring Icebergs :)
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« Reply #41 on: 03/04/2011 06:16:09 »
You know, this treadmill is indeed a marvelous invention :)
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« Reply #42 on: 03/04/2011 06:20:02 »
I'm not disputing this Geezer "They support a mass and minimize friction at the interface between the mass and some other surface." But you wrote it yourself. "They support a mass."
==
And yes, looking at it your way I can see that you might think so. But to me the inertia will constantly work against the force trying to make it move, with the tread mills band one thousand of a second behind. And the greater its mass, relative the power, the longer the stretch it will need for that takeoff. Well, as I see it.

Let us assume that the aircraft 'jumps' like blackbody radiation. from 10 to 20, to 30 km/h

10 - treadmill nul
20 - 10 (one millisecond later)

In reality both accelerations are constant ones, so the divergence will widen between the two, to the aircrafts advantage, it's just that I expect it to take a awful amount of time for that plane to ever get airborn myself, if ever.





« Last Edit: 03/04/2011 06:30:08 by yor_on »
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« Reply #43 on: 03/04/2011 06:20:48 »
Am I to understand this to mean that, according to you, there is no difference between 10 kg resting on three wheels, or 10 tons?

Ah! You are finally getting it.

Assuming the bearings were designed properly, the coefficient of friction would be the same in either case.

Put a mass of 10 kg and a mass of 10,000 kg, both on rolling bearings, on a 1% incline. Which one will have greater acceleration? Answer: Neither. They will have equal acceleration.
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« Reply #44 on: 03/04/2011 06:25:12 »
If you support a mass of 1000 kg on an air bearing, you can move it quite easily with a force of 1 kg.
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« Reply #45 on: 03/04/2011 06:31:46 »
The inertia will differ, and it won't disappear just because it starts to roll Geezer. It's there along with the mass. Even though it will lessen with speed and lift I guess. Which is a stupid statement as it must have the same inertia always, but it's the lift that confuses me here? As that will change the mass resting on the tread belt.
==

And that's my point I think.
You can assume any acceleration as inertial reactions.
There has to be one to any increase in speed.
 
But I'm not as sure any more :)
« Last Edit: 03/04/2011 06:44:03 by yor_on »
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« Reply #46 on: 03/04/2011 06:39:47 »
The inertia will differ, and it won't disappear just because it starts to roll Geezer. It's there along with the mass. Even though it will lessen with speed and lift I guess.

No it won't. The inertia of the aircraft has nothing to do with the speed of the conveyor belt.

Try again  [:D]

(You're really clutching at straws now.)
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« Reply #47 on: 03/04/2011 06:48:17 »
No, I'm not ready to throw in the towel, just yet.
Action and reaction.

The mass rest on those wheels. What mass there is, will be inertial even if the wheels are friction less. So to get it to accelerate will be as constantly making its mass to move from zero, add to that that the treadmill constantly are stealing that motion by accelerating the other way.
==

Think of those objects in space instead. To get 10 g or ten ton to move will take a different force. And an acceleration is like constantly coming back to that ten ton object and push it a little more. And the weight (on earth) is all resting on those three wheels that meet a surface rushing the other way.
==

But I'm not as sure as I was, da* :)
==

the only reason it would matter would then be the magical prop, or jet?
Take a cyclist sitting on his bike, take yourself running on a treadmill.
Exactly how do you see that differing from using a propeller to get that forward moving force?
As long as you're still on that treadmill?
==

Let us assume no prop. What would happen if the treadmill started to move. The aircraft would go with it, do you agree? So the inertia matters, as does its weight relative what it stands on, which are the wheels. To get to your point I will have to assume that the only thing that would matter was the friction the wheels had relative the aircraft and the treadmill, ignoring the aircrafts mass as well as the treadmills opposite motion.
« Last Edit: 03/04/2011 07:10:47 by yor_on »
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« Reply #48 on: 03/04/2011 07:06:33 »
You keep going back to the mass resting on the wheels as if it had some magical property. I'm sorry to disappoint you, but it doesn't!

It results in a bit of friction force that has to be overcome, but compared with the other forces involved, it's quite negligible.

If you don't believe me, put your car on a small incline, release the brake, and try to stop it.
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« Reply #49 on: 03/04/2011 07:07:36 »
The difference between one plane trying to take off on a normal runway and another trying to take off on a conveyor belt will be as insignificant as one ball rolling down a rising escalator and another ball rolling down an escalator that is turned off.

One example gets acceleration from gravity, the other from jet engines.