Puzzle for you

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Jari

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Ok, here's a question for you, imagine the following:

A plane is sitting on a huge conveyor belt. This belt will act as a runway for the said plane. Problem is that the belt is set up to match the speed of the plane, but it will run in opposite direction. Think of a treadmill, if you have trouble imagining what I mean.

Here's the question: will the plane be able to take off?
 
No. A plane must be able to lift from the ground, and that can't happen if there's no air presure under the wings (?)

but it makes you think alittle  :mrgreen:

Edit: Nice avater jari :D
 
It would take off.  The wheels under the plane are roll freely, so their isn't the same reaction as say an automobile.  Their would only be slight friction, but the forward inertia/thrust would easily compensate for that.
 
It would take off.  The wheels under the plane are roll freely, so their isn't the same reaction as say an automobile.  Their would only be slight friction, but the forward inertia/thrust would easily compensate for that.
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Look it doesn't matter if the weels are spinning at lightspeed, the plane would not take off. It's the air under the wings that creates the lift. The engines just pushes it forward to get the air pressure to be able to take off :)

The plane isn't moving at all, the conveyor belt is just meaking the weels move and will not let the plane be able to get enough lift for it to take off.

I may be wrong, but as far as my thinking, it would just stand still on the ground, with full thrust on the engines.
 
The plane may not be physically moving from the point of view of the observer, but the force of the planes kinetic energy is still acting on the free moving air.  Unless this is in some crazy vacuum.  :-D
 
Yes. if it's a Harrier, it would just take off, (but, it's rare to se a harrier actually take off verticaly, the thrusters can't handel the heat off the engines while taking off. so they are saving the "cooling" when they are going to decent verticaly.)
 
Forgive me, if I'm brief but something that happened in this thread earlier (it seems to be gone now, but it's so sad because it's the third time this person has done something like this - so I have no other option than to assume that it was an intentional attempt to ruin this) really pissed me off. For the same reason I'm cutting this thing short.

Otokoshi has the exactly right answer - it will take off just as usual, except that the wheels will be spinning twice as fast. Entirely regardless of what kind of a plane it is. Only thing that could possibly prevent it from taking off is the increased rolling resistance of the tires, and increased friction of the bearings - since they are spinning twice as fast as they normally would. This is a much smaller force than the aerodynamic drag however, so no plane should have much difficulty overcoming it.

As the plane's engines try to move it in relation to the air surrounding it (which certainly is not traveling with the belt's surface), it doesn't actually matter what the belt does - how fast it moves or which way - or what the wheels are doing, as long as they are not locked.

There is another version of the question, with a slightly different wording, and that is much, much more tricky; the question specifies that the belt is trying to match the speed of the wheels (well, the speed the wheel's road surface is traveling around and around - wheel itself of course travels just as fast as the plane), and as far as I can see it, this could lead to a case where belt's speed reaches infinity soon after the start. But since this question was talking only about the speed of the plane, it's not a problem here.

And yes, if someone is wondering, the speed of the plane and the speed of the belt's surface are both in relation to a stationary outside observer - or "real world", if you prefer. The question obviously wouldn't work, if they were for example measured in relation to each other, or something like that.

Here's one explanation and here's another (that explains the "matching the wheel speed"-issue, as well).

Reason I asked this was because this seems to generate massive amounts of debate, and even some real life pilots seem to misunderstand the simple laws physics that make their planes fly:

http://txfx.net/2005/12/08/airplane-on-a-conveyor-belt/
http://forum.physorg.com/index.php?showtopic=2417
http://www.airliners.net/discussions/tech_ops/read.main/136068/1/#1
http://www.r3vlimited.com/board/showthread.php?t=79439

:-D


PS. On the other hand, car with a wings attached to it (nevermind that such a thing could never actually really fly - without a propeller or a jet engine or some other alternative means of propulsion, since wheels can't power it when airborne) would never lift off. Simply because cars, unlike planes, use their wheels to generate motion and as such have to do it in relation to the belt's surface, not air. So, car would just sit on the belt, never moving, because the belt would simply match any attempt to go forward.
 
Ahh fair enough.  Im not into physics and whatnot so I was mainly guessing.

Also, what do you mean "something happened"?  Have there been dissapereing posts again?
 
Don't worry about it, someone just went and blurted out the right answer right at the start.

I'm going to give him the benefit of a doubt, and assume that he actually had not read the post where I first mentioned this question, told the correct answer, and that I was going to post it here... and that while it was okay for the people in the know to answer, I would not like it if they explained it - given that I had already explained it to them.

I actually do believe that there was no ill will intended, but this person has done this before - more than once - which is what pissed me off.

Anyway, no big deal. :)


EDIT: This is also why the actual voting is locked - I was more than little peeved at that point. One of the admins must have intervened afterwards, but they didn't re-open the poll.
 
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What the hell? I posted my answer in spoiler tags that apparently don't work on this board anymore. I didn't realise that at the time and I would have corrected it following your first PM however I've been away for New Year so sorry? And when have I done this before?
 
Actually, the thing might be even more clever than I originally thought.

For planes it should work exactly as described - that part is right. Plane will care very little about the belt, since it tries to move itself in relation to the air surrounding it.

But for cars... actually, I think that a car might be able to accelerate as well. See, it states that the belt will match the speed of the car (just imagine that it's a car now, instead of a plane). So, if a car is moving forward at 50 mph, the belt is moving 50 mph to the opposite direction. Thing is... that only means that the wheels of the cars are spinning at a rate that corresponds to it going 100 mph - because that's exactly what it's doing - relative to the belt.

Right?

If it was trying to match the speed of the wheels - or rather the speed of the surface of the wheel - car would never move (for a plane this might be much more trickier, like I said before...). But since it does not try to do this...


Jedimark: Oh, when have you done this before? Remember that quiz by Nori? You know, the one which she asked the replies be given as PMs, instead of writing them on the thread? And certain PM from me, that I thought was meant as private, not for people to read?

And you of course were totally incapable of both, previewing the post, or checking the thread after you posted? Both of these things would have revealed that tags indeed did not work.

I'm willing to assume that you don't do these things out of malice, but just because you type before you think. And yes, I'm mad because of it - mostly because you don't seem to learn at all. Shall we leave at that, or do you want to argue?
 
I don't want to argue with anyone but I think we should leave it there before it gets even more trivial than it already is.
 
To Jari and Otokoshi,

You have to remember that thrust is a vector quantity, with both speed and direction. If the conveyor speed is exactly matched to the aircraft speed but reversed direction, then in fact the two vectors (aircraft and conveyor) cancel each other out. Bernoulli's principal, which is what governs the "lift" of aircraft wings, states that the airflow has to be significantly different between two sides to produce the "lift" nessesary to allow an object flight. In most cases, an aircraft requires 100 (for small)- 175 (cargo/passnger) miles per hour forward air speed (FAS) to acheive enough airflow/lift to maintain flight.  However, on a conveyor system like you suggest, you cancel out the forward direction needed.  Discounting the airflow to/from the engines, the aircraft would experience no more airflow over the craft then if it were standing still, hence it could not "take off".
 
We are aware of what makes planes fly - at least I am, and I'm fairly sure that Otokoshi is, too.

But can you explain how the conveyor manages to exert force to aircraft to stop it from moving, through a set of wheels that are freewheeling under the aircraft? Other than the rather minor (compared to aerodynamic drag, for example) force exerted via the higher than normal rolling resistance of the wheels and friction at the bearings (since they are spinning at twice the speed they would on a normal takeoff).

Do note that the question does not say that the conveyor would be trying to keep the plane still - it isn't. It's merely running to the opposite direction at the same speed the plane is moving, making the plane's wheels spin twice as fast.

Yes, the plane needs to move to take off - well, in relation to air at least. And it will move, just like it would on a regular runway. It will accelerate bit slower, due to the reasons I mentioned, but it will take off.
 
To Jari and Otokoshi,

You have to remember that thrust is a vector quantity, with both speed and direction. If the conveyor speed is exactly matched to the aircraft speed but reversed direction, then in fact the two vectors (aircraft and conveyor) cancel each other out. Bernoulli's principal, which is what governs the "lift" of aircraft wings, states that the airflow has to be significantly different between two sides to produce the "lift" nessesary to allow an object flight. In most cases, an aircraft requires 100 (for small)- 175 (cargo/passnger) miles per hour forward air speed (FAS) to acheive enough airflow/lift to maintain flight.  However, on a conveyor system like you suggest, you cancel out the forward direction needed.  Discounting the airflow to/from the engines, the aircraft would experience no more airflow over the craft then if it were standing still, hence it could not "take off".
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I believe Jari might have already explained it well enough, but I'll just beat this to the ground.  :-D

As I said earlier, the plane my not be theoretically moving on the ground, but the air is still circulating as a normal takeoff procedure.  Whatever the aircraft's FAS and so on is comes down to simple physics.  As I mentioned earlier, it may not be moving on the ground, but the potential energy gained by the plane will achieve lift capable for takeoff.  This is the simplest explanation I could find.

Kinetic energy is the energy by virtue of the motion of an object. It is defined as the work needed to accelerate a body from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that velocity.
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With the free rolling wheels of the plane, the only negative work being done is the slight friction of the wheels contact with the belt.  The air would act as normal, the airflow would remain the same going to/from the engines...unless this is done in a vacuum of course.  :-D

Edit:
I hope this all makes sense, I'm still recovering from the New Years party I threw. *grabs more Aspirin*
 
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As I said earlier, the plane my not be theoretically moving on the ground, but the air is still circulating as a normal takeoff procedure.  Whatever the aircraft's FAS and so on is comes down to simple physics.  As I mentioned earlier, it may not be moving on the ground, but the potential energy gained by the plane will achieve lift capable for takeoff.  This is the simplest explanation I could find.
Click to expand...
Edit:
I hope this all makes sense, I'm still recovering from the New Years party I threw. *grabs more Aspirin*
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Heh. Actually, I do believe that it indeed is moving on the ground. Or relative to the ground - if one wants to nitpick that it's actually on a conveyor and not on ground. :P

I agree with Ooine's view:

If you were watching from the sidelines, the take-off roll would look identical to any other that you may have seen.  The only difference being the rate at which the wheels are spinning.  They would be spinning approximately twice as fast.
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And since the air is assumed to be still (relative to that very same ground) or at least not moving more than it would due to regular wind - which planes have to deal with in on daily basis, the plane is also moving through air, thus eventually gaining sufficient airspeed for takeoff.

Hmmm... perhaps it would help if I'd use numbers - all of these are relative to ground, or a stationary observer, if you wish:


  • plane moves 150 kts one way
  • belt moves 150 kts the other way
  • air is still
  • thus plane moves 150 kts relative to air
  • and 300 kts relative to the belt
 
The thing is, the kinectic force is being transfered. The aircraft is pushing down on the conveyor and forward. The conveyor is push up towards the aircraft and back. A little ASCI art might show this better:

                                      <=======Aircraft
                                                           ||
                                                           ||
                                                           \/
                                                           /\
                                                           ||
                                                           ||
                                                      Conveyor=======>

Normally, the aircraft is only dealing with the up-down forces (along with inertia) so it can direct it's energy to forward momentum. However, this shows that an opposite running conveyor would, in fact, negate that forward momentum. Unless the conveyor slows down or stops, the plane has a net sum of 0 mometum for forward motion. Therefore, it is ,in a sense, standing still in relation to it's surroundings (except the conveyor itself). It would be just like being on a ten-speed bicycle on top of a treadmill. Do you think you would feel a breeze if it was running full speed? No, because you are not moving in relation to the room, only in relation to the treadmill surface. And, in this case, the aircraft is not moving in relation to it's surroundings, merely in relation to the conveyor.
 
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