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12/07/2006 10:46:44 PM · #251 |
I came to this late and skipped over pages 3-9, but there is one thing which I haven't seen addressed regarding the friction of the wheels with the treadmill, which answers the question.
If the wheels have infinite friction with the treadmill, such that there will be no slipping, then the plane will be stationary relative to an observer not on the treadmill, there will be no air flow over/under the wing and the plane will remain on the ground.
If there is not infinite friction, then the wheels can be dragged forward by the thrust of the plane and there will presumably be enough forward movement to allow air to flow over/under the wings creating thrust allowing takeoff.
Given that infinite friction is unlikely, fasten your seatbelts for takeoff.
(To look at this in a similar situation, imagine if there were no treadmill but the wheels were locked with brakes on... the plane would still be able to take off as the tires can skid forward without rotating. also similar as to how planes with floats that take off on water work) |
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12/07/2006 10:50:17 PM · #252 |
Originally posted by Bear_Music:
I can't believe I've missed this thread through so many pages :-) I've never seen so many people getting so far off track on anything, I don't think. I haven't read every single post (I got bogged down in giggling) but of COURSE the plane can take off. If my takeoff speed is 60 mph (about right for a small plane) then when I am traveling through the air at 60 mph the treadmill would be running backwards at 60 mph and the wheels would be "spinning at 120 mph", but the plane is still moving forward at 60 mph and it will lift off. The wheels are irrelevant, they just exist to provide a (relatively) friction-free connection to the ground until lift takes over and separates me from the ground.
Right?
R. |
Yes, you are right.
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12/07/2006 10:52:28 PM · #253 |
Originally posted by asimchoudhri:
If the wheels have infinite friction with the treadmill, such that there will be no slipping, then the plane will be stationary relative to an observer not on the treadmill, there will be no air flow over/under the wing and the plane will remain on the ground.
If there is not infinite friction, then the wheels can be dragged forward by the thrust of the plane and there will presumably be enough forward movement to allow air to flow over/under the wings creating thrust allowing takeoff.
Given that infinite friction is unlikely, fasten your seatbelts for takeoff. |
If there was infinite friction, nothing would be going anywhere. And there is no reason the plane would have to drag the wheels forward, they would simply roll that way. So all you have in the end is the minimal rolling friction in the wheels which is easily overcome by the engines.
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12/07/2006 11:00:24 PM · #254 |
Originally posted by SamDoe1:
Originally posted by asimchoudhri:
If the wheels have infinite friction with the treadmill, such that there will be no slipping, then the plane will be stationary relative to an observer not on the treadmill, there will be no air flow over/under the wing and the plane will remain on the ground.
If there is not infinite friction, then the wheels can be dragged forward by the thrust of the plane and there will presumably be enough forward movement to allow air to flow over/under the wings creating thrust allowing takeoff.
Given that infinite friction is unlikely, fasten your seatbelts for takeoff. |
If there was infinite friction, nothing would be going anywhere. And there is no reason the plane would have to drag the wheels forward, they would simply roll that way. So all you have in the end is the minimal rolling friction in the wheels which is easily overcome by the engines. |
Not really.
First let me address the friction issue: you want zero friction on the axis of rotation for the wheel, but without friction between the wheel and the ground you get sliding... friction between the ground and the wheel is needed for rolling. Normally, this is what you want (the lack of friction between the ground and the wheel is what happens when your brakes lock)
In this case, the plane cannot move forward unless there is some slipping. If slipping is allowed in this scenerio (not specifically stated in the original parameters), the plane can take off. If the original scenerio was that there was no slipping and that all movement of the plane resulted in rotation of the wheels, then the treadmill can keep up and the plane cannot take off. |
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12/07/2006 11:04:13 PM · #255 |
It is entertaining to keep finding new people who don't get it, jumping in at the end of the thread ;) Funnier still that this was discussed in some of the links a couple of days ago that this keeps happening.
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12/07/2006 11:09:29 PM · #256 |
If you look at it from the opposite perspective, it is specified that the treadmill will match the speed of the plane, moving in "reverse" to the plane's movement. If it were NOT possible to take off, then by extension it's not possible for the plane to move, and therefore the treadmill will never have to spin...
R.
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12/07/2006 11:11:40 PM · #257 |
Originally posted by Gordon:
It is entertaining to keep finding new people who don't get it, jumping in at the end of the thread ;) Funnier still that this was discussed in some of the links a couple of days ago that this keeps happening. |
1- entire thread was started less then 36 hours ago, so nothing was discussed a couple of days ago
2- i've since read all of the pages and i didn't think the issue of friction between the wheels and the treadmill were clearly addressed. i will re-read all 11 pages to make sure that it wasn't addressed appropriately |
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12/07/2006 11:12:08 PM · #258 |
| I wonder if this thread could become as long as say "dpc word assosiation game"... Seems like this one is a winner. |
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12/07/2006 11:16:26 PM · #259 |
Originally posted by Bear_Music:
If you look at it from the opposite perspective, it is specified that the treadmill will match the speed of the plane, moving in "reverse" to the plane's movement. If it were NOT possible to take off, then by extension it's not possible for the plane to move, and therefore the treadmill will never have to spin...
R. |
ok, so i saw that it said the treadmill was opposite the plane's speed... in that case, the plane can take off regardless. i was approaching it that the treadmill matched the wheel rotation speed, which i why i said the plane would not take off in the absence of sliding. i still think that holds true if the treadmill was matching the rotational speed of the wheel, but that is not what the parameters are here.
sorry for the wasted ascii characters... |
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12/07/2006 11:18:44 PM · #260 |
Originally posted by asimchoudhri:
First let me address the friction issue: you want zero friction on the axis of rotation for the wheel, but without friction between the wheel and the ground you get sliding... friction between the ground and the wheel is needed for rolling. Normally, this is what you want (the lack of friction between the ground and the wheel is what happens when your brakes lock)
In this case, the plane cannot move forward unless there is some slipping. If slipping is allowed in this scenerio (not specifically stated in the original parameters), the plane can take off. If the original scenerio was that there was no slipping and that all movement of the plane resulted in rotation of the wheels, then the treadmill can keep up and the plane cannot take off. |
You are right in the first part, that's what we mean by the rolling friction. It has to be there for the wheels to roll, otherwise the treadmill would just be one huge sheet of ice.
I'm not sure what you mean by the second part of the post, could you explain it more?
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12/07/2006 11:19:55 PM · #261 |
Originally posted by asimchoudhri:
ok, so i saw that it said the treadmill was opposite the plane's speed... in that case, the plane can take off regardless. i was approaching it that the treadmill matched the wheel rotation speed, which i why i said the plane would not take off in the absence of sliding. i still think that holds true if the treadmill was matching the rotational speed of the wheel, but that is not what the parameters are here.
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No, not even then. The only scenario where the treadmill can sufficiently stop the plane from taking off is when it imparts an equal, but opposite force to counteract the acceleration of the plane (via F=ma)
as only a very small component of the treadmill movement results in backward force on the plane, the acceleration has to vastly exceed that of the plane to even start slowing things down. That is entirely different to matching the plane speed.
Message edited by author 2006-12-07 23:20:23.
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12/07/2006 11:26:50 PM · #262 |
Originally posted by SamDoe1:
[...]
I'm not sure what you mean by the second part of the post, could you explain it more? |
this is no longer relevent to this discussion as i was approaching the problem as if the treadmill speed matched the rotational speed of the wheels (angular velocity x radius) but in the opposite direction. were this the case, the only way for the plane to take off is for the planes velocity to be faster than the rotational velocity of the wheel.
for instance, if the wheel was rotating at 1 revolution per second with a circumference of 1 meter, the treadmill would be moving at negative one meter per second. if the wheel can slip, however, you could have a wheel moving at 1 revolution per second / circumfrence of one meter but the plane could be going forward at 2 meters per second.
anyways, it's easier for me to explain my thoughts with a diagram and two sentences than it would be with two pages of text... perhaps reaching this conclusion that a picture is worth a thousand words brings back the relevence to this forum. |
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12/07/2006 11:41:53 PM · #263 |
Originally posted by Gordon:
Originally posted by asimchoudhri:
ok, so i saw that it said the treadmill was opposite the plane's speed... in that case, the plane can take off regardless. i was approaching it that the treadmill matched the wheel rotation speed, which i why i said the plane would not take off in the absence of sliding. i still think that holds true if the treadmill was matching the rotational speed of the wheel, but that is not what the parameters are here.
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No, not even then. The only scenario where the treadmill can sufficiently stop the plane from taking off is when it imparts an equal, but opposite force to counteract the acceleration of the plane (via F=ma)
as only a very small component of the treadmill movement results in backward force on the plane, the acceleration has to vastly exceed that of the plane to even start slowing things down. That is entirely different to matching the plane speed. |
agreed... while i was approaching a different scenerio than was originally presented, the "infinite friction" statement was regarding an unrealistic extreme where there was no slippage... if it was matching the wheel's rotational velocity and there was no slippage, then by definition there would be no movement relative to an external observer. only movement relative to an external observer would represent air moving across wings (thus lift).
Edit: removed some confusing stuff...
ps: i'm out... again, sorry for wasting bandwidth...
Message edited by author 2006-12-07 23:54:36. |
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02/01/2008 10:34:21 AM · #264 |
Going to bump this thread cause mythbusters just did this experiment and the plane did actually take off! I thought their test methods, although a bit crude, were decent and that they did the experiment quite well.
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02/01/2008 10:54:31 AM · #265 |
The speed of the treadmill has no bearing on the thrust of the plane. The wheels do not propel an aircraft on the ground in the way that an automobile does. If a corvette where placed on a treadmill it would literally 'sit still'. Why you ask because the power, or thrust if you will, propels the vehicle via the ground ie "treadmill". The only way the plane could NOT take off is if you placed the aircraft in a wind-tunnel that could always match the, or counter-act against, the engine thrust.
Yes, the treadmill will have an effect on the plane, only in the fact that the wheels will be spinning at EXACTLY twice the speed of a normal take off.
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02/01/2008 11:29:22 AM · #266 |
THe ONLY thing an airfoil sees is airspeed. No matter how fast the wheels are turning, or how much friction, the air over and under the lift surface is what causes the aircraft to fly. The faster air over the upper surface causes higher pressure beneath the wing than above. The speeed in relation to ground, treadmill, or coconut cream pie means nothing.
I've seen a small plane takeoff from an aircraft carrier, heading into a strong wind. There was no need for catapult, in fact the aircraft was just barely rolling when it left the deck. A 15 knot vessel steaming into a 30 knot wind is just about enough. |
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02/01/2008 11:54:43 AM · #267 |
if the plane is stationery on a treadmill, wouldn't it just roll off the end when the treadmill starts?
If the jets or propellors are on then i don't see how it can take off because the thrust of the jets would be counteracted by the speed of the treadmill, so the jet remains at a standstill except with very very very hot wheels.
It's like when you do a horsepower test on a car, it sits in some traps that roll against it so it can't move but it's still generating power that would otherwise blast it through the wall. |
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02/01/2008 12:03:56 PM · #268 |
Originally posted by Tez:
...i don't see how it can take off... |
If you are truly interested in learning how, I recommend reading this thread or watching the reruns of the Mythbusters episode. Explanations abound in both locations. |
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02/01/2008 12:06:28 PM · #269 |
| I read the first 2 pages, then it seemed to get a bit repetetive. If the forces are equal then there won't be lift, it will take off (or fall off the treadmill) when one force over rides the other. |
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02/01/2008 12:06:57 PM · #270 |
| or else watch this video that demonstrates the answer! YouTube |
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02/01/2008 12:22:46 PM · #271 |
I have not read the other responses.
A car or a person on a treadmill stays in place because people and cars achieve forward thrust by pushing against the ground. Planes, on the other hand, achieve their forward thrust by pushing against the air.
Consider a car on a treadmill, in neutral, pointed downhill. Even if the treadmill matched the speed of the car, the car would still roll downhill. The wheels would just spin faster. as anyone who has driven on ice known, the speed of the wheels need not always match the speed of the car.
Now, apply that logic to a plane, ignoring the treadmill for a moment. Since the landing gear is not attached to a drive system, the landing gear is essentially always in neutral. The plane would be pushed forward by the force of the jet engines (or propeller) against the air. The wheels spin as a side effect of the plane's speed relative to the runway, and not as a means of propelling the plain forward. If the runway were to also start moving, the wheels would simply spin faster (or slower) to compensate. The added friction would have a slowing effect on the plane, but that effect would be minor.
One additional factor to consider is that with the runway moving at a speed equal and opposite the plane speed, that the landing gear would be under additional stress not normally encountered at takeoff.
With all those factors considered, my answer is that the plane would take off, assuming:
- The runway is long enough to handle a longer than normal runout distance.
- The added rotational speed of the landing gear does not cause the gear to fail due to exceeding the design speed.
~Terry
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02/01/2008 12:25:59 PM · #272 |
Originally posted by Tez:
I read the first 2 pages, then it seemed to get a bit repetetive. If the forces are equal then there won't be lift, it will take off (or fall off the treadmill) when one force over rides the other. |
Yeah, it does get repetitive. The question is a bit of a trick question in that almost everyone who weighed in that it wouldn't think of a plane like a car, except we aren't talking about a car here.
A car's forward movement is derived from turning it's wheels, essentially "pushing off" the ground so that if the ground were moving in the opposite direction at an equal speed the car wouldn't move. A plane, on the other hand, is driven by it's props (or compressor blades and combustion in the case of a jet) moving air. The wheels spin freely so the treadmill would have an almost immeasurable effect. All the asinine arguments about the drag on the bearings increasing with speed or the bearings seizing up, etc, are fatally flawed. In my Cessna I can get off the ground at around 46kts and I assure you that the wheels turn freely well beyond the 92kts they would experience on the hypothetical treadmill at liftoff.
Those who said they presume it to mean that the plane would be stationary are presuming things that weren't stated. To them I answer that *I* am presuming the plane in question is an Osprey which is capable of vertical takeoff so, yes, it would still fly. My presumption is every bit as valid as theirs.
If the plane were indeed stationary (and not a VTOL craft) then *of course* it wouldn't take off (absent a REALLY strong wind). The only thing that matters regarding the plane taking flight is it's movement relative to the air mass around it.
Jeez! :-)
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02/01/2008 12:31:54 PM · #273 |
Originally posted by ClubJuggle:
I have not read the other responses.
A car or a person on a treadmill stays in place because people and cars achieve forward thrust by pushing against the ground. Planes, on the other hand, achieve their forward thrust by pushing against the air.
One additional factor to consider is that with the runway moving at a speed equal and opposite the plane speed, that the landing gear would be under additional stress not normally encountered at takeoff.
With all those factors considered, my answer is that the plane would take off, assuming:
- The runway is long enough to handle a longer than normal runout distance.
- The added rotational speed of the landing gear does not cause the gear to fail due to exceeding the design speed.
~Terry |
Ha. Looks like we were making the same argument at the same time.
One minor flaw in your last point, though. The runway would not need to be *any* longer. The planes would travel roughly the same distance relative to the ground before achieving lift-off. In fact, the takeoff from the pilots perspective would be scarcely any different than normal. He would just feel like the ground was passing faster than normal.
Oh, and I would not get in a plane whose gear couldn't stand rotating twice the speed required for lift-off.
What a silly argument, huh? :-)
Message edited by author 2008-02-01 12:32:50.
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02/01/2008 12:45:16 PM · #274 |
[quote=good_ham] [quote=ClubJuggle] I have not read the other responses.
A car or a person on a treadmill stays in place because people and cars achieve forward thrust by pushing against the ground. Planes, on the other hand, achieve their forward thrust by pushing against the air.
Fair enough. Not beingv a pilot myself I don't know about the design specifications for landing gear, which is why I added those caveats.
I agree the difference in tolling resistance for a Cessna would be low, but what about, say, a loaded 747 with a 200 mph (320 km/h) takeoff speed? In that extreme case, design limits, rolling resistance, and runout distance might come into play. I do agree that the difference in runout distance would be marginal, but it technically would exist.
~Terry
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02/01/2008 12:54:33 PM · #275 |
Originally posted by Matthew:
The treadmill has a clever design and always matches the speed of the plane, but runs in the opposite direction.
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The key sentence. If the treadmill always matches the speed of the plane you could accelerate with a 1,000,000 pound thrust engine and never out run the treadmill. Therefore no airspeed. No flight.
It's interesting looking at all the replies and comparing some of them with their arguments on Global Warming/Climate change. |
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