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#11
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I've thought about this problem for who knows how long and I recently did a presentation which involved explaining the answer to this question. Let me attempt to give a clear answer and put this issue to rest.
The answer is: Yes. Why? Because a plane generates speed (thrust) from its engines; it does not step on the gas so that the wheels turn. As a matter of fact, the wheels surve no purpose other than to reduce friction. First, imagine this scenario: The speed of the wheels is zero. In other words, the runway doens't move backwards (it moved at speed zero) and the wheels don't move either. The plane turns its engines on. If they can provide enough force, the plane will just begin to slide on the runway until it slides fast enough to take off. From this, you can see where I'm going. If the wheel speed was, say 100 mph, the runway would move back at 100 mph. But this doens't mean that the plane itself couldn't move faster. If the plane travels at 250 mph, the wheels basically compensate for 100 mph of that and the rest is just from sliding. So, basically, yes the plane can take off if its engines are powerful enough to overcome the friction of the "still" wheels. |
#12
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Note that the way this is phrased is what causes a lot of the confusion. The conveyer belts can't move at the same speed as the wheels in this case. As stated above, the wheels on a plane don't have any power applied to them to move when taking off, they just spin.
A simple comparison would be you holding a toy car on a treadmill. Assuming friction is negligable, the treadmill can go as fast as it wants and the wheels of the car will simply spin in place. Now move the car foward, the wheels of the car go faster than the treadmill and it is impossible for the treadmill to go as fast (in theory the treadmill would be going infinitly fast since it would keep trying and never catch up). This is basically the same deal with a plane, except instead of your hand the engines on the plane provide the foward momentum so the plane will still move foward and take off. |
#13
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in this case the wind isnt traveling over the wings the wind is staionary so the plane will not rise. if the plane was held stationary and the wind was driven over the wings at liftoff speed the plane would rise verically.
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#14
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[ QUOTE ]
in this case the wind isnt traveling over the wings [/ QUOTE ] Pretty sure the jet engines would still create wind traveling over and under the wings. The motion of the converter belt would also create some mininal lift. |
#15
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As long as the entire weight of the plane is on the wheels, and the tires don’t skid, the plane cannot move forward.
In fact, the wings don’t even have to be there until some significant forward motion is achieved. |
#16
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As long as the entire weight of the plane is on the wheels, and the tires don’t skid, the plane cannot move forward. In fact, the wings don’t even have to be there until some significant forward motion is achieved. [/ QUOTE ] This would only be true if the pilot were applying the brake while attempting takeoff (which would define him as an idiot, I suppose). |
#17
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The tires will skid if the wheels are turning at a different speed than the plane is moving.
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#18
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The wheels spin on an axle with very low friction (much lower than the tires against the belt). Thus, the tires spin with [are driven by] the belt, in either direction, at any speed, but the plane is still pulled forward by the engines.
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#19
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Since the plane is attached to the wheels, for ever little increment the plane tries to move forward, the conveyor belt will speed up and pull it back, assuming the tires don’t skid.
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#20
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Since the plane is attached to the wheels, for ever little increment the plane tries to move forward, the conveyor belt will speed up and pull it back, assuming the tires don’t skid. [/ QUOTE ] Are you just [censored] with me? The tires spin against the belt, having no effect on the movement of the rest of the plane. This is so elementary I'm having a hard time believing that you don't get it. |
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