Simple solution:
If the speed of the treadmill matches the speed of the wheels, the plane can not take off.
If the speed of the treadmill matches the speed of the plane, the plan can take off.
If the speed of the treadmill matches the speed of the wheels, the plane can not take off.
If the speed of the treadmill matches the speed of the plane, the plan can take off.
Oh, I guess I missed something. I thought that the treadmill speed was supposed to directly match the plane's forward motion. The plane can take off if that's the case. Dullard's post doesn't mention anything about tethering, unless it has a meaning I don't understand.
mobobuff
Lifer
- Apr 5, 2004
- 11,099
- 1
- 81
I can see how it's going to work...
They set it up the best they can, the plane takes off every time, they explain that it's a hypothetical scenario and they can't satisfy it fully with their contraptions because they cannot accelerate the belt quickly enough. And then after all is said and done, they'll explain that the plane would still take off even if the treadmill could go as fast as it wanted.
Maybe they'll make a cool 3D demo.
They set it up the best they can, the plane takes off every time, they explain that it's a hypothetical scenario and they can't satisfy it fully with their contraptions because they cannot accelerate the belt quickly enough. And then after all is said and done, they'll explain that the plane would still take off even if the treadmill could go as fast as it wanted.
Maybe they'll make a cool 3D demo.
I don't understand.
Theoretically, if you hold a plane stationary on a treadmill and the treadmill moves backwards at 100, the wheels will be moving at 100 in the other direction. If the treadmill moves at 100 and the plane moves forward at 100, the wheels move at 200. What's wrong with that?
no actually you are right, except in the "treadmill matches the speed of the wheels" scenario, its only correct because the only time that the treadmill can match the speed of the wheels is at zero. which makes it more of a trick question then a practical one.
In the treadmill = wheels scenario, the only way for there to be a net of zero would be if the plane was held stationary by a cable or something, right? There is no way for there to be net zero motion if the planes engines move and the treadmill = wheels, correct?
It's impossible to escape this thread since it is on a treadmill that matches your forward motion exactly. Which begs the question, is stop, drop, and roll still feasible?
your assuming the wheels are perfectly frictionless?
it was a joke anyway. i was referencing the doomed thread and goading argument in the same breath
but if you insist...
if wheels have friction and belt can move infinitely fast then the plan will not move.
if they are frictionless wheels then the plan will take off regardless of whether the belt is magical or not.
however both belt and wheels being of this world, yes the plane will take off.
i never understood the problem with understanding this. it really boils down to... does the plane move, or can you stop a plane on wheels from moving by putting it on a treadmill. only in one strange outlandish situation is this possible.
Why would the conveyor belt's speed be a factor in the plane taking off? As long as the wheel/axle assembly can take the increased speed the speed of the conveyor has nothing to do with the speed of the air going over the wings and the wheels don't provide propulsion.
exactly. i'm not saying just the belt's speed, i'm saying the belt's speed ability in relation to the wheels ability to cope with that speed.
any friction the wheels have (internally axle,hub) will allow the treadmill's movement to be exerted against the plane.
just to clarify the plane will take off. barring a completely unrealistic scenario.
completely unrealistic... kind of relative when your talking about a plane on a conveyor belt trying to take off.
NOW THAT's Funnay!Originally posted by: RapidSnail
It's impossible to escape this thread since it is on a treadmill that matches your forward motion exactly. Which begs the question, is stop, drop, and roll still feasible?
Cerpin Taxt
Lifer
- Feb 23, 2005
- 11,940
- 542
- 126
Yes, but not because of some physical impossibility, but instead because of a rather trivial incoherency in the premise. In other words, there's nothing stopping the plane from moving forward, but as soon as it does, it is no longer true that "treadmill = wheels" with regard to speed. In still other words, it's impossible to create a feedback loop wherein the "speed of the treadmill = the speed of the wheels" under every circumstance.
There's nothing stopping the plane from moving forward, and thus accelerating the wheels, but once that happens you initiate an infinitely recursive self-oscillation where the treadmill must accelerate to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels, which in turn accelerates the wheels even more, which then demands that the treadmill accelerates to match the new speed of the wheels...
Get the idea?
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