I think the simple answer to the airplane on a treadmill problem is that once you understand what the problem is REALLY saying, the problem becomes uninteresting and trivial. If the problem wasn't intended to give the impression of a stationary (with relation to the surrounding air) plane taking off, no one would've cared in the first place and there would be no discussion. No one is getting confused by math or aerodynamics or physics or anything. They're getting confused by a problem intended to be confusing.
There's an obvious intuition that the author seems to be missing with the 'two aces' paradox.<p>If you have a hand with two aces, you have two chances of getting the "Ace of Spades." Having a hand with two aces doesn't make it any more likely to have "an ace" than having a hand with one ace does.<p>This pretty much explains the phenomenon.
Airplane on the treadmill will take off huh?<p>You know those big giant things sticking out from the side of the fuselage? They're called wings. Air flowing over the and under the wings creates different pressure; hence, lift. There is a magic speed for each aircraft that over which the lift will be sufficient to move the entire aircraft vertically. For a 747, this is 180 mph/290 km/h at 80,000 lbs. That's pretty fucking fast.<p>Any movement relative to the ground that is under that speed will not result in flight. This is true even if the aircraft is actually in the air.<p>If, as the problem states, the treadmill can equal the forward thrust of the engines, no lift will be produced by the wings, ergo, it will stay right where it is.<p>This is why aircraft have an airspeed and a ground speed. It's also why stalling is a major concern.<p>EDIT: the fact that finding a treadmill that can act in a manner as suggested by the problem is impossible is another issue completely.<p>EDIT 2: Think people: They have breaks on aircraft wheels do they not? (They do) You can sit on a runway with the engine revved equal to your break ability and <i>not move</i>. If the engine in the problem can outperform the treadmill, you've solved a different problem.
My issue with this problem was that it was presented rather poorly to me the first time. Instead of saying that the treadmill matched the negative of the velocity of the plane relative to the ground, the questioner instead stated that the treadmill "matched the speed of the wheel." I took this to mean that the treadmill matched velocity of the wheel as would be given by a conventional speedometer.<p>Of course, this is prima facie ridiculous. Consider, at any nonzero speed, the speed of both the treadmill and the wheel would quickly approach infinity. Given that the speedometer speed would essentially be the velocity of the wheel + the tangential velocity of the wheel touching the treadmill, the entire thing would just blow up (remember, non-slip == velocity of the wheel == velocity of treadmill). Therefore, I basically answered c) it's a stupid problem and the plane wouldn't move.<p>Of course, if you define it this way, than it is obvious that the plane would move.