The airplane/treadmill problem seems to assume that the wheels provide thrust for a plane that is trying to take off. They don't. The thrust is provided by the jets (or propellers) against the air, not against the runway. Spinning the wheels of an airplane doesn't result in air flow over the wings. A plane becomes airborne due to the Bernoulli effect of airflow over the wings, not because of spinning wheels, or any other motion with respect to the ground. The arguments that I've seen offered about this so-called problem seem to completely lack understanding of how aerodynamic lift actually works.
Because the jets exert thrust against the air, and not against the runway, the fact that the runway is a treadmill - assuming that could actually work - would be to hold the plane stationary in the air, and there would be no resulting airflow over the wings, and thus no resulting lift. The blowing of the jets (or propellers) would have all the resulting upwards lift of a box fan.
Actually, the more I think about it, I'm not at all sure what assumptions folks are making. But they don't involve any understanding of lift. If the wings don't move relative to the air (or vice versa) there will be no lift.
No fence riding. Will it take off or not?
You're right (though I'm not sure how much of a jet propulsion is rocket (F=(dm/dt)*v + m*(dv/dt)) and how much is moving air), though there are other effects to consider
1) the air at the surface of the treadmill is moving about the same speed as the treadmill, but it falls off with height (I don't know what form this falloff takes; maybe exp?) So if the plane is stationary wrt the greater air mass (i.e. moving wrt the treadmill at -v_{treadmill}} it will probably have airflow over the wings due to the air moving with the treadmill.
2) If it's stationary wrt the treadmill (i.e. moving with v_{treadmill} wrt the air) then is the velocity enough to take off? Is there a range or ranges of velocities such that the plane can fly backwards?
3) If it's moving wrt the treadmill, how much angular velocity can the wheels take before they fly apart and the airplane gets *forced* to move at v_{treadmill}? :)
I am with Rich. Why is this even a question? Do people think that the wheels of a plane are what cause it to taxi down the runway? If that were so, then the discussion would have some content. But jet planes taxi, take off, and fly because of the engines, not the wheels. They have little taxi trucks to push them around for precision movement. So, yes, the plane would take off.
Two corrections, though. Jet planes don't provide propulsion because the exhaust pushes against the air. Rather, it's a simple Newtonian action-reaction mechanism.
Also, the Bernoulli effect is less important in causing lift than most people asssume. Think about this: How can a jet plane fly upside-down? (This would be a bit awkward in a commercial jet, but fighter jets do this all the time at air shows.) The main lift is actually caused by the angle of the wings; they are not horizontal. Stick your hand out of a car window (while a passenger in a moving car) and rotate your wrist to see what happens.
Personally, I think that the only effect on the treadmill would be to turn ever so slightly in the direction of the plane's movement due to the friction of the tires as they are dragged forward by the jet engines. That assumes that the friction/resistance of turning the belt is less than the friction of the tires on the belt. If not, the belt doesn't move whatsoever.
I read another version of this that explains it much clearer:The Straight Dope: Airplane on a Conveyory belt. It actually will take off for one reason and one reason only: The conveyor belt would not succeed in keeping the plane in place - it would still move forward because the engines would provide throust moving the plane forward. The wheels would speed at approx. twice the aircraft's speed as in the problem it says the belt goes at the same speed in the opposite direction. Basically the wheels would be traveling twice the distance of the plane. It still takes off, and it still moves on down the runway, even though the runway tries to keep it from doing so.
I'm confused by your 2nd paragraph:
"Because the jets exert thrust against the air, and not against the runway, the fact that the runway is a treadmill - assuming that could actually work - would be to hold the plane stationary in the air, and there would be no resulting airflow over the wings, and thus no resulting lift. The blowing of the jets (or propellers) would have all the resulting upwards lift of a box fan."
You seem to be saying that the jet would be held stationary and therefore not take off. But clearly it would. The plane would move, treadmill or no. Both groundspeed and airspeed would increase exactly as they would if it were a normal runway. The only difference is that the wheels would be spinning faster on the treadmill than they would on a solid runway.
There's no way of keeping the jet in place as long as its wheels are free to rotate. Of course these are real wheels, so there will be small amounts of friction "pulling" the plane back, so the plane will need more than its usual thrust in order to take off, but it really shouldn't be much more.
The only way for this to work in the way that most people imagine this in their heads, is for the plane to be tied in place with a big rope to a fixed point on the sides or back of the treadmill. Then it would not be able to take off or move at all. The treadmill would move under it, but the plane would sit there and no lift would result. But once the rope is cut, the plane would start moving and eventually take off.
It's interesting that the catapults on aircraft carriers are basically the same thing but in reverse, with one critical caveat: the plane is firmly attached to the catapult, and it's not just sitting on top of it with its wheels. If it were so, the catapult could shoot off, but the plane would just sit there because the catapult rolled right under the wheels. But since the catapult is actually attached to the plane, the plane moves with the catapult/treadmill and thus takes off.
Basically the lesson here (as you seem to be saying) is that a "treadmill" is irrelevant to an airplane, despite its relevance to walkers, runners, horses, cars, trains, and other means of transportation which are motivated by pushing backward against the ground. A true "airplane treadmill" would basically be a wind tunnel. And yes, once the wind got strong enough, the plane would lift off.
Boing Boing links to this post from Cecil Adams. Pretty good I thought.
http://www.straightdope.com/columns/060203.html
This is totally bogus. The question just shows how confusing things get when you have an impossible condition.
Assuming a free spinning wheel with a frictionless axis it is not physical possible for the treadmill to go as fast as the wheel. Why is this you ask? Simple what is the speed of the wheel? It is not some random number that the treadmill equals.
Assuming a free spinning wheel with a frictionless axis on a treadmill spins at the same speed as the treadmill. If you add a force it spins the speed of the treadmill plus the velocity due the force. In our case that’s
Vw = Vt + (Fe/Mp)*t
Vw = Velocity of wheel
Vt = Velocity of tread
Fe = Force due to engins
Mp = Mass of the plane
t = Time
Our given condition is that the velocity of the treadmill equals the velocity of the wheel. So subbing in subbing in the velocity of the treadmill VTfor teh velocity of the wheel
Vt = Vt + (Fe/Mp)*t
THIS CANNOT BE TRUE such that Fe or t not equal to 0 which they are not in the question. A number plus another number cannot be equal to itself.
It is that an object in contact with the ground cannot move if the ground moves back just as fast without slipping. But despite what the question says that is not possible with a free spinning wheel on a treadmill. If there is a force applied the wheel will move faster.