Ok, so I\'m sure most of folks agree that it takes some wind velocity to generate the lift the plane needs. I think the initial question was just poorly posed. Here\'s why-
When the original post says that \"the treadmill runs in the opposite direction at the same speed the plane is moving\". Is the speed relative to the treadmill or to an inertial reference frame?
Anyone who has taken any elementary mechanics course knows that:
V(p)=V(p/t)+V(t) (vector equation)
where the V\'s are vectors, and (p) denotes absolute velocity of plane, (t) is absolute velocity of treadmill belt, and (p/t) is the velocity of the plane with respect to the belt.
If the original question implied that the speed of the plane with respect to the treadmill was the same as the speed of the treadmill belt (as viewed by an inertial observer) then the plane would not be moving in an inertial frame. Call the speed \'u\'.
That is, V(p/t)=u (in the positive x direction), and V(t)=u (in negative x direction). Think about if you were a particle stuck in the treadmill as you are \'dragged\' backwards (negative x)- the plane would appear to be moving away from you with speed u in the positive x direction.
Adding these vectors gives a vector sum of zero. Thus, as viewed by an inertial frame (someone looking at the treadmill from the ground), the plane is not moving. If the air is stagnant (again, from the inertial frame), then the plane is not moving relative to the stagnant air- thus producing no lift.
If the plane had a V(p/t) that was greater than V(t), then the vector sum would be positive, and the plane would move (as viewed by an inertial observer). Then, if the speed were great enough, you can get lift.
In conclusion- WITH RESPECT TO THE BELT, the plane would have to go it\'s normal takeoff speed (i.e. with no treadmill) PLUS the speed of the treadmill in order to get in flight.