Can a Plane on a Treadmill Take Off?

As posed by the question, I agree that the belt is a distracter. However, it
is not irrelevant as the thrust applied to achieve flying speed must
overcome not only the normal tire rotation friction but twice that. That the
plane will fly presumes that there is enough excess thrust to do just that.

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Travis
"Morgans" wrote in message
...

"alexy" wrote

Reread the stated problem:

"a conveyer belt that moves in the
opposite direction at exactly the speed that the airplane is moving
forward."

All it is, is a trick question, aimed at testing your reading and
comprehension ablility. The plane moves off in exactly the same manner as
on a regular runway. The moving belt is a distracter.
--
Jim in NC

On Sat, 04 Feb 2006 16:10:16 GMT, "Travis Marlatte"
wrote:

As posed by the question, I agree that the belt is a distracter. However, it
is not irrelevant as the thrust applied to achieve flying speed must
overcome not only the normal tire rotation friction but twice that. That the
plane will fly presumes that there is enough excess thrust to do just that.

So long as the bearings don't melt or seize, there's not much (if any)
extra "rotation friction" at double the normal takeoff speed.
Ron (EPM) (N5843Q, Mooney M20E) (CP, ASEL, ASES, IA)

However the airspeed-vs-wheel-rotation argument works out,
I think it is worth pointing out that all this time on the treadmill
should result in a _significant_ loss of weight for the aircraft,
resulting in more useful load.