Can a Plane on a Treadmill Take Off?

Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??

darthpup wrote:
Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??

Wind has nothing to do with it. The airplane will accelerate and move
down the treadmill just as it would a stationary runway. It cannot feel
the treadmill at all. The wheels can, but the wheels spin independently
of the thrust generated by an airplane.

He suggested a wind that is dynamic and tied to the speed of the conveyor
(and therefore also tied to the speed of the plane).

The plane can feel the conveyor - wheels are not frictionless. The friction
is not even insignificant. An amplified example would be trying to take off
in slushy snow. I think you will agree that the plane will feel that drag.

Back to the original puzzle - yes, the plane will accelerate and takeoff but
it will be a longer takeoff roll to overcome the increasing friction of the
wheels turning at twice the normal speed.

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Travis
"cjcampbell" wrote in message
oups.com...

darthpup wrote:
Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??

Wind has nothing to do with it. The airplane will accelerate and move
down the treadmill just as it would a stationary runway. It cannot feel
the treadmill at all. The wheels can, but the wheels spin independently
of the thrust generated by an airplane.

Interesting question. Are you suggesting a wind machine that is capable of
creating a wind that matches the conveyor?

Let's start with no wind. The plane must achieve lift off speed relative to
the stable air mass to take off.

airspeed = Vlo
Conveyor moving = - Plane motion relative to ground
Conveyor moving = -xVlo (relative to ground)

airspeed = Vlo = Plane motion relative ground + headwind = xVlo + 0Vlo;
solve for x = 1
Wheels spinning = Plane motion - conveyor motion = xVlo - -x Vlo = 2 Vlo

Plane will liftoff at an airspeed that matches its groundspeed of Vlo. The
conveyor is moving backward also at Vlo. The wheels are spinning at twice
that speed.


Now with a headwind. The plane must still achieve lift off speed relative to
the air mass which is now an accelerating headwind.

airspeed = Vlo
Conveyor moving = - Plane motion relative to ground
Conveyor moving = -x Vlo (relative to ground)
Headwind = Conveyor moving
Headwind = - x Vlo (relative to ground)

airspeed = Vlo = Plane motion relative ground - headwind = x Vlo + x
Vlo; x = 1/2
Wheels spinning = Plane motion - conveyor motion = 1/2 Vlo - -1/2 Vlo =
Vlo

The plane will be able to lift off at 1/2 the ground speed than under normal
circumstances. The wheels are spinning at twice that speed which is just
Vlo. It would be feasible to construct both such a conveyor and the wind
machine.



Now with a tailwind. The plane must still achieve lift off speed relative to
the air mass which is now an accelerating tailwind.

airspeed = Vlo
Conveyor moving = - Plane motion relative to ground
Conveyor moving = -x Vlo
Tailwind = - Converyor moving
Tailwind = x Vlo (relative to ground)

airspeed = Vlo = Plane motion relative ground - Tailwind = x Vlo - x
Vlo; ******** No solution ********

With this simple formula, the plane will never take off but will continue to
accelerate (until the wheels disintegrate) since the thrust of the engine is
acting on the air mass which is accelerating as a tailwind. Very much the
dog chasing his tail.

Assumptions:
As with most things in life, we need a point of reference. Almost any
will do but I chose the ground upon which the conveyor belt is built and we
are standing watching. It is from this reference that the conveyor belt's
velocity (backwards) is derived.

I arbitrarily chose the positive sign of velocity in the direction of
the plane motion.

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Travis
"darthpup" wrote in message
oups.com...
Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??