Can a Plane on a Treadmill Take Off?

wrote)
[snip]
The airplane works in the air whether or not the wheels are on the ground.


BUT...the AIRplane isn't in the AIR yet!

Gravity, wheels, belt, zero airspeed. Once the wings get some lift under
them, sure...

But just like a plane sitting on the tarmac, our plane is still heavy enough
to rest on those happy little spinning wheels.

It's not a frigg'n blimp!


Montblack 83.7
More from the double-diget crowd.

Once again: assume the airplane is moving at 60 miles an hour to the
north. The tread mill then by definition is turning so that the belt is
going 60 miles an hour to the south. The wheels are turning as though
the airplane was going 120 miles an hour on the runway. If 60 is enough
airspeed for this airplane to lift off, the pilot just needs to apply
back pressure to the yoke. It's a question nicely phrased to make some
of us draw the wrong inferences: that's how come this is the 197th
response, and I think it's correct but then again others who don't
agree think they are correct as well. They have a right to be wrong
:-)!

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??

BUT...the AIRplane isn't in the AIR yet!

Ok, there is some friction involved.

Let's say that the airplane will take off in 1010 instead of 1000 feet,
ok?

Piero