Can a Plane on a Treadmill Take Off?

Taxi is just nomenclature for the airplane moving along the ground.
"Takeoff run" would be more correct I guess, but in this case things
are so weird, as it is ambiguous whether the plane is going to takeoff
or not.

If the treadmill is stationary and the belt speed is equal
to the required take-off speed, the airplane will have zero
airspeed if it is "moving" in relation to the belt, the
airplane is moving, the prop has thrust and is balancing the
rearward movement of the belt. The tires are rolling, but
the airplane is stationary and there is no airspeed or lift.



--
James H. Macklin
ATP,CFI,A&P

--
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But government sees it as an obstacle to be overcome.
some support
http://www.usdoj.gov/olc/secondamendment2.htm
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"Doug" wrote in message
oups.com...
| Taxi is just nomenclature for the airplane moving along
the ground.
| "Takeoff run" would be more correct I guess, but in this
case things
| are so weird, as it is ambiguous whether the plane is
going to takeoff
| or not.
|

Jim Macklin wrote:
If the treadmill is stationary and the belt speed is equal
to the required take-off speed, the airplane will have zero
airspeed if it is "moving" in relation to the belt, the
airplane is moving, the prop has thrust and is balancing the
rearward movement of the belt. The tires are rolling, but
the airplane is stationary and there is no airspeed or lift.

Let's take this to a logical extreme. The purpose of a wheel is to
reduce friction, right? (Well, excluding steering and braking, since we
aren't using the brakes here) Anyways, let's now assume that the
airplane is sitting on the conveyor belt, and there is no friction
between it and the belt. For all intents and purposes, you now have an
antigravity device as your landing gear. Now run the engine up. If
there is no friction between the airplane and the belt (and
consequently no way to transmit force), how is the belt going to keep it
stationary? Remember, sum of forces=mass*acceleration, and the sum of
the forces in the horizontal plane is now mass*acceleration=thrust-drag
(where drag is a function of airspeed squared). No force from the
conveyor belt.

Now let's put the wheels back on. Certainly, if a wheel's purpose is to
try and reduce friction as much as possible, you aren't going to
suddenly have some wheels that drag on you with as much thrust as your
prop exerts...

if it comes down to it, I'll write a Matlab simulation of this, and show
the results to everyone.

Taken to the extreme, it might be considered ambiguous but that is only
nit-picking the puzzle. The general answer to the riddle is: if the plane
has enough excess thrust to overcome the additional drag of friction of the
wheels turning at twice the liftoff speed, the plane will fly. It will
accelerate slower and require a longer run due to the excess friction, but
it will fly.
--
-------------------------------
Travis
"Doug" wrote in message
oups.com...
Taxi is just nomenclature for the airplane moving along the ground.
"Takeoff run" would be more correct I guess, but in this case things
are so weird, as it is ambiguous whether the plane is going to takeoff
or not.