Can a Plane on a Treadmill Take Off?

"Damian" wrote in message
...
Paul...dont look now, but that airplane is NOT flying off the ground until
the AIRSPEED is up...the treadmill is only moving the TIRES, that means
diddly squat to an airplane.

Damian, don't look now but Paul is exactly right (except for his rude
nature, of course).

The treadmill is irrelevant to the airplane's motion. If the airplane is
stationary on the treadmill, it's because it has a headwind the same speed
as the treadmill and enough thrust to fly into the headwind at the same
speed as the treadmill. Of course, the wheels will be turning on the
treadmill, but only because the treadmill is rotating them against the
air-based stationary nature of the airplane.

Without a suitable headwind for the airplane to fly into, the treadmill
would just push the airplane backward. Airplanes don't use their wheels for
transmitting power to forward motion (most don't, anyway ).

Pete

"The Flying Scotsman" wrote in message
ups.com...
Look..... for all you people that think that the plane will take off.
Whats the point in having CAT Launching systems on aircraft carriers
priced at billions of dollars a piece... they could just pop down to
wallmart and buy a treadmill and the aircraft will not need to use any
runway what so ever........

No, they couldn't. Or, put another way, a scenario in which an airplane
launching from a carrier could remain stationary on a treadmill would rely
not on the treadmill, but on the presence of a suitable headwind. Of
course, in the presence of a suitable headwind, the treadmill is not needed.
But that's the whole point to this trick question: the treadmill is a red
herring.

DONT BE STUPID....

You should think a little harder before throwing the "stupid" word around.

IT CANNOT BE DONE !!! IF IT CAN BE THEN SOME CLEVER BUGGER 50 YEARS AGO
WOULD OF DONE IT BY NOW..

plus, has anyone thought what will happen to that aircaraft if i does
manage to generate enough lift..... ITS GOING TO HAVE NO AIRSPEED,
stall and fall out of the sky like a brick.

The airplane can't generate lift without airspeed. It's absurd to claim
that it would have "no airspeed". If it has enough lift to fly, then by
definition it has airspeed.

Simple physics lads...

Yes, it is. But the physics only give you the correct answer if you apply
them correctly.

Pete

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.

Exactly right.

In the end we'd have to say it's a nicely phrased question. My first
instinct was to say the ariplane had zero speed relative to the ground
the moving belt is on, but that is NOT the condition the problem
stated.

So, under the usual circumstances (not having a significant tail wind,
for example) you'd lift off assuming the wheels are not going to self
distruct turning at twice their usual takeoff speed.

"Tony" wrote in message
oups.com...
Exactly right.

In the end we'd have to say it's a nicely phrased question. My first
instinct was to say the ariplane had zero speed relative to the ground
the moving belt is on, but that is NOT the condition the problem
stated.

So, under the usual circumstances (not having a significant tail wind,
for example) you'd lift off assuming the wheels are not going to self
distruct turning at twice their usual takeoff speed.

The plane would take off from the treadmill even if there were a tail wind
equal to Vr (though in that case, the wheels would be turning at *four*
times their usual speed).

--Gary

"cjcampbell" wrote in message
oups.com...
Saw this question on "The Straight Dope" and I thought it was amusing.

http://www.straightdope.com/columns/060203.html

The question goes like this:

"An airplane on a runway sits on a conveyer belt that moves in the
opposite direction at exactly the speed that the airplane is moving
forward. Does the airplane take off?" (Assuming the tires hold out, of
course.)


AH! ...here's the problem! Are the airplane and the belt moving at equal
speeds in opposite directions
relative to the world? (-X mph for the belt & +X mph for the plane = eg.
airspeed of 100mph &
wheel speed of 200mph) If so the airplane could take off. The answer to this
question would be easy --
is the airspeed high enough or not?

......OR relative to each other? If so, there could be just enough thrust
applied to overcome frictional
forces and the airplane doesn't move relative to the world so airspeed is 0.

BUT WAIT!!! .... ANY two objects can be said to be moving (or not) at equal
speeds relative to each other. A point
on the conveyer belt moving east at 4mph and a jet moving west at 600mph
each have a relative velocity of 604
with respect to each other and there could be an observer who sees each
object moving in opposite directions
at 302mph. The only real question is how fast is the airplane moving with
respect to the air(world).

Thrust is an external force applied to the conveyer belt/airplane system.


Cecil Adams (world's smartest human being) says that it will take off
normally.


He likely had a little more information than is available in the OP.

Only if there is an 80 mph tailwind.

If the plane is rolling 80mph against a conveyor going 80mph back the
difference would be zero.

-Robert

"Robert M. Gary" wrote:

Only if there is an 80 mph tailwind.

If the plane is rolling 80mph against a conveyor going 80mph back the
difference would be zero.

True. But I was referring to the problem stated, that the conveyor was
moving the same speed and opposite direction to the speed and
direction that the plane was MOVING, not this different scenario in
which the conveyor was moving the opposite direction and same speed to
that at which the plane's tires were turing.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

"muff528" wrote in message
news:OW2Ff.179$DV2.5@trnddc07...

.....OR relative to each other? If so, there could be just enough thrust
applied to overcome frictional
forces and the airplane doesn't move relative to the world so airspeed is
0.

That would have to be either a very underpowered airplane, or wheels with a
lot of friction.

BUT WAIT!!! .... ANY two objects can be said to be moving (or not) at
equal
speeds relative to each other. A point
on the conveyer belt moving east at 4mph and a jet moving west at 600mph
each have a relative velocity of 604

But there's the trick. A treadmill belt isn't really moving at all, it's
turning.
Try this for a brain scrambler. Think about a tire on your car, driving down
the highway. At the point where the tire contacts the ground, it's speed is
zero. 180° away, at the top, it is moving forward at twice the speed of the
car.

Gary Drescher wrote:
The plane would take off from the treadmill even if there were a tail wind
equal to Vr (though in that case, the wheels would be turning at *four*
times their usual speed).

SMALL corrections:
*First of all, a plane doesn't take off at Vr but at Vlof (lift off
speed). Vr is the speed at which you lift the nosewheel from the ground
and this speed is smaller than Vlof which is the speed at which the
plane lifts off the ground. So: "The plane would take off from the
treadmill even if there were a tail wind equal to Vlof". But you
probably meant it right.
*Second, in the case of a tailwind equal to Vlof, when the plane leaves
the ground, the wheels would spin at a speed 3 times their usual speed
and not 4. Actually this entire question and solution is about adding
and substracting velocity vectors and a perfect example of Einstein's
relativity theory. It all depends on what you take as a reference (the
ground, the tredmill or the air). As some other folks said here, the
question was not clear enough and there was not enough info! So
obviously we were dealing with a communication problem here. Anyway,
since that is solved now, let me get into adding and substracting
velocity vectors to explain you the case of a tailwind.

----------(4) ----------(2) ----------(1) vectors in
reference to the conveyor belt
----------(1a) vector in ref
to the airplane
_____________________________conveyor belt

the plane moves from right to left in the above drawing and the
conveyor belt from left to right.
(1a) is the speed (let's call it "x MPH") at which the conveyor belt
moves

NO WIND CONDITION:
*Engines not running:
Assuming perfectly frictionless wheels, the plane's speed relative to
the surrounding ground (Ground Speed or GS) will be zero. Since there
is no wind, the speed relative to the air (True Air Speed or TAS) is
also 0. However, the conveyor belt moves at a speed x in reference to
the plane (vector 1a) and the wheels will spin at a speed x (vector 1)
and this is also the speed at which the plane moves forward in ref to
the belt.
Briefly:
GS=0 TAS=0
Tire speed=x not taking off!

*At takeoff thrust and the plane has reached Vlof=x MPH:
The engine thrust is pushing the aircraft away from the air behind it
to put it in simple words. In other words, we are now moving at an
airspeed (TAS) of x MPH=Vlof and since there is still no wind,
groundspeed is also x MPH BUT the plane is now moving at a speed equal
to 2x in ref to the conveyor belt. Twice the usual speed.
Briefly:
GS=x TAS=x
Tirespeed= 2x Plane lifts off!

I'll have to make an additional post since I reached max number of
characters . To be continued...