Can a Plane on a Treadmill Take Off?

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...

"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.

"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.

"Michael Ware" wrote in message
. ..

"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.

Yes!..that's why I said "COULD be just enough thrust..." More thrust than is
necessary
to overcome friction would result in the airplane moving forward relative to
the air. Then
it's only a question of how much thrust would be necessary to move the plane
forward fast
enough through the air to overcome gravity :-) A little less thrust would
result in the airplane
going backwards but not as fast as the conveyer. In any case the relative
velocities of the
plane to the conveyer would be equal to observers on either object but NOT
to an observer
standing on dirt.


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.

Again, Yes....but that's why I said "a POINT on the conveyer.." not the
conveyer
system itself.

The trick is that the original question as posted asks a question (will the
plane take off) and gives just enough info
to cause assumptions that aren't specified.

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.

Yes, but only for a very brief instant in time. And since velocity is
measured as a function of
time, is that point on the tire really moving at all at that one brief
instant when the measurement
is taken? :-)

In article ,
says...

"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.

Negative - yer forgetting centripetal force.

http://www.wordiq.com/definition/Centripetal

--
Duncan

"Dave Doe" wrote in message
. nz...
In article ,
says...
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.

Negative - yer forgetting centripetal force.

? Negative what? Talking about a point on the surface of the tire, not the
wheel as a whole. Centripital force has nothing to do with the forward
velocity of that point (how it travels in one axis).

http://www.wordiq.com/definition/Centripetal

--
Duncan

In article ,
says...

"Dave Doe" wrote in message
. nz...
In article ,
says...
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.

Negative - yer forgetting centripetal force.

? Negative what? Talking about a point on the surface of the tire, not the
wheel as a whole. Centripital force has nothing to do with the forward
velocity of that point (how it travels in one axis).

Are you talking about a round tire or not - are you then talking about a
big long flat tire of say infinite length. Sorry bud, can't make the
initial assumption that's been made - as it's on a tire, and yep, even
that point, at that time - has the centripetal force.

--
Duncan

Dave Doe wrote:


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.

Negative - yer forgetting centripetal force.

http://www.wordiq.com/definition/Centripetal

Well, I'm impressed that you know of the existence of centripetal
force. But in what possible way do you think it negates the comment
about the speeds (relative to the ground) of points at the top and
bottom of the tire on a moving car?
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

Let' say it this way. The airplane is moving forward at 60 kts. Does
that make it more clear?

The airplane is moving forward at 60, the belt, using the model in the
OP, is moving backwards at 60. The wheels are turning at 120 kts. If
it's a 172 it'll lift off into, on a calm day, 60 kts of airspeed over
the wings.

It's a nicely phrased question that caught me at first as well.
Substitute real speeds into what had been posted and the answer becomes
clear to me, although a lawyer in the group might find a (ground) loop
hole.

Cecil Adams (world's smartest human being) is correct to a first
approximation. The correct answer to a second approximation is that it will
take off normally less some small correction factor for the increased
friction of the tires, wheels, and wheel bearings.

Note the "trick" of the question. It does not say that the conveyor keeps
the AIRPLANE at zero speed relative to the real world, just that it rotates
at a speed equal to the airplane moving forward. The question itself
supposes forward velocity of the aircraft relative to the earth and the only
thing the conveyor belt does is spin the wheels twice as fast.

Jim




"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.)

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