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...