Can a Plane on a Treadmill Take Off?

"Greg Copeland" wrote:


In other words, you ASSUMED that speed was not zero
If one applies power to the plane (and yes, I am assuming that is
implied in the question of whether the plane takes off normally), and
the conveyor only moves backward at the same rate as the plane moves
forward, yes, the plane will move forward.

and Cecil ASSUMED
the runway was not driven by it's own motor.
As said before, I saw no such assumption, and in fact would assume the
opposite. I see no way for the conveyer to move without its own power.

Lots of assumptions.
What are you reading?

You then go on to talk about left turns you madel. Simply stated, it
is *all* about wheel speed and lift
Maybe for you. I didn't see anything about wheel speed.

If the wheel is turning
slower than the treadmill, then you are moving backwards.

If you are moving backwards, and the conveyer, according to the
statement of the problem, is moving backwards at the same rate as the
plane is moving forwards, what direction is it moving?

Thusly, a
delta of zero or less means NO LIFT...NO FLIGHT.

I still don't know what delta you are referring to, but I think we are
in agreement that a plane will not fly backwards. (Jokes about canards
aside).

This isn't exactly rocket science...
No, it's the wheels that are confusing folks. Make it "rocket science"
and it would become clearer. g

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

Let's do the old Einstein thought experiment. Let's presume a motorized
conveyor belt that is being rotated towards the departure end of the runway.
Let's also presume frictionless wheel bearings in an aircraft sitting at the
departure end of the runway on the conveyor belt.

What happens to the aircraft? Nothing. THe airplane remains motionless
because the aircraft wheels, which are rotating, do not impart any force to
the aircraft to make it move in any direction (F=ma). To a bystander
sitting on the taxi light at the end of the runway, the conveyor belt is
moving left to right, the wheels are spinning in a counterclockwise
direction, and the aircraft itself is motionless.

Now introduce wheel bearing friction. The aircraft will begin to slowly
move left to right as a function of how much friction there is. Fire up the
propeller and give it just enough throttle to overcome wheel bearing
friction. Again, the observer on the taxi light sees the aircraft
motionless.

Now give it full throttle. Not only do we now have enough thrust to
overcome wheel bearing friction, we have more than enough to launch the
aircraft successfully into the air.

If ya can't see this, I give up.

Jim




"BDS" wrote in message
t...

"Greg Copeland" wrote
Nitpicking aside, I suspect that everyone agrees that in order for the
plane to take off it must move
forward along the conveyor.

As a tribute to CJ Campbell for starting this wonderful thread, I propose
his official RAS and RAP call sign be "La Brea."

Jon

Which is exactly my point! If you have a motoroized conveyor which
always reduces the plane's forward movement to zero, no lift is
generated, preventing the plane from flying.

That's not what the original problem stated. And we have explained at least
ten times why the conveyor belt CANNOT prevent the plane from moving
forward. My attempt, an earlier post:

You are taking the statement 'a conveyer belt that moves in the opposite
direction at exactly the speed that the airplane is moving
forward' to mean that somehow there is a force being applied to the mass of
the aircraft, equal and opposite the thrust generated by the propellor. The
only place the treadmill can exert any force an the airplane is the only
place the treadmill is touching the airplane: the wheels. Any motion of the
treadmill belt will be translated into rotation of the wheels. This will not
prevent the aircraft from moving forward, through the air and taking off.

"Greg Copeland" wrote in message
ups.com...
Only if you make many assumptions. Otherwise, he's wrong. The only
correct answer is "unknown" because of lacking information. The only
way to get off of, "unknown", is to make assumptions, which Cecil
happily did. Therefore, if he's allowed to make assumptions, so are
the rest of us. Which means, the answer is equally, "no".

Either way, he's wrong because he made an assumption or he's wrong
because we are allowed to make assumptions in the other direction,
thusly proving he's wrong.


Bull. The only unknown is the amount of friction due to rolling resistance
of the tires and wheel bearings. Give me that and I'll give you the takeoff
distance.

RST Engineering wrote:
Let's do the old Einstein thought experiment. Let's presume a motorized
conveyor belt that is being rotated towards the departure end of the runway.
Let's also presume frictionless wheel bearings in an aircraft sitting at the
departure end of the runway on the conveyor belt.

What happens to the aircraft? Nothing. THe airplane remains motionless
because the aircraft wheels, which are rotating, do not impart any force to
the aircraft to make it move in any direction (F=ma). To a bystander
sitting on the taxi light at the end of the runway, the conveyor belt is
moving left to right, the wheels are spinning in a counterclockwise
direction, and the aircraft itself is motionless.

Actually, that isn't true. You don't need wheel bearing friction to
apply a horizontal force to the wheel at the contact point. The wheel
has inertia and accelerating the wheel will cause a reaction at the
contact point with the belt and the aircraft will begin to move along
the direction of the conveyor. This force will go to zero once the belt
reaches a steady-state speed, but the aircraft will continue to move
along with the belt.

Now if the wheels have no mass as well as no bearing friction... :-)


Matt

"Jon Woellhaf" wrote

As a tribute to CJ Campbell for starting this wonderful thread, I propose
his official RAS and RAP call sign be "La Brea."

No, I go the opposite way, and say that his name be "damned for all time!"
g

It has been somewhat entertaining, I must admit, but at what cost?

"Oh, the humanity!" :-)
--
Jim in NC

Can anyone appreciate the number of electrons that have been very
disturbed because of this thread? One could ask oneself, if an electron
was moving to the left, while someone was pulling the wire to the right
at exactly the migration speed of the electron, would the thread be
extended. One could ask that, but that might start another thread.

Travis Marlatte wrote:
The propulsion system is irrelevant as long as it is independant of the
treadmill.

No, it doesn't even have to be "independant of the treadmill." Even
if the wheels of the plane were providing the thrust, all that would
happen is that the wheels would be spinning twice as fast by the time
the plane lifted off.

The key is in the wording of the question. The people here who have
gotten it wrong have misinterpreted the riddle to imply that it means
the aircraft is being held stationary. But that's not true. That's
not what it said. It simply said the belt is moving backwards at the
same speed the plane is moving forward. If the belt were moving
backwards fast enough to keep the plane motionless, then you've just
violated the fundamental rule of the riddle. Vbelt != -(Vplane) in
that case.

Picture it this way:


--- Plane @ 100 mph
Treadmill @ 100 mph ---

Now, what is the TAS of that aircraft? 100 mph. I assure you, it
will fly.

The only braintwister is that one must realize that the WHEELS are
turning at 200 mph, rather than 100 mph.

Kevin.

Must consider the wind at time of experiment. If wind is same speed as
conveyor then real problem??