Being Airborne on A Conveyor Belt
Hi there is a question going around on a remote control plane forum that goes like this:
"Imagine a plane is sat on the beginning of a massive conveyor
belt/travelator type arrangement, as wide and as long as a runway,
and intends to take off. The conveyer belt is designed to exactly
match the speed of the wheels at any given time, moving in the
opposite direction of rotation.
There is no wind.
Can the plane take off?"
My answer is it depends on the plane's thrust. If you have for
example, a model plane that weighs 5 oz, and the motor produces 10
oz of thrust the plane should take off. My theory is based on
thefriction formula F=u R, where f= friction force, u is friction
coefficient, and R= weight. With the highest possible coeff. of
1.0, the most force the conveyor could exert is 5 oz, leaving 5 oz of
thrust left over. The plane effectively now has a 1:1
thrust:weight ratio, and applying slip & stick law, should still be able to
move forward. Is this correct?
This question was answered very well recently on
avweb.com. The link is
http://www.avweb.com/news/columns/191034-1.html. Scroll down to the
section labeled "conveyor-belt runway." (P.S. I don't know how long
this link will be maintained. It was active 1/11/06)
Yes, the plane can take off. The key is that the plane's wheels
*freewheel*, they are not driven. the conveyor belt therefore provides
NO force to the plane (OK, there's a little friction in the bearings
which could provide a couple of pounds force to a nomal GA category
airplane, but that's insignificant compared to the thrust. For an RC
plane, the situation should be similar). The prop pushes the plane
through the air until it reaches takeoff AIRspeed, which is the same no
matter how fast the wheels are going. So if you were inside watching
the instruments, you would observe that the wheels are spinning at a
GROUNDspeed different than the indicated AIRspeed, but that does not
matter, since it is AIRspeed you need in order to take off. (Caps
added for emphasis).
Your friction calculation basically assumes you have skids instead of
wheels, and would be basically correct for that case, but notice that
velocity does not appear anywhere in the equation. If you had a plane
with skids, it would take off if it had the power to get sliding with
or without a conveyor runway. It would not matter how fast the
skid-to-runway surface were moving. The friction force F = u * N is
the same at either groundspeed.
Hope this helps!
David Brandt, P.E.
Based on what was stated in the original question, the plane would not
be able to take off. I am sure you are aware that the idea of a runway
is to allow an aircraft to build forward speed with respect to the air
surrounding it not necessarily the ground. The air passing over the
wings due to the forward movement of the aircraft (caused by the
aircraft's thrust) produces lift which gets the plane off the ground.
The question states that the conveyer will negate my forward speed and
not allow the plane to move forward through the air. Thus, no lift is
generated and the plane does not fly. The friction force does not
matter in this case because no matter how much thrust the plane
produces, the aircraft does not move with respect to the air.
Another version of this would be a giant wind tunnel. Now the plane
does not have to produce any thrust. It will stay in one spot with
respect to the ground, but it will fly.
Some people just have too much time on their hands! Well, here is
The airplanes ability to fly is related to its relation to the
air. In particular here, for the production of thrust and
lift. How fast the wheels turn is irrelevant. Yes, it would take off.
Now, if you become enamored by the effect of the rolling friction of
the wheel bearings. (The static friction between the tire and the
runway is of no importance as long as the wheel turns.) In the real
world, rolling friction is so small that... yes, you would still take off.
If you were able to throttle the rolling friction on the bearings,
you could still (assuming the friction is constant or linear) take
off as long as the wheel would turn and the thrust continued to be
greater than the friction. The length of the take off run would vary.
How about if the wheels were frozen and not turning? Even with
this, I would bet an F-15 would get of the ground. You only need to
keep thrust greater than friction to allow the acceleration over the
surface until flying speed is obtained.
(Who has landed with a main gear frozen and wished he had your
conveyor belt on the runway.)
If I properly understand your travelator, the travelator moves at
exactly the speed of the airplane, but in the opposite
direction. This means the wheels rotate twice as fast as they would
on a normal runway and nothing else is different. Right?
In that case, I claim the plane would take off normally except that
the wheels would be rotating twice as fast as normally. Since the
frictional force is, as you say, f=uR, the frictional force will be
exactly the same in the two cases since v does not appear in the
equation for the frictional force. In other words, the frictional
force is independent of the speed. In that case the forces on the
plane are exactly the same whether the travelator is operating or
not and so the plane takes off the same way in the two cases.
Did I understand your question correctly?
Best, Dick Plano, Professor of Physics emeritus, Rutgers University
The only thing that matters is the motion of the wings through the
air. This is what develops lift. The force pulling the plane
forward is developed by the propeller's motion in the air and has
little to do with the wheels or the runway. Since the plane still
moves forward (due to the force of the propeller against the air) it
will still accelerate to the speed required for takeoff. When it
leaves the ground the wheels will not be rotating but otherwise it
will be a perfectly normal takeoff.
If to a stationary observer, an aircraft is moving forward at say 10 mph
and the conveyor is set at 10 mph in the opposite direction, then the
wheels, which are free turning are spinning at 20 mph. The plane will
continue to increase forward speed until it reaches take-off speed at
which point it will take off independent of the conveyor or wheel speed.
At this point the wheels will be spinning at 2 times the takeoff speed.
We could say that Speed of plane + Speed of conveyor = Speed of Wheels
Speed of Wheels - Speed of Conveyor = Speed of Plane
In the statement of question, Francisco stated:
"The conveyer belt is designed to exactly match the speed of the wheels
at any given time, moving in the opposite direction of rotation."
He did not state the speed of the aircraft, but the speed of the wheels.
Given the formula above the forward speed of the plane should always be
zero and can not generate enough lift to takeoff.
Update: June 2012