Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Being Airborne on A Conveyor Belt
Name: Francisco
Status: other
Grade: other
Location: OH
Country: NY
Date: 1/24/2006

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?

Hi, Francisco.

This question was answered very well recently on The link is 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.

Hi Francisco,

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.

Good question,

Bob Hartwell

Francisco - Some people just have too much time on their hands! Well, here is my thought.

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.

Larry Krengel

(Who has landed with a main gear frozen and wished he had your conveyor belt on the runway.)

Hi Francisco,

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.

Greg Bradburn


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.

Bob Hartwell

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory