# Plane on a conveyor belt - revised

This whole “plane on a conveyor belt” debacle had me wondering about the reverse. This may be a much simpler question entirely, but here’s my ignorance, and I hope you can fight it–

What would happen if a plane landed on a conveyor belt? As in, would its deceleration be affected? Would it be able to use a shorter runway length?

My gut instinct is to say yes-- it seems to me that a plane landing on a conveyor belt moving in the same direction as the plane’s travel will need a MUCH longer distance to stop. Therefore, the reverse should hold, right?

Let’s assume for the purposes of this question that the conveyor belt is moving opposite to the plane’s direction of travel. You can play with the speed however you want – run it backwards at the same speed as the plane is traveling, faster, slower, etc.

The initial deceleration is against the air, so it wouldn’t increase stopping distance. Does anyone know whether airplanes have brakes for low speed braking?

It should add some drag. This can happen in real life with float planes. The plane could land in a river with a current moving in the opposite direction. The moving current would add some drag to make the landing shorter. I suppose the same would happen to a plane with regular landing gear but the effect would not be as strong.

Sure they do for more modern planes. Some tail-draggers didn’t in the old days. Larger airliners have traditional brakes, reverse thrusters, flaps, and spoilers to slow down.

My uneducated layman’s guess:

The plane needs a certain length to decelerate, but it only needs this length relatively to the surface on which it is landing. If this surface is moving relatively to the ground around it (for example because it’s a conveyor belt), a shorter runway length, measured relatively to the ground, will suffice, but the distance your plane travels on the belt will remain constant.

All this is assuming you have a large enough belt moving at a sufficiently high speed, of course.

I’m going to say that no, the conveyor belt would not aid in stopping the plane - much. I’m going to speak in generalizations and I don’t want anyone jumping in to tell me that in so and so model of so and so plane this whatchamajigger was replaced by this whoozeewhatzit.

Modern aircraft have enough braking power to lock up the wheels. They will not do that on landing - even when tryin to stop in the shortest possible distance - because the coeffient of static friction is greater that that of kinetic friction. Which means that wheels do a better job of stopping the plane when the are not skidding then when they are.

The whole root of the plane on a conveyor problem is whether or not the wheels exert sufficient friction on the axles and therefore plane to make a difference on takeoff.

If the plane is using brakes, the friction between the brake pads(or shoes) and brake disc (or drum) is great enough to lock up the wheel. And for minimum rollout, the wheels are going to be almost locked up - but not quite. So, and resistence due to friction will not make a difference.

So, to qualify my answer. A plane with brakes? It won’t make a difference.
A plane without brakes - it will make a small difference.

I like you thought process. And it would be true if the limiting factor on the aircraft was the friction between the brakes and braking surface.

The limiting factor is the friction between the tire and the conveyor surface.

If you take out the factor of rotational inertia of the wheels, it won’t make much difference if you have slow the wheels down from 300mph to 150 mph while you are slowing an aircraft from 150mph to 0mph. The wheels should have about the same amount of traction and should exert about the same amount of stopping force to the plane.

The distance travelled over the belt will be greater because the aeroplane was initially travelling much faster relative to the belt (twice as fast for arguments sake.)

What if the belt starts moving after the plane has had frist ground contact?`You’d surely get problems with inertia, but the speed of the plane relative to the belt would initially be the same as over ground speed.

Let’s replace the conveyor belt scenario with a slightly more realistic one: The plane lands on an aircraft carrier. Let’s assume the plane is landing in a northward direction, while the carrier is moving southward. One would suppose that after the end of the entire landing maneuver, the plane’s geographic position would be south of the position where it would be if the carrier had not been moving. Isn’t that the same as saying that the deceleration distance, over ground, has shortened?

This thread has a very similar scenario, started by yours truly waaayyyyy before the plane on a conveyor thing ever started.

What would happen if I rode my bike onto a moving conveyor?

Same principles, I think.

Yes, the belt will have an effect will be on braking performance, but that’s it. Everything else is about reaction with the air, and we’ve been through that already. An airplane is slowed after landing by a combination of aerodynamic drag and wheel braking, in a ratio that tilts increasingly toward the brakes as the plane slows. So yes, landing roll in relation to the ground will be longer or shorter depending on the belt direction.

The difference between this question and the takeoff one is that the airplane is not driven through the wheels during takeoff, but it is braked through the wheels upon landing.

Let’s set up an improvised calculation (I have no idea if the values I use for parameters are realistic,and I guess they aren’t, but just for argument’s sake, let’s assume they are).

Our plane’s velocity at the beginning of the landing is, say, 50 meters per second. Let’s assume the deceleration is 0.5 m/s².

a = v/t -> t = v/a, so the deceleration will take 100 seconds. Since s = 0.5 a * t², the landing will require a runway of 0.5 * 0.5 m/s² * 10,000 s² = 2,500 meters (seems unrealistically high for these speeds, but let’s just face that).

Now assume the runway itself is moving at 2 m/s in the opposite direction. The velocity of the plane relative to the runway will increase to 52 m/s, the duration of the landing procedure to 104 seconds, and the deceleration distance travelled on the runway to s = 0.5 * 0.5 m/s² * 10,816 s² = 2,704 meters, an increase of 204 meters.

But during these 104 seconds, the runway itself will have moved 208 meters in the opposite direction, giving you a net decrease of 4 meters for decerlation distance over ground.