Aeroplane on a conveyor belt puzzle

[Guest Nomad]

I hope this is the right forum for this and that it hasn't been posted before.

 

Okey doke. Here's a puzzle for all you with the aero smarts:

 

Consider a (normal, serviceable) aeroplane on a runway-sized conveyor belt. The belt is controlled so that it automatically runs at the same speed as the plane, but in the opposite direction. Can the aeroplane ever take off?

[Admin]

In prac - No, in theory - Yes

 

In prac - all you are achieving is the wheels going around but there isn't any air going over the wings to generate lift

 

In theory - any movement will cause air displacement therefore if the conveyor belt could go fast enough (and we are talking faster then anything known) to generate an extreme amount of air movement at the wing of the aircraft then enough lift, in theory, could be generated BUT the wheels on the aircraft would have to be extremely huge to allow for the massive speed of the conveyor belt which would also generate drag and weight.

 

At least that would be my answer ;)

 

Oops - just re-read the question, I was assuming the prop wasn't going around but if it was then again in theory it could but the prop would have to be big enough and engine powerful enough to force enough air over the wings to generate lift - we all know that if there is enough headwind aircraft can actually hover

 

 

[Guest micgrace]

Hi

 

I'm taking it as zero displacement ie < > vectors of distance covered by the a/c and conveyor belt are equal but opposite. the a/c is actually "stationary" in relation to the air (nil wind) & surroundings so no lift or takeoff can occur.

 

Now put that conveyor belt running in the same direction super stol.

 

Micgrace

 

 

[tonymcg]

My first thought is that, assuming you have frictionless bearings in the wheels, the only effect is that the wheels will spin twice as fast.

 

 

[hihosland]

Imagine a very small very light conveyor belt

 

and place it under the wheels of a flying aircraft

 

the wheels will spin

 

the plane will fly

 

nothing changes.

 

==============

 

All the conveyor belt under the wheels would do is reduce the ground contact friction.

 

Try to take off with the brakes applied or from a very muddy field such that the plane can move across the ground but very slowly. It will eventually take off if the runway is available is long (loooooooooong ) enough.

 

The conveyor belt is, I posit, the opposite of having the brakes on.

 

The plane will still fly,

 

and take off sooner because of the reduced friction with the ground.

 

Would a plane on ice skates on a frozen lake take off quicker or slower than with wheels on a runway? I believe quicker.

 

Regardless of the conveyor belt the force applied to the air via the propeller will still accelerate the air over the wings and the plane will fly.

 

dem's my thoughts

 

Davidh

 

 

[Guest Teenie2]

Nomad,you got most poeple.Smarty pants.

 

 

[hiperlight]

The short answer is: YES, it can takeoff...quite normally. However, while the wheels are in contact with the conveyor belt the wheel bearings are taking a hammering as they are turning at twice the speed they would if the conveyor belt was stationary.

 

(Look at it this way...if the aircraft has no airspeed or groundspeed the conveyor belt is stationary...it only starts to move in the opposite direction when the aircraft moves forward.)

 

Bruce

 

 

[slartibartfast]

The way I see it, if the converyor is moving backwards as fast as the plane wants to move forward, the nett effect is the plane has no forward speed relative to the air it is sitting in (no more than it started with anyway) and no ground speed. If you look at a spot on the ground next to the conveyor belt, the plane is staying adjacent to it.

 

If the plane is not increasing its ground speed, it isn't increasing its airspeed over the wings either. So unless the wind increases to a speed above that required for flying, it's a brick with lots of potential energy.

 

Ross

 

 

[Deskpilot]

Hi guys, I have to agree with Ross. Prop wash over a relativly small part of the wings, does not generate sufficient lift for the plane to fly. If it did, we'd see this happening every time we do a full power, static, engine test. Can you imagine it. A plane tethered to a tree and hovering three feet off the ground. Only in comics my friends. As for wheel speed, bear friction, parasitic drag and any other form of speed reducing characteristics, not an issue as the conveyor belt is controlled to the aircraft air/ground speed (zero in both cases) Given sufficient head wind, then yes, the a/c might get airbourne, but that in not in this equation. Well done Nomad for raising such an interesting topic. Cheers, Doug

 

 

[hihosland]

Desk Pilot did say,

 

============

 

I have to agree with Ross. Prop wash over a relativly small part of the wings, does not generate sufficient lift for the plane to fly. If it did, we'd see this happening every time we do a full power, static, engine test. Can you imagine it.

 

=========

 

But this is not the case here.

 

There is nothing in the original question to stop the aircraft accelerating into the air.

 

The aircraft will move forward along the conveyor belt with the wheels spinning faster and faster. As in a normal take off the prop will claw itself into the air accelerating the air over the entire wing until the plane starts to fly.

 

at least that's my thinking.

 

Davidh

 

 

[slartibartfast]

This is a great hypothetical (what if there were no such thing as a hypothetical question?).

 

David, the original question does say that "The belt is controlled so that it automatically runs at the same speed as the plane, but in the opposite direction"

 

Therefore the plane will remain stationary in relation to the ground, therefore will never get an increase in air over the wings, therefore won't fly.

 

Just to clarify my thinking. Could still be wildy wrong of course.

 

Ross

 

 

[hiperlight]

Ross, Doug and Davidh,

 

I put it to you that if the aircraft has no airspeed (or groundspeed) it is not moving and thus the automatic conveyor belt is also stationary. Think about it!

 

Read the original 'quote' again.

 

The initial rotation of the wheels and the consequential movement of the conveyor belt will not happen until the aircraft starts to move forward.

 

Bruce

 

 

[slartibartfast]

But if the prop is spinning, the plane would gain groundspeed if the conveyor didn't move backwards to prevent it. The propellor adds energy to the system. It is an action for which there must be an equal and opposite reaction. If there were no conveyor, the reaction would be for the plane to start accelerating forward and build groundspeed. The conveyor cannot remain still, for that would mean there is no reaction to the propellor's action - which is against the law.

 

FYC,

 

Ross

 

 

[hihosland]

Ross says,

 

==========

 

David, the original question does say that "The belt is controlled so that it automatically runs at the same speed as the plane, but in the opposite direction"

 

Therefore the plane will remain stationary in relation to the ground, therefore will never get an increase in air over the wings, therefore won't fly.

 

============

 

As I see it as the engine goes to take off power (regardless of being a normal takeoff situation, on the conveyor belt, or a tethered aircraft ) the prop will

 

a..produce some forward thrust

 

b..force air over some of the wing giving some lift however small

 

.

 

The forward thrust will cause the aircraft to move forward along the conveyor belt regardless of how fast or slow the wheels are spinning.

 

If the aircraft accelerated by driving the wheels as in a car then the conveyor belt would stop forward motion. But it does not. The aircraft moves forward by accelerating the air past itself and will, I believe, do so on the conveyor belt.

 

.

 

This gives more air moving over the wing, which in turn gives more lift until flying is achieved.

 

.

 

Stand behind a tethered aircraft at full power and you will experience the force that will cause this hypothetical plane to move forward regardless of the speed of the conveyor belt.

 

.

 

Imagine this hypothetical aircraft at full power both tethered and on the conveyor belt. Now cut the tether.

 

IF ""Therefore the plane will remain stationary in relation to the ground, therefore will never get an increase in air over the wings, therefore won't fly."" was true nothing would happen when you cut the tether.

 

Regardless of the conveyor belt being stationary or rotating at takeoff speed I believe that the plane will rocket away when the tether is cut.

 

.

 

Supplemental question.

 

Consider a tethered aircraft sitting on scales would the scales read a lower figure with the engine developing flying power?. I believe that they would.

 

Davidh

 

 

[hihosland]

do we have an emoticon for

 

"my brain hurts"?

 

Davidh

 

 

[slartibartfast]

do we have an emoticon for "my brain hurts"?

How about this one? :yuk:

 

The original question says that the belt is controlled so that it automatically runs at the same speed as the plane - not the same speed as the plane's wheels. I read that as being the same as the tether.

 

That's what I meant about the plane being the same as a brick with lots of potential. If you suddenly stop the conveyor, it would react even more than your tethered example - it would rocket away because the wheels would suddenly stop free-wheeling and provide initial thrust forward as they bite (or explode). With a tether, the energy is being stored by stretching the tether. When cut, there would be a catapult effect by the release of that stored energy. I suspect the spinning wheels would store energy more efficiently.

 

Because it is tethered, it isn't getting air moving over the wings (except prop-wash) so it can't fly.

 

The engine doesn't provide lift, only thrust. So your tethered plane on the scales would only be lighter by the small amount of lift generated by prop-wash.

 

It all comes down to interpretation of the question - as usual.

 

Ross

 

 

[Guest micgrace]

Hi

 

reminds me of some crook physics 101 questions when I did such at the uni. All would be suprised by the number of "supposed" physics students that simply couldn't grasp the vector concept. (and that's not your heading)

 

Of course the assumption is the wheels are in contact with the conveyor without slipping. Physics still a crook subject (some do love it) but I don't.

 

Micgrace

 

Heres another question. You are flying along straight and level. Assuming you land the aircraft at exactly that speed and ignoring air resitance, you drop a bolt from your main wheel. Will it land in front of, behind or at the same point as the main wheel point where the bolt fell out (assume no rotation)?

 

 

[Guest brentc]

My answer is that the aircraft will fly.

 

Pretty simple stuff.

 

Unless your aircraft has driven wheels instead of a propellor, it will fly!

 

 

[Guest pelorus32]

Before I start let me say that :yuk: already...

 

I reckon that there is some confusion in thinking here. IF the situation was that the drive of the aircraft was delivered through the wheels then the conveyor running backwards would be a fatal impediment to flight. But that's not the case. The question then is one of the relative friction isn't it? In other words at some thrust t the aircraft will be overcoming the friction of its wheels on the backward moving conveyor and holding itself still. That value t has nothing to do with the "thrust" of the conveyor belt, simply the friction present in the wheel assembly of the a/c. Its like the roller skate rider holding themselves still on a conveyor belt. They don't have to use much arm force to do that.

 

Then at some thrust t2 the aircraft will be applying more thrust than the wheel assembly friction and it will move forward. Finally at some thrust t3 the aircraft gains enough speed to fly and off it goes - accelerating rapidly as it frees itself of the friction burden fb=(fc-fn) where fc is the friction experienced on the speeding conveyor and fn is the normal friction with the runway experienced on takeoff.

 

I reckon that the conceptual issue is that the speeding conveyor is simply turning the wheels - a la the roller skater. It's not holding the a/c still.

 

What do you reckon - a complete crock or near the mark?:;)3:

 

Ohhh :yuk: really badly.

 

Regards

 

Mike

 

 

[hiperlight]

The aircraft's tyres are not glued to the conveyor belt...come on...think about it. For a given aircraft speed the wheels will turn at twice the normal takeoff rate of rotation (until the aircraft is airborne of course)

 

Bruce

 

 

[Guest Teenie2]

Of course the a/c will takeoff ,just like normal,but the wheels will have a greater rotational speed.Nomad you are still a smarty pants.

 

 

[hihosland]

""Heres another question. You are flying along straight and level. Assuming you land the aircraft at exactly that speed and ignoring air resitance, you drop a bolt from your main wheel. Will it land in front of, behind or at the same point as the main wheel point where the bolt fell out (assume no rotation)?""

 

Bolt when released has a forward motion the same as the aircraft so they will start flying along parallel to each other. Since there is no air resistance they will continue with the same forward speed.

 

Gravity acts on the bolt to pull it down , whereas the lift from the wing counteracts gravity's effect on the aircraft (or the wheel on the ground stops the plane descending any further). Consequently the two vectors separate.

 

If bolt comes free at the precise point of touch down it will land as far forward of that point as it can travel in the time for it to fall to the ground.

 

Meanwhile the ground friction is slowing the aircraft down causing the plane to travel a lesser distance than the bolt does in the time taken for the bolt to fall to the ground.

 

The bolt will thus land in front of where the plane was when it landed and in front of where the plane is when the bolt lands.

 

Davidh

 

 

[Guest Ken deVos]

The aircraft will takeoff normally, but the wheel bearings would be cactus.

 

Look at it this way...

 

All aircraft are sitting on a moving 'runway sized conveyor belt' called the Earth. The Earth's circumference is 21,639 nautical miles and rotates at the rate of once every 24h. Therefore, the surface velocity at the equator is approximately 900 knots!

 

This being the case, has anyone observed that their aircraft takes longer to takeoff, or requires more thrust when pointing West compared to East?

 

 

[Guest micgrace]

Hi

 

A clue. How about when you are say, flying straight and level at say, 20m/s @ 9.8m (not that you would!) and the bolt drops. In one second will the bolt hit the ground foward, underneath or to the rear of the a/c (assumptions in place)

 

Micgrace

 

 

[Yenn]

My plane flies through the air, and the wheels are only there to stop the axle digging in the runway. It will still take off from the conveyor belt and the bolt will land behind due to the force of gravity reducing it's forward speed.