can an airplane takeoff on a conveyor belt?

[AceViperX]

Good feedback and discussions indeed. Let look at the aircraft carrier, the steam catapult helps accelerate aircraft by applying ALONG the direction of plane going. Right again the plane wheels are free to rotation and friction effect is not in concern. Ask, what happen if the steam catapult forcing in opposite direction as the plane going while its jet thrust continue apply? It is analog to the case of plane on conveyor?

I agree with Jasonfinn that the steam catapult cannot be equated to the conveyor belt. The catapult is physically connected to the gears while the conveyor belt isn't. In order for the conveyor belt to be the equivalent of the catapult, then the airplane wheels should not rotate and there must be zero slip between the tyre and conveyor belt.

But to answer your question, the airplane would accelerate in reverse until the thrust of the engine equals or exceeds the force of the catapult.
F=MA

Keep the discussion going!

[jansonfinn]

anyway, the all aircrafts are free wheel
the friction on the wheel and the belt + 9.81gravity
only make the aircraft take off slower
and
of course, in order for the aircraft to take off.........
the engine will counter all the friction and gravity
and in some moments the wheel of the aircraft will be in free wheel
and take off.........
for those still wanna say can not: the engine of the aircraft
are meant to lift it.......so it sure to make the wheel "free wheel"
and loss contact with the belt...........

Let me do some correction to the above. The engines of the aircraft are not meant to lift it but to provide thrust to the aircraft as well as to overcome drag forces.

It is the wings that provide the lift.

[renold]

I agree with Jasonfinn that the steam catapult cannot be equated to the conveyor belt. The catapult is physically connected to the gears while the conveyor belt isn't. In order for the conveyor belt to be the equivalent of the catapult, then the airplane wheels should not rotate and there must be zero slip between the tyre and conveyor belt.

But to answer your question, the airplane would accelerate in reverse until the thrust of the engine equals or exceeds the force of the catapult.
F=MA

Keep the discussion going!

Brother, you are correct by saying: F = ma.
But you must understand the "F" refers to SUM OF FORCES. The resultant Force. So; " Propel force by plane - Any counted force = ma "

Also in solving dynamics problems, there are too many "things" to begin with. So most people will adopt the method of making "assumptions". First, you must assume the entire system is rigid. Although parts may have some elasticity, deformation or break. These come last part in the study but begin with finding the resultant and direction.

[AceViperX]

Brother, you are correct by saying: F = ma.
But you must understand the "F" refers to SUM OF FORCES. The resultant Force. So; " Propel force by plane - Any counted force = ma "

Also in solving dynamics problems, there are too many "things" to begin with. So most people will adopt the method of making "assumptions". First, you must assume the entire system is rigid. Although parts may have some elasticity, deformation or break. These come last part in the study but begin with finding the resultant and direction.

You are right to say the assumptions have to be made ie. every parts is an ideal component meaning to say it doesn't stretch deform, etc.

So isn't it safe to say that wheels of the airplane is also ideal in that it has zero friction and drag?

Also yes you are right in saying F in F=ma is the resultant force acting on the object. So if the force of the catapult is greater than the thrust of the engines, aircraft will accelerate backwards. Let me try to type out the equation:

F-f=ma

(F-f)/m=a

But (F-f) =< 0 so 'a' will be -ve


But apply this formula into the conveyor belt and 'a' will be positive. Why? Because you have to ask how much force is required to pull the conveyor belt. Going by this formula, the airplane will not accelerate only if the force required to pull the conveyor belt is equal to the force of the airplane's thrust (not counting the friction against the ground etc). This adheres to the conservation of energy.

Looking forward to inputs!

[renold]

You are right to say the assumptions have to be made ie. every parts is an ideal component meaning to say it doesn't stretch deform, etc.

So isn't it safe to say that wheels of the airplane is also ideal in that it has zero friction and drag?

Yes, brother zero friction and drag on all wheels. Infact these are insignificant in our case. We are concern the difference between the plane and conveyor speed. Such different have accounted for friction and drag.

That TV program does not tells you the speed of that truck, plane and conveyor. Are they same? Which is faster? What you see if the plane airborned. So solely believe with your eyes is magic. Believe in principle is Physic.

[AceViperX]

Yes, brother zero friction and drag on all wheels. Infact these are insignificant in our case. We are concern the difference between the plane and conveyor speed. Such different have accounted for friction and drag.

That TV program does not tells you the speed of that truck, plane and conveyor. Are they same? Which is faster? What you see if the plane airborned. So solely believe with your eyes is magic. Believe in principle is Physic.

Yes if the wheels have zero friction and drag, then pulling the conveyor belt should be akin to pulling a table cloth out from under the cutleries.

[renold]

Also yes you are right in saying F in F=ma is the resultant force acting on the object. So if the force of the catapult is greater than the thrust of the engines, aircraft will accelerate backwards. Let me try to type out the equation:

F-f=ma

(F-f)/m=a

But (F-f) =< 0 so 'a' will be -ve


But apply this formula into the conveyor belt and 'a' will be positive. Why? Because you have to ask how much force is required to pull the conveyor belt. Going by this formula, the airplane will not accelerate only if the force required to pull the conveyor belt is equal to the force of the airplane's thrust (not counting the friction against the ground etc). This adheres to the conservation of energy.

Looking forward to inputs!

Sorry, brother have not answer your question. We can have negative 'a'. That is retardation. I mentioned F = ma is some one asked in previous thread.
Force can be expressed as F = m(v2-v1). {forgotten Newton which laws} That is change in momentums. This the case of conveyor and plane, this equation is more applicable. Think off the momentum of conveyor caused by turck and momentum of plane.

If positive 'ma' then the plane move faster. That TV program does not provide any data. What we see is the plane airbored. So in areodynamics, there must be air speed flow over the wing of that plane.

[Vortices]

Hello guys,

My take.
1) If we are only looking for the answer to 'can a plane take off from a conveyer belt'
-The answer is yes.
To understand this is simple terms, imagine all the east/west facing runway on/near the equator. The earth is spinning slowly and surely, therefore it is like all airplanes taking off to the west is on a slow conveyer belt. Can the plane take off, of course it can.

2) Can a plane take off from a very fast conveyer belt?
-The answer is, it depend on how powerful the aircraft engine is
To understand this, imagine the airplane sitting on the conveyer belt with its engine shut down, it would just be moving with the belt backwards. As the engines is started and spooled up the plane would then starts to move relative to the belt and begin to move backwards slower, as the plane continue to accelerate, it will eventually be stationary relative to space and subsequently moving fwd relative to space, when sufficient airspeed is reach it would then take off.

3) Are there any advantage to try to take off from a back moving conveyor belt?
-Absolutely none, as a matter of fact it is detrimental, it would increase both take off roll time and distance.

4) Where does this myth originate?
-I think this is where the real question lies, IMHO, I think most Joe public doesnt understand how planes fly. Looking at an airplane roll on the runway before taking off may seem to indicate, the wheels needs to be a certain speed before take off is possible. Believing that would lead to the smart idea of trying to spin up the wheels on belts to assist take offs. Just my opinion.

[zattan22]

Let me do some correction to the above. The engines of the aircraft are not meant to lift it but to provide thrust to the aircraft as well as to overcome drag forces.

It is the wings that provide the lift.

yes......plus the aero-dynamics.......

[zattan22]

as we talking so much about F=MA.....
how about acceleration.........
the plane of course will move will the belt
as object stationery will remain stationery
the belt will move the plane in the opp direction
BUT
DO NOTE THE WHEEL IS WITHOUT BRAKES AND THE AERO-DYNAMICS
OF THE PLANE..............EVEN THE ENGINE OF THE PLANES.....

THE ENGINE:
AIR INTAKE AREA IS BIGGER THAN THE OUTPUT.......THEREFORE,
THERE WILL NEVER BE A PUSH TO THE DIRECTION OF THE PLANE IN
ANY SITUATIONS.........

EVEN FOR A PROP PLANE THE ENGINE SURE TO PUSH THE AIRFLOW.....

if the air-flow is into the opp direction the engine will fail liao.

............................................
the only way, the plane can not take off is to acceleration is too fast
that the air flow through the turbine or turning the prop in the
opp direction that the engine fail.....
or
F=MA........ fixed the wheels of the plane to the belt.....

[renold]

as we talking so much about F=MA.....
how about acceleration.........

Yes, F = ma.

The "a" is the acceleration.
The "m" is the mass of the object we are looking.
The "F" is the force acting on the object.

F = ma is define as: Force required to cause a mass to accelerate.
F can also define as: the rate of change in momentum.
All the about thanks to Mr. Newton.

[xypher]

I think the aircraft will take off ultimately. Maybe it will take slightly longer, but it will still take off. IMO

The thrust of the aircraft is applied on the air, not the ground. Thus there is minimal effect on the takeoff.

Imagine I change the qstn to : Will you be able to run on a conveyor belt going the opposite direction. Most prob you will be sent flying behind or match the conveyor belts speed and stay on the spot.

My thinking is that this is because the force required to run is applied directly to the ground (conveyor belt) thus it is cancelled out.

Same applies if you manage to put a car on a conveyor belt.


Disclaimer : I have pondered on this for less than 5 mins before posting, with no calculations done. But based on the mythbusters episode + observations, I have come to this conclusion

[DL8698]

The movement of the conveyor belt has absolutely no bearing on the aircraft's ability to take off. If the wheel bearings are 100% friction free then the plane will take off in exactly the same time and distance, relative to the non moving surroundings.

Since nothing in real life is 100% friction free, the time taken to reach lift off speed will be slightly longer, but the Distance, relative to the non moving surroundings should be identical.

I think those who have difficulty catching this point are confusing how an airplane is powered through the air by it's air screw or jet, while the wheels are free wheeling; and a car, which obtain its motion via the Wheels which are rotated by the engine, and due to friction between the wheels and the road (the more the better!) the car is moved.

Therefore, for a car, if the wheels are rotating at the equivalent of 100 mph and the ground is moving Backwards at 100 mph then the car will remain absolutely stationary. (Ever observed your car being tested at Viacom where they spin the driven wheels at high speed on rotating rollers? The car does not move relative to the non moving surrounding)

To put it in a nutshell, anything that derives it motion from powered wheels, will have it's speed affected by the movement of the surface that the wheels are standing on.

A plane derives its motion, not through its wheels, but from its propellers or jet engine and will not be affected by the movement of the surface the wheels are standing on, except for a tiny amount which can be attributed to the friction in the wheel bearings.

An easy way to visualise this is, imagine the airscrew is a long cable tied to the nose of the aircraft and pulled by a winch way off in the distance, exactly like a winch launch of gliders.

The plane WILL move forward at more or less the same speed as the winch is pulling the cable in (if we totally disregard wheel bearing friction) no matter which way the ground is moving. The plane will still reach rotation speed in more or less the same time and distance.

The equivalent of the car being driven on rotating rollers and remaining stationery would be if, somehow, we can make the air that contacts the propeller only and not the wings, move backwards at 100 mph. Now even if the prop is spinning at the equivalent rpm that would normally result in airspeed of 100 mph, the plane itself will remain at 0 air speed and will not fly.

In real life, the air would of course flow through the prop as well as over the wings. Thats why we have "Air speed" and " ground speed" which differ depending on "wind speed".