Airplane on a Conveyor conundrum

[darkdezigns]

it wont work,easiest way to work it out is think of a harrier jump jet,it can hoover,but because it can move the thrust

[Apache]

Wow!

We need a Harrier here then! My missus never Hoovers!

[nero279]

GAWD this is painful

The aircraft accelerates not because the wheels are driven but because of the thrust generated by the engines, assuming the brakes are released the wheels are free to do WTF they like, the conveyor has no effect on the planes speed, just the speed of the wheels as they are not what's driving the plane. So the plane accelerates at it's normal rate, and takes off as normal.

The wheels are freewheeling, the fact that the conveyor belt is running at the opposite speed to the plane means the wheels freewheel twice as fast, but the fact that they are still freewheeling and never driven means it has no effect on the planes acceleration.

The acceleration is down to the air movement or thrust of the engines not because the wheels are being driven.

[darkdezigns]

Wow!

We need a Harrier here then! My missus never Hoovers!

lolol hover then smart ass lol

[gwh200]

I'm waiting for GWH to come along and break this discussion to down to a sub-atomic level

ah but no ome has mentioned any kinematic equations,which when applied in this case will prove beyond a reasonable doubt that without airspeed
you have no velocity and therefore no take off,im just working through the bones of it and will post my reply shortly,but i have to add....im deffo with apache on this one

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as a bye if you use v=u+1/2at^"

where v = velocity
u= initial velocity(omph in this case)
a=acceleration(ahem....again 0 in this case

and t^2 equals time squared

you get velocity = 0+1/2x0x t^2

now as we have been given no time parameter and we are told in the initial premise that the conveyor belt is long enough to accomodate a planes initial acceleration from 0 to take off speed(im assuming 1 minute)

we get finally
velocity =0 +1/2 x 0 x 1^2 minute which all conveniently equals ...err...i minute !!!

ie a time(1 minute) an initial velocity (0mph) and a final velocity (0mph)

the initial and final velocity all equal zero.removing a net effect from the plane(ie taking it off the conveyor) will alter the original lemma,so any reactive effect will be lost

cog,this CAN change according to where the ,say, fuel is and probably where all the (by now very bored)punters are sat
reason being that weight whilst staying the same ,there will be a shift of mass,relative to the moments of force.
if you observe all the conditions of the original premise,the plane will not fly.
in a real life situation you may generate enough lift from an exterior force acting on the object to achieve the net result.
boyles law has more to do with gases,thermodynamics have no bearing on this problem,bernouillis eqation shows a relationship between pressure volume and temperature,it makes use of avagdros number and has no relevance to this problem in any way shape or form
this is all about classic newtonian physics and not much else


and finally myth busters whilst being most entertaining does in no way practice good science.the only relatively reliable and representative practices are those carried out by grant immahara(spelling) ,but most of his contributions are theoretic


as you need a reaction in both directions,and one has been taken away
(the conveyor belt matching any potential motion) the net gain of energy(assuming this is an ideal,non myth buster skewed reult will always equal 0)

now bring it on !!!

[BUSHWHACKER]

Wow, there's 40 members logged on tonight!

[Slugsie]

You original premise did not state an airspeed differential, and it cant be assumed.

Broken down simply, if it has airspeed it will acheive lift. If it has groundspeed (relative to the conveyor) it need not.

Its a trick, or badly worded question.

My original premise stated that whatever air/ground speed (they're the same in still air, and by ground speed I meant the real ground, not the conveyor 'ground') the aircraft achieved, the conveyor would mirror. Now, if you truly believe that the conveyor, acting through nothing more than wheels that are able to spin freely, will be able to pull back on the aircraft with the same force as the engines are pushing it forward, then yes the airplane will stop. However, I don't see how the conveyor would be able to excerpt such a force.

[Slugsie]

ah but no ome has mentioned any kinematic equations,which when applied in this case will prove beyond a reasonable doubt that without airspeed
you have no velocity and therefore no take off,im just working through the bones of it and will post my reply shortly,but i have to add....im deffo with apache on this one

..
..

as a bye if you use v=u+1/2at^"

where v = velocity
u= initial velocity(omph in this case)
a=acceleration(ahem....again 0 in this case

and t^2 equals time squared

you get velocity = 0+1/2x0x t^2

now as we have been given no time parameter and we are told in the initial premise that the conveyor belt is long enough to accomodate a planes initial acceleration from 0 to take off speed(im assuming 1 minute)

we get finally
velocity =0 +1/2 x 0 x 1^2 minute which all conveniently equals ...err...i minute !!!

ie a time(1 minute) an initial velocity (0mph) and a final velocity (0mph)

the initial and final velocity all equal zero.removing a net effect from the plane(ie taking it off the conveyor) will alter the original lemma,so any reactive effect will be lost

That is mostly irrelevant. You start off by making an assumption that the aircrafts final velocity is zero. That is exactly what you can't do as we are trying to work out what the aircrafts final velocity will be. Because one of your initial assumptions is wrong, the entire calculation is wrong.

See my last post in reply to Apache for an explanation of why the aircrafts final velocity will be anything but zero.

[Apache]

Slugsie, Nero, this was the premise.

"You have an airplane sat on a huge conveyor belt. The conveyor belt is computer controlled such that its speed is always the same as that of the aircraft - only backwards. I.e. If the aircraft is moving forwards at 10mph, then the conveyor will move backwards at 10mph"

1) No-one stipulated air or groundspeed.

However, it is reasonable to assume that when "its speed is always the same as that of the aircraft - only backwards" that it means there is no net speed differential. Therefore, there can be NO LIFT. Therefore the f***ing aircraft WILL NOT FLY!!! Not in my universe where the laws of physics apply!

Hands up all those who have designed and bolted stuff on aircraft?

What about those who have flown aircraft of their own design (albeit models - same rules apply) and had such designs reviewed by peers and published?

No? Then shut the f*** up with your nonsense!

(PS - really no offence intended, but what a great topic after months of 'here's a spammer' and 'hi I'm a newbie should I buy a Surf' type threads!)

[Slugsie]

and finally myth busters whilst being most entertaining does in no way practice good science.the only relatively reliable and representative practices are those carried out by grant immahara(spelling) ,but most of his contributions are theoretic

Whilst I would agree that their methods are not 100% scientific, the very fact they do the experiment, and the aircraft does in fact take off must demonstrate that there is some fault in the logic of anyone who states that it will just sit there and not move.

[da SLUG man]

is the undercarriage down?

[gwh200]

for every action there is a reaction(newtons whatever law)
in an ideal scenario (which you initially pitched) any reaction forward tending from 0 towards a final velocity will require an equal and opposite force(we can call this the runway)

its the same as the ground pushing against you with the same force you are appying to it.
turn that into linear motion and net effects tend towards zero.

if the plane and the belt were not touching then there would be no reactive,negative force,but it is touching,and whether the wheels were driven or not,they connect a forward force with an inverse force(the backwards belt)

you have to overcome a reaction to say,lift a rocket into space,or even drive a car up ahill

so next time you are stuck,up to your axles in mud and the driven wheels will not get you out,strap a 747 jet to your surf and hope the ground stays where it is, and doesnt try to counter your jet propelled attempts to get free

velocity really has little to do with this as much as forces in opposite directions

[Apache]

However, I don't see how the conveyor would be able to excerpt such a force.

This is where your assumption (and mine, cos both angles are valid) is flawed.

Yes, the plane can move as fast as it likes assuming freewheeling wheels, its just that the wheels will move at the aircrafts V2 speed plus whatever speed the conveor is moving backwards, obviously! BUT the premise was that they are both moving at the same speed but in opposite directions.

I say again, its a badly worded question or a play on words designed to confuse, intentionally so, so Mythbusters can make a program about it.

Any plonker knows if there is airspeed, the plane will fly, but the way the question is written, its by no means clear whether there can be.

[BUSHWHACKER]

Bumble bees can't fly, but they do!

[da SLUG man]

hehehe!!... i just looked it up on youtube and see some fella trashing his model into a chest of drawers!!