Will it take off?

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posted on Feb, 13 2006 @ 11:47 PM
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Remeber, for every action there is an equal and opposite reaction. The thrust of the engine pushes the plane forward, no mater how fast the conveyor is going.




posted on Feb, 14 2006 @ 12:16 AM
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Originally posted by Bhadhidar
As stated before "take-off" is dependent upon lift. Lift is generated by the Bernoulli Effect resulting from air moving at higher speed over the top of the wing; thus creating an area of lower pressure Above the wing. the relatively higher pressure under the wing thus "lifts" the plane into the air.

Thrust, provided either by a propeller or jet exhaust, serves to push the wing through the ambient air at the speed required to generate sufficient lift.

If a sufficient volume of air does not flow around the airfoil of the wing with sufficient velocity (ie.: take-off speed), there will not be suffiecient lift generated to allow the plane to fly.

Speed is measurement of velocity relative to another object, generally speaking, that object is the ground. Which, although the Earth (the "ground" in this case) spins at some 3000 MPH, I believe, is generally perceived of as being stationary (Except sometimes in California!).

In the thought excercise given, the forward velocity of the plane is exactly countered by the "treadmill runway" thus negating the forward velocity of the plane. The plane would have the same velocity, therfore , if its wheels were locked and unmoving.


You were off to a good start, but the runway will not negate the speed of the aircraft, and it's speed is relative to the air.



posted on Feb, 14 2006 @ 02:14 AM
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Originally posted by Travellar

Originally posted by Bhadhidar


You were off to a good start, but the runway will not negate the speed of the aircraft, and it's speed is relative to the air.


The treadmill runway in the excercise was intended, as I read it, to negate any forward motion, speed (more precisely: ground-speed).

The ground, the plane on the ground, and the air above the ground all start with the same relative speed, 0 MPH.

The plane starts its engine. In normal operation, the plane then begins to gain speed, relative to the ground and the air above the ground, which are still at 0 MPH relative to each other. The plane's (forward) velocity through the air (still at a relative 0 MPH relative to the ground) moves a sufficient volume of air at a sufficient velocity over the wing's surface to provide adequate lift for take off.

If the surface upon which the plane is to taxi counter-acts the forward motion of the plane, which is what the theoretical "treadmill" in the excercise was designed to do, it effectively renders the plane motionless relative to the the actual ground (the Earth) and the air mass above it; no air will move over the wing, hence no lift. The plane will have no speed (0 MPH) relative to the air surrounding it.

If the plane cannot move forward, no matter how much thrust its engine is generating, because its wheels are essentially spinning in place, it cannot gernerate lift; it cannot fly.



posted on Feb, 14 2006 @ 02:44 AM
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bleh..

Your right it will take off. A conveyor type system actually wont pull the plane back at all. It will not keep it from moving. A plane relies on air to push/pull itself forward not the wheels.

Now if the brakes where on (and held) it wouldnt take off.



[edit on 14-2-2006 by Xerrog]



posted on Feb, 14 2006 @ 05:27 AM
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Xerrog,

>>
Your right it will take off. A conveyor type system actually wont pull the plane back at all. It will not keep it from moving. A plane relies on air to push/pull itself forward not the wheels.
>>

WRONG.

And the proof is simple.

Assume it's a tractor-driven prop aircraft.

Say the 'P effect' of slipstream from a propellor acts against the airfoil (and fuselage) inboard of about the 1/2 span point sufficient to create lift enough for the plane to fly. If this wash is in effect the only fluidic behavior acting on the airframe as a function of 'powered lift', why waste the weight and _drag_ of the outboard span which is NOT under P Effect?

Particularly if the jet is not designed to have added maneuver much beyond 1G level flight.

Now assume it's a jet. If the wash (had damn well better be) aft of the wing /pushing/ it forward is only sufficient to keep it motionless, WHAT is generating the powered lift (propulsive fluid force directed over) ANY PART of the wings sufficient to provide lift?

Now assume it's a helicopter, if the helicopter needs only move forward at sufficient speeds that it's rotor acts as a fixed airfoil, why the wasted weight of a transmission and anti torque system to spin the main rotor for vertical takeoff?

Because the blades independent motion _faster than the forward motion of the aircraft_ is required to generate enhanced differential pressure and thus lift effect, that's why.

People like this assume that jet=thrust and thrust=lift where in, really, lift is based on relative motion THROUGH the air to create differential pressure GREATER than thrust.

Even if you are only 'pushing' with a quarter of the total mass-offset needed to push it (straight up) directly. The air holds you up. Thrust only parts it as a function of motion.

All motion in this hypothesis begins and ends with the frictional cancellation of energy through the rotational effect OF THE WHEELS. And it is their very lack of 'traction' (which is to say more torque from a motor than drag from the conveyor) _not friction_ that prevents them from ever overcoming the equal and opposed force sufficient to generate independent airfoil movement through the air.

So long as both conveyor belt and jet are cancelling each others applied forces (Newtonian Physics people) so that THE WING stays motionless, there is NO LIFT.

And no takeoff.

No matter how fast your wheels spin.


KPl.

[edit on 14-2-2006 by ch1466]



posted on Feb, 14 2006 @ 06:08 AM
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It's a trick question. If this question involved an automobile, the conveyer would cancel forward impulse. However, it's an aircraft, with an engine that imparts motion relative to the air. You could reverse the conveyer's direction to keep the wheels from turning at all, and it still wouldn't change the aircraft's motion through the air.



posted on Feb, 14 2006 @ 06:29 AM
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Travellar,

>>
It's a trick question.
>>

As postulated, if the aircraft generates no relative motion along the length of the runway, the velocity of the airfoil _moving through the air_ will never be sufficient to take off.

If it were any other way, we would not need Harriers or Helicopters to achieve VTOL.

It needs to be stated _specifically_ that the wheels motion is not sufficient to prevent the THRUST of the aircraft from /skidding/ the airframe forward along the runway (i.e. the tires are slipping faster than the wheels can spin in counterrotation, without 'flipping' the plane completely.).

That I would believe.

But not if the airframe itself maintains a constant position relative to the viewer's perspective and the viewer is him/herself static relative to the runway's position on the ground. Because whether you 'thrust' against the air. Or along the direct contact path of the conveyor.

If you don't accelerate the total vehicle THROUGH the air so that fluid flow and differential pressure above and below the wings exceeds stall speed, the aircraft won't fly.

Period. Dot.


KPl.



posted on Feb, 14 2006 @ 11:15 AM
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Neglecting friction of the wheels (as they'd probably explode otherwise) and taking the standard measurement instrument of an aircraft... the pitot static tube - the aircraft will have no bother taking off.

It will take off... and it may do it in a shorter take off "run" than on a fixed runway if the flow is very turbulent and the boundary layer from the conveyor belt is thick enough to influence the wing.



[edit on 14-2-2006 by kilcoo316]



posted on Feb, 14 2006 @ 11:32 AM
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The question is, is the plane stationary? Obviously, if there is no lift, no forward movement, the plane will not fly. Given that, if the plane moves in the horizontal plane at a speed that lift occurs, it will fly, regardless of how fast the wheels or converyor is moving.

No horizontal movement, no lift, no fly.....period.



posted on Feb, 14 2006 @ 12:56 PM
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We all understand aerodynamics well enoug to realise the aircraft actually needs to be moving to fly. The trick to the question is it's attempt to mislead you into believing the aircraft will have no relative motion.

Aircraft moving 60 knots,
Runway velocity of -60 knots,
The wheels will be spinning at 120 knots,
but the airplane is still moving at 60 knots.



posted on Feb, 14 2006 @ 02:08 PM
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Originally posted by ch1466
Xerrog,

>>
Your right it will take off. A conveyor type system actually wont pull the plane back at all. It will not keep it from moving. A plane relies on air to push/pull itself forward not the wheels.
>>

WRONG.

And the proof is simple.

Assume it's a tractor-driven prop aircraft.

Say the 'P effect' of slipstream from a propellor acts against the airfoil (and fuselage) inboard of about the 1/2 span point sufficient to create lift enough for the plane to fly. If this wash is in effect the only fluidic behavior acting on the airframe as a function of 'powered lift', why waste the weight and _drag_ of the outboard span which is NOT under P Effect?

Particularly if the jet is not designed to have added maneuver much beyond 1G level flight.

Now assume it's a jet. If the wash (had damn well better be) aft of the wing /pushing/ it forward is only sufficient to keep it motionless, WHAT is generating the powered lift (propulsive fluid force directed over) ANY PART of the wings sufficient to provide lift?

Now assume it's a helicopter, if the helicopter needs only move forward at sufficient speeds that it's rotor acts as a fixed airfoil, why the wasted weight of a transmission and anti torque system to spin the main rotor for vertical takeoff?

Because the blades independent motion _faster than the forward motion of the aircraft_ is required to generate enhanced differential pressure and thus lift effect, that's why.

People like this assume that jet=thrust and thrust=lift where in, really, lift is based on relative motion THROUGH the air to create differential pressure GREATER than thrust.

Even if you are only 'pushing' with a quarter of the total mass-offset needed to push it (straight up) directly. The air holds you up. Thrust only parts it as a function of motion.

All motion in this hypothesis begins and ends with the frictional cancellation of energy through the rotational effect OF THE WHEELS. And it is their very lack of 'traction' (which is to say more torque from a motor than drag from the conveyor) _not friction_ that prevents them from ever overcoming the equal and opposed force sufficient to generate independent airfoil movement through the air.

So long as both conveyor belt and jet are cancelling each others applied forces (Newtonian Physics people) so that THE WING stays motionless, there is NO LIFT.

And no takeoff.

No matter how fast your wheels spin.


KPl.

[edit on 14-2-2006 by ch1466]


As smart as your post attempts to make you sound, you are completely WRONG.

The wheels on a plane are there simply to support the aircraft while it is on the ground. They do not provide the force that moves the plane forward such as what a car would do. The force to move foward is provided soley by the propellor and/or jet engine against the air mass around the aircraft. Thus, the force of the engine against the air mass is not affected by wheels that are essentially free spinning and the aircraft would take off normally with just the speed of the wheels themselves being affected.

How else would planes take off on ice???



posted on Feb, 14 2006 @ 02:10 PM
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Originally posted by ch1466
As postulated, if the aircraft generates no relative motion along the length of the runway, the velocity of the airfoil _moving through the air_ will never be sufficient to take off.




Actually that is not what is postulated. What is postulated is that the belt moves backward at the same speed that the plane is moving forward. Therefore, if the plane was truly stationary, then the belt would be also.

It is a trick question.


Let’s look at the problem another way. Let’s say that instead of an airplane you have a rocket car. That is a car that is powered by rocket exhaust. Would the car move forward to the end of the belt?

Let’s say that instead of a rocket car you took a sidewinder missile, removed the warhead and mounted the missile on a four wheel cart. Would the missile-cart reach the end of the belt?

If you fired just the missile itself, above the belt it wouldn’t make any difference at all how fast the belt was moving.

What if you had the cart, but put wings on the missile? What then?



posted on Feb, 14 2006 @ 02:35 PM
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Originally posted by ch1466

WRONG.

And the proof is simple.


All motion in this hypothesis begins and ends with the frictional cancellation of energy through the rotational effect OF THE WHEELS. And it is their very lack of 'traction' (which is to say more torque from a motor than drag from the conveyor) _not friction_ that prevents them from ever overcoming the equal and opposed force sufficient to generate independent airfoil movement through the air.

No matter how fast your wheels spin.


You are assuming the forces are balanced:

T(eng)@V1 = 2x Fr(wheels)@V1 (Since V1 is effectively doubled for the wheels)


Now, I doubt that the friction from the wheels would match the thrust from the engine until a point well beyond twice the take-off speed.

So the forces would be unbalanced (i.e. the aircraft keeps accelerating), reaches take off speed, then takes off.


If it continued to stay on the runway, you are right that eventually an equilibrium position would be reached where the wheel friction would balance the engines out and the aircraft would go no faster - but it still would have all the motion built up before the equalisation - which I believe would be well above take off speed.



posted on Feb, 14 2006 @ 02:51 PM
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The takeoff run will be exactly the same as if it was on a normal runway. It would still need so many thousand feet to take off. It wouldn't take off vertically.

I remember reading about a really old idea of using a conveyer moving in the same direction as the aircraft to help it take off. This is post WW1 era. The problem was that the takeoff speed and weight of the aircraft quickly got bigger than what the conveyer technology at the time could handle.



posted on Feb, 14 2006 @ 05:36 PM
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You can however find that technology in use on the flight decks of US Aircraft carriers.



posted on Feb, 14 2006 @ 05:50 PM
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Bhadhidar is completely and totally right.

The plane WILL not take off.

Now, I believe the better question here would be "If the plane is on a conveyor belt moving forward at 150 km/h while the conveyor belt is moving in the opposite direction of the plane at 150 km/h, is it stationary or in motion to an observer?"

If to an observor it is stationary, then no, the plane WILL not take off, but if the plane is motion to an observe despite the fact that the conveyor belt is moving against the aircraft, then it will take off, remember the landing gear does not produce lift. I think that before we can truely know the answer to this question, we have to know whether or not the if the aircraft is in motion or stationary to an observing.

Shattered OUT...



posted on Feb, 14 2006 @ 06:01 PM
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No lift, just think of a car on a dyno, the dyno says the wheel speed is 100mph, the dyno rollers are moving in the opposite direction of the car wheels, but as everyone else said the ground speed is zero, there is NO AIR going over the car on the dyno hence the need for a fan for cooling.

Same thing with a plane on your conveyor. NO AIRSPEED = NO LIFT, the amount of thrust is irrelevant since you are countering the thrust with the opposite motion of the conveyor. And yes the wheels are connected to the airframe by means of the wheel axle mount, which is connected to the gear strut which is connected to the landing gear fuselage mount therefore force is transferred, even if you replaced the wheels with skis still on the conveyor you have no positive air speed over the plane since you are equallizing the force by running the conveyor in the opposite direction.

If you reveresed the mechanisim you have what is known as a slingshot and you can impart the force directly to the airframe to create positive air speed, forward motion and lift.



posted on Feb, 14 2006 @ 08:19 PM
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Won't fly, and here's why - thrust to weight ratio.

Cut the wings off a plane and ignoring the control problem will it fly? NO! not unless it is a jet with a thrust greater than the weight. Same with the plane on the treadmill, no lift on the wings means that unless the engines are capable of lifting the plane on its own, not likely for a propeller aircraft, then it will NOT FLY period, nothing else matters.

Here's another way to put it -

You are in a aircraft waiting to take off into a 80mph head wind. If you need 100mph to take off, what speed do you need the plane to be moving forward to take off? Just 20mph. Now the same but the plane is moving with the wind. It now needs 180mph to take off. No wind under the wings means no lift.



posted on Feb, 14 2006 @ 08:39 PM
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The post Ring made reminded me of something I was taught in my class, something that all pilots are taught while in training, if any of you actually are learning to become pilots, you would already know the answer to this question because something like this is taught.(I can't think of what it's called right now, just know that I've been taught it and I do remember bits and pieces of it amongst the many other things I need to know)

Shattered OUT...



posted on Feb, 14 2006 @ 08:56 PM
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Originally posted by ShatteredSkies
Bhadhidar is completely and totally right.

If to an observor it is stationary, then no, the plane WILL not take off, but if the plane is motion to an observer despite the fact that the conveyor belt is moving against the aircraft, then it will take off, remember the landing gear does not produce lift. I think that before we can truely know the answer to this question, we have to know whether or not the if the aircraft is in motion or stationary to an observing.


Technically this is not always true. What if the conveyor and airplane are on a platform (say stretched across a few beds of 18 wheelers, with the plane facing in the direction of the motion of the trucks) and the observer is also on that platform. If the trucks were able to match the necessary speed for plane lift the observer would watch the plane take off. It's really stupid because you wouldn't even need the conveyor (the trucks would be in essence providing the "thrust") and the observer would be moving and feel the same airflow as the plane. Also it's not stated as an assumption. Just semantics, I guess.





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