HOHN

Your roller skate analogy assumes that the belt is moving the SAME direction as the wheels, or that the wheels are not moving at all.

HOwever, in the puzzle, the wheels are moving the OPPOSITE direction. That means Clockwise to a counterclockwise or vice versa.

This would negate the forward motion, and hence the lift on the wing.

jh

hemijustin

NONE..... but that doesn’t matter. why??? because of the wording used in the question. you’ll figure it out someday........ its not that the plane cant take off on a conveyer. its that the conveyer cant match the speed of a moving plane.... therefore making the results of the experiment useless....

RustyJC

Two different things:

1. You're right - wheel speed can never match conveyor speed once the plane starts its rollout since the tire surface patch will be traveling at Vc + Vp (conveyor velocity plus plane velocity) while the conveyor belt, being anchored and referenced to the ground, can only travel at Vc.

2. I'm right - with frictionless bearings and no brakes applied, conveyor and wheel velocities have no effect whatsoever on the plane's rollout acceleration, airspeed and ability to achieve Vr. Conveyor and wheel speed are red herrings.

So, since the stipulations of the original riddle (Vc will always be equal to wheel speed) can never be met once the plane begins moving, the problem as stated is invalid.

Rusty

HOHN

There's nothing in the problem stating that there is any kind of limit to the force or speed of the conveyor.

In your scenario with frictionless bearings and unbraked wheels, the belt can always and perfectly match the rotational speed of the wheels.

Say we are viewing the plane from the starboard side (right side, or passenger side in CTD-ese). Say it is pointed to our right, trying to take off.

Just before throttles are advanced, rotational velocity of both belt and wheels are ZERO-- i.e., not moving.

The INSTANT that the thrust of the engines caused the wheels to rotate (from our starboard view, the wheels will rotate clockwise), the conveyor belt MATCHES that (albeit slow) velocity IN THE OPPOSITE DIRECTION.

Thus, no actual forward movement of the plane, and no lift.


In your engineering-speak, Rusty, your vectors are equal in magnitude, and opposite in direction. Therefore, the cancel each other out perfectly, much like destructive interference affects a waveform-- generating something perfectly out of phase, etc.

jh

hemijustin

YOu nailed it!!!!!!! good job... as long as the plane dont move its Valid. but as soon as it does it becomes invalid!!!!!!!! YAY!

RustyJC

Nope, ain't buying that. The plane is moving (and accelerating) relative to the air, regardless of which way (fore or aft) the conveyor belt is moving. The speed of the wing relative to the conveyor belt is meaningless. There is no force vector applied from the conveyor through the wheel that would retard the acceleration of the plane (frictionless bearings), so the thrust would accelerate the plane relative to the atmosphere as it would on a normal takeoff, and when the wing is moving through the atmosphere at Vr, we have the ability to rotate and leave the conveyor.

Rusty

HOHN

A-HA!! I was assuming that the wheel would be able to even turn at all! WRONG! The wheels will never even turn.

But it doesn't matter, because of the equal and opposite linkage of wheelspeed and conveyor speed. These vectors will ALWAYS be exactly opposite in direction, and the problem states that the magnitudes will always cancel out.

jh

hemijustin

i told u that way long ago HOHN...

RustyJC

That's why the problem is invalid. Once the plane is moving, the surface velocity of the tire (Vc + Vp) cannot be the same as the surface velocity of the conveyor (Vc).

Rusty

hemijustin

Iv been trying to tell you guys this fovever!

HOHN

Well, we have to balance forces, now don't we? And using the 3rd Law (equal and opposite reaction for every action), we can analyze.

1)Force of thrust. Acts in the direction of the rear of the plane. Reaction? A pushing force acting to propel the plane forward. So this equal-opposite pair is thrust/propulsion.

2) The force of propulsion attempts to induce a change in position of the plane. The rate of change of this position is Velocity (fwd).

Now, the problem stated that the conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation.

So, the belt and the wheels have IDENTICAL VELOCITIES, in OPPOSITE DIRECTION.

Since the velocities are identical (change in position), and occur over the same time (in opposite directions), then the POSITION MUST BE THE SAME.




The perfect example is running on a treadmill. Your speed is completely offset by the speed of the belt in the opposite direction.

Now, to simulate the thrust of the jet engines, let's say you fart the largest fart in human history. This would normally propel you forward faster than the treadmill is moving you backward.

HOWEVER, in this case, the treadmill speeds up to absorb the thrust.

One question: is it possible to accelerate the plane relative to the air without moving the planes wheels? Well, the problem says that if the wheels don't move, neither does the conveyor (both are V=0).


In order to takeoff, the plane would HAVE to have EITHER a wheel that can spin on a motionless conveyor, OR a motionless wheel riding on an accelerating conveyor. Since both the wheel and the conveyor move (and in opposite directions), the plane cannot produce motion relative to the air.
jh

HOHN

So, we've proven this to be impossible?

The whole point is that the plane can't move AT ALL, right? If the plane moves even an inch, then the premise of the wheelspeed and beltspeed corresponding MUST be false, correct?

What a dumb problem.

RustyJC

Justin,

The wheel is imparting no forces on the plane other than a vertical force to overcome gravity. It has 100% traction against the conveyor belt but (1.) is not being braked and (2.) has frictionless bearings. Therefore, it cannot retard the acceleration of the plane. That being the case, the thrust force vector (forward) is only working against the mass (inertia) and aerodynamic drag vectors (aft) - nothing different than a normal takeoff. The plane will accelerate and lift off, even if the conveyor speed has matched the plane's velocity and the wheels are stationary. If the conveyor is moving opposite the plane and matching the plane's velocity (minus sign relative to the plane), then the wheels will be rotating at twice their normal speed, but the plane will still take off.

The key is that the velocity of the plane relative to the conveyor is meaningless. All that matters is the plane's velocity relative to the atmosphere.

Rusty

RustyJC

Yep....

Rusty

SafeHarbor

Forgive me for this, folks, but the jet engine is a reaction engine. It accelerates the air inside it to produce forward thrust. Pure jets accelerate a small amount of air a lot. Some jets, like the big turbofans on airliners, have big fan portions that impart a smaller velocity increase to a large amount of air in addition to the thrust from the core.

The important to remember is that there IS that equal and opposite reaction to the thust, and it's moving the plane forward through the air. Now, lets say that conveyor belt IS busy trying to eat up forward motion. That does not keep the plane from trying to accelerate, and from the plane's perspective, it's doing exactly that because there is a reaction.

What's neat about this is that the jet can produce thrust equivalent to that required to get it up to several hundred miles per hour at low altitude. That would have to be a HECK of a conveyor to match the acceleration and ultimate speed of the plane. (Not that the pilot would ever leave it at a takeoff power setting up to those kinds of airpspeeds.)

I think, if we ignore the wheel speeds and friction, the airplane couldn't care less about what's going on underneath it. It only cares about forward motion and air speed.

This might have been an easier experiment with a car...