think about this...

[Hiding]

[ QUOTE ]
The plane can take off. The wheels and runway will accelerate like a mofo as it does. The wheels have nothing to do with the lift.

[/ QUOTE ]

At what point does airflow start over the wings? No props, use a jet.

[Slow Play Ray]

[ QUOTE ]
Hence, if the conveyor itself isn't moving, the plane isn't moving and we have no ground speed.

[/ QUOTE ]

You were doing good right up until this part.

[Patrick del Poker Grande]

Alright, kids. I'm going to let you in on a little rocket science secret. Prop planes do not generate lift by blowing air over the wings.

[4_2_it]

[ QUOTE ]
Alright, kids. I'm going to let you in on a little rocket science secret. Prop planes do not generate lift by blowing air over the wings.

[/ QUOTE ]

Yea right! Next you are going to tell me that Santa Claus and the Easter Bunny don't exist.

[Patrick del Poker Grande]

I suggest everyone draw a Free Body Diagram. Use MS Paint - it'll be fun.

[Slow Play Ray]

[ QUOTE ]
At what point does airflow start over the wings? No props, use a jet.

[/ QUOTE ]

It starts as soon as the plane starts moving.

And why would it matter if its props or jet??

[Slow Play Ray]

[ QUOTE ]
Alright, kids. I'm going to let you in on a little rocket science secret. Prop planes do not generate lift by blowing air over the wings.

[/ QUOTE ]

way to take half the fun out of this thread!

[kevyk]

If you are saying that the runway moves backward at the speed the wheels’ center of mass translates forward, the plane is not moving forward with respect to the air, and will never take off.

If the runway moves backward at the tangential speed at which the edge of the wheels move, then the plane can achieve forward movement with respect to the air, and presumably take off.

The plane moves a distance d equal to the circumference of its wheels in time t. The time t is the period of rotation of the plane’s wheels, which is equal to 1/w, the angular speed of the wheels’ rotation.

Vplane=(2*pi*rwheel)*w

Omega is equal to the speed of the edge of the wheel divided by their radius by geometry.

Vedge=rwheel*w

Thus, Vplane=2*pi*Vedge.

So the plane speed with respect to the air = Vplane-Vrunway = Vplane-Vplane/(2*pi) = Vplane*(1-1/(2*pi)). (Vplane is the speed of the plane with respect to the runway...)

[Eurotrash]

[ QUOTE ]
I suggest everyone draw a Free Body Diagram. Use MS Paint - it'll be fun.

[/ QUOTE ]



oh, sweet. I was going to suggest drawing free body diagrams earlier in the thread. I don't know if I'm up for the challenge at this point, though

[Slow Play Ray]

[ QUOTE ]
If you are saying that the runway moves backward at the speed the wheels’ center of mass translates forward, the plane is not moving forward with respect to the air, and will never take off.

If the runway moves backward at the tangential speed at which the edge of the wheels move, then the plane can achieve forward movement with respect to the air, and presumably take off.

[/ QUOTE ]

It's really irrelevant how fast the conveyor belt is moving.