Originally posted by: KillerCharlie
Originally posted by: Evadman
Unless the wheels were bolted to the conveyor, and a speed differential between the conveyor and the air is impossible, the plane will fly. The plane only havs to make slightly more power than a regular take off to make up for the friction loss of the bearings.
Interestingly, the resistance from the wheels (to a high order of accuracy) depends just on the normal force (weight) of the aircraft, not the speed of the wheels. Given that the aircraft is not constrained in the fixed reference frame, then the conveyor speed actually has 0 effect on the airplane.
This means the ODE governing the aircraft velocity is:
m(dv/dt)=T-D-mu*(W-L)
T=thrust
D=drag
mu=coefficient of rolling friction
W=weight
L=lift
Of course thrust, drag, and lift are functions of velocity. Drag is also a function of lift, but because of ground effect the induced drag changes quite a bit, meaning you'd need a simple numerical ODE solver for this.
The wheel resistance does not depend on the speed, so the conveyor speed is irrelevant (once again, only if the airplane is not constrained in the fixed reference frame). If the aircraft is fixed in the reference frame as per some interpretations, then the thrust needed to cancel out the wheel resistance and keep the plane stationary is constant.
Where does anything in the OP imply that the aircraft is "constrained in the fixed reference frame"? There is just the airplane, and the conveyer. How can the conveyer possibly prevent the plane from taking off?