My 2 cents worth.
The engines produce thrust which propels the plane forwards (every action has a equal and opposite reaction). Therefore the ONLY answer is the plane will push itself off the conveyor belt, IRRESPECTIVE of the conveyor belt and tire speed of the aircraft. Conveyor belt speed is irrelevant.
The thrust developed by the engines will only provide lift to the immediately adjacent areas in the wash of the engine (if they are present), ie the airflow at the tip of the wing is minuscule from the engine in relation to the wing root area (wing mounted engines) or in some aircraft, simply the area adjacent to the engine. The lift is generated by the engines pushing that large amount of aluminum until
all the surfaces have enough airflow over them to generate lift.
The original question should probably have been asked like this:
"Will an aircraft that is tethered to the ground(equaling the zero ground speed of the original Q) lift off the ground under full throttle.
In this cas the answer is NO, as there is insufficient airflow over ALL the lifting surface (wing roots through to tip) to generate sufficient lift.
All the naysayers???
Explain this....
![[Linked Image]](http://www.flightsource.com/boeing727x.jpg)
Engines so far aft of the wing as to have little effect on the airflow directly on the main wing (ground staff can safely walk around the front of the aircraft with the tug attached). All they are doing is giving a "rear wheel drive" boot up the bum to the aircraft to generate enough speed to get the airflow velocity to what it needs to take off.
Short answer, aircraft on conveyor belt will advance fwd and eventually off the conveyor belt regardless of belt speed, therefore given a long enough conveyor belt, yes it will take off
![[Linked Image]](http://img82.imageshack.us/img82/7574/1201615966095pr3.jpg)
