Internet says: The plane is driven by the thrust of the engines acting on the air not the ground. It *will* accelerate forward as a result, and acquire speed relative to the air and therefore take off. This comment is wrong. It completely ignores the friction of the wheels on the ground. There are two kinds of friction that are generated by wheels. The first is the friction from the ball bearings. This type of friction is small and irrelevant to the question. The second is the friction of the wheels on the ground, or in this case, the conveyor belt. This friction needs to be great enough to make the wheels grip the road under a variety of conditions. This friction is not so great (nor insignificant) when the wheels roll, but it is dramatically greater when the wheels skid. Once the airplane’s wheels start to roll, the conveyor belt will match their speed, effectively keeping the plane in place. The rotational speed of the wheels and the speed of the conveyor belt will both increase together until the maximum rotational speed of the wheels is reached. As long as the wheels roll without skidding, the plane will go exactly nowhere. For clarity, let’s assume the maximum rotational speed of the wheels is 300mph. Once that speed is reached, if the airplane’s engine still has additional power, the wheels will start to slip. The wheel’s rotational speed, and therefore the conveyor belt’s speed, will both max out at 300mph. Let’s assume the plane needs to achieve airspeed of 100mph to take off. The engine, then, would have to have enough power to make the plane skid on its wheels down the runway at 100mph. How much power is that? If the plane were not on a treadmill, it is equivalent to locking the wheels so they cannot rotate, and accelerating the plane on skidding wheels to 100mph. That would take a LOT of power. And also quickly shred the tires. Internet is always right Adam