Will the plane take off smoothly if it glides on a conveyor belt that is wide and fast enough?

In fact, the test in this question is mainly the concept of frame of reference. There's also the concept of scrolling. For quantitative analysis, let's first introduce the hypothesis of non-slip scrolling.

You'll need it later. First of all, let's assume that the conveyor belt is not moving, and the surface is stationary. At this time, the aircraft gains a forward velocity v1 (relative stationary frame of reference).

Then relative to the conveyor belt reference system, the speed of the axle is also v1. Slip-free scrolling means that the point and surface where the wheel is in contact with the surface are always relatively stationary. Then we know that if the forward velocity of the axle relative to the surface is v1, the velocity of the lower surface of the wheel relative to the wheel axle backward is also v1.

According to the basic principle of wheel rotation, the velocity of the upper surface of the wheel relative to the axle is also v1. Since the forward velocity of the axle relative to the stationary reference frame is still v1, the forward velocity of the point on the upper surface of the wheel relative to the stationary reference frame is 2*v1. Okay, now that we're done with the foreshadowing, let's start talking about that.

The speed of the belt backwards is the same as the speed of the wheel, and this is where the controversy in this issue comes in. The reason for this is that it doesn't specify a frame of reference, and it doesn't specify what point the speed of the wheel actually refers to. If we want to discuss it more practically, I think that the velocity here should refer to the velocity of the relatively stationary frame of reference.

Otherwise, it won't make sense to mess up. The next question is what point of speed on the wheels does speed mean? The most straightforward way to understand this is that we should take the velocity of the axle relative to the stationary frame of reference.

Because the position of the axle is obviously where the center of mass of the wheel lies. To describe the speed of an object's motion, if you use the concept of particles, then it is obviously the velocity at the position of the center of mass. At this point, we can quantitatively analyze.

Suppose that the airplane obtains a forward velocity v relative to the stationary frame of reference due to the jet, then the forward velocity of the axle is the same as that of the airplane, which is also v. According to the setting, the speed of the conveyor surface relative to the stationary frame of reference is -v. Repeating the previous analysis, the velocity of the wheel axle relative to the surface of the conveyor belt is 2V, and the linear velocity of the wheel surface relative to the axle is also 2V due to the assumption of no sliding rolling.

Then the velocity of the upper surface of the wheel relative to the stationary frame of reference is 3V, and the velocity of the lower surface of the wheel relative to the stationary frame of reference is -V, the same as that of a conveyor belt. The analysis is complete. In other words, if the speed of the wheels refers to the speed of the axles under the condition of the question, then the airplane can move at any speed, but the wheels themselves will rotate twice as fast as if they were stationary.

Considering that the amount of friction is independent of the speed, the aircraft is not affected in any way. <>

Airplanes do not have the concept of maneuvering speed and non-maneuvering speed. The thrust during the take-off and taxiing phase is also derived from the aircraft engine. After the speed reaches 170km hours, the flaps are turned into 2nd gear and can take off.

Since the conveyor belt is fast enough, the aircraft reaches a relative speed with the air, and the conveyor belt can slow down the aircraft before it provides enough lift, although the rolling friction is very small, but the belt speed here is infinite, so the aircraft cannot accelerate to get rid of gravity and the resulting friction. The wheels of the airplane are pulleys, there is no power, and the problem condition itself has no result except to turn the wheels to a puncture, as long as the pilot refuels the accelerator, the plane still moves forward, and when it reaches speed, it flies. <>

Is this a physical problem? But there is no data to calculate it.

No. The take-off process of the aircraft, including the flight process, must take into account the absolute velocity of the air in the vicinity of the coordinates of the aircraft and the relative velocity between the aircraft and the air.

Conveyor belt boundary layer analysis.

If we simply follow the assumption of the infinite length of the conveyor belt in the title (i.e., the coordinate extension range of the conveyor belt is (- then the theoretical steady-state boundary layer thickness is infinite, and the air velocity at any position above the conveyor belt is the same as the conveyor belt velocity, and the force and motion process of the aircraft taking off in this physical situation are no different from those when there is no conveyor belt at rest on the ground, then the detailed explanation of this problem can be derived from the classical fluid mechanics theory of wing lift (omitted here). Here consider the take-off of an airplane on a conveyor belt that is half infinitely long, infinitely wide, and has a sufficiently fast flow rate. Semi-infinite length refers to the coordinate range covered by the conveyor belt (0,+.

Because in this physical situation, the flight process of the aircraft will be significantly different from the flight process on the stationary ground. <>