This is my bucket experiment. I use this experiment to explain lift. Am I right? 


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*Hang the bucket on a long rope. Rotate the bucket to tighten the rope. Then fill the bucket with water and close the lid (shown in the blue line). At this time, the bucket and water are in a static state, and the isobaric surface in the water is plane.

*Release the bucket and the torsion of the rope makes the bucket rotate.

*After rotating for a period of time, water also rotates because of its viscosity. At this time, the isobaric surface in water is no longer a plane, but a parabolic surface: the pressure at the wall of the bucket is high, and the pressure at the center of the bucket is low.
Why?
Because of the rotation, the water in the center of the bucket expands, so the pressure decreases, while the water in the wall of the bucket is compressed, so the pressure increases.
The lift of the wing is the same. The airflow at the top of the wing tends to be away from the wing along the normal direction, so the airflow is expanded and the pressure is reduced. The airflow at the bottom of the wing tends to approach the wing along the normal direction, so the airflow is compressed and the pressure increases. One high, one low, there is pressure difference, so it produces lift.

I use air instead of water, then fill the air into a bucket deep enough and large enough, and then rotate the bucket, the air in the bucket will form a parabolic surface. What does this mean?
 A: If there is no head space, then the density at the outer radius is slightly higher than at the center.  If $\rho_0$ is the initial density before rotation, then during rotation, the density during rotation is $$\rho=\rho_0[1+\beta (P-P_0)]$$ where P is the pressure at radial location, $P_0$ is the initial pressure, and $\beta$ is the bulk compressibility of the liquid. A differential force balance on the rotating fluid yields $$\frac{dP}{dr}=\rho \omega^2 r$$where $\omega$ is the angular velocity of rotation.  If we combine these two equations, we obtain$$\frac{dP}{dr}=\rho_0[1+\beta (P-P_0)]\omega^2r$$The solution to this equation is $$\frac{1+\beta(P-P_0)}{1+\beta(P_c-P_0)}=\exp{\left(+\frac{\beta\rho_o\omega^2r^2}{2}\right)}$$where $P_c$ is the pressure at the centerline of the bucket.  Since the bulk compressibility of water is very low, we can linearize this equation with respect to $\beta$ and obtain:  $$P=P_c+\frac{\rho_0\omega^2r^2}{2}$$And thus, $$\rho=\rho_0\left[1+\beta (P_c-P_0)+\beta\left(\frac{\rho_0\omega^2r^2}{2}\right)\right]$$ Since mass is conserved, we must have that:  $$\int_0^R{2\pi r\rho dr}=\pi R^2\rho_0$$where R is the radius of the bucket.  This enables us to determine the pressure at the centerline of the bucket:
$$P_c=P_0-\frac{\rho_0\omega^2R^2}{4}$$Substitution of this into the equation for the density yields: $$\rho=\rho_0\left[1+\beta \rho_0\omega^2\left(\frac{r^2}{2}-\frac{R^2}{4}\right)\right]$$
For a fluid with low compressibility like water, the change in density from the average original density is only very slight.
A: Your explanation of lift is basically correct,low pressur above the wing and high pressure below produces lift,but I can't see what the bucket of water has to do with it. The situation in the bucket is that the water is trying to obey Newton's 1st Law of Motion, ie an object in uniform motion will continue to travel in a straight line until it is acted on by a force. The wall of the bucket exerts a force on the H2O molecules that 'want' to travel in a straight line so they deviate from a straight line and attempt to climb the wall which is in their way. Another way of describing the same thing is to say that centrifugal force generated by the rotation of the bucket and its contents forces the H2O molecules to move to the side of the bucket, which, because it is in their way, they attempt to climb. This makes the surface parabolic.
The water doesn't actually expand, it's just drawn away from the centre by the force which compels it to move to the side. If it were a very fragile bucket and the wall collapsed (assuming that this happened on the international space station so that gravity didn't take over), the water molecules would then obey Newton's 1st Law and travel in a straight line. 
A: YES! You are correct.
You've hit as HOME RUN with this bucket.!.!.!.
Your analogy with the spinning bucket is extremely good, even though Chet M. and Michael W. do not see it. It correctly captures the conditions both below and above a wing. There is only one minor issue I describe below.
You often see people claiming that lift is due to the "turning" of air, which is correct,  and this shows it perfectly.
The air above a wing is indeed trying to go straight (due to inertia - Newton's 1st Law) and, in effect, 'pulling away' from the surface just as it is trying to move away from the center of the bucket. This is centripetal acceleration of a fluid and for this to occur, the pressure must be less on the inside of the curved path than on the outside of the curved path. On the wing, the higher pressure is atmospheric pressure far above the wing.  The curved flow is the cause of the reduction in pressure on a wing's upper surface...NOT Bernoulli.
FACT: A decrease in pressure is caused by the curved flow of the fluid in the bucket AND above a wing.
The air below a wing is also pushed into a downward curve in the very same way the bucket surface is forcing the water to follow its curved path. View a wing at a modest Angle of Attack.  As a wing moves forward, the lower surface runs into the air and pushes it down, just like the bucket wall pushes the water away from its desires straight path, into a curved path downward. On a wing, the lower pressure is atmospheric pressure far below the wing.
Fact: The increase in pressure is due to the air and wing moving toward each other.  

The minor Issue I mentioned above may cause some confusion to some readers.
It is the fact that viewing a wing next to a top-down view of he bucked puts the pressure reduction above the wing, but in the center of the bucket.  Then, the pressure increase is below the wing, but on the outside of the bucket (which may be viewed as being on top in the drawing).  This can be viewed as upside down and difficult to resolve at first.
Very good Enbin, this is a very intuitive and correct analogy.

Also, the change in density for normal flight is so small that it can be ignored and has no significant contribution to lift.


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*Regards


A: The atmosphere is open, and it has no walls of buckets. The centripetal force at the top of the wing is generated by the tendency of the airflow away from the aircraft. The centripetal force in the bucket is generated by the tendency of the flow of water away from the center. The tendency to stay away causes depression, so there is centripetal force. Why is there a tendency to stay away? In buckets, because the center is not in line with the direction of the flow. On the wing, because of the angle of attack, the center and the direction of the air flow are not collinear.

When air rotates with the bucket, isobaric surfaces are formed in the bucket. These isobaric surfaces are parabolic. The water in the bucket also forms an isobaric surface.The faster the air rotates, the lower the pressure in the center of the water.So low pressure can be produced without viscosity, and low pressure can be produced with rotation.So your theory that you have to rely on viscosity to produce low pressure is wrong.
