How can the Magnus effect be explained with Bernoulli?

On Sat, 20 Jun 2009 12:04:46 -0700 (PDT), Mikki
wrote:

Hi folks,

According to Bernoulli’s principle, an increase of speed will be
accompanied by a decrease in static pressure. However, Bernoulli’s law
only pertains to flows without any external energy being added (or
removed).
That’s why I do not understand why Bernoulli can be used to explain
the Magnus effect.
If a clockwise rotating cylinder (Flettner-Rotor) is moved through the
air from right to left (i.e. free stream direction is from left to
right), this will cause the flow on the upper site (free stream
velocity + rotation) to be faster than on the lower site (free stream
velocity - rotation). More importantly, an upward lifting force acting
on the turning cylinder can be observed.
All books / internet sites I checked so far explain the lifting effect
by Bernoulli’s law (faster flow on the upper site of the cylinder
causes a drop in static pressure and hence lift).
However, I don’t see why Bernoulli applies here as the rotation of the
cylinder means to add additional energy to the free stream and thus
should not lead to a reduced static pressure, should it?

I'm no expert but I'll have a stab at it.

at the very edge of the air where it is touching the rotor there is a
boundary layer entrained with the surface of the rotor.
when you spin the rotor you drag that boundary layer around with the
rotor.
put it in a breeze and you end up with an area just off the boundary
layer where going into wind there is an increased relative velocity
compared with the part going with the wind where there is a reduced
relative velocity.
the velocity difference creates the pressure difference which causes
the lift.

just remember that you can fly the wing through still air or you can
blow the wind over a stationary wing and still create the same lift
force.

I dont believe that it is a very large lift force generated. the
flettner rotor propelled ship was a failure in practise I believe.

Stealth Pilot