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Old June 21st 09, 06:44 PM posted to rec.aviation.piloting
Mikki
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Posts: 3
Default How can the Magnus effect be explained with Bernoulli?


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.


Why should the velocity difference experienced between the top and
bottom side of the spinning cylinder attribute to lift???
According to Bernoulli changes in stream velocity will be accompanied
by changes in pressure ONLY if no external energy is being added or
removed.
That is the reason why the static pressure can be measured in a fast
moving aircraft without being biased by velocity. Even though the
airplane may fly with several hundred kilometers per hour, i.e. feels
the air passing with several hundred km/h, the static pressure is not
changed compared to still air (as long as the orifice for measuring
the static pressure is not placed on the wings). This is because
Bernoulli's law does not apply in this scenario, as for propelling the
aircraft external energy is consumed.
As for turning the Flettner-Rotor (as well as the air in the boundary
layer of the rotor) extra energy is required, too, Bernoulli's law
should not apply in this case, either.
Therefore, I consider the explanation provided too simple. But maybe I
am wrong!

That's why still appreciate any further answers/suggestions to my
question
Mikki :-)