How can the Magnus effect be explained with Bernoulli?

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?
Therefore, I would be very grateful for any advice why the Magnus
effect can be explained with Bernoulli’s law.

Thank you
Mikki

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

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 :-)

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 :-)

On Jun 21, 1:44*pm, Mikki wrote:
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
thelift.

Why should the velocity difference experienced between the top and
bottom side of the spinning cylinder attribute tolift???

It shouldn’t, for one thing there is no velocity difference. Where
Bernoulli’s effect requires the motion of air the relative airflow or
free stream that causes aerodynamic force does not. The relative
airflow influencing the cylinder is made up by its rotation and the
breeze it is in. The motion of air (breeze) is the same on the top as
it is on the bottom although its relative airflow may have a velocity
difference on top and bottom as a result of rotation. Another thing
if you could generate an actual steady speed circular airflow around
the cylinder the air on top will have a much different velocity than
the air on bottom by virtue of their totally different directions.
This drastic change in velocity does not cause any pressure
differences as a result of Bernoulli’s effect because its speed does
not change.

For the second thing a rotating ball or cylinder moving through the
air does not generate lift at all in the real world only in the
artificial world of formula based texts. Lift and drag are very
similar forces, aircraft have circumnavigated the earth in spite of
drag while others have used drag exclusively as a means of horizontal
acceleration to fly around mother earth. Even in circumstances where
drag opposes one motion it can cause another. The drag from a prop
causes the airplane to tend to roll in a direction opposite prop
rotation. When you paddle a canoe the drag from the paddle is what the
canoe uses as a means of thrust.

When the ball or cylinder is spinning drag is reflected in a force
that opposes rotation. The surface drag on one side of the ball is in
one direction but the ball does not move in this direction because the
surface drag on the opposite side is not only in the opposite
direction it is the same amount that is on the other side. If the
amount of surface drag on one side of the ball were to become
different than the opposite side a more linear friction drag force
will be created. What makes a spinning car tire move linearly is the
non-aerodynamic drag between the tire and the ground. One way to make
the surface drag around the ball or cylinder to be uneven is to move
it through the air while it is spinning, pushing it into the air.

Why do people call this uneven surface drag lift? This requires some
intentional ignorance. If the ball generated a force perpendicular to
its flight path without spinning in would truly be lift but it don’t.
What makes the ball generate a force perpendicular to its flight path
is the fact that it is spinning. Lift is most accurately defined as
being perpendicular to the relative airflow that caused it. The
relative airflow that is influencing the ball is caused by its motion
while in the air (spinning) and its motion through the air. To derive
at the inaccurate assumption the spinning ball is generating lift you
must totally ignore the large fact that it is spinning. How many
people would believe the spinning ball is generating lift if you told
them that it was based on the false premise that the ball is not
spinning? The drag on the spinning ball opposes its linear motion as
well as its circular motion, how many directions is that? The uneven
drag around a paddle wheel causes a paddleboat to move forward. This
is a good example of drag and thrust being the same force.