Robert Ehrlich wrote:

Colin wrote:
...
1. A slipping turn can be made at a high bank angle and low airspeed.
2. Rate of turn is dependent on angle of bank and airspeed, such that
the highest rate of turn is achieved with a high bank angle and a low
airspeed.
Brian illustrated this by inviting us to compare the rate of turn
achieved by a C150 and a jet fighter at the same angle of bank.
It has to be a slipping turn or we would stall, so the maximum bank
which can be used is dependent on the amount of top rudder available.
...

Ok, if you use the lift provided by the fuselage in a slipping turn,
you can devote more of the lift provided by the wings for generating
the centripetal force, by increasing the bank angle, and so the horizontal
component of the wing lift. But you pay this by an increase in drag, i.e.
more height loss. It is not obvious if the gain in time to turn override
the increased rate of sink, but I have some argument that should show that
the answer is no.

The analysis of Dr. Rogers as well as my own one can be used in the case
of a slipping turn can be used if you consider as bank angle not the
geometric bank angle, i.e. the angle between the wing plane and the
horizontal plane, but the aerodynamic bank angle, i.e. the angle between
the total lift vector (wings + fuselage) and the vertical (well as we consider
a glide it is rather total aerodynamic force than lift, but the difference
between both angles can be neglected). In this case our common conclusion
is again that the optimal (aerodynamic) bank angle is 90 degrees, although
the geometric bank angle is higher, and we have to maximize Cl/(Cd^2) in
Dr. Rogers' analysis, minimize Vz*V in my analysis. It is difficult to determine
if you may have a higher max Cl/(Cd^2) with slip, but I have a good argument
that we can't lower Vz*V. Remember in this analysis Vz and V are the horizontal
and vertical speed at zero (aerodynamic) bank angle and the same angle of attack,
i.e. what you find in an usual glider polar. In the case of a slipped turn,
we should use a polar showing Vz versus V for a straight flight with the
same slip angle. I never saw such a polar, but I think it is obvioous that
at any speed V the vertical speed Vz is higher with slip than without it.
i.e. the polar with slip is entirely below the polar without slip. Now the
minimum of Vz*V for the polar without slip is where this polar is tangent
to one of the hyperbolas Vz*V = constant, and except for the point of
contact, the polar is entirely below this hyperbola. So the polar with
slip is also entirely below this hyperbola, and no point of it can
provide a value for Vz*V that is lower or equal to the the value on
the hyperbola, i.e. the minimum of Vz*V obtained without slip cannot
be obtained with slip.

Er, yes ...
But how does the result compare with a large tear-drop flown at normal
glide speed, or for that matter, a smaller tear-drop flown with higher
airspeed and more bank ?
The point is how to complete a rapid 180 without spinning-in, and a
full sideslip with small angle of attack precludes a spin.
The author also encourages us to go out and try it, high up to begin
with of course.
- Colin.