Eric Greenwell wrote:

In article ,
says...

"Jim" wrote in message
...
In his book Gliding, p100, Derek Piggott writes:

"In most modern gliders, the elevator power is not adequate to pull
the wing beyond the stalling angle in a steep bank and it is only just
possible to reach the pre-stall buffet with the stick right back.
This is very different from straight flight and gentle turns where a
movement right back on the stick would definitely stall the aircraft,
requiring a significant loss of height to pick up speed before full
control is regained."

If this is the case, what are the aerodynamics that account for
this? Does it have something to do with the elevator's limited
power to deal with the load factor resulting from a steep, level turn?

I'll give this one a try.

As we all know, the pitch stability/control system is like a seesaw with the
download produced by the horizontal tail balanced by the downward force of
the weight of the glider acting at the center of gravity with the center of
lift acting as the fulcrum.

In level flight the downforce at the center of gravity equals the all-up
weight of the glider and there is sufficient reserve up elevator authority
to stall the wing.

In a 60 degree bank, for example, the downforce at the CG is twice the
weight of the glider due to the centrifugal force of the turn. However, the
elevator effectiveness is the same as in level flight so it cannot overcome
the increased downforce at the CG and bring the wing to a stalling AOA.

As Derek points out, with most modern gliders in a steep turn, the wing
cannot be brought to a stalling AOA. The glider is, in effect, becoming
nose-heavy due to centrifugal force.

Since the wing is able to develop twice the gliders weight, the
elevator should also be able to develop twice the force it normally
does, shouldn't it? And this explanation would suggest the elevator is
unable to generate more than 2 g's, even in level flight, but we know
it can do that.

I believe the reason the elevator becomes less effective in circling
flight is due to the change in relative airflow between the wing and
the tail. Because the glider is turning partly in the pitch plane
(mostly in the pitch plane at 60 degrees bank), the airflow at the
tail meets the tailplane at a higher angle of attack than it does at
the wing. This higher angle of attack means more "up elevator" is
required to produce the same download. At the point the elevator
reaches it's stop, it is then producing less download than it can in
level flight, and is unable to force the wing to the stall AOA.

A quick glance at "Fundamentals of Sailplane Design" didn't find an
reference to it, but Frank Zaic described the effect 50 years ago for
model airplanes, using the term "circular airflow".
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Eric Greenwell
Richland, WA (USA)


I completely buy your explanation (which I would have posted if you didn't
do it before) and think all other explanations based on increasing forces
with increasing G are wrong. If this would be true, this would also apply
in level flight, i.e. ballasting a glider while keeping its CG in the
same place should have the same effect, and it is not the case. There is
a simple experiment that everybody can easily do showing the effect of
the change in AOA on the tailplane due to the circular airflow. Fly a
stabilized turn at e.g. 45 degrees bank (load factor = sqrt(2) ~ 1.414)
and set the trim to have a neutral stick. The fly stright and level
at the same AOA, i.e. ~ 15% slower (sqrt(1/1.414) ~ .85), without changing
the trim. You need to push on the the stick to obtain that, altough both
wing and tailplane should be at the same AOA relative to the local airflow,
the change in forces is only due to the change in speed. This shows that
the direction of the airflow changed on the tailplane.