Unintentional fully-developed spins...

I have over 2000 hours in gliders including many types of glass. I also
have extensive experience in powered and glider aerobatics. Not long after
the purchase of my first fiberglass glider (Pegasus) I was thermaling around
400' over a mountain (US east coast variety) and attempted the hang glider
maneuver (which I had also been flying a lot of lately) of horsing the
aircraft into the center of a turbulent rotor type thermal. Retrospectively
the fact that it meant cross controlling while in a 60 degree bank made it a
no brainer re what resulted.....in a heartbeat I was looking at the mountain
through the top of the canopy. My first thought was "Hmmm....I've been here
before" and made immediate corrections with little more than a couple
hundred feet and an increased heart rate to show for it. Luckily the only
other pilots were above me. Take home message.....what they teach you re
spin entry is real and aerobatic training is a definite help if you get in
that situation, or at least spin training is.

As a sidebar to this discussion I noticed one person posted that he is
constantly on the edge of stalling his glider during thermaling. I would
argue that he is flying very inefficiently if that is in fact the case. To
convince yourself try thermaling (when alone) at the buffet speed vs adding
5-10 kts at differing angles of bank and focus on the VSI and see what the
results are. Look at any polar as well. Also if you try this out here in
the turbulent wild west let me know when you go flying 'cause I don't want
to be below you!

Casey Lenox
KC
Phoenix

As a sidebar to this discussion I noticed one person posted that he is
constantly on the edge of stalling his glider during thermaling. I would
argue that he is flying very inefficiently if that is in fact the case. To
convince yourself try thermaling (when alone) at the buffet speed vs adding
5-10 kts at differing angles of bank and focus on the VSI and see what the
results are. Look at any polar as well.

Casey Lenox
KC
Phoenix

I haven't seen polars that take into effect bank angle, but
from doing the calculations of turn radius and angles of
bank, I'm convinced that in very long wing gliders at
high angles of bank and slow speeds (and ergo light weights too),
the inner wing is significantly slower than the
outer wing, and tacking on some knots is most
efficient (to keep the length of the inner wing nicely above stall)...

Mark Boyd

(Mark James Boyd) wrote in message news:401ec4d0$1@darkstar...
As a sidebar to this discussion I noticed one person posted that he is
constantly on the edge of stalling his glider during thermaling. I would
argue that he is flying very inefficiently if that is in fact the case. To
convince yourself try thermaling (when alone) at the buffet speed vs adding
5-10 kts at differing angles of bank and focus on the VSI and see what the
results are. Look at any polar as well.

Casey Lenox
KC
Phoenix

I haven't seen polars that take into effect bank angle, but
from doing the calculations of turn radius and angles of
bank, I'm convinced that in very long wing gliders at
high angles of bank and slow speeds (and ergo light weights too),
the inner wing is significantly slower than the
outer wing, and tacking on some knots is most
efficient (to keep the length of the inner wing nicely above stall)...

Mark Boyd

I don't believe your argument is correct. What determines the lift and
drag coefficients is angle of attact, NOT airspeed. The inner wing is
flying at the same angle of attack as the outer wing, think about it.
Speeding up won't make you climb better.

F1y1n wrote:
high angles of bank and slow speeds (and ergo light weights too),
the inner wing is significantly slower than the
outer wing, and tacking on some knots is most
efficient (to keep the length of the inner wing nicely above stall)...

Mark Boyd


I don't believe your argument is correct. What determines the lift and
drag coefficients is angle of attact, NOT airspeed.

This is true, and if the glider is at a certain pitch angle straight,
level, and coordinated in still air, both wings are at the
same AOA. If the two wings are at different airspeeds,
like in a turn or skid, the two wings are at different angles of attack.

The inner wing is
flying at the same angle of attack as the outer wing, think about it.

Not if the wings are at different airspeeds. This is how we
do a spin. One wing is "more" stalled than the other wing
(i.e. has a higher AOA because it is the inside wing, and has less
airspeed). A turn is similar in the sense the wings are at
different airspeeds (but the same pitch angle), but in a turn,
it isn't true that both wings are stalled (that is the difference
between a turn and a spin).

Speeding up won't make you climb better.

Err...well, we are trying to sink less. If the wingspan is very
short, the point on the polar is the same for every part of
the wing. If the wingspan is long and in a turn, different
parts along the wings are flying at different airspeeds, and
are at different "efficiencies" and points along the polar.

The goal is to minimize the average sink rate along the
wing. The best way to do this is to strongly avoid the
back side of the polar (which drops off steeply), which
is the inner wing in the turn. If we fly a little faster
than recommended IAS (from the "G" table for bank) then
a larger portion of the long inner wing is near the
min-sink point, and the outer wing, although not optimally
efficient, is just displaced a little way along the front side of
the min-sink curve (which is a little flatter).

The amount of extra speed that is optimal should be based on
wingspan, the polar, and the bank angle. The
magnitude of this speed "correction" is something I have yet
to calculate, but thank you to the folks who have corrected
some of the previous attempts to calculate it...

(Mark James Boyd) wrote in message news:40252a00$1@darkstar...
F1y1n wrote:
high angles of bank and slow speeds (and ergo light weights too),
the inner wing is significantly slower than the
outer wing, and tacking on some knots is most
efficient (to keep the length of the inner wing nicely above stall)...

Mark Boyd


I don't believe your argument is correct. What determines the lift and
drag coefficients is angle of attact, NOT airspeed.

This is true, and if the glider is at a certain pitch angle straight,
level, and coordinated in still air, both wings are at the
same AOA. If the two wings are at different airspeeds,
like in a turn or skid, the two wings are at different angles of attack.

In a coordinated turn both wings are at the same AOA (please see my
reply to the other post in this thread). In an uncoordinated turn this
is not the case. Think of the direction of the airflow over the wing -
in a coordinated turn the airflow is always from the same direction
regardless of position on the wings; in an uncoordinated turn this is
not so, hence the AOA will be different. I was talking about
coordinated turns only (and I presume you were too in your original
post).

The inner wing is
flying at the same angle of attack as the outer wing, think about it.

Not if the wings are at different airspeeds. This is how we
do a spin. One wing is "more" stalled than the other wing
(i.e. has a higher AOA because it is the inside wing, and has less
airspeed).

Yes, in a spin the inner wing is flying at a higher AOA. But spin is
not a coordinated maneuver - see above. This analogy is moot.

Speeding up won't make you climb better.

Err...well, we are trying to sink less.

Well, it's half-full for me.

F1y1n wrote:
(Mark James Boyd) wrote in message news:40252a00$1@darkstar...

In a coordinated turn both wings are at the same AOA (please see my
reply to the other post in this thread). In an uncoordinated turn this
is not the case. Think of the direction of the airflow over the wing -
in a coordinated turn the airflow is always from the same direction
regardless of position on the wings; in an uncoordinated turn this is
not so, hence the AOA will be different. I was talking about
coordinated turns only (and I presume you were too in your original
post).

Hmmm...a little history. All of this argument came up
because we were discussing spins, and a bunch of posters
were talking about skids being the cause, and some of
us (the other posters) thought it was more due to
aileron stall and different airspeeds of different
wings in steep banks.

So we're trying to calculate the magnitude of the
airspeed difference caused by bank angles in steep turns
with long wings, vs. that caused by skids. How significant
is a skid vs. bank? Is it a skid or an accelerated skid (coarse use of
rudder) that's causes these spins?

Clearly the effect of bank on precipitating a skid
is a little surprising to other posters as well. It
was interesting and novel to me as well a few days ago...

I don't believe your argument is correct. What determines the lift and
drag coefficients is angle of attact, NOT airspeed. The inner wing is
flying at the same angle of attack as the outer wing, think about it.
Speeding up won't make you climb better.

In fact, the inner wing is not flying at the same airspeed. It has the
same angular speed, but it is transribing a smaller circle than the
outer wing and thus going a shorter distance in the same amount of
time. Both wings are sinking at the same rate, therefore, since the
tangential (straight line) speed of the inner wing is lower, its angle
of attack is higher. Same thing happens at the wheels of your car,
which is why you need a differential gear, to accomodate the
difference in speed between the inside and outside wheels during a
turn. The outside wheel travels a greater distance, though both have
the same angular speed.

(Chris OCallaghan) wrote in message . com...
I don't believe your argument is correct. What determines the lift and
drag coefficients is angle of attact, NOT airspeed. The inner wing is
flying at the same angle of attack as the outer wing, think about it.
Speeding up won't make you climb better.

In fact, the inner wing is not flying at the same airspeed.

Agreed. I didn't say it did.


It has the
same angular speed, but it is transribing a smaller circle than the
outer wing and thus going a shorter distance in the same amount of
time. Both wings are sinking at the same rate, therefore, since the
tangential (straight line) speed of the inner wing is lower, its angle
of attack is higher.

This I do not agree with. The angle of attack of both wings is the
same. The air is impacting both wings from exactly the same angle -
the wings are connected by the fuselage. The inner wing is flying
slower, and thus producing less lift, while the outer wing is flying
faster, and thus producing more lift. The angle of attack and hence
the position on the polar is the same for both wings. You are correct
that both wings are sinking at the same rate, but this does imply that
the both create the same amount of lift.

Incidentally, the reason why you need some contra-aileron in a steep
turn is precisely for this reason - to correct for the net torque
trying to lift the outer, faster, wing!

You are confusing AOA with sink rate. The sink rate is the same across
the airfoil, but AOA is dependent on sink rate and forward speed, so:

If an airfoil has a forward motion of 10 and sink rate of one, then
its angle of attack can be measured -- about 5.7 degrees. If we then
slowed its forward speed to 9 while maintaining a sink rate of 1, the
angle of attack would be higher: 6.3 degrees.

We agree that the angular speed is the same across the span. We agree
"that the inner wing is flying slower." The sink rate is the same
across the span. As you've stated, this is a given: the wings are
fixed to one another. Since AOA is dependent on both sink rate and
forward speed, then the inside wingtip must have a higher AOA.

Inner wing slower, higher AOA. Outer wing faster, lower AOA. Lift is
dependent on both AOA and speed. So even though the outer wing is at a
lower angle of attack, it is moving through the air more rapidly, and
producing slightly more lift than the inner wing. With resulting
overbanking tendency.

Balance this knowledge against the sailplane's response to a turning
stall. Inner wingtip typically drops first. Why? Because it has a
higher AOA. No aggrevation from the aileron required.

The point I'm replying to is:

I'm convinced that in very long wing gliders at
high angles of bank and slow speeds (and ergo light weights too),
the inner wing is significantly slower than the
outer wing, and tacking on some knots is most
efficient (to keep the length of the inner wing nicely above stall)...

I grant you that the AOA is slightly higher for the inner wing due to
the contribution from the sink, but this is negligible. Consider a
45deg bank, 45 knots. The turn radius (at the fuselage) is about 50
meters, so for a 15-m glider the speed of the outer wingtip is about
50knots, and the speed of the inner wingtip is about 40knots. If the
sink rate in this configuration is 1.5 knots, the difference in AOA
for the two wingtips is about 0.4 degrees. You will notice that (for a
good reason!) this is much less than the typical twist of a wing. You
cannot stall the inner wingtip in a steep turn without stalling both
wing roots first! For the same reason, the inner wingtip is NEVER on
the back side of the polar when thermaling. If it was the wing roots
would already be stalled.

To answer the original question - should one speed up when thermalling
with a steep bank - the answer is no. There are too many factors that
come into play - the twist of the wing as a function of position, the
wing profile as a function of position, the drag produced by the
aileron deflection needed to correct for the overbanking tendency as a
function of speed, and so on. In the end, these effects will tend to
cancel each other: if you speed up a little to bring the wing roots to
the front side of the polar you will a) create more drag on the wing
tips and b) need more aileron input to correct for the overbanking
torque and hence create more drag. I suspect that amount by which one
should speed up or slow down to optimize the sink rate in theory will
be much smaller than the speed of the turbulent currents in the
thermal, and thus utterly irrelevant in practice. Your time will be
better spent flying cleanly and in the core of the thermal rather than
trying to nail the speed to within 0.2 knots.


(Chris OCallaghan) wrote in message . com...
You are confusing AOA with sink rate. The sink rate is the same across
the airfoil, but AOA is dependent on sink rate and forward speed, so:

If an airfoil has a forward motion of 10 and sink rate of one, then
its angle of attack can be measured -- about 5.7 degrees. If we then
slowed its forward speed to 9 while maintaining a sink rate of 1, the
angle of attack would be higher: 6.3 degrees.

We agree that the angular speed is the same across the span. We agree
"that the inner wing is flying slower." The sink rate is the same
across the span. As you've stated, this is a given: the wings are
fixed to one another. Since AOA is dependent on both sink rate and
forward speed, then the inside wingtip must have a higher AOA.

Inner wing slower, higher AOA. Outer wing faster, lower AOA. Lift is
dependent on both AOA and speed. So even though the outer wing is at a
lower angle of attack, it is moving through the air more rapidly, and
producing slightly more lift than the inner wing. With resulting
overbanking tendency.

Balance this knowledge against the sailplane's response to a turning
stall. Inner wingtip typically drops first. Why? Because it has a
higher AOA. No aggrevation from the aileron required.