Puchaz Spinning thread that might be of interest in light of the recent accident.

Dave Martin wrote:

it must be realised that the pilot caused the inadvertent
stall in the first place by inappropriate use of the
controls. He is unlikely to start making skilful or
precise movements now.

LOL. Back to the "don't stall either wing in the first place"
technique.

Beware the
unnecessary use of coarse control, particularly rudder and particularly
near the ground!

IAN STRACHAN
Lasham Gliding Society

I did a few calculations of an imaginary glider with a stall speed
of 32 knots, a min sink speed of 43 knots, and
a wingspan of 87 feet.

In a 50 degree bank at 54 knots (good thermalling speed if you
believe)

www.stolaf.edu/people/hansonr/soaring/spd2fly/

the fuse and ASI says you are at radius 180 ft circling every
7 seconds. The inner wingtip is 3/4 of that distance, and
3/4 of that airspeed, and should be stalled. The outer
wingtip is 5/4 of that distance from center, and 5/4
of that airspeed, and producing excellent lift.

Now throw in a down aileron near the wingtip, increasing the
AOA of the inner wing. Now have the student
not compensating for adverse yaw, and the instructor yelling
"get that string centered right now!"

Now have the student jam in lots of rudder, and watch the
difference in airspeed and AOA during this coarse
movement.

This is probably why coarse rudder is often used to
coarsely demonstrate a spin entry...

This is also why I fly a glider with a short wingspan and
a weak rudder... (getting a worse L/D design was faster
than getting better skill)

"Mark James Boyd" wrote in message
news:401eb7ea$1@darkstar...
A spin means both wings have too high AOA and
one wing has more AOA than the other.

If you can change the AOA of both wings so they are unstalled,
using elevator only, and the stress from the now entered spiral
doesn't make the aircraft wings twist and shatter during recovery dive,
then fine, do that.

If you can't, then it would be great to have both
wings at the same AOA, then reduce the AOA. Rudder is
a possible way to do this (make both wings have the
same AOA by making them both the same airspeed, by
countering the yawing motion). In the ensuing dive
recovery, the wings are level. In some aircraft
these stresses are different than turn/spiral stresses
and the wing structure handles them better.

I suspect this is the reasoning behind
the PARE mnemonic, where rudder is used before elevator.

Power off (for them motorglider thingies)
Aileron Neutral
Rudder Opposite
Elevator forward enough to break stall

Of course, even this mnemonic doesn't work all the
time (sometimes extra power to make the tail surfaces
more effective is better, etc.).

So results for any generalization may vary...

I did this calculation for my Nimbus 2 and found a 14 Kt. speed difference
across the 20 meter span in a normal thermalling situation with the ship
dry. (45 Kts/45 degree bank.)

Pushing the envelope a bit by slowing up and tightening the turn, I found
the typical big wing roll-off toward the low wing, but it didn't seem like a
spin departure. What I think is happening is that the inside wing is on the
back side of the polar and outside of the drag bucket, but still not
stalled. This produces a pronounced roll and yaw into the turn which
develops into a spiral dive if allowed to continue. The recovery is the
same as an incipient spin, reduce the back pressure, let the speed increase
a bit, reduce the bank and stay coordinated.

Bill Daniels

Robert John wrote in message ...
I was taught this 'pause' between full opposite rudder
and stick forward and the wind 'shadow' effect was
the reason; However, since it has been proven that
even a Puchacz, which has a low(ish) tailplane, will
recover faster without the pause (Dick Johnson) and
most gliders have 'T' tails to which it doesn't apply
at all, I for one will not be teaching the 'pause'
to my students.
Rob John
Duo 'Si' K6 '350'


Then I hope you will read the revision to the AS-K 21 POH,
which updated/changed the spin recovery protocol to include the
'pause' based on flight testing, after a spinning fatality
in the K-21.
No pause, slower recovery.
Pause, more prompt recovery.
K-21 is a T-tail.

Beware broad judgments.
Please know your POH and its recommended procedures.
If you teach/deliberately enter spins, have a predetermined exit
altitude for non-responsive behavior, or don't bother wearing the chutes.

If there was on line access for the USAF Spin Eval report for the K-21,
I would make it available... but I have no electronic source.

Cindy B
www.caracolesoaring.com

The flight manual for the AS-K21 that I flew yesterday basically says
full opposite rudder, pause, and then stick forward. The manual also
had a note that some of the manual's contents had been included due
use by the USAF.

On the other hand, the flight manual for my LS-3a states to terminate
spins by "pronounced deflection of rudder opposite to spin direction
and careful pull out". I guess that means you don't have to move the
stick forward for spin recovery! Hmmm...must be magic!

Steve

Robert John wrote in message ...
I was taught this 'pause' between full opposite rudder
and stick forward and the wind 'shadow' effect was
the reason; However, since it has been proven that
even a Puchacz, which has a low(ish) tailplane, will
recover faster without the pause (Dick Johnson) and
most gliders have 'T' tails to which it doesn't apply
at all, I for one will not be teaching the 'pause'
to my students.
Rob John
Duo 'Si' K6 '350'

In a fully developed spin the tail surfaces can see
an
airflow that has a significant component coming from
underneath the tail surfaces. If the tail surfaces
are
'conventional,' (i.e. not a T-tail), and the elevator
and
horizontal stabilizer are on the fuselage, below the
rudder,
then forward stick produces a 'shadow' in this airflow
which
can block the lower portion of the rudder near the
elevator.
This 'shadow' is reduced when the stick is back. If
you
stand below the elevator and look upward (difficult,
I know)
and move the stick forward in a 1-26, for example,
this
'shadow' effect can be seen. Thus, I was told there
are
some POH's for conventional tail aircraft that recommend
using rudder *before* forward stick in the full spin
to
maximize the effectiveness of the anti-spin rudder.


At least this is what I recall as being the explanation
received from my first flight instructor. Does anyone
else
recall this 'explanation?'


Todd Pattist - 'WH' Ventus C
(Remove DONTSPAMME from address to email reply.)

Steve Pawling wrote:
...
On the other hand, the flight manual for my LS-3a states to terminate
spins by "pronounced deflection of rudder opposite to spin direction
and careful pull out". I guess that means you don't have to move the
stick forward for spin recovery! Hmmm...must be magic!
...

The stick forward is in some way implied by the "careful pull out", if
you keep the stick at the place which caused the spin, i.e. near the
back stop, the pull out would be rather agressive if not stalled.

Mark James Boyd wrote:
...
I did a few calculations of an imaginary glider with a stall speed
of 32 knots, a min sink speed of and
a wingspan of 87 feet.

In a 50 degree bank at 54 knots (good thermalling speed if you
believe)

www.stolaf.edu/people/hansonr/soaring/spd2fly/

the fuse and ASI says you are at radius 180 ft circling every
7 seconds. The inner wingtip is 3/4 of that distance, and
3/4 of that airspeed, and should be stalled. The outer
wingtip is 5/4 of that distance from center, and 5/4
of that airspeed, and producing excellent lift.
...

I don't completely agree with your computations. I agree with
the 54 knots, i.e. 43 knots multiplied by the square root of
the load factor at 50 degree bank. However the radius I find
for this speed and bank is 66 m (sorry, I prefer to do my
calculations in metric, because I know the formulas for metric
data), i.e 216.5 ft. A wingspan of 87 feet translate into 26.5 m,
the inner wingtip is inside the circle by an amount which is
the half wingspan multiplied by the cosine of 50 degree, this is
8.5 m or 27.8 ft. The ratio of the two radii is .87 rather than
..75 and the speed at the inner wing tip is 37.4 kt.

Anyway even with your values tis doesn't implies the inner wing tip
is stalled, because stall depends on AOA rather than speed. Of course
you need an increase of AOA in order to compensate for the
lower speed in order to keep an equal lift on both wings. Some difference
in AOA between both wings is already provided by the simple fact that
the glider is sinking, i.e. both wings have the same vertical component
of velocity but different horizontal ones. The complement is provided
by aileron deflection, which change not only the AOA but the whole
airfoil shape, so that the action is an increased Cl due to both changes
in AOA and shape. The stall case would be if the needed Cl would be higher
than the maximum achievable Cl, but this can't be decided just from the
value of the speed at wing tip.

Robert Ehrlich wrote:
Mark James Boyd wrote:
...
I did a few calculations of an imaginary glider with a stall speed
of 32 knots, a min sink speed of and
a wingspan of 87 feet.

In a 50 degree bank at 54 knots (good thermalling speed if you
believe)

www.stolaf.edu/people/hansonr/soaring/spd2fly/

the fuse and ASI says you are at radius 180 ft circling every
7 seconds. The inner wingtip is 3/4 of that distance, and
3/4 of that airspeed, and should be stalled. The outer
wingtip is 5/4 of that distance from center, and 5/4
of that airspeed, and producing excellent lift.
...

I don't completely agree with your computations. I agree with
the 54 knots, i.e. 43 knots multiplied by the square root of
the load factor at 50 degree bank. However the radius I find
for this speed and bank is 66 m (sorry, I prefer to do my
calculations in metric, because I know the formulas for metric
data), i.e 216.5 ft. A wingspan of 87 feet translate into 26.5 m,
the inner wingtip is inside the circle by an amount which is
the half wingspan multiplied by the cosine of 50 degree, this is
8.5 m or 27.8 ft. The ratio of the two radii is .87 rather than
.75 and the speed at the inner wing tip is 37.4 kt.

I forgot to change the wingspan to 90 feet to make the
math easy. Sorry. I was really just trying to make the
point that the wings have different airspeeds, and this is significant
at high bank angles and low speeds with long wings. If
this is untrue please let me know.

The bigger error of mine that you pointed out
was that I did the radius calculations assuming the wings
were level. This was incorrect on my part, and resulted
in a fairly large discrepancy....


Anyway even with your values tis doesn't implies the inner wing tip
is stalled, because stall depends on AOA rather than speed. Of course
you need an increase of AOA in order to compensate for the
lower speed in order to keep an equal lift on both wings. Some difference
in AOA between both wings is already provided by the simple fact that
the glider is sinking, i.e. both wings have the same vertical component
of velocity but different horizontal ones. The complement is provided
by aileron deflection, which change not only the AOA but the whole
airfoil shape, so that the action is an increased Cl due to both changes
in AOA and shape. The stall case would be if the needed Cl would be higher
than the maximum achievable Cl, but this can't be decided just from the
value of the speed at wing tip.

Exactly why I think AOA indicators halfway+ down the wings
would be nice. I've never heard of them on any gliders.
Why is this?

On Tue, 03 Feb 2004 15:59:18 +0000, Robert Ehrlich
wrote:

Mark James Boyd wrote:
...
I did a few calculations of an imaginary glider with a stall speed
of 32 knots, a min sink speed of and
a wingspan of 87 feet.

In a 50 degree bank at 54 knots (good thermalling speed if you
believe)

www.stolaf.edu/people/hansonr/soaring/spd2fly/

the fuse and ASI says you are at radius 180 ft circling every
7 seconds. The inner wingtip is 3/4 of that distance, and
3/4 of that airspeed, and should be stalled. The outer
wingtip is 5/4 of that distance from center, and 5/4
of that airspeed, and producing excellent lift.
...

I don't completely agree with your computations. I agree with
the 54 knots, i.e. 43 knots multiplied by the square root of
the load factor at 50 degree bank. However the radius I find
for this speed and bank is 66 m (sorry, I prefer to do my
calculations in metric, because I know the formulas for metric
data), i.e 216.5 ft. A wingspan of 87 feet translate into 26.5 m,
the inner wingtip is inside the circle by an amount which is
the half wingspan multiplied by the cosine of 50 degree, this is
8.5 m or 27.8 ft. The ratio of the two radii is .87 rather than
.75 and the speed at the inner wing tip is 37.4 kt.

Anyway even with your values tis doesn't implies the inner wing tip
is stalled, because stall depends on AOA rather than speed. Of course
you need an increase of AOA in order to compensate for the
lower speed in order to keep an equal lift on both wings. Some difference
in AOA between both wings is already provided by the simple fact that
the glider is sinking, i.e. both wings have the same vertical component
of velocity but different horizontal ones.

In a descending turn, which is what gliders do in turns, it is not the
case that both wings have the same vertical component of velocity. In
a stable descending turn the inside wing is always undergoing a
downward motion relative to the outer wing. This is one cause for the
inside wing to be at a higher AOA than the outer wing, and one reason
for the resulting earlier stall than the outer wing. In an ascending
turn, power airplanes I guess, it is the outer wing that is always
undergoing a downward moovement relative to the inner wing.

I found this difficult to visualize at first, but if you try "flying"
a stable descending "turn" with your hand you will experience it
clearly.


The complement is provided
by aileron deflection, which change not only the AOA but the whole
airfoil shape, so that the action is an increased Cl due to both changes
in AOA and shape. The stall case would be if the needed Cl would be higher
than the maximum achievable Cl, but this can't be decided just from the
value of the speed at wing tip.

....in a stable descending turn the inside wing is always undergoing a
downward motion relative to the outer wing. This is one cause for the
inside wing to be at a higher AOA than the outer wing, and one reason
for the resulting earlier stall than the outer wing.

Understood! What I don't unerstand is how much washout plays into this
equation. I would suspect that it would reduce this efffect but how much?

Tony V.