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

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:
...
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.

On Tue, 03 Feb 2004 14:41:20 -0500, Tony Verhulst
wrote:


....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.

Really good question! I don't know. Since washout is, in a sense, a
relative term -- that is washout produces a lower AOA at the wing tips
compared to the AOA at the wing roots -- my guess is that in all cases
where AOA is critical the wing tip washout delays the effects we might
expect from what we see of the nose attitude of the aircraft. But
then, this is really not saying anything new!

....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?

Really good question! I don't know. Since washout is, in a sense, a
relative term -- that is washout produces a lower AOA at the wing tips
compared to the AOA at the wing roots

After thinking about this for a while, I suspect that it (washout)
doesn't matter. After all, both wings tips have an equal amount of
washout and so the net effect cancels out. The lower wing tip will still
have a higher angle of attack than the upper and will still stall first.
In this case, the effect of washout is a (wait for it :-) ) wash.

Tony V.

On Wed, 04 Feb 2004 11:14:33 -0500, Tony Verhulst
wrote:


....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?

Really good question! I don't know. Since washout is, in a sense, a
relative term -- that is washout produces a lower AOA at the wing tips
compared to the AOA at the wing roots

After thinking about this for a while, I suspect that it (washout)
doesn't matter. After all, both wings tips have an equal amount of
washout and so the net effect cancels out. The lower wing tip will still
have a higher angle of attack than the upper and will still stall first.
In this case, the effect of washout is a (wait for it :-) ) wash.

Tony V.

Makes sense to me! Thanks.

Jim

Jim wrote:
...
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.


Can't understand that. If both wingtips have a different vertical component
of velocity, the vertical distance between them should change,
increasing the bank angle if the inner wing sinks faster than the outer
one. This difference must anyway stop at 90 degrees bank. But as long
as the bank angle remains constant, both wings should have the same
vertical component of velocity.

Robert Ehrlich wrote:
Jim wrote:

...
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.



Can't understand that. If both wingtips have a different vertical component
of velocity, the vertical distance between them should change,
increasing the bank angle if the inner wing sinks faster than the outer
one. This difference must anyway stop at 90 degrees bank. But as long
as the bank angle remains constant, both wings should have the same
vertical component of velocity.
Relative to what? - is the point.

You are correct that their vertical component of velocity must be the
same because of geometry, if the bank angle remains constant. However,
because the inner wing is describing a smaller diameter spiral the
relative wind will present at a higher angle of attack on the inner wing
tip - relative to the outer wingtip. Velocity relative to the ground is
not entirely sufficient to understand what is happening in three dimensions.

In the same time the inner tip travels a smaller distance, but descends
the same vertical distance, hence the greater angle of descent, not
rate. People seem to continuously confuse rates and angles? Airflow
behavior is very dependent on angles and chord wise component of airflow
velocity...