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

On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist
wrote:

Jim wrote:
I should not have gone farther than just the
observation that the inside wing in a stable descending turn is
going down while the outside wing is going up ( and the opposite
situation in an ascending turn).

This is incorrect. Both wings are going down (and both are
going forward). They both go down at the same rate, they go
forward at different rates (inner wing slower). If you think
about it, that means that the inner wing has a higher AOA.


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

Thank you for clarifying again for me. I sure love this stuff but
I sure don't have the background to truly understand it!
But I sure love flying gliders!

Jim

On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist
wrote:

Jim wrote:
I should not have gone farther than just the
observation that the inside wing in a stable descending turn is
going down while the outside wing is going up ( and the opposite
situation in an ascending turn).

This is incorrect. Both wings are going down (and both are
going forward). They both go down at the same rate, they go
forward at different rates (inner wing slower). If you think
about it, that means that the inner wing has a higher AOA.

I was just reading the Gider Pilot Manual by Ken Steward, pages 68 &
69. There he talks about the larger angle of attack difference in
shallow turns than in steep turns, because of the larger difference in
the circumference described by both wingtips, and concludes that steep
turns are safer than shallow turns.

I believe Steward left out an important factor, intentionally or not.
Even if the the distance between the inner and the outer circles is
larger in shallow turns, they also are much wider, and that has a
large impact in the AOA difference. Otherwise, the shalowest turn
would have the largest AOA difference, which is clearly not true.

So far this thread talked about one bank angle. Would anybody care to
compute the AOA delta for the whole range of bank angles? I guess it
would be given as a function of the wing span...

In article ,
jhandl wrote:
On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist
wrote:

Jim wrote:
I should not have gone farther than just the
observation that the inside wing in a stable descending turn is
going down while the outside wing is going up ( and the opposite
situation in an ascending turn).

This is incorrect. Both wings are going down (and both are
going forward). They both go down at the same rate, they go
forward at different rates (inner wing slower). If you think
about it, that means that the inner wing has a higher AOA.

I was just reading the Gider Pilot Manual by Ken Steward, pages 68 &
69. There he talks about the larger angle of attack difference in
shallow turns than in steep turns, because of the larger difference in
the circumference described by both wingtips, and concludes that steep
turns are safer than shallow turns.

I believe Steward left out an important factor, intentionally or not.
Even if the the distance between the inner and the outer circles is
larger in shallow turns, they also are much wider, and that has a
large impact in the AOA difference. Otherwise, the shalowest turn
would have the largest AOA difference, which is clearly not true.

So far this thread talked about one bank angle. Would anybody care to
compute the AOA delta for the whole range of bank angles? I guess it
would be given as a function of the wing span...

I'm figuring bank angle, airspeed, and wingspan are the three
factors. What would be really interesting is comparing the results
against the airspeeds given in the simple bank angle
tables for G loading found in many books, POH's and manuals.

If one flies the airspeeds for the bank angles shown on these
charts, for each glider (pick some popular ones), what is
the bank angle that causes the highest ratio of
airspeed between outer wingtip and inner wingtip?

Perhaps Bob Hanson at St. Olaf can be convinced to code this up

"Gary Boggs" wrote in message ...
The paper that Rich wrote on spin training that is posted on his web site is
a must read!

Thank you very much Rich.

--
Gary Boggs
Hood River, Oregon, USA


You're welcome, Gary! BTW, I'll be giving a seminar on Stalls & Spins
at the NW Aviation Conference in Puyallup, WA on Feb. 21 (4:00 PM) if
anyone's interested...

Rich
http://www.richstowell.com

Mark James Boyd wrote:

In article ,
jhandl wrote:
On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist
wrote:

Jim wrote:
I should not have gone farther than just the
observation that the inside wing in a stable descending turn is
going down while the outside wing is going up ( and the opposite
situation in an ascending turn).

This is incorrect. Both wings are going down (and both are
going forward). They both go down at the same rate, they go
forward at different rates (inner wing slower). If you think
about it, that means that the inner wing has a higher AOA.

I was just reading the Gider Pilot Manual by Ken Steward, pages 68 &
69. There he talks about the larger angle of attack difference in
shallow turns than in steep turns, because of the larger difference in
the circumference described by both wingtips, and concludes that steep
turns are safer than shallow turns.

I believe Steward left out an important factor, intentionally or not.
Even if the the distance between the inner and the outer circles is
larger in shallow turns, they also are much wider, and that has a
large impact in the AOA difference. Otherwise, the shalowest turn
would have the largest AOA difference, which is clearly not true.

So far this thread talked about one bank angle. Would anybody care to
compute the AOA delta for the whole range of bank angles? I guess it
would be given as a function of the wing span...

I'm figuring bank angle, airspeed, and wingspan are the three
factors. What would be really interesting is comparing the results
against the airspeeds given in the simple bank angle
tables for G loading found in many books, POH's and manuals.

If one flies the airspeeds for the bank angles shown on these
charts, for each glider (pick some popular ones), what is
the bank angle that causes the highest ratio of
airspeed between outer wingtip and inner wingtip?

Perhaps Bob Hanson at St. Olaf can be convinced to code this up

As a former math teacher, I liked to do a little math about the subject.
The result was that the airspeed difference deltaV between the CG and
the wingtip as function of bank angle phi, airspeed V, half wingspan b
and gravity acceleration g is given by:
deltaV = (b*g*sin(phi))/V
i.e. speed delta constantly increases with bank angle when the airspeed
remains the same (surprising ?), and decreases when airspeed increases
at a given bank angle (not surprising).

The surprising result for the difference as function of the bank angle
should be relativized as it is not aerodynamically correct to compare
the deltas at different bank angles and the same airspeed. We know that
the load factor increases with increasing bank angle, so lift must increase
in the same way. This can be done by increasing either AOA or airspeed,
but AOA is limited by the stall angle and similar aerodynamic conditions
are better maintained by increasing airspeed. If we express deltaV as
a function of bank angle at a given AOA, which may be characterized as
the speed V1 at which the glider would fly at this AOA with load factor 1,
we get:
deltaV = (b*g*sin(phi)sqrt(cos(phi)))/V1
and this has a maximun for cos(phi) = 1/sqrt(3), equivalent to
tan(phi) = sqrt(2), or phi = .955 radians = 54.7 degrees.


For the AOA delta at bank phi, airspeed V and sink Vz, I find:
deltaA = sin(phi)*cos(phi)*b*g*Vz/(V**3)
Here again it would be more significative to consider what happens at
a given AOA, characterized by the airspeed V1 and sink Vz1 obtained
at load factor 1, but as Vz is Vz1*(n**(3/2)) and V is V1*(n**(1/2)
when the load factor is n, the effect of the load factor vanishes and
the formaula remains the same:
deltaA = sin(phi)*cos(phi)*b*g*Vz1/(V1**3)
and the maximum is obtained when phi = 45 degrees.