Hi Karel,
I do not follow your explanation.
If I carry out the same 180 degree manouver at 5000
feet, even in a 50kt wind, both I and the glider are
quite unaware of groundspeed. No change in attitude
is required or made.
The only difference in doing it at 100 feet is surely
the close view of the ground and the APPEARANCE of
changing speed which may cause me to lower or raise
the nose when I should not.
Regards
Robert
At 19:06 09 February 2004, Ir. K.P. Termaat wrote:
Hi Shawn.

Since 1978 I am an instructor myself and teach aerodynamics
to new
pilots as
well as new instructors since then. Next month we will
have a
discussion in
our instructor's team on the matter of spinning and
especially on how
to
avoid this killing phenomenon when happening at low
altitude. If you
don't
understand my wordings please let me know; I am quite
willing to
elucidate
on what I sayd. If you think my interpretation of the
Magdenburg crash
with the DG500 is wrong please explain, I am quite
willing to listen
to better theories about this. Something like 'you
need .... ' doesn't
help much Shawn.

Karel, NL


Shawn Curry wrote in message news:...
ir. K.P. Termaat wrote:

Did some simple calculations to get an idea of what
caused the spin of
the DG500.
If the glider flew initially with an IAS of 100km/h
and had a headwind
of say 25 km/h then its speed relative to the ground
is 75km/h. If
after making the 180° turn back to the airfield the
glider flew again
with an IAS of 100km/h but now with a tailwind of
25km/h, then its
speed relative to the ground is 125km/h. This means
that during the
180° turn the glider had to be accellerated from
75km/h to 125km/h
relative to the ground.

For a banking angle of 45° and an IAS of 100km/h
one finds from simple
mathematics that a 180° turn takes 8.9 secs when
properly flown. The
forward accellaration of the glider during the 180°
turn must then be
(125-75)/(3.6)/8.9=1.56m/s2 to come out at the same
speed of 100km/h.
Suppose the mass of the glider (including the pilot)
is 650kg, then
the force needed to accelarate the glider with 1.56m/s2
is 650x1.56 =
1014kgm/s2 or 1014N.

Where does this force come from. Indeed, from gravity.
The glider must
pitch down to keep its IAS up. With a glider mass
of 650kg, its weight
is 650x9.8=6370N. The pitch down angle must then
be
arc(sin)1014/6370=9.2°. Add to this a normal glide
angle of 1.4° (for
a glide ratio of 40), then the total pitch down angle
during the 180°
turn of the DG500 should have been over 10°.

If the pilot does not move his stick quite a bit
forward to achieve
this relative large pitch angle, the glider will
loose its IAS, then
stall and spin. This looks to me what happened unfortunately
with the
DG500 at Magdenburg.

Karel, NL

You need to have a good long talk with your instructor.