Spin

Looking at the windsock I'd say there was also a considerable wind gradient.
The aircraft had one wingtip within 1 span of the ground and the other a
good 15 to 20 m above that.

This little clip really is a classic, so many things wrong in one 10 second
period!

Ian




"ir. K.P. Termaat" wrote in message
om...
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

At 10:00 08 February 2004, 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.

That old red herring again!

The glider is flying in an airmass which is moving
over the ground at a constant rate. No additional
acceleration is required apart from that normally needed
in a turn to supply the turning force.

There may be some effect caused by descending/putting
the lower wing down through any wind gradient but this
actually improves the situation as the air is moving
'away' from the path of the glider more slowly and
will consequently cause some increase in airspeed.
(You can try the opposite of that effect by pulling
up from a downwind racing finish through a strong wind
gradient; watch the airspeed decay at an alarming rate).

The biggest problem is that the apparent speed over
the ground in say a 15kt wind jumps by 30kts and results
in people trying to reduce the ground rush by raising
the nose with no reference to the ASI.

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.

In article , Janos Bauer
writes:

What are the standards altitudes for such incident? Here are the list
I learnt: 50 straight landing, 50&100 one 180 degree turn, 180 two
turns or small circle. Of course in strong wind I would increase these
values.


Best rules, and those now taught in the UK are, if you can land safely ahead,
then you should do so.

Heights are not a good guide as they can mislead you into all sorts of trouble.
The higher the wind, the greater the land ahead opportunities up to heights
well above those where it is clearly safe to do a reasonably normal short
circuit. The tricky situation on a short run is a hot, nil wind day when the
margin between landing ahead and adequate height for a short circuit is very
narrow.

However, the GOLDEN RULE is first fly the aeroplane. i.e if both options are
marginal, recover from nose high and get into stable flight at correct speed,
then assess your options. there is usually ample time.

Barney
UK

There may be some effect caused by descending/putting
the lower wing down through any wind gradient but this
actually improves the situation as the air is moving
'away' from the path of the glider more slowly and
will consequently cause some increase in airspeed.
(You can try the opposite of that effect by pulling
up from a downwind racing finish through a strong wind
gradient; watch the airspeed decay at an alarming rate).

The biggest problem is that the apparent speed over
the ground in say a 15kt wind jumps by 30kts and results
in people trying to reduce the ground rush by raising
the nose with no reference to the ASI.

There is also the potential of poor turn coordination caused by the
perceptual changes that occur when maneuvering below pivotal altitude,
although that didn't appear to be the case in the video. The
suddenness of the departure was a bit surprising, but it seemed to be
triggered by the opening of the spoilers rather than by any obvious
lack of coordination.

- Rich Carr

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 hlink.net...
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.

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.

I'm certainly not speaking for Shawn, but what feels wrong-headed to
me about your description of acceleration RELATIVE TO THE GROUND,
and the need then to find forces to account for that apparent
acceleration, is that, in my view of things, no such acceleration of
the glider is occurring at all and thus no forces need be conjured.

The changes in groundspeed are not accompanied by changes in airspeed
and are a result of the movement over the ground of the air in which
the glider is turning.



On 9 Feb 2004 11:01:42 -0800, (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 hlink.net...
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.

ir. K.P. Termaat wrote:

Shawn Curry wrote in message hlink.net...

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.


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


Your description of the situation sounded naive. If you were
considering wind shear (decreasing wind velocity with altitude) you
didn't note that. As Robert noted in another response a turn is a turn,
as far as the aircraft is concerned, whether at 80 meters or 2000 unless
the wind changes during the turn.
I assumed you did not understand this, and figured a "usenet education"
is a poor (perhaps deadly) substitute for time spent with an instructor.
Thus my response.

Shawn

Robert John wrote in message ...
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

Hi Robert,

You are right Robert. The glider is unaware of groundspeed.
Looked several times at the short film of the crash where it is
obvious that the DG500 is flying to slow relative to the fast moving
air rather then to slow relative to the ground while having a lot of
tailwind (which is not very fast either).
During standard circling no accelleration forces in the longitudinal
direction of the glider are required to keep the IAS constant when the
glider makes perfect circles relative to the moving layer of air. From
the ground this looks quite different of course. But that is indeed
irrelevant.

Regards and thanks for your comment,
Karel