Nimbus 4 Accident

Michael Clarke wrote:

Derek, my Nimbus 4 flight manual says:

a) Apply opposite rudder.
b) Hold ailerons neutral.
c) Ease control sick forward until roation ceases and
the airflow is restored.
d) Centralise rudder and pull gently out of the dive.

No surprize: JAR22 explicitely demands "standard spin recovery". And it
must demonstrated from a spin of "at least 5 turns" and with "the most
unfavorable configuration". (Cited from memory, so the exact wording may
be different.)

Stefan

Just for general clarification, Section 4.5.6 of my
Nimbus 4T flight manual says:

“Note:

In order to achieve a high maneuverability, a favourable
c/g position (when using the fin tank) and a maximum
in ground clearance of the wing tips on take-off and
landing, it is always recommended to fill the inboard
water tanks first.”

I know it is the opposite for the Nimbus 3, I assume
the wings are just that much stronger on the 4. In
my experience it does make take off less fraught and
handling a bit nicer on the 4 than the 3.

Mike


At 18:30 11 July 2005, Roy Bourgeois wrote:
I think what Bert says is technically correct (wing
load doesn't change
when you open the air brakes) - but the distribution
does change a lot -
especially on a ship like the N3 & N4 where the brakes
are located inboard
on the inner panels. Stated differently, when the
brakes are opened the
outer panels are being asked to do more work supporting
the fuselage (and
non flying portions or the inner panels) than before
the dive brakes were
opened. The Nimbus 3 and 4 are placarded against
carrying water ballast
in the inner panel tanks with the outer panel tanks
empty for structural
reasons. You also must dump the inner tanks first.
The same structural
problem occurs when the dive brakes are open and that
part of the inner
panel becomes 'dead weight'. So - while the brakes
should be used to
prevent the glider getting to extreme speeds - we need
to be cautious about
suggesting that nothing bad is going to happen if you
open them at extreme
speeds.

Roy B. (Nimbus 3 # 65)

At 20:18 13 July 2005, T O D D P A T T I S T wrote:
Don Johnstone
wrote:

If you have to open the brakes, do so before Vne is
reached

Absolutely correct, carve that in stone.

The implication here is that if you find yourself extremely
nose down, but at an initially low speed, opening the
brakes
is a desirable action. I disagree.

Opening the brakes might be advisable if brakes were
able to
produce large amounts of drag - sufficient to limit
you to
speeds below Vne - but that is generally not true.
So, your
only other option is to get the nose back up to stop
the
acceleration. The only way to do that is to apply
the
maximum force possible in a direction perpendicular
to the
downwardly angled path of the aircraft to curve it
back
towards level.

A force perpendicular to the path of the aircraft is
called
'lift' and by opening the brakes you prevent much of
the
wing from producing the lift you desperately need to
bring
you back to level flight. This delays the critical
recovery. Up until you reach Va, you can operate the
wings
at maximum lift coefficient and produce maximum lift
without
risk of structural damage, and that's exactly what
you want
to do to get the nose back up.

In addition, by opening the brakes, you seriously increase
the risk that the pilot will overstress the aircraft
at
higher speeds. With brakes open, the max G load for
many
gliders is so low that the pilot simply does not think
he's
about to break anything. His built in warning system
does
not begin to go off until much higher G loads are felt.

Finally, if you open the brakes, you increase the altitude
loss significantly, a potentially critical factor in
a low
altitude recovery.

The proposal to use a tail chute does not suffer from
these
problems, as a tail chute does not decrease lift or
max G
limits. It also has the advantage of allowing recovery
from
otherwise unrecoverable spin modes.

I accept what you say my original reponse was to the
whole paragraph 'My main original point was that the
first action in
any sort of loss of control situation in a flapped
glider must be to select neutral or negative flap.
If you have to open the brakes, do so before Vne is
reached'

I standby that. It is important to select a non positive
flap setting and if the brakes are going to be used
it should be before VNE is reached.

At 04:48 16 August 2005, Gerhard Wesp wrote:
Ian wrote:
If you are in a situation (perhaps recovering from
a spin) with your nose
pointing 60 deg down below the horizontal and the
ASI reading 180 km/h and
accelerating rapidly. You could try:

[ leave A/B closed and pull 4G ...]

I have not worked out the maths of the two options
but I think that in
a clean glass ship, you would have a better chanced
completing the first
manoeuvre without exceeding VNE than the second.

I couldn't resist the temptation and did the maths.
I kind of lost the
habit, so it took me 20 minutes. The result is: My
intuition was
right: Do *not* use the airbrakes.

Even in an ideal ship (zero drag) and with the nose
pointing vertically
down initially, the peak speed is only 240km/h under
your assumptions
(4g pull-up and 180km/h initially). In a real ship
and with only 60
degrees down, the peak speed would of course be lower.

Here's the script with the phugoid equation (Mathematica).
Feel free to
modify the initial conditions and parameters and play
around. The
unknowns are v1, v2, the horizontal and vertical component
of the speed
vector.

kmh = 1 / 3.6
Ng = 4
g = 9.81
v0 = 180 kmh
tmax = 8
dt = .1

v =
NDSolve[
{
v1'[ t ] == Ng g * v2[ t ] / Sqrt[ v1[ t ]^2 +
v2[ t ]^2 ] ,
v2'[ t ] == Ng g * -v1[ t ] / Sqrt[ v1[ t ]^2 +
v2[ t ]^2 ] - g ,
v1[ 0 ] == 0 ,
v2[ 0 ] == -v0
} ,
{ v1 , v2 } ,
{ t , 0 , tmax }
][[ 1 ]]

Table[ ( Sqrt[ v1[ t ]^2 + v2[ t ]^2 ] /. v ) / kmh
, { t , 0 , tmax , dt } ]

Regards
-Gerhard

I think the ground might get in the way before I have
done the sums :-)

since induced drag increase due to airbrakes is=20
high=20

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Don't you mean "form" drag? I thought induced drag resulted (crudely)
from the wingtip vortices.

Also, are you sure that drag is proportional to lift?

Rgds,

Derrick Steed

Derrick Steed a écrit :
since induced drag increase due to airbrakes is=20
high=20

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Don't you mean "form" drag? I thought induced drag resulted (crudely)
from the wingtip vortices.

No. Wingtip vortices are only the visible part of the iceberg ;-)

Induced drag is the result of lift itself and is lower when the span is
higher and when the spanwise lift repartition is close to elliptic. And
with the airbrakes out, the lift repartition is very bad, closer to 3
wings of 5 m of span separated by no lift (at the airbrakes) than to 1
27 m wing !

Also, are you sure that drag is proportional to lift?

Induced drag is proportionnal to square of lift, while form drag don't
vary much with lift. Total drag definitely increases with lift, but to
quantify it that's where we need the proportion of induced drag...



--
Denis

R. Parce que ça rompt le cours normal de la conversation !!!
Q. Pourquoi ne faut-il pas répondre au-dessus de la question ?

In article ,
Denis wrote:

Derrick Steed a écrit :
since induced drag increase due to airbrakes is=20
high=20

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Don't you mean "form" drag? I thought induced drag resulted (crudely)
from the wingtip vortices.

No. Wingtip vortices are only the visible part of the iceberg ;-)

Induced drag is the result of lift itself and is lower when the span is
higher and when the spanwise lift repartition is close to elliptic. And
with the airbrakes out, the lift repartition is very bad, closer to 3
wings of 5 m of span separated by no lift (at the airbrakes) than to 1
27 m wing !

This appears to be confused.


Also, are you sure that drag is proportional to lift?

Induced drag is proportionnal to square of lift, while form drag don't
vary much with lift. Total drag definitely increases with lift, but to
quantify it that's where we need the proportion of induced drag...

And this is clearly totally mistaken, or misspoken. Two gliders flying
have twice as much lift as one glider, but they have four times as much
induced drag as one glider? And ten gliders have one hundred times as
much induced ddrag as one?

--
Bruce | 41.1670S | \ spoken | -+-
Hoult | 174.8263E | /\ here. | ----------O----------

Bruce Hoult a écrit :

Also, are you sure that drag is proportional to lift?

Induced drag is proportionnal to square of lift, while form drag don't
vary much with lift. Total drag definitely increases with lift, but to
quantify it that's where we need the proportion of induced drag...


And this is clearly totally mistaken, or misspoken.

.... or misunderstood ;-)

Two gliders flying
have twice as much lift as one glider, but they have four times as much
induced drag as one glider? And ten gliders have one hundred times as
much induced ddrag as one?

in this thread it is question of drag vs load factor - i.e. lift of one
glider, not of ten !


--
Denis

R. Parce que ça rompt le cours normal de la conversation !!!
Q. Pourquoi ne faut-il pas répondre au-dessus de la question ?