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-   -   AF 447 Search to resume (part2) (https://www.pprune.org/tech-log/449639-af-447-search-resume-part2.html)

bearfoil 8th May 2011 18:01

Machinbird

Ballistic agglomeration: Descent in the Sea. I have never argued that there was an inflight breakup. Loss of the odd bit? PERHAPS

Chris Scott 8th May 2011 18:17

bearfoil, quote:
"If the decision had been made to turn around, why not a maximum rate turn to port off the airway at some point past LKP ??"

We hope soon to have an indication from the CVR of the point at which this flight ceased to be "ops-normal" from the pilots' viewpoint. Not sure what you mean by a "maximum-rate turn", but I cannot think of any circumstance where a crew already experiencing severe turbulence in a Cb or line squall would attempt to turn back. In heavy precipitation, the radar tends to be fairly useless until you come through it. The time to do a big turn is well before entry.

Once you are in it, you are going to avoid anything but the most gentle of turns. Prior to that, you can try to plan a zig-zag course between cells, staying clear of the worst stuff, but there's always the chance of a weaker/smaller cell eclipsing the big, nasty bugger beyond.

Doubt you would find any pilot who has actually decided to abandon a passage across the ITF (sorry, ITCZ) for lack of alleyways. In any case, the analysis suggests that this crew would have been anticipating a routine one. And evidence suggests that they had not deemed it necessary to secure the cabin in advance of this event.

gums
and Machinbird,
As you both tend towards the deep-stall scenario, maybe resulting from a sudden pitch-up, wouldn't it be interesting to see how far the high-altitude, low-speed part of the envelope was explored during certification?

takata 8th May 2011 18:24

Hi Gums,

Originally Posted by gums
The wreckage data ( distribution) we already have has a wealth of data that will be combined with the recorded parameters, IMHO. So we should get a very good replay of impact.

Right, we'll certainly know everything if the recorders are fully readable.


Originally Posted by gums
The big question is how in the hell the plane got there, ya think?

As far as the horizontal distance versus vertical distance goes, it is not betond my personal experience. As a light plane pilot previously recalled, if the plane is in a fully-developed stall, it can descend at a great rate without much forward velocity. In my own experience, one jet I flew had exceptional directional control in a full stall ( rudder versus aileron for maintaining a heading and picking up the low wing). Just release back pressure and it came right out. I could descend at 10,000 feet per minute or more ( the vertical velocity meter was pegged) with an indicated speed of about 60 knots. So I could go one mile and lose 10,000 feet or so.

We already tried to put some numbers together in the previous thread around this stage: http://www.pprune.org/tech-log/39510...ml#post6356297

My point was that it looks not likely that an early loss of control (0210-0210:20) happened to AF447 based on those published data I'll repost below.

From nine cases studied, we'll find an altitude loss in the range between 12,000 and 32,000 feet per minute, and a ground covered distance range between 3 to 17 Nautical Miles, while no event lasted more than 210 seconds overall after the LOC event:
__________________
9 unrecovered upsets from cruise level (data Metron Inc., BEA study):
Date.........ft/mn..Level..Dur....Dist..Type....-> Cause
07.12.1995..32,000..FL310...57s...8 NM..TU-154B -> Spiral / roll upset
19.12.1997..29,000..FL350...75s...5 NM..B-737.. -> Unknown
21.12.2002..27,000..FL180...40s...2 NM..ATR 72. -> Stall / icing
19.11.2001..26,000..FL260...59s...4 NM..IL-18V. -> Spiral / diving
01.01.2007..20,000..FL350..105s...9 NM..B-737.. -> Roll upset
15.07.2009..16,000..FL240...90s...5 NM..TU-154M -> Spiral / roll upset
22.08.2006..14,000..FL390..166s...3 NM..TU-154M -> Stall-spin
23.03.1994..12,000..FL310..156s...3 NM..A-310.. -> Spiral-spin / roll upset
16.08.2005..12,000..FL310..210s..17 NM..MD-82.. -> Stall
__________________
AF447 - timed cases for one upset from cruise level:
01.06.2009..35,000..FL350...60s..~5 NM..A-330 (1 mn)
01.06.2009..23,000..FL350...90s..~5 NM..A-330 (1.5 mn)
01.06.2009..17,500..FL350..120s..~5 NM..A-330 (2 mn)
01.06.2009..12,000..FL350..180s..~5 NM..A-330 (3 mn)
01.06.2009...9,000..FL350..240s..~5 NM..A-330 (4 mn)
01.06.2009...7,000..FL350..300s..~5 NM..A-330 (5 mn)
01.06.2009...6,000..FL350..360s..~5 NM..A-330 (6 mn)
__________________
1 Recovered upset: B747 China Airlines (17.02.1985)
1. 1014:50 -> 40,442 ft -> roll upset (4.8 G pull up)
2. 1015:23 -> 30,132 ft -> unreliable attitude data (5.1 G pull up)
3. 1017:15 -> 9,577 ft -> recovery & climb

Total upset duration: 145 sec.
1..10,310 ft lost...33 sec..18,700 ft/mn
2..21,155 ft lost..112 sec..11,300 ft/mn
3..31,465 ft lost..145 sec..13,000 ft/mn (average)

__________________
"Upset" definition (from “2008 - Airplane Upset Recovery Training Aid Revision 2”):
An airplane upset is defined as an airplane in flight unintentionally exceeding the parameters normally experienced in line operations or training. In other words, the airplane is not doing what it was commanded to do and is approaching unsafe parameters.

While specific values may vary among airplane models, the following unintentional conditions generally describe an airplane upset:
• Pitch attitude greater than 25 deg, nose up.
• Pitch attitude greater than 10 deg, nose down.
• Bank angle greater than 45 deg.
• Within the above parameters, but flying at airspeeds inappropriate for the conditions.

Mr Optimistic 8th May 2011 18:29

Turbine D, thanks. Read the report. Suppose it makes sense that the set of pitots, if exposed to the same conditions, would behave the same way. I wonder if modern systems would spot the short period of disagreement, as the icing was taking place before the data became consistently erroneous. Alternatively, disagreement would occur when the condition was recovering after the period when the system was listening to lies.

bearfoil 8th May 2011 18:45

takata

hello.

"Upset" definition (from “2008 - Airplane Upset Recovery Training Aid Revision 2”):
An airplane upset is defined as an airplane in flight unintentionally exceeding the parameters normally experienced in line operations or training. In other words, the airplane is not doing what it was commanded to do and is approaching unsafe parameters.


"While specific values may vary among airplane models, the following unintentional conditions generally describe an airplane upset:
• Pitch attitude greater than 25 deg, nose up.
• Pitch attitude greater than 10 deg, nose down.
• Bank angle greater than 45 deg.
• Within the above parameters, but flying at airspeeds inappropriate for the conditions.
"....

These are the limits of the autopilot supplied on the A330, as I understand. So the a/c will fly on auto pilot within this range, and beyond it (upset) the Pilots are to assume control?? The a/c is already 'upset'........ So this would qualify as a challenge, one would think.

captplaystation 8th May 2011 18:57

woodvale,

You may not post often, but good post when you do !

My personal suspicion is that this will all pan out to have been a cockpit light on /Capt snoozing/read the papers/wx radar brightness dim/ don't understand tilt or gain so try to fly over it / CB encounter , followed by loss of control, either partially or not at all excacerbated by frozen pitots & Airbus gubbins.

But, the problem in any case,I fear, originated with US ,gentlemen, the pilots.

Hope not, but I fear it will be so.

RetiredF4 8th May 2011 19:01

A test pilot’s point of view
 
Concerning definition of upsets, stalls and their recovery some words from year 2000 from airbus chief testpilot captain william wainwright.

Download PDF Airplane upset recovery

takata 8th May 2011 19:15

Hi Bearfoil,


Originally Posted by Bearfoil
These are the limits of the autopilot supplied on the A330, as I understand. So the a/c will fly on auto pilot within this range, and beyond it (upset) the Pilots are to assume control?? The a/c is already 'upset'........ So this would qualify as a challenge, one would think.

Come on Bear!
Those figures are certainly not A330 system related. They are taken from the whole Industry standards and issued in the courses of general training as a specific "upset" definition.

Each word/expression should have a clear definition (here boundaries to qualify as "an upset"). It's like your "en ligne de vol" stuff... or you will only be making some noise around the fuzz.

henra 8th May 2011 19:31


Originally Posted by EDLB (Post 6436719)
So it came down at a steeper than 45 degree angle in about 4 minutes. Anyone familiar with what type of spin that could have been?
Since it did not disintegrate to small pieces at impact I assume that the speed at impact had to be relative benign i.e. sub 200kn.


I'm with Chris Scott on this one.
I do not see any possibility to arrive at that point after 4mins without some kind of detour.
I had calculated before as did Chris Scott that this would require ~60kts ground speed and ~60kts vertical speed.
Given the Lift/drag dimension of an airliner and the available thrust I cannot see any way to achieve this (Btw. none of the planes in the russian study took longer than 2,5 minutes to arrive at the scene). This would be helicopter (/ light GA aircraft) territory !
Moreover you first would have to decellerate from ~470kts at 2:10:00 to less than 60kts (final speed would have to be even slower due to the higher speed at the beginning) in 4 mins without traveling further than 4nm.

My theory was a massive wing drop as initiating event for the turn.
Could have been intentional as well, however. We'll see.
hopefully.

mm43 8th May 2011 19:33

RetiredF4;

Thanks for that "Airplane Upset Recovery" link.

Wainwright dealt in general terms with the issues, but seems to have ignored some of the FBW limitations that are inserted into the equation when not in Normal Law - see below.

Machinbird;

... trying to break a stable stall by putting in a bootful of rudder and holding it.
The boot wouldn't have helped that much! Low IAS and only 7.9° of rudder allowed by the RTLU!

RR_NDB 8th May 2011 19:40

Parameters and data to FDR
 
What kind of info from WX radar (if any) goes to the data recorder?


But, the problem in any case,I fear, originated with US ,gentlemen, the pilots.

In my opinion one of the worst “advances” in aviation has been the reduction in the ability of the pilot to use or improve his technical skills, specifically in this instance, in radar interpretation.
Question:

The only information on the crew "decision making" to go through WX is the one that (may) is recorded in CVR or QAR?

henra 8th May 2011 19:42


Originally Posted by gums (Post 6436746)
Salute!


The big question is how in the hell the plane got there, ya think?

As far as the horizontal distance versus vertical distance goes, it is not betond my personal experience. As a light plane pilot previously recalled, if the plane is in a fully-developed stall, it can descend at a great rate without much forward velocity. In my own experience, one jet I flew had exceptional directional control in a full stall ( rudder versus aileron for maintaining a heading and picking up the low wing). Just release back pressure and it came right out. I could descend at 10,000 feet per minute or more ( the vertical velocity meter was pegged) with an indicated speed of about 60 knots. So I could go one mile and lose 10,000 feet or so.

Hi gums,

were it for the horizontal distance alone I would agree.
The problem is that the combination of horizontal speed and vertical speed does not fit at all for an A330 @210t coming down from 35kft.

The average RoD of 6000 ft/min is the typical spin RoD of a GA light aircraft. As I calculated earlier for an A330 @210 t I would expect RoD of ~20.000 ft /min at altitude slowing down to ~14.000 ft/min at S/L as terminal velocity e.g. in a flat spin.

That is the problem I have with the 4nm and 4,5 mins. I simply cannot see it descending with 60kts at a rate of 6.000 ft/min. At 60kts (~30kts IAS at altitude) the required lift for that RoD is not there.

electric-chris 8th May 2011 19:51

Forgive me if this is an ignorant question (I'm not a pilot), but it seems to be a strong possibility that the pitots iced rapidly due to supercooled water. If that is the case, could ice have also formed elsewhere on the aircraft in a similar manner and could that have potentially played a role in making the situation worse?

mm43 8th May 2011 20:14

takata posted the following summary of an A310 LOC.
Date.............ft/mn....Level....Dur....Dist..Type...-> Cause
23.03.1994..12,000..FL310..156s...3 NM..A310...-> Spiral-spin / roll upset

From what I understand, the aircraft departed FL310 cruise altitude into a gentle right-hand turn which then became a high bank roll and dive with the aircraft turning through 180 degrees before making a partial recovery. Control was finally lost at around FL130 and the aircraft did a combo spiral dive and finished up spinning in.

The point I am making is that any number of manoeuvres, controlled or otherwise, could conspire to place the aircraft close to the point that LOC occurred.

Machinbird 8th May 2011 20:25


That is the problem I have with the 4nm and 4,5 mins. I simply cannot see it descending with 60kts at a rate of 6.000 ft/min. At 60kts (~30kts IAS at altitude) the required lift for that RoD is not there.
Henra, did you forget the vertical component of velocity? They add vectorially. And as the velocity vector is largely pointed down, drag (which is substantial) helps support much of the weight of the aircraft.

HazelNuts39 8th May 2011 20:31


Originally Posted by electric chris
it seems to be a strong possibility that the pitots iced rapidly due to supercooled water.

Two fragments taken from BEA Report #1; Appendix 1; Meteorological study undertaken by Météo France: Meteorological situation in the tropical atlantic on 1st june 2009 - :


The conditions that are the most favourable to icing (presence of super-cooled water) are generally located in the lower central part of the cumulonimbus "tower", in an altitude range where the temperatures are between 0 and -25 °C. However, the icing conditions can persist down to -40 °C or less, that’s to say up to around flight level FL350, but ice crystals are encountered at this altitude.

the presence of super-cooled water at FL350 was unlikely and would necessarily have been limited to small quantities.
To this can be added that transport category airplanes are designed and demonstrated to be able to cope with large concentrations of supercooled water as can be found at lower altitudes.

takata 8th May 2011 20:43


Originally Posted by Machinbird
Henra, did you forget the vertical component of velocity? They add vectorially. And as the velocity vector is largely pointed down, drag (which is substantial) helps support much of the weight of the aircraft.

Right.
But, more vertical component means also much less duration (a faster rate of descent). You can't have both. Now, we are facing the opposite: a very long duration (the longuest ever recorded if LOC occured so early) but a very short distance from departure point. This will translate into a lot of horizontal component but in a very tiny airspace. Those figures should be somewhat in the range of what is already known or something very new happened... or is just wrong with the basic assumption.

grity 8th May 2011 20:55


Moreover you first would have to decellerate from ~470kts at 2:10:00 to less than 60kts (final speed would have to be even slower due to the higher speed at the beginning) in 4 mins without traveling further than 4nm.
henra theoretical: if the bird is without any control but starts with an extrem climb mayby 60 deg, this (~balistic) path will reach a maximal high of ~12700m (full change the kinetic horizontal energie into potencial energie), while reducing the speed to very low untill it will deap stall and if it then fall with a stable AOA of 50-60 deg. back to 10500m and afterwards down to zero.....

this path could need ~100 sec longer than yours and can end within a range of 8000m

Machinbird 8th May 2011 21:00


Right.
But, more vertical component means also much less duration (a faster rate of descent). You can't have both. Now, we are facing the opposite: a very long duration (the longuest ever recorded if LOC occured so early) but a very short distance from departure point. This will translate into a lot of horizontal component but in a very tiny airspace. Those figures should be somewhat in the range of what is already known or something very new happened... or is just wrong with the basic assumption.
Good Point Olivier
The reason for a very long LOC event is fairly obvious, the Airbus FBW system was still in control. But how was the airspeed lost so quickly? Logically the aircraft departed controlled flight with a strong upward vector.

Now if someone can match up these endpoints, we will likely come very close to the cause of the loss of AF447 without even using the recorder data.

Looks like great minds think alike.:} Grity is right on target.

henra 8th May 2011 21:01


Originally Posted by Machinbird (Post 6437088)
Henra, did you forget the vertical component of velocity? They add vectorially. And as the velocity vector is largely pointed down, drag (which is substantial) helps support much of the weight of the aircraft.


No, I did not, at least not intentionally (may be I was a bit unclear?!)
I said horizontal component was 60kts. Vertical was ~60kts as well.
But even if you combine both you end up with ~85 kts along the trajectory.
And these 60kts and 85kts is even TAS not IAS, so when entering it into a L/D diagram, you would have to use IAS which is roughly TAS divided by 1,8 at altitude, i.e. ~35kts horizontally or ~50 along the trajectory, increasing to 60/85kts at S/L.
That's not really much.
You would have to enter the L/D diagram of the A330 at an Alpha of ~45° with 85 kts and see if the combined Lift and drag higher than the 2100 kN (+ thrust of the engines).
That would mean ~1500kN along the trajectory, if no thrust of the engines available (idle, stalled) and ~ 1800kN at Cruise Thrust along the 45° path.
The projected surface at 45° being roughly 520 sqm, at 44m/s you get a resisting force of 620 kN at Cd = 1.
This means the Fuselage + wing would have to have a drag coefficient of ~ 2,5 - 3.
That's definitely too much.
I would expect something around 1.

At Cd = 1 it would mean ~80m/s => ~160kts along the trajectory. =>
~110kts horizontally + 110 kts vertically

On top of all that you would additionally have to shed the initial kinetic energy of it flying @470kts.


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