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# AF 447 Search to resume

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# AF 447 Search to resume

21st Aug 2010, 15:14

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HazelNuts39:
The more I think about the proportion of energy dissipation provided by altitude vs. air speed, with a relationsship of 4:1 for altitude, the more a high altitude stall seems probable to me.
If we stall it first, we would have to have only ~ 0,8m/s^2 to decelerate horizontally to 50kts.
The remaining energy would be dissipated by the vertical drag while falling belly first.
Vertical drag should be massive. Assuming a Cw ~1 in the Vertical plane (which should roughly fit), an A330 would arrive at a terminal velocity of ~150kts.
That combined with a forward 50 - 100 kts wouldn't be too far off the attitude which BEA concluded for the impact.

21st Aug 2010, 15:39

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henra;

I can't quite follow your numbers. To get kinetic energy, you need to add horizontal and vertical components of speed vectorially. If the airplane attained a speed of 50 kt at any time, it would need accelerate to a much higher speed (of the order of 180 - 200 kt) to be able to impact in nearly level attitude, whether stalled or not.

Drag is vertical only if there is no horizontal speed. The terminal velocity for cw=1 is 189 kCAS at sealevel.

regards,
HN39

21st Aug 2010, 16:13

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HazelNuts:
The terminal velocity for cw=1 is 189 kCAS at sealevel
Hmmm, what area did you take for that calculation ?
I took Wing area 362m^2
Fuselage 55m x 5,64m.
Weight 210t
=> I get 140 kts
Or do you have real data ?

Regarding the other numbers:
I was just referring to the forward speed vector. I assumed the forward component was just 50 kts upon impact, just a wild guess based on BEA's description of the liekly impact attitude.
For the pure horizontal deceleration that would be OK, but not for the overall energy state.
Regarding overall energy loss, you are absolutely right: I should have considered the vertical component as well. So the overall energy loss was rather equivalent to 43000 ft than to 45000 ft. Equivalent speed would be 980 kts then.
Thanks for that correction !

Doesn't change the overall picture, though.
This wasn't meant as a scientific analysis, rather a rough calculation, to get an idea about the general proportions

Last edited by henra; 21st Aug 2010 at 17:13.

21st Aug 2010, 17:30

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henra;

I was just considering the wing. The cw of a circular cylinder moving perpendicular to its axis at high Re is of the order of 0.4. But as you say,fair enough for a rough calculation. Shall we split the difference?

Originally Posted by henra
I assumed the forward component was just 50 kts upon impact, just a wild guess based on BEA's description of the liekly impact attitude.
The BEA speaks about attitude and vertical speed. The latter is based on examination of elements recovered from the cabin, and on 'arm 36g'. Regarding horizontal speed, bear in mind BEA's description of the separation of the fin under longitudinal (horizontal) inertia loads.
regards,
HN39

Last edited by HazelNuts39; 21st Aug 2010 at 18:28. Reason: consideration of horizontal component

23rd Aug 2010, 05:03

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Thoughts

The longer I look at the AF447 loss, the more it looks like the aircraft promptly entered a stall and then descended to the water. The question I asked earlier regarding how protection "collisions" are handled applies. When one protection commands an action and it conflicts with the limits of another protection, how is that handled?? Is it always in fact the logical prioritization?
The latest thought relates to the Vmo/Mmo protection in normal law. If the protection commands an additional amount of g to get the nose moving up and it "thinks" the aircraft is at Vmo, how hard does it pull g. If 1.65 g is commanded and only 1.6 g is actually available, what happens then? It would seem that the protection should apply g as mitigated by the envelope protection software, however if this software had an erroneous CAS input, it might be possible to pull so much g that a departure would result. Ok, you ask, what about Alpha Protect limits.
Well first we have a limits collision in a little explored part of the software (erroneous airspeed input). Does it behave the way we think it should?
Second, I am not entirely sure that Airbus is using actual angle of attack data in all of their angle of attack calculations. They may possibly be substituting inferred angle of attack based upon aircraft configuration data and airspeed.
And finally, it appears that if bogus airspeed data tripped the Vmo/Mmo protections, ACARS messages infer that the aircraft then shifted in a few seconds into Alt2 flight laws where it is possible to stall the aircraft. Has anyone run that kind of scenario in a SIM to see if there are additional implications?
When you see indications of anomalous behavior, it is time to look for a path to how the anomaly could be created. Maybe it is time for BEA to do some forensic software analysis. It might cost less than the actual search for the boxes and be more productive.

24th Aug 2010, 09:42

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Bogus Vmo/Mmo exceedance

Machinbird,

If the pitot intake and drains are blocked, the pressure rise due to heating depends on the volume of trapped air, which includes that in the tubing up to the pressure transducer in the ADM, and for the STBY pitot that in the pneumatic line up to the STBY ASI on the instrument panel. Wouldn't that difference result in less pressure rise in the STBY system?

If the airplane really reached the buffet boundary at FL384 with a deck angle of 27.8 degrees (see discussion 29th june), I wonder if the alpha protection authority (4 deg of elevator in QF72 report) would suffice to prevent stalling in normal law?

regards,
HN39

Last edited by HazelNuts39; 24th Aug 2010 at 11:39.

25th Aug 2010, 01:51

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HN39
If the pitot intake and drains are blocked, the pressure rise due to heating depends on the volume of trapped air, which includes that in the tubing up to the pressure transducer in the ADM, and for the STBY pitot that in the pneumatic line up to the STBY ASI on the instrument panel. Wouldn't that difference result in less pressure rise in the STBY system?
The same question bothered me, and you are completely correct when the air trapped in the pitot tubes piping is dry air.
When you are at ~FL350 and you have air and moisture trapped in the pitot tubes in contact with heated metal, the moisture will flash to steam and the steam will try to hold a pressure dependent on the temperature of the trapped water. Now this isn't hot steam that you are acquainted with from your tea kettle. It is really just cool water vapor. The situation is very analogous to the newer steam catapults with the boiling water type steam receivers. As the launch valves open and begins driving the catapult pistons, water flashes to steam and tends to hold the pressure constant.
With 3 identical pitot tubes that have just experienced common mode plugging due to weather, you are going to have 3 identical steam generators that will attempt to keep the pitot tube interior at a pressure characteristic of the trapped water temperature. As long as there is water to flash to steam, the pressure in all 3 pitot systems should be relatively constant. The transition from open to plugged pitot tubes should look like a step function on a pressure graph with the system with the greatest volume lagging behind the other two systems by a second or two and all stabilizing at the same (higher) absolute pressure value.

I see the calculation in the QF72 report. I'd like to do some homework before I comment on the second part.

Last edited by Machinbird; 25th Aug 2010 at 02:36. Reason: Mention second point.

25th Aug 2010, 09:40

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With 3 identical pitot tubes that have just experienced common mode plugging due to weather, you are going to have 3 identical steam generators that will attempt to keep the pitot tube interior at a pressure characteristic of the trapped water temperature. As long as there is water to flash to steam, the pressure in all 3 pitot systems should be relatively constant.
The problem is that it would require the average water temperature throughout the whole tubing to be identical.
Also the steam generators may be identical, but if the volume of the systems is signifcantly different the time to achieve equilibrium will be different between them.
With a long tubing in the stdby system it should take considerablư longer to achieve the same average temperature and thus steam pressure as in the systems with less volume.
Regarding flash boilers:
The constant pressure is achieved by the moving of the catapult in combination with a constant evaporation rate. If the piston wouldn't move, the pressure should rise.

I wouldn't rule out this theory though we may have to think further on how this could potentially happen.

25th Aug 2010, 17:12

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The problem is that it would require the average water temperature throughout the whole tubing to be identical.
Also the steam generators may be identical, but if the volume of the systems is signifcantly different the time to achieve equilibrium will be different between them.
With a long tubing in the stdby system it should take considerablư longer to achieve the same average temperature and thus steam pressure as in the systems with less volume.
Regarding flash boilers:
The constant pressure is achieved by the moving of the catapult in combination with a constant evaporation rate. If the piston wouldn't move, the pressure should rise.
Hi Henra,
The key temperature would be the peak water temperature in the systems. That would control the vapor pressure. See the chart below.In the case of a pitot tube, I would expect the water trap area of the tube to be the warmest spot on a frozen up pitot tube.
The flash boiler doesn't build pressure indefinitely because it is sealed in. There is a characteristic equilibrium pressure for each temperature. In the case of the catapult steam receiver, the water temperature generates about 550 psi steam pressure at equilibrium. (I haven't personally operated any of the newer catapults so I don't know the exact number).
The below chart gives pressure in Torr versus water temperature. 1 Torr= 1 mmHG.
50 Torr=1.96850366457 inches of mercury
100 Torr=3.93700732914 inches of mercury
200 Torr=7.87401465828 inches of mercury

25th Aug 2010, 18:14

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Terminal Velocity

HazelNuts, henra, Machinbird...

In your terminal velocity calcs, can you include a time variable -- how much time required to fall, say, 35,000 ft.?

With the leaning tower and the feather and stone experiment in mind, seems that a stalled A/C would descend more rapidly at higher altitudes, slowing gradually (smooth asymptotic curve) as air density (drag) increases until sea level. Perhaps 200t of mass overwhelms the influence of fluid density, but theoretically, I should think Terminal Velocity (where drag=weight) would vary by altitude...

Without simulations, hard to tell how much residual aerodynamic lift might also be at play in (what kind of) a stall. Not enough lift to keep the A/C flying, of course, but possibly not zero, either, thus a factor to include when computing Terminal Velocity and estimating duration of fall.

GB

25th Aug 2010, 18:16

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The key temperature would be the peak water temperature in the systems. That would control the vapor pressure.
Hi Machinbird,
The peak temperature would indeed define the local max pressure. But if the evaporated volume of water is small compared to the overall volume, the overall presssure throughout the system would be much lower, more a function of the average temperature in the system.

Regarding the Flash Boiler:
Yes indeed, it won't generate indefinite pressure, but in order to ensure the rapid expansion I would expect the temperature (and thus the local /instantaneous pressure) to be much higher than the average required for expansion and moving of the catapult. So if the catapult won't move the pressure should rise to some extent until the temperature of the whole steam in the system is identical to the boiler temperature. Then there would be equilibrium.
I don't think these catapults work at equilibrium of the whole system as in that case the force exerted on the catapult would decrease wgeb expanding/moving

25th Aug 2010, 18:39

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Perhaps 200t of mass overwhelms the influence of fluid density, but theoretically, I should think Terminal Velocity (where drag=weight) would vary by altitude...
Hi GreatBear,
The Terminal velocity should be roughtly constant, if yuu consider it being IAS/CAS and not the absolute value.

At 35000 ft this would be roughly 1.8 times higher.
So let's assume for a very crude first guess a factor of 1.4 - 1.5 for the average speed.
That would give you (again very roughly) somewhere between 200 - 250 kts.
Which would mean ~20000 ft/min.
So it would take close to two minutes for that vertical free fall.
But this is indeed an extremely crude calculation, not considering any aerodynamic or even local Mach effects.
To achieve terminal velocity at altitude (~280 - 300kts) would only take about 15sec. amd a drop of ~3000ft.
Edit:
This is pure theoretical as it would only apply for a pure vertical drop.
In reality this will be a transition and the plane will not immediately start to fall like a rock at the point of stall.
If you look at a wing polar you will notice that at the stall point the lift drops quite a bit but directly at the point of stall there is still quite some Cl, then decreasing with increasing AoA
/Edit

It has to be considered though that the initial altitude may have been higher, as this is one of the more likely sceanrios to get quickly into a stall in the given circumstances.
The initiating altitude may have been as much as 38000ft considering the initial Energy potential of ~45000ft (see posts above) and a stall speed of 220 kts IAS (equivalent altitude of 220kts ~7000ft. Alternatively: Speed loss 270kts to 220kts gives an altitude difference of ~3000ft).

Again, all numbers very crude, just to give an idea about the orders of magnitude we are talking

Last edited by henra; 25th Aug 2010 at 19:24. Reason: Edit block added

25th Aug 2010, 20:16

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Thanks, henra,

What is "known" or can be reasonably inferred is

AF447 was in serious upset at 0210.

Af447 was in the water by 0214 or 0215.

Henra, your calculation estimates ~15 secs for decel to stall in the high AOA scenario (ALT Law, no alpha prot) with a gain of altitude, staying under 3g (per JD-EE), and puts the A/C in the water soonest in ~2 minutes from stall.

Once stalled, no control surfaces work. Kick the rudder; nothing happens. Push forward on the stick; nothing happens; aerilons useless. Try thrust to the stalled wing's engine? Flameout? Try this; try that? A/C in a horizontal plane. High g rotational forces in cockpit. A/C pretty much descending vertically. Time running short. All the while, automatic ACARS messages are being sent as the situation evolves...

For the "where," I do hope they mount a Phase 4 effort and look near the LKP. For the "why," I hope they successfully recover the recorders.

So far, there are no answers and a great many plausible scenarios

GB

25th Aug 2010, 20:40

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AF447 was in serious upset at 0210.
Not necessarily.
From a pure timing perspective, it would be sufficient to change the flight path 3,5° up. That would mean according to my rough calculation that about one minute later they would be at 38000 ft and 220kts if Throttles remained unchanged (admittedly that is not 100% correct as with the speed decrease the drag would reduce quite a bit so the engines would have some SET, so energy conservation is not 100% constant which it was at 270kts with level attitude for the given N1/EPR). OK, let's take 5° up. (Alternatively, cut the throttle back)
Then they would hit the stall.
As I mentioned the drop would not be immediately like a brick, but after some decelleration the plane might be dropping at vertical velocities towards terminal velocity. lets add another 30 secs for this transition phase. Add 30 secs for the initiation of the pull up.
And then 2 minutes down.
And here we go: 4 minutes.
Just pure speculation though, but at least it could fit.

But even if this was the case, the most important question would be:
WHY ?? And that's why we need the black boxes.
Edit:
What also remains a mystery to me in this scenario, even if it was possible from a timing point of view is the question if a Conventional Tail Airliner can really deep stall straight down. So far I always thought this was a 'privilege' of T- Tails.
The only way I thought this would be possible with Conventional Tail was with elevator or at least Trim full up.
If this assumption would be correct the question would be again: Why ?
/Edit

Last edited by henra; 25th Aug 2010 at 21:01. Reason: edit Block added

25th Aug 2010, 21:20
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It is also quite possible the a/c was out of control, never to regain it, well ahead of a/p out. Sanchez shows the a/c exiting the cell. Prior to entering the cell, a/p had control. Entering the ring around the column could involve upwards of 7k fpm up. From cruise, the a/p pulls back some throttle and inputs Nose Down, to cage speed and altitude. So far so good. Maintaining altitude, throttled back, the a/c exits the up column, and encounters neutral or even down air. Now shes "losing" 7k fpm, (indicated, read a/p), but the a/c is too out of trim for the automatics in Pitch and Power, the very thing the book says to fly when a/p is lost. The autopilot trips out because the controls require beyond her installed limits, (expressed as total deflection, or "excursion", the pilots might be surprised because they are trained for additive a/p input, not "recover" reset" she rapidly Stalls, and the floor falls out, loss of a/p, u/a/s ensues (unusual attitudes), and the pilots do......what? Is it time to unfocus our steely stare at Thales?

The Vertical Stabilizer did not pop off the fuselage at impact. There is simply not enough damage to the root or to the assembly itself due to tumbling through the concrete of an upset sea at in excess of 100 knots (at the very least). The damage to the base of the Rudder is consistent with being dragged across a tailcone with ram and Rapid inputs at altitude and highspeed.

Last edited by bearfoil; 25th Aug 2010 at 22:02.

25th Aug 2010, 21:29

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@ GreatBear:
Just curious: Did Airbus actually stall one of these aricraft during original operational test to see what it would do? The reason I ask is that flat spins are a beast to get out of, if they can be, even for fighter aircraft that can enter them. Did the manufacturer have enough data to decide that going into full stall was not necessary to prove a point, given the operational environment expected of the aircraft? (Air Transport of persons).
Once stalled, no control surfaces work. Kick the rudder; nothing happens. Push forward on the stick; nothing happens; aerilons useless.
Flaps any help at this point (fully developed stall), or is relative wind too vertical for any airfoil?
Try thrust to the stalled wing's engine? Flameout?
I may be confusing this from another thread's discussion, but would not added thrust create a pitching moment (down?) Or do I have that arse backward, and thrust would induce a pitch up moment?
I don't have a clear mental picture of the moment arm the force of the engine would be acting upon in re the CG, but I do have an idea that airflow not roughly parallel to fuselage and engine center of thrust could lead to engine stall/flameout ... maybe a pull to idle would somewhat drop the nose ... but up near the coffin corner, oops, that may not be the help you need. (Ah, I see bearfoil and I have cross posted, he answers a small bit of my question on what to do with throttles ... )
Try this; try that? A/C in a horizontal plane.
I suppose that in a Hollywood film, they'd try to drop the gear, but I don't see that as much help in a fully developed stall in real life.
High g rotational forces in cockpit.
G manifested as negative g/eyeballs out, if I understand the scenario you are painting. Do I have that right?
A/C pretty much descending vertically. Time running short. All the while, automatic ACARS messages are being sent as the situation evolves...
Interesting, that in what looks like a flat spin, the geometry for the message center signal link could be maintainted ...

I presume you paint this scenario regarding a fully developed stall, which would go through first initial stall, then post stall gyration to get into something roughly "steady state" (albeit a rough ride as described).

This takes me to a question about recognizing stall and incipient spin, and
anti-stall inputs being called for before the stall / spin gets fully developed.

Is there training for that for line pilots? I noticed some pages back that a stall/upset of this sort could sneak up on you up near the coffin corner ... catch you by surprise.

About 20 years ago, the Navy lost a trainer (T-34C) to what was eventually figured out as a spiral, not a spin. One won't get out of a spiral with anti-spin inputs. Ten years later, two more were lost near Pensacola from similar cause. Two fatalities each event, six dead in three wrecks.

In the meantime, on the wet wear side, the training philosophy focused on stall and spin prevention, even though a spin was a standard training maneuver. (And a fairly benign one at that, all things considered). What was lacking somewhat was training on recognizing the difference between a spiral and a spin.

One could, with a poorly executed stall to spin entry by the student, end up in a spiral ... it can sneak up on the IP if he misses that roll input at spin entry.

So with the idea of the stall/upset/spin sneaking up on you, you paint a picture of there being a point of no return in a fully developed stall/upset/spin.

Are there anti stall/recovery inputs for this model aircraft? I realize it's built as a tranpsport, not a trainer or fighter aircraft. Some of the other threads here (and on tech discussion forum) indicate that in some heavies, there is a way out of such upset, though one needs quite a bit of altitude.

Yet even with a recovery via flight controls, it seems that a full throw on the vertical stab (rudder opposite spin?) might, if the rudder bites, lead to material failure whilst one is in the middle of getting out of that rotation! At this point one is short at least one flight control surface and still in upset ...

Taking that thought a step further, and I think this has been raised a few dozen pages ago, I wonder if your described scenario doesn't suggest a plane breaking up some time interval before impact with the ocean, a prospect that considerably challenges the search effort.

Where did what bit land? The front bit would not necessarily be oriented in the direction of flight if things began to fall apart waaaay up there during the flat rotation.

25th Aug 2010, 21:34
bearfoil
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The Bonanza makes a very iffy training platform, what with ADs and all.(Weren't they all lost via broken main spar One for the AD and the next two chasing the correct strength?

The VS was lost in a Rudder reversal, with 7.9 degrees available and 16 total, rapid and non consonant deflection at M.82, I doubt the VS lasted beyond the first out of phase reversal. Antenna considerations place the VS/separation well into the upset, when in a Stall that is fully developed, the controls are not being operated by cool customers. Can we agree that the pilots were at least, out of phase with the upset? Hence no recovery. Cabin Pressure alert likely happened at nearly cruise altitude, but no where near 9,000 Feet. The BEA should open a shop selling two-legged stools. So that's one for ACARS done at ~35k.

Last edited by bearfoil; 25th Aug 2010 at 22:06.

25th Aug 2010, 21:40
bearfoil
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Lonewolf50

It was a very long time ago, and a very unpopular position to take, at the time.

Knew you'd arrive. Pax were lost at altitude, there is no way to explain floating (assumed to be belted but unbelted at discovery) unbelted dead people, with the precise but off target injuries inventory proposed by BEA. Like the VS at impact separation pronouncement, the injuries have a glaring blunder.

I would welcome a discussion of maneuveringVS loss at altitude.....

Last edited by bearfoil; 25th Aug 2010 at 21:52.

25th Aug 2010, 22:17

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The following quote reporting on the cause of a BA1-11 prototype test flight crash in 1963 caught may attention - more to do with the wording than anything else.
The aircraft was evaluating stall characteristics at varying center of gravity locations when the flight crew found the flight controls unresponsive after entering a stable stall and the aircraft struck the ground at a wings level attitude with a high rate of descent and little forward speed.
Bearing in mind that that BA1-11 was an aft powered T-Tail, and the circumstances of its crash identified the "deep stall" condition, I couldn't resist being drawn to the "center of gravity" comment and the possible implications of that in the case of AF447 if in fact it entered a stable stall with a +20°nose-up attitude and low IAS. The later part of the quote replicates the BEA's description of the crash.

I believe the original crash report determined that G-ASHG descended from FL180 in 80 seconds - 13,850 fpm. How accurate that figure is or how it was determined, I have no idea.

Some important questions spring to mind in respect of AF447 -
(a) Was it also in a flat spin, i.e. tail rotating to port on impact?
(b) Likely engine flame-outs?
(c) Why no ACARS WNG from the FADEC on engine(s) status?
(d) Relevance of RTLU WNG. Was the rudder booted?
mm43

25th Aug 2010, 22:21
bearfoil
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It most likely had six thousand pounds of fuel in the tail (trim) tank adding a bit of impetus to any noseup control. No need for Warning from FADEC, the engines soldiered on. The auto pilot trip limits are rather impressive, I doubt any control was had after a/p left the loop. RTLU was set at 16 degrees sweep, per BEA. Without a/s or statics, the game is up. The ACARS was a snitch after the fact. I think no one was alive to feel the impact. The litany of ACARS are all we have. Wanting to "fly" is the design consideration for all commercial a/c, and she did what she could, alone, and witless.

It seems intuitive, with all respect.

1. A "Born Free" autopilot

2. Auto flight in trash Wx

3. No positive control (by definition) at A/P OFF

If the a/c was nosedown and behind on power leaving the unstable (VS read), Mach Limit, Yes? (ACARS). Recovering with full power and nose up, that's the ranch, Hank. Especially if the pilots started a "recovery" before the a/p quit, and inputs were ADDITIVE. With an emphatic Pitch Up, the a/p quits, that is unlikely but possible. It's very quiet, must be post last call.

bearfoil

Last edited by bearfoil; 25th Aug 2010 at 22:47.