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# AF 447 Thread No. 5

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# AF 447 Thread No. 5

25th Jul 2011, 19:43

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it was said before that the change of the direktion horizontal vers vertical cost an extra portion energie.... so the calculation was not much/exact valid it was under a limit value view, and yes it is possible to climb with an bird and change Ekin into Epot... (in detail the pressure on the water-surface deep under the bird has to rise a little bit, and the earth will change her position on her way round the sun also a liiiiitle bit)

a calculation just to control the plausibility of datas

and due to the lots of lessknown faktors (AoA, Cl, Cd, Cm, temp. density, spezific weight, reynold.....) it is just a practician desicion of the calculator if he will more talk over the marginal conditions or over the basic applications

and yes the used calculations of Ekin has some mistake in turbulence conditions with changes in the airspeed, this mostly cost a second portion of energie....

25th Jul 2011, 19:59

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One last admission from my side to end this energy discussion:
When I simply enter the 275 kts into the simplified formula of an E6B calculator I get 467,5 kts as TAS.
When calculating considering the Mach values I get between 475 kts (M0,80) and 490kts (M0,82) for an ISA temperature of -54,3° (ISA standard atmosphere at 35000 feet).

Depending on the real temperature this value can vary significantly and therefore the TAS can probably have been easily been 460kts as well as 500kts. We shouldn't read too much into it without knowing OAT.

Mach dependent TAS:
a0*M*sqrt(T/T0)
where
a0: speed of sound at S/L
M: actual Mach number
T0: ISA temperature at sea level in Kelvin
T: actual temperature.

25th Jul 2011, 20:08

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Originally Posted by henra
... when you are flying at 100kts and suddenly a tailwind of 100kts hits you....
henra, you're making a mistaek....
Your aircraft has mass/inertia, so it will not be instantly dragged along by your 100 kts tailwind. Your airspeed will now be 0 kts. Not a healthy situation....

25th Jul 2011, 20:17

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Originally Posted by ChristiaanJ
henra, you're making a mistaek....
Your aircraft has mass/inertia, so it will not be instantly dragged along by your 100 kts tailwind. Your airspeed will now be 0 kts. Not a healthy situation....

That is EXACTLY the point I was trying to make.

It was my point about change of state of the reference system in reply to his remark:
Yeah, suppose the windspeed goes from 100 knots back to zero, does this change the Ekin of the airplane ? If not, then you will need to take the groundspeed as speed reference.
My point being that for the problem of loss of effective Ekin to trade for altitude it makes no difference if you have a headwind of 100kts while flying 100kts, i.e. being at a standstill with regard to the ground and the wind suddenly stopping or
you are flying at 100 kts and are suddenly hit by a 100kts tailwind. In the latter case you have no airspeed left to fly after being hit by the wind either.

Seems my explanation was a bit brief last time..

Edit:
To give another example:
You have a Microlight tied to the ground and the wind is blowing at 70kts. What will happen if you cut the ties?
It will lift off and start to fly backwards with regard to the ground. You have traded speed for altitude at that moment. And it had kinetic energy with regard to the air at that moment although with regard to the ground it didn't. It's all about relevant reference systems....

Last edited by henra; 25th Jul 2011 at 20:33.

25th Jul 2011, 20:32

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henra,
Sorry, I dropped into the Ekin discussion halfway. I agree with your point of view.

25th Jul 2011, 21:29

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At post #657 HN39 made the following "tongue in cheek" comment -
As a student I piloted gliders. These routinely fly tight circles to stay in thermals. In a strong wind there is no airspeed loss turning downwind, nor gain turning upwind. I've never quite figured that out, may be as a physicist you can.
I don't think that everyone picked up on what he said, but the discussion over KE principles has been interesting and informative.

henra;
I believe the forecast temperature was -46°C at FL350, and I wouldn't be surprised if it was even warmer where AF447 was, e.g. -42/-43°C.

Last edited by mm43; 25th Jul 2011 at 22:09.

25th Jul 2011, 21:39

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Originally Posted by mm43
At post #657 HN39 made the following "tongue in cheek" comment -I don't think that everyone picked up on what he said, but the discussion over KE principles has been interesting and informative.

henera;
I believe the forecast temperature was -46°C at FL350, and I wouldn't be surprised if it was even warmer where AF447 was, e.g. -42/-43°C.
Thanks mm43 !

That would give us a TAS of 489kts at M0,80 and 500kts at M,82.
Looking at the setting of the RTLU at the time of the beginning of UAS a Mach speed of roughly M0,80 seems likely. So we would be back somewhere around the 490kts and ~4000ft.

25th Jul 2011, 21:44

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Originally Posted by CONF iture
• Should not we have expected a relevant ACARS message ?
• If 2 ADRs self auto detect at fault, should not we get SPD and ALT flags accordingly ?
a) PROBE-PITOT fault seems to be the relevant ACARS, by logic priority, over ADR faults which should have been compiled under the same ATA 341.
b) ADR fault is not self detected (there is no ADR internal failure) ; monitored channels (airspeed, Mach,...) are declared faulty by FMGCs and FCDCs, but they are still displayed.

25th Jul 2011, 21:53

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@HazelNuts39
Originally Posted by Dutch M
I'm not a pilot, though do have a degree in Applied Physics

As a student I piloted gliders. These routinely fly tight circles to stay in thermals. In a strong wind there is no airspeed loss turning downwind, nor gain turning upwind. I've never quite figured that out, may be as a physicist you can.
Yes, that is also one of the aspects I considered, though did not have enough time to dive into. Though 2 aspects might be relevant in this:
- Speeds are relatively low, glider speed and "jetstream".
- An important control target is the constant airspeed. Small rudder changes and/or differences in thermal-absorbing might hide the effects in speed changes.
- An experiment might be to circle around and measure the pitch angle and rudder usage during the circles. I would expect a somewhat sinus shaped change in rudder/pitch position.

Let me change the challenge a little:
- Land your glider with a airspeed of 30.25 m/s and a headwind of 30 m/s (just enough to move a little bit related to the ground).
- The presumed TAS based Ekin would be 0.5xmx915 =458 m J
- The moment the glider does touch down, everything becomes ground speed based and the ground speed based Ekin would be 0.5xmx0.25x0.25 = 0.03 m J.

Where did the Ekin of (458 - 0.03) m J go ?????

25th Jul 2011, 21:59

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Originally Posted by Dutch M
Let me change the challenge a little:
- Land your glider with a airspeed of 30.25 m/s and a headwind of 30 m/s (just enough to move a little bit related to the ground).
- The presumed TAS based Ekin would be 0.5xmx915 =458 m J
- The moment the glider does touch down, everything becomes ground speed based and the ground speed based Ekin would be 0.5xmx0.25x0.25 = 0.03 m J.

Where did the Ekin of (458 - 0.03) m J go ?????

It got nowhere. The only energy that got converted into heat was the 1/2*m*(0,25m/s)^2 during braking on the ground.
With regard to the air the remaining 1/2*m*(30m/s)^2 is still in the glider. Being fixed on the ground it is effectively being towed through the air at 30m/s by the ground with regard to the air. See my example of the microlight.
If the wind breezes up by 1m/s to 31 m/s it will lift off again. Or push it forward at 1m/s and it will take off.
If the air is still you will have to push at 31m/s. Big difference for you, no difference for the glider for its ability to lift off and fly. The difference for the glider is how it will move in relation to the 'other' reference system, i.e. the ground.
That's why aircraft prefer to take off and land against the wind. Mother Earth gives a certain amount of kinetic energy to fly to the Aircraft for free. For the same Aircraft you need less energy (thrust*distance) to leave the ground. And upon landing you have to dissipate less energy in your brakes. Still in between you were equally able to fly as if you took off and landed with the wind.

The trick is that you need to understand that for an Aircraft once it has left the ground, the movement of the ground below is totally irrelevant for the ability to stay in the air. And it stays irrelevant until it contacts the ground again. In between only its relative movement through the air is relevant.

Last edited by henra; 25th Jul 2011 at 22:27.

25th Jul 2011, 22:05

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Originally Posted by mm43
At post #657 HN39 made the following "tongue in cheek" comment -I don't think that everyone picked up on what he said....
I did, but I don't have all the answers, and formulas, and figures, at my fingertips any more...

To stir up the discussion, what about a 'breakdown' of the Ekin of a Fieseler Storch (or similar) flying at about 50kts in a 50kts headwind ?

25th Jul 2011, 22:29

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@CONF iture

Ground speed is affected, but not airspeed.
Once airborne, the glider is in the air mass that is moving at the speed of the wind, but for the glider itself regarding that air mass, there is no notion of tail or head wind.
Sure, but the remark is, the airspeed should change when the inertial speed stays the same, whereas the airspeed is perceived to be constant (which in effect isn't, but this is masked by other factors).

25th Jul 2011, 22:35

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@rudderrudderrat

Simply use the air mass as your frame of reference. It may be moving horizontally over the earth's surface, but the conservation of KE + PE still holds.
IAS is just a measure of dynamic air pressure, it must be converted to TAS.
Unfortunately it isn't. See my example about the landing glider, a few posts back. The airframe can't be the reference point for conservation of energy.

25th Jul 2011, 22:42

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New report

The BEA to publish another report on Friday?

25th Jul 2011, 22:43

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@henra

It got nowhere. The only energy that got converted into heat was the 1/2*m*(0,25m/s)^2 during braking on the ground.
With regard to the air the remaining 1/2*m*(30m/s)^2 is still in the glider. Being fixed on the ground it is effectively being towed through the air at 30m/s by the ground with regard to the air. See my example of the microlight.
If the wind breezes up by 1m/s to 31 m/s it will lift off again. Or push it forward at 1m/s and it will take off.
If the air is still you will have to push at 31m/s. Big difference for you, no difference for the glider for its ability to lift off and fly. The difference for the glider is how it will move in relation to the 'other' reference system, i.e. the ground.
That's why aircraft prefer to take off and land against the wind. Mother Earth gives a certain amount of kinetic energy to fly to the Aircraft for free. For the same Aircraft you need less energy (thrust*distance) to leave the ground. And upon landing you have to dissipate less energy in your brakes. Still in between you were equally able to fly as if you took off and landed with the wind.

The trick is that you need to understand that for an Aircraft once it has left the ground, the movement of the ground below is totally irrelevant for the ability to stay in the air. And it stays irrelevant until it contacts the ground again. In between only its relative movement through the air is relevant.
Yes, the ability to stay in the air is a function of airspeed.
The inertial Kinetic energy is not a function of airspeed, a very subtle difference......

So the amount of speed which can be bled of is related to the min. airspeed to stay in the air. The actual amount of Ekin released is depending on the absolute speed compared to the fixed reference. The reference should at least be earth (-surface). (And when you get convinced of that, you will also understand the earth rotation stuff).

25th Jul 2011, 22:47

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@ChristiaanJ

I did, but I don't have all the answers, and formulas, and figures, at my fingertips any more...

To stir up the discussion, what about a 'breakdown' of the Ekin of a Fieseler Storch (or similar) flying at about 50kts in a 50kts headwind ?
Let's extend this: Headwind 49.99 kts and the Storch is having a collide with a tower, best would be just above a platform.

What would happen: A very gentle little bump against the tower and then the subtle drop/land on the towers' platform. Nothing bent.

25th Jul 2011, 22:51

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How to enter a deep stall

I am amazed at all the kinetic energy and OAT and such discussion here for last two days.

The profile of AF447, as we know it, and regardless of control laws or pilot inputs, resembles the classic manner of getting the Viper into a deep stall. You simply climb at a fairly steep attitude, fairly level roll attitude, and at low AoA until you run outta energy, then sit there and watch the jet try to nose over too late. AoA increases rapidly, and with little or no "nose down" pitch moment available from the flight controls, you are there!!

The jet's "protections" (or limits, as I prefer) are fooled. We simply fly past the jet's control authority to provide the so-called "protections". Worse, and in the case of the Airbus, we have a myriad of reversion "laws" that could cause the crew to do something worse than just sit there and hold attitude/power. The overspeed warning is what I am concerned about, as that could explain either pilot or computer commands, or both.

I refuse to believe that the Airbus is a poorly-designed jet from the aerodynamic aspect. I truly believe you could exceed the mach "protections" until reaching maybe 0.95M or so with no ill effects. I truly believe that you could fly the jet at 10 or 15 degrees AoA. I truly believe you could pull 3 gees without the wings falling off. I truly believe the jet has exceptional lateral stability, or we would not see a proflile with a slow rotation versus a tendency to enter a spin.

What I do see is an embedded "autopilot" influence that changes control laws depending upon flight phase ( and Viper had some of those, but not to the extent of the Airbus). I see confusing cockpit warning/caution indications. I see no firm "hang your hat on the jet's capabilities" control law that the human crew can use when things go to hell in a handbasket. Worst of all, I see no aspect of the system that acknowledges loss of air data and simply reverts to a basic control law while the crew and HAL figure things out.

25th Jul 2011, 23:02

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Originally Posted by Dutch M
An experiment might be to circle around and measure the pitch angle and rudder usage during the circles. I would expect a somewhat sinus shaped change in rudder/pitch position.
You'd be disappointed. The point is - if the wind speed is horizontal and constant in magnitude and direction, there is absolutely no effect on the airplane, except that the center of the circles moves over the earth surface at the speed of the wind. Except for that, the airplane flies perfect circles at constant airspeed, bank angle, pitch angle, angle of attack, and control surface positions.

Another matter entirely is a vertical wind component, or changes of wind speed or direction, but let's keep that for next year ...

25th Jul 2011, 23:03

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the Old Chestnut ,wind effect when circling

That Old Chestnut

I had it beaten out of of me when learning pattern B instrument flying on a Harvard "circling upwind or downwind has no effect on your airspeed as you are only flying/ circling with respect to the airmass"
I was convinced I had to keep adjusting power as I went round the 360 degrees maintaining an exact height
However we were also warned to be careful turning downwind near the ground for fear of stalling due to visual illusions of speed
In gliding, circling on the ridge, if one correctly maintains airspeed there seems to be a need to speed up/nose down turning downwind, and there is surge of lift as you turn back into the headwind .Though the detail effects are distorted by the changes of updraft and windspeeds near the hill .
Nevertheless there is a loss of height turning downwind, and a regain height as you turn into wind.
It is also interesting how the angle of bank changes as you fly elliptically over the ground while performing a supposed perfect circle in the airmass?
Swirlyflier

25th Jul 2011, 23:23

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Hello henra
Originally Posted by henra
It got nowhere. The only energy that got converted into heat was the 1/2*m*(0,25m/s)^2 during braking on the ground.
And that was the only Kinetic Energy that it had left, when it crossed the virtual border between the two reference systems - air, and ground.

Originally Posted by henra
With regard to the air the remaining 1/2*m*(30m/s)^2 is still in the glider.
Being fixed on the ground it is effectively being towed through the air at 30m/s by the ground with regard to the air. See my example of the microlight.
Then, when I am standing watching kids playing in the yard, under a breeze of 0.5m/s, I must have a 0.5m/s corresponding Kinetic Energy.... even though I don't move...

The glider at this time does not move, so it has 0 kinetic energy. (relative to the ground, just to be accurate).

Kinetic Energy goes beyond Aerodynamics, and thus, similar examples are abundant in various other fields of the Dynamics, some involving wheels and legs, instead of air, and wings....

For instance the example of a train at a certain speed, and a passenger walking on the train: if the passenger hits an object on the train, his pain is going to be relatively little, when compared with the pain he would have if he hit somehow, an object on the ground - classic comedy films with guys walking on the top of the train, while the train goes under a very low overpass bridge, come to mind.

The passenger Kinetic Energy relative to the train is given by his walking speed, measured by the little pain of hitting an object on the train, while his Kinetic Energy relative to the ground, is augmented by the train's Kinetic Energy, measured by considerable more pain, and possible destruction in the second case..

With that, back to the glider example, the Kinetic Energy corresponding to the 30m/s speed wind, is the air's Kinetic Energy, not the glider's.

The glider on the ground, is like the passenger off the train - no Kinetic Energy from the air, none from the train. Put the glider on the air - push it to make it fly - it's like having the passenger back on the moving train.

All of these examples contain momentum, and inertia aspects, which may add to the fun, or to the confusion....

Originally Posted by henra
The trick is that you need to understand that for an Aircraft once it has left the ground, the movement of the ground below is totally irrelevant for the ability to stay in the air. And it stays irrelevant until it contacts the ground again. In between only its relative movement through the air is relevant.
After these examples it is also clear that it's good to keep the Energy Conservation equation pure, in its General and Universal aspects, which transgress the specifics of the Dynamics fields, like Aerodynamics.