As a ‘newbie’ here I have been reluctant to say too much in the presenc e of so much expertise, but I do think that I may be able to make a positive contribution towards understanding the aerodynamics – at least I hope so! I should preface any remarks by explaining that I was once an aerodynamicist, but anything I write is based on general principles and standard methods, not any particular knowledge of the A330, although I have of course used such data as is publically available. JT gave a lucid explanation of how the stability changes as CG is moved back, but before anyone goes off down one of those typical thread tangents about how the A330 with relaxed stability must be dangerous, it should be recognised that the CG where one arrives at static instability on the A330 is way back around 45~50% mac (it depends on Mach Number). Compare that with Takata’s CG envelope and you will see that it is nowhere near static instability even at the aft CG limit.
If I have understood correctly, the A330 in Normal, Alt and Alt2 laws operates under some form of C* control, and it is only in Direct law that this would come into consideration anyway. Also, AI would have had to show, for certification, that the aircraft was flyable in all states, including Direct law, without requiring exceptional piloting skill.
The requirements BTW, say that a pull must be required to obtain and maintain speeds below the specified trim speed and a push for speeds above that. The speed (in cruise) must return to within 7.5% of the original trim speed when the control force is slowly released but it is acceptable for the aeroplane, without control forces, to stabilise on speeds above or below the desired trim speeds if “exceptional attention on the part of the pilot is not required to return to and maintain the desired trim speed and altitude”. Not my words, not AI words; JAR words.
Retired F4 posed a lot of questions, most of which he answered by “We don’t know”. In this of course he is entirely correct, but that doesn’t stop us making sensible engineering estimates.
I suspect that most peoples’ mental image of a stall is a sudden loss of lift and a nose down pitching moment, typical of the straight, unswept wings on which they learned to fly (I am not a pilot BTW). However, a modern swept airliner wing does not stall like that.
The sweep, taper, camber and twist of a typical modern design will result in the maximum local lift coefficient occurring first at somewhere between half and two thirds semispan. This is generally a little way aft of the CG, so the first effect will be a gentle pitch up. What happens next will depend on how the wing is twisted. If the outer wing is more heavily loaded there will be a steadily increasing, but still relatively modest pitch up. When the inner (and forward of the CG) part of the wing stalls there will be a compensating pitch down. Throughout all of this process the wing lift coefficient is departing from the linear relationship with AoA that it had before stall, and eventually it will end up as a fully stalled wing with a more or less constant normal force coefficient. The actual ‘lift’ and ‘drag’ will then be the component of this normal force resolved into axes parallel and normal to the incoming airflow (i.e. to the FPA usually). When it gets to this point I think the centre of lift/pressure will be fairly close to the centre of area of the exposed wing. For the A330, this is about 70% mac. This means that when fully stalled the pitching moment from the wing will be a nose down value which does not vary much with AoA since the PM will be the normal force coefficient times the moment arm of this 70% to the CG at 29%, and the normal force coefficient will be nearly constant.
Net result of all this is that the actual process of stalling can be quite a gentle affair as Gums has suggested from time to time.
A couple of pages ago I posted an explanation of the mechanics of aerodynamic equilibrium in deep stall conditions, and nobody has yet said they think this was nonsense, so I am sticking with it. When you look at this, it is clear that the question to be asked is not will there be enough down elevator power to give a ND recovery moment, but rather will there be enough elevator power to get and hold 60 deg AoA. To recover all that would be necessary would be to remove the up elevator, although of course some down elevator wouldn’t hurt – so long as you don’t stall the THS.
OK, is forward or aft CG better/worse for stall/recovery?
The wing will stall at the same AoA regardless of CG position. With a forward CG the aerodynamic moment about the CG when approaching stall will be more ND because of the increased moment arm. So more up elevator will be required to approach and maintain a given AoA, the negative tail lift will oppose the wing and the overall lift coefficient will be lower with forward CG than aft and the aircraft will stall at a higher airspeed – but the same AoA!.
When the wing is stalled, for the reasons described above, the moment (a ND moment remember) will be nearly constant, so any variability in speed of recover will come from the elevator’s ability to provide ND pitch acceleration. This is going to be bigger with the moment arm from a forward CG, so my vote goes to forward CG as being more favourable.
Although it has been said that we cannot know what the post stall conditions are, we can calculate them for at least one point where we have ground speed, attitude, and descent rate; namely the point of impact. Taking the quoted values as ‘gospel’, one can calculate the airspeed as about 151 kts, the AoA as about 61 deg and the FPA as -45 deg. In addition it is believed that the engines were at Flight Idle at this time. For the sort of engine we are looking at this is as near zero net thrust as makes no difference. Assuming (OK, it is an assumption) that this was a stabilised state, then you can calculate what the lift and drag coefficients would have had to be to match – both lift and drag coefficients at about 1.07.
Hope this does eventually help and not spread more confusion!