RetiredF4,
To begin with your last point: If the nose-down pitch rate is maintained until the stall warning stops, that would occur at t=21 seconds and 27 degrees ND pitch. Level flight would be achieved at t=42.5 seconds, FL262, 324 kCAS. I suspect that Owain Glyndwr did not show that scenario in his graph because he found it difficult to imagine that an ATP would deliberately do that in an airliner.
Owain explained the 'Performance Model' in his post #87:
OTOH, with HN39's help (but any errors are mine!) I have been able to construct a means of calculating performance at and beyond the stall, and this in turn allows a systematic study of recovery possibilities - something that
AFAIK has not been presented here before, so perhaps it qualifies as "new data"?
On the diagram, the yellow line is the actual AF447 flight path derived from BEA traces of inertial ground speed and barometric altitude, corrected for wind effects wherever possible.
The purple line is a theoretical calculation of that flight path using the developed method with actual flight pitch and power (N1) as input data. Given the fact that the aircraft was gyrating with six degrees of freedom and the model is restricted to three, I think it a reasonable match which gives some credence to any recovery calculations. Others may differ.
The 'Performance Model' consists of lift, drag and thrust data as functions of relevant parameters, and the trajectory then follows from Flight Mechanics, i.e. Newton's laws. It is driven by the specified variation of pitch attitude versus time. As opinions differ regarding the the achievable rate of pitch change, Owain's graph shows trajectories for pitch rates of 1, 2 and 3 degrees/second. These rates bracket the values observed by PJ2 in the simulator, and the responses of the airplane to elevator movement observed in the DFDR traces of AF447.
"Wasn´t the initial AOA according BEA higher than 40°?" The recorded AoA at 02:11:45 was just about 40°. In my graph I chose to stay with the recorded pitch attitude and calculated flight path angle, which result in AoA=36° for symmetrical flight in still air. The difference between that value and BEA's is attributable to the bank angle of 30° existing at that moment.