Originally Posted by
airtren
Your rephrasing of my text is still in line with what I meant
Thanks for confirming it
Originally Posted by
airtren
So my point again is that the change of AoA affecting the CI alone, may not be sufficient to cause the exit from Stall.
As for the maths: I'm sorry, I'm not fond of maths, I must say. Logic is enough to me
For that reason, I won't comment on your formula.
You may be unstalled (i.e. your wing is flying, producing
lift & allowing control) but not have a CI big enough to support your weight. Then, you're descending. But no more stalled.
Originally Posted by
airtren
You say, the "gain of speed is a means to lower AOA".
I don't see how the "gain of speed" has necessarily a "lower AoA" consequence. Perhaps your elaborating would help.
I wrote : "the gain of speed is a mean to lower AoA (
all other parameters equals)" (my underlining)
Reverse: For the same velocity vector, your AoA is lower if you fly quick than if you fly slow.
That's why I wrote that if you fly quicker, you'll lower your AoA. Note that it was the spirit of the (now discarted IIRC) "approach to stall" procedure: To escape "falling" into an actual stall with more speed ; Without loosing altitude (<= maintaining the (level) velocity vector)
My point is : Speed is a consequence. AoA is what matters.
If stalled (or nearly stalled), you must lower AoA, be it by altering flight path (velocity vector => push the damn stick) or by gaining more speed (*)
The former is more efficient, as pitch control is a direct AoA control. That doesn't mean a bit more speed won't help: it will. But what you seeks, ultimately, is not more speed but less AoA.
(*) which, as shown above, will allow you a lesser AoA for the same flight path, or a higher flight path with the same AoA.