Originally Posted by
BOAC
Posters talking about '30 secs sustained ND or even more' need to remember that
a) This a/c was descending at around 10,000 fpm ie from 10k, one minute to impact
b) A pitch change of around 30-40 degrees nose down would have been required to initiate unstall, which would probably have raised the r o d to around 20k fpm
c) Now pull out at xxx g?
I agree it is not really important in the given case,
just out of curiosity I had again a rough look at the physics behind.
We are probably talking about somewhere between 10 and 20kFt.
I appologise upfront for re- doing the calc.
[Assuming stall speed of ~180kts @200t, no flaps, we need 250kts IAS for a 2g recovery. At 10 kft that corresponds to ~300kts IAS. At an angle of 45 ° this would produce ~22kfpm (300kts x sin(45°)). If we assume linear acceleration (which I admit is not 100% true) that would give us an average 16kfpm.
In 30s that would mean a drop of 8k.
Edit : At 30° - 35° RoD would be 16kfpm. Simplified average 13kfpm.
This would give us 6500ft drop.
Now the 2 g recovery:
R = v^2/a: with a = 9,81 m/s (Out of the 2g maneuver capability @300kts TAS (=250kts IAS) one g is required for Sir Isaac Newton). That gives us a radius of
2430m. At an angle of 45° the corresponding lost altitude would be (1- cos(45°)) x Radius. That would gives us an additional 710m (2370ft)].
Here I err on the conservative side, as recovery in the given example would have been below 3kft and thus 250kts IAS wouldn't have meant 300kts TAS but rather 260kts.
(Edit: 1500ft at 35°)
So we have lost somewhere between 6,5 and 10kft for acceleration to a speed where recovery could be executed and between 1,5 kft and 2,5kft for the recovery itself. However it would be easy to start to early with the recovery and run into the next stall.
However, from a purely theoretical PoV somewhere close to 10.000ft might have been possible.
Apologies again for beating this poor dead horse again.
But I fully agree this has no practical meaning in the given case.