Originally Posted by
SteinarN
I did some calculation.
Starting at 6000 feet/2000 meter, the time in free fall to ground is 20,2 sec. This aircraft took about 18 sec from 6000 feet to ground. This means the G throughout the dive, from start to end, would have to be close to zero or slightly negative. Incorrect
This seems to imply that the elevator/h.stabilizer must have been at average at approximately zero degrees angle of attack thus producing zero force. Incorrect
I do wonder, if the jack screw for the stabilizer breaks loose, in which position will the aerodynamic forces put the stabilizer into? Would it "free float" into zero degrees AoA?
There is some flawed thinking here.
An object released to freefall will experience a 1G (gravitational) acceleration until atmospheric friction/drag will stabilize its speed at terminal velocity.
Depending on surface area and drag coefficient this terminal velocity may be significantly less then a dive while still under engine power.
https://www.grc.nasa.gov/www/k-12/airplane/termv.html
Recovery attempt from a steep dive can cause very high G-forces on the airframe
An example of this is here where they descended under power at appr. 12000fpm (!)
https://en.wikipedia.org/wiki/China_Airlines_Flight_006
The fact that the time of freefall more or less equals the time to descent is nothing but coincidence.