I am sorry to belabor this point, but hopefully I will understand it better
if you have the patience to help me.

I understand that gravity's acceleration is indistinguishable from any other
acceleration in a frame of reference, and that it is 1g in a constant
descent or climb. I also understand that a turn is accelerating to the
center of the arc whether it is descending, level, or climbing.

Are you saying the force vector sums are equivalent when comparing level
turns at the same bank angle to constant rate descent? In other words, that
a 2g load factor occurs in a 60 degree bank regardless of whether remaining
level or a constant rate of descent and that the stall speeds are identical?
The engine power required is plainly different.

Thanks.



The center just becomes a line in that case, doesn't it? You end up
with the same centripetal force needed towards that center line in
order to turn.

Nosing over into a descent does temporarily reduce load factor, but as
soon as you're established in a constant descent, you're back at the
same 1g load. Gravity is *acceleration*, not *velocity*.

--
James Carlson, Solaris Networking