Turn to Final - Keeping Ball Centered

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

wrote:


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?

Yes. That's exactly the case.

Stalls are determined by the critical angle of attack. The so-called
stall speed is the minimum speed required to maintain (vertical)
unaccellerated flight without exceeding the critical angle. It doesn't
matter if it is "level" or any other constant rate.


The engine power required is plainly different.

That is true, but engine power has nothing to do with stalling.

In article xQUBj.19725$TT4.4490@attbi_s22, wrote:

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.

In the case of descending flight, gravity is supplying some of the power
required to maintain flight. With wings level, you are at 1.0 g whether
climbing, level or descending. The same rules apply to turning flight.

--
Remove _'s from email address to talk to me.

Orval, I can't thank you enough (missing energy found!). That was the
missing piece, even though it is so obvious after you point it out. I
haven't seen that part explained or referred to in all of the usual
discussions.

Thank you very much.


"Orval Fairbairn" wrote in message

In the case of descending flight, gravity is supplying some of the power
required to maintain flight. With wings level, you are at 1.0 g whether
climbing, level or descending. The same rules apply to turning flight.