But the rate of descent quickly becomes constant or close enough to it that you're sitting at 1G. You need to PULL to send the lift horizontal so you're going to need to do 2Gs to get it around at the rate you're saying even though you're descending.
Well sure you're still sending lift horizontal. But your nose is going to fall. You're going to rapidly build airspeed. And you're not going to turn at much of a rate.
By this logic, I can get to stall speed, initiate a spin, stop rotation, bank to 60 degrees (while still stalled???) and pull 1.5g and continue the downward spiral.
Where does the stall break? Does it break from the reduction in .5G?
accelerating downwards so that kinda threw me off. I FELT (and who knows how accurate that is) like I had one G going through the bottom of the seat from the rotation and one G going through the back of the seat from the acceleration... which would add up to 2Gs but we need physicist now I think. What I lost though was more than 500feet of altitude so probably better to hold closer to best glide speed it seemed very inefficient. I suppose you would get some altitude back by pulling up to bring the speed down but not all of it. So yes you can pull 180+ degree turn without going from best glide to over Va but for the cost in altitude your probably better off holding a speed you know is safely above stall speed in a sixty degree bank.Initially it goes down. Then it goes up. If it doesn't stabilize at 1g very quickly, you won't survive the impact.
Then we're probably pulling about 1.5-2 g's on the g-meter (with vectors of 1 g perpendicular to the earth's surface and about 1.5 g's parallel to it) throughout the turning portion of the maneuver at the probable 45-60 degree bank for least altitude loss.
I'd say the entire stabilized portion of the turn (i.e., all but rolling in and rolling out). The only time you'll be under 1g will be when you initially unload to get to the target speed for the maneuver, as you'll probably be well below it before you get your brain unscrambled after the engine quits at 7-10 degrees nose up.
Yes, it is stabilized in terms of velocity during the turning portion, just like a constant-speed level turn. If it weren't, your airspeed and vertical velocity (VSI) would be running away (which would create a lot of other problems), and if executed per Prof. Rogers' guidelines, they aren't. If you hold constant airspeed and bank angle during the turn (as Prof. Rogers suggests), you really will be in a stabilized condition during the turn. The only time it isn't stabilized is during the initial acceleration to target speed, roll-in/roll-out, and during deceleration to landing speed at the bottom.
Correct, but an engine with out oil pressure isn't likely to windmill very long.
Not many -- no time for keeping the coffee in the cups.
Huh? Unless you're flying a V-22, engine thrust vector is always pretty much right down the longitudinal axis, while gravity is always pulling straight down towards the earth. Unless you're going through the vertical (which you shouldn't be on this maneuver), the forces of gravity and engine thrust will always be different vectors -- and it won't matter if the steep turn is level or ascending/descending, nor whether a level turn is steep or shallow.
It seems you are overthinking. It's ok, since we all express oursleves differently. Where you replied to Ron that it wasn't stabilized I was thinking yes it is because, other than roll-in/roll out, it is a contstant bank, constant pitch turn.
Hen I really don't follow your point at all.