Originally Posted by
n5296s
I don't think so - though I'd be happy to see why I'm wrong.
If VS is constant - whether zero or 20,000 ft/min - then by definition acceleration is zero and hence net force is zero, i.e. downward force due to gravity (mg) equals upward force due to lift. And hence lift perpendicular to the wing surface = mg/cos(angle of bank).
The angle of pitch has exactly the same effect as the AoB (if normally quite a lot smaller).
Put your elbow on the desk and hold your hand straight up - this will be the lift vector and the height of your hand will be the vertical component (which must equal the weight of the aircraft if it is not changing vertical velocity).
Now roll your arm left and right and note how the hand is closer to the desk (i.e. reduced vertical component). This works exactly as described in many earlier posts.
Now move your arm forward and back. This is what happens when you change pitch (you can see the reduced vertical component in this as well).
Finally if you rotate the arm and move it for/aft you have even less vertical component.
Normally we ignore this because even a 15 degree deck angle (which is going to be going down at 1600 fpm in a 60 knot turn (giving only 19 seconds to get the turn in from 500 feet) is only going to add 5% or so to the stall speed due to reduced vertical component of lift.
I would suggest in this discussion it can be reasonably ignored as well