Pilot DAR...
Yes, for a given angle of bank, which corresponds to a given G load, there will be a stall speed increase.
Correct. Just to clarify that it's a given angle of bank in a level turn that will correspond to a given G load.
Ignoring changes in configuration (flaps, slats, ice) the angle of attack at which that airfoil stalls is proportional to the lift it must create (weight X G), but nothing else.
Yes, 1G stall speed is directly related to 1G weight. As you increase your G loading, the stall speed will change. Vs2=Vs1 x sq.rt(load factor) --- Vs1=Vs x sq.rt(current weight/max gross weight)
Vs1 will be the new stall speed for the reduced weight. Plug that into the Vs2 equation and that will give you the stall speed at a particular load factor.
If for a given condition of flight, the pilot increases the angle of attack, the pilot takes that wing closer to stalling, but the speed at which the stall occurs is not changing because the angle of attack is changing. If the change of angle of attack (with no other configuration changes) resulted in a change of the stall speed, the angle of attack indicator would not really be a useful indicator of the approach to the stall.
If you're doing this all at 1G, then that's correct, ie: you're increasing your angle of attack by slowing down. The stall speed won't change because you're still at 1G. If you increase your G loading, your stall speed will increase according to the Vs2 equation above.
The section of airfoil at any spanwise point along the wing (and in co-ordinated flight) does not know if it is in level or banked flight. It just knows it is reacting a given load (weight X G). It knows its angle of attack, and how close it is getting to its Cl max, but it will wait to get to that Cl max before it stalls. Increasing angle of attack takes it closer to that Cl max and the stall. but does not change Cl max, so the stall speed is not being changed.
In level flight at 1G that's correct. But in a turn that's not correct. Just imagine a highly maneuverable fighter, for example, in level flight at 300 kts with a stall speed of 100 kts. If the pilot yanked back on the stick, the aircraft would pitch up abruptly and would be able to stall at that very high speed. He would feel a high G loading and that G loading would be what causes the stall speed to increase. All these equations go back to the Lift equation.
When you "go ballistic" over the top of a push over, it is possible that you are not reducing the angle of attack as the pitch attitude of the aircraft (relative to earth) reduces. You have, however dramatically reduced the stall speed, as you have reduced G, and the demand for lift.
Exactly! In the push-overs, you will have a lower stall speed than published and, like you pointed out, it's because of reduced load factor (G). The same relationship still happens when you have positive load factor - the stall speed will now increase.
It seems like you understand that G directly affects stall speed but don't see how angle of attack plays into it. Going back to the lift equation -- Lift = 0.5 rho Vsquared S Cl
That can be rewritten for our purposes as -- Weight = Vsquared AoA
So, at a higher G, I essentially 'weight' more and need proportionally more lift. To balance the equation I can either increase velocity or increase the AoA. If I keep the same AoA and increase the speed, I'll increase the lift to balance the weight. If I keep the same speed and increase the AoA, I'll increase the lift to balance the weight. I could do any combination of either and come out with an increase in lift to balance the weight. In a steep turn you have to add power because to maintain level, you're increasing your angle of attack. That increases drag which must be offset by thrust or you will slow down.
If I wanted to accelerate in level flight, I add power to accelerate but I decrease the AoA so that I don't get any net increase in upwards force.