We show that accelerated stall speed is a function of load factor and straight-and-level stall speed only. Accelerated stall speed does not depend on bank angle directly; it only depends on bank angle in turns, if one insists on level flight (or on turns with constant descent rate).
Straight-and-Level Stall Speed
First let us remind ourselves of the expression for straight-and-level stall speed. The equations for lift and weight are given as follows:
$$L=\frac{1}{2}\rho V^2SC_L$$
$$W=mg$$
In straight-and-level flight lift equals weight
$$L=W$$
Inserting the above expressions into the above equation and solving for \(V\), we obtain:
$$V=\sqrt{\frac{2mg}{\rho S C_L}}.$$
Stall is defined as \(C_L\) being maximal, and therefore the corresponding straight-and-level stall speed is:
$$V_S=\sqrt{\frac{2mg}{\rho S C_{L_{max}}}}.$$
Load Factor and Accelerated Stall Speed
The load factor \(n\) is defined as
$$n:=\frac{L}{W}$$
(Note that load factor does not necessary correspond to the g-force the airplane experiences. An airplane flying vertically up and accelerating may experience many g’s, but zero load factor, because it is operating at a zero-lift angle of attack and the wings are not producing any lift.)
With the above equation, we can do the same calculation as above to arrive at the equation for the accelerated stall speed \(V_{Sacc}\):
$$V_{Sacc}=\sqrt{\frac{2mgn}{\rho S C_{L_{max}}}} = \sqrt{n}V_S$$
The only difference to the expression for straight-and-level stall speed derived earlier is the factor of \(n\). This shows our initial claim that the accelerated stall speed is a function only of load factor and stall speed. Note that no additional assumptions entered the calculation, only the definition of load factor.
A conclusion is, if you know the load factor in a given situation, you know the stall speed.
Accelerated Stall Speed in Level Turns and in Turns with Constant Descent Rate (Apparent Bank Angle Dependence)
It follows from the above that knowing the bank angle \(\mu\) in a given situation does not tell you the stall speed. In the above equation, the bank angle is nowhere to be found. An additional assumption will have to be made which relates bank angle to load factor. In that sense, this is a special case, which does not hold universally.
The assumption we shall make now is that we make a level turn, i.e. one where we hold altitude, or a descending turn with constant descent rate. In both those cases, the downward velocity component is constant, i.e. does not change with time, and thus the vertical component of lift, \(L_v\), must equal weight as an additional requirement, since otherwise we would experience a downward acceleration. This leads to:
$$L_v = L\cos\mu = W$$
Comparison to the definition of load factor we saw earlier,
$$L=nW,$$
yields
$$n=\frac{1}{\cos\mu}.$$
So for a level turn and for a turn with constant descent rate (which is an additional assumption we had to put in here), accelerated stall speed does depend on bank angle via
$$V_{Sacc}=\frac{1}{\sqrt{\cos\mu}}V_S.$$
But this is true for level turns only and for turns with constant descent rate. The pilot may choose to fly at a steep bank angle well below this speed without stalling, if she/he is willing to start losing altitude (faster) and does not pull on the stick to increase lift by increasing angle of attack, and thereby also does not increase the load factor.