Originally Posted by
Pugilistic Animus
(p-pinfinity)
/(1/2rhoV^2)/Cpi=1/[1-M^2]^.5+ [M^2/1+{1-M^2}^.5]*(cpi/2)
or maybe just 1/[1-M^2]^.5
no further comments
Actually, no, mot just the "standard" Mach effect. Stall AOA ends up being a powerful function of freestream Mach, due to transonic effects in the (very accelerated) boundary layer. As a result the freestream Mach is not the Mach directly having the effect.
What you're alluding to is fine at more 'normal' AOAs.