Well, if L = W.Nz = 0.5 rho TAS^2 S CL
CL = a.AoA
I'd get rid of rho.TAS^2 and instead use rho_0.EAS^2 which is the same thing
So, W.Nz = 0.5 rho_0.EAS^2 S a.AoA
Which will give you AoA = (2.W.Nz)/(rho_0.EAS^2 . S .a)
So, you need:
W = aircraft weight, from operational plan and fuel state
Nz = normal acceleration, from INS, g-meter, etc.
EAS = equivalent airspeed, from FMS, or derived from IAS via CAS
S= wing area
a = lift curve slope, hopefully available from wind tunnel data for the whole aeroplane (DO NOT just use aerofoil data, because the tailplane and fuselage have a significant effect). You might be able to come up with an estimate of this, if you don't have wind tunnel data, using ESDU data sheets - but that can only be an approximation.
Obvious cautions to this:
- CL=a.AoA assumes that CL=0 when AoA=0, if it's more complex than that, which it may well be, you need to modify the formulae.
- It'll only work in steady state conditions (so Nz, EAS, bank, sHp are all constant for any given data point)
- It'll still be an approximation, and no substitute for flying an external air data sensor in clear air.
G
(Edited 18/5/11 because I realised I'd missed a Rho_0 out of the formulae)