I'm trying to figure out the lift coefficient of an A320.

What I'm doing is the following: during cruise, the lift is equal to the weight of the aircraft ($mg$) so that I used the cruise table taken from the FCOM to get the speed, the wing area is known ($122.6 m^2$) and I calculated using a tool online the air density at FL290 (29,000 ft).

The result is that the lift coefficient is $0.03029$. Is it right for the assumptions given above?

Next, if I want to relate the $C_L$ with the Angle of Attack, what is the formula and the procedure?

  • 2
    $\begingroup$ Well.... I doubt that any current airliner will be able to fly with just 0.03 of $C_L$, did you mulitplied the mass by the gravity? $\endgroup$ May 4, 2015 at 10:11
  • $\begingroup$ 50000kg is the "weight" so already multiplied by g, that's not the mass. $\endgroup$
    – Afe
    May 4, 2015 at 11:46
  • $\begingroup$ Well, first hint for knowing that the gravity shall be used is that "weight" is measured in Kg instead of N. Second is having a so low $C_L$, having actually lower $C_L$ than usual "C_D$ values. Another warning message is that if 50000 is actual "weight" the mass will be 5000Kg, which means that for an A320 being an airplane of more than 100 passenguers of around 75Kg (in mass) having a total mass of 7500Kg we have a negative mass of the structure of -2500Kg for reaching the the 50000/9,8 mass value. In that sense, I think it is more likely that "weight" is actually "mass" than other option $\endgroup$ May 4, 2015 at 13:03
  • $\begingroup$ It's the same as when I say for example: "My weight is 60Kg". This means that my mass is 60/9.8. $\endgroup$
    – Afe
    May 4, 2015 at 15:34
  • $\begingroup$ My point is that, although it is said "weight" it actually means "mass". Having a $C_L$ of ust 0.03 is unphysical. An airplane of 100 passenguers with a actualy weight of 50kN is not possible .... $\endgroup$ May 4, 2015 at 15:37

1 Answer 1


I found a different number for $C_L$, please check your calculation:

$C_L=\frac{2 m g}{\rho V^2 S}$

substitute: $ m = 50000 \textrm{ kg}\\ g = 9.81 \textrm{ m}/\textrm{s}^2\\ V = 462 \textrm{ KTAS} = 237.67\textrm{ m/s}\\ S = 122.6 \textrm{ m}^2\\ \rho = 0.475 \textrm{ kg}/\textrm{m}^3 $

This will give: $C_L = 0.298$

Note that this lift coefficient is for the total aircraft. It includes the lift of the wing, the fuselage lift and also the negative lift of the vertical stabilizer. There is also a small vertical component of thrust that is neglected.

Now that we have an estimate of the lift coefficient, we can estimate the angle of attack. Normally the lift coefficient is assumed to vary linearly with the angle of attack:

$C_L = \frac{\textrm{d}C_L}{\textrm{d}\alpha}\cdot (\alpha-\alpha_0)$

$\alpha_0$ is the zero-lift angle of attack, the angle of attack at which the wing does not generate any lift. For symmetric airfoils, $\alpha_0 = 0$, for chambered airfoils $\alpha_0 < 0$.

I don't know what that angle is for the A320, it will be difficult to obtain. Let's assume for now it is -1.2 degrees.

For infinite long wings in incompressible flow the lift slope $\frac{\textrm{d}C_L}{\textrm{d}\alpha}$ is $2\pi$.

$(\alpha-\alpha_0) = \frac{C_L}{2\pi} =0.0474 \textrm{ rad} \approx 2.72^\circ$

This would result in an angle of attack of $\alpha \approx 1.52^\circ$

Because the wing of the Airbus A320 is not infinitely long but has a span of about 30 meters (excluding the fuselage) we need to correct for that. The reason we need to correct for the finiteness of the wing is that the circulation will cause the local angle of attack of the wing to be lower than the free stream angle of attack. The effective angle of attack $\alpha_{eff} = \alpha - \alpha _i$

The induced angle of attack $\alpha_i$ is given by:

$\alpha_i = \frac{C_L}{\pi AR} = \frac{C_LS}{\pi b^2}$

substitute: $ S = 122.6 \textrm{ m}^2\\b = 30 \textrm{ m} $

gives $\alpha_i = 0.0129 \textrm{ rad} \approx 0.74^\circ$

Adding this to our earlier angle of attack results in:

$ 1.52 + 0.74 = 2.26^\circ$

This may not be very accurate as

a) the zero-lift angle of attack may be very different,

b) the lift slope may be flatter due to compressibility effects

  • $\begingroup$ The two densities used are slightly different. I think this could explain the 1.6% difference. $\endgroup$ May 4, 2015 at 8:42
  • $\begingroup$ @MikeFoxtrot the difference is almost 90%, or 883% depending on which CL you take as denominator... $\endgroup$
    – DeltaLima
    May 4, 2015 at 9:00
  • $\begingroup$ took the liberty of fixing a comma-decimal dot typo $\endgroup$
    – Federico
    May 4, 2015 at 11:08
  • $\begingroup$ @Federico thnx! $\endgroup$
    – DeltaLima
    May 4, 2015 at 11:08
  • $\begingroup$ @DeltaLima Just spotted the decimal points was off so indeed :P The online tool gives a density of 0.47545 as compared to 0.475 that you use. $\endgroup$ May 4, 2015 at 11:19

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