Is there an equation relating AoA to lift coefficient?

I have been searching for a while: there are plenty of discussions about the relation between AoA and Lift, but few of them give an equation relating them.

I know that for small AoA, the relation is linear, but is there an equation that can model the relation accurately for large AoA as well? (so that we can see at what AoA stall occurs)

I am not looking for a very complicated equation. Can anyone just give me a simple model that is easy to understand?

(Of course, if it has to be complicated, then please give me a complicated equation)

  • $\begingroup$ Are you asking about a 2D airfoil or a full 3D wing? $\endgroup$
    – MikeY
    Commented May 19, 2019 at 1:42
  • $\begingroup$ @MikeY 3D is good, but 2D is OK $\endgroup$ Commented May 19, 2019 at 3:56
  • 1
    $\begingroup$ For a 3D wing, you can tailor the chord distribution, sweep, dihedral, twist, wing airfoil selection, and other parameters to get any number of different behaviors of lift versus angle of attack. So your question is just too general. $\endgroup$
    – MikeY
    Commented May 19, 2019 at 13:30

5 Answers 5


In the post-stall regime, airflow around the wing can be modelled as an elastic collision with the wing's lower surface, like a tennis ball striking a flat plate at an angle. Lift and drag are thus:

$$c_L = sin(2\alpha)$$ $$c_D = 1-cos(2\alpha)$$

I superimposed those (blue line) with measured data for a symmetric NACA-0015 airfoil and it matches fairly well. I don't know how well it works for cambered airfoils.

cl-vs-angle cd-vs-angle


A bit late, but building on top of what Rainer P. commented above I approached the shape with a piecewise-defined function. Not perfect, but a good approximation for simple use cases.

$$ C_L = \left\{ \begin{align*} \sin(6 \alpha) ,\ \alpha &\in \left\{0\ <\ \alpha\ <\ \frac{\pi}{8},\ \frac{7\pi}{8}\ <\ \alpha\ <\ \pi\right\} \\ \sin\left(2\alpha\right) ,\ \alpha &\in \left\{\ \frac{\pi}{8}\le\ \alpha\ \le\frac{7\pi}{8}\right\} \end{align*} \right. $$ Note that I'm using radians to avoid messing the formula with many fractional numbers. The post-stall regime starts at 15 degrees ($\pi/12$).

Plot of the lift coefficient formula


No, there's no simple equation for the relationship.

Here's an example lift coefficient graph:

A lift coefficient graph for the NACA 0015 airfoil

(Image taken from http://www.aerospaceweb.org/question/airfoils/q0150b.shtml.)

This is actually three graphs overlaid on top of each other, for three different Reynolds numbers. I'll describe the graph for a Reynolds number of 360,000.

We see that the coefficient is 0 for an angle of attack of 0, then increases to about 1.05 at about 13 degrees (the stall angle of attack). From here, it quickly decreases to about 0.62 at about 16 degrees. Then it decreases slowly to 0.6 at 20 degrees, then increases slowly to 1.04 at 45 degrees, then all the way down to -0.97 at 140, then...

Well, in short, the behavior is pretty complex. The most accurate and easy-to-understand model is the graph itself.

  • $\begingroup$ "there's no simple equation". Could you give me a complicated equation to model it? $\endgroup$ Commented May 19, 2019 at 4:53
  • 1
    $\begingroup$ You could take the graph and do an interpolating fit to use in your code. $\endgroup$
    – MikeY
    Commented May 19, 2019 at 13:28
  • $\begingroup$ @Holding Arthur, the relationship of AOA and Coefficient of Lift is generally linear up to stall. So for an air craft wing you are using the range of 0 to about 13 degrees (the stall angle of attack) for normal flight. There is an interesting second maxima at 45 degrees, but here drag is off the charts. This is why coefficient of lift and drag graphs are frequently published together, $\endgroup$ Commented May 19, 2019 at 14:33
  • $\begingroup$ @HoldingArthur Perhaps. What are you planning to use the equation for? I don't want to give you an equation that turns out to be useless for what you're planning to use it for. $\endgroup$ Commented May 20, 2019 at 0:55
  • $\begingroup$ Great graph and source (there's also a graph for drag), but the conclusion seems wrong. Based on those graphs, lift is clearly proportional to sin(2α) in the post-stall region. $\endgroup$
    – Rainer P.
    Commented May 21, 2019 at 20:27

The lift coefficient is linear under the potential flow assumptions. So just a linear equation can be used where potential flow is reasonable.

Potential flow solvers like XFoil can be used to calculate it for a given 2D section. Or for 3D wings, lifting-line, vortex-lattice or vortex panel methods can be used (e.g. using XFLR5).

When the potential flow assumptions are not valid, more capable solvers are required.

XFoil has a very good boundary layer solver, which you can use to fit your "simple" model to (e.g. a spline approximation). And I believe XFLR5 has a non-linear lifting line solver based on XFoil results.

For 3D wings, you'll need to figure out which methods apply to your flow conditions. Possible candidates are: experimental data, non-linear lifting line, vortex panel methods with boundary layer solver, steady/unsteady RANS solvers, ...


You mention wanting a simple model that is easy to understand. Always a noble goal. You then relax your request to allow a complicated equation to model it.

@ruben3d suggests one fairly simple approach that can recover behavior to some extent. However one could argue that it does not 'model' anything.

I.e. the arbitrary functions drawn that happen to resemble the observed behavior do not have any explanatory value. You wanted something simple to understand -- @ruben3d's model does not advance understanding.

@ranier-p's approach uses a Newtonian flow model to explain behavior across a wide range of fully separated angle of attack. That does a lot to advance understanding.

For most aircraft use, we are most interested in the well behaved attached potential flow region (say +-8 deg or so). Much study and theory have gone into understanding what happens here. Another ASE question also asks for an equation for lift.

Is there a formula for calculating lift coefficient based on the NACA airfoil?

In this limited range, we can have complex equations (that lead to a simple linear model). These are based on formal derivations from the appropriate physics and math (thin airfoil theory). They are complicated and difficult to understand -- but if you eventually understand them, they have much more value than an arbitrary curve that happens to lie near some observations.

On the other hand, using computational fluid dynamics (CFD), engineers can model the entire curve with relatively good confidence.

From one perspective, CFD is very simple -- we solve the conservation of mass, momentum, and energy (along with an equation of state) for a control volume surrounding the airfoil. We divide that volume into many smaller volumes (or elements, or points) and then we solve the conservation equations on each tiny part -- until the whole thing converges. I.e. we subject the problem to a great deal computational brute force.

This is a very powerful technique capable of modeling very complex flows -- and the fundamental equations and approach are pretty simple -- but it doesn't always provide very satisfying understanding because we lose a lot of transparency in the computational brute force.

I.e. It could be argued that that the Navier Stokes equations are the simple equations that answer your question. But that probably isn't the answer you are looking for.

There is no simple answer to your question. Instead, there is the fascinating field of aerodynamics.

  • $\begingroup$ Ehm, the asker would like to see "an equation relating AoA to lift coefficient", I don't see any in your answer... $\endgroup$
    – sophit
    Commented Feb 23, 2023 at 10:29
  • $\begingroup$ @sophit that is because there is no such thing. I try to make the point that just because you can draw a curve to match observation, you do not advance understanding unless that model is based on the physics. I also try to make the point that just because a simple equation is not possible does not mean that it is impossible to understand or calculate. $\endgroup$ Commented Feb 23, 2023 at 20:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .