# How do you calculate the lift coefficient of an airfoil at zero angle of attack?

Thin airfoil theory gives $C = C_o + 2\pi\alpha$, where $C_o$ is the lift coefficient at $\alpha = 0$. However, I couldn't find any equation to calculate what $C_o$ is which must be some function of the airfoil shape. In other words how do you extend thin airfoil theory to cambered airfoils without having to use experimental data?

This is my own attempt, I made this airfoil model of the lift coefficient of the airfoil at zero angle of attack for a project I am working on. It's derived from a Joukowsky transform. it seems to work but how is $C_o$ actually calculated?

Because it is derived from the Joukowsky transform of the inviscid potential flow around a cylinder it's more accurate at high Reynolds numbers.

• I marked this as "unclear what you're asking", but only for lack of a better label. The problem, in just my opinion, with this question is that the only possible answer is 'yes' because we cannot disprove a negative. Ie. if you've discovered a new way, then no one can realistically be expected to comprehensively prove no one has ever done it before. – Ryan Mortensen Aug 31 '18 at 21:56
• The linear term ($C_0=4\pi \cdot p$) can be found in many books, while the extra terms affect the result only in single digit percentages. This is the same order of magnitude as the error, as well as smaller than the magnitude of other factors you're not taking into account such as the chordwise position of max camber. That's probably a reason why the formula is not commonly shown the way you wrote it down: the same way we use $C_l=C_{l\alpha}\cdot \alpha$ instead of $C_l=C_{l\alpha}\cdot \sin(\alpha)$, "$1-\sqrt(4p^2+1)$" makes the calculation more complex without improving the accuracy. – Gypaets Sep 3 '18 at 6:30
• I see you created two accounts. If you wish to join them, please follow the instructions here: aviation.stackexchange.com/help/merging-accounts – Federico Sep 3 '18 at 7:05
• also, please consider typing the equations here, rather than posting an image, it makes answering and searching a bit easier – Federico Sep 3 '18 at 7:06