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Are there any formulas that can be used to calculate this?

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    $\begingroup$ There are, but it quickly gets a lot more complex. What are you really trying to do? There are many NACA airfoil families, which are you interested in? Two-dimensional characteristics of an airfoil only get you so far -- the switch to 3D gets much more complex and there are far fewer closed form solutions. $\endgroup$ Nov 27, 2022 at 6:44
  • $\begingroup$ I'm interested in two dimensional NACA 4 digit airfoil $\endgroup$
    – Martreb
    Nov 27, 2022 at 14:35
  • $\begingroup$ 5-digit and 6 digit there are number telling their coefficient of lift, but not 4 digit. For the 5-digit: the 1st number multiplied by 3/2 then divided by 10, or 1st number multiplied by 0.15. For 6-digit: the 3rd number divided by 10. youtube.com/watch?v=y6D3qhss7vc $\endgroup$ Nov 29, 2022 at 3:55
  • $\begingroup$ To my knowledge the 4-digit NACA airfoils have been specifically conceived during a project to systematically measure the effects of camber, thickness, etc. rather than just relying on "a formula" (thin airfloil theory or such). The results of are in NACA Report No. 460 (publicly accessible), there the results are also compared to theoretical values. $\endgroup$
    – Apfelsaft
    Dec 30, 2022 at 4:31
  • $\begingroup$ Is there any reason you have not accepted my answer below? Do you need any further clarification? $\endgroup$ Dec 30, 2022 at 5:51

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Thin airfoil theory can be used to calculate lift based on the shape of the mean camber line. This involves some integrals, but these have been worked out ahead of time for certain camber line families like the 4-digit one you are interested in.

$c_l = 2 \pi \alpha + \pi (A_1-2 A_0)$

$A_0 = \frac{1}{\pi} \int_0^\pi \frac{dy}{dx} d\theta$

$A_1 = \frac{2}{\pi} \int_0^\pi \frac{dy}{dx} \cos(\theta) d\theta$

$\frac{dy}{dx}$ Is the slope of the camber line.

We use the transformation:

$x = \frac{1}{2} (1-\cos( \theta))$

These integrals have been worked out for a 4-digit mean camber line (Houghton and Carpenter, 4th Ed, pp. 252).

$A_0 = \frac{m}{\pi p^2} [(2 p - 1) \theta_p + \sin(\theta_p) ]\\ + \frac{m}{\pi (1-p)^2} [(2 p - 1)(\pi - \theta_p) + \sin(\theta_p) ]$

$A_1 = \frac{2 m}{\pi p^2} [(2 p - 1) \sin(\theta_p) + \frac{1}{4} \sin(2 \theta_p) + \frac{\theta_p}{2}]\\ - \frac{2 m}{\pi (1-p)^2} [(2 p - 1) \sin(\theta_p) + \frac{1}{4} \sin(2 \theta_p) - \frac{\pi - \theta_p}{2}]$

For a 4412, $m=0.04$ and $p=0.4$. $\theta_p=\cos^{-1}(1-2*p)=1.3694$ radians. $A_0 = 0.00811$ and $A_1 = 0.1632$.

Pulling it all together gives:

$c_l = 0.46175 + 2 \pi \alpha$

Finite thickness and viscous effects will reduce the true value a bit from this, but it is pretty good. To do better than this, things get substantially more complex.

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    $\begingroup$ Where or how did you get this (CL = 0.512 + 2 * pi * alpha_rad) formula? As far as I know, the first digit is representing its maximum camber in chord line (C), i.e., 4% of its C, the second digit is representing its location of its maximum camber x 10%, i.e., 4x10x% of C, and the last two digits representing maximum thickness in regard to its C. Nothing about coefficient of lift. $\endgroup$ Nov 29, 2022 at 3:48
  • $\begingroup$ I missed a term in my exposition above corresponding to the angle of attack for the design lift coefficient -- it is not zero. I will edit my answer and also give a reference to a textbook with this example worked out. $\endgroup$ Nov 30, 2022 at 0:02

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