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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?

enter image description here

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

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  • $\begingroup$ 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. $\endgroup$
    – Ryan Mortensen
    Aug 31, 2018 at 21:56
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    $\begingroup$ 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. $\endgroup$
    – Gypaets
    Sep 3, 2018 at 6:30
  • $\begingroup$ I see you created two accounts. If you wish to join them, please follow the instructions here: aviation.stackexchange.com/help/merging-accounts $\endgroup$
    – Federico
    Sep 3, 2018 at 7:05
  • $\begingroup$ also, please consider typing the equations here, rather than posting an image, it makes answering and searching a bit easier $\endgroup$
    – Federico
    Sep 3, 2018 at 7:06

2 Answers 2

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The solution and the approximate solution to zero-lift angle of attack from thin airfoil theory can be found in ESDU 98011. The derivation of an analytical solution can also be found in Anderson, Fundamentals of Aerodynamics. $C_{l_0}$ can then be found by multipling it with $2\pi$.

Edited to add classical thin-airfoil result:

$$C_{l_0}=2\int_0^\pi{\frac{dz}{dx}(cos\theta-1)d\theta}$$ $$x=\frac{1}{2}(1-cos\theta)$$

$x$ is the chord line coordinate from 0 to 1; $z$ is the mean camber line height at coordinate $x$, normalized to the chord length; $\frac{dz}{dx}$ is the curvature of the camber line.

Now you just have to numerically integrate to get the lift at zero incidence.

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  • $\begingroup$ This could be improved by citing the relevant parts of those sources. $\endgroup$
    – AEhere supports Monica
    Jul 26, 2019 at 7:24
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An approach for finding $C_l$ of a cambered wing is to look at the angle of the tail section, and take this as if it was the $\alpha$ of a flat plate

enter image description here

In this example the max camber is pretty far forward, which makes a good illustration of the principle. It works best if the outflow has had ample opportunity to follow the aft end of the profile.

As an aside, $2\pi \alpha$ in an approximation at small angles of $2\pi \cdot sin (\alpha)$

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