0
$\begingroup$

Thin airfoil theory says that lift coefficient is directly proportional to the angle of attack in radians. I haven't been able to find any limit, short of stall, for applying this theory. It would give us a huge lift coefficient at 90 degrees, where lift is actually zero, and double even that at 180 degrees, where again the lift is zero. This is basic reasoning. So, what am I missing?

$\endgroup$
  • $\begingroup$ What makes you think there should be another limit than stall? That seems like a pretty definitive limit to me. $\endgroup$ – Sanchises Jun 7 at 12:08
2
$\begingroup$

Of course not. Thin Airfoil Theory produces a beautiful result that relates the zero-AOA lift to the mean camber of the airfoil, as well as the lift slope vs AOA of any thin airfoil is $2\pi$. But that's assuming:

  1. Airfoil is asymptotically thin: such that the mean camber line represents the flow path
  2. Camber is close to the chord: such that the vortex sheet is assumed to lie on the chord while resulting in the flow path that lies on the camber line
  3. AOA and camber are small: such that all angles are linearized ($\sin\theta \approx \tan\theta \approx \theta$); obviously at large angles, this assumption breaks down and the results are worthless
  4. Potential and incompressible flow: by the virtue of irrotationality, stall will not be modeled.

So in all, the result is only valid for small range of AOA on an idealized thin airfoil. For engineering purposes on general airfoils, you'd want to use panel method + boundary layer equations. But even then, extrapolating results beyond stall lead to ridiculous results (certainly at 90deg or 180deg).

More details on Thin Airfoil Theory can be found in Drela, Flight Vehicle Aerodynamics or Anderson, Fundamentals of Aerodynamics.

| improve this answer | |
$\endgroup$
  • $\begingroup$ AOA and camber are small thanks! So that explains why my simulated plane is climbing straight up! $\endgroup$ – Abdullah Jun 7 at 13:15
  • $\begingroup$ @Abdullah which software do u use for that simulation? $\endgroup$ – Auberron Jun 7 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.