Question

Using thin airfoil theory, find the lift coefficient of a NACA 0012 airfoil at \(\alpha=5^\circ\) in uniform flow (rounded to 2 decimal places).

Hint:

For small angles, remember the thin-airfoil slope \(2\pi\) per radian (\(\approx 0.11\) per degree). Thus \(5^\circ \times 0.11 \approx 0.55\) is a quick mental check.

Answer 1

Step 1: Thin airfoil theory relation.
For a symmetric airfoil (NACA 0012), \(\alpha_{L0}=0\). Thin airfoil theory gives \[ C_L = 2\pi\,\alpha \quad (\alpha \text{ in radians}). \] Step 2: Convert angle to radians and compute.
\[ \alpha = 5^\circ = 5\left(\frac{\pi}{180}\right) \text{ rad} = \frac{\pi}{36} \approx 0.087266. \] \[ C_L = 2\pi \times 0.087266 \approx 6.283185 \times 0.087266 \approx 0.5483 \approx 0.55\ \text{(to 2 d.p.)}. \] \[ \boxed{C_L \approx 0.55} \]