Thin airfoil theory predicts the zero-lift angle of attack \(\alpha_{L=0}\) of NACA 2412 airfoil as \(-2.1^{\circ}\). The corresponding prediction of \(\alpha_{L=0}\) for NACA 5410 airfoil is _______ degrees (rounded off to 1 decimal place).
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Correct Answer: -5.3
Solution and Explanation
In the NACA 4-digit system XYZZ, X is the maximum camber as a percentage of chord (\(m\)), Y is the location of maximum camber in tenths of chord (\(p\)), and ZZ is the thickness as a percentage of chord.
NACA 2412: \(m = 2\%\), \(p = 0.4c\), thickness \(= 12\%\).
NACA 5410: \(m = 5\%\), \(p = 0.4c\), thickness \(= 10\%\).
Step 2: Recall what thin airfoil theory actually depends on.
Thin airfoil theory predicts lift and \(\alpha_{L=0}\) purely from the mean camber line's slope \(dz/dx\); thickness has no effect on this prediction. So the 12% versus 10% thickness difference between the two airfoils is irrelevant here.
Step 3: Use the linear scaling of the camber-line slope with m.
The NACA 4-digit camber line (both its forward and aft parabolic-arc segments) has a slope \(dz/dx\) that is directly proportional to \(m\) for a fixed camber location \(p\). Since both airfoils share the same \(p = 0.4c\), the entire camber-line slope of NACA 5410 is exactly \((5/2)\) times that of NACA 2412 at every point along the chord.
\[ \alpha_{L=0} = -\frac{1}{\pi}\int_0^{\pi} \frac{dz}{dx}(\cos\theta_0 - 1)\,d\theta_0 \]
Because this integral is linear in \(dz/dx\), scaling \(dz/dx\) by a constant factor scales \(\alpha_{L=0}\) by the same factor:
\[ \alpha_{L=0}(5410) = \frac{5}{2}\,\alpha_{L=0}(2412) \]
Step 4: Substitute.
\[ \alpha_{L=0}(5410) = 2.5 \times (-2.1^{\circ}) = -5.25^{\circ} \]
Final Answer:
\[ \boxed{\alpha_{L=0} \approx -5.3^{\circ}} \]
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