Question

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).

Hint:

In the NACA 4-digit naming, the first digit gives the max camber (percent chord) and the second gives its location (tenths of chord). Thin airfoil theory's \(\alpha_{L=0}\) is a linear function of the camber-line slope, so with the same camber location it scales directly with camber.

Answer 1

Correct Answer: -5.3

Step 1: Write the piecewise NACA 4-digit camber line slope formula.
For $0 \le x/c \le p$:
\[ \frac{dz}{dx} = \frac{2m}{p^2}\left(p - \frac{x}{c}\right) \]
For $p \le x/c \le 1$:
\[ \frac{dz}{dx} = \frac{2m}{(1-p)^2}\left(p - \frac{x}{c}\right) \]
Both pieces have the factor $m$ multiplying an expression that depends only on $p$ and $x/c$. Since $p=0.4$ is identical for NACA 2412 and NACA 5410, only $m$ differs between the two, and it appears as a simple multiplier out front in both pieces.

Step 2: Compare the two airfoils term by term.
For NACA 2412, $m_1 = 0.02$. For NACA 5410, $m_2 = 0.05$. At any chordwise station $x/c$,
\[ \frac{(dz/dx)_{5410}}{(dz/dx)_{2412}} = \frac{m_2}{m_1} = \frac{0.05}{0.02} = 2.5 \]
so the entire slope function for 5410 is just the 2412 slope function multiplied by 2.5, everywhere along the chord.

Step 3: Substitute into the zero-lift angle integral.
Since $\alpha_{L=0} = -\frac{1}{\pi}\int_0^{\pi}\frac{dz}{dx}(\cos\theta_0-1)\,d\theta_0$ is linear in $dz/dx$, multiplying $dz/dx$ everywhere by 2.5 multiplies the whole integral, and hence $\alpha_{L=0}$, by exactly 2.5:
\[ \alpha_{L=0}(5410) = 2.5\,\alpha_{L=0}(2412) = 2.5(-2.1^{\circ}) \]

Final Answer:
\[ \boxed{\alpha_{L=0} \approx -5.3^{\circ}} \]