Subtopic: Free & Forced Vibrations of Continuous Beams
Question Statement:
Thin airfoil theory predicts the zero-lift angle of attack 𝛼L=0 of NACA 2412 airfoil
as -2.1°. The corresponding prediction of 𝛼L=0 for NACA 5410 airfoil is
________ degrees (rounded off to 1 decimal place).
Official Answer Key:
-5.25
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.63)**
**1. Intuitive Physical Setup & Given Parameters**:
From thin airfoil theory:
- NACA 2412 zero-lift angle of attack: $\alpha_{L=0} = -2.1^\circ$.
- We need to find the predicted zero-lift angle of attack $\alpha_{L=0}$ for a **NACA 5410** airfoil in degrees.
**2. Step-by-Step Aerodynamic Derivation**:
1. **NACA 4-Digit Numbering Convention**:
In NACA $M P X X$:
- First digit ($M$): Maximum camber as a percentage of chord ($m = M / 100$).
- Second digit ($P$): Location of maximum camber as tenths of chord ($p = P / 10$).
- Last two digits ($XX$): Maximum thickness as a percentage of chord.
2. **Airfoil Parameters**:
- For **NACA 2412**: $M = 2 \implies m = 0.02$, $P = 4 \implies p = 0.4$.
- For **NACA 5410**: $M = 5 \implies m = 0.05$, $P = 4 \implies p = 0.4$.
3. **Linearity of Thin Airfoil Theory**:
- In thin airfoil theory, the mean camber line equation $z(x)$ is directly proportional to maximum camber $m$:
$z(x) = m \cdot f(x/c, p)$
- The zero-lift angle of attack is given by:
$\alpha_{L=0} = -\frac{1}{\pi} \int_0^\pi \frac{dz}{dx} (\cos\theta - 1) d\theta$
- Since both airfoils share the exact same camber location parameter ($p = 0.4$), the camber distribution function $f(x/c, p)$ is identical.
- Therefore, $\alpha_{L=0}$ scales **linearly** with maximum camber $m$:
$\alpha_{L=0} \propto m$
4. **Calculate for NACA 5410**:
$\alpha_{L=0}(\text{NACA 5410}) = \alpha_{L=0}(\text{NACA 2412}) \times \left(\frac{m_{5410}}{m_{2412}}\right) = -2.1^\circ \times \left(\frac{5}{2}\right) = -2.1^\circ \times 2.5 = -5.25^\circ$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $-5.25^\circ$.
- **Rounding Requirement**: One decimal place $\implies \mathbf{-5.2}$ or $\mathbf{-5.3}$ (nominal **-5.25**).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **-5.25** (Acceptable Range: **-5.4 to -5.1**)
- **Exam Takeaway**: In thin airfoil theory, when maximum camber location $p$ is identical, zero-lift angle of attack is directly proportional to maximum camber: $\alpha_{L=0} \propto m$.
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