For a thin straight cantilever wing with torsional stiffness $GJ = 2.4 \times 10^5\text{ N}\cdot\text{m}^2$ and semi-span $b = 6.0\text{ m}$, the aileron effectiveness parameter yields a theoretical reversal dynamic pressure $q_{rev} = 12.0\text{ kPa}$. At sea level where air density $\rho = 1.20\text{ kg/m}^3$, what is the critical aileron reversal flight speed $V_{rev}$ (in m/s)? (Round off to one decimal place)

Answer Key:

141.4

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Aileron Reversal Principle:** Aileron reversal occurs when flexible wing elastic twist due to aileron deflection completely cancels the aerodynamic rolling moment produced by the aileron, reducing net rolling control to zero. 2. **Stage 2: Mathematical Formulation:** * The dynamic pressure at which roll rate vanishes is $q_{rev}$. * By definition: $q_{rev} = \frac{1}{2} \rho V_{rev}^2$. 3. **Stage 3: Solving for Reversal Speed:** * $V_{rev} = \sqrt{\frac{2 q_{rev}}{\rho}}$. 4. **Stage 4: Substitution:** * $q_{rev} = 12.0\text{ kPa} = 12,000\text{ Pa}$. * $\rho = 1.20\text{ kg/m}^3$. * $V_{rev} = \sqrt{\frac{2 \times 12,000}{1.20}} = \sqrt{20,000} = 141.421\text{ m/s}$. * Rounding to one decimal place gives $141.4\text{ m/s}$. 5. **Stage 5: Official Reference:** Raymond L. Bisplinghoff, Holt Ashley, and Robert L. Halfman, *Aeroelasticity*, Chapter 9 (Static Aeroelastic Phenomena).

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