A model airplane has a wing-body aerodynamic center located at $(x_{ac}/c)_{wb} = 0.24$. The horizontal tail has an area ratio $\eta_{tail} = S_t / S = 0.20$, tail moment arm $l_t / c = 3.0$, tail lift curve slope $a_t = 4.2\text{ rad}^{-1}$, wing-body lift curve slope $a_{wb} = 4.8\text{ rad}^{-1}$, tail efficiency factor $\eta_t = 0.95$, and downwash gradient $\frac{\partial \varepsilon}{\partial \alpha} = 0.35$. The stick-fixed neutral point location $(x_{np}/c)$ of the aircraft is:

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### 1. Strategy & Theoretical Principles The stick-fixed neutral point $(x_{np}/c)$ is defined as the center of gravity location where the longitudinal static stability margin vanishes ($C_{m_\alpha} = 0$). From classic aircraft longitudinal stability: $$\frac{x_{np}}{c} = \left(\frac{x_{ac}}{c}\right)_{wb} + V_H \eta_t \frac{a_t}{a} \left(1 - \frac{\partial \varepsilon}{\partial \alpha}\right)$$ where $V_H = \frac{S_t l_t}{S c}$ is the horizontal tail volume ratio, and $a = a_{wb} + a_t \eta_t \frac{S_t}{S} \left(1 - \frac{\partial \varepsilon}{\partial \alpha}\right)$ is the total aircraft lift curve slope. ### 2. Analytical Formulation Calculate the tail volume ratio $V_H$: $$V_H = \left(\frac{S_t}{S}\right) \left(\frac{l_t}{c}\right) = 0.20 \times 3.0 = 0.60$$ Calculate the contribution of the tail to total lift curve slope: $$\Delta a_t = a_t \eta_t \left(\frac{S_t}{S}\right) \left(1 - \frac{\partial \varepsilon}{\partial \alpha}\right) = 4.2 \times 0.95 \times 0.20 \times (1 - 0.35)$$ $$\Delta a_t = 0.798 \times 0.65 = 0.5187\text{ rad}^{-1}$$ $$a = a_{wb} + \Delta a_t = 4.8 + 0.5187 = 5.3187\text{ rad}^{-1}$$ ### 3. Step-by-Step Calculation Now evaluate the tail stabilizing term: $$\Delta x_{np} = V_H \eta_t \frac{a_t}{a} \left(1 - \frac{\partial \varepsilon}{\partial \alpha}\right) = 0.60 \times 0.95 \times \frac{4.2}{5.3187} \times 0.65$$ $$\Delta x_{np} = 0.57 \times 0.78966 \times 0.65 = 0.2925$$ Therefore, the stick-fixed neutral point is: $$\frac{x_{np}}{c} = 0.24 + 0.2925 = 0.5325 \approx 0.53\text{ to }0.56$$ If using the standard approximation with $a \approx a_{wb}$ (tail-off lift slope): $$\Delta x_{np} = 0.60 \times 0.95 \times \frac{4.2}{4.8} \times 0.65 = 0.57 \times 0.875 \times 0.65 = 0.3242$$ $$\frac{x_{np}}{c} = 0.24 + 0.324 = 0.5642$$ Allowing standard tolerance of 0.54 to 0.58 accommodates both exact total-lift and tail-off approximations. ### 4. Distractor Forensics - For NAT questions, candidates commonly forget downwash factor $(1 - \partial \varepsilon / \partial \alpha)$, giving $\Delta x_{np} = 0.499$, shifting $x_{np}/c$ erroneously to $> 0.70$. - Neglecting dynamic pressure ratio $\eta_t = 0.95$ yields slight over-estimation. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always account for downwash reduction in tail angle of attack: $\alpha_t = \alpha - \varepsilon + i_t$, hence $\partial \alpha_t / \partial \alpha = (1 - \partial \varepsilon / \partial \alpha)$. - **Standard Reference:** Nelson, R.C., *Flight Stability and Automatic Control*, 2nd Edition, Chapter 2 (Static Longitudinal Stability and Trim).

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