An aircraft of gross weight $W = 60{,}000\text{ N}$ has a wing planform area $S = 30\text{ m}^2$ and maximum lift coefficient $C_{L,\max} = 1.6$. Flying in standard sea-level air density ($\rho = 1.225\text{ kg/m}^3$), the positive structural design limit load factor is $n_{\max} = +6.0$. The corner velocity (maneuver speed $V^*$ in $\text{m/s}$) at which the aircraft achieves the maximum limit load factor right at aerodynamic stall is closest to:

(A) 78.2 m/s

(B) 101.0 m/s

(C) 41.2 m/s

(D) 123.7 m/s

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles On the $V-n$ flight envelope (maneuver diagram), the corner velocity (or maneuver speed $V^*$) defines the precise intersection between the aerodynamic stall boundary and the positive structural limit load factor $n_{\max}$. At this flight speed, the wing produces exactly the maximum allowable normal load factor $n_{\max}$ when operating at its peak maximum lift coefficient $C_{L,\max}$: $$n_{\max} = \frac{L}{W} = \frac{\frac{1}{2} \rho (V^*)^2 S C_{L,\max}}{W}$$ ### 2. Analytical Formulation Rearranging directly for the corner velocity $V^*$: $$(V^*)^2 = \frac{2 n_{\max} W}{\rho S C_{L,\max}} \implies V^* = \sqrt{\frac{2 n_{\max} W}{\rho S C_{L,\max}}}$$ ### 3. Step-by-Step Calculation Substitute the known flight conditions: - $W = 60{,}000\text{ N}$ - $n_{\max} = 6.0$ - $\rho = 1.225\text{ kg/m}^3$ - $S = 30\text{ m}^2$ - $C_{L,\max} = 1.6$ Numerator of radical: $$2 \times 6.0 \times 60{,}000 = 720{,}000\text{ N}$$ Denominator of radical: $$\rho S C_{L,\max} = 1.225 \times 30 \times 1.6 = 58.8\text{ kg/m}$$ Evaluating velocity: $$(V^*)^2 = \frac{720{,}000}{58.8} \approx 12{,}244.898\text{ m}^2/\text{s}^2$$ $$V^* = \sqrt{12{,}244.898} \approx 101.02\text{ m/s}$$ ### 4. Distractor Forensics - **Option A ($78.2\text{ m/s}$):** Computed without multiplying by 2 in dynamic pressure formulation or misapplying load factor. - **Option B ($101.0\text{ m/s}$):** Correct standard solution. - **Option C ($41.2\text{ m/s}$):** Corresponds to unaccelerated 1g stall speed $V_{s1} = \sqrt{2W/(\rho S C_{L,\max})} = \sqrt{120000 / 58.8} = 45.18\text{ m/s}$ evaluated with errors. - **Option D ($123.7\text{ m/s}$):** Erroneously used ultimate load factor ($1.5 \times 6 = 9$) instead of limit load factor $n_{\max} = 6.0$. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always ensure whether the problem specifies *limit* load factor or *ultimate* load factor. Structural limit $n_{\max}$ governs the corner speed $V^*$. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, 8th Edition, McGraw-Hill, Chapter 6 (Aircraft Performance: The Maneuver $V-n$ Diagram).

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