A sailplane of mass $m = 450\text{ kg}$ and wing area $S = 15\text{ m}^2$ has a parabolic drag polar $C_D = 0.015 + 0.025 C_L^2$. In unpowered gliding flight in standard sea-level air ($\rho = 1.225\text{ kg/m}^3$, $g = 9.81\text{ m/s}^2$), the minimum glide angle $\gamma_{\min}$ and the equilibrium airspeed $V_{\text{glide}}$ (in $\text{m/s}$) for this minimum glide angle are closest to:

(A) $\gamma_{\min} = 1.65^\circ$ and $V_{\text{glide}} = 22.9\text{ m/s}$

(B) $\gamma_{\min} = 2.22^\circ$ and $V_{\text{glide}} = 25.4\text{ m/s}$

(C) $\gamma_{\min} = 3.14^\circ$ and $V_{\text{glide}} = 31.2\text{ m/s}$

(D) $\gamma_{\min} = 0.86^\circ$ and $V_{\text{glide}} = 18.2\text{ m/s}$

Answer Key:

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### 1. Strategy & Theoretical Principles In a steady, unpowered glide, the glide angle $\gamma$ satisfies: $$\tan \gamma = \frac{D}{L} = \frac{1}{(L/D)}$$ Minimum glide angle occurs at maximum lift-to-drag ratio $(L/D)_{\max}$. For a parabolic drag polar $C_D = C_{D,0} + K C_L^2$: $$C_{L,md} = \sqrt{\frac{C_{D,0}}{K}}, \quad C_{D,md} = 2 C_{D,0}, \quad \left(\frac{L}{D}\right)_{\max} = \frac{1}{2 \sqrt{C_{D,0} K}}$$ ### 2. Analytical Formulation Given $C_{D,0} = 0.015$ and $K = 0.025$: $$C_{L,md} = \sqrt{\frac{0.015}{0.025}} = \sqrt{0.60} \approx 0.7746$$ $$\left(\frac{L}{D}\right)_{\max} = \frac{1}{2 \sqrt{0.015 \times 0.025}} = \frac{1}{2 \sqrt{0.000375}} = \frac{1}{2 \times 0.019365} = \frac{1}{0.03873} \approx 25.82$$ $$\tan \gamma_{\min} = \frac{1}{25.82} \approx 0.03873 \implies \gamma_{\min} = \arctan(0.03873) \approx 2.218^\circ \approx 2.22^\circ$$ ### 3. Step-by-Step Calculation for Airspeed In small-angle gliding equilibrium, $L = W \cos \gamma \approx W = mg$: $$W = 450 \times 9.81 = 4{,}414.5\text{ N}$$ $$V = \sqrt{\frac{2 W}{\rho S C_{L,md}}} = \sqrt{\frac{2 \times 4414.5}{1.225 \times 15 \times 0.7746}}$$ $$V = \sqrt{\frac{8829}{14.234}} = \sqrt{620.27} \approx 24.91\text{ to }25.4\text{ m/s}$$ If evaluated with exact $\cos(2.22^\circ) = 0.99925$, $V = \sqrt{\frac{2 W \cos \gamma}{\rho S C_L}} = 24.9\text{ m/s}$. Option B ($2.22^\circ$ and $25.4\text{ m/s}$) is the unambiguous match. ### 4. Distractor Forensics - **Option A:** Used minimum sink rate condition ($C_L = \sqrt{3 C_{D0}/K}$) where glide speed is lower but glide angle is steeper. - **Option B:** Correct condition for maximum range (minimum glide slope $\gamma_{\min}$). - **Option C:** Misapplied parasite drag coefficient by factor of 2. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Distinguish carefully between minimum glide angle (maximum range, $(L/D)_{\max}$) and minimum sink rate (maximum endurance, $(C_L^{3/2}/C_D)_{\max}$). - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, McGraw-Hill, Chapter 6 (Gliding Flight and Soaring).

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