# Problem Overview We are given a tapered wing with the following parameters: - **Span (b):** 36 ft - **Root chord (Cr):** 6 ft - **Tip chord (Ct):** ft (not a practical design, but for calculation) - **Airfoil:** NASA 4412 - **Aircraft weight:** 300 lb - **Altitude:** 200 m (~656 ft) - **Speed:** 200 mph --- # Step 1: Convert Units ### Convert speed from mph to ft/sec: \[ v = 200\, \text{mph} \times \frac{528\, \text{ft}}{1\, \text{mile}} \times \frac{1\, \text{hr}}{360\, \text{sec}} \] \[ v = 200 \times \frac{528}{360} \approx 200 \times 1.467 \approx 293.4\, \text{ft/sec} \] ### Convert altitude to feet: \[ h = 200\, \text{m} \times 3.281\, \text{ft/m} \approx 656\, \text{ft} \] --- # Step 2: Determine Air Density (\( \rho \)) At 200 m (~656 ft), the air density is approximately: \[ \rho \approx .002377\, \text{slugs/ft}^3 \] *(Standard sea level density is about .0023769, so this is close)* --- # Step 3: Calculate Wing Area (S) Since the wing is tapered from root to tip: \[ S = \frac{b}{2} (Cr + Ct) \] Given: \[ b = 36\, \text{ft},\quad Cr=6\, \text{ft},\quad Ct=\, \text{ft} \] \[ S = \frac{36}{2} (6 + ) = 18 \times 6 = 108\, \text{ft}^2 \] --- # Step 4: Determine Lift Required The aircraft weight is balanced by lift: \[ L = W = 300\, \text{lb} \] Lift coefficient (\( C_L \)) at level flight: \[ L = \frac{1}{2} \rho v^2 S C_L \] Solve for \( C_L \): \[ C_L = \frac{2L}{\rho v^2 S} \] Plugging in the numbers: \[ C_L = \frac{2 \times 300}{.002377 \times (293.4)^2 \times 108} \] Calculate denominator: \[ .002377 \times 293.4^2 \times 108 \] \[ 293.4^2 \approx 86,058 \] \[ .002377 \times 86,058 \approx 204.4 \] \[ 204.4 \times 108 \approx 22,095 \] Now compute \( C_L \): \[ C_L = \frac{600}{22,095} \approx .271 \] **Result:** \[ \boxed{ C_{L} \approx .271 } \] --- # Step 5: Estimate Drag Coefficient \( C_D \) ### 5.1: Find the Airfoil Data NASA 4412 is a typical airfoil with: - **Drag coefficient at \( C_{L} = .271 \):** Approximate \( C_{D_} \) (parasitic drag) and \( C_{D_i} \) (induced drag). From typical data: | Parameter | Approximate Value | |---|---| | \( C_{D_} \) (Zero-lift drag) | .02 (typical for clean airfoil) | | \( k \) (Induced drag factor) | .055 (approximate for elliptical lift distribution) | --- ### 5.2: Calculate \( C_D \) \[ C_D = C_{D_} + C_{D_i} \] \[ C_{D_i} = \frac{C_L^2}{\pi e AR} \] Where: - \( e \) = Oswald efficiency factor (~.8) - \( AR \) = Aspect ratio = \( \frac{b^2}{S} \) Calculate \( AR \): \[ AR = \frac{b^2}{S} = \frac{36^2}{108} = \frac{1296}{108} = 12 \] Calculate \( C_{D_i} \): \[ C_{D_i} = \frac{C_L^2}{\pi e AR} = \frac{(.271)^2}{\pi \times .8 \times 12} \] \[ (.271)^2 \approx .0734 \] \[ \pi \times .8 \times 12 \approx 3.1416 \times .8 \times 12 \approx 3.1416 \times 9.6 \approx 30.17 \] \[ C_{D_i} = \frac{.0734}{30.17} \approx .00243 \] --- ### 5.3: Total \( C_D \): \[ C_D = C_{D_} + C_{D_i} \approx .02 + .00243 = .02243 \] **Final Total Drag Coefficient:** \[ \boxed{ \textbf{(a) } C_{D} \approx .0224 } \] --- # Step 6: Wing Angle of Attack for Level Flight Since lift equals weight: \[ L = \frac{1}{2} \rho v^2 S C_L \] \[ \Rightarrow C_L = .271 \quad (\text{from earlier}) \] Lift coefficient relates to angle of attack (\( \alpha \)): \[ C_L = C_{L_\alpha} \times \alpha \] Where: - \( C_{L_\alpha} \) is the lift curve slope (~.11 per degree for NASA 4412). Rearranged: \[ \alpha = \frac{C_L}{C_{L_\alpha}} \] \[ \alpha = \frac{.271}{.11} \approx 2.46^\circ \] **Final answer:** \[ \boxed{ \textbf{(b) } \text{Angle of attack } \alpha \approx 2.46^\circ } \] --- # **Summary** | Quantity | Result | |---|---| | **(a) Total drag coefficient \( C_D \)** | **.0224** | | **(b) Wing angle of attack \( \alpha \)** | **2.46°** | This completes the step-by-step solution. Let me know if you'd like further clarification!