Let's break down the problem step by step. Given: - Downwash velocity relation: \[ w = \frac{L}{2\rho V_0 b^2} \] Where: - \( L \) = Lift - \( \rho \) = Air density - \( V_0 \) = Free-stream velocity - \( b \) = Span --- ### **Part a: How does \( C_{D_i} \) depend on \( C_L \)?** #### **Step 1: Express Lift \( L \) in terms of \( C_L \):** \[ L = C_L \frac{1}{2} \rho V_0^2 S \] where \( S \) is the wing area. #### **Step 2: Substitute \( L \) into the downwash equation:** \[ w = \frac{C_L \frac{1}{2} \rho V_0^2 S}{2\rho V_0 b^2} = \frac{C_L V_0 S}{4 b^2} \] #### **Step 3: Induced Drag Coefficient \( C_{D_i} \):** The induced drag is given by: \[ D_i = L \sin \alpha_i \approx L \alpha_i \] where the induced angle of attack \( \alpha_i \) is: \[ \alpha_i \approx \frac{w}{V_0} \] So, \[ D_i = L \frac{w}{V_0} \] The induced drag coefficient: \[ C_{D_i} = \frac{D_i}{\frac{1}{2} \rho V_0^2 S} = \frac{L \frac{w}{V_0}}{\frac{1}{2} \rho V_0^2 S} = \frac{L w}{\frac{1}{2} \rho V_0^3 S} \] Now, substitute \( L \) and \( w \) from above: \[ C_{D_i} = \frac{C_L \frac{1}{2} \rho V_0^2 S \cdot \frac{C_L V_0 S}{4 b^2}}{\frac{1}{2} \rho V_0^3 S} = \frac{C_L \cdot \frac{C_L V_0 S}{4 b^2}}{V_0 S} = \frac{C_L^2}{4 b^2} \] So, \[ \boxed{C_{D_i} \propto C_L^2} \] with the proportionality constant \( \frac{1}{4b^2} \). --- ### **Part b: Wing \( C_L(\alpha) \) Relation and \( \frac{dC_L}{d\alpha} \) Compared to 2D Value** #### **Step 1: Effective Angle of Attack** The effective angle of attack is: \[ \alpha_\text{eff} = \alpha - \alpha_i \] where \( \alpha_i \approx \frac{w}{V_0} \). #### **Step 2: 2D Lift Slope** For a 2D airfoil: \[ C_L = 2\pi \alpha_\text{eff} \] #### **Step 3: Substitute \( \alpha_\text{eff} \)** \[ C_L = 2\pi (\alpha - \alpha_i) \] But \( \alpha_i = \frac{w}{V_0} = \frac{C_L S}{4b^2} \) (from earlier). So, \[ C_L = 2\pi \left[\alpha - \frac{C_L S}{4 b^2}\right] \] #### **Step 4: Solve for \( C_L \) in terms of \( \alpha \):** \[ C_L = 2\pi \alpha - 2\pi \frac{C_L S}{4 b^2} \] \[ C_L + \frac{2\pi S}{4 b^2} C_L = 2\pi \alpha \] \[ C_L \left[1 + \frac{2\pi S}{4 b^2}\right] = 2\pi \alpha \] \[ C_L = \frac{2\pi \alpha}{1 + \frac{2\pi S}{4 b^2}} \] #### **Step 5: Compute \( \frac{dC_L}{d\alpha} \):** \[ \frac{dC_L}{d\alpha} = \frac{2\pi}{1 + \frac{2\pi S}{4 b^2}} \] #### **Step 6: Compare to 2D Value** The 2D lift slope is \( 2\pi \). The finite span wing's lift slope is **less** than \( 2\pi \) due to induced effects. --- ### **Summary Table** | Quantity | 2D (Airfoil) | 3D (Finite Wing) | |----------|--------------|------------------| | \( C_{D_i} \) | - | \( \propto C_L^2 \) | | \( \frac{dC_L}{d\alpha} \) | \( 2\pi \) | \( \frac{2\pi}{1 + \frac{2\pi S}{4 b^2}} \) | --- If you need a more detailed explanation or further simplification, let me know!