A thin cambered airfoil has a parabolic camber line defined by $z(x) = 4 h \frac{x}{c} \left(1 - \frac{x}{c}\right)$, where $h$ is the maximum camber at mid-chord. In thin airfoil theory with coordinate transformation $x = \frac{c}{2}(1 - \cos\theta)$, the camber slope $\frac{dz}{dx}$ expressed in terms of $\theta$ is:

(A) $\frac{dz}{dx} = \frac{4h}{c} \cos\theta$

(B) $\frac{dz}{dx} = \frac{4h}{c} \sin\theta$

(C) $\frac{dz}{dx} = \frac{4h}{c} (1 - \cos\theta)$

(D) $\frac{dz}{dx} = \frac{8h}{c} \cos\theta$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Given camber profile: $$z(x) = 4h \left(\frac{x}{c} - \frac{x^2}{c^2}\right)$$ Differentiating with respect to $x$: $$\frac{dz}{dx} = 4h \left(\frac{1}{c} - \frac{2x}{c^2}\right) = \frac{4h}{c} \left(1 - \frac{2x}{c}\right)$$ ### 2. Coordinate Transformation In Glauert's transformation for thin airfoils: $$x = \frac{c}{2} (1 - \cos\theta) \implies \frac{2x}{c} = 1 - \cos\theta$$ Substitute into the slope expression: $$\frac{dz}{dx} = \frac{4h}{c} \left[1 - (1 - \cos\theta)\right] = \frac{4h}{c} \cos\theta$$ ### 3. Step-by-Step Verification - At the leading edge ($x = 0, \theta = 0$): $\frac{dz}{dx} = \frac{4h}{c}$ (maximum positive slope). - At the mid-chord ($x = c/2, \theta = \pi/2$): $\frac{dz}{dx} = 0$ (peak camber summit). - At the trailing edge ($x = c, \theta = \pi$): $\frac{dz}{dx} = -\frac{4h}{c}$ (maximum negative slope). ### 4. Distractor Forensics - **Option A:** Correct trigonometric transformation. - **Option B:** Uses $\sin\theta$, which incorrectly gives zero slope at the leading edge. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Note that because $dz/dx \propto \cos\theta$, the Fourier series expansion of circulation has only the $A_1$ term ($A_1 = \frac{4h}{c}$, with $A_0 = A_2 = A_3 = \dots = 0$). - **Standard Reference:** Anderson, J.D., *Fundamentals of Aerodynamics*, Chapter 4.

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