Finite Wing Theory Questions and Answers

Answer: B

The induced drag coefficient is \(\dfrac{C_{L}^{2}}{\pi AR}\).

An elliptical lift distribution produces a uniform downwash along the span, with induced angle \(\alpha_{i} = C_{L}/(\pi AR)\). The lift vector is tilted back by this angle, and its rearward component is the induced drag:

\[ C_{D,i} = C_{L}\alpha_{i} = \frac{C_{L}^{2}}{\pi AR} \]

\(C_{L}/(\pi AR)\) is the induced angle itself, not the drag coefficient. The forms with \(2\pi AR\) or \(AR^{2}\) in the denominator do not come from lifting-line theory; induced drag falls with the first power of aspect ratio.

Answer: B

The lift coefficient is \(\pi^{2}/12\).

For an elliptic wing the induced angle of attack is \(\alpha_{i} = C_{L}/(\pi AR)\), so with the thin-airfoil section slope \(a_{0} = 2\pi\) the wing lift slope is \(a = \frac{a_{0}}{1 + a_{0}/(\pi AR)} = \frac{2\pi}{1 + 2/AR}\). The sections are symmetric, so lift is zero at zero angle of attack.

\[ \begin{aligned} a &= \frac{2\pi}{1 + 2/6} = \frac{3\pi}{2} \\ \alpha &= 10^\circ = \frac{\pi}{18}\ \text{rad} \\ C_{L} &= \frac{3\pi}{2} \cdot \frac{\pi}{18} = \frac{\pi^{2}}{12} \end{aligned} \]

Ignoring the finite span gives \(2\pi \cdot \frac{\pi}{18} = \frac{\pi^{2}}{9}\), which is not offered; \(\pi^{2}/16\) and \(\pi^{2}/8\) correspond to other finite-span factors than \(AR/(AR + 2) = 3/4\), and \(\pi/2 \approx 1.57\) is more than even the two-dimensional value of about 1.10 at this angle.

Answer: B

The answer is \(L = 0\) and \(D_{i} \neq 0\).

With \(y = -\frac{b}{2}\cos\theta\), \(dy = \frac{b}{2}\sin\theta\,d\theta\) and \(\theta\) runs from 0 to \(\pi\). The lift is \(\rho V_{\infty}\int\Gamma\,dy\):

\[ \begin{aligned} L &= \rho V_{\infty}A\tfrac{b}{2}\int_{0}^{\pi}\sin 3\theta\sin\theta\,d\theta \\ &= 0 \end{aligned} \]

because \(\sin 3\theta\) and \(\sin\theta\) are orthogonal on \([0, \pi]\). The induced drag is \(\rho\int w\,\Gamma\,dy\), and the \(\sin\theta\) in the downwash cancels the one in dy:

\[ \begin{aligned} D_{i} &= \rho V_{\infty}\tfrac{3A^{2}b}{2}\int_{0}^{\pi}\sin^{2}3\theta\,d\theta \\ &= \rho V_{\infty}\tfrac{3\pi A^{2}b}{4} \neq 0 \end{aligned} \]

In lifting-line terms only the first Fourier coefficient produces lift, but every coefficient adds induced drag. A distribution made only of the third harmonic gives no net lift yet still costs drag, which rules out the other three options.