Finite Wing Theory Questions and Answers

Answer: A

The paradox states that a closed body in steady inviscid flow experiences no drag, but its derivation assumes no wake and no vorticity left behind. A lifting wing of finite span violates that assumption: it trails vorticity indefinitely downstream. Induced drag is therefore a genuine inviscid drag and no contradiction arises.

Answer: B

Winglets spread and displace the tip vorticity so the wing induces less downwash on itself than its geometric span alone would suggest. Aerodynamically it behaves as a longer wing, and the effective aspect ratio, which is what enters the induced drag expression, is raised above the geometric value.

Answer: C

The vortex pair induces downwash between the tips but upwash outside them. A wing sitting in that upwash meets a relative wind tilted upward, so its lift vector tilts forward and its induced drag falls. Birds in a V and military formation flights both exploit the effect, and measured savings of ten per cent or more have been recorded.

Answer: D

At the tip the upper and lower surfaces open to the same body of air, so no pressure difference can be maintained and the local lift, and with it the circulation, must vanish. This holds whatever the tip chord happens to be. It is the boundary condition that the sine series in lifting-line theory is built to satisfy automatically.

Answer: A

Lower aspect ratio means larger induced angle for the same lift coefficient, so more geometric angle is needed to leave the section with the effective angle it requires. The lift curve is flattened and its stalling angle pushed higher, which is why slender delta wings land at such conspicuously nose-high attitudes.

Answer: B

Using \( C_{D_i} = C_L^2 / (\pi e AR) \), the numerator is \( 0.8^2 = 0.64 \) and the denominator is \( \pi \times 0.9 \times 8 = 22.62 \). Dividing gives \( 0.64 / 22.62 = 0.0283 \). Note that the whole of the induced drag comes from the lift, so this figure would have to be added to the zero-lift drag coefficient to obtain the total.

Answer: C

An elliptical loading means \( e = 1 \), so \( C_{D_i} = C_L^2 / (\pi AR) \). Here that is \( 1.0 / (\pi \times 10) = 1 / 31.42 = 0.0318 \). It is worth remembering that this is the lowest induced drag any planar wing of that span can achieve at that lift coefficient.

Answer: D

For elliptical loading \( \alpha_i = C_L / (\pi AR) \) in radians, which gives \( 0.9 / (\pi \times 6) = 0.9 / 18.85 = 0.0478 \) radians. Multiplying by \( 57.3 \) converts this to \( 2.74 \) degrees. Each section therefore meets the flow at nearly three degrees less than the geometric angle of attack.

Answer: A

The correction is \( a = a_0 / (1 + a_0 / (\pi AR)) \). Here \( a_0 / (\pi AR) = 6.283 / 25.13 = 0.25 \), so the denominator is \( 1.25 \) and \( a = 6.283 / 1.25 = 5.03 \) per radian. The finite wing has lost a fifth of its section's lift curve slope purely to downwash.

Answer: B

Aspect ratio is \( b^2 / S \), so \( 10^2 / 20 = 100 / 20 = 5 \). The tempting wrong answer is span divided by area, which gives 0.5 and is dimensionally meaningless. For a rectangular wing the same result could be reached as span divided by chord, since the chord here would be 2 m.