Question bank — Finite wing theory — induced drag, Prandtl's lifting line

True — with no tips there is no trailing sheet, no downwash, no lift tilt; this is d'Alembert's paradox (all drag vanishes in ideal 2D flow).

False — induced drag is the kinetic energy left spinning in the trailing vortices, a pressure/lift effect that survives with zero viscosity; only friction drag needs viscosity.

False — it minimizes drag for a given lift and span by making downwash uniform ($\delta =0$); it does not maximize lift, it minimizes the wake energy cost.

False — since $C_{D,i}\propto C_L^{2}S/(\pi b^2)$, at fixed span growing area (chord) raises induced drag; only span helps.

False — orthogonality of $\sin n\theta$ leaves only $A_1$: $C_L=\pi AR{A}_1$; higher harmonics add drag but no lift.

True — induced downwash "eats" part of the geometric angle, so you need more $\alpha$ per unit $C_L$: $a=a_0/(1+a_0/\pi AR)$.

False — $e=1/(1+\delta )$ with $\delta \ge 0$, so $e\le 1$ with equality only for the elliptic (single-harmonic) case.

True — $C_{D,i}\propto 1/AR$, so $AR\to 2AR$ gives half the induced drag; this is why gliders are long and thin (see Aspect ratio & wing design).

False — they exist because $\Gamma$ varies along span; Helmholtz forbids a filament ending in fluid, so vorticity $-d\Gamma /dy$ must shed wherever circulation changes.

The trailing filaments are semi-infinite (they start at the wing and run to infinity), so it is $\Gamma /(4\pi r)$ — exactly the factor in the downwash integral. The $2\pi$ formula is for a doubly-infinite line.

$A_2$ contributes zero to lift by orthogonality; it only raises $\delta$, adding induced drag. Only $A_1$ sets $C_L$.

Induced drag is inviscid, lift-coupled, and lives in the wake's swirl energy — completely separate from the surface skin-friction (profile) drag.

It subtracts: ${\alpha }_{\text{eff}}=\alpha -{\alpha }_i$, because downwash tilts the local flow downward, reducing the angle the section actually feels.

It scales as $C_L^2$, so at zero lift ($C_L=0$) induced drag vanishes — it is entirely a drag-due-to-lift; the point is only that it needs no viscosity.

Every term $\sin n\theta$ vanishes at $\theta =0,\pi$ (the tips), forcing $\Gamma =0$ there — this is built in and physically required, since a finite tip circulation would demand an infinite shed vortex.

Lift is defined perpendicular to the local relative wind; downwash rotates that wind downward by ${\alpha }_i$, so the still-perpendicular lift rotates back by ${\alpha }_i$, giving it a rearward component (from Thin airfoil theory's perpendicular-lift rule).

Lift integrates $\Gamma \cdot \sin \theta$, and ${\int }_0^{\pi }\sin n\theta \sin \theta d\theta =0$ unless $n=1$; drag integrates $\Gamma \cdot {\alpha }_i$ which brings in $n{A}_n^2$ from every mode via Biot–Savart downwash.

Constant downwash means every strip pays the same induced-angle "tax," which is the Cauchy–Schwarz optimum for a fixed total lift; any spanwise variation ($\delta >0$) is wasted wake energy.

The trailing vorticity is spread over a longer span, so the downwash per unit lift is weaker; $C_{D,i}\propto 1/AR=S/b^2$, and it is span $b$ (squared) that dominates, not area.

$\delta ={\sum }_{n\ge 2}n(A_n/A_1{)}^2$ is a sum of squares, so it is $\ge 0$; every departure from the pure-$A_1$ elliptic shape only adds drag, captured by $e=1/(1+\delta)$.

Part of the geometric angle is "used up" cancelling the induced downwash angle ${\alpha }_i$, so the section's effective angle is reduced; you must add back ${\alpha }_i$ geometrically, which is why the 3D slope $a<a_0$.

A constant $\Gamma$ that must drop to zero at the tips means $d\Gamma /dy\to -\infty$ there, shedding an infinitely strong tip vortex — physically impossible, so real loading tapers to zero (elliptic being the smooth optimum).

$C_{D,i}=C_L^2/(\pi AR)\to 0$ and $a=a_0/(1+a_0/\pi AR)\to a_0=2\pi$ — recovering the 2D drag-free, full-slope limit, consistent with d'Alembert's paradox.

No — $A_1=0$ so $C_L=0$, and $C_{D,i}=C_L^2/(\pi eAR)=0$; with no net lift there is no spanwise loading to shed lift-carrying vortices.

No — $C_{D,i}=\pi AR\sum n{A}_n^2$ depends on $A_n^2$, so it stays positive; drag cannot be recovered as thrust from the wake, only $C_L$ flips sign.

Lift is zero (no $A_1$), yet $\delta ={\sum }_{n\ge 2}n{A}_n^2/A_1^2$ blows up / $C_{D,i}=\pi AR(2A_2^2)>0$ — pure induced drag with no lift, the worst possible efficiency.

Identical — induced drag depends on span and total lift, not chord: $C_{D,i}=L^2/(\frac{1}{2}\rho V^2\pi b^{2}q\dots )$ reduces to depending on $b$ and lift only, so equal span ⇒ equal induced drag.