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

The winner is the elliptical distribution — the oval-shaped $\Gamma$ curve in the right panel of the figure above, i.e. $\Gamma \propto \sin \theta$ where $\theta$ is the spanwise relabelling $y=-\frac{b}{2}\cos \theta$ (only the first Fourier coefficient $A_1$ nonzero). All higher harmonics $A_2,A_3,\dots$ add drag but no lift, so switching them off is optimal. The result is $C_{D,i}=\frac{C_L^2}{\pi AR}.$

For elliptic loading $e=1$ — it is the best possible, so $e$ can never exceed it. A value $e<1$ means the lift is shaped worse than an oval, so the wing pays extra induced drag: the same lift costs $1/e$ times the elliptic minimum.

Elliptic ⇒ $e=1$, use the clean formula. $C_{D,i}=\frac{C_L^2}{\pi AR}=\frac{0.9^2}{\pi \cdot 6}=\frac{0.81}{18.849}=\mathrm{0.04297.}$ So $C_{D,i}\approx 0.0430$.

First the aspect ratio (definition $AR=b^2/S$): $AR=\frac{15^2}{9}=\frac{225}{9}=25.$ Then, elliptic so $e=1$: $C_{D,i}=\frac{1.0^2}{\pi \cdot 25}=\frac{1}{78.54}=\mathrm{0.01273.}$ $C_{D,i}\approx 0.0127$ — tiny, which is exactly why gliders use these long thin wings.

Use $a=\frac{a_0}{1+a_0/(\pi AR)}$. Here $a_0/(\pi AR)=2\pi /(10\pi )=0.2$. $a=\frac{2\pi }{1+0.2}=\frac{6.2832}{1.2}=5.236\text{/rad}.$ Compared to the 2D value $6.283$/rad the finite wing is "softer": some geometric angle is eaten by downwash.

Recall the $A_n$ are the numbers weighting each harmonic in $\Gamma (\theta )=2b{V}_{\infty }{\sum }_n{A}_n\sin n\theta$. The penalty sums the weighted higher harmonics: $\delta ={\sum }_{n\ge 2}n(A_n/A_1{)}^2$. $\delta =2(0.08{)}^2+3(0.06{)}^2=2(0.0064)+3(0.0036)=0.0128+0.0108=\mathrm{0.0236.}$ Efficiency $e=\frac{1}{1+\delta }=\frac{1}{1.0236}=0.9769$. The drag is $(1+\delta )=1.0236$ times the elliptic minimum, i.e. $\approx 2.36\%$ above it.

Rewrite induced drag coefficient with $AR=b^2/S$: $C_{D,i}=\frac{C_L^2}{\pi AR}=\frac{C_L^{2}S}{\pi b^2}.$ Here $q\equiv \frac{1}{2}{\rho }_{\infty }{V}_{\infty }^2$ is the dynamic pressure (defined above): the same for both wings since they fly at the same speed and altitude. Lift $L=qS{C}_L$ is fixed, so $S{C}_L$ is fixed ⇒ $C_L\propto 1/S$. Then $C_{D,i}\propto C_L^{2}S\propto \frac{1}{S^2}\cdot S=\frac{1}{S}.$ So the smaller-area wing X has the larger $C_{D,i}$. The induced-drag force is $D_i=qS{C}_{D,i}\propto S\cdot (1/S)=\text{const}$ — actually equal! Let's confirm the cleaner route: $D_i=\frac{L^2}{q\pi b^2}$, which depends only on lift, speed and span — not area at all. So both wings have the same induced drag force. Area affects the coefficient, span (through $b^2$) fixes the force. Takeaway: span, not area, governs the induced drag force.

Elliptic reference: $C_{D,i}^{\text{ell}}=\frac{0.6^2}{\pi \cdot 7}=\frac{0.36}{21.99}=0.016368$. Real rectangular wing: divide by $e$: $C_{D,i}=\frac{C_{D,i}^{\text{ell}}}{e}=\frac{0.016368}{0.9}=\mathrm{0.018187.}$ So the rectangular wing's induced drag is $1/0.9=1.111$ times, i.e. $\approx 11.1\%$ higher than the elliptic ideal at the same lift and aspect ratio.

Induced part (elliptic): $C_{D,i}=\frac{0.5^2}{\pi \cdot 9}=\frac{0.25}{28.274}=\mathrm{0.008842.}$ Total: $C_D=0.008+0.008842=\mathrm{0.016842.}$ Ratio $C_{D,i}/C_{D,0}=0.008842/0.008=1.105$. At this cruise condition induced drag slightly exceeds profile drag — a reminder that at moderate-to-high $C_L$, drag-due-to-lift dominates the budget.

Maximum $L/D$ occurs when induced drag equals profile drag (a standard result for parabolic drag polars). Let $k=\frac{1}{\pi AR}=\frac{1}{28.274}=0.035368$. Set $C_{D,0}=k{C}_L^{\star 2}$: $C_L^{\star }=\sqrt{\frac{C_{D,0}}{k}}=\sqrt{C_{D,0}\pi AR}=\sqrt{0.008\cdot 28.274}=\sqrt{0.226195}=\mathrm{0.47560.}$ At this point $C_D=2C_{D,0}=0.016$, so ${\left(\frac{L}{D}\right)}_{\max }=\frac{C_L^{\star }}{2C_{D,0}}=\frac{0.47560}{0.016}=\mathrm{29.725.}$ Interpretation: best glide happens exactly where "lift-tax" and "friction-tax" are balanced — push $C_L$ higher and induced drag explodes; lower and you fly too fast with profile drag dominating.

Step 1 — change variable. From $y=-\frac{b}{2}\cos \theta$ we get $dy=\frac{b}{2}\sin \theta d\theta$; as $y$ runs $-b/2\to b/2$, $\theta$ runs $0\to \pi$. Why this substitution? It maps the finite span onto $[0,\pi ]$ where sine series are natural and the tips ($\theta =0,\pi$) automatically give $\Gamma =0$.

Step 2 — insert the series. $L={\rho }_{\infty }{V}_{\infty }{\int }_0^{\pi }\left(2b{V}_{\infty }{\sum }_n{A}_n\sin n\theta \right)\frac{b}{2}\sin \theta d\theta ={\rho }_{\infty }{V}_{\infty }^2{b}^2{\sum }_n{A}_n{\int }_0^{\pi }\sin n\theta \sin \theta d\theta .$

Step 3 — orthogonality. The integral ${\int }_0^{\pi }\sin n\theta \sin \theta d\theta$ equals $\frac{\pi }{2}$ if $n=1$ and $0$ otherwise. This is the key fact: different sine harmonics are "perpendicular", so every term except $n=1$ integrates to nothing. $L={\rho }_{\infty }{V}_{\infty }^2{b}^2{A}_1\cdot \frac{\pi }{2}.$

Step 4 — form the coefficient. By definition $C_L=\frac{L}{\frac{1}{2}{\rho }_{\infty }{V}_{\infty }^{2}S}=\frac{L}{qS}$ (with $q$ the dynamic pressure): $C_L=\frac{{\rho }_{\infty }{V}_{\infty }^2{b}^2{A}_1\frac{\pi }{2}}{\frac{1}{2}{\rho }_{\infty }{V}_{\infty }^{2}S}=\frac{\pi b^2{A}_1}{S}=\pi A_1\frac{b^2}{S}=\boxed{\pi AR{A}_1}.$ Only $A_1$ appears — the higher harmonics are "lift-invisible". $■$

Step 1 — factor out the first term. $C_{D,i}=\pi AR(A_1^2+{\sum }_{n\ge 2}n{A}_n^2)=\pi AR{A}_1^2(1+{\sum }_{n\ge 2}n\frac{A_n^2}{A_1^2})=\pi AR{A}_1^2(1+\delta ).$

Step 2 — replace $A_1$ using lift. From $C_L=\pi AR{A}_1$ we have $A_1=\frac{C_L}{\pi AR}$, so $A_1^2=\frac{C_L^2}{{\pi }^{2}A{R}^2}$. Substitute: $C_{D,i}=\pi AR\cdot \frac{C_L^2}{{\pi }^{2}A{R}^2}(1+\delta )=\boxed{\frac{C_L^2}{\pi AR}(1+\delta )}.$

Step 3 — sign of $\delta$. Each term $n(A_n/A_1{)}^2$ has $n\ge 2>0$ and a squared factor $\ge 0$, so every term is $\ge 0$ and thus $\delta \ge 0$. Equality $\delta =0$ demands every term vanish, i.e. $A_2=A_3=\cdots =0$: only $A_1$ survives, which is precisely $\Gamma (\theta )=2b{V}_{\infty }{A}_1\sin \theta$ — the elliptic distribution. Hence elliptic loading is the unique minimiser. $■$

Step 1 — the two definitions of lift slope. The section (2D) airfoil obeys $C_L=a_0({\alpha }_{\text{eff}}-{\alpha }_{L=0})$: lift responds to the local, tilted flow measured against the section's zero-lift angle ${\alpha }_{L=0}$. The wing (3D) obeys $C_L=a(\alpha -{\alpha }_{L=0})$: lift measured against the geometric angle we set with the stick.

Step 2 — link the two angles. Effective = geometric − induced: ${\alpha }_{\text{eff}}=\alpha -{\alpha }_i$. Substitute into the section law: $C_L=a_0(\alpha -{\alpha }_i-{\alpha }_{L=0}).$

Step 3 — insert the elliptic downwash. For elliptic loading ${\alpha }_i=\frac{C_L}{\pi AR}$ (constant across span — the beautiful special feature). Then $C_L=a_0(\alpha -{\alpha }_{L=0}-\frac{C_L}{\pi AR})=a_0(\alpha -{\alpha }_{L=0})-\frac{a_0}{\pi AR}{C}_L.$

Step 4 — solve for $C_L$. Collect the $C_L$ terms: $C_L(1+\frac{a_0}{\pi AR})=a_0(\alpha -{\alpha }_{L=0})\Rightarrow C_L=\frac{a_0}{1+a_0/(\pi AR)}(\alpha -{\alpha }_{L=0}).$ Comparing with $C_L=a(\alpha -{\alpha }_{L=0})$ reads off $\boxed{a=\frac{a_0}{1+a_0/(\pi AR)}}.$ With $a_0=2\pi$ this is $a=\frac{2\pi }{1+2/AR}$. As $AR\to \infty$, $a\to a_0$: the infinite wing recovers the 2D slope. $■$

Numeric check (same $AR=10$ as L2.3): $a=\frac{2\pi }{1+2/10}=\frac{2\pi }{1.2}=5.236$/rad — matching the value we computed in L2.3, so the two routes agree.