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

Winner elliptical distribution hai — figure ke right panel mein oval-shaped $\Gamma$ curve, yaani $\Gamma \propto \sin \theta$ jahan $\theta$ spanwise relabelling $y=-\frac{b}{2}\cos \theta$ hai (sirf pehla Fourier coefficient $A_1$ nonzero). Saare higher harmonics $A_2,A_3,\dots$ drag add karte hain lekin koi lift nahi, isliye unhe switch off karna optimal hai. Result hai $C_{D,i}=\frac{C_L^2}{\pi AR}.$

Elliptic loading ke liye $e=1$ — yeh best possible hai, isliye $e$ kabhi isse zyada nahi ho sakta. $e<1$ ka matlab hai lift oval se worse shaped hai, isliye wing extra induced drag pay karta hai: usi lift ke liye elliptic minimum ka $1/e$ guna lagta hai.

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

Pehle aspect ratio (definition $AR=b^2/S$): $AR=\frac{15^2}{9}=\frac{225}{9}=25.$ Phir, elliptic toh $e=1$: $C_{D,i}=\frac{1.0^2}{\pi \cdot 25}=\frac{1}{78.54}=\mathrm{0.01273.}$ $C_{D,i}\approx 0.0127$ — bahut chhota, aur yahi reason hai kyun gliders in lambi patli wings ka use karte hain.

$a=\frac{a_0}{1+a_0/(\pi AR)}$ use karo. Yahan $a_0/(\pi AR)=2\pi /(10\pi )=0.2$. $a=\frac{2\pi }{1+0.2}=\frac{6.2832}{1.2}=5.236\text{/rad}.$ 2D value $6.283$/rad se compare karte hue finite wing "softer" hai: kuch geometric angle downwash kha jaata hai.

Yaad karo ki $A_n$ woh numbers hain jo $\Gamma (\theta )=2b{V}_{\infty }{\sum }_n{A}_n\sin n\theta$ mein har harmonic ko weight karte hain. Penalty higher harmonics ko weighted sum karta hai: $\delta ={\sum }_{n\ge 2}n(A_n/A_1{)}^2$. $\delta =2(0.08{)}^2+3(0.06{)}^2=2(0.0064)+3(0.0036)=0.0128+0.0108=\mathrm{0.0236.}$ Efficiency $e=\frac{1}{1+\delta }=\frac{1}{1.0236}=0.9769$. Drag elliptic minimum ka $(1+\delta )=1.0236$ guna hai, yaani $\approx 2.36\%$ usse upar.

Induced drag coefficient ko $AR=b^2/S$ se rewrite karo: $C_{D,i}=\frac{C_L^2}{\pi AR}=\frac{C_L^{2}S}{\pi b^2}.$ Yahan $q\equiv \frac{1}{2}{\rho }_{\infty }{V}_{\infty }^2$ dynamic pressure hai (upar define kiya gaya): dono wings ke liye same hai kyunki woh same speed aur altitude par fly karte hain. Lift $L=qS{C}_L$ fixed hai, isliye $S{C}_L$ fixed hai ⇒ $C_L\propto 1/S$. Tab $C_{D,i}\propto C_L^{2}S\propto \frac{1}{S^2}\cdot S=\frac{1}{S}.$ Toh chhhote-area wala wing X ka $C_{D,i}$ bada hai. Induced-drag force $D_i=qS{C}_{D,i}\propto S\cdot (1/S)=\text{const}$ hai — actually equal! Cleaner route se confirm karte hain: $D_i=\frac{L^2}{q\pi b^2}$, jo sirf lift, speed aur span par depend karta hai — area par bilkul nahi. Isliye dono wings ki induced drag force same hai. Area coefficient ko affect karta hai, span ($b^2$ ke zariye) force fix karta hai. Takeaway: span, area nahi, induced drag force control karta hai.

Elliptic reference: $C_{D,i}^{\text{ell}}=\frac{0.6^2}{\pi \cdot 7}=\frac{0.36}{21.99}=0.016368$. Real rectangular wing: $e$ se divide karo: $C_{D,i}=\frac{C_{D,i}^{\text{ell}}}{e}=\frac{0.016368}{0.9}=\mathrm{0.018187.}$ Toh rectangular wing ka induced drag $1/0.9=1.111$ guna hai, yaani elliptic ideal se same lift aur aspect ratio par $\approx 11.1\%$ zyada.

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 $C_{D,i}/C_{D,0}=0.008842/0.008=1.105$. Is cruise condition par induced drag profile drag se thoda zyada hai — ek reminder ki moderate-to-high $C_L$ par, drag-due-to-lift budget dominate karta hai.

Maximum $L/D$ tab hota hai jab induced drag equals profile drag (parabolic drag polars ke liye ek standard result). $k=\frac{1}{\pi AR}=\frac{1}{28.274}=0.035368$ maano. $C_{D,0}=k{C}_L^{\star 2}$ set karo: $C_L^{\star }=\sqrt{\frac{C_{D,0}}{k}}=\sqrt{C_{D,0}\pi AR}=\sqrt{0.008\cdot 28.274}=\sqrt{0.226195}=\mathrm{0.47560.}$ Is point par $C_D=2C_{D,0}=0.016$, isliye ${\left(\frac{L}{D}\right)}_{\max }=\frac{C_L^{\star }}{2C_{D,0}}=\frac{0.47560}{0.016}=\mathrm{29.725.}$ Interpretation: best glide theek wahan hoti hai jahan "lift-tax" aur "friction-tax" balanced hain — $C_L$ zyada push karo toh induced drag explode karta hai; kam karo toh aap bahut tez fly karte ho profile drag dominate karte hue.

Step 1 — variable change. $y=-\frac{b}{2}\cos \theta$ se $dy=\frac{b}{2}\sin \theta d\theta$ milta hai; jaise $y$ run karta hai $-b/2\to b/2$, $\theta$ run karta hai $0\to \pi$. Ye substitution kyun? Ye finite span ko $[0,\pi ]$ par map karta hai jahan sine series natural hain aur tips ($\theta =0,\pi$) automatically $\Gamma =0$ dete hain.

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

Step 3 — orthogonality. Integral ${\int }_0^{\pi }\sin n\theta \sin \theta d\theta$ equals $\frac{\pi }{2}$ if $n=1$ aur $0$ otherwise. Yahi key fact hai: alag sine harmonics "perpendicular" hain, isliye $n=1$ ke alawa har term integrate hokar kuch nahi deta. $L={\rho }_{\infty }{V}_{\infty }^2{b}^2{A}_1\cdot \frac{\pi }{2}.$

Step 4 — coefficient form karo. Definition se $C_L=\frac{L}{\frac{1}{2}{\rho }_{\infty }{V}_{\infty }^{2}S}=\frac{L}{qS}$ ($q$ dynamic pressure ke saath): $C_L=\frac{{\rho }_{\infty }{V}_{\infty }^2{b}^2{A}_1\frac{\pi }{2}}{\frac{1}{2}{\rho }_{\infty }{V}_{\infty }^{2}S}=\frac{\pi b^2{A}_1}{S}=\pi A_1\frac{b^2}{S}=\boxed{\pi AR{A}_1}.$ Sirf $A_1$ appear karta hai — higher harmonics "lift-invisible" hain. $■$

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

Step 2 — $A_1$ ko lift se replace karo. $C_L=\pi AR{A}_1$ se $A_1=\frac{C_L}{\pi AR}$ milta hai, isliye $A_1^2=\frac{C_L^2}{{\pi }^{2}A{R}^2}$. Substitute karo: $C_{D,i}=\pi AR\cdot \frac{C_L^2}{{\pi }^{2}A{R}^2}(1+\delta )=\boxed{\frac{C_L^2}{\pi AR}(1+\delta )}.$

Step 3 — $\delta$ ka sign. Har term $n(A_n/A_1{)}^2$ mein $n\ge 2>0$ hai aur ek squared factor $\ge 0$ hai, isliye har term $\ge 0$ hai aur thus $\delta \ge 0$. Equality $\delta =0$ ke liye har term ka vanish hona zaroori hai, yaani $A_2=A_3=\cdots =0$: sirf $A_1$ bachta hai, jo exactly $\Gamma (\theta )=2b{V}_{\infty }{A}_1\sin \theta$ hai — elliptic distribution. Isliye elliptic loading unique minimiser hai. $■$

Step 1 — lift slope ki do definitions. Section (2D) airfoil $C_L=a_0({\alpha }_{\text{eff}}-{\alpha }_{L=0})$ follow karta hai: lift local, tilted flow ke response mein, section ke zero-lift angle ${\alpha }_{L=0}$ ke against measure kiya. Wing (3D) $C_L=a(\alpha -{\alpha }_{L=0})$ follow karta hai: lift us geometric angle ke against measure kiya jo hum stick se set karte hain.

Step 2 — dono angles link karo. Effective = geometric − induced: ${\alpha }_{\text{eff}}=\alpha -{\alpha }_i$. Section law mein substitute karo: $C_L=a_0(\alpha -{\alpha }_i-{\alpha }_{L=0}).$

Step 3 — elliptic downwash insert karo. Elliptic loading ke liye ${\alpha }_i=\frac{C_L}{\pi AR}$ (span ke across constant — yeh khubsoorat special feature hai). Tab $C_L=a_0(\alpha -{\alpha }_{L=0}-\frac{C_L}{\pi AR})=a_0(\alpha -{\alpha }_{L=0})-\frac{a_0}{\pi AR}{C}_L.$

Step 4 — $C_L$ ke liye solve karo. $C_L$ terms collect karo: $C_L(1+\frac{a_0}{\pi AR})=a_0(\alpha -{\alpha }_{L=0})\Rightarrow C_L=\frac{a_0}{1+a_0/(\pi AR)}(\alpha -{\alpha }_{L=0}).$ $C_L=a(\alpha -{\alpha }_{L=0})$ se compare karte hue $\boxed{a=\frac{a_0}{1+a_0/(\pi AR)}}.$ $a_0=2\pi$ ke saath yeh $a=\frac{2\pi }{1+2/AR}$ hai. Jab $AR\to \infty$, $a\to a_0$: infinite wing 2D slope recover kar leta hai. $■$

Numeric check (L2.3 wala same $AR=10$): $a=\frac{2\pi }{1+2/10}=\frac{2\pi }{1.2}=5.236$/rad — L2.3 mein compute ki gayi value se match karta hai, isliye dono routes agree karte hain.