Finite wing theory — induced drag, Prandtl's lifting line: Visual walkthrough

3.1.22 · D2 · Physics › Compressible Flow & Aerodynamics › Finite wing theory — induced drag, Prandtl's lifting line

Step 1 — Ek real wing ko vortices kyun shed karne padte hain

KYA. Hum sirf ek cheez se shuru karte hain jo parent se maani hui hai: ek wing circulation carry karti hai, jise $Γ$ (Greek capital "gamma") likhte hain. $Γ$ ko socho kitni strongly air wing ke around spin ho rahi hai — bada $Γ$ = strong spin = bahut zyada lift. Ek finite wing par yeh spin har jagah same nahi hoti: beech mein strong, tips par zero (tip ke baad koi wing nahi hoti jo koi spin hold kar sake).

KYUN. Spinning "tube" of air (ek vortex filament) Helmholtz ki ek sakht rule follow karta hai: ek vortex line fluid ke andar simply band nahi ho sakti. Toh agar bound spin $Γ$ bahar ki taraf kamzor hoti jaaye, toh "lost" spin gayab nahi ho sakti — use peel off hoke wing ke peeche trail karna hi padega.

PICTURE. Neeche: span left–right chalta hai. Blue curve hai $Γ (y)$ , centre par sabse bada, dono tips par zero tak girta hua. Jahan bhi $Γ$ gir raha hai, wahan chhote orange trailing vortices downstream ki taraf peel off ho rahe hain.

Trailing vorticity ki strength jo per unit span shed hoti hai exactly $- \frac{d Γ}{d y}$ hai. Minus sign kehta hai: jahan $Γ$ girta hai, wahan positive trailing vorticity appear hoti hai.

Step 2 — Ek trailing filament ka kick: Biot–Savart

KYA. Ek single trailing vortex pick karo jo station $η$ (Greek "eta" — bas ek aur spanwise position, vortex ka source) par peel off hota hai. Hum poochte hain: yeh kisi doosre station $y$ par (observer) kitna downward air-speed create karta hai?

KYUN Biot–Savart, aur $4 π$ kyun. Biot–Savart law woh tool hai jo "distance $r$ par strength $Γ$ ki ek spin line" ko "woh point par jo speed induce hoti hai" mein convert karta hai. Ek full infinite straight vortex line ke liye yeh speed deta hai $\frac{Γ}{2 π r}$ . Lekin ek trailing filament semi-infinite hai: yeh wing par shuru hota hai aur sirf downstream infinity tak chalta hai — yeh exactly ek infinite line ka aadha hai. Aadhi line, aadhi speed: $\frac{Γ}{2 π r} half \frac{Γ}{4 π r} .$

PICTURE. Filament $η$ se shuru hoke right (downstream) ko infinity tak jaata hai. Lifting line par observation point $y$ par, induced velocity neeche ki taraf point karti hai. Perpendicular distance hai $r = ∣ y - η ∣$ .

Step 3 — Saare filaments ko add karo: downwash integral

KYA. Real trailing sheet in filaments ki ek continuous row hai, $η$ ke har value par tip se tip tak ek filament hai. Station $y$ par total downwash $w (y)$ paane ke liye, hum un sabke chhote kicks add karte hain — integral sign $\int$ ka yahi matlab hai: "infinitely many slices ka sum."

KYUN integrate karo. Har filament at $η$ ne $d w$ diya. Har $η$ par ek filament hai. Un sabko sum (integrate) karna hi combined effect paane ka honest tarika hai.

PICTURE. Span ke across bahut saare orange filaments; observer par black arrow $w (y)$ total downward push hai, saare chhote $d w$ arrows ka pile-up.

Step 4 — Downwash flow ko tilt karti hai: induced angle

KYA. Wing aage $V_{\infty}$ speed se fly kar rahi hai (" $\infty$ " matlab far upstream, undisturbed). Locally use downwash $w$ bhi feel hoti hai jo air ko neeche push karti hai. In dono ko arrows ki tarah add karo: air jo wing actually "feel" karti hai woh neeche ki taraf ek chhote angle se tilt hoti hai jise hum $α_{i}$ kehte hain, induced angle ( $α$ Greek "alpha" = angle; subscript $i$ = induced).

KYUN tangent, aur kyun hum ise drop kar sakte hain. Dono velocities ek right triangle banate hain: forward leg $V_{\infty}$ , downward leg $w$ . Resultant aur forward direction ke beech ke angle ka $tan α_{i} = \frac{w}{V _{\infty}}$ hota hai — kyunki tangent = opposite over adjacent, aur yahan "opposite" downward $w$ hai, "adjacent" forward $V_{\infty}$ hai. Kyunki $w ≪ V_{\infty}$ (downwash flight speed ke comparison mein tiny hai), angle chhota hai aur $tan α_{i} \approx α_{i}$ (small-angle rule). Isliye hum likh sakte hain: $α_{i} (y) = \frac{w ( y )}{V _{\infty}} .$

PICTURE. Velocity triangle, aur — sabse important — lift arrow. Lift hamesha us flow ke perpendicular hoti hai jo wing actually feel karti hai. Felt flow ko $α_{i}$ se neeche tilt karo, toh poora lift arrow $α_{i}$ se peeche tilt ho jaata hai. Uska chhota rearward shadow induced drag hai.

Step 5 — Loop band karo: lifting-line equation

KYA. Ab hamare paas ek hi local lift ke baare mein do independent facts hain, aur hum unhe equal set karte hain.

Aerodynamics side (Thin airfoil theory): strip ka lift coefficient hai $c_{ℓ} = 2 π (α_{eff} - α_{L = 0})$ , jahan effective angle woh hai jo airfoil actually feel karta hai downwash ke $α_{i}$ churaane ke baad: $α_{eff} = α - α_{i}$ .
Circulation side (Kutta–Joukowski theorem): usi strip ka $c_{ℓ} = \frac{2Γ}{V _{\infty} c ( y )}$ hai, jahan $c (y)$ local chord (front-to-back width) hai.

KYUN. Dono ek hi physical strip describe karte hain, isliye equal hain. $α_{i}$ ki jagah downwash integral substitute karne se yeh ek aisa equation ban jaata hai jiska single unknown sirf shape $Γ (y)$ hai.

PICTURE. Ek single spanwise strip, uska geometric angle $α$ , chura hua $α_{i}$ , aur jo effective angle bachi — ek strip par angles ki "accounting."

Step 6 — Solve karne ki trick: ek Fourier sine series

KYA. Hum coordinate $y$ se ek angle $θ$ mein change karte hain by $y = - \frac{b}{2} cos θ$ . Jab $θ$ sweep karta hai $0 \to π$ , tip-value $y$ sweep karta hai $- b /2 \to + b /2$ . Phir hum guess karte hain ki $Γ$ sine waves ka sum hai: $Γ (θ) = 2 b V_{\infty} \sum_{n = 1}^{\infty} A_{n} sin (n θ) .$

KYUN sines, kyun yeh $θ$ . Tips par ( $θ = 0$ aur $θ = π$ ) har $sin (n θ) = 0$ automatically — toh yeh guess "tips par zero spin" rule kabhi nahi todega, chahe coefficients $A_{n}$ kuch bhi hon. Sines orthogonal bhi hote hain (Step 7), jo lift aur drag ko clean formulas mein collapse kar dega.

PICTURE. Pehle kuch building blocks: $sin θ$ ek clean hump hai (elliptic shape), $sin 2 θ$ ek wiggle hai jo ek side par positive aur doosri par negative hai, $sin 3 θ$ aur bhi wigglier. Real $Γ$ = in sabka weighted stack.

Step 7 — Lift aur drag harvest karo: sirf $A_{1}$ lift karta hai, sab drag dete hain

KYA. Series ko total-lift integral $L = ρ V_{\infty} \int Γ d y$ aur induced-drag integral $D_{i} = ρ V_{\infty} \int Γ α_{i} d y$ mein feed karo. Do clean results milte hain: $C_{L} = π A R A_{1}, C_{D, i} = π A R \sum_{n = 1}^{\infty} n A_{n}^{2} .$

KYUN sirf $A_{1}$ lift mein. Lift ke integral mein $\int_{0}^{π} sin (n θ) sin θ d θ$ hai. Orthogonality se yeh har $n \neq = 1$ ke liye zero hota hai — wiggly harmonics span par khud ko cancel kar dete hain, lift mein exactly kuch contribute nahi karte. Lekin drag integral mein har harmonic khud se pair hota hai, giving $n A_{n}^{2} \geq 0$ — har harmonic pay karta hai.

PICTURE. Do bar charts: lift bars sirf $A_{1}$ ke liye non-zero hain; drag bars sab $n$ ke liye non-zero hain. "Lift loves A-one; drag pays for the rest."

Step 8 — Sabse best possible wing (aur edge cases)

KYA. Drag minimize karo. Kyunki $δ \geq 0$ aur har extra harmonic sirf drag add karta hai jabki lift kuch nahi add karta, minimum tab milta hai jab sab ko kill karo: $A_{2} = A_{3} = \dots = 0$ , sirf pure $Γ \propto sin θ$ bachta hai — ek elliptical lift distribution. Tab: $C_{D, i} = \frac{C _{L}^{2}}{π A R}, C_{D, i} = \frac{C _{L}^{2}}{π e A R} general case mein.$

KYUN yeh floor hai, aur edge cases:

Elliptic ( $δ = 0$ , $e = 1$ ): downwash poore span par constant hoti hai — kahin bhi wasted swirl nahi. Minimum drag.
Koi bhi non-elliptic ( $δ > 0$ , $e < 1$ ): same lift ke liye strictly zyada drag.
$A R \to \infty$ (2D limit): $C_{D, i} \to 0$ — hum infinite wing ka zero drag recover karte hain (d'Alembert's paradox). Finite-wing lift slope $a = \frac{2 π}{1 + 2 π / ( π A R )} \to 2 π$ , 2D value.
$C_{L} = 0$ (no lift): $C_{D, i} = 0$ — induced drag purely drag-due-to-lift hai; lift na karo, induced penalty na bharo.

PICTURE. Span ke across downwash: elliptic ke liye flat blue line, non-elliptic ke liye dipping/humping orange (tips ke paas zyada downwash = wasted energy).

Ek-picture summary

Poori chain ek frame mein: varying $Γ$ → shed trailing sheet (Helmholtz) → downwash (Biot–Savart, $4 π$ ) → tilted lift → induced drag; solve hoti hai ek sine series se jahan sirf $A_{1}$ lift karta hai aur elliptic case floor set karta hai $C_{L}^{2} / (π A R)$ .

Recall Feynman retelling — poori walk plain words mein

Ek wing lift banane ke liye air ko spin karti hai, aur woh beech mein sabse zyada spin karti hai, tips par zero tak fade hoti hai. Woh fading spin kahin na kahin jaani hai, toh woh back edge se trailing whirlpools ki tarah peel off hoti hai (Helmholtz ka rule: ek whirlpool mid-air mein band nahi ho sakti). Woh whirlpools bilkul wahin air ko gently neeche blow karte hain jahan wing fly kar rahi hai — yahi downwash hai. Kyunki wing ab apni khud ki hawa ko thoda neeche tilted feel karti hai, lift arrow (hamesha felt wind ke square) thoda peeche tip ho jaata hai, aur woh backward lean ek brand-new drag hai — zero friction ke saath bhi. Spin ki exact shape predict karne ke liye hum ise sine-wave humps ke stack ke roop mein likhte hain; tips automatically zero aate hain, aur ek khoobsurat cancellation ka matlab hai ki sirf single clean hump lift karta hai jabki har extra wiggle sirf drag karta hai. Toh sabse clever wing sirf clean hump use karti hai — oval-shaped lift — constant, minimal downwash aur physically possible sabse chhoti drag deti hai: $C_{L}^{2} / (π A R)$ . Lamba aur patla jeetta hai.

Connections

Finite wing theory — induced drag, Prandtl's lifting line — woh parent jise yeh page unpack karta hai.
Biot–Savart law · Helmholtz vortex theorems · Kutta–Joukowski theorem · Thin airfoil theory — step by step use kiye gaye chaar tools.
d'Alembert's paradox · Aspect ratio & wing design · Oswald efficiency factor — jahan limits aur results connect hote hain.