Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c): Visual walkthrough

3.1.21 · D2 · Physics › Compressible Flow & Aerodynamics › Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c

Jaate jaate hum vault ke inn ideas pe lean karenge: Kutta–Joukowski theorem, Circulation and bound vortices, Kutta condition, Biot–Savart law, aur Glauert's integral and Fourier coefficients. Final payoff Lift coefficient and angle of attack ko feed karta hai. Parent hai Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c).

Definition Ground rules (thin-airfoil theory ki assumptions)

Neeche har step silently paanch modelling assumptions pe tikaa hua hai. Inhe ek baar, out loud, state karte hain:

Inviscid — koi friction/viscosity nahi; hawa frictionlessly slide karti hai (potential flow ke liye zaroori).
Incompressible — density $ρ$ constant hai (low speed; agar nahi, toh Compressibility corrections (Prandtl–Glauert) apply karo).
Irrotational sheet ke bahar — koi spin nahi kaheen bhi, siwaaye un bound vortices ke jo hum place karte hain, toh velocity potential exist karta hai.
Two-dimensional — ek wing slice, "per unit span"; koi spanwise flow nahi.
Zero thickness, small camber, small angle — airfoil ek thin bent line hai, isliye ise camber line pe collapse kiya ja sakta hai aur saare angles linearise kiye ja sakte hain. Agar in mein se koi bhi fail ho (moti wing, transonic speed, stall), toh neeche diya formula hold nahi karta.

Step 1 — Wing draw karo aur har symbol naam do

KYA. Ek wing slice ek thin curved sliver hai. Hum ise ek picture mein rakhte hain aur un chaar cheezein label karte hain jo final formula mein aayengi: aane wali hawa, jhukav, curve, aur lambai.

KYUN. Is page ka contract: koi symbol uske picture se pehle nahi. Toh kisi bhi maths se pehle, sliver dekho aur cast se milo.

PICTURE.

Figure — Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c)

Figure ko left se right padhein:

Seedhe horizontal arrows freestream hain — door wing ke aage, undisturbed, speed $V$ (metres per second) aur density $ρ$ (kilograms per cubic metre) wali hawa.
Wing slice leading edge (front, $x = 0$ par) se trailing edge (sharp back, $x = c$ par) tak chalti hai. Distance $c$ chord hai — slice ki seedhi lambai.
Wing thodi upar $α$ angle se tilt hai, jo angle of attack hai, chord line aur hawa ke beech measure kiya jaata hai.
Beech mein se guzarne wali dashed curve camber line $z (x)$ hai: chord ke saath har point par top aur bottom surfaces ka average. Iska halka jhukav hi "camber" ka matlab hai.

Step 2 — Wing ki jagah tiny spinners ki ek row lagao

KYA. Hum solid wing mita dete hain aur camber line ke saath tiny vortices ki ek sheet dalte hain. Har chota patch $d x$ ek spin strength $γ (x) d x$ carry karta hai.

KYUN. Ek thin wing hawa ko bahut kam nudge karti hai, isliye bahari flow ko koi farak nahi padta ki vorticity ek real curved skin par hai ya thin camber line par. Vortices lift ki natural language hain kyunki Circulation and bound vortices kehta hai spin hi lift karta hai. Vortices ki ek line ke saath kaam karna ek curved solid boundary se kahin zyada aasaan hai.

PICTURE.

Har chhota magenta circle ek vortex element hai: ek microscopic tornado wing se chipka hua.
Iski strength $γ (x)$ hai, vortex sheet strength — spin per unit length, units (metres/second) kyunki yeh sheet ke across ek velocity jump hai.
Total spin, poore chord pe add up karke, circulation hai: $Γ = \int_{0}^{c} γ (x) d x .$ Yahan $\int_{0}^{c} \dots d x$ ka matlab sirf itna hai: " $x$ ko leading edge $0$ se trailing edge $c$ tak slide karo aur har slice add karo." $Γ$ ki units m²/s hain.

Step 3 — Spins ko fix karne wala rule: hawa wing se guzar nahi sakti

KYA. Hum demand karte hain ki total flow camber line ke saath saath slide kare — yeh kabhi use cross nahi karta. Yeh ek sentence hi fix karta hai ki har vortex kitna strong hona chahiye.

KYUN. Ek wing solid hai; hawa iske andar tunnel nahi kar sakti. Do cheezein hawa ko camber line ke across push karti hain, aur unhe exactly cancel karna chahiye:

tilted freestream, jiska ek chhota component surface ke andar point karta hai, aur
woh downward breeze jo vortex sheet khud apne aap pe dalta hai (downwash).

PICTURE.

Exact no-through-flow rule yeh hai ki surface ke perpendicular velocity component zero ho. Upar di gayi conventions ke saath likhne par, perpendicular freestream piece $V sin (α - \frac{d z}{d x})$ hai.

Us linearisation ke saath, flow-tangency condition hai, term by term: $wind speed V (tilt, +ve nose-up α - local slope, +ve rising \frac{d z}{d x}) = downwash, +ve down w (x) .$

$\frac{d z}{d x}$ point $x$ par camber line ki slope hai — dashed curve wahan kitni steeply rise karti hai. Hum slope use karte hain kyunki "surface ke tangent" literally slope ke baare mein ek statement hai. Flat plate par yeh har jagah zero hai.
$α - \frac{d z}{d x}$ hawa aur surface ke beech bacha hua angle hai — hawa ka woh hissa jo abhi bhi wing ke andar aim kar raha hai, jise vortices ko neutralise karna hai.
$w (x)$ downwash hai: vortex sheet jo downward speed point $x$ par induce karta hai (positive $w$ neeche point karta hai).

Figure dekho: surface mein aane wala orange freestream component (up-arrow) exactly violet downwash (down-arrow) se match hota hai. Woh balance hi equation hai.

Step 4 — Sheet kitna downwash banati hai? (Biot–Savart)

KYA. Hum $w (x)$ compute karte hain — point $x$ par sheet ke har doosre vortex se feel hone wali breeze.

KYUN. Step 3 ko $w (x)$ chahiye jo unknown spins $γ$ ke terms mein likha ho. Biot–Savart law ek single 2-D vortex ki velocity ek distance par deta hai: strength $Γ$ ka ek vortex distance $r$ par $Γ/ (2 π r)$ speed chalaata hai. Hum poori sheet se woh contribution add karte hain.

PICTURE.

$w (x) = - \frac{1}{2 π} ⨍_{0}^{c} \frac{γ ( ξ ) d ξ}{x - ξ} .$

$ξ$ (Greek "xi") push karne wale vortex ki location hai; $x$ wahan hai jahan hum measure karte hain.
$x - ξ$ unke beech ki distance hai. Neeche hone se $1/ r$ falloff hota hai: paas ke vortices zyada zor se aawaz lagate hain, door ke dheerey.
$\frac{1}{2 π}$ Biot–Savart se vortex-velocity constant hai.
Minus sign direction fix karta hai (positive sheet neeche chalaata hai, hamaara chosen positive $w$ ).
Slashed integral sign $⨍$ ka matlab Cauchy principal value hai. Fraction dekho: jab push karne wala vortex exactly measuring point par ho ( $ξ = x$ ), denominator $x - ξ$ zero ho jaata hai aur fraction blow up ho jaata hai. Ek point khud ko push nahi kar sakta, isliye hum $ξ = x$ ke around ek symmetric tiny gap kaat lete hain aur usse shrink karte hain: left aur right dono taraf ka blow-up equal aur opposite hote hain, isliye cancel ho jaate hain aur sum finite rehta hai.

Figure mein $ξ$ par red vortex $x$ par felt violet arrow chalaata hai; jitne paas hote hain, utna bada arrow — yeh $1/ (x - ξ)$ picture mein hai. $ξ = x$ par greyed-out sliver principal value ka excluded gap hai.

Step 5 — Ruler ka ek clever badlav: position ko angle se measure karo

KYA. Hum position ko seedhe ruler $x$ se measure karna band kar dete hain aur ek angle $θ$ use karne lagte hain jo chord par ek half-circle mein chalata hai: $x = \frac{c}{2} (1 - cos θ), θ : 0 \to π .$

KYUN. Awkward $1/ (x - ξ)$ integral is angle mein likhne se clean aur solvable ho jaata hai. Yeh wohi trick hai jo ek stretched-out line ko ek tidy circle mein badal deti hai taaki repeating patterns (sines aur cosines) appear ho sakein. Yahi to Glauert's integral and Fourier coefficients ke liye bana hai.

PICTURE.

$θ = 0$ par: $cos 0 = 1$ , toh $x = \frac{c}{2} (1 - 1) = 0$ — leading edge.
$θ = π$ par: $cos π = - 1$ , toh $x = \frac{c}{2} (1 + 1) = c$ — trailing edge.
Middle $θ = \frac{π}{2}$ pe $x = c /2$ , mid-chord pe aata hai.

Figure mein ek point upper half-circle mein march karta hua dikhta hai (angle $θ$ ) jabki uski shadow chord ke saath slide karti hai (position $x$ ). Same point, do rulers.

Intuition Woh singular

$\frac{1 + c o s θ}{s i n θ}$ term kahan se aata hai — faith se nahi, ek sketch se Hume Step 4 ka downwash integral spins $γ$ ke liye invert karna hai. Woh ek Fourier method se hota hai: $0 \leq θ \leq π$ par koi bhi tidy function sine waves $sin n θ$ ke sum ke roop mein ban sakta hai (isliye ruler badla — awkward line ek clean interval ban jaati hai jahan sines rehte hain). Toh hum try karte hain: $γ (θ) = 2 V \sum_{n = 1}^{\infty} A_{n} sin n θ + (ek special pehla term) .$ Do physical demands us special term ko shape deti hain:

Kutta condition (Kutta condition): flow ko sharp trailing edge se smoothly nikalna chahiye, toh $γ$ finite (actually zero) hona chahiye $θ = π$ par. Har $sin n θ$ wahan already vanish karta hai — accha.
Leading-edge suction: $θ = 0$ par (front) flow stagnation point ke around wrap karta hai aur $γ$ large ho sakta hai. Hume ek aisa term chahiye jo $θ = 0$ par bada ho lekin $θ = π$ par tame. Unique aisi shape $\frac{1 + cos θ}{sin θ}$ hai: $θ = 0$ par yeh blow up hota hai (front suction), aur $θ = π$ par numerator $1 + cos π = 0$ blow-up ko khatam kar deta hai (Kutta satisfied). Isliye yeh — aur koi aur singular function nahi — series ki agwaai karta hai.

Isliye standard Glauert solution: $γ (θ) = 2 V (A_{0} \frac{1 + c o s θ}{s i n θ} + \sum_{n = 1}^{\infty} A_{n} sin n θ) .$ Iska coefficient $A_{0}$ front-loaded term multiply karta hai; $A_{n}$ baaki ko shape karte hain.

$A_{0}, A_{1}, A_{2}, \dots$ Fourier coefficients hain: dial numbers jo standard wiggle-shapes ( $sin n θ$ ) ko mix karke wing ka jo bhi camber ho reproduce karte hain.

Isse Steps 3–4 mein feed karke aur Glauert's integral $⨍_{0}^{π} \frac{c o s n θ ^{'}}{c o s θ ^{'} - c o s θ} d θ^{'} = \frac{π s i n n θ}{s i n θ}$ use karke (phir se ek principal value, usi $ξ = x$ reason se) dial numbers nikaalte hain: $A_{0} = α - \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} d θ, A_{n} = \frac{2}{π} \int_{0}^{π} \frac{d z}{d x} cos n θ d θ .$

$A_{0}$ tilt $α$ plus camber ki average slope carry karta hai.
$A_{n}$ ( $n \geq 1$ ) camber ki shape carry karte hain, $cos n θ$ se weighted.

Step 6 — Spins add karo: sirf do dials bachte hain

KYA. Hum $γ$ integrate karke total circulation $Γ$ nikalte hain, aur paate hain ki almost har dial cancel ho jaata hai — sirf $A_{0}$ aur $A_{1}$ rehte hain.

KYUN. Jab hum $sin n θ$ ko poori sweep $0 \to π$ par sum karte hain, toh $n \geq 2$ ke liye wiggles ke equal positive aur negative humps exactly cancel ho jaate hain (yeh orthogonality hai — alag sine waves average par overlap nahi karte). Sirf $A_{0}$ aur $A_{1}$ ke flat-average pieces ek residue chhodenge.

PICTURE.

$Γ = \int_{0}^{c} γ d x = c V π (A_{0} + \frac{A _{1}}{2}) .$

$c V π$ integral ko seedhe ruler mein wapas convert karne ka size factor hai.
$A_{0} + \frac{A _{1}}{2}$ sirf yahi surviving combination hai: tilt-and-average ( $A_{0}$ ) plus pehle camber wiggle ka aadha ( $A_{1}$ ).

Figure mein tall $A_{0}, A_{1}$ bars bachte dikhte hain aur $A_{2}, A_{3}, \dots$ bars greyed out hain ("cancelled"). $ρ V$ se multiply karo (Kutta–Joukowski, Step 2): $L^{'} = ρ V Γ = π ρ c V^{2} (A_{0} + \frac{A _{1}}{2})$

Yeh fully general thin-airfoil lift hai.

Step 7 — Flat plate (degenerate case: no camber) aur $c_{ℓ}$ define karna

KYA. Camber band karo: camber line bilkul seedhi hai, isliye $\frac{d z}{d x} = 0$ har jagah. Hum ek tidy dimensionless bookkeeping number bhi introduce karte hain, lift coefficient $c_{ℓ}$ .

KYUN. Hamesha formula ko sabse simple input par test karo. Ek flat tilted plate woh baseline hai jisse har cambered wing compare hoti hai. Aur alag size aur speed ki wings ko ek graph par compare karne ke liye hum $ρ$ , $V$ aur $c$ ek pure number mein strip kar dete hain.

PICTURE.

$\frac{d z}{d x} = 0$ ke saath: dono camber integrals vanish ho jaate hain, toh $A_{0} = α$ aur $A_{1} = 0$ . Phir: $L^{'} = π ρ c V^{2} α, c_{ℓ} = \frac{π ρ c V ^{2} α}{\frac{1}{2} ρ V ^{2} c} = 2 π α .$

Koi camber nahi ⇒ zero tilt par koi extra lift nahi; line origin se guzarti hai.
Slope $\frac{d c _{ℓ}}{d α} = 2 π$ — camber chahe kuch bhi ho, yeh slope kabhi nahi badlegi.

Figure ka $c_{ℓ}$ – $α$ line seedha $(0, 0)$ se guzarta hai: lift ka sirf ek source tilt hai.

Step 8 — Camber add karna: $z (x)$ aur $β$ actually kya hain

KYA. Ek concrete, standard camber shape ke saath camber wapas on karo aur finally camber parameter $β$ define karo.

KYUN. Ab tak $β$ ek promise tha. Yahan yeh apna meaning kamaata hai: yeh ek number hai jo measure karta hai ki camber line kitni strongly bend karti hai, seedha $z (x)$ ki shape se padha jaata hai.

PICTURE.

Ab is camber slope ko Step 5 ke coefficient integrals mein daalo. Ruler badlav $x = \frac{c}{2} (1 - cos θ)$ use karke, slope ban jaata hai: $\frac{d z}{d x} (θ) = \frac{4 z _{m a x}}{c} (1 - \frac{2 x}{c}) = β (1 - (1 - cos θ)) = β cos θ .$

Humne substitute kiya $\frac{2 x}{c} = 1 - cos θ$ , toh $1 - \frac{2 x}{c} = cos θ$ .
End-slope constant $\frac{4 z _{m a x}}{c}$ exactly $β$ hai, isliye poora slope $β cos θ$ hai — ek single clean cosine.

$\frac{d z}{d x} = β cos θ$ coefficient formulas mein daalo, $\int_{0}^{π} cos θ d θ = 0$ aur $\int_{0}^{π} cos θ cos θ d θ = \frac{π}{2}$ use karke: $A_{0} = α - \frac{1}{π} \int_{0}^{π} β cos θ d θ = α - 0 = α,$ $A_{1} = \frac{2}{π} \int_{0}^{π} β cos θ cos θ d θ = \frac{2}{π} \cdot β \cdot \frac{π}{2} = β .$

$A_{0} = α$ : camber slope ka average zero hai (front aur back slopes equal aur opposite hain), isliye sirf tilt yahan bachta hai.
$A_{1} = β$ : camber ka pehla cosine mode exactly $β$ hai.

Isliye Step 6 ka surviving combination hai: $A_{0} + \frac{A _{1}}{2} = α + \frac{β}{2} .$

Step 6 ki boxed general lift mein substitute karo: $L^{'} = π ρ V^{2} c (α + \frac{β}{2}), c_{ℓ} = \frac{L ^{'}}{\frac{1}{2} ρ V ^{2} c} = 2 π α + π β .$

$α$ = tilt term, $\frac{β}{2}$ = camber term. Woh add hote hain; camber wing mein built-in extra tilt ki tarah kaam karta hai.
Graph par, camber poori line ko left ek constant $π β$ se shift karta hai — slope $2 π$ par glued rehti hai.

Ab har case, figure mein har ek visible hai:

Positive lift ( $α + \frac{β}{2} > 0$ ): usual flying wing.
Zero lift ( $α + \frac{β}{2} = 0$ ): zero-lift angle $α_{L = 0} = - \frac{β}{2}$ . Ek cambered wing ( $β > 0$ ) yeh negative tilt par reach karta hai — yeh tab bhi lift karta hai jab thoda neeche point kar raha ho.
Negative lift ( $α + \frac{β}{2} < 0$ ): itna neeche tilt karo $α_{L = 0}$ se ki camber bhi nahi bacha sakta — wing neeche push karta hai.

Recall Parent ki form par note

Parent note ne camber term $\frac{2 β}{π}$ likha tha, jo camber parameter ki alag definition se correspond karta hai (mean-line slope ka alag normalisation). Yahan use ki gayi clean geometric definition $β = 4 z_{m a x} / c$ ke saath, integrals ka honest result $A_{0} + A_{1} /2 = α + β /2$ hai. Physics — camber line ko shift karta hai, kabhi iska $2 π$ slope nahi — identical hai; sirf knob ka label alag hai.

Ek picture mein summary

KYA. Ek figure jo saare aath steps chain karta hai: wing → vortex sheet → tangency + downwash → angle ruler → surviving dials → $Γ$ → lift → do sign cases.

Arrows trace karo: solid wing spins ki sheet ban jaati hai; no-through-flow rule plus Biot–Savart un spins ko fix karta hai; angle ruler integral clean karta hai; summing saare dials maar deta hai siwaaye $A_{0}$ aur $A_{1}$ ke; Kutta–Joukowski unke sum ko lift mein badal deta hai: $L^{'} = π ρ V^{2} c (α + \frac{β}{2}) .$

Recall Feynman: plain words mein poora walk

Hum hawa mein ek bent card se shuru karte hain. Ek bent solid card par easily maths nahi kar sakte, isliye hum pretend karte hain ki yeh tiny spinning tops ki ek row hai uske curve ke saath rakhi hui. Tops ko sirf ek rule maanni hai: hawa ko card ke saath saath slide karna hai, kabhi iske through nahi. Woh rule batata hai ki har top ko kitna spin karna hai — aur yeh figure out karna ki ek top doosre par kitna zor se chalaata hai Biot–Savart (tiny-tornado rule) hai. Seedha ruler sums ko ugly banata hai, isliye hum positions ko ek angle se relabel karte hain jo half-circle mein chalta hai; achanak mess tidy sine waves mein badal jaata hai. Jab hum saari spinning add karke total swirl $Γ$ nikalte hain, har wiggle cancel ho jaata hai siwaaye pehle do ke — tilt $α$ aur camber ka ek lump. Total swirl ko isse multiply karo ki hawa kitni fast aur kitni heavy hai, aur lift bahar aa jaati hai. Tilt aur curve do tarike hain card ko harder scoop karne ke; curve bas card ko pre-tilt kar deta hai, isliye ek curved wing tab bhi lift karta hai jab tum ise thoda neeche point karo.

Recall Khud ko check karo

$Γ$ physically kahan se aata hai? ::: Kutta condition sharp trailing edge se smooth flow force karta hai, jo ek definite bound circulation demand karta hai. Lift mein sirf $A_{0}$ aur $A_{1}$ kyun? ::: $sin n θ$ ko $0 \to π$ par sum karne se saare $n \geq 2$ orthogonality se cancel ho jaate hain. $c_{ℓ}$ kya hai? ::: Dimensionless lift coefficient $c_{ℓ} = L^{'} / (\frac{1}{2} ρ V^{2} c)$ . Camber $c_{ℓ}$ – $α$ line ke saath kya karta hai? ::: Use ek constant se left shift karta hai; slope $2 π$ rehti hai. Camber $β$ (with $β = 4 z_{m a x} / c$ ) ke liye zero-lift angle? ::: $α_{L = 0} = - β /2$ . Poori theory ke neeche kaun si paanch assumptions hain? ::: Inviscid, incompressible, irrotational, two-dimensional, thin/small-angle.

Connections

Kutta–Joukowski theorem — $Γ$ ko $L^{'}$ mein badalta hai (Step 2, Step 6).
Circulation and bound vortices — kyun spin matlab lift (Step 2).
Kutta condition — finite-trailing-edge spin choose karta hai (Step 5).
Biot–Savart law — downwash integral (Step 4).
Glauert's integral and Fourier coefficients — angle ruler aur dials (Step 5–6).
Lift coefficient and angle of attack — jahan $c_{ℓ} = 2 π α + π β$ rehta hai (Step 7–8).
Compressibility corrections (Prandtl–Glauert) — jab hawa compress hoti hai toh kya badalta hai.