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

3.1.21 · D1 · Physics › Compressible Flow & Aerodynamics › Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c

Is page mein assume kiya gaya hai ki tumne kuch nahi dekha. Hum har symbol ko ek picture se build karte hain, equation mein aane se pehle. End tak tum parent note Thin airfoil theory ko line by line padh sakte ho.

0 · "Wing side se dekhne par" kaisi dikhti hai?

Kisi bhi symbol se pehle, picture apne dimag mein fix karo. Ek wing ko seedha across kaat do, kati hui face ko dekho. Tumhe ek patli si shape milti hai — aage se thodi moti, peeche sharp point par tapering. Yeh 2-D cross-section hi poora game hai. Parent page ka sab kuch is flat-ish sliver ke baare mein hai jo ek wind mein baitha hai.

Figure s01 — ek side-on wing slice. Lavender outline physical skin hai; dashed slate line straight chord hai leading edge (LE, front) se trailing edge (TE, sharp back) tak. Yeh picture define karti hai ki "2-D airfoil", LE, TE aur chord length $c$ ka matlab kya hai — kisi algebra se pehle.

1 · Air: density $ρ$ aur speed $V$

Parent page hamesha $V^{2}$ (V squared) kyun likhta hai, plain $V$ nahi? Kyunki lift pressure se aati hai, aur moving fluid jo pressure exert kar sakti hai woh speed ke square ke saath badhti hai (speed double karo, punch char guna hoga). Hum yeh square §7 mein $\frac{1}{2} ρ V^{2}$ ke andar lift coefficient mein phir milenge.

2 · Angle of attack $α$ — tilt knob

Figure s02 — mint arrows freestream wind $V$ hain; lavender bar chord hai. Coral arc angle of attack $α$ hai, wind se chord tak sweep hoti hai. Curved coral rotation arrow positive (nose-up about the LE) sense mark karta hai. Visual point: $α$ purely wing ka wind ke relative tilt hai, aur positive matlab nose-up hai.

3 · Camber $z (x)$ , uski slope $d z / d x$ , aur $β$

Ek flat card tilt hone par lift kar sakti hai. Lekin real wings gently curved hoti hain, tilt na hone par bhi — jaise ek shallow smile. Yeh curve camber hai, aur yeh wing ko air neeche scoop karne deti hai freely.

Figure s03 — lavender arch camber line $z (x)$ hai; dashed line chord hai ( $z = 0$ ). Do coral tangent segments local slope $d z / d x$ dikhate hain: positive (uphill) front ke paas, negative (downhill) back ke paas. Yeh derivative $d z / d x$ ke peeche ki picture hai.

Exactly yeh tool (derivative) kyun? Parent page ki boundary condition kehti hai air ko surface ke saath flow karna chahiye. Har jagah surface kis direction mein point karti hai yeh jaanne ke liye, tumhe uska local slope chahiye — aur "ek point par curve ka slope" ka matlab precisely derivative $d z / d x$ hota hai. Koi aur tool jawaab nahi deta "yahan skin ka rukh kidhar hai?"

Angle variable $θ$ — aur yeh $x$ ko replace kyun karta hai

$β$ likhne se pehle, hamen ek naya bookkeeping variable $θ$ milna chahiye jo theory chord ke saath chalne ke liye use karti hai.

Definition

$θ$ — chord-angle variable, aur change of variable == $θ$ == ("theta") ek naya coordinate hai jo chord ko re-label karta hai: seedhi ruler position $x$ ki jagah, hum set karte hain $x = \frac{c}{2} (1 - cos θ), θ : 0 \to π .$ $θ = 0$ par: $cos 0 = 1$ , toh $x = 0$ — leading edge. $θ = π$ par: $cos π = - 1$ , toh $x = c$ — trailing edge. Jaise $θ$ $0 \to π$ sweep karta hai, point $x$ smoothly LE→TE chalta hai. Kyun takleef uthao? Yeh chord ko ek half-circle par map karta hai, jo exactly woh shape hai jo vortex-sheet integrals ko solvable banati hai (dekho Glauert's integral and Fourier coefficients). Yeh sharp edges ke paas sample points crowd karta hai jahan flow sabse tezi se vary karti hai. Jacobian (kaise $d x$ $d θ$ banta hai). Substitution differentiate karo: $\frac{d x}{d θ} = \frac{c}{2} sin θ ⟹ d x = \frac{c}{2} sin θ d θ .$ Toh chord ke saath koi bhi integral $\int_{0}^{c} (\dots) d x$ ban jaata hai $\int_{0}^{π} (\dots) \frac{c}{2} sin θ d θ$ . Yahi rule hai jo neeche har " $d θ$ " integral ko uska meaning deta hai: yeh ordinary chord integrals hain, $θ$ costume mein re-dress kiye gaye hain aur factor $\frac{c}{2} sin θ$ theory ki algebra mein already absorb ho chuka hai.

Figure s05 — change of variable $x = \frac{c}{2} (1 - cos θ)$ . $θ$ mein constant speed se lavender half-circle (upar) ke around move karta point chord (neeche) par project (coral dashed drop-lines) hota hai. Notice karo kaise $θ$ mein equal steps LE aur TE ke paas crowd karte hain — yahi "edges ke paas bunching" hai jo theory chahti hai.

Definition

$β$ — camber parameter, properly define kiya gaya Ab $θ$ haath mein hai, theory ke Fourier coefficients camber slope $d z / d x$ ke weighted averages hain (har $d z / d x$ us $x$ par evaluate hoti hai jo us $θ$ se correspond karti hai): $A_{0} = α - \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} d θ, A_{1} = \frac{2}{π} \int_{0}^{π} \frac{d z}{d x} cos θ d θ .$ Only combination jo lift mein enter karti hai woh hai $A_{0} + \frac{A _{1}}{2}$ . Uski camber part collect karna == $β$ == define karta hai: $\frac{2 β}{π} \equiv - \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} d θ + \frac{1}{2} \cdot \frac{2}{π} \int_{0}^{π} \frac{d z}{d x} cos θ d θ$ Toh $β$ chord ke around camber-line slope $d z / d x$ ka ek specific weighted average hai — poori shape $z (x)$ se nikaala gaya ek number. Flat plate mein $d z / d x = 0$ har jagah ⇒ $β = 0$ . Downward-scooping arc $β > 0$ deta hai. Picture: $β$ arch ki "smile strength" hai — lekin ab tum jaante ho ise $z (x)$ se exactly kaise compute karte hain.

Intuition Exactly

$A_{0} + A_{1} /2$ kyun, aur woh weightings kyun? Tum formulas accept karne ki jagah reason maangne mein bilkul sahi ho. Yeh honest sketch hai (poori machinery Glauert's integral and Fourier coefficients mein hai):

Theory unknown sheet strength $γ (θ)$ ko ek Fourier series ke roop mein likhti hai — simple wiggles ka sum $A_{0} (term) + A_{1} sin θ + A_{2} sin 2 θ + \dots$ . Numbers $A_{0}, A_{1}, A_{2}, \dots$ bas "kitna har wiggle present hai" hain.
Har $A_{n}$ find karne ke liye tum boundary condition ko matching wiggle se multiply karo aur $θ$ par integrate karo. Yahi "multiply by $cos n θ$ and integrate" exactly kyun hai ki $A_{0}$ mein $\int \frac{d z}{d x} d θ$ (weight $1$ ) aata hai aur $A_{1}$ mein $\int \frac{d z}{d x} cos θ d θ$ (weight $cos θ$ ) aata hai: har coefficient camber slope aur us particular wiggle ka overlap hai.
Total circulation $Γ = \int_{0}^{c} γ d x$ poori series integrate karta hai. $n \geq 2$ ke liye sine wiggles $sin n θ$ average to zero ek full chord par (woh utna hi upar wiggle karte hain jitna neeche) — yeh orthogonality hai. Sirf $A_{0}$ term aur $A_{1}$ term non-zero total dete hain, aur un do surviving integrals ki arithmetic precisely combination $A_{0} + \frac{A _{1}}{2}$ produce karti hai. Ek line mein: higher wiggles khud ko cancel kar dete hain jab poori wing add up karte ho, toh lift sirf pehle do Fourier voices, $A_{0}$ aur $A_{1}$ , ko "sunta" hai.

Recall Camber extra angle of attack ki tarah kyun kaam karta hai?

Kyunki back mein curved-down line pehle se trailing flow ko neeche point karti hai, tab bhi jab chord level ho — geometrically same hai jaise flat plate tilt karna. Toh camber aur tilt add hote hain: $α + \frac{2 β}{π}$ . ::: Woh add hote hain kyunki dono ek hi kaam karte hain — air ko neeche deflect karna — toh unke effects ek bracket ke andar sum hote hain.

4 · Circulation $Γ$ , aur Kutta condition jo ise pin down karti hai

Yahan dil hai. Imagine karo sliver ke poore around ek bada loop walk karo aur, chalte chalte, ek running tally rakho: meri path ke saath air mujhe forward help kar rahi hai ya lad rahi hai? (air-speed-along-my-step) × (step ki length) poore round mein add karo. Yeh total circulation hai.

Figure s04 — dashed slate loop woh path hai jo tum walk karte ho. Coral arrows (upar) fast air dikhate hain, mint arrows (neeche) slow air; loop clockwise walk karte hue, fast top run dominate karta hai, positive tally $Γ > 0$ deta hai. Yeh picture $Γ$ ka meaning aur sign dono fix karti hai.

Definition Kutta condition — yeh

sahi amount of swirl select karta hai Mathematically bahut se alag-alag amounts of swirl $Γ$ "air surface ke saath flow kare" satisfy kar sakte hain. Nature exactly ek choose karta hai: Kutta condition kehti hai air ko sharp trailing edge se smoothly leave karna chahiye, infinite speed par corner ke around whip kiye bina. Vortex-sheet language mein yeh force karta hai ki sheet strength trailing edge par finite (actually zero) ho, $γ (TE) = 0$ . Kyun chahiye: is rule ke bina maths under-determined hai — yeh nahi keh sakta ki $Γ$ kitna strong hai. Kutta condition woh physical extra fact hai (ek real sharp edge infinitely fast turn-around flow support nahi kar sakta) jo unique $Γ$ , hence unique lift, lock in karta hai. Yeh poori theory ki cornerstone hai. Dekho Kutta condition. Picture: sharp back point par, top stream aur bottom stream milte hain aur cleanly saath mein peel off hote hain — unmen se koi tip ke around violently curl nahi karta.

Figure s06 — trailing edge par do candidate flows. Left (coral, ✗): flow sharp corner ke around huge speed par whip karta hai — physically impossible. Right (mint, ✓): Kutta condition, flow TE se smoothly aur parallel leave karta hai. Sirf right wala hota hai, aur ise demand karna circulation $Γ$ fix karta hai.

5 · Prime $L^{'}$ — lift per unit span (aur "span" ka matlab kya hai)

Per-slice kyun kaam karte hain? Kyunki haari poori picture ek 2-D cross-section hai. Hum jo bhi compute karte hain woh sliver ke liye hai, toh sab answers naturally "per metre of span" nikalte hain.

6 · Vortex sheet $γ (x)$ — model trick

Real wings curved solid skin hoti hain. Yeh mushkil hai. Clever move: wing ko chord ke saath laid infinitely many tiny spinning points ki line se replace karo, har ek swirl ka ek sliver contribute karte hue.

Un sab tiny spins ko add karo aur poori circulation recover hoti hai: $Γ = \int_{0}^{c} γ (x) d x$

Parent note ke Steps 3–5 sab machinery hai is $γ (x)$ ko find karne ki — Kutta condition trailing edge par iska value fix karta hai, Biot–Savart law woh downwash deta hai jo yeh induce karta hai, aur Glauert's integral and Fourier coefficients poori sheet solve karta hai — phir ise $Γ$ mein integrate karo, phir $Γ$ ko $L^{'} = ρ V Γ$ mein feed karo. Yahi poora pipeline hai.

7 · Lift coefficient $c_{ℓ}$ — ek units-free score

Parent page lift ko ek dimensionless number ke roop mein bhi report karta hai. Ise abhi milo taki yeh stranger na lage.

Prerequisite map

Map ko words mein padhna (jo bhi render nahi kar sakta uske liye): Left par do air properties se shuru karo — air density $ρ$ aur freestream speed $V$ — dono seedha Kutta–Joukowski ( $L^{'} = ρ V Γ$ ) mein feed hote hain. Kutta–Joukowski ka teesra input circulation $Γ$ hai, aur Kutta–Joukowski output deta hai lift per unit span $L^{'}$ . $Γ$ kahan se aata hai? Kai streams isme flow karte hain: angle of attack $α$ , camber parameter $β$ , vortex sheet $γ (x)$ , aur — crucially — Kutta condition jo iska size fix karti hai. Camber parameter $β$ khud camber line $z (x)$ aur uski slope aur chord-angle variable $θ$ se bana hai, jabki vortex sheet chord $c$ ke along laid hai. Finally $L^{'}$ dono lift coefficient $c_{ℓ}$ aur poori thin-airfoil-theory topic ko feed karta hai. Toh dependency chain hai: geometry knobs ( $α$ , $β$ $z (x)$ aur $θ$ se, $c$ ) + Kutta condition → circulation $Γ$ → ( $ρ$ , $V$ ke saath) Kutta–Joukowski → lift $L^{'}$ → $c_{ℓ}$ aur topic.

Equipment checklist

Self-test: right side cover karo aur reveal se pehle answer do.

$L^{'}$ mein prime ka matlab kya hai? :::

Symbol

c

kya measure karta hai, aur kis units mein?

Chord length (LE se TE), metres mein.

ρ

kya hai aur iska sea-level value?

Air density, approximately

1.225 kg/m^{3}

Lift

V^{2}

kyun use karta hai,

V

nahi?

Fluid pressure/dynamic push speed ke square ke saath badhta hai.

α

kya hai aur iska sign rule?

Angle of attack;

α > 0

jab chord leading edge ke baare mein nose-up ho.

α

radians mein kyun hona chahiye?

Theory ne

sin α \approx α

use ki, sirf radians mein valid.

α

ke kis range ke liye thin-airfoil theory trustworthy hai?

Roughly

∣ α ∣ ≲ 0.17

rad (

\sim 10°

); usse aage linearisation fail hoti hai aur flow stall ho jaata hai.

6°

ko radians mein convert karo.

6 \times π /180 = 0.1047

rad.

z (x)

kya describe karta hai?

Position

x

par chord ke upar mean camber line ki height.

d z / d x

physically kya mean karta hai?

Camber line ka local slope (facing direction).

Variable

θ

kya hai aur

d x

d θ

se kya relation hai?

x = \frac{c}{2} (1 - cos θ)

se chord-angle;

d x = \frac{c}{2} sin θ d θ

Shape se

β

kaise define hota hai?

Weighted slope average ke roop mein

\frac{2 β}{π} = - \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} d θ + \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} cos θ d θ

Lift mein sirf

A_{0}

aur

A_{1}

kyun enter karte hain?

Higher sine wiggles

sin n θ

(

n \geq 2

) chord par average to zero hote hain (orthogonality).

Ek sentence mein, circulation

Γ

kya hai?

Wing ke around air ka net swirl, yaani speed-along-a-loop uske around sum kiya gaya.

Positive

Γ

ka sign convention kya hai?

Clockwise loop (wind left-to-right, lift upar); positive

Γ

= top par faster air = upward lift.

Kutta condition kya require karta hai, aur kyun?

Sharp trailing edge se smooth flow (

γ (TE) = 0

); yeh uniquely

Γ

fix karta hai.

Kutta–Joukowski state karo.

L^{'} = ρ V Γ

"Span" ka matlab kya hai?

Wingtip-to-wingtip doori, chord slice ke perpendicular.