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

3.1.21 · D3 · Physics › Compressible Flow & Aerodynamics › Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c

Yeh page parent note ka formula har tarah ke input par drill karta hai — positive aur negative angles, positive aur negative camber, zero camber, zero angle, degenerate flat plate, ek stall-warning limit, ek real-world word problem, aur ek sneaky exam twist. Koi bhi number daalne se pehle, hum saare cases ka ek map banate hain taaki koi bhi scenario aise na aaye jo humne dikhayi na ho.

Sab kuch parent ke do boxed results par tika hua hai, jo plain words mein dobara bataye gaye hain:

Definition Dynamic pressure — chalti hawa ka "dhakka"

Example 5 mein use karne se pehle, quantity $q = \frac{1}{2} ρ V^{2}$ se milo, jise dynamic pressure kehte hain. Plain words mein: speed $V$ se chalti hawa ki kinetic energy $\frac{1}{2} m v^{2}$ per unit mass hoti hai; $q$ woh energy density hai, aur yeh exactly woh extra pressure hai jo hawa kisi surface par maarti hai toh daal sakti hai. Iski units hain Pa = N/m². $q$ ko area se multiply karo toh force milta hai. Ek wing section ke liye, span ke har metre ka reference area sirf chord $c$ hai (ek $c \times 1$ strip), toh $q c$ ki units hain (N/m²)(m) = N/m — span per force. Lift coefficient $c_{ℓ}$ ki definition woh dimensionless ratio hai jo is reference push ko actual lift mein badalta hai: $c_{ℓ} \equiv \frac{L ^{'}}{\frac{1}{2} ρ V ^{2} c} ⟹ L^{'} = q \frac{1}{2} ρ V^{2} c c_{ℓ} .$ Yeh koi nayi physics formula nahi hai — yeh $c_{ℓ}$ ki definition hai, rearranged. Hum ise Example 5 mein use karenge $c_{ℓ}$ se force tak jaane ke liye.

Scenario matrix

Is formula se banne wala har problem inhi cells mein se ek mein aata hai. Neeche ke worked examples us cell ke saath tagged hain jo wo cover karta hai, aur saath mein yeh sab unhe hit karte hain.

Cell	Case class	Kya special hai	Example
O	Flat plate, zero angle	$β = 0, α = 0 \Rightarrow L^{'} = 0$ (trivial boundary)	Ex 0
A	Flat plate, positive $α$	$β = 0$ : sirf tilt lift karta hai	Ex 1
B	Cambered wing, positive $α$	Dono knobs add hote hain	Ex 2
C	Negative angle of attack	Sign flip hoti hai — lift neeche ja sakti hai	Ex 3
D	Zero-lift angle (degenerate $L^{'} = 0$ )	Terms exactly cancel hote hain (flat plate ⇒ $α = 0$ ; cambered ⇒ shifted)	Ex 4
E	Zero angle, pure camber	$α = 0$ : sirf curve se lift	Ex 5
E−	Negative camber ( $β < 0$ )	Curve reversed — line ko doosri taraf shift karta hai	Ex 5b
F	Limiting / stall-warning behaviour	Formula bina bound ke badhta hai; physics kahan quit karti hai	Ex 6
G	Real-world word problem	Total lift, kitna mass support kar sakta hai	Ex 7
H	Exam twist: ulta solve karo	$L^{'}$ diya hua hai, required $α$ nikalo	Ex 8

Sign story cells O–E− ko ek saath bandhti hai — compute karne se pehle ek picture dekhne layak hai.

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

Figure s01 — Master sign-map: flat plate (blue) aur cambered wing (green) ke liye $c_{ℓ}$ – $α$ lines. Alt text: slope $2 π$ ki do parallel seedhi lines; blue line origin se jaati hai, green line $4 β$ se upar-baayein shift hai, Cell O (origin), Cell D (green ka negative angle par zero-crossing), aur Cell E (vertical axis par green ki value) ke markers ke saath.

Figure ko dhyan se padhiye — yeh cells O se E− ke liye master map hai:

Horizontal axis angle of attack $α$ hai degrees mein; vertical axis lift coefficient $c_{ℓ}$ hai.
Blue line flat plate hai, $c_{ℓ} = 2 π α$ : yeh exactly origin se jaati hai, toh $α = 0$ par yeh $c_{ℓ} = 0$ deti hai (woh hai Cell O, trivial boundary).
Green line ek positively cambered wing ( $β = 0.05$ ) hai, $c_{ℓ} = 2 π α + 4 β$ : same slope $2 π$ lekin constant $4 β$ se upar aur baayein khiskayi hui hai.
Red dot jahan green line zero cross karti hai woh cambered zero-lift angle hai (Cell D), jo ek chhoti negative $α$ par baitha hai.
Orange square vertical axis par green line ki value hai $α = 0$ par — bina tilt ke pure-camber lift (Cell E).
Kisi bhi line ke zero-crossing ke baayein sab kuch axis ke neeche hai: negative (downward) lift (Cell C). Dhyan do dono lines parallel hain: camber shift karta hai, kabhi tilt nahi karta. Negative camber ( $β < 0$ , Cell E−) line ko ulti taraf, neeche aur daayein, slide kar dega.

Worked examples

Zyaadatar examples ke liye baseline flow: $ρ = 1.225$ kg/m³, $V = 50$ m/s, $c = 1.5$ m. Tab constant lump $π ρ V^{2} c = π (1.225) (5 0^{2}) (1.5) = 14432 N/m$ baar baar aata hai — ise $K = 14432$ N/m bolo, toh $L^{'} = K (α + \frac{2 β}{π})$ .

Cell O — flat plate, zero angle (trivial boundary)

Worked example Example 0 — kuch tila nahi, kuch curved nahi

Flat plate ( $β = 0$ ) bilkul level hawa mein baitha hai ( $α = 0$ ). $L^{'}$ nikalo.

Forecast: koi bhi lift hogi?

Dono knobs off hain: $α = 0$ aur $β = 0$ , toh bracket $(α + \frac{2 β}{π}) = 0$ . Yeh step kyun? Yeh woh boundary hai jahan dono lift sources khatam ho jaate hain — woh reference "kuch na karo" state jisse har doosra case measure hota hai.
$L^{'} = K \times 0 = 0 N/m$ . Yeh step kyun? Constant $K$ ko zero bracket se multiply karna formula complete karta hai aur physical expectation confirm karta hai — hum multiplication explicitly carry karte hain taaki zero derived ho, sirf claim na ho.

Verify: figure par yeh origin hai, jahan blue line zero cross karti hai ✓. Flow ke saath aligned flat plate hawa ko net turn nahi karta, isliye koi lift nahi — symmetry se consistent (upar aur neeche mein koi fark nahi) ✓.

Cell A — flat plate, positive angle

Cell B — cambered wing, positive angle

Worked example Example 2 — dono knobs on

Same flow, ab $β = 0.05$ rad, abhi bhi $α = 5^{\circ}$ .

Forecast: Ex 1 se zyaada ya kam, aur roughly kitna?

Convert: $α = 5 \times π /180 = 0.08727$ rad. Yeh step kyun? Same radian rule jaise Ex 1 mein — linearisation $sin α \approx α$ radians maangta hai, toh hum kisi bhi arithmetic se pehle convert karte hain.
Camber term: $\frac{2 β}{π} = \frac{2 ( 0.05 )}{π} = 0.03183$ . Yeh step kyun? Camber ek effective extra angle add karta hai, slope change nahi.
Bracket ke andar sum: $0.08727 + 0.03183 = 0.11910$ . Yeh step kyun? Hum tilt aur camber contributions pehle add karte hain kyunki formula $K$ ko unke sum se multiply karta hai; inhe ab combine karna final multiplication ko ek clean step bana deta hai.
$L^{'} = 14432 \times 0.11910 = 1719 N/m$ . Yeh step kyun? Benchmark constant $K$ ko combined bracket se multiply karna cambered wing ki total lift produce karta hai — woh number jo hum flat plate se compare karte hain.

Verify: flat plate se ratio $= 1719/1259 = 1.37$ , ek $\sim 37%$ gain — parent ke " $\sim 36%$ " se match karta hai ✓. Gain ek shift hai, steeper slope nahi (Cell A aur B same $2 π$ slope share karte hain).

Cell C — negative angle of attack

Worked example Example 3 — naak neeche ki taraf

Flat plate ( $β = 0$ ), $α = - 3^{\circ}$ . $L^{'}$ aur uski direction nikalo.

Forecast: lift ka sign kya hoga?

$α = - 3 \times π /180 = - 0.05236$ rad. Yeh step kyun? Negative tilt formula mein genuinely negative number hai, aur yeh abhi bhi radians mein hona chahiye taaki linearisation hold kare.
$L^{'} = K α = 14432 \times (- 0.05236) = - 755.6 N/m$ . Yeh step kyun? Kuch special nahi — linear formula bas negative $α$ mein run ho jaata hai.
Minus sign ka matlab aerodynamic force neeche point karti hai (ek "down-force"). Yeh step kyun? Hum negative numeric result ko physically interpret karte hain — sign padhna is case ka poora point hai, kyunki yeh batata hai wing neeche push karta hai, upar nahi.

Verify: magnitude Ex 1 ka $3/5$ scale hai: $1259 \times (3/5) = 755$ N/m ✓. Figure par yeh point flat-plate (blue) line par origin ke baayein baitha hai, axis ke neeche ✓.

Cell D — zero-lift angle (exact cancellation)

Worked example Example 4 — balance point (flat-plate aur cambered dono case)

(a) Flat plate ( $β = 0$ ): kis $α$ par $L^{'} = 0$ hoga? (b) Cambered wing $β = 0.05$ rad: kis $α$ par $L^{'} = 0$ hoga?

Forecast: har ek ke liye, positive, zero, ya negative angle?

$L^{'} = 0$ set karo: kyunki $K \neq = 0$ , humein chahiye $α + \frac{2 β}{π} = 0$ . Yeh step kyun? Product tab zero hota hai jab bracket zero ho.
(a) Flat plate: $β = 0$ ke saath bracket bas $α$ hai, toh $α_{L = 0} = 0$ . Yeh step kyun? Ek symmetric flat plate mein koi built-in camber offset nahi hota, toh uski zero-lift line origin se jaati hai — Cell O se consistent.
(b) Cambered: $α_{L = 0} = - \frac{2 β}{π} = - \frac{2 ( 0.05 )}{π} = - 0.03183$ rad $= - 1.82 4^{\circ}$ . Yeh step kyun? Hum $α$ ke liye zero-bracket condition solve karte hain aur phir degrees mein convert karte hain, kyunki "kya angle" question ka answer degrees mein quote karna zyaada useful hai bhale hi woh radians mein derive hua ho.

Verify: (a) $α = 0, β = 0$ plug karo: bracket $= 0$ ⇒ $L^{'} = 0$ ✓ (Cell O se match). (b) wapas plug karo — $α + 2 β / π = - 0.03183 + 0.03183 = 0$ ✓. Toh zero-lift boundary flat plate ke liye $α = 0$ hai aur camber ki amount se exactly negative shift hoti hai — ek consistent rule, do cases. Figure par (a) blue line ka origin crossing hai aur (b) green line par red dot hai.

Cell E — zero angle, pure camber

Worked example Example 5 — bina tilt ke lift

Cambered wing $β = 0.05$ rad, $α = 0$ . $c_{ℓ}$ aur $L^{'}$ nikalo.

Forecast: kya ek wing bilkul level hawa mein lift kar sakta hai?

$c_{ℓ} = 2 π (0) + 4 β = 4 (0.05) = 0.2$ . Yeh step kyun? $α = 0$ ke saath $c_{ℓ} = 2 π α + 4 β$ mein sirf camber term bachta hai.
$L^{'} = \frac{1}{2} ρ V^{2} c c_{ℓ} = 0.5 (1.225) (2500) (1.5) (0.2) = 459.375 N/m$ . Yeh step kyun? Hum upar callout se $c_{ℓ}$ ki definition use karte hain, $L^{'} = q c c_{ℓ}$ mein rearrange karke jahan $q = \frac{1}{2} ρ V^{2}$ dynamic pressure hai — yeh dimensionless lift number ko actual force per span mein convert karta hai.
Direct form se cross-check: $L^{'} = K \cdot \frac{2 β}{π} = 14432 (0.03183) = 459.4$ N/m ✓. Yeh step kyun? Same lift ko doosre route se recompute karna ( $c_{ℓ}$ definition ki jagah direct $K$ -formula) ek self-consistency test hai — agreement confirm karta hai koi slip nahi hua.

Verify: do independent routes same $\approx 459$ N/m dete hain ✓. Haan — camber akela zero tilt par lift karta hai; yehi reason hai real wings curved hoti hain. Poori curve ke liye Lift coefficient and angle of attack dekhein.

Cell E− — negative camber

Worked example Example 5b — reflex camber (Cell E ka mirror)

Ek tail surface jisme negative camber $β = - 0.05$ rad hai, bilkul level baitha hai ( $α = 0$ ). $c_{ℓ}$ aur $L^{'}$ nikalo.

Forecast: agar positive camber upar lift kiya, toh negative camber kya karega?

$c_{ℓ} = 2 π (0) + 4 β = 4 (- 0.05) = - 0.2$ . Yeh step kyun? $β$ linearly enter karta hai, toh curve par sign flip uski lift contribution ka sign flip kar deta hai.
$L^{'} = \frac{1}{2} ρ V^{2} c c_{ℓ} = 0.5 (1.225) (2500) (1.5) (- 0.2) = - 459.375 N/m$ . Yeh step kyun? Same $c_{ℓ}$ -definition jaise Ex 5 mein; negative $c_{ℓ}$ seedha negative (downward) force tak jaata hai.

Verify: Ex 5 ka exact mirror: same magnitude, opposite sign ✓. Figure par yeh green line ko origin se reflect karega — neeche aur daayein slide karega, ek positive zero-lift angle dega. Reflex/negative camber tailless aircraft par trim ke liye use hota hai (rear ko neeche push karta hai), jo dikhata hai formula poori $β$ sign range cover karta hai.

Cell F — limiting behaviour / jahan theory quit karti hai

Worked example Example 6 —

$α$ ko stall ki taraf push karna Flat plate. Formula kehta hai $c_{ℓ} = 2 π α$ hamesha ke liye. $c_{ℓ}$ compute karo $α = 5^{\circ}, 1 5^{\circ}, 2 5^{\circ}$ par aur mark karo jahan physics disagree kare.

Forecast: kya lift sach mein $9 0^{\circ}$ tak linearly badhti rehti hai?

$c_{ℓ} (5^{\circ}) = 2 π (0.08727) = 0.548$ . Yeh step kyun? $5^{\circ}$ ek chhota angle hai jahan linear theory trustworthy hai — stall ke neeche yeh hamara reliable anchor point hai.
$c_{ℓ} (1 5^{\circ}) = 2 π (0.26180) = 1.645$ . Yeh step kyun? $1 5^{\circ}$ real stall region ke bilkul edge par baitha hai — theory yahan itni lift predict karti hai ( $1.645$ ) jo ek real airfoil rakh hi nahi sakta ( $\approx 1.5$ ), toh yeh number pehla hai jis par hume shak karna chahiye.
$c_{ℓ} (2 5^{\circ}) = 2 π (0.43633) = 2.742$ . Yeh step kyun? Stall ke kaafi baad linear theory ka koi upper bound nahi hai — yeh baar baar badhti lift predict karta hai, jo physically impossible hai.
Lekin real airfoils stall karte hain $c_{ℓ} \approx 1.5$ ke paas (around $12 - 1 5^{\circ}$ ): flow separate ho jaata hai, Kutta condition assumption toot jaata hai, aur measured lift collapse kar jaata hai. Yeh step kyun? Derivation ne small angles aur attached flow assume kiya; stall ke baad woh assumptions fail ho jaate hain.

Verify: numbers linearly badhte hain (ratio $0.548 : 1.645 : 2.742 = 1 : 3 : 5$ , $5 : 15 : 25$ se match karta hai) ✓. Yaad rakhne wali limit: straight line par sirf chhote $α$ ke liye trust karo; physical ceiling stall hai, formula nahi.

Figure s02 — Cell F stall warning: seedhi linear-theory line (blue) bina limit ke badhti hai, jabki real airfoil (red dashed) $c_{ℓ} \approx 1.5$ ke paas peak karta hai aur stall par gir jaata hai. Alt text: ek badhti seedhi blue line aur ek red dashed curve jo uske saath $\approx 1 3^{\circ}$ tak jaati hai, phir mur jaati hai aur gir jaati hai; stall ke baad shaded gap dikhata hai jahan formula lift over-predict karta hai.

Figure dikhata hai seedhi theory line real (dashed) curve ko rocket ki tarah cross karti hai, jo stall par peak karke gir jaata hai — shaded region woh jagah hai jahan formula jhooth bolta hai.

Cell G — real-world word problem

Worked example Example 7 — yeh wing kitna hold kar sakti hai?

Ek light aircraft wing: span $b = 10$ m, chord $c = 1.5$ m, cruising $V = 50$ m/s par $ρ = 1.225$ kg/m³ ki hawa mein, angle of attack $α = 4^{\circ}$ , camber $β = 0.04$ rad. Total lift aur mass jo yeh support karta hai ( $g = 9.81$ m/s²) nikalo.

Forecast: ek car lift karne ke liye kaafi? ek insaan? ek plane?

Convert: $α = 4 \times π /180 = 0.06981$ rad. Yeh step kyun? Jaise har example mein, linearised formula angle radians mein maangta hai pehle ki hum use camber term mein add karein.
Camber term $\frac{2 β}{π} = \frac{2 ( 0.04 )}{π} = 0.02546$ . Yeh step kyun? $β$ pehle se radians mein diya hua hai (ek slope, dimensionless), toh koi conversion nahi — lekin add karne se pehle $α$ ki units se match karne ke liye hume $2 β / π$ form karna hoga.
Per-span lift $L^{'} = K (α + 2 β / π) = 14432 (0.06981 + 0.02546) = 14432 (0.09527) = 1375 N/m$ . Yeh step kyun? Dono knobs combine karna aur $K$ se multiply karna har metre of span par force deta hai — woh per-strip lift jo hum span ke across replicate karenge.
Span par total lift: $L = L^{'} \times b = 1375 \times 10 = 13750$ N. Yeh step kyun? $L^{'}$ force per metre hai; span $b = 10$ m se multiply karna saare das $1$ -metre strips add karta hai (strip theory) poori wing ka force pane ke liye.
Supported mass $m = L / g = 13 750/9.81 = 1402 kg$ . Yeh step kyun? Level flight mein lift weight $m g$ balance karta hai, toh lift force ko $g$ se divide karna $L = m g$ invert karta hai us mass ko reveal karne ke liye jo wing hawa mein rakh sakta hai.

Verify: units — (N/m)(m) = N ✓, N ÷ (m/s²) = kg ✓. Lagbhag 1.4 tonnes — ek plausible small aircraft ✓. (Yeh strip theory use karta hai; true 3-D wings tip vortices se kuch lift lete hain — Circulation and bound vortices dekhein.)

Cell H — exam twist: ulta solve karo

Worked example Example 8 — required angle nikalo

Baseline flow ( $K = 14432$ N/m), flat plate ( $β = 0$ ). Designer ko exactly $L^{'} = 2000$ N/m chahiye. Konsa angle of attack (degrees mein) chahiye?

Forecast: $5^{\circ}$ se bada ya chhota jo 1259 N/m diya?

$L^{'} = K α$ ko $α$ ke liye rearrange karo: $α = L^{'} / K = 2000/14 432 = 0.13858$ rad. Yeh step kyun? Hum formula invert karte hain kyunki unknown ab angle hai, force nahi — target lift ko constant $K$ se divide karna $α$ isolate karta hai.
Convert: $α = 0.13858 \times 180/ π = 7.9 4^{\circ}$ . Yeh step kyun? Tilt ke baare mein answers degrees mein quote hote hain, toh hum radian result ko bilkul end mein convert karte hain ek human-readable design angle ke liye.

Verify: $α = 0.13858$ rad wapas plug karo: $L^{'} = 14432 \times 0.13858 = 2000$ N/m ✓. $5^{\circ}$ se bada (expected, kyunki hume zyaada lift chahiye) aur Cell F se stall ke neeche comfortably hai ✓.

Recall Saare cells par rapid self-test

Ex 0 flat plate zero angle par ::: $0$ N/m Ex 1 5° par flat-plate lift ::: $1259$ N/m Ex 2 cambered lift ::: $1719$ N/m ( $\sim 37%$ zyaada) Ex 3 $α = - 3^{\circ}$ par lift ::: $- 756$ N/m (down-force) Ex 4a flat-plate zero-lift angle ::: $0^{\circ}$ Ex 4b $β = 0.05$ ke liye cambered zero-lift angle ::: $- 1.8 2^{\circ}$ Ex 5 $α = 0$ par pure-camber lift ::: $459$ N/m, $c_{ℓ} = 0.2$ Ex 5b negative camber ::: $- 459$ N/m, $c_{ℓ} = - 0.2$ Ex 6 line kahan jhooth bolta hai ::: stall ke baad ( $\sim 12$ – $1 5^{\circ}$ ) Ex 7 supported mass ::: $\approx 1402$ kg Ex 8 2000 N/m ke liye angle ::: $7.9 4^{\circ}$

Connections

Kutta–Joukowski theorem — woh $L^{'} = ρ V Γ$ jinse yeh numbers aakhir mein aate hain.
Lift coefficient and angle of attack — woh $c_{ℓ}$ – $α$ line jis par Cells C–E rehte hain.
Kutta condition — woh assumption jo Cell F (stall) mein fail ho jaati hai.
Circulation and bound vortices — kyun 3-D span multiplication (Cell G) sirf approximate hai.
Compressibility corrections (Prandtl–Glauert) — high $V$ par yeh numbers kaise change hote hain.