Angle of attack, lift coefficient, drag coefficient: Worked examples

3.1.20 · D3 · Physics › Compressible Flow & Aerodynamics › Angle of attack, lift coefficient, drag coefficient

Yeh deep dive parent topic note ko drill karta hai — lift/drag machinery ke har tarah ke case ke through walk karte hue. Pehle hum poora territory map karte hain, phir har cell ko ek full worked example se cover karte hain.

Yahan sab kuch parent ke teen formulas pe tikaa hai, jinhein hum dobara likh rahe hain taaki koi bhi symbol "unearned" na lage:

Formula Teen tools jo hum baar baar use karenge

$q = \frac{1}{2} ρ V^{2} (dynamic pressure, in pascals)$ $L = C_{L} q S, D = C_{D} q S$ $C_{L} = a_{0} (α - α_{L = 0}), C_{D} = C_{D, 0} + \frac{C _{L}^{2}}{π e AR}$

Letters ka meaning (sab parent mein defined hain): $ρ$ = air density (kg/m³), $V$ = speed (m/s), $S$ = wing area (m²), $α$ = angle of attack, $a_{0}$ = lift-curve slope (yahan per degree), $α_{L = 0}$ = woh angle jo zero lift deta hai, $C_{D, 0}$ = profile drag, $AR$ = aspect ratio (span²/area), $e$ = Oswald efficiency (0 aur 1 ke beech ki ek number).

Stall case ke liye do aur symbols chahiye, yahan define kar lete hain taaki kuch bhi "unearned" na rahe:

$α_{stall}$ = stall angle — woh angle of attack jahan lift apne peak pe pahunche aur flow separate hone lage; iske upar linear law use karna illegal hai.
$C_{L, m a x}$ = maximum lift coefficient — airfoil ka sabse zyada $C_{L}$ jo ho sakta hai, jo exactly $α_{stall}$ pe milta hai.

The scenario matrix

Har problem ko is grid ke ek cell mein rakhke socho. Neeche ke examples tag karte hain ki woh kaunsi cell hit kar rahe hain, aur mil ke poori table cover karte hain.

Cell	Case class	Kya tricky hai	Example
A	Positive $α$ , linear region	"textbook" case	Ex 1
B	$α$ below $α_{L = 0}$ → negative $C_{L}$	sign flip, downward lift	Ex 2
C	$α = α_{L = 0}$ (zero-lift)	degenerate: $C_{L} = 0$	Ex 3
D	Stall / beyond linear region	formula breaks — use karna NAHI chahiye	Ex 4
E	Coefficient → actual newtons (full force)	units, $\frac{1}{2}$ ka factor	Ex 5
F	Drag split + best glide ( $L / D$ )	two-term $C_{D}$ , optimum	Ex 6
G	Limiting input: $V \to 0$ , aur $AR \to \infty$	degenerate/boundary behaviour	Ex 7
H	Real-world word problem (steady climb)	words ko equations mein translate karo	Ex 8
I	Exam twist: backwards solve for $α$	linear law ko invert karo	Ex 9
J	Drag coefficient → force ( $D = C_{D} q S$ )	drag side ki practice	Ex 10

Ex 1 — Cell A: textbook positive angle

Forecast: compute karne se pehle andaza lagao — kya $C_{L}$ $0.5$ , $0.9$ , ya $1$ se upar hoga?

Effective angle nikalo $α - α_{L = 0} = 6 - (- 3) = 9^{\circ}$ . Yeh step kyun? Lift ka generation chord se nahi balki zero-lift line se tilt measure karke hota hai — airfoil ki camber already $α = 0$ pe lift bana rahi hoti hai.
Slope se multiply karo: $C_{L} = 0.10 \times 9 = 0.90$ . Yeh step kyun? Linear region mein har extra degree ek fixed $a_{0}$ lift coefficient add karta hai.

Verify: $α = α_{L = 0} = - 3^{\circ}$ pe formula deta hai $0.10 \times 0 = 0$ ✓ (zero-lift angle ke meaning se consistent). $0.90$ typical stall value $C_{L, m a x} \approx 1.3$ – $1.6$ se comfortably neeche hai, toh hum valid linear zone mein hain. ✓

Ex 2 — Cell B: zero-lift se neeche → negative $C_{L}$

Forecast: kya lift upar jayegi ya neeche?

Effective angle: $α - α_{L = 0} = - 8 - (- 3) = - 5^{\circ}$ . Yeh step kyun? Ab hum zero-lift line se bhi zyada nose-down hain, isliye effective angle negative hai.
Coefficient: $C_{L} = 0.10 \times (- 5) = - 0.50$ . Yeh step kyun? Linear law symmetric hai; negative effective angle negative coefficient deta hai.
Sign ka matlab samjho: negative $C_{L}$ matlab aerodynamic "lift" neeche point kar raha hai (wing ke belly ki taraf). Yahi cheez ek inverted race-car wing ya diving stunt use karta hai.

Figure — Angle of attack, lift coefficient, drag coefficient

Verify: picture sign convention dikhata hai — zero-lift line ke upar arrow upar hai, neeche arrow neeche flip ho jaata hai $C_{L} = 0$ crossing ke through. Magnitude check: crossing se same $5^{\circ}$ door, jaise $α = + 2^{\circ}$ deta $+ 0.5$ , mirror image ✓. Units: $C_{L}$ dimensionless hai, toh bare number sahi hai. ✓

Ex 3 — Cell C: degenerate zero-lift point

Forecast: airfoil nose-down tilt pe hai. Kya phir bhi lift banega?

Effective angle: $α - α_{L = 0} = - 4 - (- 4) = 0^{\circ}$ . Yeh step kyun? Hum zero-lift line pe hain — woh ek angle jahan camber ka upward bias aur nose-down tilt exactly cancel ho jaate hain.
Coefficient: $C_{L} = a_{0} \times 0 = 0$ , kisi bhi slope $a_{0}$ ke liye. Yeh step kyun? Zero effective angle → zero coefficient, slope ki value chahe kuch bhi ho.
Lift force: $L = C_{L} q S = 0 \times 5000 \times 15 = 0 N$ . Yeh step kyun? Lift force = coefficient × dynamic pressure × area; $C_{L} = 0$ hone se poora product zero ho jaata hai chahe $q$ aur $S$ kitne bhi bade hon.

Verify: key insight — ek cambered wing $α = 0$ (chord level) pe zero lift pe nahi hoti; use saari lift khatam karne ke liye $- 4^{\circ}$ tak nose neeche karna padta hai. Isliye $α_{L = 0}$ cambered airfoils ke liye negative hota hai. $0 \times kuch bhi = 0$ ✓, units N·(dimensionless) = N ✓.

Ex 4 — Cell D: stall ke baad (jahan formula illegal hai)

Forecast: kya lift $α = 2 0^{\circ}$ tak badhti rehti hai?

Naive plug-in: $C_{L} = 0.10 (20 - (- 3)) = 0.10 \times 23 = 2.3$ . Yeh step kyun? Hum yeh jaanbujhkar karte hain trap expose karne ke liye — linear law ko stall ka koi ilm nahi hota.
$C_{L, m a x}$ se reality check: $2.3 > 1.4$ , jo impossible hai. $1 5^{\circ}$ ke baad boundary layer separate ho jaati hai (dekho Boundary Layer Separation & Stall) aur $C_{L}$ girta hai — ek real airfoil $2 0^{\circ}$ pe shayad sirf $C_{L} \approx 1.0$ de. Yeh step kyun? Linear formula sirf $α < α_{stall}$ mein valid hai; iske baad hume experimental data use karna padta hai, formula nahi.

Verify: figure dikhata hai curve linearly rise karta hai, $(1 5^{\circ}, 1.4)$ pe peak karta hai, phir collapse ho jaata hai — amber "forbidden extrapolation" dashed line $2.3$ tak shoot karti hai jabki true curve neeche mud jaata hai. Lift stall pe maximum hoti hai, monotonic nahi. ✓

Ex 5 — Cell E: coefficient → newtons ( $\frac{1}{2}$ ka dhyan rakho)

Forecast: ek 10-tonne aircraft ke weight ( $\approx 98000$ N) se zyada ya kam?

Dynamic pressure: $q = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.15) (80)^{2} = 0.5 \times 1.15 \times 6400 = 3680 Pa$ . Yeh step kyun? $q$ airstream ki kinetic-energy density hai (dekho Bernoulli's Equation); $\frac{1}{2}$ optional nahi hai — isko drop karne se answer double ho jaata hai.
Lift: $L = C_{L} q S = 0.6 \times 3680 \times 25 = 55200 N$ . Yeh step kyun? Force = coefficient × dynamic pressure × area — coefficient shape/angle ki saari physics carry karta hai.

Verify: units — $Pa \cdot m^{2} = \frac{N}{m ^{2}} \cdot m^{2} = N$ ✓. Supported mass $= 55200/9.81 \approx 5627 kg$ , ek plausible small aircraft — aur 10 tonnes se kam jaise expect tha. ✓

Ex 6 — Cell F: drag split aur lift-to-drag ratio

Forecast: kya $L / D$ $5$ , $18$ , ya $40$ ke kareeb hoga?

Induced drag: $C_{D, i} = \frac{C _{L}^{2}}{π e AR} = \frac{0. 8 ^{2}}{π ( 0.85 ) ( 10 )} = \frac{0.64}{26.70} = 0.02397$ . Yeh step kyun? Lift generate karne se tip vortices bante hain (Induced Drag & Wingtip Vortices); unka cost $C_{L}^{2}$ ke saath badhta hai, lamba patla wing (bada $AR$ ) se cheaper.
Total drag coefficient: $C_{D} = 0.018 + 0.02397 = 0.04197$ . Yeh step kyun? Profile drag (friction+form) aur induced drag simply add ho jaate hain.
Glide ratio: $L / D = C_{L} / C_{D} = 0.8/0.04197 = 19.06$ . Yeh step kyun? Same $q$ aur $S$ pe areas aur pressures cancel ho jaate hain, toh $L / D = C_{L} / C_{D}$ — pure aerodynamic efficiency.

Verify: $L / D \approx 19$ expected $5$ – $40$ band mein hai ✓. Sanity: agar $AR$ half hokar $5$ ho jaaye, $C_{D, i}$ double ho jaayega, $L / D$ giregi — lamba wing zyada door glide karta hai, gliders ke bade spans se match karta hai. ✓

Ex 7 — Cell G: limiting / degenerate inputs

Forecast: kya zero speed pe wing kuch bhi lift karta hai? Kya infinitely lamba wing induced drag banata hai?

Part (a): $L = C_{L} \frac{1}{2} ρ V^{2} S$ . Jab $V \to 0$ , $V^{2} \to 0$ , toh $L \to 0$ . Yeh step kyun? $C_{L}$ $α$ se fix hai, lekin koi airstream nahi push karne ke liye — lift speed ke square ke saath vanish ho jaati hai. Isliye aircraft ko takeoff run chahiye.
Part (b): $C_{D, i} = \frac{C _{L}^{2}}{π e AR} \to 0$ jab $AR \to \infty$ . Yeh step kyun? Infinitely lamba wing ke "tips hi nahi" hain, toh tip vortices nahi, toh induced drag nahi — yeh thin-airfoil theory ka ideal 2-D airfoil hai.

Verify: dono limits physically sensible boundary cases hain. (a) parent ki baat confirm karta hai " $C_{L}$ $V$ pe depend nahi karta, lekin $L$ karta hai" — $V = 0$ pe, $C_{L}$ abhi bhi jo $α$ deta hai wahi hai, lekin $L = 0$ . (b) explain karta hai kyun gliders huge aspect ratios ke peeche bhaagte hain. ✓

Ex 8 — Cell H: real-world word problem (steady level flight)

Forecast: level flight mein lift kaneq kya equal honi chahiye?

Words translate karo: "straight and level" ⇒ lift weight balance karta hai, toh $L = m g$ . Yeh step kyun? Koi vertical acceleration nahi matlab vertical forces cancel ho rahe hain — yeh physics us phrase mein chhupa hua hai.
Required lift compute karo: $L = m g = 1200 \times 9.81 = 11772 N$ . Yeh step kyun? Weight woh target lift hai jo wing ko produce karni hai.
Dynamic pressure: $q = \frac{1}{2} (1.10) (55)^{2} = 0.5 \times 1.10 \times 3025 = 1663.75 Pa$ . Yeh step kyun? Har lift/drag formula $q$ pe built hai, isliye coefficient extract karne se pehle airstream ki kinetic-energy density compute karni padti hai.
Coefficient ke liye solve karo: $C_{L} = \frac{L}{q S} = \frac{11772}{1663.75 \times 16} = \frac{11772}{26620} = 0.4422$ . Yeh step kyun? $L = C_{L} q S$ ko invert karo woh coefficient nikaalane ke liye jo pilot $α$ ke through "dial in" karta hai.

Verify: $C_{L} \approx 0.44$ ek typical cruise value hai (stall se bahut neeche) ✓. Cross-check: $L = 0.4422 \times 1663.75 \times 16 = 11772 N = m g$ ✓. Units: $\frac{N}{Pa \cdot m ^{2}}$ dimensionless ✓.

Ex 9 — Cell I: exam twist — $α$ ke liye backwards solve karo

Forecast: ek chhota sa nose-up angle, ya stall ke kareeb kuch dramatic?

Linear law se shuru karo $C_{L} = a_{0} (α - α_{L = 0})$ aur usse invert karo: $α = α_{L = 0} + \frac{C _{L}}{a _{0}} .$ Yeh step kyun? Hume coefficient pata hai aur cause chahiye — toh formula ko ulta chalate hain.
Effective angle: $C_{L} / a_{0} = 0.4422/0.11 = 4.0 2^{\circ}$ . Yeh step kyun? Yeh zero-lift line se upar tilt hai.
Offset add karo: $α = - 2 + 4.02 = 2.0 2^{\circ}$ . Yeh step kyun? Zero-lift line $- 2^{\circ}$ pe baithti hai, toh true angle wahan se measure hota hai.

Verify: forward check — $C_{L} = 0.11 (2.02 - (- 2)) = 0.11 \times 4.02 = 0.4422$ ✓. Aur $2.0 2^{\circ}$ $\sim 1 5^{\circ}$ stall angle se bahut neeche hai, toh linear formula yahan legal hai (Ex 4 ke unlike). ✓

Ex 10 — Cell J: drag coefficient → drag force

Forecast: $C_{D}$ typical $C_{L}$ se lagbhag $23 \times$ chhota hai, toh kya $D$ kuch sau ya kuch hazaar newtons hogi?

Dynamic pressure: $q = \frac{1}{2} ρ V^{2} = 0.5 \times 1.10 \times 6 0^{2} = 0.5 \times 1.10 \times 3600 = 1980 Pa$ . Yeh step kyun? Drag, lift ki tarah, $C \cdot q \cdot S$ hai; pehle airstream ki kinetic-energy density banana padta hai.
Drag force: $D = C_{D} q S = 0.04197 \times 1980 \times 18 = 1495.8 N$ . Yeh step kyun? Drag equation lift equation ka exact twin hai — $D = C_{D} q S$ mein $C_{L} \to C_{D}$ swap karo; yeh matrix ke drag side ki practice karata hai.

Verify: units $Pa \cdot m^{2} = N$ ✓. Cross-check glide ratio se: same wing $C_{L} = 0.8$ pe lift dega $L = 0.8 \times 1980 \times 18 = 28512 N$ , aur $L / D = 28512/1495.8 = 19.06$ — exactly Ex 6 ka $L / D$ , confirm karta hai ki dono forces consistent hain. ✓

Recall Quick self-test

Kaunsi cell linear $C_{L}$ formula use karna forbid karti hai? ::: Cell D — stall angle ke baad flow separate ho jaata hai aur $C_{L}$ line se nahi, data se aana chahiye. Level flight mein $C_{L}$ kya satisfy karta hai? ::: $C_{L} = m g / (q S)$ , kyunki lift weight balance karta hai. Fixed $α$ ke saath $V \to 0$ hone par $C_{L}$ aur $L$ ka kya hota hai? ::: $C_{L}$ unchanged rehta hai (woh $α$ pe depend karta hai), lekin $L \to 0$ kyunki $L \propto V^{2}$ . Cambered airfoil ke liye $α_{L = 0}$ negative kyun hota hai? ::: Camber $α = 0$ pe bhi lift banata hai, toh zero lift tak pahunchne ke liye nose neeche karna padta hai. Drag coefficient ko drag force mein kaise convert karte hain? ::: $D = C_{D} q S$ — exactly $L = C_{L} q S$ ka twin.

Connections

Angle of attack, lift coefficient, drag coefficient — woh parent topic jise yeh examples drill karte hain.
Bernoulli's Equation — $q = \frac{1}{2} ρ V^{2}$ ka source, jo har force example mein use hota hai.
Boundary Layer Separation & Stall — kyun Ex 4 ka formula stall ke baad fail ho jaata hai.
Induced Drag & Wingtip Vortices — Ex 6 mein $C_{L}^{2}$ term aur Ex 7 mein limit.
Kutta-Joukowski Theorem — lift-curve slope ka origin jo poore mein use hota hai.
Reynolds Number & Mach Number — secondary parameters jo yeh curves shift karte hain.
Compressible Flow — $a_{0}$ ke liye high-speed corrections.