Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions of angle of attack, Mach: Worked examples

3.1.29 · D3 · Physics › Compressible Flow & Aerodynamics › Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions o

Yeh page parent topic ki drill-ground hai. Yahan hum koi naya theory introduce nahi karenge — balki har case class ko cover karenge jo coefficient relations pose kar sakti hai, ek worked example per cell, taaki koi bhi exam scenario surprise na lage.

Yahan sab kuch parent note ke char building blocks par based hai. Kuch bhi compute karne se pehle, har symbol ko plain words mein re-earn karte hain taaki pehli baar padhne wala reader kabhi bhi confused na ho.

Definition Jo symbols hum use karenge (sab plain words mein defined)

$C_{N}$ = normal-force coefficient: chord line (wing section ke nose se tail tak ki seedhi line) ke perpendicular push, ek pure number ke roop mein.
$C_{A}$ = axial-force coefficient: chord ke saath, aft direction mein (tail ki taraf) push, ek pure number ke roop mein.
$V_{\infty}$ = freestream velocity: wing se door, aage se aane wali hawa ki speed aur direction, wing ke disturb karne se pehle. Yeh woh arrow hai jiske around "wind frame" bana hai.
$C_{L}, C_{D}$ = lift and drag coefficients: wahi total push, lekin us aane wali hawa $V_{\infty}$ ke perpendicular aur along measure kiya gaya.
$α$ = angle of attack: chord line aur wind $V_{\infty}$ ke beech ka angle. Chota $α$ = wing lagbhag wind ke saath aligned; bada $α$ = wing upar ki taraf tilt.
$M_{\infty}$ = freestream Mach number: flight speed $V_{\infty}$ ko door undisturbed hawa mein local speed of sound se divide kiya. $M_{\infty} < 1$ = subsonic, $M_{\infty} > 1$ = supersonic. Is page par hum Mach number hamesha apne subscript ke saath $M_{\infty}$ likhte hain taaki yeh moment se kabhi confuse na ho.
$C_{m}$ = pitching-moment coefficient: ek reference point ke baare mein twisting (nose-up / nose-down) moment, ek extra chord length se non-dimensionalized.

Dono frames ke beech ka bridge (parent note mein ek vector ko rotate karke prove kiya gaya): $C_{L} = C_{N} cos α - C_{A} sin α, C_{D} = C_{N} sin α + C_{A} cos α$

Baaki tools — thin-airfoil lift, drag polar, Prandtl–Glauert, Ackeret — har ek ko us example ke andar restate kiya gaya hai jise pehli baar unki zaroorat padti hai, taaki koi bhi formula aapko cold na mile.

Scenario matrix

Is topic ke har problem ka ek cell mein placement hota hai. Sabse dayi column us example ka naam batati hai jo use cover karti hai.

Cell	Case class	Isme kya tricky hai	Covered by
A	Chota $α$ , ordinary signs	baseline conversion, lift ≈ normal	Ex 1
B	$α = 0$ (degenerate)	$sin α = 0$ rotation ko collapse kar deta hai	Ex 2
C	Bada $α$ (near/at stall)	lift aur normal force mein bahut difference	Ex 3
D	Negative $α$ (sign flip)	kaunse coefficients sign change karte hain?	Ex 4
E	Inverse direction (wind → body)	rotation ko undo karo	Ex 5
F	Subsonic compressibility limit	Prandtl–Glauert as $M_{\infty} \to 1$	Ex 6
G	Supersonic (new physics: wave drag)	alag square root, extra drag	Ex 7
H	Real-world word problem	words strip karo, cell chuno	Ex 8
I	Exam twist (stability sign, $C_{m}$ )	$d C_{m} / d α$ ka matlab, sirf plug-in nahi	Ex 9

Neeche wala figure Cells A–E ka master picture hai: ek resultant force $R$ (woh red arrow) jo do axis systems mein dekha gaya hai jo $α$ se rotate hue hain. Har rotation example bas yahi diagram hai alag numbers ke saath.

Figure — Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions of angle of attack, Mach

Example 1 — Cell A: chota angle, ordinary signs

Worked example Body → wind at

$α = 6^{\circ}$ Given $C_{N} = 0.80$ , $C_{A} = 0.04$ , $α = 6^{\circ}$ . $C_{L}$ aur $C_{D}$ nikalo.

Forecast: compute karne se pehle guess karo — kya $C_{L}$ , $C_{N}$ se bada hoga ya chota? Kya $C_{D}$ tiny hoga ya bada?

Angle convert karo: $cos 6^{\circ} = 0.9945$ , $sin 6^{\circ} = 0.1045$ . Yeh step kyun? Rotation formulas ko $α$ ki cosine aur sine chahiye, degree value nahi — trig hi ek angle ko projection ratio mein convert karti hai. Upar wale master figure mein, $α$ chota hai, toh dono frames lagbhag overlap karte hain.
Lift: $C_{L} = 0.80 (0.9945) - 0.04 (0.1045) = 0.7956 - 0.0042 = 0.7914$ . Yeh step kyun? $N$ lift direction par $N cos α$ ke roop mein project hota hai; aft-pointing $A$ thoda lift ke khilaf tilt karta hai, isliye minus.
Drag: $C_{D} = 0.80 (0.1045) + 0.04 (0.9945) = 0.0836 + 0.0398 = 0.1234$ . Yeh step kyun? Dono forces ka ek piece downstream point karta hai; woh add hote hain.

Verify: $C_{L} \approx C_{N} = 0.80$ jaise small-angle sanity check predict karta hai (lift ≈ normal at low $α$ ). ✓ Units: charon dimensionless hain, consistent. ✓

Example 2 — Cell B: degenerate case $α = 0$

Worked example Wing perfectly aligned with the wind

Given $C_{N} = 0.30$ , $C_{A} = 0.02$ , $α = 0^{\circ}$ . $C_{L}$ aur $C_{D}$ nikalo.

Forecast: chord wind ke parallel hone par, kaunsi body force ban jaati hai lift, aur kaunsi ban jaati hai drag?

$α = 0$ par: $cos 0 = 1$ , $sin 0 = 0$ . Yeh step kyun? Yeh degenerate input hai — poora rotation switch off ho jaata hai. Master figure mein, imagine karo wind axes ko wapas tab tak rotate karo jab tak woh body axes par exactly na baith jaayein.
$C_{L} = 0.30 (1) - 0.02 (0) = 0.30$ . Yeh step kyun? Trig values plug in karne par: $sin$ -term zero se kill ho jaata hai, toh sirf normal force lift mein survive karta hai. Yeh "frames coincide" idea ko abstract ki jagah concrete banata hai.
$C_{D} = 0.30 (0) + 0.02 (1) = 0.02$ . Kyun: chord wind ke saath hone par, "perpendicular to chord" hi hai "perpendicular to wind," isliye $C_{L} = C_{N}$ exactly; "along chord" hi hai "along wind," isliye $C_{D} = C_{A}$ exactly.

Verify: $α = 0$ par dono frames ek hi frame hain, isliye lift normal ke barabar aur drag axial ke barabar honi chahiye. Dono exactly hold karte hain. ✓ Yeh woh anchor hai jo batata hai ki nonzero $α$ sirf in numbers ko rotate karta hai.

Example 3 — Cell C: bada angle (near stall)

Worked example High angle of attack,

$α = 3 0^{\circ}$ Given $C_{N} = 1.20$ , $C_{A} = 0.25$ , $α = 3 0^{\circ}$ . $C_{L}$ , $C_{D}$ , aur ratio $C_{L} / C_{N}$ nikalo.

Forecast: $3 0^{\circ}$ par, kya lift abhi bhi ≈ normal force hai, ya gap khul gaya hai?

$cos 3 0^{\circ} = 0.8660$ , $sin 3 0^{\circ} = 0.5000$ . Yeh step kyun? Bade angle par sine ab tiny nahi raha (poora $0.5$ hai), isliye iska exact value chahiye — yeh precisely woh regime hai jahan $sin α$ drop karne se galat answer aata. Master figure mein, wind axes ko puri $3 0^{\circ}$ body axes se swung imagine karo.
$C_{L} = 1.20 (0.8660) - 0.25 (0.5000) = 1.0392 - 0.1250 = 0.9142$ . Yeh step kyun? Ab $sin α$ bada hai, isliye axial term ek real chunk subtract karta hai — "lift ≈ normal" shortcut ab kaam nahi karta.
$C_{D} = 1.20 (0.5000) + 0.25 (0.8660) = 0.6000 + 0.2165 = 0.8165$ . Kyun: normal force ka bada hissa ( $N sin α = 0.6$ ) ab downstream point karta hai — high $α$ par yeh dominant drag hai.
Ratio $C_{L} / C_{N} = 0.9142/1.20 = 0.762$ . Yeh step kyun? Ratio woh single number hai jo expose karta hai ki lift normal force se kitni dur drift hui hai — is cell ka headline lesson.

Verify: ratio ~0.99 (Ex 1) se 0.76 tak gir gaya — exactly parent note ki woh warning ki "lift = normal" high $α$ par fail hoti hai. ✓ Aur $C_{D}$ ab $C_{L}$ ke comparable hai: high- $α$ flight bahut draggy hoti hai, jaise stall ke paas expect kiya jaata hai. ✓

Example 4 — Cell D: negative angle of attack

Neeche wala figure exactly dikhata hai ki "negative $α$ " geometrically kya karta hai. Black arrow resultant force $R$ hai positive angle par; red arrow wahi force hai negative angle par. Notice karo ki unki length same hai aur horizontal reach same hai (cosine unchanged hai), lekin unke vertical parts opposite ways point karte hain (sine flip hota hai). Yeh picture woh reason hai jis se cosine terms sign change survive karte hain jabki sine terms reverse hote hain — step 1 padhte waqt yeh yaad rakhna.

Worked example Nose-down attitude,

$α = - 1 0^{\circ}$ Given $C_{N} = 0.55$ , $C_{A} = 0.03$ , $α = - 1 0^{\circ}$ . $C_{L}$ aur $C_{D}$ nikalo.

Forecast: kaunsa flip sign karta hai — drag, lift, ya dono?

Negative angle: $cos (- 1 0^{\circ}) = cos 1 0^{\circ} = 0.9848$ (cosine even hai — sign se unchanged), aur $sin (- 1 0^{\circ}) = - sin 1 0^{\circ} = - 0.1736$ (sine odd hai — yeh flip karta hai). Yeh step kyun? Yahi sign case ka poora point hai: cosine $α$ ke sign ko ignore karta hai, sine use track karta hai. Upar wala figure dekho — red vector $R$ black wale jitna lamba hai, bas doosri taraf tilt hai.
$C_{L} = 0.55 (0.9848) - 0.03 (- 0.1736) = 0.5416 + 0.0052 = 0.5468$ . Kyun: axial term ab lift mein add hota hai (double negative), Example 1 se ulta.
$C_{D} = 0.55 (- 0.1736) + 0.03 (0.9848) = - 0.0955 + 0.0295 = - 0.0660$ . Yeh step kyun? Ek negative $C_{D}$ ? Yeh real drag nahi hai — yeh bata raha hai ki is contrived (positive) $C_{N}$ ke saath negative $α$ par, normal force ka streamwise projection upstream point karta hai. Physically, negative $α$ par ek real wing ka $C_{N}$ bhi negative hoga, aur drag positive aayegi. Math faithfully warn kar raha hai ki input combination unphysical hai.

Verify: Example 1 ( $+ 6^{\circ}$ -style) se compare karo: $α$ ka sign flip karna $cos$ terms ko rakhta hai aur $sin$ terms ko flip karta hai — exactly wahi hum dekhte hain (lift barely changed, $sin$ -driven parts reversed). ✓

Example 5 — Cell E: inverse rotation (wind → body)

Worked example

$C_{L}, C_{D}$ measured hain — $C_{N}, C_{A}$ recover karo Ek wind-tunnel run $C_{L} = 0.90$ , $C_{D} = 0.12$ at $α = 1 0^{\circ}$ deta hai. $C_{N}$ aur $C_{A}$ nikalo.

Forecast: hum un-rotate kar rahe hain. Kya hum $sin$ terms ke signs swap karte hain?

Parent note se inverse relations use karo: $C_{N} = C_{L} cos α + C_{D} sin α, C_{A} = - C_{L} sin α + C_{D} cos α .$ Yeh step kyun? $- α$ se wapas rotate karne par har $sin$ ke andar sign flip ho jaata hai; yeh wahi rotation matrix transposed hai. Master figure mein, ab hum wind-frame arrow jaante hain aur uska body-frame shadow chahiye.
$cos 1 0^{\circ} = 0.9848$ , $sin 1 0^{\circ} = 0.1736$ . Yeh step kyun? Inverse relations ko wahi do trig values chahiye — un-rotating wahi angle $α$ use karta hai, bas formula mein swapped signs ke saath wired, isliye ek baar yahan cosine aur sine compute karte hain.
$C_{N} = 0.90 (0.9848) + 0.12 (0.1736) = 0.8863 + 0.0208 = 0.9071$ . Yeh step kyun? Dono wind-frame forces ka ek piece body-normal direction mein hai; yahan woh add hote hain (plus note karo, forward formula ke minus ka mirror image).
$C_{A} = - 0.90 (0.1736) + 0.12 (0.9848) = - 0.1562 + 0.1182 = - 0.0380$ . Kyun: negative $C_{A}$ matlab chordwise force actually forward point karti hai — yeh plausible hai jab leading edge ke paas ek suction peak section ko chord ke along pull karta hai (leading-edge thrust).

Verify (round-trip): $C_{N} = 0.9071, C_{A} = - 0.0380$ ko forward relations mein wapas plug karo: $C_{L} = 0.9071 (0.9848) - (- 0.0380) (0.1736) = 0.8933 + 0.0066 = 0.90$ ✓ aur $C_{D} = 0.9071 (0.1736) + (- 0.0380) (0.9848) = 0.1575 - 0.0374 = 0.120$ ✓. Forward aur inverse maps ek doosre ko undo karte hain.

Example 6 — Cell F: subsonic compressibility, limit tak push karna

Neeche wala figure Prandtl–Glauert factor plot karta hai: jaise $M_{\infty}$ 1 ki taraf badhta hai, $C_{L}$ ki red curve upar rocket karti hai. Woh runaway is example mein numeric banta hai.

Worked example Prandtl–Glauert from

$M_{\infty} = 0.3$ to $M_{\infty} = 0.85$ Ek thin airfoil ka incompressible $C_{L, incomp} = 0.50$ hai. Restated tool: $C_{L} = \frac{C _{L, incomp}}{1 - M _{\infty}^{2}}, β = 1 - M_{\infty}^{2}$ ( $β$ compressibility factor hai: subsonic air flow ko kitna "stiffen" karti hai.) $M_{\infty} = 0.3$ , $0.6$ , aur $0.85$ par $C_{L}$ nikalo.

Forecast: jaise $M_{\infty}$ 1 ki taraf badhta hai, kya lift slowly badhti hai ya runaway ho jaati hai?

$M_{\infty} = 0.3$ : $β = 1 - 0.09 = 0.91 = 0.9539$ , toh $C_{L} = 0.50/0.9539 = 0.5242$ . Yeh step kyun? Isliye hum low-speed flow ko "incompressible enough" kehte hain — sirf 5% correction.
$M_{\infty} = 0.6$ : $β = 1 - 0.36 = 0.64 = 0.80$ , toh $C_{L} = 0.50/0.80 = 0.625$ . Yeh step kyun? Hum $β$ ko higher Mach par recompute karte hain correction ko grow hote dekhne ke liye — factor $1/ β$ ab $1.25$ hai, solid 25% boost, ab negligible nahi.
$M_{\infty} = 0.85$ : $β = 1 - 0.7225 = 0.2775 = 0.5268$ , toh $C_{L} = 0.50/0.5268 = 0.9492$ . Yeh step kyun? $M_{\infty}$ ko validity ke edge tak push karne se denominator zero ki taraf collapse hota dikhta hai, isliye $C_{L}$ lagbhag double ho gaya — yeh woh runaway growth hai $M_{\infty} \to 1$ ke saath jo hum dekhna chahte the.

Verify (limiting behaviour): teeno answers $M_{\infty}$ ke saath monotonically increase karte hain, aur jump accelerate hota hai ( $+ 0.024$ , phir $+ 0.10$ , phir $+ 0.32$ ). ✓ Caution flag: $M_{\infty} = 0.85$ par hum parent note ki validity band ( $M_{\infty} ≲ 0.7$ ) se bahar hain; formula compute toh karta hai lekin real shocks aa chuke hote. Faithful math, physically stretched. Dekho Prandtl-Glauert Compressibility Correction.

Example 7 — Cell G: supersonic, new physics (wave drag)

Neeche wala figure ek axis par dono square-root factors contrast karta hai: Mach 1 se neeche factor $1/ 1 - M_{\infty}^{2}$ lift ko amplify karta hai, Mach 1 se upar factor $1/ M_{\infty}^{2} - 1$ use shrink karta hai, aur ek brand-new red wave-drag term appear hota hai.

Worked example Ackeret flat plate at

$M_{\infty} = 2$ Restated tool (Supersonic Linearized (Ackeret) Theory, 2D flat plate): $C_{L} = \frac{4 α}{M _{\infty}^{2} - 1}, C_{D, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1}$ $α$ radians mein hona chahiye. $M_{\infty} = 2$ , $α = 3^{\circ}$ lo. $C_{L}$ aur $C_{D, wave}$ nikalo, aur ratio $C_{D} / C_{L}$ bhi.

Forecast: notice karo ki square root $M_{\infty}^{2} - 1$ ho gaya. Mach 1 se upar, zyada $α$ kya drag linearly cost karta hai ya quadratically?

Radians mein convert karo: $α = 3^{\circ} \times π /180 = 0.05236$ rad. Yeh step kyun? Ackeret formulas $α$ small in radians ke saath derive kiye gaye hain; degrees se nonsense aata.
Denominator: $2^{2} - 1 = 3 = 1.7321$ . Yeh step kyun? Supersonically compressibility factor $M_{\infty}^{2} - 1$ ban jaata hai — subsonic jaisa wahi square root lekin root ke neeche sign flip hoke.
$C_{L} = 4 (0.05236) /1.7321 = 0.20944/1.7321 = 0.1209$ . Yeh step kyun? Radian angle aur denominator ko lift formula mein plug karne se actual supersonic lift milti hai — same $α$ par ek subsonic wing se choti, kyunki $M_{\infty}^{2} - 1$ factor ab upar ki jagah neeche divide karta hai.
$C_{D, wave} = 4 (0.05236)^{2} /1.7321 = 4 (0.002742) /1.7321 = 0.010967/1.7321 = 0.006332$ . Yeh step kyun? Yeh drag sirf supersonically exist karti hai — yeh shock losses se aati hai, friction se nahi. Subsonic inviscid flow mein zero wave drag hoti hai.
Ratio $C_{D} / C_{L} = 0.006332/0.1209 = 0.05236 = α$ (radians mein!). Kyun: dono formulas divide karne par $α$ ka ek factor chhodkar sab cancel ho jaata hai — ek elegant check.

Verify: ratio $C_{D, wave} / C_{L}$ exactly $α$ radians mein nikla (0.05236). Woh algebraic identity confirm karti hai ki dono numbers consistent hain. ✓ Aur wave drag $α$ mein quadratic hai (squared term), parent note ke us statement se match karta hai ki wave drag $\propto α^{2}$ . ✓

Example 8 — Cell H: real-world word problem

Worked example Ek flight-test report padhna

"Ek UAV $M_{\infty} = 0.6$ par cruise karta hai. Uska 2-D airfoil section cruise angle par incompressible lift coefficient $0.45$ produce karta hai. Usi waqt, sensors section force ko $C_{N} = 0.60$ body normal ke along aur $C_{A} = 0.05$ axial resolve karte hain, $α = 4^{\circ}$ ke saath. (a) Compressibility kaunsa lift coefficient deti hai? (b) Body-frame numbers ko wind-frame lift mein convert karke cross-check karo."

Forecast: do independent routes roughly same lift par land karni chahiye — har sentence ka cell pehchanno.

Part (a) Cell F hai. $β = 1 - 0.36 = 0.80$ , toh $C_{L} = 0.45/0.80 = 0.5625$ . Yeh step kyun? "Incompressible ... at $M_{\infty} = 0.6$ " phrase Prandtl–Glauert ka exact trigger hai.
Part (b) Cell A hai. $cos 4^{\circ} = 0.99756$ , $sin 4^{\circ} = 0.06976$ . Yeh step kyun? Part (b) seedha body-to-wind rotation hai, isliye forces project karne se pehle diye gaye $α = 4^{\circ}$ ka trig chahiye.
$C_{L} = 0.60 (0.99756) - 0.05 (0.06976) = 0.59854 - 0.00349 = 0.5951$ . Yeh step kyun? Body-frame reading measured, already-compressible force hai, isliye yahan koi PG factor apply nahi hota — yeh direct rotation hai.

Verify: dono routes $0.5625$ aur $0.5951$ dete hain — same ballpark (~6% ke andar), chota gap real 3-D aur measurement effects ki wajah se. Word problem decode ho gaya: sentence 1 → compressibility cell, sentence 2 → rotation cell. ✓

Example 9 — Cell I: exam twist (stability sign, $C_{m}$ )

Neeche wala figure dono wings ke liye $C_{m}$ versus $α$ plot karta hai. Red line neeche slope karti hai (stable: koi bhi nudge wapas push hoti hai); black line upar slope karti hai (unstable: koi bhi nudge badhti hai). Line ki height nahi, slope ka sign padho.

Worked example Kaunsa wing statically stable hai?

Do configurations apne center of gravity ke baare mein test ki gayi hain ( $α$ radians mein, $C_{m}$ = pitching-moment coefficient):

Wing P: $C_{m} = 0.04 - 0.90 α$
Wing Q: $C_{m} = - 0.02 + 0.30 α$

Kaunsa statically stable hai, aur kaunse $α$ par har ek trim karta hai (yaani $C_{m} = 0$ )?

Forecast: stability slope ke sign ke baare mein hai, $C_{m}$ ki value ke baare mein nahi. Compute karne se pehle guess karo kaunsa wing kaisa hai.

Har line se slope $d C_{m} / d α$ padho: Wing P $\to - 0.90$ , Wing Q $\to + 0.30$ . Yeh step kyun? $d C_{m} / d α < 0$ stability condition hai (parent note): ek chota pitch-up nose-down restoring moment generate karna chahiye. $α$ multiply karne wala coefficient wahi slope hai, toh hum ise figure ki line steepness se directly padhte hain.
Wing P: slope $- 0.90 < 0$ → statically stable. Wing Q: slope $+ 0.30 > 0$ → unstable (equilibrium se aur door pitch karta hai). Dekho Static Longitudinal Stability. Yeh step kyun? Hum har slope ko zero ke stability threshold se compare karte hain — yahi "twist" hai: answer ek sign par hinge karta hai, magnitude par nahi.
Wing P ka trim angle ( $C_{m} = 0$ set karo): $0 = 0.04 - 0.90 α \Rightarrow α = 0.04/0.90 = 0.04444$ rad $= 2.54 6^{\circ}$ . Yeh step kyun? Trim wahan hai jahan net pitching moment vanish hoti hai — woh equilibrium jahan aircraft actually fly karega. Hum $α$ ke liye linear equation solve karte hain jo $C_{m}$ zero karta hai; figure mein yeh wahan hai jahan red line zero cross karti hai.
Wing Q ka trim angle ( $C_{m} = 0$ set karo): $0 = - 0.02 + 0.30 α \Rightarrow α = 0.02/0.30 = 0.06667$ rad $= 3.82 0^{\circ}$ . Yeh step kyun? Wing Q ke liye same algebra — iska bhi ek trim point hai (black line ki zero-crossing), lekin jaise step 2 ne dikhaya yeh hold nahi kar sakta, isliye trim angle ek paper equilibrium hai jisse aircraft diverge ho jaata.

Answer: Wing P statically stable hai (negative slope) aur $α = 0.0444$ rad ( $2.5 5^{\circ}$ ) par trim karta hai; Wing Q unstable hai (positive slope) aur, halanki $α = 0.0667$ rad ( $3.8 2^{\circ}$ ) par trim karta hai, woh equilibrium maintain nahi kar sakta.

Verify: Wing P ko $α = 0.05$ par nudge karo (uske trim $0.0444$ se thoda upar): $C_{m} = 0.04 - 0.90 (0.05) = - 0.005 < 0$ → nose-down, trim ki taraf wapas push. ✓ Ab Wing Q ko uske trim se upar nudge karo, $α = 0.08 > 0.0667$ par: $C_{m} = - 0.02 + 0.30 (0.08) = + 0.004 > 0$ → nose-up, $α$ ko aur upar push karta hai, trim se door → instability confirm hoti hai. ✓ Twist: identical-looking algebra, opposite physics, poori tarah ek sign se decide. Dekho Aerodynamic Center vs Center of Pressure.

Recall Yeh kaunsa cell hai? (self-test)

"Given $C_{L} = 0.7$ , $C_{D} = 0.1$ , $α = 1 2^{\circ}$ , $C_{N}$ nikalo." ::: Cell E (inverse rotation, wind → body). " $C_{L} = 0.6$ incompressible, $M_{\infty} = 0.7$ , compressible $C_{L}$ nikalo." ::: Cell F (Prandtl–Glauert). " $M_{\infty} = 2.5$ flat plate, wave drag nikalo." ::: Cell G (Ackeret, new physics). " $d C_{m} / d α = + 0.2$ — stable?" ::: Cell I: nahi, positive slope unstable hai.