Aerodynamic coefficients — CA (axial force), CN (normal force), Cm (pitching moment): Worked examples

3.4.7 · D3 · Physics › Rocket Flight Mechanics › Aerodynamic coefficients — CA (axial force), CN (normal forc

Shuru karne se pehle, ek vaada: neeche har symbol parent note mein earn kiya gaya tha. Safe rehne ke liye, ye sab ek jagah hain.

Definition Is page ke har symbol (expand karne ke liye tap karo)

$q = \frac{1}{2} ρ V^{2}$ — dynamic pressure, wo "push per unit area" jo hawa carry karti hai. Units: pascals ( $Pa = N/ m^{2}$ ). Dekho Dynamic pressure and Bernoulli.
$ρ$ — air density (kg/m³). $V$ — flight speed (m/s).
$S$ — reference area (m²), rocket ke liye choose kiya gaya ek fixed number (aksar body tube ka cross-section).
$d$ — reference length (m), usually body diameter — sirf moments ke liye chahiye.
$A$ — axial force: hawa sidha nose–tail line ke along peeche dhakelta hai.
$N$ — normal force: hawa sideways dhakelta hai, body ke perpendicular.
$M$ — pitching moment: hawa rocket ko nose-up (+) ya nose-down (−) twist karta hai.
$α$ — angle of attack: nose kitna door point kar raha hai us jagah se jahan rocket actually ja raha hai.
$D$ — drag: force flight path ke along, velocity ko oppose karta hai. $L$ — lift: force flight path ke perpendicular. Ye wind frame mein rehte hain (dekho Drag and Lift in wind axes), jabki $A$ aur $N$ body frame mein rehte hain.
$C_{A} = \frac{A}{q S}$ , $C_{N} = \frac{N}{q S}$ , $C_{m} = \frac{M}{q S d}$ , $C_{D} = \frac{D}{q S}$ , $C_{L} = \frac{L}{q S}$ — pure-number "shape factors".
$C_{N α} = \frac{\partial C _{N}}{\partial α}_{0}$ — normal-force slope, per radian. Dekho Fin design and normal-force slope $C_{N\alpha}$.
$C_{m α} = \frac{\partial C _{m}}{\partial α}_{0}$ — moment slope (per radian): twisting coefficient kitni tezi se badhta hai jab nose tilt hota hai. Iska sign stability decide karta hai (negative = restoring = stable). Dekho Static and dynamic stability of rockets.
static margin $= \frac{x _{c p} - x _{c g}}{d}$ — centre of pressure aur centre of gravity ke beech gap diameters mein.
$x_{c p}$ , $x_{c g}$ — nose se centre of pressure aur centre of gravity ki distances (m).
Mach number $Ma = \frac{V}{a}$ — flight speed divided by local speed of sound $a$ . "Mach 2" ka matlab hai $V = 2 a$ . Ye wo number hai jo model aur full-size rocket ke beech match karna chahiye taaki coefficients transfer ho sakein. Dekho Reynolds and Mach scaling.

Body forces se wind forces tak (wo rotation jo hum baar baar use karte rahenge)

Neeche do examples (Ex 1 aur Ex 6) drag $C_{D}$ aur lift $C_{L}$ maangti hain, body forces $C_{A}$ aur $C_{N}$ nahi. Toh pehle hume do frames ke beech ka bridge derive karna hoga — use karne se pehle banao, kabhi bhi andar nahi.

Figure — Aerodynamic coefficients — CA (axial force), CN (normal force), Cm (pitching moment)

Intuition DO frames alag kyun hain, aur unhe ek rotation kyun connect karta hai

Figure dekho. Body axis (chalk-white) nose–tail line ke along point karta hai. Wind axis (blue) flight velocity $V$ ke along point karta hai. Jab rocket angle of attack $α$ par fly karta hai, ye do directions exactly $α$ se misaligned hote hain.

Axial force $A$ body ke along backward point karta hai. Wind direction par uska shadow (projection) $A cos α$ hai — wo part $V$ ko oppose karta hai, isliye ye drag hai.
Normal force $N$ body ke perpendicular point karta hai. Body ko $α$ se upar tilt karo aur $N$ ka ek part, yaani $N sin α$ , ab wind ke along backward bhi point karta hai (figure mein pink dashed drop) — extra drag. Do drag contributions ko add karo, aur lift ke liye wind ke perpendicular same bookkeeping karo:

$D = A cos α + N sin α, L = N cos α - A sin α .$

Lift ko $+ N cos α$ milta hai ( $N$ ka zyada hissa) lekin $- A sin α$ bhi ( $a x ia l f or ce t h o d a l i f t d i r ec t i o nk e * a g ain s t * j h u k t ahai) . H a r t er mk o$ qS$ se divide karne par forces coefficients ban jaate hain, jo do formulas deta hai jinhe hum baar baar use karenge:

Pitching-moment law (doosra tool jo hum baar baar use karenge)

Kuch examples neeche moment coefficient $C_{m}$ chahti hain, aur jaise humne drag/lift rotation use karne se pehle derive ki, usi tarah hum yahan moment law banate hain — kabhi bhi example ke andar nahi.

Scenario matrix

Is topic ka har problem in cells mein se ek hai. Neeche ke examples [Cell n] se tagged hain taaki tum dekh sako ki poori grid cover hai.

#	Cell class	Isme tricky kya hai	Covered by
1	Force from coefficient, $α = 0$	pure plug-in; $C_{D} = C_{A}$ yahan	Ex 1
2	Positive $α$ ( $α > 0$ )	$C_{N} > 0$ , nose-up angle	Ex 2
3	Negative $α$ ( $α < 0$ )	har sign flip hota hai — track karna padega	Ex 3
4	Zero / degenerate: symmetric rocket at $α = 0$ , ya CP = CG	$C_{N} = 0$ , $C_{m} = 0$ , neutral stability	Ex 4
5	Unstable rocket ( $C_{m α} > 0$ )	wrong-sign moment, diverge karta hai	Ex 5
6	Large $α$ (body vs wind axes)	$cos α, sin α$ ab $\approx 1, α$ nahi raha	Ex 6
7	Real-world word problem	tumhe khud $q$ , $S$ , $α$ pick out karne honge	Ex 7
8	Exam twist: static margin ke liye backwards solve karo	moment equation rearrange karo	Ex 8

Hum ek "base rocket" reuse karenge taaki numbers comparable rahein: $ρ = 0.9 kg/ m^{3}, V = 300 m/s, S = 0.05 m^{2}, d = 0.25 m .$ Toh uska dynamic pressure hamesha hai $q = \frac{1}{2} (0.9) (300)^{2} = 40, 500 Pa .$

Example 1 — Zero angle par Plug-in · [Cell 1]

Worked example Bilkul seedha fly karte hue axial force

Base rocket ka $C_{A} = 0.35$ hai $α = 0$ par. Axial force $A$ nikalo, aur drag coefficient $C_{D}$ batao.

Forecast: andaza lagao ki $A$ 100 N, 700 N, ya 7000 N ke kareeb hai — aur kya $C_{D}$ , $C_{A}$ ke equal hai yahan.

Step 1. $q = 40, 500 Pa$ (base numbers se). Ye step kyun? Coefficient akela Newtons mein kuch nahi batata; $q$ mein saari speed-and-density scaling hoti hai. Pehle ise compute karo, hamesha.

Step 2. $A = C_{A} q S = 0.35 \times 40, 500 \times 0.05 = 708.75 N$ . Ye step kyun? Bas $C_{A} = A / (q S)$ ko rearrange kiya. Shape number ko physical scale $q S$ par wapas multiply karo.

Step 3. $α = 0$ par: $C_{D} = C_{A} cos 0 + C_{N} sin 0 = C_{A} \cdot 1 + 0 = C_{A} = 0.35$ . Ye step kyun? Upar derive ki gayi rotation use karo. Body axis aur wind axis coincide karte hain jab nose exactly wahaan point karta hai jahan rocket fly karta hai, isliye "body ke along" hi "wind ke along" hai. Ye sirf wahi case hai jahan $C_{D} = C_{A}$ .

Verify: units — $[Pa] \cdot [m^{2}] = N/ m^{2} \cdot m^{2} = N$ . ✔ Magnitude hundreds of newtons hai, jo "700 N ke kareeb" wale guess se match karta hai.

Example 2 — Positive angle of attack · [Cell 2]

Worked example Nose-up angle se Normal force

Base rocket, $C_{N α} = 2.0 ra d^{- 1}$ , angle of attack $α = + 4^{\circ}$ . $C_{N}$ aur normal force $N$ nikalo.

Forecast: kya $N$ , Ex 1 ke axial force se bada ya chhota hoga?

Step 1. Convert karo: $α = 4^{\circ} \times \frac{π}{180} = 0.06981 rad$ . Ye step kyun? Slope $C_{N α}$ per radian defined hai. Degrees feed karne par 57 ka factor blunder dega.

Step 2. $C_{N} = C_{N α} α = 2.0 \times 0.06981 = 0.13963$ . Ye step kyun? Chhote $α$ ke liye, $C_{N}$ linearly badhta hai — ye linearized rule $C_{N} \approx C_{N α} α$ hai.

Step 3. $N = C_{N} q S = 0.13963 \times 40, 500 \times 0.05 = 282.72 N$ . Ye step kyun? Same $q S$ jaise Ex 1 mein — coefficients ki yahi khoobsurti hai, har force ke liye ek scale factor.

Verify: $N = 282.72 N$ , $A = 708.75 N$ se thoda chhota hai. Sensible — gentle $4^{\circ}$ tilt par sideways dhakka real hai par head-on push se chhota. Sign check: $α > 0 \Rightarrow C_{N} > 0$ (nose-up se "upward" normal force milta hai). ✔

Example 3 — Negative angle of attack · [Cell 3]

Worked example Har sign flip hota hai

Same rocket aur slope, lekin ab $α = - 4^{\circ}$ (nose flight path ke neeche point karta hai). $C_{N}$ , $N$ , aur pitching-moment sign nikalo, static margin $= 1.5$ ke saath.

Forecast: compute karne se pehle $C_{N}$ aur $C_{m}$ ka sign predict karo.

Step 1. $α = - 0.06981 rad$ . Ye step kyun? Convert karo, minus sign rakhte hue — tilt ki direction puri tarah us sign mein rehti hai.

Step 2. $C_{N} = 2.0 \times (- 0.06981) = - 0.13963$ , isliye $N = - 282.72 N$ . Ye step kyun? Linearity sign ko seedha carry karta hai: nose-down ⇒ negative normal force (hawa ab nose region ko doosri taraf push karti hai).

Step 3. $C_{m} = - C_{N} \times margin = - (- 0.13963) (1.5) = + 0.20944$ . Ye step kyun? Upar banaya gaya pitching-moment law $C_{m} = - C_{N} \cdot (static margin)$ use karo. Negative $C_{N}$ se $C_{m}$ positive banta hai — ek nose-up twist.

Verify: kya ye restore karta hai? Nose ko neeche push kiya gaya tha ( $α < 0$ ); moment positive = nose-up nikla, use zero ki taraf wapas dhakelta hai. ✔ Ek stable rocket ko yahi karna chahiye — restoring twist hamesha disturbance ko oppose karta hai. Parent ke Ex 3 se compare karo (positive $α$ ne negative $C_{m}$ diya): mirror image, jaisa hona chahiye.

Example 4 — Zero aur degenerate cases · [Cell 4]

Worked example Symmetric rocket, aur neutral rocket

(a) Ek fin-symmetric rocket exactly $α = 0$ par fly karta hai. $C_{N}$ aur $C_{m}$ kya hain? (b) Ek rocket ka centre of pressure exactly uske centre of gravity par hai, $x_{c p} = x_{c g}$ , $α = + 4^{\circ}$ par. $C_{m}$ kya hai? Kya ye stable hai?

Forecast: padhne se pehle dono moments guess karo.

Step 1 (a). Symmetry se, $α = 0$ par koi sideways preference nahi: $C_{N} = C_{N α} \cdot 0 = 0$ , hence $N = 0$ aur $C_{m} = - C_{N} (\dots) = 0$ . Ye step kyun? Seedha fly karta hua ek symmetric body kisi bhi particular sideways direction mein push ya twist karne ki koi wajah nahi rakhta — dono sides cancel ho jaate hain. Sirf $C_{A}$ bachta hai.

Step 2 (b). static margin $= \frac{x _{c p} - x _{c g}}{d} = \frac{0}{d} = 0$ . Tab $C_{m} = - C_{N} \times 0 = 0$ bhi, chahe $C_{N} = 0.13963 \neq = 0$ ho. Ye step kyun? Lever arm wo distance hai CG se jahan force act karta hai. Zero distance ⇒ zero torque, force chahे kitna bhi bada ho.

Verify: (a) mein dono zero hain kyunki cause ( $α$ ) zero hai. (b) mein $C_{m} = 0$ hai kyunki lever arm zero hai. "No twist" tak do alag raaste — aur case (b) neutrally stable hai: ise dhakelo aur ye na wapas aata hai na diverge karta hai, kyunki $C_{m α} = C_{m} / α = 0$ . Ye stable Ex 3 aur unstable Ex 5 ke beech exact borderline hai. ✔

Example 5 — Unstable rocket · [Cell 5]

Worked example CG, CP ke peeche — ek tumbler

Ek rocket ka CG uske CP ke peeche rakha gaya hai: static margin $= - 0.8$ (negative!). $α = + 4^{\circ}$ par, $C_{N} = 0.13963$ . $C_{m}$ aur $C_{m α}$ nikalo, aur stability decide karo.

Forecast: nose-up disturbance ke liye, kya moment nose ko aur upar dhakela ga (bura) ya wapas neeche (achha)?

Step 1. $C_{m} = - C_{N} \times margin = - (0.13963) (- 0.8) = + 0.11170$ . Ye step kyun? Same moment law; negative margin $C_{m}$ ka usual sign flip kar deta hai.

Step 2. $C_{m α} = \frac{C _{m}}{α} = \frac{0.11170}{0.06981} = + 1.6 ra d^{- 1}$ . Ye step kyun? Stability ka faisla puri tarah moment slope ke sign par hota hai, uske size par nahi.

Verify: $α > 0$ (nose up) ne $C_{m} > 0$ diya (nose up twist) — disturbance amplify ho raha hai. Kyunki $C_{m α} = + 1.6 > 0$ , rocket statically unstable hai aur tumble karta hai. Ye Ex 3 ki mirror-warning hai: positive margin restore karta hai, negative margin diverge karta hai, zero margin (Ex 4b) fence par baitha hai. Dekho Static and dynamic stability of rockets. ✔

Example 6 — Bada angle: body axes vs wind axes matter karte hain · [Cell 6]

Step 1. $cos 3 0^{\circ} = 0.86603$ , $sin 3 0^{\circ} = 0.5$ . Ye step kyun? $3 0^{\circ}$ par small-angle shortcuts ( $cos \approx 1$ , $sin \approx α$ ) das percent se zyada galat hain — humhe real trig chahiye. Figure body axis ko wind axis se $3 0^{\circ}$ tilted dikha raha hai; wind axis par projections exactly ye cosines aur sines hain.

Step 2. $C_{D} = C_{A} cos α + C_{N} sin α = 0.35 (0.86603) + 0.90 (0.5) = 0.30311 + 0.45 = 0.75311$ . Ye step kyun? Drag wind ke along measure hota hai, isliye hum upar derive ki gayi body-to-wind rotation apply karte hain. Axial force wind par $cos α$ se project karta hai; normal force, ek baar body tilt karo, drag direction mein bhi ek component $sin α$ spill karta hai (figure mein pink arrow dekho).

Step 3. $C_{L} = C_{N} cos α - C_{A} sin α = 0.90 (0.86603) - 0.35 (0.5) = 0.77942 - 0.175 = 0.60442$ . Ye step kyun? Lift wind ke perpendicular hai. Yahan normal force $cos α$ se project karta hai, lekin axial force $A sin α$ subtract karta hai (ye thoda lift direction ke against jhukta hai).

Verify: $C_{D} = 0.753$ , $C_{A} = 0.35$ se double se bhi zyada hai — isliye " $C_{D} = C_{A}$ " $3 0^{\circ}$ par bilkul galat hai; ye sirf $α = 0$ par hold karta hai (Ex 1). Consistency check: total force magnitude frame-independent honi chahiye. $C_{A}^{2} + C_{N}^{2} = 0.3 5^{2} + 0.9 0^{2} = 0.9656$ aur $C_{D}^{2} + C_{L}^{2} = 0.75 3^{2} + 0.60 4^{2} = 0.9656$ — identical, kyunki axes rotate karne se vector ki length kabhi nahi badlti. ✔ Dekho Drag and Lift in wind axes.

Example 7 — Real-world word problem · [Cell 7]

Worked example Wind-tunnel model ko full scale par transfer karna

Mach 2 par ek model rocket ka wind-tunnel test $C_{A} = 0.40$ measure karta hai. Full-size rocket same Mach 2 par altitude par fly karta hai jahan $ρ = 0.30 kg/ m^{3}$ aur local speed of sound $a = 340 m/s$ hai; uska reference area $S = 0.20 m^{2}$ hai. Full-size rocket ko kitna axial force feel hota hai?

Forecast: kya model ka $C_{A}$ bade rocket par unchanged apply hota hai? (Haan — isliye coefficients exist karte hain. Force kN mein guess karo.)

Step 1. Mach number se speed recover karo: $Ma = V / a$ , isliye $V = Ma \times a = 2 \times 340 = 680 m/s$ . Ye step kyun? Problem speed sirf "Mach 2" ke roop mein deti hai, jiska definition hai $V = 2 a$ . $q$ banane se pehle ise m/s mein turn karna hoga. Model aur full scale ke beech Mach number match karna hi allow karta hai model ka $C_{A}$ borrow karna (dekho Reynolds and Mach scaling).

Step 2. $q = \frac{1}{2} ρ V^{2} = 0.5 (0.30) (680)^{2} = 0.5 \times 0.30 \times 462, 400 = 69, 360 Pa$ . Ye step kyun? $q$ ke liye same recipe; sirf $ρ$ aur $V$ local flight values hain, wind tunnel ke nahi. $q$ saari scaling carry karta hai jo model aur full scale ke beech differ karti hai.

Step 3. $A = C_{A} q S = 0.40 \times 69, 360 \times 0.20 = 5548.8 N \approx 5.5 kN$ . Ye step kyun? Humne chhote model se dimensionless $C_{A}$ borrow kiya lekin full-size $q$ aur $S$ apply kiye. Yahi coefficients ka point hai — shape number transfer hota hai, scale factor local hai.

Verify: units newtons dete hain ✔; magnitude ~5.5 kN ek launch vehicle ke liye high dynamic pressure par reasonable hai. Agar hum (galti se) tunnel ka chhota $q$ ya model ka chhota $S$ reuse karte, toh ek hassi waala chhota force milta — transfer sirf coefficient ke through kaam karta hai, kabhi raw newtons se nahi. ✔

Example 8 — Exam twist: static margin ke liye backwards solve karo · [Cell 8]

Worked example Measured moment diya hai, pata karo CP kahan hai

Base rocket ( $q = 40, 500$ , $S = 0.05$ , $d = 0.25$ ) par flight telemetry $α = + 4^{\circ}$ par pitching moment $M = - 70 N m$ record karti hai, $C_{N α} = 2.0 ra d^{- 1}$ ke saath. Static margin nikalo aur batao ki CP CG se kitne diameters peeche hai.

Forecast: moment nose-up $α$ par negative hai — compute karne se pehle, kya ye rocket stable hai?

Step 1. $C_{m} = \frac{M}{q S d} = \frac{- 70}{40 , 500 \times 0.05 \times 0.25} = \frac{- 70}{506.25} = - 0.13827$ . Ye step kyun? Moment definition $C_{m} = M / (q S d)$ undo karo dimensionless $C_{m}$ get karne ke liye — hume coefficients compare karne hain, raw newton-metres nahi.

Step 2. $C_{N} = C_{N α} α = 2.0 \times 0.06981 = 0.13963$ (jaise Ex 2 mein). Ye step kyun? Margin formula $C_{m} = - C_{N} \cdot margin$ ko $C_{N}$ chahiye; hum ise angle se build karte hain.

Step 3. Geometry ke liye moment law rearrange karo: $margin = - \frac{C _{m}}{C _{N}} = - \frac{- 0.13827}{0.13963} = 0.99026 \approx 0.99$ diameters. Ye step kyun? $C_{m} = - C_{N} \cdot (static margin)$ ko moment ke bajaye margin ke liye solve karna — exam ka favourite inversion.

Verify: margin $\approx + 0.99 > 0$ ⇒ CP lagbhag ek diameter CG ke peeche hai ⇒ statically stable, nose-up $α$ par negative moment se match karta hai. ✔ Sanity: wapas plug karo, $C_{m} = - 0.13963 \times 0.99026 = - 0.13827$ ✔, aur $M = C_{m} q S d = - 0.13827 \times 506.25 = - 70.0 N m$ ✔. Toh ek stable rocket ko nose-up $α$ par zaroor ek negative moment dikhana chahiye, exactly jaisa record hua.

Recall

Recall Kaun sa cell poora

$cos α, sin α$ use karta hai aur kyun? Large-angle problems (Cell 6). ~ $5^{\circ}$ se neeche tum $cos \approx 1, sin \approx α$ use kar sakte ho, lekin $3 0^{\circ}$ par body-to-wind rotation $C_{D}$ ko $0.35$ se $0.75$ badal deta hai — real trig se project karna padega. ::: Cell 6.

Ek stable rocket ke liye $C_{m α}$ ka sign hai ::: negative ( $C_{m α} < 0$ ): ek nose-up disturbance ko ek nose-down (negative) restoring moment produce karna chahiye.

Zero moment do tarahon se aa sakta hai — unhe name karo ::: ya toh cause zero hai ( $α = 0 \Rightarrow C_{N} = 0$ ) ya lever arm zero hai (CP = CG ⇒ margin $= 0$ ).

Ex 7 mein model ka $C_{A}$ full-size rocket par transfer kyun hota hai? ::: Kyunki $C_{A}$ dimensionless hai aur dono same Mach number par fly karte hain; sirf $q$ aur $S$ (local scale factors) differ karte hain.