Dynamic stability — pitch - yaw damping derivatives: Worked examples

3.4.11 · D3 · Physics › Rocket Flight Mechanics › Dynamic stability — pitch - yaw damping derivatives

Shuru karne se pehle, ek line ka reminder har us symbol ka jo hum use karte hain (har ek ko parent-linked topic mein build kiya gaya tha — Static Stability — Center of Pressure & Margin aur Barrowman Equations aero background dete hain):

Definition Symbol dictionary (plain words mein)

$x$ ::: distance peeche (aft of) center of gravity se measure ki gayi, metres mein. Positive = CG ke peeche.
$α$ (alpha) ::: angle of attack — body axis aur aane wali hawa ke beech ka angle (upar define kiya, Figure s03), radians mein.
$q$ ::: pitch rate — kitni tezi se naak upar ya neeche swing kar rahi hai, radians per second mein.
$V$ ::: body axis ke along flight speed, m/s.
$ρ$ ::: air density, kg/m³ (altitude ke saath girta hai — dekho Atmospheric Density vs Altitude).
$S$ ::: ek fixed reference area (usually body cross-section), m². Bas ek bookkeeping constant hai.
$d$ ::: reference length (body diameter), m.
$c_{n_{α}} A$ ::: "kitni sideways force per unit angle of attack" ek surface banata hai, m²/rad. Bade fins → bada number.
$C_{m}$ ::: pitching-moment coefficient — naak-upar/naak-neeche torque ko $\frac{1}{2} ρ V^{2} S d$ se dimensionless banaya gaya. Yeh "torque, pure-number form mein" hai taaki alag-alag rockets compare ho sakein.
$\overset{q}{^}$ ::: non-dimensional pitch rate, $\overset{q}{^} = q d / (2 V)$ — raw rate $q$ ko us time $d / (2 V)$ se scale kiya gaya jitne mein hawa body cross karti hai. Yeh "rad/s" ko pure number mein badal deta hai taaki $C_{m_{q}}$ bhi pure number ho sake.
$C_{m_{q}}$ ::: pitch-damping derivative, $C_{m_{q}} = \partial C_{m} / \partial \overset{q}{^}$ — moment coefficient kitna badalta hai non-dimensional pitch rate ke har unit par. Ek pure number jo batata hai rocket pitch rate ko kitni strongly resist karta hai.
$C_{m_{α}}$ ::: static-stability derivative, $C_{m_{α}} = \partial C_{m} / \partial α$ — moment coefficient kitna badalta hai angle of attack $α$ ke har unit par. Yeh spring hai (wind ki taraf restore karta hai); static stability ke liye yeh negative hona chahiye (dekho Static Stability — Center of Pressure & Margin). Ise damping derivative $C_{m_{q}}$ se confuse mat karo: ek angle pe react karta hai, doosra rate pe.
$I_{y}$ ::: pitch axis ke baare mein moment of inertia (dekho Moments of Inertia of a Rocket), kg·m².
$ζ$ (zeta) ::: damping ratio — 0 matlab hamesha ring karta rahega, ≥1 matlab bilkul oscillation nahi.

Do master formulas jinhe hum hammer karenge (dono parent se):

Scenario matrix

Is topic ka har problem in cells mein se ek (ya blend) hota hai. Har row ek "tarah ke input" ka naam deta hai aur physical question jo woh test karta hai.

#	Cell class	Input mein kya special hai	Worked in
A	Baseline sign	ordinary fins CG ke peeche, $x_{f} > 0$	Ex 1
B	$x^{2}$ scaling	same fins, distance change kiya	Ex 2
C	Distributed area	area ek length par spread, integrate karna padega	Ex 3
D	Degenerate: zero rate	$q = 0$ — kya koi damping moment hai?	Ex 4
E	Sign flip (danger)	fins CG ke aage, static instability	Ex 5
F	Limiting: altitude	$ρ \to$ chhota, damping fade hoti hai	Ex 6
G	Real-world word problem	geometry + mass se poora $ζ$	Ex 7
H	Exam twist	do fin sets, superposition, kaunsa dominate karta hai	Ex 8
I	Straddling the CG	$x < 0$ AND $x > 0$ ke saath loading, sign-aware integral	Ex 9

Neeche diye figures un cases ke liye geometry carry karte hain jahan picture argument karta hai. Figure s01 Cell A/B illustrate karta hai ( $x^{2}$ weighting jo Ex 1–3 mein use hoti hai); Figure s02 Cell E illustrate karta hai (Ex 5); Figure s03 angle of attack $α$ define karta hai jo throughout use hota hai.

Figure — Dynamic stability — pitch - yaw damping derivatives

Figure s01 mein red bar sabse outer surface hai $x = 2.0$ m par: uska $x^{2}$ contribution near stations par tower karta hai. Ise dekho ki kyun fins ko peeche le jaana (Ex 2) damping quadruple karta hai aur kyun distributed load ka far end (Ex 3) dominate karta hai.

Figure s03 symbol $α$ ke peeche ki picture hai: red arrow aane wali hawa hai, body axis se $α$ se tilt ki hui.

Worked examples

Worked example Ex 1 — Cell A: baseline sign & magnitude

Fins $x_{f} = 0.6$ m CG ke peeche lumped, $c_{n_{α}} A_{f} = 8$ m²/rad, $S = 0.008$ m², $d = 0.1$ m. $C_{m_{q}}$ find karo.

Forecast: pehle sign guess karo — $+$ hoga ya $-$ ? Aur roughly, tens ya tens-of-thousands?

Lumped formula likho: $C_{m_{q}} = - \frac{2}{S d ^{2}} (c_{n_{α}} A_{f}) x_{f}^{2}$ . Yeh step kyun? Saari area ek station par hai, toh integral value × $x_{f}^{2}$ mein collapse ho jaata hai. Figure s01 mein red bar dekho — yeh woh ek dominant station hai.
Denominator compute karo $S d^{2} = 0.008 \times (0.1)^{2} = 8 \times 1 0^{- 5}$ . Yeh step kyun? Yeh fixed non-dimensionalising constant hai; ise ek baar get karo.
Numerator: $2 \times 8 \times 0.36 = 5.76$ . Toh $C_{m_{q}} = - 5.76/ (8 \times 1 0^{- 5}) = - 7.2 \times 1 0^{4}$ . Yeh step kyun? $x_{f}^{2} = 0.36$ ; minus physics se aata hai — force rotation ko oppose karta hai.

Verify: sign negative hai ✔ (damping, as required). Units: $C_{m_{q}}$ dimensionless hai — check: $\frac{m ^{2}}{m ^{2} \cdot m ^{2}} \cdot m^{2} = 1$ ✔.

Worked example Ex 3 — Cell C: distributed area (integrate karna zaroori)

Ek point ki jagah, aerodynamic area uniformly aft body par $x = 0.4$ m se $x = 1.0$ m tak spread hai, constant per-length loading $c_{n_{α}} \frac{d A}{d x} = k = 5$ m/rad ke saath. Same $S, d$ . $C_{m_{q}}$ find karo.

Forecast: far end dominate karna chahiye — kya answer near limit ya far limit ke contribution ke kareeb baithega?

Set up karo: $C_{m_{q}} = - \frac{2}{S d ^{2}} \int_{0.4}^{1.0} k x^{2} d x$ . Yeh step kyun? Area ab ek station par nahi hai, toh $x^{2}$ -weighted integral properly karna padega (yahi Barrowman-style integration hai jo Barrowman Equations mein hai). Figure s01 ke bars exactly woh $x^{2}$ weights hain jo sum ho rahe hain.
Integrate karo: $\int_{0.4}^{1.0} x^{2} d x = \frac{x ^{3}}{3}_{0.4}^{1.0} = \frac{1.0 - 0.064}{3} = \frac{0.936}{3} = 0.312$ . Yeh step kyun? $\int x^{2} = x^{3} /3$ ; $x^{3}$ outer end (1.0 m) ko inner (0.4 m) se kaafi zyada contribute karwata hai.
Toh $C_{m_{q}} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 5 \times 0.312 = - \frac{3.12}{8 \times 1 0 ^{- 5}} = - 3.9 \times 1 0^{4}$ . Yeh step kyun? Constant, loading $k$ , aur geometric integral assemble karo.

Verify: sanity bound ke roop mein, agar saari area far end $x = 1.0$ par hoti: total $c_{n_{α}} A = k \times 0.6 = 3$ , deta $- \frac{2}{8 \times 1 0 ^{- 5}} \times 3 \times 1 = - 7.5 \times 1 0^{4}$ ; near end $x = 0.4$ par: $- \frac{2}{8 \times 1 0 ^{- 5}} \times 3 \times 0.16 = - 1.2 \times 1 0^{4}$ . Hamara $- 3.9 \times 1 0^{4}$ beech mein hai ✔ aur far end ki taraf lean karta hai jaise expected ✔.

Worked example Ex 4 — Cell D: degenerate zero pitch rate

Rocket $q = 0$ ke saath fly kar raha hai (abhi rotate nahi kar raha). Abhi damping moment kya hai, aur $C_{m_{q}}$ kya hai?

Forecast: kya $C_{m_{q}}$ zero hai kyunki kuch rotate nahi ho raha? Padhne se pehle guess karo.

Physical damping moment, $C_{m} = M / (\frac{1}{2} ρ V^{2} S d)$ aur $\overset{q}{^} = q d / (2 V)$ ke consistent, $M_{damp} = \frac{1}{2} ρ V^{2} S d C_{m_{q}} \overset{q}{^} = \frac{1}{4} ρ V S d^{2} C_{m_{q}} q .$ $q = 0$ par, $\overset{q}{^} = 0$ , toh $M_{damp} = 0$ . Yeh step kyun? Damping ek rate effect hai — koi rate nahi, koi damping torque nahi. Jaise ek shock absorber jo kuch push nahi karta jab wheel move nahi kar rahi. (Note karo ki $\overset{q}{^}$ ke neeche $V^{2}$ ek single $V$ mein collapse ho jaata hai jab $\overset{q}{^} = q d / (2 V)$ substitute karte hain — yeh exactly $ζ$ formula ke $\frac{1}{4} ρ V S d^{2} C_{m_{q}}$ factor se match karta hai.)
Lekin $C_{m_{q}} = \partial C_{m} / \partial \overset{q}{^}$ moment-coefficient curve ka ek slope hai, value nahi. Origin par line ka slope kahin bhi same hai. Toh $C_{m_{q}}$ unchanged hai (e.g. Ex 1 se abhi bhi $- 7.2 \times 1 0^{4}$ ). Yeh step kyun? Derivative poori curve ki property hai, current operating point se independent.

Verify: moment $\propto q$ , aur $q = 0 \Rightarrow M = 0$ ✔; ek straight line ka derivative constant hai, toh $C_{m_{q}} \neq = 0$ ✔. $M_{damp} = \frac{1}{4} ρ V S d^{2} C_{m_{q}} q$ par units check: $\frac{kg}{m ^{3}} \cdot \frac{m}{s} \cdot m^{2} \cdot m^{2} \cdot \frac{1}{s} = \frac{kg \cdot m ^{2}}{s ^{2}} = N\cdotm$ ✔ (ek torque). Do quantities ( $M$ aur $C_{m_{q}}$ ) different objects hain — yeh ek classic trap hai.

Worked example Ex 5 — Cell E: sign flip, fins CG ke aage (danger case)

Ek build error fins ko CG ke aage $x_{f} = - 0.5$ m par rakh deta hai. $c_{n_{α}} A_{f} = 8$ , same $S, d$ lo. $C_{m_{q}}$ kya hai, aur kya rocket safe hai?

Forecast: formula mein $x_{f}^{2}$ hai. Kya negative $x_{f}$ $C_{m_{q}}$ ka sign change karta hai?

$C_{m_{q}} = - \frac{2}{S d ^{2}} (c_{n_{α}} A_{f}) x_{f}^{2} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 8 \times (- 0.5)^{2}$ . Yeh step kyun? $x_{f}$ squared hai, toh $(- 0.5)^{2} = 0.25$ — damping term negative rahta hai chahe fins CG ke kisi bhi side hon.
$= - \frac{2 \times 8 \times 0.25}{8 \times 1 0 ^{- 5}} = - 5 \times 1 0^{4}$ . Damping abhi bhi "good" lagti hai. Yeh step kyun? Numerically pitch-rate damping abhi bhi stabilising hai.
Lekin static stability fail ho jaati hai: CG ke aage fins $C_{m_{α}} > 0$ dete hain, toh $ω_{n}^{2} \propto - C_{m_{α}} < 0$ . Oscillation frequency imaginary ho jaati hai → motion diverge karti hai (badhti hai), kabhi oscillate nahi karti. Yeh step kyun? Damping sirf ek oscillation se energy remove karta hai. Agar koi restoring spring nahi hai, toh koi oscillation nahi hai jise damp karo — naak bas bhaag jaati hai. Dekho Damped Harmonic Oscillator kyun $ω_{n}^{2} < 0$ matlab exponential growth hai.

Verify: $C_{m_{q}} = - 5 \times 1 0^{4} < 0$ (damping term theek) ✔; lekin $C_{m_{α}} > 0 \Rightarrow ω_{n}^{2} < 0 \Rightarrow$ divergence ✔. Lesson: ek achha $C_{m_{q}}$ statically unstable rocket ko rescue nahi kar sakta.

Figure s02 dono bodies dikhata hai: red fins CG ke peeche hain ( $x > 0$ , safe, $ω_{n}^{2} > 0$ ); black fins aage hain ( $x < 0$ ) aur naak diverge karti hai. Note karo dono negative $C_{m_{q}}$ dete hain — picture clearly dikhata hai ki sirf damping sign se safety ke baare mein kuch nahi pata chalta.

Worked example Ex 6 — Cell F: altitude ke saath limiting behaviour

Ex 1 rocket ka damping ratio $ζ_{0} = 0.06$ ground ke paas hai ( $ρ_{0} = 1.225$ kg/m³). 20 km altitude par $ρ = 0.089$ kg/m³ aur speed unchanged hai. Wahan $ζ$ estimate karo (quick estimate ke liye $ω_{n}$ roughly constant rakhho).

Forecast: kya $ζ$ thoda girta hai ya bahut zyada?

Formula se, physical damping term $\propto ρ V$ hai; $V$ fixed ke saath, $ζ \propto ρ$ (estimate ke liye $ω_{n}$ ko slowly varying maante hue). Yeh step kyun? $C_{m_{q}}$ geometry hai (fixed), lekin jो moment yeh produce karta hai usme $ρ$ hota hai — patli hawa, kamzor shock absorber.
$ζ = ζ_{0} \times \frac{ρ}{ρ _{0}} = 0.06 \times \frac{0.089}{1.225} = 0.06 \times 0.0727 = 0.00436$ . Yeh step kyun? Ground value ko density ratio se scale karo.

Verify: $0.00436 ≪ 0.05$ ✔ — comfort line se neeche, toh oscillations upar zyada decay nahi karti. Yeh exactly parent ki warning hai: $ζ$ har altitude par evaluate karo. Density ratio $0.089/1.225 \approx 0.073$ ✔.

Worked example Ex 7 — Cell G: scratch se poora real-world

$ζ$ Ek hobby rocket: $C_{m_{q}} = - 7.2 \times 1 0^{4}$ (Ex 1), $C_{m_{α}} = - 12$ /rad, $ρ = 1.225$ , $V = 80$ m/s, $S = 0.008$ m², $d = 0.1$ m, $I_{y} = 0.05$ kg·m². $ω_{n}$ aur $ζ$ find karo. Kya yeh calm ho jaata hai?

Forecast: guess karo $ζ$ 0.05 comfort line ke upar ya neeche land karta hai.

$ω_{n} = \frac{- \frac{1}{2} ρ V ^{2} S d C _{m_{α}}}{I _{y}}$ . Top $= - \frac{1}{2} (1.225) (8 0^{2}) (0.008) (0.1) (- 12)$ . Yeh step kyun? $ω_{n}$ ko static derivative $C_{m_{α}} < 0$ chahiye; do minus signs ise positive banate hain.
$\frac{1}{2} (1.225) (6400) = 3920$ ; $\times 0.008 = 31.36$ ; $\times 0.1 = 3.136$ ; $\times (-) (- 12) = 37.63$ . $I_{y}$ se divide: $37.63/0.05 = 752.6$ . $ω_{n} = 752.6 = 27.4$ rad/s. Yeh step kyun? Constants ko carefully chain karo; $ω_{n}$ hai kitni tezi se yeh undamped ring karta.
$ζ = \frac{- \frac{1}{4} ρ V S d ^{2} C _{m_{q}}}{2 I _{y} ω _{n}}$ . Top $= - \frac{1}{4} (1.225) (80) (0.008) (0.01) (- 7.2 \times 1 0^{4})$ . Yeh step kyun? Yeh dashpot term hai; $C_{m_{q}} < 0$ ise positive banata hai.
$\frac{1}{4} (1.225) = 0.30625$ ; $\times 80 = 24.5$ ; $\times 0.008 = 0.196$ ; $\times 0.01 = 0.00196$ ; $\times 7.2 \times 1 0^{4} = 141.1$ . Bottom $= 2 (0.05) (27.4) = 2.744$ . $ζ = 141.1/2.744 = 51.4$ . Yeh step kyun? Ratio ke dono parts assemble karo.

Verify: $ζ = 51.4$ . Yeh $≫ 1$ hai — yeh configuration bahut zyada over-damped hai (ek bahut stubby, heavily finned rocket bina ring kiye trim par wapas aata hai). $ζ > 0$ ✔ aur 0.05 se kaafi upar ✔ — yeh definitely calm ho jaata hai. $ω_{n} = 27.4$ rad/s ✔.

Worked example Ex 8 — Cell H: exam twist, do fin sets (superposition)

Ek rocket mein canards $x_{1} = - 0.3$ m par $c_{n_{α}} A_{1} = 2$ ke saath aur tail fins $x_{2} = + 0.9$ m par $c_{n_{α}} A_{2} = 6$ ke saath hain. Same $S, d$ . Damping par kaunsa set dominate karta hai, aur total $C_{m_{q}}$ kya hai?

Forecast: tail 3× area hai aur zyada door hai — lekin woh kitne se jeetta hai?

Damping additive hai (integral split ho jaata hai): $C_{m_{q}} = - \frac{2}{S d ^{2}} [(c_{n_{α}} A_{1}) x_{1}^{2} + (c_{n_{α}} A_{2}) x_{2}^{2}]$ . Yeh step kyun? Har surface $x^{2}$ -weighted integral mein independently contribute karti hai; superpose karo.
Canard term: $2 \times (- 0.3)^{2} = 2 \times 0.09 = 0.18$ . Tail term: $6 \times (0.9)^{2} = 6 \times 0.81 = 4.86$ . Yeh step kyun? Unka weight compare karne ke liye har $c_{n_{α}} A \cdot x^{2}$ alag compute karo.
Sum $= 5.04$ . Toh $C_{m_{q}} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 5.04 = - 1.26 \times 1 0^{5}$ . Yeh step kyun? Pooled second-moment ko constant se multiply karo.

Verify: tail $4.86/5.04 = 96.4%$ damping contribute karta hai ✔ — far-back set overwhelmingly dominate karta hai $x^{2}$ weighting ki wajah se, chahe woh sirf 3× area ho (woh $3 \times 3^{2} = 27$ -ish se jeetta hai jab arm bhi 3× lamba hai). Total sign negative ✔.

Worked example Ex 9 — Cell I: CG ko straddle karne wali loading (sign-aware integral)

Ek body mein aerodynamic loading uniformly $x = - 0.4$ m (CG ke aage) se $x = + 0.8$ m (CG ke peeche) tak spread hai, constant $c_{n_{α}} \frac{d A}{d x} = k = 5$ m/rad. Same $S, d$ . $C_{m_{q}}$ find karo. Kya CG-ke-aage-wala part damping se subtract karta hai?

Forecast: aage ka part $x < 0$ hai — kya uska contribution negative hoga (damping weakening) ya positive (phir bhi help karta)?

Same integral set up karo, ab ek range par jo zero cross karta hai: $C_{m_{q}} = - \frac{2}{S d ^{2}} \int_{- 0.4}^{0.8} k x^{2} d x$ . Yeh step kyun? Formula $x^{2}$ se weight karta hai, jo kabhi negative nahi hota. Toh har strip — aage ya peeche — second moment mein positively add karta hai. Damping CG ke aas-paas cancel nahi hoti.
Integrate karo: $\int_{- 0.4}^{0.8} x^{2} d x = \frac{x ^{3}}{3}_{- 0.4}^{0.8} = \frac{0.512 - ( - 0.064 )}{3} = \frac{0.576}{3} = 0.192$ . Yeh step kyun? $(- 0.4)^{3} = - 0.064$ , aur ek negative subtract karna add karta hai — yeh sign-aware part hai. Front section abhi bhi contribute karta hai kyunki $x^{2}$ usse unsigned kar deta hai.
$C_{m_{q}} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 5 \times 0.192 = - \frac{1.92}{8 \times 1 0 ^{- 5}} = - 2.4 \times 1 0^{4}$ . Yeh step kyun? Pehle ki tarah constant × loading × integral assemble karo.

Verify: front strip ka $x^{2}$ genuinely positive hai: $\int_{- 0.4}^{0} x^{2} d x = 0.02133 > 0$ ✔, toh woh damping mein add karta hai, subtract nahi. Total $- 2.4 \times 1 0^{4} < 0$ ✔. Moral: damping ke liye CG side irrelevant hai (sirf $∣ x ∣$ matter karta hai); static stability ( $C_{m_{α}}$ , jo $x^{1}$ se weight karta hai) ke liye sign kaafi matter karta.

Recall Matrix ke aas-paas quick self-test

Kaunsa cell puchta hai kya $C_{m_{q}}$ zero hai jab rocket rotate nahi kar raha? ::: Cell D — aur answer hai nahi; $C_{m_{q}}$ ek slope hai, sirf moment zero hai. Fins CG ke aage: kya $C_{m_{q}}$ abhi bhi negative hai? ::: Haan ( $x^{2}$ ki wajah se), lekin rocket phir bhi diverge karta hai kyunki $C_{m_{α}} > 0$ (Cell E). Do fin sets — kya unke damping contributions add hote hain? ::: Haan, integral linear hai, toh $c_{n_{α}} A x^{2}$ terms superpose karo (Cell H). CG ko straddle karti loading — kya aage ka part peeche ko cancel karta hai? ::: Nahi; $x^{2} \geq 0$ toh har strip damping mein add karta hai chahe kisi bhi side ho (Cell I). Altitude par $ζ$ kyun collapse ho jaata hai? ::: Damping moment mein $ρ$ hota hai, jo steeply girta hai — $ζ \propto ρ$ (Cell F).

Related: Fin Design & Sizing · Moments of Inertia of a Rocket · Atmospheric Density vs Altitude