Exercises — Dynamic stability — pitch - yaw damping derivatives

Damped. Parent note ka rule: ek stable, damped rocket ka $C_{m_q}<0$ hota hai, matlab aerodynamic moment hamesha pitch rate ko oppose karta hai — jaise ek shock absorber jo swing se energy nikal leta hai. Positive $C_{m_q}$ ka matlab hoga ki moment pitch rate mein add ho raha hai — yeh oscillation ko feed karta hai, swings ko badhata hai. Yeh hai anti-damping (aerodynamically unstable in rate).

• $C_{m_{\alpha }}$ → (a) spring: angle of attack $\alpha$ per moment. Static stability ke liye negative hona chahiye.
• $C_{m_q}$ → (b) dashpot: (scaled) pitch rate per moment. Damping ke liye negative hona chahiye.
• $\hat{q}$ → (c) pitch rate $q$ ko dimensionless banana ka tarika, $\hat{q}=qd/(2V)$.
• $\zeta$ → (d) damping ratio: 0 = kabhi khatam nahi hota, 1 = critically damped, >1 = koi oscillation nahi.

$\hat{q}=\frac{qd}{2V}=\frac{3\times 0.1}{2\times 200}=\frac{0.3}{400}=7.5\times 10^{-4}.$ Bahut chhota — expected hai: real rockets pitch slowly karte hain iske comparison mein ki hawa unke paas se kitni tez badhti hai. $\hat{q}$ essentially "yeh hai ki hawa half diameter travel karne mein nose kitna turn karta hai."

Area ek station par lumped hai, toh integral $\int c_{n_{\alpha }}{x}^{2}dA$ sirf $(c_{n_{\alpha }}{A}_f)x_f^2$ ban jaata hai:

Pehle $\hat{q}=\frac{2\times 0.12}{2\times 180}=\frac{0.24}{360}=6.667\times 10^{-4}$. Dynamic-pressure block: $\frac{1}{2}\rho V^2=0.5\times 1.0\times 180^2=16200\text{ Pa}$. Phir $\frac{1}{2}\rho V^{2}Sd=16200\times 0.01\times 0.12=19.44$. $M=19.44\times (-4.69\times 10^4)\times 6.667\times 10^{-4}\approx -607.7\text{ N⋅m}.$ Negative → moment positive pitch rate ko oppose karta hai. Yahi shock absorber ka pushback hai.

$C_{n_r}=C_{m_q}\approx -4.69\times 10^4$. Pitch aur yaw ek body of revolution ke liye same geometry hai, bas 90° rotate — hawa ko koi fark nahi padta. (Dekho Fin Design & Sizing ki yeh odd number of fins ke saath kaise break hota hai.)

Integral lumps ka sum ban jaata hai: $\int c_{n_{\alpha }}{x}^{2}dA=(1.5)(0.3{)}^2+(5)(0.9{)}^2=0.135+4.05=\mathrm{4.185.}$ $C_{m_q}=-\frac{2}{0.01\times 0.0144}\times 4.185=-\frac{8.37}{1.44\times 10^{-4}}\approx -5.81\times 10^4.$ Fin fraction: $4.05/4.185=0.968$, yaani ≈96.8% damping fins se aa rahi hai. Interpretation: bhale hi tail cone ka area fins ke 30% hai, $x^2$ weighting ($0.9^2/0.3^2=9\times$ leverage) fins ko completely dominate karati hai. Isliye damping design = fin placement design.

$C_{m_q}=-\frac{2}{S{d}^2}(c_{n_{\alpha }}{A}_f)x_f^2$ ko invert karo: $x_f^2=\frac{-C_{m_q}S{d}^2}{2(c_{n_{\alpha }}{A}_f)}=\frac{8.0\times 10^4\times 1.44\times 10^{-4}}{2\times 5}=\frac{11.52}{10}=\mathrm{1.152.}$ $x_f=\sqrt{1.152}\approx 1.074\text{ m}.$ Sirf positive root physical hai (fins CG ke peechhe hone chahiye). Negative $x_f$ unhe aage rakh deta, jo hum aage address karte hain.

Damping: $x_f^2=(-0.5{)}^2=0.25$ — square sign khatam kar deta hai, toh $C_{m_q}=-\frac{2}{1.44\times 10^{-4}}\times 5\times 0.25\approx -1.74\times 10^4<0.$ Numerically abhi bhi "damped." Lekin static stability $C_{m_{\alpha }}$ par depend karti hai, jiska sign area ke first moment se aata hai — CG ke aage fins $C_{m_{\alpha }}>0$ dete hain (destabilising). Toh ${\omega }_n^2<0$: motion pure divergence hai, oscillation nahi, aur damping ke liye kuch hai hi nahi. Damping sirf ek statically stable body ki help karta hai.

Natural frequency (spring, static stability se): ${\omega }_n=\sqrt{\frac{-\frac{1}{2}\rho V^{2}Sd{C}_{m_{\alpha }}}{I_y}}.$ $\frac{1}{2}\rho V^2=0.5\times 1.225\times 150^2=13781.25$. Times $Sd=0.01\times 0.12=1.2\times 10^{-3}$ deta hai $16.5375$. Times $-C_{m_{\alpha }}=6.0$: $99.225$. Divide by $I_y=0.9$: $110.25$. ${\omega }_n=\sqrt{110.25}=10.5\text{ rad/s}.$ Damping ratio (dashpot): $\zeta =\frac{-\frac{1}{4}\rho VS{d}^2{C}_{m_q}}{2I_y{\omega }_n}.$ Numerator: $\frac{1}{4}\rho VS{d}^2=0.25\times 1.225\times 150\times 0.01\times 0.0144=6.615\times 10^{-3}$. Times $-C_{m_q}=4.7\times 10^4$: $310.905$. Denominator: $2\times 0.9\times 10.5=18.9$. $\zeta =\frac{310.905}{18.9}\approx \mathrm{16.4.}$ Yeh bahut zyada overdamped hai ($\zeta >1$) — bilkul koi oscillation nahi, bas dheere wapas aata hai. (Realistic $C_{m_q}$ magnitudes aur inertias usually $\zeta$ $0.05$–$0.3$ ke around dete hain; yeh illustrative set deliberately stiff hai.)

${\omega }_d=10.5\sqrt{1-0.08^2}=10.5\sqrt{0.9936}=10.5\times 0.99679\approx 10.47\text{ rad/s}.$ Envelope $e^{-\zeta {\omega }_{n}t}$ ki tarah decay karta hai, toh $1/e$ time hai $\tau =\frac{1}{\zeta {\omega }_n}=\frac{1}{0.08\times 10.5}=\frac{1}{0.84}\approx 1.19\text{ s}.$ Toh har bada swing frequency almost nahi badlta (chhote $\zeta$ ke liye damped $\approx$ undamped), lekin amplitude har $1.19$ s mein $\sim 63\%$ shrink ho jaata hai.

Dekho ki har piece $\rho$ ke saath kaise scale karta hai. Formulas se, ${\omega }_n\propto \sqrt{\rho }$ aur damping-ratio numerator $\propto \rho$, denominator $\propto {\omega }_n\propto \sqrt{\rho }$. Toh $\zeta \propto \frac{\rho }{\sqrt{\rho }}=\sqrt{\rho }.$ Isliye ${\zeta }_{\text{alt}}={\zeta }_0\sqrt{\frac{\rho }{{\rho }_0}}=0.08\sqrt{\frac{0.30}{1.225}}=0.08\sqrt{0.2449}=0.08\times 0.4949\approx \mathrm{0.0396.}$ $0.0396<0.05$ → threshold fail. Rocket ab altitude par dangerously lightly damped hai bhale hi launch par sab theek tha. Yeh exactly woh third parent-note mistake hai jo concrete ban gayi. (Dekho Atmospheric Density vs Altitude.)

(a) ${\omega }_n$: $\frac{1}{2}\rho V^2=0.5\times 0.40\times 250^2=12500$. Times $Sd=1.2\times 10^{-3}$: $15.0$. Times $-C_{m_{\alpha }}=8$: $120$. Divide by $I_y=1.2$: $100$. ${\omega }_n=\sqrt{100}=10.0\text{ rad/s}.$ (b) Required $C_{m_q}$: $\zeta =\frac{-\frac{1}{4}\rho VS{d}^2{C}_{m_q}}{2I_y{\omega }_n}$ ko rearrange karo: $C_{m_q}=-\frac{2I_y{\omega }_n\zeta }{\frac{1}{4}\rho VS{d}^2}.$ Numerator: $2\times 1.2\times 10.0\times 0.10=2.4$. Denominator: $0.25\times 0.40\times 250\times 0.01\times 0.0144=3.6\times 10^{-3}$. $C_{m_q}=-\frac{2.4}{3.6\times 10^{-3}}=-\mathrm{666.67.}$ (c) $x_f$: $C_{m_q}=-\frac{2}{S{d}^2}(c_{n_{\alpha }}{A}_f)x_f^2$ ko invert karo: $x_f^2=\frac{-C_{m_q}S{d}^2}{2(c_{n_{\alpha }}{A}_f)}=\frac{666.67\times 1.44\times 10^{-4}}{2\times 5}=\frac{0.096}{10}=9.6\times 10^{-3}.$ $x_f=\sqrt{9.6\times 10^{-3}}\approx 0.098\text{ m}.$ Toh fins ko CG ke ~9.8 cm peechhe rakhne se worst flight point par target milta hai. Aur peechhe rakhne par margin badhta hai. (Inertia cross-check karo against Moments of Inertia of a Rocket.)

$C_{m_q}$ aur hence $\zeta$ dono $c_{n_{\alpha }}{A}_f$ mein linear hain ($x_f$, $\rho$, $V$ fixed ke saath). Toh ${\zeta }_{\text{new}}=0.10\times \frac{4.25}{5}=0.10\times 0.85=\mathrm{0.085.}$ $0.085>0.05$ → abhi bhi safety threshold ke upar, bhale hi margin design $\zeta =0.10$ se kam hua. Ek robust design ko pehle se zyada target karna chahiye tha (jaise $\zeta =0.12$) taaki 15% build tolerance use edge ke paas nahi le jaaye. Yahi mastery ka lesson hai: tolerance-worst case ke liye design karo, nominal ke liye nahi.

$C_{m_q}$ ko $-\frac{2}{S{d}^2}\int c_{n_{\alpha }}{x}^{2}dA$ se compute karo ($x^2$ dhyan se), ise $\zeta \propto \sqrt{\rho }$ mein convert karo, aur design karo taaki altitude aur build tolerance mein minimum $\zeta$ abhi bhi ~0.05 se zyada ho.