Question bank — Dynamic stability — pitch - yaw damping derivatives

False. Static stability ($C_{m_{\alpha }}<0$) sirf ye guarantee karta hai ki woh wapas mudna shuru karta hai; damping ($C_{m_q}<0$) ke bina woh hamesha overshoot karta rahega ya, kam $\rho$ par, bahut mushkil se shant hoga. Dono chahiye — dekho Static Stability — Center of Pressure & Margin.

True. Damping moment ko oscillation se energy nikaalane ke liye pitch rate ka virodh karna chahiye, aur woh sign convention $C_{m_q}<0$ ke roop mein dikhta hai.

False. Coefficient geometry se fixed hai, lekin physical moment $\propto \rho V{d}^2{C}_{m_q}$ hai; jaise rocket upar jaata hai, $\rho$ girta hai (dekho Atmospheric Density vs Altitude) aur $V$ badalti hai, isliye real damping dramatically kamzor ho jaati hai.

True. Pitch aur yaw ek revolution ke body par geometrically identical planes sweep karte hain, isliye second-moment-of-area integral dono planes mein same hota hai.

False. Yeh doori ke square ke proportion mein badhta hai: $x$ doori par rakha fin teji se sweep karta hai (speed $qx$) aur lambe lever arm par push karta hai ($x$), isliye torque $\propto x\cdot x=x^2$.

False. CG ke aage fins $C_{m_{\alpha }}>0$ banate hain (statically unstable): ${\omega }_n^2<0$, jo pure divergence deta hai, oscillation nahi — aur damping ek unstable body ko nahi bacha sakti.

True. Damping moment rate $q$ ke proportional hai; jis instant rocket momentarily rotate nahi kar raha (jaise maximum overshoot angle par), $q=0$ hota hai isliye damping contribution vanish ho jaata hai — wahan sirf static spring kaam karta hai.

True ideal linear model mein — bina damping ke Damped Harmonic Oscillator pure oscillator ban jaata hai; reality mein doosre losses eventually energy bleed karte hain, lekin design us par rely nahi kar sakta.

Galat derivative hai — woh $C_{m_{\alpha }}$ (static) hai. $C_{m_q}=\partial C_m/\partial \hat{q}$ hai, moment ka rate ke saath change, angle ke saath nahi.

Yeh velocity $\dot{\theta }$ ka virodh karta hai, displacement ka nahi. Static restoring moment $\theta$ ka virodh karta hai; damping rate ka virodh karta hai, isliye woh equation of motion mein $\dot{\theta }$ term ke roop mein aata hai.

Arm double karne se damping chaar guna ho jaati hai kyunki $C_{m_q}\propto x^2$. Factor of 2 expect karna is baat ko ignore karta hai ki local sweep speed $qx$ bhi $x$ ke saath scale karta hai.

Factor of 2 missing hai: non-dimensional pitch rate $\hat{q}=qd/(2V)$ hai, pitch rate ko ek chord-time (half-diameter over speed) se scale karta hai.

Magnitude $x^2$ ke zariye damping drive karta hai, lekin nose surfaces (CG ke aage) simultaneously statically destabilise karti hain. Saaf damping ki kahani ye assume karti hai ki aerodynamic area CG ke peeche baithti hai; nose area stability liability hai, damping bonus nahi.

Ulta hai. Damping moment $\propto V$ (aur $\propto \rho$) hai, isliye apogee ke paas kam speed aur pataali hawa par woh sabse kamzor hoti hai, jo exactly woh waqt hai jab tumbling ka risk sabse zyada hota hai.

Nahi — $\zeta \propto 1/(I_y{\omega }_n)$ aur ${\omega }_n\propto 1/\sqrt{I_y}$ hai, isliye $\zeta \propto 1/\sqrt{I_y}$ hai. Ek bhaari, lamba body (bada $I_y$) actually $\zeta$ ko ghata deta hai. Dekho Moments of Inertia of a Rocket.

$x$ ka ek factor transverse sweep speed $qx$ hai jo local angle of attack banata hai; $x$ ka doosra factor woh moment arm hai jo us force ko torque mein convert karta hai. Unka product $x^2$ hai — second moment of aerodynamic area.

Neeche sweep karta tail ek upar ka extra angle of attack dekhta hai, isliye hawa ise wapas upar push karti hai — resulting moment us rotation ka virodh karta hai jisne ise cause kiya, jo mathematically $\hat{q}$ ke respect mein $C_m$ mein negative contribution hai.

$C_{m_{\alpha }}<0$ real natural frequency ${\omega }_n$ deta hai (woh spring jo ise wapas modta hai), jabki $C_{m_q}<0$ positive damping ratio $\zeta$ deta hai (woh dashpot jo har swing ko chota karta hai). Ek ke bina doosra ya toh permanent oscillator hai ya divergence.

Kyunki damping $x^2$ ke roop mein scale karta hai: aerodynamic centre ki thodi si pichward movement squared payoff deti hai, jo fin area ya airspeed mein changes se kahin aage hai. Dekho Fin Design & Sizing.

Kyunki dynamic stability physical moment $\propto \rho V$ par aur $I_y$ par depend karti hai — alag flight speeds, altitudes, ya mass distributions ${\omega }_n$ aur $\zeta$ ko badal dete hain chahe coefficient fixed ho.

Chhote angles ke liye static moment $\theta$ mein linear hai (spring) aur damping moment $\dot{\theta }$ mein linear hai (dashpot); un dono terms wala linear second-order ODE hi damped oscillator hai, isliye poora Damped Harmonic Oscillator toolbox seedha apply hota hai.

Damping moment $\propto \rho$ zero ho jaata hai kyunki swept fins par push karne ke liye koi hawa nahi hoti; aerodynamic damping simply switch off ho jaati hai, aur attitude control thrust vectoring ya spin stabilisation se aana chahiye.

Kuch nahi. $x=0$ ke saath sweep speed $qx$ aur lever arm dono vanish ho jaate hain, isliye integral mein iska $x^2$ contribution zero hai — yeh na sweep karta hai na lever karta hai.

Peak angle par rate $\dot{\theta }=0$ hota hai, isliye static restoring moment (yahan sabse bada) saara kaam karta hai jabki damping moment momentarily zero hota hai; damping apna kaam swing ke dauran karta hai, turnaround par nahi.

Bilkul bhi oscillation nahi. $C_{m_{\alpha }}>0$ se ${\omega }_n^2<0$ hota hai, isliye roots real aur positive hain: disturbance exponentially badata hai (divergence). Damping terms woh oscillation create nahi kar sakti jo stability ke liye chahiye hogi.

Technically stable lekin marginal — oscillations itni slowly decay karti hain ki gusts ya thrust misalignment unhe baar baar re-excite kar sakte hain. Designers typically $\zeta \gtrsim 0.05$–$0.1$ require karte hain taaki swings actually kaafi cycles mein shrink ho sakein.

Practical structure aur static-margin limits: bahut zyada rearward area desired static margin ko overshoot kar sakta hai (rocket ko over-stable aur gust-sensitive banata hai) aur tail mass badhata hai $I_y$ badhaata hai, jo $\zeta$ ko phir se ghatata hai — isliye $x^2$ gain in penalties se compete karta hai.