Exercises — Static stability — weather-cocking tendency

Har ek ke liye $\ell =X_{cp}-X_{cg}$ nikalo aur rule padhlo: $\ell >0$ stable, $\ell <0$ unstable, $\ell =0$ neutral.

• A: $\ell =0.55-0.40=+0.15\text{m}>0$ → CP, CG ke peeche → stable ✅
• B: $\ell =0.60-0.60=0$ → CP bilkul CG par → neutrally stable (kisi bhi direction mein koi restoring moment nahi).
• C: $\ell =0.62-0.70=-0.08\text{m}<0$ → CP, CG ke aage → unstable ❌

Humne kya kiya: teen position pairs ko ek ek signed lever arm mein badla. Kyun: CP aur CG ki ordering, jo $\ell$ ke sign se capture hoti hai, static stability ki poori kahani hai.

(a) Weather-cocking (weather-vaning). (b) Yeh sabit karta hai ki rocket statically stable hai — aerodynamic force ne ek restoring moment produce kiya.

Step 1: $\ell =0.92-0.80=0.12\text{m}$. Kyun: positive ⇒ CP, CG ke peeche ⇒ principle mein stable. Step 2: $\text{SM}=\ell /d=0.12/0.04=3.0$ calibers. Kyun $d$ se divide karo: ek caliber = ek diameter, jo ek size-independent number deta hai. Result: $\text{SM}=3.0$ — stable lekin over-stable (2 se kaafi upar). Crosswinds mein aggressive weather-cocking expect karo.

Pehle tilt convert karo: $\alpha =4^{\circ }=4\times \pi /180=0.06981\text{rad}$. Kyun: $C_{N_{\alpha }}$ "per radian" hai, isliye $\alpha$ radians mein hona chahiye. (a) $q=\frac{1}{2}\rho V^2=\frac{1}{2}(1.2)(80^2)=\frac{1}{2}(1.2)(6400)=3840\text{Pa}$. Kyun: dynamic pressure force ka scale set karta hai. (b) $N=qA{C}_{N_{\alpha }}\alpha =3840(1.5\times 10^{-3})(2.5)(0.06981)=1.005\text{N}$. (c) $M=-N\ell =-1.005(0.10)=-0.1005\text{N⋅m}$. Restore karta hai? Haan — minus sign matlab moment positive tilt ko oppose karta hai, $\alpha$ ko zero par wapas push karta hai.

Reasoning: $M=-\frac{1}{2}\rho V^{2}A{C}_{N_{\alpha }}\ell \alpha$. Sirf $V$ badla, aur $M\propto V^2$. $V$ double karne se $M$ $2^2=4$ se multiply hota hai. Predict: $M_{\text{new}}=4\times (-0.1005)=-0.402\text{N⋅m}$. Verify: $q_{\text{new}}=\frac{1}{2}(1.2)(160^2)=15360\text{Pa}$; $N=15360(1.5\times 10^{-3})(2.5)(0.06981)=4.021\text{N}$; $M=-4.021(0.10)=-0.402\text{N⋅m}$. ✔ Insight: stability authority speed ke square ke saath badhti hai — pad se abhi-abhi utha slow rocket ka restoring moment sabse kamzor hota hai, isliye low launch speed woh dangerous moment hai.

Step 1: $\ell =\text{SM}\times d=1.5\times 0.05=0.075\text{m}$. Kyun: $\text{SM}=\ell /d\Rightarrow \ell =\text{SM}\cdot d$. Step 2: $\ell =X_{cp}-X_{cg}\Rightarrow X_{cg}=X_{cp}-\ell =1.00-0.075=0.925\text{m}$. Kyun: CG, CP se lever arm jitna aage hota hai. Result: $X_{cg}=0.925\text{m}$ nose se.

Step 1 — target lever arm: chahiye $\ell =\text{SM}\times d=1.5\times 0.04=0.06\text{m}$. Step 2 — required CG: $\ell =X_{cp}-X_{cg}\Rightarrow X_{cg}=X_{cp}-\ell =0.58-0.06=0.52\text{m}$. Step 3 — sanity: original $X_{cg}=0.60$, target $0.52$ ⇒ CG ko aage $0.08\text{m}$ move karna hoga. Nose ballast CG ko aage move karta hai, toh yeh achievable hai. ✔ Result: CG ko $X_{cg}=0.52\text{m}$ par le jaao. Ab $\ell =+0.06\text{m}>0$ aur $\text{SM}=1.5$. Iske alternative ke liye, CP ko peeche move karne ke liye Fin Design and Sizing dekho.

Step 1: chahiye $\ell =1.5\times 0.04=0.06\text{m}$. Step 2: $\ell =X_{cp}-X_{cg}\Rightarrow X_{cp}=X_{cg}+\ell =0.60+0.06=0.66\text{m}$. Result: CP ko $X_{cp}=0.66\text{m}$ par le jaao (Barrowman Equations for CP location ke according bade/aft fins). Same margin, opposite lever — aage CG ya peeche CP dono kaam karte hain kyunki dono $\ell$ badhate hain.

Margins ko lever arms mein convert karo: $\ell =\text{SM}\cdot d$, isliye $\ell \in [1.0\times 0.05,2.0\times 0.05]=[0.05,0.10]\text{m}$. CG positions mein badlo: $X_{cg}=X_{cp}-\ell$.

• $\ell =0.05$ par: $X_{cg}=1.10-0.05=1.05\text{m}$ (yeh sabse peeche allowed CG hai, $\text{SM}=1$).
• $\ell =0.10$ par: $X_{cg}=1.10-0.10=1.00\text{m}$ (sabse aage allowed CG, $\text{SM}=2$). Result: allowed band $X_{cg}\in [1.00,1.05]\text{m}$. Bada $\ell$ (zyada aage CG) ⇒ zyada margin ⇒ over-stable ke kareeb; chhota $\ell$ ⇒ neutral ke kareeb.

Pad par: ${\ell }_0=1.00-0.90=0.10\text{m}$, ${\text{SM}}_0=0.10/0.05=2.0$ calibers — comfortably stable. Flight mein: ${\ell }_1=0.93-0.85=0.08\text{m}$, ${\text{SM}}_1=0.08/0.05=1.6$ calibers — abhi bhi positive, abhi bhi stable. Comment: aage CP drift ne margin khaaya, lekin fuel burn se simultaneous aage CG shift ne kuch wapas diya. Net result $1$–$2$ calibers ke andar rehta hai, isliye poori flight mein stability survive karti hai. Lesson (dekho Thrust Vectoring vs Passive Stability): margin ko worst instant par check karna zaroori hai, sirf pad par nahi, kyunki CP aur CG dono move karte hain.

Lever arm: $\ell =0.70-0.70=0$. Static margin: $\text{SM}=0/d=0$ calibers. Moment: $M=-\frac{1}{2}\rho V^{2}A{C}_{N_{\alpha }}\ell \alpha =-(\cdots )\times 0\times \alpha =0$ har $\alpha$ ke liye. Physically: zero lever arm ke saath air force pivot se guzarti hai, isliye koi torque produce nahi hota. Rocket neutrally stable hai: gust use tilt karta hai aur woh simply naye angle par reh jaata hai — na correct karta hai, na diverge. Woh course se bhatak jaayega. Isliye designers ek strictly positive margin maangte hain, sirf non-negative nahi.