Static stability — weather-cocking tendency: Worked examples

3.4.10 · D3 · Physics › Rocket Flight Mechanics › Static stability — weather-cocking tendency

Yeh page static stability ka drill floor hai. Parent note ne ideas build kiye; yahan hum har case class ke through grind karte hain jo yeh topic exam mein throw kar sakta hai, taaki koi bhi exam scenario naya na lage.

Shuru karne se pehle, do symbols jo aap har line mein use karenge — parent note se mile hue:

Scenario matrix

Is topic mein jo bhi pucha ja sakta hai woh in cells mein se kisi ek mein hai. Neeche har worked example un cell(s) ke saath tagged hai jo woh cover karta hai.

Cell	Case class	Kya khaas hai	Example
A	$ℓ > 0$ , normal margin (1–2 cal)	Healthy design	Ex 1
B	$ℓ > 0$ lekin over-stable ( $SM ≫ 2$ )	Bahut zyada margin	Ex 2
C	$ℓ < 0$ (CP, CG ke aage)	Unstable — aur kaise fix karein	Ex 3
D	$ℓ = 0$ exactly	Degenerate: neutral stability	Ex 4
E	Actual restoring moment compute karo (signs & units)	Force → torque chain	Ex 5
F	Zero / limiting inputs ( $α = 0$ , $V \to 0$ , $V \to$ large)	Kya vanish hota hai, kya blow up hota hai	Ex 6
G	Real-world word problem (crosswind launch)	Story ko $ℓ, SM$ mein convert karo	Ex 7
H	Exam twist: CP flight mein move karta hai	Stability flight mein kho sakti hai	Ex 8

Jo tools hum reuse karte hain (parent note mein build kiye): static margin $SM = ℓ / d$ calibers mein measure kiya (ek caliber = ek body diameter $d$ ), aur moment $M = - \frac{1}{2} ρ V^{2} A C_{N_{α}} ℓ α$ . Dynamic-pressure factor $\frac{1}{2} ρ V^{2}$ ke liye dekho Dynamic Pressure and Aerodynamic Coefficients; $α$ ke liye dekho Angle of Attack and Aerodynamic Forces.

Figure — Static stability — weather-cocking tendency

Figure dekho: har cell mein wahi rocket outline hai, CG (teal dot), CP (orange dot) aur resulting swing arrow ke saath. Yeh poora matrix ek nazar mein hai — padhte waqt iske paas wapas refer karo.

Cell A — healthy design

Worked example Ex 1 — Kya yeh stable hai, aur kitna stable?

Ek model rocket ka $X_{c g} = 0.60 m$ , $X_{c p} = 0.74 m$ nose se, body diameter $d = 0.08 m$ hai.

Forecast: Pehle guess karo — kya CP, CG ke peeche hai? Roughly kitne calibers ka margin?

Step 1. Lever arm compute karo: $ℓ = X_{c p} - X_{c g} = 0.74 - 0.60 = 0.14 m$ . Yeh step kyun? $ℓ$ ka sign hi decide karta hai stability — hum ise hamesha pehle nikalte hain.

Step 2. Kyunki $ℓ = + 0.14 > 0$ hai, CP CG ke peeche hai → restoring torque exist karta hai → stable. Yeh step kyun? Positive lever arm ka matlab hai hawa tail ko pakadti hai aur nose ko wapas hawa mein swing karti hai (weather-cocking).

Step 3. Ek comparable number mein convert karo: $SM = ℓ / d = 0.14/0.08 = 1.75$ calibers. Yeh step kyun? Diameter se divide karne par metres universal "calibers" scale mein aa jaate hain taaki kisi bhi rocket size ko usi 1–2 rule par judge kiya ja sake.

Result: $SM = 1.75$ calibers, 1–2 window ke andar → well-behaved stable rocket.

Verify: Units: $m / m$ = dimensionless ✓. Sanity: $1.75$ ek aur do ke beech hai jaisa ek acche design mein hona chahiye ✓.

Cell B — ek acchi cheez bahut zyada ho gayi

Worked example Ex 2 — Over-stable rocket

Wahi diameter $d = 0.08 m$ , lekin ab $X_{c g} = 0.50 m$ aur $X_{c p} = 0.82 m$ hai.

Forecast: CG aur CP ke beech bada gap — kya yeh zyada desirable hoga ya problem?

Step 1. $ℓ = 0.82 - 0.50 = 0.32 m$ . Kyun? Pehle sign: positive → principle mein abhi bhi stable.

Step 2. $SM = 0.32/0.08 = 4.0$ calibers. Kyun? Rule of thumb ke against compare karo.

Step 3. $4.0 ≫ 2$ → over-stable. Yeh kyun matter karta hai: ek over-stable rocket choti-choti gusts par bhi bahut strong restoring torque generate karta hai, isliye yeh apna nose crosswind mein steeply swing karta hai aur launch pad ke downwind mein fly karta hai.

Result: Statically stable lekin over-stable; aggressive weather-cocking expect karo. Dekho Fin Design and Sizing — fix hai smaller fins, CP ko aage move karo.

Verify: $ℓ / d = 0.32/0.08 = 4.0$ ✓, aur $4.0 > 2$ over-stable flag confirm karta hai ✓.

Cell C — unstable, aur do fixes

Worked example Ex 3 — Diagnose aur repair karo

Ek rocket ka $X_{c g} = 0.55 m$ , $X_{c p} = 0.50 m$ , $d = 0.05 m$ hai.

Forecast: CP number, CG number se chhota hai — yeh kya kehta hai?

Step 1. $ℓ = 0.50 - 0.55 = - 0.05 m$ . Kyun? Negative lever arm ⇒ CP CG ke aage hai.

Step 2. $SM = - 0.05/0.05 = - 1.0$ caliber. Kyun? Negative margin ka matlab moment kisi bhi disturbance ko add karta hai → rocket tumble karta hai.

Step 3 — Fix A (fins). Bade/rearward fins add karo taaki $X_{c p}$ ko, manlo, $0.62 m$ tak push kiya ja sake. Naya $ℓ = 0.62 - 0.55 = + 0.07 m$ , $SM = 0.07/0.05 = 1.4$ cal → stable. Kyun? Fins pressure point ko move karte hain, Barrowman Equations for CP location ke anusaar.

Step 4 — Fix B (ballast). Iske bajaye nose ballast add karo taaki $X_{c g}$ ko $0.44 m$ tak aage move kiya ja sake. Naya $ℓ = 0.50 - 0.44 = + 0.06 m$ , $SM = 0.06/0.05 = 1.2$ cal → stable. Kyun? Ballast mass point ko move karta hai; dono fixes $ℓ$ ko positive banate hain.

Result: Original $SM = - 1.0$ (unstable); koi bhi fix positive margin restore karta hai (1.4 ya 1.2 cal).

Verify: $- 0.05/0.05 = - 1.0$ ✓; Fix A $0.07/0.05 = 1.4$ ✓; Fix B $0.06/0.05 = 1.2$ ✓.

Cell D — degenerate case $ℓ = 0$

Cell E — poori force→torque chain signs ke saath

Worked example Ex 5 — Restoring moment ka size

$ρ = 1.2 kg/m^{3}$ , $V = 100 m/s$ , $A = 2 \times 1 0^{- 3} m^{2}$ , $C_{N_{α}} = 2 rad^{- 1}$ , $ℓ = 0.12 m$ , disturbance $α = 5^{\circ}$ hai.

Forecast: Compute karne se pehle $M$ ka sign guess karo — restoring ka matlab hai...?

Step 1 — Angle to radians. $α = 5^{\circ} = 5 \times π /180 = 0.08727 rad$ . Kyun? $C_{N_{α}}$ radian per hai, isliye units cancel karne ke liye $α$ radians mein hona chahiye.

Step 2 — Dynamic pressure. $q = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.2) (10 0^{2}) = 6000 Pa$ . Kyun? Yeh har aerodynamic force ka scale set karta hai — moving air jo pressure exert kar sakti hai.

Step 3 — Normal force. $N = q A C_{N_{α}} α = 6000 (0.002) (2) (0.08727) = 2.094 N$ . Kyun? Pressure × area × (tilt kitni strongly sideways force banata hai) × tilt.

Step 4 — Moment about CG. $M = - N ℓ = - (2.094) (0.12) = - 0.2513 N\cdotm$ . Minus kyun? Positive $ℓ$ + positive $α$ ko ek negative (opposing) torque dena chahiye taaki $α$ wapas zero par push ho.

Result: $M \approx - 0.25 N\cdotm$ , negative ⇒ restoring, forecast se match karta hai.

Verify: Units: $Pa \cdot m^{2} \cdot rad^{- 1} \cdot rad \cdot m = N\cdotm$ ✓. Magnitude $\approx 0.251$ ✓.

Cell F — zero aur limiting inputs

Worked example Ex 6 — Kya vanish hota hai, kya survive karta hai

Ex 5 wala rocket use karo ( $ℓ = 0.12$ , stable). $M = - q A C_{N_{α}} ℓ α$ ke teen limits examine karo.

Forecast: Inme se kaun-se restoring torque ko khatam karte hain, aur kaun-se nahi?

Step 1 — Zero angle of attack ( $α = 0$ ). $M = - q A C_{N_{α}} ℓ \cdot 0 = 0$ . Kyun? Koi tilt nahi ⇒ koi sideways force nahi ⇒ koi correction ki zarurat nahi. Stability disturbance ka response hai, seedha fly karte waqt silent rehti hai.

Step 2 — Standstill ( $V \to 0$ ). $q = \frac{1}{2} ρ V^{2} \to 0$ , isliye $M \to 0$ . Yeh kyun matters: launch rail par ya bahut slowly move karte rocket mein almost koi aerodynamic stability nahi hoti — yahi wajah hai launch rails/rods exist karte hain, direction pakadne ke liye jab tak $V$ itna bada na ho jaye. Yeh woh limiting case hai jo beginners bhool jaate hain.

Step 3 — High speed ( $V$ large). $M$ , $V^{2}$ ki tarah badhta hai. Kyun? Restoring torque speed ke square ke saath strong hota jaata hai, isliye rocket accelerate karne par stiffer hota jaata hai — lekin agar high speed par CP aage move kar jaye (Cell H), toh margin phir bhi shrink ho sakta hai.

Result: $α = 0$ aur $V \to 0$ dono $M = 0$ dete hain; large $V$ mein $M \propto V^{2}$ (tezi se badhta hua).

Verify: $α = 0$ par: $M = 0$ ✓. $V$ ko 100 se 200 double karne par $M$ chaar guna hota hai: check $M (200) / M (100) = 4$ ✓.

Cell G — real-world crosswind word problem

Worked example Ex 7 — Launch day ki story

"Aap ek rocket launch karte ho jiska $X_{c g} = 0.90 m$ , $X_{c p} = 1.14 m$ , $d = 0.10 m$ hai, ek steady side-wind mein. Ek judge kehta hai ki landing ko pad se zyada door jaane se rokne ke liye margin 2 calibers se kam hona chahiye. Kya yeh pass karta hai, aur yeh kis taraf lean karega?"

Forecast: Story se margin aur lean direction guess karo.

Step 1 — Story ko translate karo. "Side-wind" ⇒ rocket airflow ke relative ek chhota angle of attack $α > 0$ par fly karta hai. Kyun? Weather-cocking tabhi act karta hai jab correct karne ke liye koi tilt ho.

Step 2 — Lever arm. $ℓ = 1.14 - 0.90 = 0.24 m > 0$ ⇒ stable, aur yeh apna nose hawa mein ghumaega (weather-cock). Kyun? CP, CG ke peeche hai.

Step 3 — Margin vs judge ka rule. $SM = 0.24/0.10 = 2.4$ calibers $> 2$ . Kyun? Stated limit se compare karo.

Result: Judge ke rule mein fail — 2.4 cal par over-stable ⇒ yeh hawa mein hard lean karega aur pad ke downwind mein drift karega. Recommend karo fin area trim karna (CP ko aage move karta hai) taaki ~1.5 cal tak pahuncha ja sake.

Verify: $0.24/0.10 = 2.4$ ✓; $2.4 > 2$ ⇒ fails ✓.

Cell H — exam twist: CP flight mein shift karta hai

Worked example Ex 8 — Launch par stable, speed par unstable

Low speed par ek rocket ka $X_{c p} = 0.70 m$ , $X_{c g} = 0.60 m$ , $d = 0.05 m$ hai. Jaise-jaise yeh transonic hota hai CP aage move karta hai $X_{c p} = 0.58 m$ par (CP fixed nahi hai — yeh Mach aur $α$ par depend karta hai).

Forecast: Launch par stable, sure — lekin jab CP aage khiskata hai toh kya hota hai?

Step 1 — Launch margin. $ℓ_{1} = 0.70 - 0.60 = 0.10 m$ , $SM_{1} = 0.10/0.05 = 2.0$ cal → stable. Kyun? Initial condition par standard check.

Step 2 — Transonic margin. $ℓ_{2} = 0.58 - 0.60 = - 0.02 m$ , $SM_{2} = - 0.02/0.05 = - 0.4$ cal. Yeh step kyun? Naye CP ke saath recompute karo; CG yahan zyada nahi move hua, isliye sirf $X_{c p}$ badla.

Step 3 — Interpret karo. $SM$ $+ 2.0$ se $- 0.4$ par aa gaya: rocket mid-flight mein stability kho deta hai aur high speed par tumble kar sakta hai. Exam ise kyun pasand karta hai: yeh assumption ko punish karta hai ki "CP, CG ki tarah fixed hai".

Result: Launch par stable (2.0 cal), transonic par unstable ( $- 0.4$ cal). Fix: bade/further-aft fins taaki CP saare Mach numbers par CG ke peeche rahe.

Verify: $0.10/0.05 = 2.0$ ✓; $- 0.02/0.05 = - 0.4$ ✓; sign positive→negative flip karta hai ✓.

Recall Quick self-test (answers reveal karo)

Kis cell mein har $α$ ke liye $M = 0$ hota hai? ::: Cell D, $ℓ = 0$ (neutral) — lever arm poore torque ko multiply karta hai. Launch rail par rocket mein almost koi aerodynamic stability kyun nahi hoti? ::: $V \approx 0$ ⇒ $q = \frac{1}{2} ρ V^{2} \approx 0$ ⇒ $M \approx 0$ (Cell F). 2.4 calibers ka margin matlab hai...? ::: Stable lekin over-stable — strong weather-cocking, hawa mein drift (Cells B, G). Agar CP high speed par CG se aage move kar jaye, toh rocket kya ban jaata hai? ::: Unstable — margin mid-flight mein negative ho jaata hai (Cell H).

Related: Thrust Vectoring vs Passive Stability (jab passive margin impossible ho toh kya karein), aur Hinglish walkthrough 3.4.10 Static stability — weather-cocking tendency (Hinglish).