Static margin = (XCP − XCG) - d — must be positive (at least 1 caliber): Exercises

3.4.9 · D4 · Physics › Rocket Flight Mechanics › Static margin = (XCP − XCG) - d — must be positive (at least

Yeh page parent topic ke liye ek self-testing workout hai. Yahan use hone wala har symbol us note mein explain kiya gaya hai; agar koi term unfamiliar lage, toh link follow karo. Hum paanch levels chadhte hain — L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery. Har problem pehle pen aur paper se try karo, phir solution kholo.

Intuition Koi bhi algebra se pehle yeh picture padho

Is page ka har number ek hi ruler pe measure hota hai: ek line jo rocket ke saath laid hai, nose tip se shuru hoti hai (yahi hamara zero hai, datum) aur tail ki taraf badhti hai (aft is positive). Neeche wali figure wohi ruler draw karti hai aur uspe do khaas points mark karti hai — yellow dot balance point CG hai, blue square air-push point CP hai. Chhota bracket jis par $d$ likha hai, woh ek caliber hai (body ki width). Static margin literally yeh hai ki "CG se CP tak ke gap mein kitne aisa brackets fit hote hain."

Apni aankh se trace karo: nose tip left side pe $x = 0$ par hai; dono points wahaan se measure hote hain; gap $X_{C P} - X_{C G}$ ek double arrow ke roop mein drawn hai; caliber bracket $d$ scale ke saath dikhaya gaya hai taaki tum division dekh sako. Har problem ke liye yeh drawing apne dimag mein rakho.

Master formula ki yaad dahaani (neeche sab kuch isi pe based hai):

L1 · Recognition

Yeh test karte hain ki tum definition padh sako aur signs identify kar sako.

Problem 1.1

Ek rocket mein $X_{C G} = 50 cm$ , $X_{C P} = 65 cm$ , $d = 5 cm$ hai. Static margin compute karo aur stable / unstable batao.

Recall Solution 1.1

WHAT: Formula mein plug karo. $SM = \frac{65 - 50}{5} = \frac{15}{5} = 3 cal$ WHY: Numerator $65 - 50 = 15 cm$ dono points ke beech ka physical gap hai (upar wali datum figure mein double arrow); $d = 5$ se divide karne par count milta hai ki us gap mein kitne caliber-brackets fit hote hain. Kyunki $SM = 3$ calibers $> 1$ hai: stable, comfortable margin ke saath.

Problem 1.2

Bina poora number compute kiye batao ki har rocket stable hai, unstable hai, ya neutral:

(a) CP, CG ke peeche hai.
(b) CP bilkul CG ke upar hai.
(c) CP, CG ke aage (saamne) hai.

Recall Solution 1.2

$X_{C P} - X_{C G}$ ka sign sab decide karta hai, aur yahan physical WHY hai (sirf algebra nahi). Jab koi gust rocket ko angle of attack se tilt karta hai, toh hawa CP par push karti hai; rocket CG ke baare mein pivot karta hai. Force $\times$ lever = moment, aur lever exactly gap $X_{C P} - X_{C G}$ hai:

(a) CP behind CG → gap $> 0$ → air-push pivot ke peeche baith ti hai, toh woh tail ko ghumati hai aur nose ko hawa mein wapas le jaati hai — yeh ek restoring moment hai. $SM > 0$ → stable ✅
(b) CP on CG → gap $= 0$ → zero lever → koi moment nahi → kuch bhi tilt correct nahi karta. $SM = 0$ → neutral ⚠️
(c) CP ahead of CG → gap $< 0$ → air-push pivot ke aage baith ti hai, toh woh nose ko aur door swing karti hai — yeh ek amplifying moment hai. $SM < 0$ → unstable / tumbles ❌

Positive lever arm (CP aft of CG) exactly wahi hai jo restoring moment produce karta hai — isliye Rocket Stability Criterion demand karta hai $SM > 0$ .

L2 · Application

Ab tum formula ko realistic inputs par chalate ho aur units convert karte ho.

Problem 2.1

$X_{C G} = 1.20 m$ , $X_{C P} = 1.35 m$ , body diameter $d = 10 cm$ . SM calibers mein nikalo.

Recall Solution 2.1

Pehle WHAT: dono lengths ko same unit mein rakho. $1.20 m = 120 cm$ , $1.35 m = 135 cm$ . WHY: formula tabhi clean caliber count deta hai jab numerator aur $d$ same units share karein. $SM = \frac{135 - 120}{10} = \frac{15}{10} = 1.5 cal$ $SM = 1.5$ calibers $> 1$ → stable, recommended $1$ – $2$ caliber sweet spot mein.

Problem 2.2

Ek rocket ko exactly $SM = 2.0$ cal tak pahunchna hai. Uska $X_{C P} = 90 cm$ aur $d = 6 cm$ hai. CG kahaan hona chahiye?

Recall Solution 2.2

WHAT: formula ko $X_{C G}$ ke liye solve karo. $2.0 = \frac{90 - X _{C G}}{6} \Rightarrow 90 - X_{C G} = 12 \Rightarrow X_{C G} = 78 cm$ WHY: hum target margin aur $X_{C P}$ jaante hain; sirf balance point unknown hai. Multiply out karo aur isolate karo. CG nose se 78 cm par hona chahiye.

L3 · Analysis

Yahan hum shifts interpret karte hain — points move karte hain, aur tum consequence ke baare mein sochte ho. Neeche wali figure dikhati hai ki CP rocket ki speed badhne par kaise ghoomta hai — problems se pehle ise study karo.

Problem 3.1

Ek rocket $X_{C G} = 100 cm$ , $X_{C P} = 112 cm$ , $d = 8 cm$ se shuru hota hai. Jaise propellant jalta hai, CG aft mein $108 cm$ tak move karta hai (burning mass CG ke forward mein tha). $X_{C P}$ wahan ka wahan rehta hai. Burn se pehle aur baad SM compute karo. Kya rocket poore time safe hai?

Recall Solution 3.1

Burn se pehle: $SM_{i} = \frac{112 - 100}{8} = \frac{12}{8} = 1.5 cal$ Burn ke baad (CG at 108): $SM_{f} = \frac{112 - 108}{8} = \frac{4}{8} = 0.5 cal$ WHY yeh gira: CG ne CP ko rearward chase kiya, gap $X_{C P} - X_{C G}$ ko $12$ cm se $4$ cm tak shrink kar diya. Verdict: woh shuru mein safe hai ( $1.5$ calibers) lekin end mein $1$ -caliber minimum se neeche hai ( $0.5$ ). Poore time safe nahi — yahi wajah hai ki tum Rocket Stability Criterion ko poore burn mein check karo, sirf liftoff par nahi.

Problem 3.2

Usi rocket ka $X_{C P}$ actually migrate bhi karta hai. Transonic regime mein $X_{C P}$ aft $3$ cm (to $115 cm$ ) move karta hai jabki CG apni mid-burn value $104 cm$ par hai, $d = 8 cm$ . Kya aft CP shift stability ko yahan help karta hai ya hurt?

Recall Solution 3.2

$SM = \frac{115 - 104}{8} = \frac{11}{8} = 1.375 cal$ Same instant se compare karo agar CP $112$ par rehta: $SM = (112 - 104) /8 = 1.0$ cal. WHY yeh help karta hai: CP ko aft move karna lever $X_{C P} - X_{C G}$ ko badata hai (migration figure mein blue curve upar chadh rahi hai), toh yeh particular transonic shift margin badhata hai ( $1.0 \to 1.375$ ). Dekho Transonic Aerodynamics — lekin khayal rakho, supersonic flight mein $X_{C P}$ often forward slide karta hai (curve gir rahi hai), jo margin shrink karta hai. Shift ki direction hi matter karti hai.

L4 · Synthesis

Multiple effects combine karo aur fix design karo.

Problem 4.1

Apni worst flight condition par ek rocket mein $X_{C P} = 140 cm$ , $X_{C G} = 133 cm$ , $d = 7 cm$ hai. (a) Margin compute karo. (b) Yeh $1$ -caliber minimum se neeche hai — tum CG ko forward kheenchne ke liye nose ballast add karne ka decide karte ho. Kaun sa naya $X_{C G}$ exactly $SM = 1.5$ cal tak pahunche ga? (c) CG ko kitne centimetres move karna hoga?

Recall Solution 4.1

(a) $SM = \frac{140 - 133}{7} = \frac{7}{7} = 1.0 cal$ Bilkul edge par — koi safety cushion nahi. (b) $X_{C P} = 140$ fixed rakho, new CG ke liye solve karo: $1.5 = \frac{140 - X _{C G}^{'}}{7} \Rightarrow 140 - X_{C G}^{'} = 10.5 \Rightarrow X_{C G}^{'} = 129.5 cm$ (c) Movement $= 133 - 129.5 = 3.5 cm$ nose ki taraf. WHY nose ballast kaam karta hai: yeh mass balance point ko forward shift karta hai, $X_{C P} - X_{C G}$ ko $7$ cm se $10.5$ cm tak enlarge karta hai.

Problem 4.2

Alternatively, CG ko $133$ cm par rakho aur instead fins bado karke CP ko aft move karo. $SM = 1.5$ cal ( $d = 7$ ) hit karne ke liye $X_{C P}$ kahaan move karna hoga?

Recall Solution 4.2

$1.5 = \frac{X _{C P}^{'} - 133}{7} \Rightarrow X_{C P}^{'} - 133 = 10.5 \Rightarrow X_{C P}^{'} = 143.5 cm$ CP ko $140$ se 143.5 cm tak aft move karna hoga, yani $3.5$ cm rearward. WHY fins yeh karte hain: bade tail fins rear mein aerodynamic load add karte hain, Center of Pressure ko peeche kheenchte hain — dekho Fin Design. Datum figure (page ke upar) mein pink arrows dono levers dikhate hain: nose ballast yellow CG dot ko forward (left) slide karta hai, bade fins blue CP square ko aft (right) slide karte hain. Dono opposite ends se ek hi gap ko kholte hain.

Figure — Static margin = (XCP − XCG) - d — must be positive (at least 1 caliber)

L5 · Mastery

Full flight-envelope reasoning — yahi real engineering task hai.

Problem 5.1

Ek rocket, $d = 8 cm$ , ko chaar flight instants par analyse kiya gaya hai. Har jagah SM compute karo, phir minimum margin batao aur kaho ki design $\geq 1$ caliber rule har jagah pass karta hai ya nahi.

Instant	$X_{C P}$ (cm)	$X_{C G}$ (cm)
Liftoff (subsonic)	150	138
Max-Q (transonic)	158	144
Burnout (supersonic)	148	141
Coast (supersonic)	146	140

Recall Solution 5.1

Har row par $SM = (X_{C P} - X_{C G}) /8$ apply karo:

Liftoff: $(150 - 138) /8 = 12/8 = 1.5$ cal
Max-Q: $(158 - 144) /8 = 14/8 = 1.75$ cal
Burnout: $(148 - 141) /8 = 7/8 = 0.875$ cal
Coast: $(146 - 140) /8 = 6/8 = 0.75$ cal

Minimum margin $= 0.75$ caliber, coast (supersonic) instant par. WHY worst case liftoff nahi hai: supersonic flight mein $X_{C P}$ forward (nose ki taraf) slide kar gaya jabki burnt-out CG rearward raha — gap shrink ho gaya. Design FAIL karta hai: do supersonic points $1$ caliber se neeche hain. Fix idea: fins bado karo ya CG forward karo jabtaik worst instant ( $0.75$ ) $\geq 1$ tak na pahunche. Neeche wali margin-versus-flight figure dikhati hai ki curve flight ke baad mein danger line ke neeche dip karta hai, launch par nahi.

Problem 5.2

Problem 5.1 ke rocket ke liye, tum fins add karte ho jo har $X_{C P}$ value ko ek constant $Δ$ cm aft shift karti hain. Sabse chhota $Δ$ (nearest $0.1$ cm tak) nikalo jo worst-case margin ko exactly $1.0$ cal banaye. ( $d = 8$ .)

Recall Solution 5.2

WHAT: worst case coast instant hai, $X_{C P} = 146$ , $X_{C G} = 140$ . CP ko aft $Δ$ se shift karne par $X_{C P} = 146 + Δ$ milta hai. $1.0 = \frac{( 146 + Δ ) - 140}{8} \Rightarrow 6 + Δ = 8 \Rightarrow Δ = 2.0 cm$ WHY worst case $Δ$ set karta hai: ek constant shift saare margins ko equally raise karta hai, toh jab minimum $1.0$ tak pahunche, baaki har instant already usse upar hai. $Δ = 2.0$ cm ka aft CP shift add karne se design har jagah pass karta hai. New minimum check karo: coast $1.0$ ho jaata hai, burnout $(148 + 2 - 141) /8 = 9/8 = 1.125$ ho jaata hai — dono ab $\geq 1$ caliber hain. ✅

Wrap-up recall

Recall One-line takeaways (answers chhupao)

Static margin ka formula ::: $SM = (X_{C P} - X_{C G}) / d$ , nose tip se aft-positive.
Negative SM ka matlab ::: CP ahead of CG → tumbling.
Margin badhane ke do tarike ::: CP aft karo (bigger fins) ya CG forward karo (nose weight).
Margin maximise kyun nahi karte ::: over-stability weathercocking aur altitude loss cause karti hai.
Margin usually worst kahaan hota hai ::: often supersonic/transonic mein, liftoff par NAHI — poora envelope check karo.
Safety rule ::: $SM \geq 1$ caliber har flight instant par.