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

3.4.9 · D3 · Physics › Rocket Flight Mechanics › Static margin = (XCP − XCG) - d — must be positive (at least

Yeh page static-margin ke har tarah ke problem ka complete tour hai jo tum kabhi bhi encounter kar sakte ho. Pehle hum ek map banate hain (called the scenario matrix), phir har cell ke liye ek example work out karte hain taaki baad mein koi bhi case tumhe surprise na kare.

Yahan sab kuch parent note par based hai: the topic note. Agar koi symbol naya lage, wahan uski poori picture milegi. Plain words mein quick reminders:

$X_{C P}$ = nose tip se Center of Pressure tak ki distance (woh single point jahan saari hawa ki push karti hai).
$X_{C G}$ = nose tip se Center of Gravity tak ki distance (mass ka balance point).
$d$ = body ka sabse zyada wide diameter, jise one caliber kehte hain.
Static margin batata hai ki CP, CG ke peeche kitne calibers hai: $SM = \frac{X _{C P} - X _{C G}}{d}$

Hume chahiye $SM > 0$ (CP, CG ke peeche) aur safety ke liye $SM \geq 1$ , per the Rocket Stability Criterion.

The scenario matrix

Har static-margin problem in mein se kisi ek cell mein aati hai. Neeche diye gaye examples har cell ko cover karte hain.

Cell	Kya special hai	Sign / regime	Example
A	Textbook stable, dono points known	$X_{C P} > X_{C G}$ , SM $> 1$	Ex 1
B	Unstable: CP, CG ke aage	$X_{C P} < X_{C G}$ , SM $< 0$	Ex 2
C	Degenerate: CP exactly CG par	$X_{C P} = X_{C G}$ , SM $= 0$	Ex 3
D	Barely stable lekin 1 caliber se neeche	$0 < SM < 1$	Ex 4
D′	Exact borderline: one-caliber safety edge	$SM = 1$ exactly	Ex 4b
E	Inverse problem: SM given hai, ek unknown position (CG/ballast) solve karni hai	formula rearrange karo	Ex 5
F	Over-stable: SM bahut bada → weathercocking	SM $≫ 1$	Ex 6
G	Limiting behaviour: kya hota hai jab $d \to 0$ ya $d \to \infty$	SM $\to \pm \infty$ ya $0$	Ex 7
H	Real-world word problem: CG burn ke dauran shift karta hai	time-varying	Ex 8
I	Exam twist: flight mein worst margin (CP migrate karta hai)	envelope check	Ex 9

Example 1 — Cell A: clean stable case

Worked example Basic caliber count

Ek rocket mein $X_{C G} = 62 cm$ , $X_{C P} = 80 cm$ , diameter $d = 6 cm$ hai. Static margin nikalo aur classify karo.

Forecast: CP clearly CG ke peeche hai, toh positive answer expect karo — lekin kya yeh 1 caliber se upar hai? Ek second sochlo.

Step 1 — Lever separation nikalo. $X_{C P} - X_{C G} = 80 - 62 = 18 cm$ Yeh step kyun? Yeh raw distance (cm mein) air-push point aur balance point ke beech ka physical gap hai. Yehi restoring torque create karta hai.

Step 2 — Calibers mein convert karo. $SM = \frac{18}{6} = 3 cal$ Yeh step kyun? $d$ se divide karne par cm-gap ek dimensionless count ban jata hai body-widths ka, taaki hum kisi bhi size ke rockets compare kar sakein.

Step 3 — Classify karo. $SM = 3 > 1$ → stable with a healthy margin. Yeh step kyun? Ek akela number answer nahi hai — hume ise safety rule ( $SM \geq 1$ ) ke against map karna hoga taaki yeh bol sakein ki rocket actually seedha fly karta hai ya nahi.

Verify: Numerator rebuild karo: $3 \times 6 = 18$ cm, jo $80 - 62$ se match karta hai. Units: $cm / cm$ = dimensionless. ✓

Example 2 — Cell B: unstable, CP ahead of CG

Worked example Danger pakdo

$X_{C G} = 90 cm$ , $X_{C P} = 78 cm$ , $d = 4 cm$ .

Forecast: Yahan $X_{C P} < X_{C G}$ — air-push balance point ke aage hai. Kya tum positive ya negative number expect karte ho? Rocket physically kya karega?

Step 1 — Lever separation (sign rakho!). $X_{C P} - X_{C G} = 78 - 90 = - 12 cm$ Yeh step kyun? Hume absolute values nahi lene. Sign hi poora message hai: negative matlab CP aage hai.

Step 2 — Convert karo. $SM = \frac{- 12}{4} = - 3 cal$ Yeh step kyun? Same caliber conversion; negative survive karta hai, instability flag karta hai.

Step 3 — Physical meaning. Negative SM → koi bhi tilt badhta hai → rocket end over end tumble karta hai. Fix: fins add karo taaki CP peeche aaye (Fin Design) ya nose weight add karo taaki CG aage aaye. Yeh step kyun? SM ka sign sirf isliye matter karta hai kyunki woh aerodynamic torque ka direction decide karta hai; " $- 3$ " ko "tumbles" mein translate karna hi number ko physically meaningful banata hai. Yahan "tilt" ka matlab hai angle of attack $α$ — rocket ki body axis aur incoming air ke beech ka angle, positive nose-up liya gaya (dekho Angle of Attack). Negative SM matlab ek chhota sa positive $α$ ek aisa torque produce karta hai jo $α$ ko aur badhata hai, toh tilt runaway ho jati hai.

Verify: $- 3 \times 4 = - 12 = 78 - 90$ . ✓ Sign negative hai → unstable, Rocket Stability Criterion se match karta hai.

Example 3 — Cell C: degenerate knife-edge

Worked example CP exactly CG par

$X_{C G} = 70 cm$ , $X_{C P} = 70 cm$ , $d = 5 cm$ .

Forecast: Dono points ek jagah hain. Zero divided by kuch bhi... kya hoga? Aur neutral stability kaisi feel hoti hai?

Step 1 — Lever separation. $X_{C P} - X_{C G} = 70 - 70 = 0 cm$ Yeh step kyun? Koi gap nahi matlab restoring torque ke liye koi lever arm nahi.

Step 2 — Convert karo. $SM = \frac{0}{5} = 0 cal$ Yeh step kyun? Zero ko (nonzero, kyunki $d > 0$ ) diameter se divide karna safe hai aur exactly zero deta hai.

Step 3 — Physical meaning. $SM = 0$ → neutrally stable. Ek tilt koi restoring torque produce nahi karta, toh rocket na khud correct karta hai na tumble — woh wahan drift karta hai jahan koi gust isse chodh de. Practically unusable; hamesha ek cushion chahiye. Yeh step kyun? Knife-edge value $0$ ko "theek hai, negative toh nahi" samajhna galat hai — hum clearly bolte hain ki zero torque matlab zero self-correction, isliye yeh case sirf marginal nahi balki unusable hai.

Verify: Koi bhi nonzero $d$ deta hai $0/ d = 0$ . Neutral case confirmed. ✓

Example 4 — Cell D: positive lekin 1 caliber se neeche

Worked example Technically stable, practically risky

$X_{C G} = 85 cm$ , $X_{C P} = 88 cm$ , $d = 5 cm$ .

Forecast: CP, CG ke peeche hai → positive SM. Lekin calibers mein kitna — safe ya borderline?

Step 1 — Lever separation. $88 - 85 = 3 cm$ Yeh step kyun? Yeh cm-gap CP aur CG ke beech raw lever arm hai; hum pehle ise compute karte hain kyunki baad ka sab kuch (caliber count aur safety judgement) isi par based hai.

Step 2 — Convert karo. $SM = \frac{3}{5} = 0.6 cal$ Yeh step kyun? Number positive hai (strict sense mein stable) lekin 1-caliber safety rule se neeche hai.

Step 3 — Judgement. $0 < 0.6 < 1$ : rocket weakly self-correct karta hai, lekin CP migration (Transonic Aerodynamics) ya burn ke dauran CG shift ke liye koi cushion nahi hai. Redesign needed. Yeh step kyun? Positive hona bare stability test pass karta hai lekin safety test nahi; hum $\geq 1$ rule ke against compare karte hain taaki real-flight dangers (CP moves, CG shifts) pakad sakein jo raw positive number hide karta hai.

Verify: $0.6 \times 5 = 3 = 88 - 85$ . ✓ Yeh interval $(0, 1)$ mein hai → "stable but under-margined." ✓

Example 4b — Cell D′: exact one-caliber edge

Worked example Safety line par exactly baitha hua

$X_{C G} = 84 cm$ , $X_{C P} = 89 cm$ , $d = 5 cm$ . Kya yeh design acceptable hai?

Forecast: Gap exactly ek diameter wide hai. Kya $SM = 1$ " $\geq 1$ caliber" rule pass karta hai, aur kya tum actually ise fly karoge?

Step 1 — Lever separation. $89 - 84 = 5 cm$ Yeh step kyun? Pehle raw cm-gap nikalo; yahan yeh exactly diameter ke barabar hai, jo is borderline case ka poora point hai.

Step 2 — Convert karo. $SM = \frac{5}{5} = 1 cal$ Yeh step kyun? $d$ se divide karne par pata chalta hai ki gap exactly one caliber hai — safety rule ka knife-edge.

Step 3 — Judgement. $SM = 1$ equality se $\geq 1$ satisfy karta hai, isliye paper par yeh minimum acceptable design hai. Lekin yeh boundary par hai: flight mein CP ka koi bhi forward migration ya CG ka aft shift ise 1 se neeche le jaayega. Prudent designers 1 se upar cushion rakhte hain rather than line par parking karne ke. Yeh step kyun? Rule $\geq$ use karta hai, toh equality technically pass karta hai — lekin ek judgement step mein real-flight drift bhi weigh karni chahiye, jo exactly boundary design ko edge ke upar le jaati hai.

Verify: $1 \times 5 = 5 = 89 - 84$ , aur $SM = 1$ satisfy karta hai $SM \geq 1$ (by equality) lekin $SM > 1$ nahi. ✓

Example 5 — Cell E: ballast ke liye backwards solve karo

Worked example CG move karke fix design karo

Ek rocket mein $X_{C P} = 90 cm$ (shape se fixed), $d = 5 cm$ , aur current $X_{C G} = 88 cm$ hai. Hume target $SM = 1.5$ cal chahiye. Naya CG nikalo aur kitna move karna hoga.

Forecast: $X_{C P}$ fixed rakhte hue margin badhane ke liye, kya CG nose ki taraf move karna hoga ya tail ki taraf? Compute karne se pehle guess karo.

Step 1 — Target equation likho. $1.5 = \frac{90 - X _{C G}^{'}}{5}$ Yeh step kyun? Hum SM ko known treat karte hain aur $X_{C G}^{'}$ ko unknown — usual direction invert karke.

Step 2 — $X_{C G}^{'}$ solve karo. $90 - X_{C G}^{'} = (1.5 cal) \times (5 cm/cal) = 7.5 cm \Rightarrow X_{C G}^{'} = 90 - 7.5 = 82.5 cm$ Yeh step kyun? Multiply out karo, phir CG position isolate karo. Units dhyan se: SM dimensionless hai calibers mein, aur ek caliber hai $d = 5$ cm, isliye $1.5 cal \times 5 cm/cal = 7.5$ cm — calibers cancel ho jaate hain aur hum centimetres mein wapas aa jaate hain, length $90 - X_{C G}^{'}$ se match karta hai.

Step 3 — Shift nikalo. $88 - 82.5 = 5.5 cm nose ki taraf$ Yeh step kyun? CG ko forward move karna (nose tip se distance chhota karna) $X_{C P} - X_{C G}$ badhata hai, margin raise karta hai. Toh hum nose ballast add karte hain.

Verify: Plug back karo: $(90 - 82.5) /5 = 7.5/5 = 1.5$ cal. ✓ CG forward move hua, "add nose weight" intuition se match karta hai.

Example 6 — Cell F: over-stable aur weathercocking

Worked example Zyada achhi cheez bhi problem hoti hai

$X_{C G} = 50 cm$ , $X_{C P} = 95 cm$ , $d = 5 cm$ . Margin par comment karo.

Forecast: Huge CP-CG gap → large positive SM. Kya bada hamesha better hota hai? Weathercocking warning yaad karo.

Step 1 — Lever separation. $95 - 50 = 45 cm$ Yeh step kyun? Pehle raw cm-gap compute karte hain kyunki yeh physical lever arm hai; is gap ki badi size hi over-stable danger ko flag karegi jab hum ise convert karenge.

Step 2 — Convert karo. $SM = \frac{45}{5} = 9 cal$ Yeh step kyun? Positive aur bahut bada.

Step 3 — Judgement. $9$ cal over-stable hai. Restoring torque itna strong hai ki rocket directly relative wind ki taraf snap karta hai. Crosswind mein yeh weathercocks — sharply upwind arc karta hai aur altitude loose karta hai. Sweet spot roughly $1 - 2$ cal hai; yahan hum CP forward laane ke liye fin area reduce karenge. Yeh step kyun? Large positive SM "safest" lagta hai lekin physics doosri taraf se kaata hai — sirf number ko $1 - 2$ cal sweet spot ke against interpret karne par hi weathercocking failure mode dikhti hai jo raw value hide karta hai.

Verify: $9 \times 5 = 45 = 95 - 50$ . ✓ Value 2 se bahut upar → over-stable regime confirmed.

Example 7 — Cell G: $d$ ka limiting behaviour

Worked example Agar diameter shrink ya grow kare toh?

Physical gap fix karo $X_{C P} - X_{C G} = 10 cm$ . $SM$ ko explore karo jab $d$ vary kare: $d = 20 cm$ , $d = 2 cm$ , aur limits $d \to \infty$ , $d \to 0^{+}$ .

Forecast: SM $= 10/ d$ . Jab rocket fatter hota hai, caliber-count badhta hai ya ghatata hai? Dono limits guess karo.

Step 1 — Fat rocket, $d = 20$ cm. $SM = \frac{10}{20} = 0.5 cal$ Yeh step kyun? Fat body matlab ek caliber ek badi distance hai, isliye same 10 cm kam calibers khareedta hai → less margin.

Step 2 — Thin rocket, $d = 2$ cm. $SM = \frac{10}{2} = 5 cal$ Yeh step kyun? Skinny body mein har caliber chhota hota hai, isliye same 10 cm zyada calibers count karta hai → more margin.

Step 3 — The limits. $lim_{d \to \infty} \frac{10}{d} = 0 lim_{d \to 0^{+}} \frac{10}{d} = + \infty$ Yeh step kyun? Pancake-wide rocket ( $d \to \infty$ ) ka margin vanishing ho jaata hai; infinitely thin needle ( $d \to 0^{+}$ ) ka margin unbounded ho jaata hai. Hum $0$ ke paas sirf upar se aate hain kyunki $d > 0$ required hai — $d = 0$ khud undefined hai. Isliye hum $d$ se normalize karte hain: same 10 cm ka matlab alag-alag shaped rockets ke liye alag stability hai.

Verify: $10/20 = 0.5$ , $10/2 = 5$ ; aur $10/ d$ monotonically 0 ki taraf decrease karta hai jab $d$ badhta hai. ✓

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

Intuition Figure mein kya dekhna hai

Blue curve hai $SM = 10/ d$ — woh margin jo tumhe ek fixed 10 cm gap se milta hai jab diameter change hoti hai. Isse left se right trace karo: yeh steeply girta hai phir flatten ho jaata hai, ek hyperbola hai. Dono marked dots notice karo — green dot ( $d = 2$ , SM $= 5$ ) steep part par upar hai, orange dot ( $d = 20$ , SM $= 0.5$ ) tail par neeche hai. Curve ko right edge ki taraf follow karo: yeh axis ke paas jaata hai lekin kabhi touch nahi karta — yeh $d \to \infty$ limit hai, $SM \to 0$ . Ise upar-left ki taraf follow karo: yeh bina bound ke top ki taraf shoot karta hai — yeh $d \to 0^{+}$ hai, $SM \to + \infty$ . Sabse zaroori baat, curve red dashed 1-caliber safety line ko cross karta hai $d = 10$ cm ke paas: us se zyada fat hone par same 10 cm gap safe nahi raha. Woh single picture yehi batati hai ki hum kabhi margin sirf cm mein quote nahi karte — cm-gap tabhi meaningful hoti hai jab $d$ se divide kiya jaaye.

Example 8 — Cell H: burn ke dauran CG shift karta hai

Worked example Real-world word problem

Ek rocket liftoff par $X_{C G} = 84 cm$ se shuru karta hai. Uska propellant CG ke aft mein hai, isliye use burn karne se balance point forward $X_{C G} = 76 cm$ par aa jaata hai burnout tak. Poore time $X_{C P} = 88 cm$ aur $d = 6 cm$ rahta hai. Liftoff aur burnout par static margin nikalo.

Forecast: Jab CG ke peeche ka propellant burn hota hai, kya margin badhta hai ya ghatata hai? Numbers crunch karne se pehle reason karo.

Step 1 — Liftoff margin. $SM_{lift} = \frac{88 - 84}{6} = \frac{4}{6} \approx 0.667 cal$ Yeh step kyun? Starting CG use karta hai. Note karo ki akela yeh 1 caliber se neeche hai — liftoff par risky.

Step 2 — Burnout margin. $SM_{burn} = \frac{88 - 76}{6} = \frac{12}{6} = 2 cal$ Yeh step kyun? Burnout par CG forward move ho gaya hai, CP–CG gap wide ho gaya hai, isliye margin badhta hai.

Step 3 — Interpretation. Kyunki propellant CG ke aft mein tha, use burn karne se stability improve hui (liftoff $0.67 \to$ burnout $2$ cal). Worst moment yahan liftoff hai. Lekin yeh design-dependent hai — agar propellant CG ke aage hota toh margin burn ke dauran shrink karta. Yeh step kyun? Do isolated numbers answer nahi hain; design question yeh hai ki rocket kab sabse weak hota hai? Dono margins compare karna aur trend ko propellant placement se jodna hi arithmetic ko engineering decision mein turn karta hai.

Verify: $(88 - 84) /6 = 0.6 \overline{6}$ aur $(88 - 76) /6 = 2$ . Burnout $>$ liftoff → margin increased. ✓

Example 9 — Cell I: exam twist, flight mein worst margin

Worked example Puri envelope check karo

Same rocket geometry $X_{C G} = 80 cm$ aur $d = 5 cm$ constant rakho, lekin $X_{C P}$ migrate karta hai Mach number ke saath Transonic Aerodynamics ki wajah se:

Subsonic (Mach 0.3): $X_{C P} = 92 cm$
Transonic (Mach 1.0): $X_{C P} = 88 cm$ (peeche jaata hai phir is design ka minimum-gap point)
Supersonic (Mach 2.0): $X_{C P} = 84 cm$ (high Mach par CP forward move karta hai)

Har speed par static margin nikalo aur design-limiting (worst) value batao.

Forecast: Kaun sa Mach number sabse chhota margin deta hai? Bahut se students liftoff guess karte hain — resist karo.

Step 1 — Subsonic. $SM = \frac{92 - 80}{5} = \frac{12}{5} = 2.4 cal$ Yeh step kyun? Pehle subsonic CP position par margin evaluate karte hain; kyunki $X_{C P}$ Mach par depend karta hai, har speed ka apna calculation hoga aur yeh baseline low-speed case hai.

Step 2 — Transonic. $SM = \frac{88 - 80}{5} = \frac{8}{5} = 1.6 cal$ Yeh step kyun? Transonic CP ke saath formula repeat karte hain jo shift ho gaya hai; margin recompute karna hoga kyunki moved CP numerator change karta hai.

Step 3 — Supersonic. $SM = \frac{84 - 80}{5} = \frac{4}{5} = 0.8 cal$ Yeh step kyun? Supersonic CP par ek baar aur recompute karte hain jo forward move ho gaya hai, numerator aur isliye margin shrink ho gaya hai — is case ka khatra yehi hai.

Step 4 — Worst pick karo. Minimum $0.8$ cal hai Mach 2.0 (supersonic) par, jo 1-caliber rule se neeche hai. Design envelope check fail karta hai bhale hi subsonically theek lagta ho. Fix: fins bade karo taaki high Mach par CP aft rahe.

Verify: $12/5 = 2.4$ , $8/5 = 1.6$ , $4/5 = 0.8$ ; minimum $0.8 < 1$ hai → envelope check fails. Worst case liftoff nahi hai. ✓

Active recall

Recall Scenario matrix test karo (answers hide karo)

SM negative aaya — kaun sa cell, aur kya hota hai? ::: Cell B; rocket tumble karta hai.
SM $= 0$ exactly — stable? ::: Cell C; neutrally stable, drift karta hai, unusable.
SM $= 1$ exactly — safety rule pass hota hai? ::: Cell D′; haan equality se ( $\geq 1$ ), lekin yeh bare minimum hai — cushion rakho.
Target SM aur $X_{C P}$ pata hai; CG solve karo — kaun sa cell? ::: Cell E (unknown position ke liye formula rearrange karo).
Fixed cm-gap ke saath $d \to 0^{+}$ par SM kya karta hai? ::: Cell G; SM $\to + \infty$ .
Worst margin aksar liftoff par kyun nahi hoti? ::: Cell I; $X_{C P}$ supersonically forward migrate karta hai, SM shrink karta hai.
Bada SM jaise 9 cal — problem? ::: Cell F; over-stable → weathercocking.