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

This page is a self-testing workout for the parent topic. Every symbol used here is built in that note; if a term feels unfamiliar, follow the link. We climb five levels — L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery. Try each problem with pen and paper first, then open the solution.

Reminder of the master formula (everything below rests on it):

L1 · Recognition

These test whether you can read the definition and identify signs.

Problem 1.1

A rocket has $X_{C G} = 50 cm$ , $X_{C P} = 65 cm$ , $d = 5 cm$ . Compute the static margin and state stable / unstable.

Recall Solution 1.1

WHAT: Plug into the formula. $SM = \frac{65 - 50}{5} = \frac{15}{5} = 3 cal$ WHY: The numerator $65 - 50 = 15 cm$ is the physical gap between the two points (the double arrow in the datum figure above); dividing by $d = 5$ counts how many caliber-brackets fit in that gap. Since $SM = 3$ calibers $> 1$ : stable, with comfortable margin.

Problem 1.2

State, without computing a full number, whether each rocket is stable, unstable, or neutral:

(a) CP is behind CG.
(b) CP is exactly on top of CG.
(c) CP is ahead of (in front of) CG.

Recall Solution 1.2

The sign of $X_{C P} - X_{C G}$ decides everything, and here is the physical WHY (not just the algebra). When a gust tilts the rocket by an angle of attack, the air pushes at the CP; the rocket pivots about the CG. Force $\times$ lever = moment, and the lever is exactly the gap $X_{C P} - X_{C G}$ :

(a) CP behind CG → gap $> 0$ → the air-push sits behind the pivot, so it swings the tail around and the nose back into the wind — a restoring moment. $SM > 0$ → stable ✅
(b) CP on CG → gap $= 0$ → zero lever → no moment → nothing corrects the tilt. $SM = 0$ → neutral ⚠️
(c) CP ahead of CG → gap $< 0$ → the air-push sits in front of the pivot, so it swings the nose further off — an amplifying moment. $SM < 0$ → unstable / tumbles ❌

A positive lever arm (CP aft of CG) is precisely what produces the restoring moment — that is why the Rocket Stability Criterion demands $SM > 0$ .

L2 · Application

Now you run the formula on realistic inputs and convert units.

Problem 2.1

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

Recall Solution 2.1

WHAT first: put both lengths in the same unit. $1.20 m = 120 cm$ , $1.35 m = 135 cm$ . WHY: the formula only gives a clean caliber count if numerator and $d$ share units. $SM = \frac{135 - 120}{10} = \frac{15}{10} = 1.5 cal$ $SM = 1.5$ calibers $> 1$ → stable, in the recommended $1$ – $2$ caliber sweet spot.

Problem 2.2

A rocket must reach exactly $SM = 2.0$ cal. Its $X_{C P} = 90 cm$ and $d = 6 cm$ . Where must the CG sit?

Recall Solution 2.2

WHAT: solve the formula for $X_{C G}$ . $2.0 = \frac{90 - X _{C G}}{6} \Rightarrow 90 - X_{C G} = 12 \Rightarrow X_{C G} = 78 cm$ WHY: we know the target margin and $X_{C P}$ ; the only unknown is the balance point. Multiply out and isolate. The CG must sit at 78 cm from the nose.

L3 · Analysis

Here we interpret shifts — points move, and you reason about the consequence. The figure below shows how the CP wanders as the rocket speeds up — study it before the problems.

Problem 3.1

A rocket starts with $X_{C G} = 100 cm$ , $X_{C P} = 112 cm$ , $d = 8 cm$ . As propellant burns, the CG moves aft to $108 cm$ (the burning mass sat forward of the CG). $X_{C P}$ stays put. Compute SM before and after. Is the rocket safe throughout?

Recall Solution 3.1

Before burn: $SM_{i} = \frac{112 - 100}{8} = \frac{12}{8} = 1.5 cal$ After burn (CG at 108): $SM_{f} = \frac{112 - 108}{8} = \frac{4}{8} = 0.5 cal$ WHY it dropped: the CG chased the CP rearward, shrinking the gap $X_{C P} - X_{C G}$ from $12$ cm to $4$ cm. Verdict: it starts safe ( $1.5$ calibers) but ends below the $1$ -caliber minimum ( $0.5$ ). Not safe throughout — this is exactly why you must check the Rocket Stability Criterion across the whole burn, not just at liftoff.

Problem 3.2

The same rocket's $X_{C P}$ actually migrates too. In the transonic regime $X_{C P}$ moves aft by $3$ cm (to $115 cm$ ) while the CG is at its mid-burn value $104 cm$ , $d = 8 cm$ . Does the aft CP shift help or hurt stability here?

Recall Solution 3.2

$SM = \frac{115 - 104}{8} = \frac{11}{8} = 1.375 cal$ Compare to the same instant if CP had stayed at $112$ : $SM = (112 - 104) /8 = 1.0$ cal. WHY it helps: moving CP aft enlarges the lever $X_{C P} - X_{C G}$ (the blue curve climbing in the migration figure), so this particular transonic shift increases margin ( $1.0 \to 1.375$ ). See Transonic Aerodynamics — but beware, in supersonic flight $X_{C P}$ often slides forward (the curve falling), which shrinks margin. Direction of the shift is what matters.

L4 · Synthesis

Combine multiple effects and design a fix.

Problem 4.1

A rocket at its worst flight condition has $X_{C P} = 140 cm$ , $X_{C G} = 133 cm$ , $d = 7 cm$ . (a) Compute the margin. (b) It's below the $1$ -caliber minimum — you decide to add nose ballast to pull the CG forward. What new $X_{C G}$ reaches exactly $SM = 1.5$ cal? (c) By how many centimetres must the CG move?

Recall Solution 4.1

(a) $SM = \frac{140 - 133}{7} = \frac{7}{7} = 1.0 cal$ Right on the edge — no safety cushion. (b) Hold $X_{C P} = 140$ fixed, solve for the new CG: $1.5 = \frac{140 - X _{C G}^{'}}{7} \Rightarrow 140 - X_{C G}^{'} = 10.5 \Rightarrow X_{C G}^{'} = 129.5 cm$ (c) Movement $= 133 - 129.5 = 3.5 cm$ toward the nose. WHY nose ballast works: it shifts the mass balance point forward, enlarging $X_{C P} - X_{C G}$ from $7$ cm to $10.5$ cm.

Problem 4.2

Alternatively, keep the CG at $133$ cm and instead enlarge the fins so the CP moves aft. To hit $SM = 1.5$ cal ( $d = 7$ ), where must $X_{C P}$ move?

Recall Solution 4.2

$1.5 = \frac{X _{C P}^{'} - 133}{7} \Rightarrow X_{C P}^{'} - 133 = 10.5 \Rightarrow X_{C P}^{'} = 143.5 cm$ The CP must move aft from $140$ to 143.5 cm, i.e. $3.5$ cm rearward. WHY fins do this: bigger tail fins add aerodynamic load at the rear, dragging the Center of Pressure backward — see Fin Design. In the datum figure (top of page) the pink arrows show the two levers: nose ballast slides the yellow CG dot forward (left), bigger fins slide the blue CP square aft (right). Both open the same gap from opposite ends.

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

L5 · Mastery

Full flight-envelope reasoning — the real engineering task.

Problem 5.1

A rocket, $d = 8 cm$ , is analysed at four flight instants. Compute SM at each, then state the minimum margin and whether the design passes the $\geq 1$ caliber rule everywhere.

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

Apply $SM = (X_{C P} - X_{C G}) /8$ at each row:

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, at the coast (supersonic) instant. WHY the worst case is NOT liftoff: in supersonic flight $X_{C P}$ slid forward (toward the nose) while the burnt-out CG stayed rearward — the gap shrank. The design FAILS: two supersonic points fall below $1$ caliber. Fix idea: enlarge fins or move CG forward until the worst instant ( $0.75$ ) reaches $\geq 1$ . The margin-versus-flight figure below shows the curve dipping below the danger line late in flight, not at launch.

Problem 5.2

For the rocket of 5.1, you add fins that shift every $X_{C P}$ value aft by a constant $Δ$ cm. Find the smallest $Δ$ (to the nearest $0.1$ cm) that makes the worst-case margin exactly $1.0$ cal. ( $d = 8$ .)

Recall Solution 5.2

WHAT: the worst case is the coast instant, $X_{C P} = 146$ , $X_{C G} = 140$ . Shifting CP aft by $Δ$ gives $X_{C P} = 146 + Δ$ . $1.0 = \frac{( 146 + Δ ) - 140}{8} \Rightarrow 6 + Δ = 8 \Rightarrow Δ = 2.0 cm$ WHY the worst case sets $Δ$ : a constant shift raises all margins equally, so once the minimum reaches $1.0$ , every other instant is already above it. Adding $Δ = 2.0$ cm of aft CP shift makes the design pass everywhere. Check the new minimum: coast becomes $1.0$ , burnout becomes $(148 + 2 - 141) /8 = 9/8 = 1.125$ — both now $\geq 1$ caliber. ✅

Wrap-up recall

Recall One-line takeaways (hide answers)

Formula for static margin ::: $SM = (X_{C P} - X_{C G}) / d$ , aft-positive from nose tip.
Negative SM means ::: CP ahead of CG → tumbling.
Two ways to increase margin ::: move CP aft (bigger fins) or CG forward (nose weight).
Why not maximise margin ::: over-stability causes weathercocking, altitude loss.
Where is margin usually worst ::: often supersonic/transonic, NOT liftoff — check whole envelope.
The safety rule ::: $SM \geq 1$ caliber at every flight instant.