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

This page is a complete tour of every kind of static-margin problem you can meet. We build a map first (the scenario matrix), then work one example for each cell so no case surprises you later.

Everything here rests on the parent note: the topic note. If any symbol feels new, that's where its full picture lives. Quick reminders in plain words:

$X_{C P}$ = distance from the nose tip to the Center of Pressure (the single point where all the air-push acts).
$X_{C G}$ = distance from the nose tip to the Center of Gravity (the balance point of the mass).
$d$ = the body's widest diameter, called one caliber.
Static margin counts how many calibers the CP trails behind the CG: $SM = \frac{X _{C P} - X _{C G}}{d}$

We need $SM > 0$ (CP behind CG) and want $SM \geq 1$ for safety, per the Rocket Stability Criterion.

The scenario matrix

Every static-margin problem is really one of these cells. The examples below are each tagged with the cell they cover.

Cell	What makes it special	Sign / regime	Example
A	Textbook stable, both points known	$X_{C P} > X_{C G}$ , SM $> 1$	Ex 1
B	Unstable: CP ahead of CG	$X_{C P} < X_{C G}$ , SM $< 0$	Ex 2
C	Degenerate: CP exactly on CG	$X_{C P} = X_{C G}$ , SM $= 0$	Ex 3
D	Barely stable but below 1 caliber	$0 < SM < 1$	Ex 4
D′	Exact borderline: the one-caliber safety edge	$SM = 1$ exactly	Ex 4b
E	Inverse problem: SM is given, an unknown position (CG/ballast) is solved for	rearrange the formula	Ex 5
F	Over-stable: SM too big → weathercocking	SM $≫ 1$	Ex 6
G	Limiting behaviour: what happens as $d \to 0$ or $d \to \infty$	SM $\to \pm \infty$ or $0$	Ex 7
H	Real-world word problem: CG shifts during burn	time-varying	Ex 8
I	Exam twist: worst margin across the flight (CP migrates)	envelope check	Ex 9

Example 1 — Cell A: the clean stable case

Worked example Basic caliber count

A rocket measures $X_{C G} = 62 cm$ , $X_{C P} = 80 cm$ , diameter $d = 6 cm$ . Find the static margin and classify it.

Forecast: CP is clearly behind CG, so guess a positive answer — but is it above 1 caliber? Take a moment.

Step 1 — Find the lever separation. $X_{C P} - X_{C G} = 80 - 62 = 18 cm$ Why this step? This raw distance (in cm) is the physical gap between the air-push point and the balance point. It is the thing that creates the restoring torque.

Step 2 — Convert to calibers. $SM = \frac{18}{6} = 3 cal$ Why this step? Dividing by $d$ turns the cm-gap into a dimensionless count of body-widths, so we can compare rockets of any size.

Step 3 — Classify. $SM = 3 > 1$ → stable with a healthy margin. Why this step? A number alone isn't an answer — we must map it back to the safety rule ( $SM \geq 1$ ) to say whether the rocket actually flies straight.

Verify: Rebuild the numerator: $3 \times 6 = 18$ cm, which matches $80 - 62$ . Units: $cm / cm$ = dimensionless. ✓

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

Worked example Spot the danger

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

Forecast: Here $X_{C P} < X_{C G}$ — the air-push is in front of the balance point. Do you expect a positive or negative number? What will the rocket physically do?

Step 1 — Lever separation (keep the sign!). $X_{C P} - X_{C G} = 78 - 90 = - 12 cm$ Why this step? We must not take absolute values. The sign is the whole message: negative means CP is ahead.

Step 2 — Convert. $SM = \frac{- 12}{4} = - 3 cal$ Why this step? Same caliber conversion; the negative survives, flagging instability.

Step 3 — Physical meaning. Negative SM → any tilt grows → the rocket tumbles end over end. Fix: add fins to drag CP aft (Fin Design) or add nose weight to pull CG forward. Why this step? The sign of SM only matters because it decides the direction of the aerodynamic torque; translating " $- 3$ " into "tumbles" is what makes the number physically meaningful. Here "tilt" means the angle of attack $α$ — the angle between the rocket's body axis and the oncoming air, taken positive nose-up (see Angle of Attack). A negative SM means a small positive $α$ produces a torque that increases $α$ further, so the tilt runs away.

Verify: $- 3 \times 4 = - 12 = 78 - 90$ . ✓ Sign is negative → unstable, matching the Rocket Stability Criterion.

Example 3 — Cell C: the degenerate knife-edge

Worked example CP exactly on CG

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

Forecast: The two points coincide. Zero divided by anything is... what? And what does neutral stability feel like?

Step 1 — Lever separation. $X_{C P} - X_{C G} = 70 - 70 = 0 cm$ Why this step? No gap means no lever arm for the restoring torque.

Step 2 — Convert. $SM = \frac{0}{5} = 0 cal$ Why this step? Dividing zero by the (nonzero, since $d > 0$ ) diameter is safe and gives exactly zero.

Step 3 — Physical meaning. $SM = 0$ → neutrally stable. A tilt produces no restoring torque, so the rocket neither corrects nor tumbles on its own — it drifts wherever a gust leaves it. Unusable in practice; you always want a cushion. Why this step? The knife-edge value $0$ is easy to misread as "fine, it's not negative" — we spell out that zero torque means zero self-correction, which is why this case is unusable rather than merely marginal.

Verify: Any nonzero $d$ gives $0/ d = 0$ . Neutral case confirmed. ✓

Example 4 — Cell D: positive but below 1 caliber

Worked example Technically stable, practically risky

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

Forecast: CP is behind CG → positive SM. But by how much in calibers — safe or borderline?

Step 1 — Lever separation. $88 - 85 = 3 cm$ Why this step? This cm-gap between CP and CG is the raw lever arm; we compute it first because everything downstream (the caliber count and the safety judgement) is built from it.

Step 2 — Convert. $SM = \frac{3}{5} = 0.6 cal$ Why this step? The number is positive (stable in the strict sense) but below the 1-caliber safety rule.

Step 3 — Judgement. $0 < 0.6 < 1$ : the rocket self-corrects weakly, but leaves no cushion for CP migration (Transonic Aerodynamics) or CG shift during burn. Redesign needed. Why this step? Being positive passes the bare stability test but not the safety test; we compare against the $\geq 1$ rule to catch the real-flight dangers (CP moves, CG shifts) that a raw positive number hides.

Verify: $0.6 \times 5 = 3 = 88 - 85$ . ✓ It lands in the interval $(0, 1)$ → "stable but under-margined." ✓

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

Worked example Sitting right on the safety line

$X_{C G} = 84 cm$ , $X_{C P} = 89 cm$ , $d = 5 cm$ . Is this design acceptable?

Forecast: The gap is exactly one diameter wide. Does $SM = 1$ pass the " $\geq 1$ caliber" rule, and would you actually fly it?

Step 1 — Lever separation. $89 - 84 = 5 cm$ Why this step? We find the raw cm-gap first; here it happens to equal the diameter exactly, which is the whole point of this borderline case.

Step 2 — Convert. $SM = \frac{5}{5} = 1 cal$ Why this step? Dividing by $d$ shows the gap is exactly one caliber — the knife-edge of the safety rule.

Step 3 — Judgement. $SM = 1$ satisfies $\geq 1$ by equality, so it is the minimum acceptable design on paper. But it is right on the boundary: any forward migration of CP or aft shift of CG during flight drops it below 1. Prudent designers keep a cushion above 1 rather than parking on the line. Why this step? The rule uses $\geq$ , so equality technically passes — but a judgement step must also weigh real-flight drift, which is exactly what pushes a boundary design over the edge.

Verify: $1 \times 5 = 5 = 89 - 84$ , and $SM = 1$ satisfies $SM \geq 1$ (by equality) but not $SM > 1$ . ✓

Example 5 — Cell E: solve backwards for ballast

Worked example Design a fix by moving the CG

A rocket has $X_{C P} = 90 cm$ (fixed by its shape), $d = 5 cm$ , and current $X_{C G} = 88 cm$ . We want a target $SM = 1.5$ cal. Find the new CG and how far it must move.

Forecast: To raise the margin with $X_{C P}$ fixed, must the CG move toward the nose or toward the tail? Guess before computing.

Step 1 — Write the target equation. $1.5 = \frac{90 - X _{C G}^{'}}{5}$ Why this step? We treat SM as known and $X_{C G}^{'}$ as the unknown — inverting the usual direction.

Step 2 — Solve for $X_{C G}^{'}$ . $90 - X_{C G}^{'} = (1.5 cal) \times (5 cm/cal) = 7.5 cm \Rightarrow X_{C G}^{'} = 90 - 7.5 = 82.5 cm$ Why this step? Multiply out, then isolate the CG position. Note the units carefully: SM is dimensionless in calibers, and one caliber is $d = 5$ cm, so $1.5 cal \times 5 cm/cal = 7.5$ cm — the calibers cancel and we are back in centimetres, matching the length $90 - X_{C G}^{'}$ .

Step 3 — Find the shift. $88 - 82.5 = 5.5 cm toward the nose$ Why this step? Moving CG forward (smaller distance from nose tip) increases $X_{C P} - X_{C G}$ , raising the margin. So we add nose ballast.

Verify: Plug back: $(90 - 82.5) /5 = 7.5/5 = 1.5$ cal. ✓ CG moved forward, matching the "add nose weight" intuition.

Example 6 — Cell F: over-stable and weathercocking

Worked example Too much of a good thing

$X_{C G} = 50 cm$ , $X_{C P} = 95 cm$ , $d = 5 cm$ . Comment on the margin.

Forecast: Huge CP-CG gap → large positive SM. Is bigger always better? Recall the Weathercocking warning.

Step 1 — Lever separation. $95 - 50 = 45 cm$ Why this step? We compute the raw cm-gap first because it is the physical lever arm; the large size of this gap is precisely what will flag the over-stable danger once we convert it.

Step 2 — Convert. $SM = \frac{45}{5} = 9 cal$ Why this step? Positive and very large.

Step 3 — Judgement. $9$ cal is over-stable. The restoring torque is so strong the rocket snaps to point directly into the relative wind. In a crosswind it weathercocks — arcing sharply upwind and losing altitude. Sweet spot is roughly $1 - 2$ cal; here we'd reduce fin area to bring CP forward. Why this step? A large positive SM looks "safest" but the physics bites the other way — only by interpreting the number against the $1 - 2$ cal sweet spot do we see the weathercocking failure mode a raw value hides.

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

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

Worked example What if the diameter shrinks or grows?

Fix the physical gap $X_{C P} - X_{C G} = 10 cm$ . Explore $SM$ as $d$ varies: $d = 20 cm$ , $d = 2 cm$ , and the limits $d \to \infty$ , $d \to 0^{+}$ .

Forecast: SM $= 10/ d$ . As the rocket gets fatter, does the caliber-count grow or shrink? Guess the two limits.

Step 1 — Fat rocket, $d = 20$ cm. $SM = \frac{10}{20} = 0.5 cal$ Why this step? A fat body means one caliber is a big distance, so the same 10 cm buys fewer calibers → less margin.

Step 2 — Thin rocket, $d = 2$ cm. $SM = \frac{10}{2} = 5 cal$ Why this step? A skinny body makes each caliber small, so the same 10 cm counts as many calibers → more margin.

Step 3 — The limits. $lim_{d \to \infty} \frac{10}{d} = 0 lim_{d \to 0^{+}} \frac{10}{d} = + \infty$ Why this step? A pancake-wide rocket ( $d \to \infty$ ) has vanishing margin; an infinitely thin needle ( $d \to 0^{+}$ ) has unbounded margin. We approach $0$ only from above because $d > 0$ is required — $d = 0$ itself is undefined. This is why we normalise by $d$ : the same 10 cm means different stability for different-shaped rockets.

Verify: $10/20 = 0.5$ , $10/2 = 5$ ; and $10/ d$ decreases monotonically toward 0 as $d$ grows. ✓

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

Intuition What to see in the figure

The blue curve is $SM = 10/ d$ — the margin you get from a fixed 10 cm gap as the diameter changes. Trace it left to right: it falls steeply then flattens, a hyperbola. Notice the two marked dots — the green dot ( $d = 2$ , SM $= 5$ ) sits high on the steep part, the orange dot ( $d = 20$ , SM $= 0.5$ ) sits low on the tail. Follow the curve toward the right edge: it hugs the axis but never touches it — that is the $d \to \infty$ limit, $SM \to 0$ . Follow it up-left: it shoots toward the top without bound — that is $d \to 0^{+}$ , $SM \to + \infty$ . Crucially, the curve crosses the red dashed 1-caliber safety line near $d = 10$ cm: any fatter than that and the same 10 cm gap is no longer safe. That single picture is why we never quote a margin in cm alone — the cm-gap only becomes meaningful once divided by $d$ .

Example 8 — Cell H: CG shifts during the burn

Worked example Real-world word problem

A rocket starts with $X_{C G} = 84 cm$ at liftoff. Its propellant sits aft of the CG, so burning it pulls the balance point forward to $X_{C G} = 76 cm$ at burnout. Throughout, $X_{C P} = 88 cm$ and $d = 6 cm$ . Find the static margin at liftoff and at burnout.

Forecast: As propellant behind the CG burns off, does the margin rise or fall? Reason it through before crunching numbers.

Step 1 — Liftoff margin. $SM_{lift} = \frac{88 - 84}{6} = \frac{4}{6} \approx 0.667 cal$ Why this step? Uses the starting CG. Note this alone is below 1 caliber — risky at liftoff.

Step 2 — Burnout margin. $SM_{burn} = \frac{88 - 76}{6} = \frac{12}{6} = 2 cal$ Why this step? At burnout the CG has moved forward, widening the CP–CG gap, so the margin grows.

Step 3 — Interpretation. Because the propellant was aft of the CG, burning it improved stability (liftoff $0.67 \to$ burnout $2$ cal). The worst moment here is liftoff. But this is design-dependent — if propellant sat ahead of the CG the margin would shrink during burn. Why this step? Two isolated numbers are not the answer; the design question is when is the rocket weakest? Comparing the two margins and tying the trend to propellant placement is what turns the arithmetic into an engineering decision.

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

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

Worked example Check the whole envelope

Same rocket geometry with $X_{C G} = 80 cm$ and $d = 5 cm$ held constant, but $X_{C P}$ migrates with Mach number because of Transonic Aerodynamics:

Subsonic (Mach 0.3): $X_{C P} = 92 cm$
Transonic (Mach 1.0): $X_{C P} = 88 cm$ (moves aft then this design's minimum-gap point)
Supersonic (Mach 2.0): $X_{C P} = 84 cm$ (CP moves forward at high Mach)

Find the static margin at each speed and state the design-limiting (worst) value.

Forecast: Which Mach number gives the smallest margin? Many students guess liftoff — resist that.

Step 1 — Subsonic. $SM = \frac{92 - 80}{5} = \frac{12}{5} = 2.4 cal$ Why this step? We evaluate the margin at the subsonic CP position first; because $X_{C P}$ depends on Mach, each speed needs its own calculation and this is the baseline low-speed case.

Step 2 — Transonic. $SM = \frac{88 - 80}{5} = \frac{8}{5} = 1.6 cal$ Why this step? We repeat the formula with the transonic CP, which has shifted; the margin must be recomputed because a moved CP changes the numerator.

Step 3 — Supersonic. $SM = \frac{84 - 80}{5} = \frac{4}{5} = 0.8 cal$ Why this step? We recompute once more at the supersonic CP, which has moved forward, shrinking the numerator and hence the margin — the whole reason this case is dangerous.

Step 4 — Pick the worst. The minimum is $0.8$ cal at Mach 2.0 (supersonic), which is below the 1-caliber rule. The design fails the envelope check even though it looks fine subsonically. Fix: enlarge fins so CP stays aft at high Mach.

Verify: $12/5 = 2.4$ , $8/5 = 1.6$ , $4/5 = 0.8$ ; minimum is $0.8 < 1$ → envelope check fails. The worst case is not liftoff. ✓

Active recall

Recall Test the scenario matrix (hide answers)

SM comes out negative — which cell, and what happens? ::: Cell B; the rocket tumbles.
SM $= 0$ exactly — stable? ::: Cell C; neutrally stable, drifts, unusable.
SM $= 1$ exactly — does it pass the safety rule? ::: Cell D′; yes by equality ( $\geq 1$ ), but it's the bare minimum — keep a cushion.
You know the target SM and $X_{C P}$ ; solve for CG — which cell? ::: Cell E (rearrange the formula for the unknown position).
As $d \to 0^{+}$ with a fixed cm-gap, SM does what? ::: Cell G; SM $\to + \infty$ .
Why is the worst margin often NOT at liftoff? ::: Cell I; $X_{C P}$ migrates forward supersonically, shrinking SM.
A big SM like 9 cal — problem? ::: Cell F; over-stable → weathercocking.