Static stability — weather-cocking tendency: Worked examples

This page is the drill floor for static stability. The parent note built the ideas; here we grind through every case class the topic can throw at you, so that no exam scenario is new.

Before we start, two symbols you will use in every line — earned from the parent note:

The scenario matrix

Everything this topic can ask lives in one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	Case class	What is special	Example
A	$ℓ > 0$ , normal margin (1–2 cal)	The healthy design	Ex 1
B	$ℓ > 0$ but over-stable ( $SM ≫ 2$ )	Too much margin	Ex 2
C	$ℓ < 0$ (CP ahead of CG)	Unstable — and how to fix	Ex 3
D	$ℓ = 0$ exactly	Degenerate: neutral stability	Ex 4
E	Compute the actual restoring moment (signs & units)	Force → torque chain	Ex 5
F	Zero / limiting inputs ( $α = 0$ , $V \to 0$ , $V \to$ large)	What vanishes, what blows up	Ex 6
G	Real-world word problem (crosswind launch)	Turn a story into $ℓ, SM$	Ex 7
H	Exam twist: CP moves mid-flight	Stability can be lost in flight	Ex 8

The tools we reuse (built in the parent note): the static margin $SM = ℓ / d$ measured in calibers (one caliber = one body diameter $d$ ), and the moment $M = - \frac{1}{2} ρ V^{2} A C_{N_{α}} ℓ α$ . For the dynamic-pressure factor $\frac{1}{2} ρ V^{2}$ see Dynamic Pressure and Aerodynamic Coefficients; for $α$ see Angle of Attack and Aerodynamic Forces.

Figure — Static stability — weather-cocking tendency

Look at the figure: the same rocket outline appears in each cell, with the CG (teal dot), CP (orange dot) and the resulting swing arrow. This is the whole matrix in one glance — refer back to it as you read.

Cell A — the healthy design

Worked example Ex 1 — Is it stable, and how stable?

A model rocket has $X_{c g} = 0.60 m$ , $X_{c p} = 0.74 m$ from the nose, body diameter $d = 0.08 m$ .

Forecast: Guess first — is CP behind CG? Roughly how many calibers of margin?

Step 1. Compute the lever arm: $ℓ = X_{c p} - X_{c g} = 0.74 - 0.60 = 0.14 m$ . Why this step? The sign of $ℓ$ decides stability at all — we always find it first.

Step 2. Since $ℓ = + 0.14 > 0$ , the CP is behind the CG → restoring torque exists → stable. Why this step? A positive lever arm means the air grabs the tail and swings the nose back into the wind (weather-cocking).

Step 3. Convert to a comparable number: $SM = ℓ / d = 0.14/0.08 = 1.75$ calibers. Why this step? Dividing by the diameter turns metres into the universal "calibers" scale so any rocket size is judged on the same 1–2 rule.

Result: $SM = 1.75$ calibers, inside the 1–2 window → a well-behaved stable rocket.

Verify: Units: $m / m$ = dimensionless ✓. Sanity: $1.75$ sits between 1 and 2 as a good design should ✓.

Cell B — too much of a good thing

Worked example Ex 2 — Over-stable rocket

Same diameter $d = 0.08 m$ , but now $X_{c g} = 0.50 m$ and $X_{c p} = 0.82 m$ .

Forecast: Bigger gap between CG and CP — will this be more desirable or a problem?

Step 1. $ℓ = 0.82 - 0.50 = 0.32 m$ . Why? Sign first: positive → still stable in principle.

Step 2. $SM = 0.32/0.08 = 4.0$ calibers. Why? Compare against the rule of thumb.

Step 3. $4.0 ≫ 2$ → over-stable. Why this matters: an over-stable rocket generates a very strong restoring torque even for tiny gusts, so it swings its nose steeply into a crosswind and flies off downwind of the launch pad.

Result: Statically stable but over-stable; expect aggressive weather-cocking. See Fin Design and Sizing — the fix is smaller fins, moving CP forward.

Verify: $ℓ / d = 0.32/0.08 = 4.0$ ✓, and $4.0 > 2$ confirms the over-stable flag ✓.

Cell C — unstable, and the two fixes

Worked example Ex 3 — Diagnose and repair

A rocket has $X_{c g} = 0.55 m$ , $X_{c p} = 0.50 m$ , $d = 0.05 m$ .

Forecast: CP number is smaller than CG number — what does that say?

Step 1. $ℓ = 0.50 - 0.55 = - 0.05 m$ . Why? Negative lever arm ⇒ CP ahead of CG.

Step 2. $SM = - 0.05/0.05 = - 1.0$ caliber. Why? A negative margin means the moment adds to any disturbance → the rocket tumbles.

Step 3 — Fix A (fins). Add larger/rearward fins to push $X_{c p}$ back to, say, $0.62 m$ . New $ℓ = 0.62 - 0.55 = + 0.07 m$ , $SM = 0.07/0.05 = 1.4$ cal → stable. Why? Fins move the pressure point, per Barrowman Equations for CP location.

Step 4 — Fix B (ballast). Instead add nose ballast to move $X_{c g}$ forward to $0.44 m$ . New $ℓ = 0.50 - 0.44 = + 0.06 m$ , $SM = 0.06/0.05 = 1.2$ cal → stable. Why? Ballast moves the mass point; both fixes make $ℓ$ positive.

Result: Original $SM = - 1.0$ (unstable); either fix restores a positive margin (1.4 or 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 — the degenerate case $ℓ = 0$

Cell E — the full force→torque chain with signs

Worked example Ex 5 — Size of the restoring moment

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

Forecast: Guess the sign of $M$ before computing — restoring means...?

Step 1 — Angle to radians. $α = 5^{\circ} = 5 \times π /180 = 0.08727 rad$ . Why? $C_{N_{α}}$ is per radian, so $α$ must be in radians for units to cancel.

Step 2 — Dynamic pressure. $q = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.2) (10 0^{2}) = 6000 Pa$ . Why? This sets the scale of every aerodynamic force — pressure the moving air can exert.

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

Step 4 — Moment about CG. $M = - N ℓ = - (2.094) (0.12) = - 0.2513 N\cdotm$ . Why the minus? Positive $ℓ$ + positive $α$ must give a negative (opposing) torque so it pushes $α$ back to zero.

Result: $M \approx - 0.25 N\cdotm$ , negative ⇒ restoring, matching the forecast.

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

Cell F — zero and limiting inputs

Worked example Ex 6 — What vanishes, what survives

Use the Ex 5 rocket ( $ℓ = 0.12$ , stable). Examine three limits of $M = - q A C_{N_{α}} ℓ α$ .

Forecast: Which of these kill the restoring torque, and which do not?

Step 1 — Zero angle of attack ( $α = 0$ ). $M = - q A C_{N_{α}} ℓ \cdot 0 = 0$ . Why? No tilt ⇒ no sideways force ⇒ no correction needed. Stability is a response to disturbance, silent when flying straight.

Step 2 — Standstill ( $V \to 0$ ). $q = \frac{1}{2} ρ V^{2} \to 0$ , so $M \to 0$ . Why this matters: a rocket on the launch rail or moving very slowly has almost no aerodynamic stability — that is exactly why launch rails/rods exist, to hold direction until $V$ is large enough. This is the limiting case beginners forget.

Step 3 — High speed ( $V$ large). $M$ grows like $V^{2}$ . Why? Restoring torque strengthens with the square of speed, so the rocket becomes stiffer as it accelerates — but if the CP moves forward at high speed (Cell H), margin can still shrink.

Result: $α = 0$ and $V \to 0$ both give $M = 0$ ; large $V$ gives $M \propto V^{2}$ (rapidly rising).

Verify: At $α = 0$ : $M = 0$ ✓. Doubling $V$ from 100 to 200 quadruples $M$ : check $M (200) / M (100) = 4$ ✓.

Cell G — real-world crosswind word problem

Worked example Ex 7 — Launch day story

"You launch a rocket with $X_{c g} = 0.90 m$ , $X_{c p} = 1.14 m$ , $d = 0.10 m$ into a steady side-wind. A judge says it should have margin under 2 calibers to avoid landing far downwind. Does it pass, and which way does it lean?"

Forecast: Guess the margin and the lean direction from the story.

Step 1 — Translate the story. "Side-wind" ⇒ the rocket flies at a small angle of attack $α > 0$ relative to airflow. Why? Weather-cocking only acts when there is a tilt to correct.

Step 2 — Lever arm. $ℓ = 1.14 - 0.90 = 0.24 m > 0$ ⇒ stable, and it will turn its nose into the wind (weather-cock). Why? CP behind CG.

Step 3 — Margin vs the judge's rule. $SM = 0.24/0.10 = 2.4$ calibers $> 2$ . Why? Compare to the stated limit.

Result: Fails the judge's rule — over-stable at 2.4 cal ⇒ it will lean hard into the wind and drift downwind of the pad. Recommend trimming fin area (moves CP forward) to reach ~1.5 cal.

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

Cell H — exam twist: CP shifts in flight

Worked example Ex 8 — Stable at launch, unstable at speed

At low speed a rocket has $X_{c p} = 0.70 m$ , $X_{c g} = 0.60 m$ , $d = 0.05 m$ . As it goes transonic the CP moves forward to $X_{c p} = 0.58 m$ (CP is not fixed — it depends on Mach and $α$ ).

Forecast: Stable at launch, sure — but what happens when the CP creeps forward?

Step 1 — Launch margin. $ℓ_{1} = 0.70 - 0.60 = 0.10 m$ , $SM_{1} = 0.10/0.05 = 2.0$ cal → stable. Why? Standard check at initial condition.

Step 2 — Transonic margin. $ℓ_{2} = 0.58 - 0.60 = - 0.02 m$ , $SM_{2} = - 0.02/0.05 = - 0.4$ cal. Why this step? Recompute with the new CP; the CG has not moved much here, so only $X_{c p}$ changed.

Step 3 — Interpret. $SM$ went from $+ 2.0$ to $- 0.4$ : the rocket loses stability mid-flight and can tumble at high speed. Why the exam loves this: it punishes the assumption "CP is fixed like CG".

Result: Stable at launch (2.0 cal), unstable transonic ( $- 0.4$ cal). Fix: larger/further-aft fins so CP stays behind CG across all Mach numbers.

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

Recall Quick self-test (reveal answers)

Which cell has $M = 0$ for every $α$ ? ::: Cell D, $ℓ = 0$ (neutral) — the lever arm multiplies the whole torque. Why does a rocket have almost no aerodynamic stability on the launch rail? ::: $V \approx 0$ ⇒ $q = \frac{1}{2} ρ V^{2} \approx 0$ ⇒ $M \approx 0$ (Cell F). A margin of 2.4 calibers means...? ::: Stable but over-stable — strong weather-cocking, drifts into the wind (Cells B, G). If CP moves forward past CG at high speed, the rocket becomes...? ::: Unstable — margin goes negative mid-flight (Cell H).

Related: Thrust Vectoring vs Passive Stability (what to do when passive margin is impossible), and the Hinglish walkthrough 3.4.10 Static stability — weather-cocking tendency (Hinglish).