Dynamic stability — pitch - yaw damping derivatives: Worked examples

Before we start, a one-line reminder of every symbol we use (each was built in the parent-linked topic — Static Stability — Center of Pressure & Margin and Barrowman Equations give the aero background):

Definition Symbol dictionary (plain words)

$x$ ::: distance measured behind (aft of) the center of gravity, in metres. Positive = behind CG.
$α$ (alpha) ::: angle of attack — the angle between the body axis and the oncoming air (defined above, Figure s03), in radians.
$q$ ::: pitch rate — how fast the nose is swinging up or down, in radians per second.
$V$ ::: flight speed along the body axis, m/s.
$ρ$ ::: air density, kg/m³ (falls with altitude — see Atmospheric Density vs Altitude).
$S$ ::: a fixed reference area (usually the body cross-section), m². Just a bookkeeping constant.
$d$ ::: reference length (the body diameter), m.
$c_{n_{α}} A$ ::: "how much sideways force per unit angle of attack" a surface makes, m²/rad. Big fins → big number.
$C_{m}$ ::: the pitching-moment coefficient — the nose-up/nose-down torque made dimensionless by $\frac{1}{2} ρ V^{2} S d$ . It's the "torque, in pure-number form" so different rockets can be compared.
$\overset{q}{^}$ ::: the non-dimensional pitch rate, $\overset{q}{^} = q d / (2 V)$ — the raw rate $q$ scaled by the time $d / (2 V)$ air takes to cross the body. This turns "rad/s" into a pure number so $C_{m_{q}}$ can be a pure number too.
$C_{m_{q}}$ ::: the pitch-damping derivative, $C_{m_{q}} = \partial C_{m} / \partial \overset{q}{^}$ — how much the moment coefficient changes per unit of non-dimensional pitch rate. A pure number telling us how strongly the rocket resists a pitch rate.
$C_{m_{α}}$ ::: the static-stability derivative, $C_{m_{α}} = \partial C_{m} / \partial α$ — how much the moment coefficient changes per unit angle of attack $α$ . This is the spring (restoring toward the wind); it must be negative for static stability (see Static Stability — Center of Pressure & Margin). Do not confuse it with the damping derivative $C_{m_{q}}$ : one reacts to angle, the other to rate.
$I_{y}$ ::: moment of inertia about the pitch axis (see Moments of Inertia of a Rocket), kg·m².
$ζ$ (zeta) ::: damping ratio — 0 means it rings forever, ≥1 means no oscillation at all.

The two master formulas we'll hammer on (both from the parent):

The scenario matrix

Every problem this topic can throw is one (or a blend) of these cells. Each row names a "kind of input" and the physical question it tests.

#	Cell class	What's special about the input	Worked in
A	Baseline sign	ordinary fins behind CG, $x_{f} > 0$	Ex 1
B	$x^{2}$ scaling	same fins, distance changed	Ex 2
C	Distributed area	area spread over a length, must integrate	Ex 3
D	Degenerate: zero rate	$q = 0$ — is there any damping moment?	Ex 4
E	Sign flip (danger)	fins ahead of CG, static instability	Ex 5
F	Limiting: altitude	$ρ \to$ small, damping fades	Ex 6
G	Real-world word problem	full $ζ$ from geometry + mass	Ex 7
H	Exam twist	two fin sets, superposition, which dominates	Ex 8
I	Straddling the CG	loading with $x < 0$ AND $x > 0$ , sign-aware integral	Ex 9

The figures below carry the geometry for the cases where a picture does the arguing. Figure s01 illustrates Cell A/B (the $x^{2}$ weighting used in Ex 1–3); Figure s02 illustrates Cell E (Ex 5); Figure s03 defines the angle of attack $α$ used throughout.

Figure — Dynamic stability — pitch - yaw damping derivatives

The red bar in Figure s01 is the outermost surface at $x = 2.0$ m: its $x^{2}$ contribution towers over the near stations. Use it to see why moving fins back (Ex 2) quadruples damping and why the far end of a distributed load (Ex 3) dominates.

Figure s03 is the picture behind the symbol $α$ : the red arrow is the oncoming air, tilted by $α$ from the body axis.

Worked examples

Worked example Ex 1 — Cell A: baseline sign & magnitude

Fins lumped at $x_{f} = 0.6$ m behind CG, $c_{n_{α}} A_{f} = 8$ m²/rad, $S = 0.008$ m², $d = 0.1$ m. Find $C_{m_{q}}$ .

Forecast: guess the sign first — will it be $+$ or $-$ ? And roughly, tens or tens-of-thousands?

Write the lumped formula: $C_{m_{q}} = - \frac{2}{S d ^{2}} (c_{n_{α}} A_{f}) x_{f}^{2}$ . Why this step? All area sits at one station, so the integral collapses to value × $x_{f}^{2}$ . Look at the red bar in Figure s01 — this is that single dominant station.
Compute the denominator $S d^{2} = 0.008 \times (0.1)^{2} = 8 \times 1 0^{- 5}$ . Why this step? This is the fixed non-dimensionalising constant; get it once.
Numerator: $2 \times 8 \times 0.36 = 5.76$ . So $C_{m_{q}} = - 5.76/ (8 \times 1 0^{- 5}) = - 7.2 \times 1 0^{4}$ . Why this step? $x_{f}^{2} = 0.36$ ; the minus comes from the physics — the force opposes the rotation.

Verify: sign is negative ✔ (damping, as required). Units: $C_{m_{q}}$ is dimensionless — check: $\frac{m ^{2}}{m ^{2} \cdot m ^{2}} \cdot m^{2} = 1$ ✔.

Worked example Ex 3 — Cell C: distributed area (must integrate)

Instead of a point, the aerodynamic area is spread uniformly over the aft body from $x = 0.4$ m to $x = 1.0$ m, with constant per-length loading $c_{n_{α}} \frac{d A}{d x} = k = 5$ m/rad. Same $S, d$ . Find $C_{m_{q}}$ .

Forecast: the far end should dominate — will the answer sit nearer the near limit or the far limit's contribution?

Set up: $C_{m_{q}} = - \frac{2}{S d ^{2}} \int_{0.4}^{1.0} k x^{2} d x$ . Why this step? Area is no longer at one station, so the $x^{2}$ -weighted integral must be done properly (this is what Barrowman-style integration in Barrowman Equations does). Figure s01's bars are exactly the $x^{2}$ weights being summed.
Integrate: $\int_{0.4}^{1.0} x^{2} d x = \frac{x ^{3}}{3}_{0.4}^{1.0} = \frac{1.0 - 0.064}{3} = \frac{0.936}{3} = 0.312$ . Why this step? $\int x^{2} = x^{3} /3$ ; the $x^{3}$ makes the outer end (1.0 m) contribute far more than the inner (0.4 m).
So $C_{m_{q}} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 5 \times 0.312 = - \frac{3.12}{8 \times 1 0 ^{- 5}} = - 3.9 \times 1 0^{4}$ . Why this step? Assemble the constant, the loading $k$ , and the geometric integral.

Verify: as a sanity bound, if all that area sat at the far end $x = 1.0$ : total $c_{n_{α}} A = k \times 0.6 = 3$ , giving $- \frac{2}{8 \times 1 0 ^{- 5}} \times 3 \times 1 = - 7.5 \times 1 0^{4}$ ; at the near end $x = 0.4$ : $- \frac{2}{8 \times 1 0 ^{- 5}} \times 3 \times 0.16 = - 1.2 \times 1 0^{4}$ . Our $- 3.9 \times 1 0^{4}$ lies between ✔ and leans toward the far end as expected ✔.

Worked example Ex 4 — Cell D: degenerate zero pitch rate

The rocket is flying with $q = 0$ (not currently rotating). What is the damping moment right now, and what is $C_{m_{q}}$ ?

Forecast: is $C_{m_{q}}$ zero because nothing is rotating? Guess before reading.

The physical damping moment, consistent with $C_{m} = M / (\frac{1}{2} ρ V^{2} S d)$ and $\overset{q}{^} = q d / (2 V)$ , is $M_{damp} = \frac{1}{2} ρ V^{2} S d C_{m_{q}} \overset{q}{^} = \frac{1}{4} ρ V S d^{2} C_{m_{q}} q .$ At $q = 0$ , $\overset{q}{^} = 0$ , so $M_{damp} = 0$ . Why this step? Damping is a rate effect — no rate, no damping torque. Like a shock absorber that pushes nothing when the wheel isn't moving. (Note the $V^{2}$ under $\overset{q}{^}$ collapses to a single $V$ once $\overset{q}{^} = q d / (2 V)$ is substituted — this matches the $ζ$ formula's $\frac{1}{4} ρ V S d^{2} C_{m_{q}}$ factor exactly.)
But $C_{m_{q}} = \partial C_{m} / \partial \overset{q}{^}$ is a slope of the moment-coefficient curve, not a value. The slope of the line at the origin is the same as anywhere. So $C_{m_{q}}$ is unchanged (e.g. still $- 7.2 \times 1 0^{4}$ from Ex 1). Why this step? A derivative is a property of the whole curve, independent of the current operating point.

Verify: moment $\propto q$ , and $q = 0 \Rightarrow M = 0$ ✔; derivative of a straight line is constant, so $C_{m_{q}} \neq = 0$ ✔. Units check on $M_{damp} = \frac{1}{4} ρ V S d^{2} C_{m_{q}} q$ : $\frac{kg}{m ^{3}} \cdot \frac{m}{s} \cdot m^{2} \cdot m^{2} \cdot \frac{1}{s} = \frac{kg \cdot m ^{2}}{s ^{2}} = N\cdotm$ ✔ (a torque). The two quantities ( $M$ and $C_{m_{q}}$ ) are different objects — a classic trap.

Worked example Ex 5 — Cell E: sign flip, fins ahead of CG (danger case)

A build error puts the fins ahead of the CG at $x_{f} = - 0.5$ m. Take $c_{n_{α}} A_{f} = 8$ , same $S, d$ . What is $C_{m_{q}}$ , and is the rocket safe?

Forecast: the formula has $x_{f}^{2}$ . Does a negative $x_{f}$ change the sign of $C_{m_{q}}$ ?

$C_{m_{q}} = - \frac{2}{S d ^{2}} (c_{n_{α}} A_{f}) x_{f}^{2} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 8 \times (- 0.5)^{2}$ . Why this step? $x_{f}$ is squared, so $(- 0.5)^{2} = 0.25$ — the damping term stays negative regardless of which side of the CG the fins sit.
$= - \frac{2 \times 8 \times 0.25}{8 \times 1 0 ^{- 5}} = - 5 \times 1 0^{4}$ . Damping still looks "good". Why this step? Numerically the pitch-rate damping is still stabilising.
But static stability fails: fins ahead of CG give $C_{m_{α}} > 0$ , so $ω_{n}^{2} \propto - C_{m_{α}} < 0$ . The oscillation frequency is imaginary → the motion diverges (grows), it never oscillates. Why this step? Damping only removes energy from an oscillation. If there's no restoring spring, there's no oscillation to damp — the nose just runs away. See Damped Harmonic Oscillator for why $ω_{n}^{2} < 0$ means exponential growth.

Verify: $C_{m_{q}} = - 5 \times 1 0^{4} < 0$ (damping term fine) ✔; but $C_{m_{α}} > 0 \Rightarrow ω_{n}^{2} < 0 \Rightarrow$ divergence ✔. Lesson: a good $C_{m_{q}}$ cannot rescue a statically unstable rocket.

Figure s02 shows both bodies: the red fins sit behind the CG ( $x > 0$ , safe, $ω_{n}^{2} > 0$ ); the black fins sit ahead ( $x < 0$ ) and the nose diverges. Note both give a negative $C_{m_{q}}$ — the picture makes clear that damping sign alone tells you nothing about safety.

Worked example Ex 6 — Cell F: limiting behaviour with altitude

The Ex 1 rocket has damping ratio $ζ_{0} = 0.06$ near the ground ( $ρ_{0} = 1.225$ kg/m³). At 20 km altitude $ρ = 0.089$ kg/m³ and speed is unchanged. Estimate $ζ$ there (hold $ω_{n}$ roughly constant for a quick estimate).

Forecast: does $ζ$ drop a little or a lot?

From the formula, the physical damping term $\propto ρ V$ ; with $V$ fixed, $ζ \propto ρ$ (treating $ω_{n}$ as slowly varying for the estimate). Why this step? $C_{m_{q}}$ is geometry (fixed), but the moment it produces carries a $ρ$ — thin air, weak shock absorber.
$ζ = ζ_{0} \times \frac{ρ}{ρ _{0}} = 0.06 \times \frac{0.089}{1.225} = 0.06 \times 0.0727 = 0.00436$ . Why this step? Scale the ground value by the density ratio.

Verify: $0.00436 ≪ 0.05$ ✔ — below the comfort line defined above, so oscillations barely decay up high. This is exactly the parent's warning: evaluate $ζ$ at every altitude. Density ratio $0.089/1.225 \approx 0.073$ ✔.

Worked example Ex 7 — Cell G: full real-world

$ζ$ from scratch A hobby rocket: $C_{m_{q}} = - 7.2 \times 1 0^{4}$ (Ex 1), $C_{m_{α}} = - 12$ /rad, $ρ = 1.225$ , $V = 80$ m/s, $S = 0.008$ m², $d = 0.1$ m, $I_{y} = 0.05$ kg·m². Find $ω_{n}$ and $ζ$ . Does it calm down?

Forecast: guess whether $ζ$ lands above or below the 0.05 comfort line.

$ω_{n} = \frac{- \frac{1}{2} ρ V ^{2} S d C _{m_{α}}}{I _{y}}$ . Top $= - \frac{1}{2} (1.225) (8 0^{2}) (0.008) (0.1) (- 12)$ . Why this step? $ω_{n}$ needs the static derivative $C_{m_{α}} < 0$ ; the two minus signs make it positive.
$\frac{1}{2} (1.225) (6400) = 3920$ ; $\times 0.008 = 31.36$ ; $\times 0.1 = 3.136$ ; $\times (-) (- 12) = 37.63$ . Divide by $I_{y}$ : $37.63/0.05 = 752.6$ . $ω_{n} = 752.6 = 27.4$ rad/s. Why this step? Chain the constants carefully; $ω_{n}$ is how fast it would ring undamped.
$ζ = \frac{- \frac{1}{4} ρ V S d ^{2} C _{m_{q}}}{2 I _{y} ω _{n}}$ . Top $= - \frac{1}{4} (1.225) (80) (0.008) (0.01) (- 7.2 \times 1 0^{4})$ . Why this step? This is the dashpot term; $C_{m_{q}} < 0$ makes it positive.
$\frac{1}{4} (1.225) = 0.30625$ ; $\times 80 = 24.5$ ; $\times 0.008 = 0.196$ ; $\times 0.01 = 0.00196$ ; $\times 7.2 \times 1 0^{4} = 141.1$ . Bottom $= 2 (0.05) (27.4) = 2.744$ . $ζ = 141.1/2.744 = 51.4$ . Why this step? Assemble both parts of the ratio.

Verify: $ζ = 51.4$ . That's $≫ 1$ — this configuration is hugely over-damped (a very stubby, heavily finned rocket returns to trim without ringing). $ζ > 0$ ✔ and well above 0.05 ✔ — it definitely calms down. $ω_{n} = 27.4$ rad/s ✔.

Worked example Ex 8 — Cell H: exam twist, two fin sets (superposition)

A rocket has canards at $x_{1} = - 0.3$ m with $c_{n_{α}} A_{1} = 2$ and tail fins at $x_{2} = + 0.9$ m with $c_{n_{α}} A_{2} = 6$ . Same $S, d$ . Which set dominates the damping, and what is total $C_{m_{q}}$ ?

Forecast: the tail is 3× the area and farther out — but by how much does it win?

Damping is additive (the integral splits): $C_{m_{q}} = - \frac{2}{S d ^{2}} [(c_{n_{α}} A_{1}) x_{1}^{2} + (c_{n_{α}} A_{2}) x_{2}^{2}]$ . Why this step? Each surface contributes independently to the $x^{2}$ -weighted integral; superpose.
Canard term: $2 \times (- 0.3)^{2} = 2 \times 0.09 = 0.18$ . Tail term: $6 \times (0.9)^{2} = 6 \times 0.81 = 4.86$ . Why this step? Compute each $c_{n_{α}} A \cdot x^{2}$ separately to compare their weights.
Sum $= 5.04$ . So $C_{m_{q}} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 5.04 = - 1.26 \times 1 0^{5}$ . Why this step? Multiply the pooled second-moment by the constant.

Verify: tail contributes $4.86/5.04 = 96.4%$ of the damping ✔ — the far-back set dominates overwhelmingly because of the $x^{2}$ weighting, even though it's only 3× the area (it wins by $3 \times 3^{2} = 27$ -ish once the arm is also 3× longer). Total sign negative ✔.

Worked example Ex 9 — Cell I: loading that straddles the CG (sign-aware integral)

A body has aerodynamic loading spread uniformly from $x = - 0.4$ m (ahead of CG) to $x = + 0.8$ m (behind CG), constant $c_{n_{α}} \frac{d A}{d x} = k = 5$ m/rad. Same $S, d$ . Find $C_{m_{q}}$ . Does the ahead-of-CG part subtract from damping?

Forecast: the front part has $x < 0$ — will its contribution be negative (weakening damping) or positive (still helping)?

Set up the same integral, now across a range that crosses zero: $C_{m_{q}} = - \frac{2}{S d ^{2}} \int_{- 0.4}^{0.8} k x^{2} d x$ . Why this step? The formula weights by $x^{2}$ , which is never negative. So every strip — front or back — adds positively to the second moment. Damping does not cancel across the CG.
Integrate: $\int_{- 0.4}^{0.8} x^{2} d x = \frac{x ^{3}}{3}_{- 0.4}^{0.8} = \frac{0.512 - ( - 0.064 )}{3} = \frac{0.576}{3} = 0.192$ . Why this step? $(- 0.4)^{3} = - 0.064$ , and subtracting a negative adds it — this is the sign-aware part. The front section still contributes because $x^{2}$ un-signs it.
$C_{m_{q}} = - \frac{2}{8 \times 1 0 ^{- 5}} \times 5 \times 0.192 = - \frac{1.92}{8 \times 1 0 ^{- 5}} = - 2.4 \times 1 0^{4}$ . Why this step? Assemble constant × loading × integral as before.

Verify: the front strip's $x^{2}$ is genuinely positive: $\int_{- 0.4}^{0} x^{2} d x = 0.02133 > 0$ ✔, so it adds to damping, it does not subtract. Total $- 2.4 \times 1 0^{4} < 0$ ✔. Moral: for damping the CG side is irrelevant (only $∣ x ∣$ matters); for static stability ( $C_{m_{α}}$ , which weights by $x^{1}$ ) the sign would matter a lot.

Recall Quick self-test across the matrix

Which cell asks whether $C_{m_{q}}$ is zero when the rocket isn't rotating? ::: Cell D — and the answer is no; $C_{m_{q}}$ is a slope, only the moment is zero. Fins ahead of the CG: is $C_{m_{q}}$ still negative? ::: Yes ( $x^{2}$ ), but the rocket still diverges because $C_{m_{α}} > 0$ (Cell E). Two fin sets — do their damping contributions add? ::: Yes, the integral is linear, so superpose the $c_{n_{α}} A x^{2}$ terms (Cell H). Loading straddling the CG — does the front part cancel the rear? ::: No; $x^{2} \geq 0$ so every strip adds to damping regardless of side (Cell I). Why does $ζ$ collapse at altitude? ::: The damping moment carries $ρ$ , which falls steeply — $ζ \propto ρ$ (Cell F).

Related: Fin Design & Sizing · Moments of Inertia of a Rocket · Atmospheric Density vs Altitude