Exercises — Dynamic stability — pitch - yaw damping derivatives

Damped. The rule from the parent note: a stable, damped rocket has $C_{m_q}<0$, meaning the aerodynamic moment always opposes the pitch rate — a shock absorber removing energy from the swing. A positive $C_{m_q}$ would mean the moment adds to the pitch rate — it feeds the oscillation, making swings grow. That is anti-damping (aerodynamically unstable in rate).

• $C_{m_{\alpha }}$ → (a) spring: moment per angle of attack $\alpha$. Must be negative for static stability.
• $C_{m_q}$ → (b) dashpot: moment per (scaled) pitch rate. Must be negative for damping.
• $\hat{q}$ → (c) the pitch rate $q$ scaled to be dimensionless, $\hat{q}=qd/(2V)$.
• $\zeta$ → (d) damping ratio: 0 = never dies, 1 = critically damped, >1 = no oscillation.

$\hat{q}=\frac{qd}{2V}=\frac{3\times 0.1}{2\times 200}=\frac{0.3}{400}=7.5\times 10^{-4}.$ Tiny — as expected: real rockets pitch slowly compared to how fast air flows past them. $\hat{q}$ is essentially "how far the nose turns in the time air travels half a diameter."

With area lumped at one station, the integral $\int c_{n_{\alpha }}{x}^{2}dA$ becomes just $(c_{n_{\alpha }}{A}_f)x_f^2$:

First $\hat{q}=\frac{2\times 0.12}{2\times 180}=\frac{0.24}{360}=6.667\times 10^{-4}$. Dynamic-pressure block: $\frac{1}{2}\rho V^2=0.5\times 1.0\times 180^2=16200\text{ Pa}$. Then $\frac{1}{2}\rho V^{2}Sd=16200\times 0.01\times 0.12=19.44$. $M=19.44\times (-4.69\times 10^4)\times 6.667\times 10^{-4}\approx -607.7\text{ N⋅m}.$ Negative → the moment opposes the positive pitch rate. That is the shock absorber pushing back.

$C_{n_r}=C_{m_q}\approx -4.69\times 10^4$. Pitch and yaw are the same geometry rotated 90° for a body of revolution — the air can't tell the difference. (See Fin Design & Sizing for how this breaks with an odd number of fins.)

The integral becomes a sum over lumps: $\int c_{n_{\alpha }}{x}^{2}dA=(1.5)(0.3{)}^2+(5)(0.9{)}^2=0.135+4.05=\mathrm{4.185.}$ $C_{m_q}=-\frac{2}{0.01\times 0.0144}\times 4.185=-\frac{8.37}{1.44\times 10^{-4}}\approx -5.81\times 10^4.$ Fin fraction: $4.05/4.185=0.968$, i.e. ≈96.8% of the damping is from the fins. Interpretation: even though the tail cone has 30% of the fins' area, the $x^2$ weighting ($0.9^2/0.3^2=9\times$ leverage) makes the fins utterly dominate. This is why damping design = fin placement design.

Invert $C_{m_q}=-\frac{2}{S{d}^2}(c_{n_{\alpha }}{A}_f)x_f^2$: $x_f^2=\frac{-C_{m_q}S{d}^2}{2(c_{n_{\alpha }}{A}_f)}=\frac{8.0\times 10^4\times 1.44\times 10^{-4}}{2\times 5}=\frac{11.52}{10}=\mathrm{1.152.}$ $x_f=\sqrt{1.152}\approx 1.074\text{ m}.$ Only the positive root is physical (fins must be behind the CG). A negative $x_f$ would put them ahead, which we address next.

Damping: $x_f^2=(-0.5{)}^2=0.25$ — the square kills the sign, so $C_{m_q}=-\frac{2}{1.44\times 10^{-4}}\times 5\times 0.25\approx -1.74\times 10^4<0.$ Numerically still "damped." But static stability depends on $C_{m_{\alpha }}$, whose sign follows the first moment of area — fins ahead of CG give $C_{m_{\alpha }}>0$ (destabilising). So ${\omega }_n^2<0$: the motion is pure divergence, not oscillation, and there is nothing for the damping to damp. Damping only helps a statically stable body.

Natural frequency (the spring, from static stability): ${\omega }_n=\sqrt{\frac{-\frac{1}{2}\rho V^{2}Sd{C}_{m_{\alpha }}}{I_y}}.$ $\frac{1}{2}\rho V^2=0.5\times 1.225\times 150^2=13781.25$. Times $Sd=0.01\times 0.12=1.2\times 10^{-3}$ gives $16.5375$. Times $-C_{m_{\alpha }}=6.0$: $99.225$. Divide by $I_y=0.9$: $110.25$. ${\omega }_n=\sqrt{110.25}=10.5\text{ rad/s}.$ Damping ratio (the dashpot): $\zeta =\frac{-\frac{1}{4}\rho VS{d}^2{C}_{m_q}}{2I_y{\omega }_n}.$ Numerator: $\frac{1}{4}\rho VS{d}^2=0.25\times 1.225\times 150\times 0.01\times 0.0144=6.615\times 10^{-3}$. Times $-C_{m_q}=4.7\times 10^4$: $310.905$. Denominator: $2\times 0.9\times 10.5=18.9$. $\zeta =\frac{310.905}{18.9}\approx \mathrm{16.4.}$ That is hugely overdamped ($\zeta >1$) — no oscillation at all, it just eases back. (Realistic $C_{m_q}$ magnitudes and inertias usually give $\zeta$ around $0.05$–$0.3$; this illustrative set is deliberately stiff.)

${\omega }_d=10.5\sqrt{1-0.08^2}=10.5\sqrt{0.9936}=10.5\times 0.99679\approx 10.47\text{ rad/s}.$ The envelope decays as $e^{-\zeta {\omega }_{n}t}$, so the $1/e$ time is $\tau =\frac{1}{\zeta {\omega }_n}=\frac{1}{0.08\times 10.5}=\frac{1}{0.84}\approx 1.19\text{ s}.$ So each big swing barely changes frequency (damped $\approx$ undamped for small $\zeta$), but the amplitude shrinks by $\sim 63\%$ every $1.19$ s.

Look at how each piece scales with $\rho$. From the formulas, ${\omega }_n\propto \sqrt{\rho }$ and the damping-ratio numerator $\propto \rho$, denominator $\propto {\omega }_n\propto \sqrt{\rho }$. So $\zeta \propto \frac{\rho }{\sqrt{\rho }}=\sqrt{\rho }.$ Therefore ${\zeta }_{\text{alt}}={\zeta }_0\sqrt{\frac{\rho }{{\rho }_0}}=0.08\sqrt{\frac{0.30}{1.225}}=0.08\sqrt{0.2449}=0.08\times 0.4949\approx \mathrm{0.0396.}$ $0.0396<0.05$ → fails the threshold. The rocket is now dangerously lightly damped at altitude even though it was fine at launch. This is exactly the third parent-note mistake made concrete. (See Atmospheric Density vs Altitude.)

(a) ${\omega }_n$: $\frac{1}{2}\rho V^2=0.5\times 0.40\times 250^2=12500$. Times $Sd=1.2\times 10^{-3}$: $15.0$. Times $-C_{m_{\alpha }}=8$: $120$. Divide by $I_y=1.2$: $100$. ${\omega }_n=\sqrt{100}=10.0\text{ rad/s}.$ (b) Required $C_{m_q}$: rearrange $\zeta =\frac{-\frac{1}{4}\rho VS{d}^2{C}_{m_q}}{2I_y{\omega }_n}$: $C_{m_q}=-\frac{2I_y{\omega }_n\zeta }{\frac{1}{4}\rho VS{d}^2}.$ Numerator: $2\times 1.2\times 10.0\times 0.10=2.4$. Denominator: $0.25\times 0.40\times 250\times 0.01\times 0.0144=3.6\times 10^{-3}$. $C_{m_q}=-\frac{2.4}{3.6\times 10^{-3}}=-\mathrm{666.67.}$ (c) $x_f$: invert $C_{m_q}=-\frac{2}{S{d}^2}(c_{n_{\alpha }}{A}_f)x_f^2$: $x_f^2=\frac{-C_{m_q}S{d}^2}{2(c_{n_{\alpha }}{A}_f)}=\frac{666.67\times 1.44\times 10^{-4}}{2\times 5}=\frac{0.096}{10}=9.6\times 10^{-3}.$ $x_f=\sqrt{9.6\times 10^{-3}}\approx 0.098\text{ m}.$ So placing the fins ~9.8 cm behind CG meets the target at the worst flight point. Anything farther back gives margin. (Cross-check inertia against Moments of Inertia of a Rocket.)

Both $C_{m_q}$ and hence $\zeta$ are linear in $c_{n_{\alpha }}{A}_f$ (with $x_f$, $\rho$, $V$ fixed). So ${\zeta }_{\text{new}}=0.10\times \frac{4.25}{5}=0.10\times 0.85=\mathrm{0.085.}$ $0.085>0.05$ → still above the safety threshold, though margin shrank from a design $\zeta =0.10$. A robust design should have targeted higher (e.g. $\zeta =0.12$) so a 15% build tolerance can't push it near the edge. This is the mastery lesson: design to the tolerance-worst case, not the nominal.

Compute $C_{m_q}$ from $-\frac{2}{S{d}^2}\int c_{n_{\alpha }}{x}^{2}dA$ (mind the $x^2$), turn it into $\zeta \propto \sqrt{\rho }$, and design so the minimum $\zeta$ over altitude and build tolerance still beats ~0.05.