Question bank — Dynamic stability — pitch - yaw damping derivatives

False. Static stability ($C_{m_{\alpha }}<0$) only guarantees it starts turning back; without damping ($C_{m_q}<0$) it overshoots forever or, at low $\rho$, barely calms. You need both — see Static Stability — Center of Pressure & Margin.

True. The damping moment must oppose the pitch rate to remove energy from the oscillation, and that sign convention shows up as $C_{m_q}<0$.

False. The coefficient is geometry-fixed, but the physical moment $\propto \rho V{d}^2{C}_{m_q}$; as the rocket climbs, $\rho$ drops (see Atmospheric Density vs Altitude) and $V$ changes, so real damping weakens dramatically.

True. Pitch and yaw sweep geometrically identical planes on a body of revolution, so the second-moment-of-area integral is the same in both planes.

False. It grows as the square of the distance: a fin at distance $x$ both sweeps faster (speed $qx$) and pushes on a longer lever arm ($x$), so torque $\propto x\cdot x=x^2$.

False. Fins ahead of CG make $C_{m_{\alpha }}>0$ (statically unstable): ${\omega }_n^2<0$, giving pure divergence, not an oscillation — and damping cannot rescue an unstable body.

True. The damping moment is proportional to the rate $q$; at the instant the rocket is momentarily not rotating (e.g. at maximum overshoot angle), $q=0$ so the damping contribution vanishes — only the static spring acts there.

True in the ideal linear model — with no damping the Damped Harmonic Oscillator reduces to a pure oscillator; in reality other losses eventually bleed energy, but the design cannot rely on that.

Wrong derivative — that's $C_{m_{\alpha }}$ (static). $C_{m_q}=\partial C_m/\partial \hat{q}$, the change of moment with rate, not with angle.

It opposes the velocity $\dot{\theta }$, not the displacement. The static restoring moment opposes $\theta$; damping opposes the rate, which is why it appears as a $\dot{\theta }$ term in the equation of motion.

Doubling the arm quadruples damping because $C_{m_q}\propto x^2$. Expecting a factor of 2 ignores that the local sweep speed $qx$ also scales with $x$.

Missing the factor of 2: the non-dimensional pitch rate is $\hat{q}=qd/(2V)$, scaling the pitch rate by a chord-time (half-diameter over speed).

Magnitude does drive damping via $x^2$, but nose surfaces (ahead of CG) simultaneously destabilise statically. The clean damping story assumes the aerodynamic area sits behind the CG; nose area is a stability liability, not a damping bonus.

Backwards. The damping moment $\propto V$ (and $\propto \rho$), so at the low speed and thin air near apogee it is weakest, which is exactly when tumbling risk is highest.

No — $\zeta \propto 1/(I_y{\omega }_n)$ and ${\omega }_n\propto 1/\sqrt{I_y}$, so $\zeta \propto 1/\sqrt{I_y}$. A heavier, longer body (larger $I_y$) actually lowers $\zeta$. See Moments of Inertia of a Rocket.

One factor of $x$ is the transverse sweep speed $qx$ that creates the local angle of attack; the second factor of $x$ is the moment arm turning that force into torque. Their product is $x^2$ — the second moment of aerodynamic area.

A tail sweeping downward sees an upward extra angle of attack, so the air pushes it back up — the resulting moment opposes the very rotation that caused it, which mathematically is a negative contribution to $C_m$ with respect to $\hat{q}$.

$C_{m_{\alpha }}<0$ gives a real natural frequency ${\omega }_n$ (the spring that turns it back), while $C_{m_q}<0$ gives a positive damping ratio $\zeta$ (the dashpot that shrinks each swing). One without the other is either a permanent oscillator or a divergence.

Because damping scales as $x^2$: modest rearward movement of the aerodynamic centre yields a squared payoff, far outpacing changes to fin area or airspeed. See Fin Design & Sizing.

Because dynamic stability depends on the physical moment $\propto \rho V$ and on $I_y$ — different flight speeds, altitudes, or mass distributions change ${\omega }_n$ and $\zeta$ even when the coefficient is fixed.

For small angles the static moment is linear in $\theta$ (spring) and the damping moment is linear in $\dot{\theta }$ (dashpot); a linear second-order ODE with those two terms is the damped oscillator, so the whole Damped Harmonic Oscillator toolbox applies directly.

The damping moment $\propto \rho$ goes to zero because there is no air to push on the swept fins; aerodynamic damping simply switches off, and attitude control must come from thrust vectoring or spin stabilisation instead.

None. With $x=0$ both the sweep speed $qx$ and the lever arm vanish, so its $x^2$ contribution to the integral is zero — it neither sweeps nor levers.

At peak angle the rate $\dot{\theta }=0$, so the static restoring moment (largest here) does all the work while the damping moment is momentarily zero; damping does its work during the swing, not at the turnaround.

No oscillation at all. $C_{m_{\alpha }}>0$ makes ${\omega }_n^2<0$, so the roots are real and positive: the disturbance grows exponentially (divergence). Damping terms cannot create the oscillation that stability would need.

Technically stable but marginal — oscillations decay so slowly that gusts or thrust misalignment can keep re-exciting them. Designers typically require $\zeta \gtrsim 0.05$–$0.1$ so swings actually shrink within a few cycles.

Practical structure and static-margin limits: too much rearward area can overshoot the desired static margin (making the rocket over-stable and gust-sensitive) and adds tail mass raising $I_y$, which lowers $\zeta$ again — so the $x^2$ gain competes with these penalties.