Visual walkthrough — Dynamic stability — pitch - yaw damping derivatives

---

Step 1 — What is "pitch rate", and what does it do to the body?

WHAT. Picture a rocket flying to the right. Its long axis is horizontal. Now imagine the nose slowly tipping up while the tail dips down — the whole rigid body rotating about a point called the center of gravity (CG), the balance point of the rocket.

The speed of that rotation is the pitch rate, written $q$, measured in radians per second (rad/s). One radian per second means the body turns through about $57^{\circ }$ every second.

WHY we start here. Damping is a force that fights rotation. So before anything, we must describe the rotation itself and — crucially — what velocity it gives to each part of the body.

PICTURE.

Set up a coordinate: put the origin at the CG, and let $x$ measure distance toward the tail (aft). So a fin near the tail sits at some positive $x$; a point on the nose sits at negative $x$.

Here is the key fact from rigid-body motion. When a rigid body spins at rate $q$ about the CG, a point a distance $x$ from the CG moves sideways (perpendicular to the axis) at a speed

$w_{\text{local}}=qx.$

Read it term-by-term:

• $q$ — the spin rate (bigger spin, faster sideways motion),
• $x$ — the lever distance (points farther out move faster, like the rim of a merry-go-round moves faster than the center),
• $w_{\text{local}}$ — the resulting sideways (here, downward) velocity of that point relative to the air.

The tail ($x>0$, big) swings down fast. A point right at the CG ($x=0$) doesn't move sideways at all. Points on the nose ($x<0$) swing up — the opposite direction. That sign flip matters later.

---

Step 2 — That sideways motion tilts the wind: a local angle of attack

WHAT. The rocket flies forward at speed $V$. From the point of view of a tail element that is also sliding downward at $w_{\text{local}}=qx$, the oncoming air no longer arrives straight along the axis — it arrives slanted, seeming to come from slightly below. That slant is an extra angle of attack, the angle between the local airflow and the body axis.

WHY. Aerodynamic forces are produced by the angle at which air meets a surface. The rotation didn't change the wind — it changed how the wind looks to each moving element. We convert "sideways velocity" into "extra angle" because forces respond to angle, not to velocity directly.

PICTURE.

Look at the little velocity triangle. The axial leg is $V$ (forward flight). The transverse leg is $w_{\text{local}}=qx$ (the sweep from Step 1). The airflow the element feels is the diagonal — the sum of the two.

The angle of that diagonal off the axis is

$\Delta \alpha (x)=\arctan \left(\frac{w_{\text{local}}}{V}\right)\approx \frac{w_{\text{local}}}{V}=\frac{qx}{V}.$

Why is $\arctan$ the right tool? $\arctan$ answers the question "which angle has this ratio of opposite-over-adjacent?" — exactly what we need to turn the triangle's two legs into an angle. And why can we drop it to just $\frac{qx}{V}$? Because in real flight $qx$ is tiny next to $V$, so the triangle is a thin sliver, and for a thin sliver the angle (in radians) equals its tangent. This is the small-angle approximation.

Term-by-term:

• $qx$ — sideways velocity (numerator, "opposite"),
• $V$ — forward speed (denominator, "adjacent"),
• $\Delta \alpha (x)$ — extra tilt of the wind at station $x$, in radians.

Notice: $\Delta \alpha$ grows linearly with $x$. Far-back elements see the biggest wind-tilt.

---

Step 3 — A tilted wind pushes: extra normal force

WHAT. A surface meeting the air at an angle $\Delta \alpha$ feels a force perpendicular to the body axis — a normal force. More tilt → more force, in proportion (for small angles).

WHY. This is the basic linear model of aerodynamics: force = (how hard the air pushes) × (how much area) × (how sensitive that surface is to angle) × (the angle). We need the force before we can get a torque.

PICTURE.

For a small patch of surface of area $dA$ at station $x$, the incremental normal force is

$dN(x)=\underset{\text{dynamic pressure}}{\underbrace{\frac{1}{2}\rho V^2}}\underset{\text{angle-sensitivity}}{\underbrace{c_{n_{\alpha }}(x)}}\underset{\text{the tilt}}{\underbrace{\Delta \alpha (x)}}\underset{\text{patch area}}{\underbrace{dA}}.$

Term-by-term:

• $\frac{1}{2}\rho V^2$ — the dynamic pressure, the "punch" of the air. $\rho$ is air density (kg/m³), $V$ the flight speed. Faster or denser air pushes harder.
• $c_{n_{\alpha }}(x)$ — the local lift-curve slope: how many units of normal-force-coefficient you get per radian of angle for that shape. Fins have big $c_{n_{\alpha }}$; a smooth body tube has small $c_{n_{\alpha }}$.
• $\Delta \alpha (x)=\frac{qx}{V}$ — carried straight in from Step 2.
• $dA$ — the tiny patch area.

Substituting $\Delta \alpha =qx/V$:

$dN(x)=\frac{1}{2}\rho V^2{c}_{n_{\alpha }}(x)\frac{qx}{V}dA=\frac{1}{2}\rho Vq{c}_{n_{\alpha }}(x)xdA.$

One $V$ cancelled. The force on each patch is proportional to $x$ (through $\Delta \alpha$). Remember that single power of $x$ — a second one is about to appear.

---

Step 4 — Each force acts on a lever: torque, and the birth of $x^2$

WHAT. A force $dN$ acting a distance $x$ from the CG produces a torque (a twisting moment) about the CG: torque = force × lever arm = $xdN$. We add up all patches to get the total damping moment $M$.

WHY. We care about the rotation, and rotation is driven by torque, not by force. A force close to the CG barely twists; the same force far out twists a lot. The lever arm $x$ is what converts force into twist.

PICTURE.

The moment from one patch is $xdN$. Summing (integrating) over the whole rocket:

$M=-\int xdN=-\frac{1}{2}\rho Vq\int c_{n_{\alpha }}(x)\underset{x^2}{\underbrace{x\cdot x}}dA=-\frac{1}{2}\rho Vq\int c_{n_{\alpha }}(x)x^{2}dA.$

Here is the heart of the whole page. Two separate powers of $x$ multiplied together:

• one power of $x$ came from Step 1–2 (far elements sweep the air faster, so bigger $\Delta \alpha$, bigger force),
• the second power of $x$ came from Step 4 (far elements act on a longer lever).

Together: torque $\propto x^2$. This is why doubling a fin's distance from the CG quadruples its damping — the famous $x^2$ that the parent note kept insisting on.

The minus sign. Why the leading $-$? Trace the physics: nose pitches up ($q>0$), the tail swings down, so the tail meets air coming from below, which pushes the tail up. Pushing the tail up rotates the nose down — opposite to the original pitch-up. The moment fights the motion. Any moment that always opposes the rotation must carry the opposite sign to $q$, hence the $-$.

Because $\int c_{n_{\alpha }}{x}^{2}dA$ is a sum of squares (always positive), $M$ has the opposite sign to $q$ for every possible rocket. It can never accidentally drive the rotation. That is what makes it damping and not the reverse.

---

Step 5 — Cleaning up: the non-dimensional derivative

WHAT. Engineers don't quote a moment in newton-metres that changes with speed and altitude. They quote a coefficient — a pure number that captures only the shape. We divide out the messy physical factors.

WHY. We want a single geometry number that lives in a table and applies at any speed. Non-dimensionalising strips away $\rho$, $V$ and the reference size so the shape stands alone.

PICTURE.

Two standard definitions:

• The moment coefficient $C_m=\frac{M}{\frac{1}{2}\rho V^{2}Sd}$, where $S$ is a reference area and $d$ the reference length (diameter). This turns the moment into a pure number.
• The non-dimensional pitch rate $\hat{q}=\frac{qd}{2V}$ — the pitch rate scaled by a natural time, "how far the body rotates in the time air travels one radius".

The damping derivative is the slope of $C_m$ against $\hat{q}$:

$C_{m_q}\equiv \frac{\partial C_m}{\partial \hat{q}}.$

Put $M=-\frac{1}{2}\rho Vq\int c_{n_{\alpha }}{x}^{2}dA$ into $C_m$, then differentiate with respect to $\hat{q}$ (which just means replacing $q$ by $2V\hat{q}/d$ and reading off the coefficient of $\hat{q}$). All the $\rho$, $V$ cancel, and out drops:

$\boxed{C_{m_q}=-\frac{2}{S{d}^2}\int c_{n_{\alpha }}(x)x^{2}dA<0}$

Term-by-term:

• $-$ — always opposes rotation (from Step 4).
• $\frac{2}{S{d}^2}$ — the normalising factor from the two definitions above.
• $\int c_{n_{\alpha }}{x}^{2}dA$ — the second moment of aerodynamic area about the CG: the geometry, and only the geometry.

By the same argument turned $90^{\circ }$ (nose swinging left/right instead of up/down), yaw gives $C_{n_r}=C_{m_q}$ for an axisymmetric rocket — the two planes are geometrically identical.

---

Step 6 — The edge cases (never leave a scenario unshown)

WHAT. Let's test the formula at its extremes so you're never surprised.

PICTURE.

• A surface right at the CG ($x=0$). Then $\Delta \alpha =q\cdot 0/V=0$: no sweep, no extra force, no torque. Contribution $\propto 0^2=0$. Surfaces near the balance point do nothing for damping.
• A surface ahead of the CG ($x<0$, the nose). Its sweep flips sign (moves up when the tail moves down), and its lever arm flips sign — two sign flips. But $x^2=(-x{)}^2>0$: it still contributes positive to the integral, i.e. still adds damping. Distance-squared doesn't care which side.
• Air density $\rho \to 0$ (high altitude). The coefficient $C_{m_q}$ is unchanged — it's pure geometry. But the physical moment $M=-\frac{1}{2}\rho V{d}^{2}S{C}_{m_q}\hat{q}$ carries $\rho$ out front. As $\rho \downarrow$, real damping fades. The rocket's shock absorber weakens as it climbs, even though the table number never moves.
• Zero spin ($q=0$). No rotation, no sweep, no damping moment — damping only exists while the body is rotating. It is a response to motion, not a static force.

---

The one-picture summary

Read the chain left to right: spin $q$ → sideways sweep $qx$ (one $x$) → wind-tilt $\Delta \alpha =qx/V$ → force $dN$ → torque on lever $x$ (second $x$) → total $\propto x^2$, opposing the spin → the negative $C_{m_q}$.

Recall Feynman retelling — the whole walkthrough in plain words

Spin the rocket a little: the nose goes up, the tail goes down. Any part of the body away from the balance point gets dragged sideways through the air — and the farther out it is, the faster it's dragged (Step 1). To that part, the wind now seems to come at a slant (Step 2). A slanted wind pushes on the surface (Step 3), and a push out on a long arm makes a big twist (Step 4). The twist always pushes the nose back down — it fights the spin, that's the minus sign. And because "far out" both means "swept faster" and "longer arm", the effect grows with distance squared — move the fins twice as far back, get four times the calming. Strip away the air density and speed to leave only the shape, and you've got $C_{m_q}$, the rocket's built-in shock absorber (Step 5). It's zero at the balance point, works from either end, and quietly weakens as the air thins out on the way up (Step 6).

Recall Quick self-check

Where do the two powers of $x$ come from? ::: One from the sweep velocity $qx$ (far points sweep faster → bigger angle → bigger force); one from the lever arm $x$ (torque = force × distance). Why is $C_{m_q}$ always negative? ::: The integral $\int c_{n_{\alpha }}{x}^{2}dA$ is a sum of squares (positive), with a leading minus — so the moment always opposes $q$. What does a surface exactly at the CG contribute? ::: Nothing — $x=0$ gives zero sweep and zero lever, so $x^2=0$. Does the geometry number $C_{m_q}$ change with altitude? ::: No, but the physical damping moment does, because it scales with $\rho V$.

---

Prerequisites & neighbours: Parent topic · Static Stability — Center of Pressure & Margin · Damped Harmonic Oscillator · Fin Design & Sizing · Atmospheric Density vs Altitude · Barrowman Equations · Moments of Inertia of a Rocket