Visual walkthrough — Quaternion kinematics — q̇ = ½ Ξ(q) ω

Before anything: three words, three pictures.

We link the prerequisite pictures as we need them: Angular velocity and the body frame, Rotation matrices and SO(3), Rodrigues rotation formula.

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Step 1 — A rotation is an axis and an angle

WHAT. Any turn of a rigid body, no matter how complicated it looks, is one spin: pick a straight line (the axis $\hat{n}$, a unit arrow) and turn by some angle $\theta$ around it. That is the whole content of a rotation.

WHY. We need the simplest honest description of "orientation" before we can ask how it changes. Euler's rotation theorem promises this axis–angle pair always exists, so we start there instead of with nine messy matrix entries (see Rotation matrices and SO(3)).

PICTURE.

The red arrow is the axis $\hat{n}$ (length 1). The orange sweep is the angle $\theta$. Everything below is just a clever way to store these two things as four numbers.

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Step 2 — Store the spin as four numbers (why half the angle)

WHAT. We pack the axis $\hat{n}$ and angle $\theta$ into a quaternion $q$, a stack of four numbers:

$$q=\underset{\begin{array}{c}\text{one scalar}\\ q_0\end{array}}{\underbrace{\cos \frac{\theta }{2}}},\underset{\begin{array}{c}\text{three numbers}\\ {\mathbf{q}}_v=(q_1,q_2,q_3)\end{array}}{\underbrace{\hat{n}\sin \frac{\theta }{2}}}$$

Read it term by term: $q_0$ is a lone number tracking how much we turned; ${\mathbf{q}}_v$ is the axis, shrunk by $\sin \frac{\theta }{2}$, tracking which way.

WHY the $\frac{\theta }{2}$ and not $\theta$? This is the seed of the famous $\frac{1}{2}$. Storing the half-angle is what makes quaternion multiplication compose rotations correctly (it is the "double cover": $q$ and $-q$ name the same physical orientation). Remember this half — it comes back to bite us in Step 6.

PICTURE.

As the true angle $\theta$ (orange) sweeps, the four stored numbers ride cosine and sine curves of $\frac{\theta }{2}$ (teal, plum). Notice they move at half the speed of the real turn — that is the half-angle, drawn.

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Step 3 — Two rotations combine by multiplying quaternions

WHAT. "Do rotation $q$, then a further tiny rotation $\delta q$" is written as a single product $q\otimes \delta q$. The symbol $\otimes$ is the Hamilton product:

$$p\otimes q=\left[\begin{array}{c}\underset{\text{new scalar part}}{\underbrace{p_0{q}_0-{\mathbf{p}}_v\cdot {\mathbf{q}}_v}}\\ \underset{\text{axes blended}}{\underbrace{p_0{\mathbf{q}}_v+q_0{\mathbf{p}}_v}}+\underset{\text{order matters}}{\underbrace{{\mathbf{p}}_v\times {\mathbf{q}}_v}}\end{array}\right]$$

Term by term: the dot product ${\mathbf{p}}_v\cdot {\mathbf{q}}_v$ (a single number measuring alignment) feeds the scalar; the cross product ${\mathbf{p}}_v\times {\mathbf{q}}_v$ (a new arrow, perpendicular to both) is the fingerprint of non-commutativity.

WHY. Composing rotations does not commute — turning then rolling differs from rolling then turning. Only a product with that $\times$ term can copy this behaviour. This is the entire reason quaternions exist.

PICTURE.

Left: rotate-then-flip. Right: flip-then-rotate. Same two moves, different final arrow — the gap between them is exactly the cross-product term.

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Step 4 — The attitude lives on a 4D ball; velocity must stay tangent

WHAT. Because $\Vert q\Vert =1$ always, the point $q$ can never leave the surface where "sum of four squares $=1$". That surface is the unit 3-sphere $S^3$. We cannot draw four dimensions, so we draw the honest 2D analogy: a point on a circle.

WHY. If a rotation is always a legal orientation, then $q(t)$ must slide along this surface for all time — it can never bulge outward or sink inward. That single geometric demand will nail down the $\frac{1}{2}$ and the matrix $\Xi (q)$ in Steps 6–7.

PICTURE.

The point $q$ sits on the circle. Its velocity $\dot{q}$ (green) is drawn tangent — it points along the surface, at a right angle to $q$ itself. Any velocity with a radial (outward) piece would grow $\Vert q\Vert$ and is forbidden.

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Step 5 — A tiny spin over a tiny time

WHAT. In a short slice of time $\Delta t$, the body turns by a small angle $\Delta \theta =\Vert \omega \Vert \Delta t$ about the current axis $\hat{\omega }=\omega /\Vert \omega \Vert$. From Step 2, that tiny extra rotation is its own quaternion:

$$\delta q=\left[\begin{array}{c}\cos \frac{\Delta \theta }{2}\\ \hat{\omega }\sin \frac{\Delta \theta }{2}\end{array}\right]$$

Now shrink $\Delta t\to 0$. For a tiny angle, $\cos \frac{\Delta \theta }{2}\approx 1$ and $\sin \frac{\Delta \theta }{2}\approx \frac{\Delta \theta }{2}$:

$$\delta q\approx \left[\begin{array}{c}1\\ \frac{1}{2}\omega \Delta t\end{array}\right]=\underset{\text{do nothing}}{\underbrace{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]}}+\frac{\Delta t}{2}\underset{\text{the nudge}}{\underbrace{\left[\begin{array}{c}0\\ \omega \end{array}\right]}}$$

WHY the approximations? Near angle zero, cosine flattens to 1 and sine becomes a straight line through the origin — the "small-angle" facts you can read straight off Step 2's curves. This turns the exact-but-messy $\delta q$ into "identity plus a small linear nudge", which we can differentiate.

WHY the $\frac{1}{2}$ again? It rode in on $\sin \frac{\Delta \theta }{2}$ — the half-angle from Step 2. This is literally where the famous factor of one-half is born.

PICTURE.

Zoomed onto the circle near $q$: the huge orange curve of $\sin$ is replaced by its straight teal tangent line. Over the tiny step, the true spin and its linear stand-in are indistinguishable — and the slope of that line carries the $\frac{1}{2}$.

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Step 6 — From "tiny step" to a genuine derivative

WHAT. Compose the current attitude with the tiny nudge (Step 3), subtract where we started, and divide by the time slice:

$$\frac{q(t+\Delta t)-q(t)}{\Delta t}=\frac{q\otimes \delta q-q}{\Delta t}=\frac{q\otimes \left(\frac{\Delta t}{2}\left[\begin{array}{c}0\\ \omega \end{array}\right]\right)}{\Delta t}=\frac{1}{2}q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]$$

Let $\Delta t\to 0$ and the left side becomes the derivative $\dot{q}$ — the instantaneous rate, the green tangent arrow of Step 4:

$$\boxed{\dot{q}=\frac{1}{2}q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]}$$

Term by term: $\frac{1}{2}$ is the half-angle survivor; $q\otimes$ says "spin measured in the body frame, so it multiplies on the right"; and $\left[\begin{array}{c}0\\ \omega \end{array}\right]$ is the angular-velocity quaternion — a quaternion with zero scalar part, because $\omega$ is a pure spin with no "amount-so-far" of its own.

WHY. The identity terms $q\otimes \left[\begin{array}{c}1\\ 0\end{array}\right]=q$ cancel exactly, leaving only the nudge. The definition of a derivative (limit of a difference quotient) turns "compose a small rotation each tick" into a smooth flow. That is the whole trick — see Numerical integration RK4 for how we later step this flow on a computer.

PICTURE.

The secant line from $q(t)$ to $q(t+\Delta t)$ (dotted) pivots down onto the tangent (solid green) as $\Delta t$ shrinks — the picture of taking the limit.

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Step 7 — Package the product as a matrix $\Xi (q)$

WHAT. The Hamilton product $q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]$ is linear in $\omega$ — doubling $\omega$ doubles the output, with no leftover terms. Anything linear can be written as a matrix times $\omega$. Grinding through Step 3's formula with $p=q$ and $q_{\text{here}}=(0,\omega )$:

$$q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]=\left[\begin{array}{c}-{\mathbf{q}}_v\cdot \omega \\ q_0\omega +{\mathbf{q}}_v\times \omega \end{array}\right]=\Xi (q)\omega$$

so

$$\dot{q}=\frac{1}{2}\Xi (q)\omega ,\Xi (q)=\left[\begin{array}{ccc}-q_1& -q_2& -q_3\\ q_0& -q_3& q_2\\ q_3& q_0& -q_1\\ -q_2& q_1& q_0\end{array}\right]$$

Term by term: the top row $-{\mathbf{q}}_v^{\top }$ came from the dot product (alignment of axis with spin). The bottom $3\times 3$ block is $q_{0}I+[{\mathbf{q}}_v\times ]$ — the scalar $q_0$ scaling $\omega$ plus the skew matrix that performs ${\mathbf{q}}_v\times \omega$.

WHY. GNC estimators like the Extended Kalman Filter for attitude estimation need a clean linear map $\dot{q}=A\omega$ they can drop into matrix code. $\Xi (q)$ is that map — the Hamilton product rewritten so a computer can multiply it.

PICTURE.

The $4\times 3$ grid of $\Xi (q)$ with its top row (dot-product) shaded plum and its lower block (scale + cross) shaded teal — showing exactly which entries came from which piece of Step 3.

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Step 8 — Sanity check the picture: the norm cannot drift

WHAT. We claimed in Step 4 that $\dot{q}$ must be tangent, i.e. $\dot{q}\perp q$, so $\Vert q\Vert$ stays 1. Check it:

$$\frac{d}{dt}\Vert q{\Vert }^2=2q^{\top }\dot{q}=2q^{\top }\left(\frac{1}{2}\Xi (q)\omega \right)=(q^{\top }\Xi (q))\omega$$

and the key fact $q^{\top }\Xi (q)=0^{\top }$:

$$q^{\top }\Xi (q)=\underset{\text{scalar times top row}}{\underbrace{q_0(-{\mathbf{q}}_v^{\top })}}+\underset{\text{vector part times block}}{\underbrace{{\mathbf{q}}_v^{\top }(q_{0}I+[{\mathbf{q}}_v\times ])}}=-q_0{\mathbf{q}}_v^{\top }+q_0{\mathbf{q}}_v^{\top }+{\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ]=0^{\top }$$

The two $q_0{\mathbf{q}}_v^{\top }$ pieces cancel, and ${\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ]=-({\mathbf{q}}_v\times {\mathbf{q}}_v{)}^{\top }=0$ because any vector crossed with itself is zero.

WHY. This is the whole promise of Step 4, cashed out in algebra: whatever $\omega$ you feed in, the rate is perpendicular to $q$. $\Vert q\Vert$ never changes. The equation cannot send the attitude off the ball.

PICTURE.

$q$ (blue radial) and $\dot{q}$ (green tangent) meet at a perfect right angle; the little box at the corner is the $q^{\top }\dot{q}=0$ we just proved.

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Step 9 — The edge and degenerate cases (nothing left uncovered)

WHAT & WHY — walk every corner:

1. Zero spin, $\omega =0$. Then $\Xi (q)0=0$, so $\dot{q}=0$. The point sits still on the ball — a non-rotating body keeps its attitude. ✔
2. At the identity, $q=(1,0,0,0)$. The top row $-{\mathbf{q}}_v^{\top }=0^{\top }$, so ${\dot{q}}_0=0$; only the vector part turns on. First motion is purely into $q_1,q_2,q_3$ — exactly the tangent direction. ✔
3. Spin along a stored axis (e.g. $\omega \parallel \hat{n}$): the cross term ${\mathbf{q}}_v\times \omega =0$ vanishes, so the body simply keeps turning about the same axis — no wobble, clean single-axis spin. ✔
4. $-q$ instead of $q$ (double cover): flip every sign in $q$ and $\Xi (q)$ flips sign too, giving $-\dot{q}$ — the same physical rotation traced by the mirror-labelled point. No contradiction; both name one orientation.
5. Big $\omega$ over a finite step. The $\frac{1}{2}$ derivation is exact in continuous time, but a discrete integrator (e.g. Numerical integration RK4) drifts slightly off the ball — so real flight code renormalizes $q\leftarrow q/\Vert q\Vert$ each tick. Contrast Euler angles and gimbal lock, where no amount of renormalizing saves you from the singularity.

PICTURE.

Four mini-circles: (a) $\omega =0$ frozen dot; (b) identity with tangent-only motion; (c) axis-aligned spin, no wobble; (d) discrete steps drifting outward, with a plum arrow snapping them back to the surface (renormalization).

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The one-picture summary

One frame carries the whole derivation: the point $q$ on the sphere; the tiny orange nudge $\delta q\approx 1+\frac{\Delta t}{2}(0,\omega )$ pulling it forward; the green tangent $\dot{q}=\frac{1}{2}\Xi (q)\omega$ that results; and the right angle proving $\Vert q\Vert$ stays 1. Half-angle in, tangent out.

Recall Feynman retelling — say it in plain words

A spinning thing has a "where am I pointing" state, and we cleverly store it as four numbers whose squares add to one — so the state is stuck on the surface of a round ball. Turning a little more means multiplying by a small turn-quaternion, and because we stored the half-angle, that small turn is "do-nothing plus one-half of the spin". Subtract the do-nothing, divide by the time, and you get the rate: one-half, times the current state, times the spin. Rewrite the multiply as a tidy matrix $\Xi (q)$ so a computer can use it. Finally, the beautiful check: the rate always points sideways along the ball, never outward, so the state can never fall off — which is exactly what "still a valid rotation" means. The whole formula is just "slide along the ball at half-speed."

Recall Quick self-test

Where does the factor $\frac{1}{2}$ come from? ::: From storing the half-angle $\sin \frac{\theta }{2}$ in the quaternion (Step 2); its slope at zero is $\frac{1}{2}$ (Step 5). Why must $\dot{q}\perp q$? ::: Because $\Vert q\Vert =1$ forever, so the velocity can only slide along the sphere, never radially outward (Step 4 / Step 8). Why is $\Xi (q)$ a matrix and not a nonlinear map? ::: Because the Hamilton product $q\otimes (0,\omega )$ is linear in $\omega$ (Step 7) — doubling $\omega$ doubles $\dot{q}$. What is $\dot{q}$ when $\omega =0$? ::: Zero — a non-spinning body holds its attitude (Step 9, case 1).

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