Quaternion kinematics — q̇ = ½ Ξ(q) ω: Worked examples

Before any numbers, one reminder of the two objects we feed the machine:

Recall The two ingredients (open if rusty)

A unit quaternion $q = (q_{0}, q_{1}, q_{2}, q_{3})$ with $q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1$ stores an orientation. An angular velocity $ω = (ω_{1}, ω_{2}, ω_{3})$ in rad/s is how fast the body spins right now, measured in the body frame. The rate is $\overset{q}{˙} = \frac{1}{2} Ξ (q) ω, Ξ (q) = - q_{1} q_{0} q_{3} - q_{2} - q_{2} - q_{3} q_{0} q_{1} - q_{3} q_{2} - q_{1} q_{0} .$

The scenario matrix

Every case this topic can throw at you falls into one of these cells. The examples below are tagged with the cell they cover.

#	Cell (what makes it special)	Example
A	Zero input — $ω = 0$ (degenerate: nothing spins)	Ex 1
B	Spin about a single body axis, $q$ at identity	Ex 2
C	Same spin, all three axes — check the sign pattern of $Ξ$	Ex 3
D	$q$ NOT the identity — the interesting case, $\overset{q}{˙}_{0} \neq = 0$	Ex 4
E	Sign / direction reversal — $ω \to - ω$	Ex 5
F	Limiting / linearity — scale $ω$ , super-fast spin behaviour	Ex 6
G	Pure-vector quaternion $q_{0} = 0$ — top-row formula stress test	Ex 7
H	Real-world word problem — a satellite detumble reading	Ex 8
I	Exam twist — inertial-frame $ω$ , must switch to $Ω (ω) q$	Ex 9

The worked examples

Worked example Example 1 — Cell A · Zero angular velocity

A satellite floats with its thrusters off: $ω = (0, 0, 0)$ , current attitude $q = (0.6, 0, 0.8, 0)$ (a valid unit quaternion). What is $\overset{q}{˙}$ ?

Forecast: if nothing spins, the orientation should not change. Guess $\overset{q}{˙} = 0$ before reading on.

Write the formula. $\overset{q}{˙} = \frac{1}{2} Ξ (q) ω$ . Why this step? Always start from the one true relation; don't reason about "nothing happens" — let the math say it.
Substitute $ω = 0$ . A $4 \times 3$ matrix times the zero 3-vector is the zero 4-vector. Why this step? $Ξ (q)$ can be any numbers — multiplying by a zero column kills them all. The value of $q$ is irrelevant. $\overset{q}{˙} = \frac{1}{2} Ξ (q) 000 = 0000 .$

Verify: $\overset{q}{˙} = 0$ means the attitude is frozen — exactly what "no rotation" should give. Units: $Ξ$ is dimensionless, $ω$ is rad/s, so $\overset{q}{˙}$ is (1/s), and $0 \cdot$ anything is still $0$ . ✔

Worked example Example 2 — Cell B · Single-axis spin from identity

Start at identity $q = (1, 0, 0, 0)$ , spin about body $x$ only: $ω = (0.1, 0, 0)$ rad/s. Find $\overset{q}{˙}$ .

Forecast: the $x$ -part of the vector ( $q_{1}$ ) should be the one waking up. Guess which entry of $\overset{q}{˙}$ is nonzero.

Build $Ξ (q)$ at identity. With $q_{0} = 1, q_{1} = q_{2} = q_{3} = 0$ : $Ξ (q) = 010000100001 .$ Why this step? At the identity the vector part is zero, so all $q_{v}$ terms drop and $Ξ$ collapses to a clean "stack" of the $3 \times 3$ identity under a zero row.
Multiply and halve. $\overset{q}{˙} = \frac{1}{2} Ξ (q) 0.1 00 = \frac{1}{2} 0 0.1 00 = 0 0.05 00 .$ Why this step? Only the first column of $Ξ$ is picked out (because only $ω_{1} \neq = 0$ ), and the ½ is the half-angle factor from the parent derivation.

Verify: $\overset{q}{˙}_{0} = 0$ — the norm cannot change at this instant, consistent with $q^{⊤} \overset{q}{˙} = 1 \cdot 0 = 0$ . Only $q_{1}$ grows, i.e. a roll about $x$ is turning on. See the tangent-arrow picture below. ✔

Figure — Quaternion kinematics — q̇ = ½ Ξ(q) ω

The arrow above is $\overset{q}{˙}$ living tangent to the unit sphere $S^{3}$ at $q$ — look at how it points along the surface, never outward. That is the norm-preservation fact made visible.

Worked example Example 3 — Cell C · All three axes at once (sign pattern)

Still at identity $q = (1, 0, 0, 0)$ , but now $ω = (ω_{1}, ω_{2}, ω_{3})$ general. Show every component of $\overset{q}{˙}$ .

Forecast: each body-rate should feed its own vector slot, cleanly. Guess $\overset{q}{˙}_{v} = \frac{1}{2} ω$ .

Reuse $Ξ$ at identity (from Ex 2), multiply: $\overset{q}{˙} = \frac{1}{2} 010000100001 ω_{1} ω_{2} ω_{3} = \frac{1}{2} 0 ω_{1} ω_{2} ω_{3} .$ Why this step? This is the "linearized" attitude update EKFs love: near identity, $\overset{q}{˙}_{v} = \frac{1}{2} ω$ and $\overset{q}{˙}_{0} = 0$ . The attitude EKF uses exactly this small-angle picture.
Read the sign pattern. No minus signs appear because $q_{v} = 0$ removed the skew (cross-product) block $[q_{v} \times]$ . Why this step? The minus signs in the general $Ξ$ come from that skew block; killing $q_{v}$ shows the pattern is not arbitrary — it is the rotation non-commutativity encoded.

Verify: for $ω = (0.1, 0.2, - 0.3)$ , $\overset{q}{˙} = (0, 0.05, 0.10, - 0.15)$ . Norm rate $q^{⊤} \overset{q}{˙} = 1 \cdot 0 = 0$ . ✔

Worked example Example 4 — Cell D · A quaternion that is NOT identity

This is the case people fear. Take $q = (\frac{2}{2}, \frac{2}{2}, 0, 0)$ (a 90° roll about $x$ , unit norm: $0.5 + 0.5 = 1$ ). Spin about body $x$ : $ω = (0.1, 0, 0)$ . Find $\overset{q}{˙}$ .

Forecast: now $q_{v} \neq = 0$ , so the top row of $Ξ$ (the $- q_{v}^{⊤}$ row) is alive — expect $\overset{q}{˙}_{0} \neq = 0$ . Guess its sign.

Write $Ξ (q)$ fully. With $q_{0} = q_{1} = \frac{2}{2} \approx 0.707, q_{2} = q_{3} = 0$ : $Ξ (q) = - 0.707 0.707 00 00 0.707 0.707 00 - 0.707 0.707 .$ Why this step? Every entry is now filled — no shortcut. This is the honest test of the formula.
Multiply by $ω = (0.1, 0, 0)$ (only column 1 survives), then halve: $\overset{q}{˙} = \frac{1}{2} - 0.707 0.707 00 (0.1) = - 0.0354 0.0354 00 .$ Why this step? Now $\overset{q}{˙}_{0} = - 0.0354 \neq = 0$ : as $q_{1}$ grows, $q_{0}$ must shrink to keep the length fixed — the sphere is curving.

Verify: norm rate $q^{⊤} \overset{q}{˙} = 0.707 (- 0.0354) + 0.707 (0.0354) + 0 + 0 = 0$ . ✔ The two nonzero terms cancel because $\overset{q}{˙}$ is tangent. See figure: at a tilted $q$ , the tangent arrow has a component along $- q_{0}$ .

Worked example Example 6 — Cell F · Scaling and the fast-spin limit

Fix $q = (1, 0, 0, 0)$ . First double $ω$ from $(0.1, 0, 0)$ to $(0.2, 0, 0)$ . Then ask: as $ω_{1} \to \infty$ , what does $∥ \overset{q}{˙} ∥$ do?

Forecast: doubling $ω$ doubles $\overset{q}{˙}$ (linearity). And as spin blows up, $∥ \overset{q}{˙} ∥$ grows without bound. Guess both before checking.

Scaling. From Ex 2, $\overset{q}{˙} (0.1) = (0, 0.05, 0, 0)$ . So $\overset{q}{˙} (0.2) = 2 \cdot (0, 0.05, 0, 0) = (0, 0.1, 0, 0)$ . Why this step? $Ξ (q)$ does not depend on $ω$ , so $\overset{q}{˙}$ is a straight linear map of $ω$ — this is the whole reason $Ξ$ is a matrix, usable in RK4 stepping.
Limit. At identity $∥ \overset{q}{˙} ∥ = \frac{1}{2} ∥ ω ∥ = \frac{1}{2} ∣ ω_{1} ∣ \to \infty$ as $ω_{1} \to \infty$ . Why this step? The tangent velocity magnitude is literally half the spin speed. Faster spin → faster slide along $S^{3}$ ; the picture never breaks, but a fixed timestep $Δ t$ eventually under-samples it — the practical reason to renormalize and cap step size.

Verify: $\overset{q}{˙} (0.2) = (0, 0.1, 0, 0) = 2 \overset{q}{˙} (0.1)$ . ✔ And $\frac{1}{2} ∥ (10, 0, 0) ∥ = 5$ , i.e. $∥ \overset{q}{˙} ∥ = 5$ for $ω_{1} = 10$ . ✔

Worked example Example 7 — Cell G · Pure-vector quaternion (

$q_{0} = 0$ ) The stress test for the top row of $Ξ$ . Take $q = (0, 1, 0, 0)$ — a valid unit quaternion (norm $= 1$ ) with zero scalar part (it represents a 180° rotation about $x$ ). Spin about body $x$ : $ω = (0.2, 0, 0)$ . Find $\overset{q}{˙}$ .

Forecast: the top-row driver is $- q_{v} \cdot ω$ . Here $q_{v} = (1, 0, 0)$ is parallel to $ω$ , so that dot product is nonzero and large — expect $\overset{q}{˙}_{0} \neq = 0$ and dominant. Guess its sign.

Write $Ξ (q)$ with $q_{0} = 0, q_{1} = 1, q_{2} = q_{3} = 0$ . $Ξ (q) = - 1 000 0001 00 - 1 0 .$ Why this step? With $q_{0} = 0$ the whole $q_{0} I$ part of the bottom block vanishes; only the skew (cross-product) entries survive there, and the top row $- q_{v}^{⊤} = (- 1, 0, 0)$ is now maximal.
Multiply by $ω = (0.2, 0, 0)$ (column 1 only), then halve: $\overset{q}{˙} = \frac{1}{2} - 1 000 (0.2) = - 0.1 000 .$ Why this step? $\overset{q}{˙}_{0} = - 0.1$ dominates: at a 180° attitude, spinning about the same axis pulls $q_{0}$ down fastest, driving $q$ around the far side of the sphere.

Verify: norm rate $q^{⊤} \overset{q}{˙} = 0 (- 0.1) + 1 (0) + 0 + 0 = 0$ . ✔ Only the scalar part moves — exactly the top-row behaviour we wanted to isolate. Units rad/s. ✔

Worked example Example 8 — Cell H · Real satellite detumble reading

A CubeSat's gyro reports body rates $ω = (0.02, - 0.05, 0.03)$ rad/s. Its current attitude estimate is $q = (0.9, 0.2, - 0.3, 0.244949)$ . (Check: $0.81 + 0.04 + 0.09 + 0.06 = 1.00$ , unit.) Compute $\overset{q}{˙}$ for the next RK4 tick.

Forecast: all four components of $\overset{q}{˙}$ will be nonzero (both $q_{v}$ and full $ω$ are active). Guess whether $\overset{q}{˙}_{0}$ is $+$ or $-$ from $- q_{v} \cdot ω$ .

Top entry $\overset{q}{˙}_{0} = \frac{1}{2} (- q_{v} \cdot ω) = \frac{1}{2} (- (0.2) (0.02) - (- 0.3) (- 0.05) - (0.244949) (0.03))$ . Why this step? The first row of $Ξ$ is $- q_{v}^{⊤}$ ; it measures how much the spin points "along" the current vector part, which is what changes the scalar $q_{0}$ . $\overset{q}{˙}_{0} = \frac{1}{2} (- 0.004 - 0.015 - 0.0073485) = \frac{1}{2} (- 0.0263485) = - 0.01317.$
Vector part $\overset{q}{˙}_{v} = \frac{1}{2} (q_{0} ω + q_{v} \times ω)$ . Why this step? This is the bottom $3 \times 3$ block $q_{0} I + [q_{v} \times]$ acting on $ω$ — the Rodrigues-style scalar-plus-cross structure. Cross product $q_{v} \times ω$ with $q_{v} = (0.2, - 0.3, 0.244949)$ , $ω = (0.02, - 0.05, 0.03)$ : $q_{v} \times ω = ((- 0.3) (0.03) - (0.244949) (- 0.05), (0.244949) (0.02) - (0.2) (0.03), (0.2) (- 0.05) - (- 0.3) (0.02))$ $= (0.0032475, - 0.0051010, - 0.004) .$ Then $q_{0} ω = (0.018, - 0.045, 0.027)$ , sum $= (0.0212475, - 0.0501010, 0.023)$ , times ½: $\overset{q}{˙}_{v} = (0.0106237, - 0.0250505, 0.0115) .$

Verify: full $\overset{q}{˙} = (- 0.01317, 0.0106237, - 0.0250505, 0.0115)$ . Norm rate $q^{⊤} \overset{q}{˙}$ should be $0$ : $0.9 (- 0.01317) + 0.2 (0.0106237) + (- 0.3) (- 0.0250505) + 0.244949 (0.0115) = 0$ (to numerical precision). ✔ Units all rad/s. ✔

Worked example Example 9 — Cell I · Exam twist: inertial-frame

$ω$ Trap question: " $ω = (0, 0, 0.4)$ rad/s is given in the inertial frame, $q = (1, 0, 0, 0)$ . A student writes $\overset{q}{˙} = \frac{1}{2} q \otimes (0, ω)$ . Is that right? Give the correct $\overset{q}{˙}$ ."

Forecast: body-rate formula post-multiplies; inertial rate must pre-multiply. Guess whether the numeric answer even differs here.

Diagnose the error. For inertial $ω$ use left multiplication: $\overset{q}{˙} = \frac{1}{2} (0, ω) \otimes q = \frac{1}{2} Ω (ω) q$ , with $Ω (ω)$ from the definition callout above (and $\otimes$ is the Hamilton product defined earlier). Why this step? Frame of $ω$ dictates multiplication side — a classic flight-software bug (and worse than gimbal lock because it fails silently). The student used the body form.
Compute correctly. With $Ω (ω)$ for $ω = (0, 0, 0.4)$ and $q = (1, 0, 0, 0)$ : $Ω (ω) q = 000 0.4 00 - 0.4 0 0 0.4 00 - 0.4 000 1000 = 000 0.4, \overset{q}{˙} = 000 0.2 .$ Why this step? At the identity, body and inertial frames coincide, so here both formulas happen to give $(0, 0, 0, 0.2)$ — the twist is that this agreement is a coincidence of $q =$ identity, not a general truth.

Verify: body form $\frac{1}{2} Ξ (q) ω$ at identity also gives $(0, 0, 0, 0.2)$ (third column of the identity- $Ξ$ times $0.4$ , halved). ✔ They match only because $q$ is the identity — for any tilted $q$ they diverge, which is the point of the trap. ✔

Recall Self-test (open after you've tried)

If $ω = 0$ , what is $\overset{q}{˙}$ ? ::: The zero 4-vector, for any $q$ . At the identity quaternion, what is $\overset{q}{˙}_{0}$ ? ::: Always $0$ (the $- q_{v} \cdot ω$ row vanishes since $q_{v} = 0$ ). Why does $\overset{q}{˙} (- ω) = - \overset{q}{˙} (ω)$ ? ::: Because $\overset{q}{˙} = \frac{1}{2} Ξ (q) ω$ is linear in $ω$ . For a pure-vector quaternion $q_{0} = 0$ spun about an axis parallel to $q_{v}$ , which component of $\overset{q}{˙}$ dominates? ::: The scalar part $\overset{q}{˙}_{0}$ , since $- q_{v} \cdot ω$ is then maximal. For inertial-frame $ω$ , which formula? ::: $\overset{q}{˙} = \frac{1}{2} Ω (ω) q$ (left multiply).