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

$\Xi (q)$ is $==4\times 3==$. Why: it eats the 3-vector $\omega$ (3 columns) and produces the 4-vector $\dot{q}$ (4 rows). A matrix that turns a 3-thing into a 4-thing must be $4\times 3$. Count the entries in the formula above to confirm: 4 rows, 3 columns.

It is a pure quaternion (also called the angular-velocity quaternion). A pure quaternion has zero scalar part; its three vector slots hold $\omega =({\omega }_1,{\omega }_2,{\omega }_3)$. It is not a unit quaternion and does not represent a rotation — it is only a convenient package so we can multiply $\omega$ into $q$.

(b). A quaternion stores $\cos \frac{\theta }{2}$ and $\sin \frac{\theta }{2}$ — the half-angle. When you differentiate $\sin \frac{\theta }{2}$ in time, the chain rule pulls out a $\frac{1}{2}$: $\frac{d}{dt}\sin \frac{\theta }{2}=\frac{1}{2}\dot{\theta }\cos \frac{\theta }{2}$. That $\frac{1}{2}$ is the whole story — it is the price of the double-cover, where both $q$ and $-q$ mean the same physical rotation.

At $q=(1,0,0,0)$: $q_0=1,q_1=q_2=q_3=0$, so $\Xi (q)=\left[\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].$ Then $\dot{q}=\frac{1}{2}\Xi (q)\left[\begin{array}{c}0\\ 0.2\\ 0\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}0\\ 0\\ 0.2\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0.1\\ 0\end{array}\right].$ What it means: the $q_2$ slot turns on first — that is a pitch about the body $y$-axis beginning. And ${\dot{q}}_0=0$: at identity the norm cannot yet change ($q^{\top }\dot{q}=0$).

Here $q_0=0.7071,q_1=0.7071,q_2=0,q_3=0$. Build $\Xi (q)$: $\Xi (q)=\left[\begin{array}{ccc}-0.7071& 0& 0\\ 0.7071& 0& 0\\ 0& 0.7071& -0.7071\\ 0& 0.7071& 0.7071\end{array}\right].$ Only ${\omega }_3=1$ is nonzero, so we take the third column of $\Xi (q)$ and halve it: $\dot{q}=\frac{1}{2}\left[\begin{array}{c}0\\ 0\\ -0.7071\\ 0.7071\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ -0.35355\\ 0.35355\end{array}\right].$ Sanity check: $q^{\top }\dot{q}=0.7071\cdot 0+0.7071\cdot 0+0\cdot (-0.35355)+0\cdot 0.35355=0$. Tangent to the sphere. ✔

Left side ($\Xi (q)\omega$, before the $\frac{1}{2}$): from 2.1 it is $(0,0,0.2,0)$. Right side ($\Omega (\omega )q$): with $\omega =(0,0.2,0)$, $\Omega (\omega )=\left[\begin{array}{cccc}0& 0& -0.2& 0\\ 0& 0& 0& -0.2\\ 0.2& 0& 0& 0\\ 0& 0.2& 0& 0\end{array}\right],\Omega (\omega )\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0.2\\ 0\end{array}\right].$ Both give $(0,0,0.2,0)$. ✔ Same physics, two bookkeeping choices — one factors out $q$, the other $\omega$.

Start from $\dot{q}=\frac{1}{2}\Xi (q)\omega$, so $q^{\top }\dot{q}=\frac{1}{2}(q^{\top }\Xi (q))\omega .$ Using the block form (restated in the definition callout above) $\Xi (q)=\left[\begin{array}{c}-{\mathbf{q}}_v^{\top }\\ q_{0}I+[{\mathbf{q}}_v\times ]\end{array}\right]$ and $q=(q_0,{\mathbf{q}}_v)$, we split $q^{\top }=\left[q_0{\mathbf{q}}_v^{\top }\right]$ and multiply block-by-block: $q^{\top }\Xi (q)=q_0(-{\mathbf{q}}_v^{\top })+{\mathbf{q}}_v^{\top }(q_{0}I+[{\mathbf{q}}_v\times ])=-q_0{\mathbf{q}}_v^{\top }+q_0{\mathbf{q}}_v^{\top }+{\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ].$ The last term uses $[{\mathbf{q}}_v\times {]}^{\top }=-[{\mathbf{q}}_v\times ]$ (skew) and ${\mathbf{q}}_v\times {\mathbf{q}}_v=0$: ${\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ]=-({\mathbf{q}}_v\times {\mathbf{q}}_v{)}^{\top }=0^{\top }$. So $q^{\top }\Xi (q)=0^{\top }$ and hence $q^{\top }\dot{q}=0$ for every $\omega$. Geometrically (see figure s01 above): the black arrow $q$ points radially out to the sphere; the red arrow $\dot{q}$ meets it at the marked right angle, lying flat against the surface. That right angle is the algebra $q^{\top }\dot{q}=0$ made visible — the motion slides along $S^3$, never off it, so $\Vert q\Vert$ stays exactly 1.

Differentiate the first equation and substitute the second: ${\ddot{q}}_0=-\frac{1}{2}{\omega }_z{\dot{q}}_3=-\frac{1}{2}{\omega }_z\left(\frac{1}{2}{\omega }_z{q}_0\right)=-(\frac{{\omega }_z}{2}{)}^2{q}_0.$ This is simple harmonic motion with angular frequency $==\frac{{\omega }_z}{2}==$. With $q_0(0)=1,q_3(0)=0$: $q_0(t)=\cos \frac{{\omega }_{z}t}{2},q_3(t)=\sin \frac{{\omega }_{z}t}{2}.$ Why half: the physical body turns through angle $\theta ={\omega }_{z}t$, but the quaternion stores $\cos \frac{\theta }{2},\sin \frac{\theta }{2}$. So the quaternion components oscillate at half the body's rotation rate. A full $360^{\circ }$ body turn ($\theta =2\pi$) only advances the quaternion by $\pi$ in half-angle — and indeed $q$ returns to $-q$ after one full turn, the famous double cover.

Columns of $\Xi (q)$: $c_1=\left[\begin{array}{c}-q_1\\ q_0\\ q_3\\ -q_2\end{array}\right],c_2=\left[\begin{array}{c}-q_2\\ -q_3\\ q_0\\ q_1\end{array}\right],c_3=\left[\begin{array}{c}-q_3\\ q_2\\ -q_1\\ q_0\end{array}\right].$ $c_1\cdot c_2=(-q_1)(-q_2)+q_0(-q_3)+q_3{q}_0+(-q_2)(q_1)=q_1{q}_2-q_0{q}_3+q_0{q}_3-q_1{q}_2=0.$ Orthogonal. ✔ And $c_1\cdot c_1=q_1^2+q_0^2+q_3^2+q_2^2=\Vert q{\Vert }^2=1$. So the three columns form an orthonormal set (when $q$ is unit). Meaning: $\Xi (q)$ is an isometry from $\omega$-space into the tangent plane — it doesn't stretch $\omega$, it just rotates it into the right 3D slice of the 4D tangent space. That is why $\Vert \dot{q}\Vert =\frac{1}{2}\Vert \omega \Vert$ exactly.

From the parent note Example 2, $\dot{q}=\frac{1}{2}\Xi (q)\omega =(0,0.05,0,0)$. Euler step: $q^{\text{new}}=(1,0,0,0)+0.5(0,0.05,0,0)=(1,0.025,0,0)$. Norm: $\Vert q^{\text{new}}\Vert =\sqrt{1^2+0.025^2}=\sqrt{1.000625}=1.0003125\dots$ — slightly bigger than 1. Renormalize: divide by the norm: $q^{\text{norm}}=\frac{1}{1.0003125}(1,0.025,0,0)=(0.99968762,0.02499219,0,0).$ Why needed: the tangency proof (3.1) is exact only in continuous time. A finite Euler step moves straight along the tangent line, which leaves the curved sphere, so $\Vert q\Vert$ creeps above 1. Renormalizing snaps it back onto $S^3$. Higher-order integrators like RK4 (four rate-samples per step) drift less but still require periodic renormalization.

Exact: $q_0=\cos \frac{{\omega }_{z}t}{2},q_3=\sin \frac{{\omega }_{z}t}{2}$. With ${\omega }_{z}t=\pi \cdot 1=\pi$: $q_0=\cos \frac{\pi }{2}=0,q_3=\sin \frac{\pi }{2}=1\Rightarrow q(1)=(0,0,0,1).$ Physical angle: $q_0=\cos \frac{\theta }{2}=0\Rightarrow \frac{\theta }{2}=\frac{\pi }{2}\Rightarrow \theta =\pi =180^{\circ }$ about the $z$-axis. Consistent: rate $\pi$ rad/s for 1 s turns the body by $\pi$ radians. ✔

$q$ maps a vector expressed in the body frame to the inertial frame. A tiny extra spin measured by gyros on the body is naturally described in body coordinates, so it acts first, in the body frame — and "first" for a map is the rightmost factor: $q\otimes \delta q$. Hence body rates post-multiply. If instead $\omega$ is given in the fixed inertial frame, the extra spin acts after $q$ has done its job, in inertial coordinates — the leftmost factor: $\delta q\otimes q$. Hence inertial rates pre-multiply. Because $\otimes$ is non-commutative, swapping these silently integrates the wrong attitude — a classic flight-software bug. This links to composition order in Rotation matrices and SO(3).

Input Jacobian $\partial \dot{q}/\partial \omega$. Write $\dot{q}=\frac{1}{2}\Xi (q)\omega$. This is linear in $\omega$ (each component of $\dot{q}$ is a fixed linear combination of ${\omega }_1,{\omega }_2,{\omega }_3$), and the derivative of a linear map $M\omega$ with respect to $\omega$ is just $M$. Hence $\partial \dot{q}/\partial \omega =\frac{1}{2}\Xi (q)$ — a $4\times 3$ matrix. Explicit entry: row 3, column 2 of $\dot{q}$'s Jacobian is $\frac{1}{2}$ times entry $(3,2)$ of $\Xi (q)$. At $q=(1,0,0,0)$ that entry is $q_0=1$, so $\frac{\partial {\dot{q}}_3}{\partial {\omega }_2}=\frac{1}{2}\cdot 1=\frac{1}{2}=\mathrm{0.5.}$ State Jacobian $\partial \dot{q}/\partial q$. Use the other form $\dot{q}=\frac{1}{2}\Omega (\omega )q$, which is linear in $q$, so $\partial \dot{q}/\partial q=\frac{1}{2}\Omega (\omega )$ — a $4\times 4$ matrix. Explicit entry: row 3, column 1. Entry $(3,1)$ of $\Omega (\omega )$ is ${\omega }_2$; with $\omega =(0,0.2,0)$ that is $0.2$, so $\frac{\partial {\dot{q}}_3}{\partial q_0}=\frac{1}{2}\cdot 0.2=\mathrm{0.1.}$ Cross-check with Exercise 2.1: there $q=(1,0,0,0)$ has only $q_0\ne 0$, and ${\dot{q}}_3=0.1$. That entire value flows through the single Jacobian path $\frac{\partial {\dot{q}}_3}{\partial q_0}\cdot q_0=0.1\cdot 1=0.1$. ✔ Both Jacobians agree with the direct computation, which is exactly why quaternion kinematics slot cleanly into an EKF's propagation and covariance-update steps.

From Exercise 3.3, the columns of $\Xi (q)$ are orthonormal for unit $q$, so $\Xi (q)$ preserves length: $\Vert \Xi (q)\omega \Vert =\Vert \omega \Vert$. Therefore $\Vert \dot{q}\Vert =\Vert \frac{1}{2}\Xi (q)\omega \Vert =\frac{1}{2}\Vert \Xi (q)\omega \Vert =\frac{1}{2}\Vert \omega \Vert .$ Number: $\Vert \dot{q}\Vert =\frac{1}{2}(0.05)=0.025$ (quaternion-units per second). Interpretation: the point $q$ slides along $S^3$ at half the body's angular speed — the geometric echo of the half-angle. Arc-length on the unit sphere equals angle, so $q$ sweeps $0.025$ rad of great-circle arc per second, and after $t=\frac{2\pi }{0.05}=125.66$ s (one full physical turn, $\Vert \omega \Vert t=2\pi$) it has traced $\pi$ of arc — landing at $-q$, the double cover again. (See figure s02.)

For constant $\omega$, $\dot{q}=\frac{1}{2}\Omega (\omega )q$ is a linear ODE with constant coefficients, whose exact solution is the matrix exponential $q(t+h)=\exp (\frac{1}{2}\Omega (\omega )h)q(t)$. Because $\Omega (\omega )$ is skew-symmetric (${\Omega }^{\top }=-\Omega$), its exponential is orthogonal, so it preserves norm exactly: $\Vert q(t+h)\Vert =\Vert q(t)\Vert$ — no drift, no renormalization needed. The closed form is $\exp (\frac{1}{2}\Omega (\omega )h)=\cos \frac{\Vert \omega \Vert h}{2}I+\frac{\sin \frac{\Vert \omega \Vert h}{2}}{\Vert \omega \Vert }\Omega (\omega ),$ which is precisely a rotation quaternion of angle $\Vert \omega \Vert h$ about $\hat{\omega }$ — the quaternion analogue of the Rodrigues rotation formula $R=I+\sin \theta [\hat{n}\times ]+(1-\cos \theta )[\hat{n}\times {]}^2$. Both split "rotate by an angle about an axis" into a $\cos$ (stay) plus a $\sin$ (turn) part. The exponential map is the bridge between the rate description ($\Omega ,\Xi$) and the finite rotation description (Rodrigues, $R\in SO(3)$).

The half comes from ::: the half-angle $\sin \frac{\theta }{2}$ stored in the quaternion (chain rule). $\dot{q}$ is orthogonal to $q$ because ::: $q^{\top }\Xi (q)=0^{\top }$, keeping motion tangent to $S^3$. $\Vert \dot{q}\Vert$ equals ::: $\frac{1}{2}\Vert \omega \Vert$ (columns of $\Xi (q)$ are orthonormal).