Exercises — DCM kinematics — Ċ = −[ω×]C

WHAT: apply the vee-map. WHY: the pattern of $[\vec{w}\times ]$ tells us exactly where each component hides.

• Entry $(3,2)=w_1\Rightarrow w_1=3$
• Entry $(1,3)=w_2\Rightarrow w_2=2$
• Entry $(2,1)=w_3\Rightarrow w_3=5$

So $\vec{\omega }=[3,2,5{]}^T$. Sanity check the top row: $-w_3=-5$ ✔, $w_2=2$ ✔.

(b) is correct. WHY: $C$ carries inertial vectors into body coordinates. When the body spins forward, the fixed inertial vectors appear to spin backward in the body frame — the minus sign records that reversal. (Choice (a) is the equation for the transpose $C^T:\mathcal{B}\to \mathcal{N}$.)

WHAT: plug into $\dot{C}=-[\vec{\omega }\times ]C$. WHY $C=I$ helps: the multiply collapses to just $-[\vec{\omega }\times ]$. $[\vec{\omega }\times ]=\left[\begin{array}{ccc}0& -4& 0\\ 4& 0& 0\\ 0& 0& 0\end{array}\right]\Rightarrow \dot{C}=-\left[\begin{array}{ccc}0& -4& 0\\ 4& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 4& 0\\ -4& 0& 0\\ 0& 0& 0\end{array}\right].$

WHAT: add $\dot{C}\Delta t$ to $I$. $C(0.01)\approx \left[\begin{array}{ccc}1& 0.04& 0\\ -0.04& 1& 0\\ 0& 0& 1\end{array}\right].$ Orthogonality check: take the top-left $2\times 2$ block. Its first column has length $\sqrt{1^2+0.04^2}=\sqrt{1.0016}\approx 1.0008\ne 1$. Conclusion: it drifted off $SO(3)$ by $\approx 8\times 10^{-4}$. The exact solution would have length exactly 1; Euler over-shoots. This is why real code renormalizes — see Matrix exponential and rotation about a fixed axis for the exact step.

WHAT: the parent note gives the exact exponential solution. $C(t)=\left[\begin{array}{ccc}\cos \omega t& \sin \omega t& 0\\ -\sin \omega t& \cos \omega t& 0\\ 0& 0& 1\end{array}\right].$ At $\omega t=\pi /2$: $\cos =0,\sin =1$, so $C=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right].$ Check: $\det =0\cdot 0-1\cdot (-1)\cdot 1=1$ ✔ (proper rotation), and each column is a unit vector ✔.

WHAT: differentiate, multiply by $C^T$, read off with vee. WHY $-\dot{C}{C}^T$: from $\dot{C}=-[\omega \times ]C$, right-multiply by $C^T$ and use $C{C}^T=I$. $\dot{C}=\left[\begin{array}{ccc}-2\sin 2t& 2\cos 2t& 0\\ -2\cos 2t& -2\sin 2t& 0\\ 0& 0& 0\end{array}\right],C^T=\left[\begin{array}{ccc}\cos 2t& -\sin 2t& 0\\ \sin 2t& \cos 2t& 0\\ 0& 0& 1\end{array}\right].$ Multiply and negate (the trig collapses via ${\cos }^2+{\sin }^2=1$): $-\dot{C}{C}^T=\left[\begin{array}{ccc}0& -2& 0\\ 2& 0& 0\\ 0& 0& 0\end{array}\right]=[\vec{\omega }\times ].$ Vee-map: ${\vec{\omega }}_{\mathcal{B}}=[0,0,2{]}^T$ rad/s — a constant spin of $2$ rad/s about body-$z$. See Attitude propagation & determination (TRIAD, QUEST) for where such measured histories come from.

WHY: $C$ maps $\mathcal{N}\to \mathcal{B}$. The stored solution $C(t)=e^{-[\omega \times ]t}$ has a built-in minus. So the physical body spin $+\omega$ shows up as a rotation by $-\omega t$ inside $C$. When we invert with $-\dot{C}{C}^T$, that minus is undone and we recover the true, positive body rate $+2$. The bookkeeping is self-consistent: minus in $\dot{C}$, minus in recovery, they cancel.

WHAT: because $\vec{\omega }$ is constant, $C(t)=e^{-[\omega \times ]t}$, a rotation by angle $\theta =|\vec{\omega }|t=t$ (since $|\vec{\omega }|=1$) about the axis $\hat{a}=[1,1,1{]}^T/\sqrt{3}$, negated (the minus in the kinematic equation). WHY exponential: $e^{-[\omega \times ]t}$ is the solution of a constant-coefficient linear matrix ODE; this is exactly Matrix exponential and rotation about a fixed axis. Using Rodrigues' formula with angle $-t$ about $\hat{a}$: $C(t)=I-\sin t[\hat{a}\times ]+(1-\cos t)[\hat{a}\times {]}^2,$ where the $-\sin t$ (not $+\sin t$) carries the minus sign. Here $[\hat{a}\times ]=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}0& -1& 1\\ 1& 0& -1\\ -1& 1& 0\end{array}\right]$. Check at $t=0$: $C(0)=I$ ✔. Check the axis is fixed: $C(t)\hat{a}=\hat{a}$ for all $t$ because rotating about $\hat{a}$ leaves $\hat{a}$ untouched — verify $[\hat{a}\times ]\hat{a}=\vec{0}$ ✔.

WHAT: differentiate $D=RC$ with $R$ constant ($\dot{R}=0$). $\dot{D}=R\dot{C}=R(-[{\vec{\omega }}_{\mathcal{B}}\times ]C)=-R[{\vec{\omega }}_{\mathcal{B}}\times ]R^{T}RC.$ WHY insert $R^{T}R=I$: to expose a skew operator acting on $D=RC$. Use the rotation identity $R[\vec{w}\times ]R^T=[(R\vec{w})\times ]$ (proved in Rotation group SO(3) and Lie algebra so(3)). With ${\vec{\omega }}_{\mathcal{S}}=R{\vec{\omega }}_{\mathcal{B}}$: $\boxed{\dot{D}=-[{\vec{\omega }}_{\mathcal{S}}\times ]D.}$ Meaning: the kinematic equation keeps its form — you just express $\vec{\omega }$ in the sensor frame. That form-invariance is exactly why the equation is so beloved.

WHAT/WHY: Jacobi's formula gives $\frac{d}{dt}\det C=\det C\cdot \mathrm{tr}(C^{-1}\dot{C})$. It converts a determinant rate into a trace — much easier. Substitute $\dot{C}=-[\vec{\omega }\times ]C$ and $C^{-1}=C^T$: $\mathrm{tr}(C^{-1}\dot{C})=\mathrm{tr}(C^T(-[\vec{\omega }\times ]C))=-\mathrm{tr}([\vec{\omega }\times ]C{C}^T)=-\mathrm{tr}[\vec{\omega }\times ].$ WHY the last trace is zero: a skew-symmetric matrix has all-zero diagonal, so its trace is $0$. Therefore $\frac{d}{dt}\det C=\det C\cdot 0=0$. $\det C$ is constant. $■$ This is the determinant-level companion to the length-preservation proof; together they say $C$ stays a proper rotation forever. Connects to the Lie-algebra fact $\mathrm{tr}(\mathfrak{so}(3))=0$ in Rotation group SO(3) and Lie algebra so(3).

DCM side: from L2.3, $C(t)$ is a rotation whose angle about $z$ is $\omega t$ (magnitude). Quaternion side: constant $\vec{\omega }$ gives $q(t)=[\cos \frac{\omega t}{2},0,0,\sin \frac{\omega t}{2}]$ (scalar-first). A unit quaternion $[\cos \frac{\varphi }{2},\hat{a}\sin \frac{\varphi }{2}]$ encodes a rotation by angle $\varphi$ about $\hat{a}$. Match: here $\varphi =\omega t$ about $\hat{a}=\hat{z}$ — identical to the DCM. WHY the half-angle: quaternions double-cover $SO(3)$, so they carry $\varphi /2$ internally but represent the same physical $\varphi$. Numeric check at $\omega t=\pi /2$: quaternion $=[\cos 45^{\circ },0,0,\sin 45^{\circ }]=[0.7071,0,0,0.7071]$, which rebuilds exactly the $C$ found in L2.3. ✔

L1 read/choose the operator → L2 build & step $\dot{C}$ → L3 invert to get $\omega$ → L4 change frames keeping the form → L5 prove the invariants and bridge to quaternions. Every rung is the same equation $\dot{C}=-[\omega \times ]C$ seen from a new angle.