Exercises — Modified Rodrigues parameters — singularity-free, compact

WHAT: The definition is $\mathbf{\sigma }=\hat{e}\tan (\Phi /4)$. WHY: The direction is the rotation axis itself; the length stores the angle, squeezed through $\tan (\Phi /4)$. At $\Phi =180^{\circ }$: quarter-angle $=45^{\circ }$, and $\tan 45^{\circ }=1$. So $\Vert \mathbf{\sigma }\Vert =1$. On figure s01: this is the single point where the blue curve meets the orange switch line — the boundary case, where (per the policy above) you do not yet switch.

Singularity at $\Phi =360^{\circ }$. Then $q_0=\cos (\Phi /2)=\cos 180^{\circ }=-1$, making the denominator $1+q_0=0$. Dividing by zero → blow-up. On figure s01 this is the red vertical wall the blue curve races toward.

False. MRPs have one singularity, at $\Phi =360^{\circ }$. What makes them practical is that this singularity is avoidable via the shadow set. Only quaternions and the DCM are globally nonsingular.

Step 1 — magnitude. $\Vert \mathbf{\sigma }\Vert =\tan (90^{\circ }/4)=\tan 22.5^{\circ }\approx 0.4142$. Why: magnitude is the quarter-angle tangent. Step 2 — direction. Attach that length to the axis: $\mathbf{\sigma }=(0,0,0.4142)$. Step 3 — safety check. $0.4142<1$, so we are left of the orange line on figure s01 — safe, no switch.

WHY invert: magnitude tells us the quarter-angle. $\Vert \mathbf{\sigma }\Vert =0.2679$. WHAT: $\Phi /4=\arctan (0.2679)=15^{\circ }$, so $\Phi =60^{\circ }$. The axis is $\hat{x}=(1,0,0)$.

Quaternion route. $q_0=\cos 60^{\circ }=0.5$; $\mathbf{q}=\hat{e}\sin 60^{\circ }=(0,0,0.8660)$. $\mathbf{\sigma }=\mathbf{q}/(1+q_0)=(0,0,0.8660)/1.5=(0,0,0.5774)$. Axis–angle route. $\tan (120^{\circ }/4)=\tan 30^{\circ }=0.5774$. ✅ Same answer — the two definitions are one and the same.

(a) $\Vert \mathbf{\sigma }\Vert =\tan (270^{\circ }/4)=\tan 67.5^{\circ }\approx 2.4142$. So $\mathbf{\sigma }=(0,2.4142,0)$. (b) $2.4142>1$ → right of the orange line, heading toward the red wall. Switch. (c) ${\mathbf{\sigma }}^S=-\mathbf{\sigma }/({\mathbf{\sigma }}^{\top }\mathbf{\sigma })=-\mathbf{\sigma }/(2.4142{)}^2$. Component: $-2.4142/5.8284=-0.4142$, so ${\mathbf{\sigma }}^S=(0,-0.4142,0)$. Equivalent angle: $\Vert {\mathbf{\sigma }}^S\Vert =0.4142=\tan 22.5^{\circ }$, so ${\Phi }^{\prime }=90^{\circ }=360^{\circ }-270^{\circ }$. ✅ Same attitude, tiny magnitude.

WHAT the skew matrix is: for any 3-vector $\mathbf{a}$, the matrix $[{\mathbf{a}}^{\times }]$ is built so that $[{\mathbf{a}}^{\times }]\mathbf{b}=\mathbf{a}\times \mathbf{b}$ — it turns a cross product into a plain matrix multiply. WHY we want it: the kinematics matrix $B$ needs cross-product coupling between axes, and a matrix is what we can multiply and invert. $[{\mathbf{\sigma }}^{\times }]=\left(\begin{array}{ccc}0& -{\sigma }_3& {\sigma }_2\\ {\sigma }_3& 0& -{\sigma }_1\\ -{\sigma }_2& {\sigma }_1& 0\end{array}\right)$ For $\mathbf{\sigma }=(0,0,0.5)$: $[{\mathbf{\sigma }}^{\times }]=\left(\begin{array}{ccc}0& -0.5& 0\\ 0.5& 0& 0\\ 0& 0& 0\end{array}\right)$ Notice the diagonal is always zero and $[{\mathbf{\sigma }}^{\times }{]}^{\top }=-[{\mathbf{\sigma }}^{\times }]$ — that is what "skew-symmetric" means.

Step 1 — build $B$. ${\sigma }^2=0.25$, so $(1-{\sigma }^2)=0.75$.

• Identity term: $0.75\mathbb{I}$.
• Skew term: $2[{\mathbf{\sigma }}^{\times }]$ from L3.2.
• Outer-product term: $2\mathbf{\sigma }{\mathbf{\sigma }}^{\top }$ has only the $(3,3)$ entry nonzero: $2(0.5)(0.5)=0.5$. Step 2 — apply to $\mathbf{\omega }=(0,0,0.1)$. Since $\mathbf{\omega }$ points along $\hat{z}$ and $\mathbf{\sigma }$ also points along $\hat{z}$, the skew term (a cross product of parallel vectors) gives zero. Only the third component survives: $B\mathbf{\omega }$ third entry $=(0.75+0.5)(0.1)=1.25\times 0.1=0.125$. Step 3 — the quarter factor. $\dot{\mathbf{\sigma }}=\frac{1}{4}B\mathbf{\omega }=(0,0,0.03125)$ per second.

WHY the inverse form: control laws prescribe how $\mathbf{\sigma }$ should move; we must back out the $\mathbf{\omega }$ command. The parent note gives $\mathbf{\omega }=4B^{-1}\dot{\mathbf{\sigma }}$ with $B^{-1}=B^{\top }/(1+{\sigma }^2{)}^2$. Step 1. ${\sigma }^2=0.09+0.16=0.25$. So $(1+{\sigma }^2{)}^2=1.25^2=1.5625$. Step 2 — build $B$. $(1-{\sigma }^2)=0.75$. $B=0.75\mathbb{I}+2\left(\begin{array}{ccc}0& 0& -0.4\\ 0& 0& -0.3\\ 0.4& 0.3& 0\end{array}\right)+2\left(\begin{array}{ccc}0.09& -0.12& 0\\ -0.12& 0.16& 0\\ 0& 0& 0\end{array}\right)$ $B=\left(\begin{array}{ccc}0.93& -0.24& -0.8\\ -0.24& 1.07& -0.6\\ 0.8& 0.6& 0.75\end{array}\right)$ Step 3 — inverse-multiply. $\mathbf{\omega }=4\frac{B^{\top }}{1.5625}\dot{\mathbf{\sigma }}$. Why the next move is legal — the matrix–vector rule. A matrix $M$ times a vector $\mathbf{v}$ produces a new vector whose entries are $M$'s columns weighted by $\mathbf{v}$'s entries: $M\mathbf{v}=v_1(\text{col }1)+v_2(\text{col }2)+v_3(\text{col }3)$. Here $\dot{\mathbf{\sigma }}=(0.05,0,0)$ has only its first entry nonzero, so only the first column of $B^{\top }$ survives, scaled by $0.05$. And the first column of $B^{\top }$ is by definition the first row of $B$, namely $(0.93,-0.24,-0.8)$. Hence: $\mathbf{\omega }=\frac{4\times 0.05}{1.5625}(0.93,-0.24,-0.8)=0.128(0.93,-0.24,-0.8)$ $\mathbf{\omega }\approx (0.11904,-0.03072,-0.1024)\text{rad/s}.$

Start from ${\mathbf{\sigma }}^S=-\mathbf{\sigma }/{\sigma }^2$ where ${\sigma }^2=\Vert \mathbf{\sigma }{\Vert }^2$. Take the norm: $\Vert {\mathbf{\sigma }}^S\Vert =\Vert \mathbf{\sigma }\Vert /{\sigma }^2=\Vert \mathbf{\sigma }\Vert /\Vert \mathbf{\sigma }{\Vert }^2=1/\Vert \mathbf{\sigma }\Vert$. Interpretation: the switch is a reciprocal map on magnitude. If $\Vert \mathbf{\sigma }\Vert =2.41$, the shadow has $\Vert {\mathbf{\sigma }}^S\Vert =0.414<1$. The two magnitudes multiply to exactly $1$, and $\Vert \mathbf{\sigma }\Vert =1$ is the fixed point where original and shadow meet — exactly the orange line on figure s01, and exactly why the boundary policy leaves that point alone.

Compute $\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)$ at each waypoint:

• $\Phi =170^{\circ }$: $\tan 42.5^{\circ }=0.9163$ → safe.
• $\Phi =175^{\circ }$: $\tan 43.75^{\circ }=0.9601$ → safe.
• $\Phi =185^{\circ }$: $\tan 46.25^{\circ }=1.0416$ → exceeds 1 → switch here.
• $\Phi =190^{\circ }$: $\tan 47.5^{\circ }=1.0913$ → would be even larger; already switched. Trigger: the switch fires the first sample where $\Vert \mathbf{\sigma }\Vert >1$ (strict inequality — the boundary $\Vert \mathbf{\sigma }\Vert =1$ does not trigger, per the policy up top), i.e. as $\Phi$ crosses $180^{\circ }$ — between the $175^{\circ }$ and $185^{\circ }$ samples. Stored magnitude after switch (at $185^{\circ }$): $\Vert {\mathbf{\sigma }}^S\Vert =1/1.0416=0.9601$, which equals $\tan ((360-185)/4{)}^{\circ }=\tan 43.75^{\circ }$. The estimator now carries a small vector describing the equivalent $175^{\circ }$ short-way rotation — smoothly, no blow-up.

Read figure s02 like this: it zooms into the $150^{\circ }$–$200^{\circ }$ window of the same curve. The two green dots on the left are the safe samples ($170^{\circ },175^{\circ }$); the two red dots on the right ($185^{\circ },190^{\circ }$) show where the raw magnitude would keep climbing above the orange $\Vert \mathbf{\sigma }\Vert =1$ line. The gray arrows are the reciprocal hop the switch performs — dropping each red point down to its green diamond shadow, which sits safely back below $1$. That downward hop is the singularity avoidance, made visible.

MRP without switch: as $\Phi \to 360^{\circ }$, $\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)\to \tan 90^{\circ }\to \infty$. The state vector's components explode; the $B$ matrix entries (quadratic in $\mathbf{\sigma }$) explode too, and numerical integration diverges. But this is a removable failure — one reciprocal switch keeps $\Vert \mathbf{\sigma }\Vert \le 1$ forever. Euler angles: gimbal lock at $\pm 90^{\circ }$ is a structural rank loss — two rotation axes align, a degree of freedom vanishes, and no coordinate relabeling within Euler angles removes it. You must change parameterization entirely. Which is worse: gimbal lock is worse operationally, for two reasons. First, $90^{\circ }$ pitch is a routine, frequently visited attitude, whereas a full $360^{\circ }$ turn is a rare extreme. Second, the Euler failure is unavoidable inside the representation — the singularity is baked into the geometry — while the MRP failure is dodgeable by a single dot-product test and a reciprocal switch that cost almost nothing. Why MRPs win overall: they combine the minimum parameter count (3, tying Euler angles) with a singularity relocated to $360^{\circ }$ (far from normal operation) and rendered unreachable by the shadow set. You get the compactness of Euler angles without gimbal lock, and nearly the robustness of quaternions without the redundant fourth number or its sign ambiguity. That balance — compact, smooth over $\pm 180^{\circ }$, singularity avoidable for free — is exactly why MRPs are the preferred small attitude state for many modern control laws.