Question bank — Modified Rodrigues parameters — singularity-free, compact

False. They have one genuine singularity at $\Phi =360^{\circ }$ (where the scalar part $q_0=-1$); the point is only that it is avoidable by switching sets, whereas quaternions and DCMs have no singularity at all.

False. That is the classical Gibbs vector. MRPs use the quarter-angle, so $\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)$ — this extra halving is precisely why the blow-up moves from $180^{\circ }$ out to $360^{\circ }$.

False. Every attitude has two MRP sets (the original $\mathbf{\sigma }$ and its shadow ${\mathbf{\sigma }}^S$), because $q$ and $-q$ describe the same rotation and each maps to its own MRP.

False. The switch is a pure change of coordinates: $\mathbf{\sigma }$ and ${\mathbf{\sigma }}^S$ encode the identical physical rotation, just from the two quaternion sign branches. Nothing physical moves.

False. It is $\frac{1}{4}$. Quaternions carry the half-angle ($\frac{1}{2}$); MRPs carry the quarter-angle, so a second halving appears in $\dot{\mathbf{\sigma }}=\frac{1}{4}B(\mathbf{\sigma })\mathbf{\omega }$.

True. MRPs use 3 numbers versus the DCM's 9 (with 6 constraints) and the quaternion's 4 (with a unit-norm constraint), making MRPs the most compact smooth-over-a-full-turn description.

True. At $\mathbf{\sigma }=0$ the identity term's factor $(1-{\sigma }^2)=1$, and both the cross-product term $2[{\mathbf{\sigma }}^{\times }]$ and the outer-product term $2\mathbf{\sigma }{\mathbf{\sigma }}^{\top }$ vanish (they carry a factor of $\mathbf{\sigma }=0$), leaving $B={\mathbb{I}}_3$.

True. $\Vert \mathbf{\sigma }\Vert =1$ means $\Phi =180^{\circ }$, exactly halfway to the $360^{\circ }$ blow-up; staying at or below $1$ (by switching sets) keeps you in the well-behaved region.

True. Because ${\mathbf{\sigma }}^S$ has magnitude $\tan ((360^{\circ }-\Phi )/4)$, an attitude at $\Phi =300^{\circ }$ is re-described with the small rotation $60^{\circ }$.

Error: at $\Vert \mathbf{\sigma }\Vert =10$ you are at $\Phi =4\arctan (10)\approx 337^{\circ }$, dangerously close to the $360^{\circ }$ singularity and to numerical blow-up. You should have switched to the shadow set once $\Vert \mathbf{\sigma }\Vert$ exceeded $1$.

Error: the Gibbs vector is $\hat{e}\tan (\Phi /2)$ (singular at $180^{\circ }$); the MRP is $\hat{e}\tan (\Phi /4)$ (singular at $360^{\circ }$). Different half-angle, different singularity — see Classical Rodrigues parameters (Gibbs vector).

Error: $B$ is nearly orthogonal, $B^{\top }B=(1+{\sigma }^2{)}^2\mathbb{I}$, so $B^{-1}=B^{\top }/(1+{\sigma }^2{)}^2$ — a closed-form transpose-and-scale, no numerical matrix inversion needed.

Error: $\Vert \mathbf{\sigma }\Vert >1$ simply means $\Phi >180^{\circ }$ — a perfectly valid rotation. It is a cue to switch sets, not a sign of bad data.

Error: Euler angles hit gimbal lock at a middle angle of $\pm 90^{\circ }$, right inside the normal operating range. MRPs push their (avoidable) singularity all the way to $360^{\circ }$ — a much better deal.

Error: they cover all rotations except $\Phi =360^{\circ }$, and the covering is two-to-one (original plus shadow), not one-to-one. "Three unconstrained numbers" does not imply a global chart.

Dividing by the scalar part $q_0$ gives the Gibbs vector, which blows up at $\Phi =180^{\circ }$ where $q_0=0$. Using $(1+q_0)$ only vanishes at $q_0=-1$, i.e. $\Phi =360^{\circ }$, pushing the trouble twice as far away.

The tangent blows up when its argument reaches $90^{\circ }$. For $\tan (\Phi /2)$ that is $\Phi =180^{\circ }$; for $\tan (\Phi /4)$ it is $\Phi =360^{\circ }$. Halving the argument doubles the angle that reaches the blow-up.

It comes from feeding $-\mathbf{q}$ (the other quaternion branch) through the MRP definition, which algebraically yields $-\mathbf{\sigma }/({\mathbf{\sigma }}^{\top }\mathbf{\sigma })$. Plain $-\mathbf{\sigma }$ would be a genuinely different rotation.

At $\Vert \mathbf{\sigma }\Vert =1$ the original and shadow magnitudes are equal, so switching there keeps you strictly in the region $\Vert \cdot \Vert \le 1$ with maximum margin from the $360^{\circ }$ blow-up.

MRP space is curved; away from identity a raw $\mathbf{\omega }$ must be rotated and scaled into the local MRP coordinates. $B$ does exactly that — the simple $\frac{1}{4}\mathbf{\omega }$ only holds near $\mathbf{\sigma }=0$.

They are minimal (3 numbers, no constraint to enforce), smooth over the full $\pm 180^{\circ }$ working range, and their kinematics invert in closed form — so control errors and feedback laws stay simple and singularity-free in practice.

The two signs give the two MRP sets; instead of being an annoyance, this pair is exactly what lets you hop to the shadow set and dodge the singularity.

$\mathbf{\sigma }=0$, since $\tan (0/4)=0$; the axis is undefined but irrelevant because it is multiplied by zero.

$\Vert \mathbf{\sigma }\Vert =\tan (45^{\circ })=1$ (finite and well-behaved), whereas the Gibbs vector $\tan (90^{\circ })$ blows up — the case where MRPs win clearly.

$\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)\to \infty$ (the singularity), while the shadow $\Vert {\mathbf{\sigma }}^S\Vert =\tan ((360^{\circ }-\Phi )/4)\to 0$ — so you can always describe the attitude by the finite shadow instead.

No — $\Phi =360^{\circ }$ is the same as $\Phi =0^{\circ }$ (a full turn is the identity), so the shadow lands at ${\mathbf{\sigma }}^S=0$, perfectly finite. In practice you never let the original reach the singularity anyway.

It is undefined (division by $\Vert \mathbf{\sigma }{\Vert }^2=0$), corresponding to the scalar part $q_0=-1$ — the singular attitude of the shadow branch. This is harmless because you only ever switch when $\Vert \mathbf{\sigma }\Vert >1$, never at $0$.

It reduces to ${\mathbb{I}}_3$, so $\dot{\mathbf{\sigma }}\approx \frac{1}{4}\mathbf{\omega }$ — the linearized, small-angle limit matching the quarter-angle intuition.

They are shadow partners describing the same physical attitude; comparing raw numbers would wrongly suggest a large error, so a controller must recognise this pairing.