Exercises — Gimbal lock — problem with Euler angles at θ = ±90°

WHAT: We recall the defining configuration. WHY: Gimbal lock is the pitch value that swings the roll axis onto the yaw axis. The middle rotation $R_y(\theta )$ tilts the last ($x$) axis relative to the first ($z$) axis. It fully overlaps them when the tilt is a quarter turn: $\theta =+90°$ or $\theta =-90°$. Answer: $\theta =\pm 90°$. The yaw $\psi$ and roll $\varphi$ become redundant (they combine into one effective angle); pitch $\theta$ is still meaningful.

WHAT: Read the kinematic matrix from the parent note. WHY: The entries feeding $\dot{\varphi }$ and $\dot{\psi }$ carry the singular functions. The dangerous terms are $\tan \theta$ and $\frac{1}{\cos \theta }$. As $\theta \to 90°$, $\cos \theta \to 0^{+}$, so both $\to +\infty$. Answer: $\tan \theta \to +\infty$ and $1/\cos \theta \to +\infty$.

WHAT: Plug into the yaw-rate row of the kinematic map. WHY: We want to feel how fast the coordinate rate diverges near lock. The relevant row is $\dot{\psi }=\frac{\sin \varphi }{\cos \theta }q+\frac{\cos \varphi }{\cos \theta }r$. With $\varphi =0$: $\sin \varphi =0,\cos \varphi =1$, so $\dot{\psi }=\frac{r}{\cos \theta }.$ At $89.9°$: $\cos (89.9°)=1.7453\times 10^{-3}$, so $\dot{\psi }=0.01/1.7453\times 10^{-3}\approx 5.73\text{rad/s}$. At $89.99°$: $\cos (89.99°)=1.7453\times 10^{-4}$, so $\dot{\psi }=0.01/1.7453\times 10^{-4}\approx 57.3\text{rad/s}$. Answer: $\approx 5.73$ then $\approx 57.3\text{rad/s}$; a 10× jump for a 10× closer approach to $90°$. A finite $r$ demands an exploding $\dot{\psi }$.

WHAT: Use the collapse rule "$R$ depends only on $\varphi -\psi$ at $\theta =+90°$." WHY: Any pair with the same difference $\varphi -\psi$ yields the same matrix. Here $\varphi -\psi =40°-10°=30°$. Pick any $(\psi ,\varphi )$ with difference $30°$, e.g. $(\psi ,\varphi )=(0°,30°)$ or $(50°,80°)$. Answer: e.g. $(\psi ,\varphi )=(0°,30°)$ — same $R$ as $(10°,40°)$.

WHAT: Set $c_{\theta }=0,s_{\theta }=-1$ and multiply $R_z(\psi )R_y(-90°)R_x(\varphi )$. WHY: We test the other singular pitch directly, rather than trusting symmetry blindly. First, $R_y(-90°)=\left(\begin{array}{ccc}0& 0& -1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right)$. Multiplying out (same procedure as the parent's $+90°$ case) gives

WHAT IT LOOKS LIKE: every nonzero off-column entry is a function of the single combination $\varphi +\psi$. Answer: Increasing $\psi$ by $\delta$ and decreasing $\varphi$ by $\delta$ leaves $R$ unchanged ⇒ effective DOF is $\varphi +\psi$. Confirmed.

WHAT: Compare where the roll axis lands after $\pm 90°$ of pitch. WHY: The sign of the combination is set by whether the roll axis lands parallel or anti-parallel to the yaw axis. WHAT IT LOOKS LIKE: In the figure, pitching $+90°$ tips the body $x$-axis to point along $+z$, so a yaw and a roll of equal size add up as co-rotations about the same directed axis — but yaw is applied before the tilt and roll after, and the intervening flip reverses one sense, leaving the difference $\varphi -\psi$. Pitching $-90°$ tips the $x$-axis to point along $-z$ (anti-parallel), reversing that intervening sense again, so now the sum $\varphi +\psi$ survives. Answer: The middle tilt sends the roll axis to $\pm z$; the sign of that landing direction flips the relative sense of the two remaining rotations, turning $\varphi -\psi$ into $\varphi +\psi$.

WHAT: Relate singularity of a map to the vanishing of a determinant. WHY (which tool and why this one): A linear map loses invertibility ⇔ its matrix has zero determinant. This is the algebraic detector of "a direction got squashed to nothing" — exactly the DOF loss we're chasing. We use the determinant because it is a single scalar whose zeros pin down all singular configurations at once. Since $\det (E)=1/\det (E^{-1})=1/\cos \theta$, the forward map $E$ blows up precisely where $\det (E^{-1})=\cos \theta =0$. Over the pitch range $\theta \in (-90°,+90°]$ used for a chart, $\cos \theta =0$ only at $\theta =\pm 90°$. WHAT IT LOOKS LIKE: the determinant is a smooth cosine bump, $=1$ at $\theta =0$ (perfectly invertible) and pinching to $0$ at the two poles. Answer: Singularities of the forward Euler-rate map are exactly $\theta =\pm 90°$; the map is well-conditioned only away from those poles, worst near them.

WHAT: Read off the single entry that encodes pitch. WHY: In the general ZYX matrix the $(3,1)$ entry equals $-\sin \theta$, isolating pitch from the other two angles. Lock is $\theta =\pm 90°$ ⇔ $\sin \theta =\pm 1$ ⇔ $R_{31}=\mp 1$. So flag lock when $|R_{31}|$ is within a tolerance of $1$, e.g. $|R_{31}|>0.9998$ (that's $\theta$ within $\approx 1.1°$ of a pole). WHY it breaks: the usual formulas $\psi =atan2(R_{21},R_{11})$ and $\varphi =atan2(R_{32},R_{33})$ read entries that have collapsed to zero at lock (see the parent's $\theta =90°$ matrix: $R_{11}=R_{21}=R_{32}=R_{33}=0$). You get $atan2(0,0)$ — undefined. Only the combination $\varphi \mp \psi$ is recoverable, from the surviving entries $\sin (\varphi \mp \psi ),\cos (\varphi \mp \psi )$. Answer: Detect lock by $|R_{31}|\to 1$; near lock, split the ambiguity by convention (e.g. set $\psi =0$ and solve $\varphi =atan2(R_{12},R_{22})$), because only $\varphi \mp \psi$ is uniquely defined.

WHAT: Contrast the algebraic form of the quaternion→$SO(3)$ map with the Euler map. WHY (which tool and why): Singularities of a coordinate map appear as divisions by a vanishing quantity or a rank-dropping Jacobian. So we inspect the map's formula for any denominator that can hit zero. The rotation from a unit quaternion is, e.g. for the first row, $R=\left(\begin{array}{ccc}1-2(y^2+z^2)& 2(xy-wz)& 2(xz+wy)\\ \cdots & \cdots & \cdots \end{array}\right)$ — pure polynomials, no $1/\cos \theta$, no denominator anywhere. There is no value of $(w,x,y,z)$ on the unit sphere that makes an entry blow up. The redundancy: $q$ and $-q$ give the identical $R$ (both rows contain products like $wz$, $wy$, $wx$ and squares, all invariant under flipping every sign). This is a two-to-one cover $S^3\to SO(3)$ — but it never loses a direction: the map's Jacobian stays full rank everywhere on $S^3$. WHAT IT LOOKS LIKE: picture $S^3$ wrapping smoothly and doubly over the rotation group with no pinch points — versus the Euler chart, which pinches shut at the two poles. Answer: No singular orientation (no lost DOF); the only quirk is the harmless sign ambiguity $q\equiv -q$. See Quaternions and Singularities of coordinate charts.

WHAT: Apply the "hairy sphere"-style obstruction: $SO(3)$ is not globally chartable by 3 parameters without a singularity. WHY: $SO(3)$ has a curved global topology (it is ${\mathbb{RP}}^3$); any single 3-angle chart is like a flat map of the round Earth — it must tear or pinch somewhere. You can move the singularity (a different axis order puts lock at a different pitch) but never delete it. The escape: Don't insist on a 3-parameter chart. Use a 4-parameter, one-constraint representation (quaternions: 4 numbers, $|q|=1$) or the $9$-number rotation matrix ($SO(3)$ with 6 constraints). Extra parameters buy you a smooth global description at the cost of a redundancy — but redundancy $\ne$ lost DOF. Answer: No — every 3-angle set has a lock somewhere; the fix is to leave the "three-angle" family entirely for internal computation (quaternions / rotation matrices), converting to Euler angles only for human display, exactly as the parent note's engineer's fix prescribes. See Rotation matrices SO(3) and Attitude determination and control.