Gimbal Lock

Never really encountered it, but I know it’s one of the reasons we use Rodrigues Rotation Formula instead of Euler angles. Also related to Coordinate Systems, Coordinate Frame, and Understanding rotational frames in 3D.

Euler angles present a failure: the so-called Gimbal Lock

• the loss of one degree of freedom in a three-dimensional, three-gimbal mechanism.
• occurs when the axes of two of the three gimbals are driven into a parallel configuration, “locking” the system into rotation in a degenerate two-dimensional space.
• no actual locking happens, the word ‘lock’ is a bit misleading.

Resource: - GuerrillaCC on yt

4D Space: Quaternions operate in 4D space and encode rotation as a single operation, not as sequential rotations around individual axes. This intrinsic nature avoids the conditions that lead to gimbal lock.

GIMBAL LOCK, PROOF

$R=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & -\sin \alpha \\ 0& \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{ccc}\cos \beta & 0& \sin \beta \\ 0& 1& 0\\ -\sin \beta & 0& \cos \beta \end{array}\right]\left[\begin{array}{ccc}\cos \gamma & -\sin \gamma & 0\\ \sin \gamma & \cos \gamma & 0\\ 0& 0& 1\end{array}\right]$

• For $\beta =\pi /2$, we have $\cos \pi /2=0,\sin \pi /2=1$

$$R=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & -\sin \alpha \\ 0& \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ -1& 0& 0\end{array}\right]\left[\begin{array}{ccc}\cos \gamma & -\sin \gamma & 0\\ \sin \gamma & \cos \gamma & 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 1\\ \sin \alpha & \cos \alpha & 0\\ -\cos \alpha & \sin \alpha & 0\end{array}\right]\left[\begin{array}{ccc}\cos \gamma & -\sin \gamma & 0\\ \sin \gamma & \cos \gamma & 0\\ 0& 0& 1\end{array}\right]$$ $$R=\left[\begin{array}{ccc}0& 0& 1\\ \sin \alpha \cos \gamma +\cos \alpha \sin \gamma & -\sin \alpha \sin \gamma +\cos \alpha \cos \gamma & 0\\ -\cos \alpha \cos \gamma +\sin \alpha \sin \gamma & \cos \alpha \sin \gamma +\sin \alpha \cos \gamma & 0\end{array}\right]$$

• We know the trigonometric formula: $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha$

$$R=\left[\begin{array}{ccc}0& 0& 1\\ \sin (\alpha +\gamma )& \cos (\alpha +\gamma )& 0\\ -\cos (\alpha +\gamma )& \sin (\alpha +\gamma )& 0\end{array}\right]$$

We can notice that only $\alpha$ and $\gamma$ appear as $(\alpha +\gamma )$ — they’re coupled together. This means:

• We’ve lost one degree of freedom
• Rotating $\alpha$ and $\gamma$ by opposite amounts produces the same result
• We cannot independently control rotation around the x and z axes
• The gimbal is “locked” - two rotation axes have become aligned