# Gimbal lock

Euler angles have a major deficiency, and that is, that it is possible, in some rotation sequences, to reach a situation where two of the three Euler angles cause rotation around the same axis of the object. In the case below, rotation around the $$x$$ axis becomes indistinguishable in its effect from rotation around the $$z$$ axis, so the $$z$$ and $$x$$ axis angles collapse into one transformation, and the rotation reduces from three degrees of freedom to two.

Imagine that we are using the Euler angle convention of starting with a rotation around the $$x$$ axis, followed by the $$y$$ axis, followed by the $$z$$ axis.

Here we see a Spitfire aircraft, flying across the screen. The $$x$$ axis is left to right (tail to nose), the $$y$$ axis is from the left wing tip to the right wing tip (going away from the screen), and the $$z$$ axis is from bottom to top:

Imagine we wanted to do a slight roll with the left wing tilting down (rotation about $$x$$) like this:

followed by a violent pitch so we are pointing straight up (rotation around $$y$$ axis):

Now we’d like to do a turn of the nose towards the viewer (and the tail away from the viewer):

But, wait, let’s go back over that again. Look at the result of the rotation around the $$y$$ axis. Notice that the $$x$$ axis, as was, is now aligned with the $$z$$ axis, as it is now. Rotating around the $$z$$ axis will have exactly the same effect as adding an extra rotation around the $$x$$ axis at the beginning. That means that, when there is a $$y$$ axis rotation that rotates the $$x$$ axis onto the $$z$$ axis (a rotation of $$\pm\pi/2$$ around the $$y$$ axis) - the $$x$$ and $$y$$ axes are “locked” together.

## Mathematics of gimbal lock

We see gimbal lock for this type of Euler axis convention, when $$\cos(\beta) = 0$$, where $$\beta$$ is the angle of rotation around the $$y$$ axis. By “this type of convention” we mean using rotation around all 3 of the $$x$$, $$y$$ and $$z$$ axes, rather than using the same axis twice - e.g. the physics convention of $$z$$ followed by $$x$$ followed by $$z$$ axis rotation (the physics convention has different properties to its gimbal lock).

We can show how gimbal lock works by creating a rotation matrix for the three component rotations. Recall that, for a rotation of $$\alpha$$ radians around $$x$$, followed by a rotation $$\beta$$ around $$y$$, followed by rotation $$\gamma$$ around $$z$$, the rotation matrix $$R$$ is:

$\begin{split}R = \left(\begin{smallmatrix}\operatorname{cos}\left(\beta\right) \operatorname{cos}\left(\gamma\right) & - \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) + \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\beta\right) & \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) + \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\beta\right)\\\operatorname{cos}\left(\beta\right) \operatorname{sin}\left(\gamma\right) & \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right) + \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\beta\right) \operatorname{sin}\left(\gamma\right) &- \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right) + \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\beta\right) \operatorname{sin}\left(\gamma\right)\\- \operatorname{sin}\left(\beta\right) & \operatorname{cos}\left(\beta\right) \operatorname{sin}\left(\alpha\right) & \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\beta\right)\end{smallmatrix}\right)\end{split}$

When $$\cos(\beta) = 0$$, $$\sin(\beta) = \pm1$$ and $$R$$ simplifies to:

$\begin{split}R = \left(\begin{smallmatrix}0 & - \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) + \pm{1} \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right) & \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) + \pm{1} \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right)\\0 & \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right) + \pm{1} \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) & - \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right) + \pm{1} \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\gamma\right)\\- \pm{1} & 0 & 0\end{smallmatrix}\right)\end{split}$

When $$\sin(\beta) = 1$$:

$\begin{split}R = \left(\begin{smallmatrix}0 & \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right) - \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) & \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right) + \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\gamma\right)\\0 & \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right) + \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) & \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) - \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right)\\-1 & 0 & 0\end{smallmatrix}\right)\end{split}$

From the angle sum and difference identities (see also geometric proof, Mathworld treatment) we remind ourselves that, for any two angles $$\alpha$$ and $$\beta$$:

\begin{align}\begin{aligned}\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,\\\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta\end{aligned}\end{align}

We can rewrite $$R$$ as:

$\begin{split}R = \left(\begin{smallmatrix}0 & V_{1} & V_{2}\\0 & V_{2} & - V_{1}\\-1 & 0 & 0\end{smallmatrix}\right)\end{split}$

where:

\begin{align}\begin{aligned}V_1 = \operatorname{cos}\left(\gamma\right) \operatorname{sin}\left(\alpha\right) - \operatorname{cos}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) = \sin(\alpha - \gamma) \,\\V_2 = \operatorname{cos}\left(\alpha\right) \operatorname{cos}\left(\gamma\right) + \operatorname{sin}\left(\alpha\right) \operatorname{sin}\left(\gamma\right) = \cos(\alpha - \gamma)\end{aligned}\end{align}

We immediately see that $$\alpha$$ and $$\gamma$$ are going to lead the same transformation - the mathematical expression of the observation on the spitfire above, that rotation around the $$x$$ axis is equivalent to rotation about the $$z$$ axis.

It’s easy to do the same set of reductions, with the same conclusion, for the case where $$\sin(\beta) = -1$$ - see http://www.gregslabaugh.name/publications/euler.pdf.