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 $\pm π / 2$ 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 $\cos (β) = 0$ , where $β$ 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 $α$ radians around $x$ , followed by a rotation $β$ around $y$ , followed by rotation $γ$ around $z$ , the rotation matrix $R$ is:

\begin{array}{r} R = (\begin{array}{c} \cos (β) \cos (γ) & - \cos (α) \sin (γ) + \cos (γ) \sin (α) \sin (β) & \sin (α) \sin (γ) + \cos (α) \cos (γ) \sin (β) \\ \cos (β) \sin (γ) & \cos (α) \cos (γ) + \sin (α) \sin (β) \sin (γ) & - \cos (γ) \sin (α) + \cos (α) \sin (β) \sin (γ) \\ - \sin (β) & \cos (β) \sin (α) & \cos (α) \cos (β) \end{array}) \end{array}

When $\cos (β) = 0$ , $\sin (β) = \pm 1$ and $R$ simplifies to:

\begin{array}{r} R = (\begin{array}{c} 0 & - \cos (α) \sin (γ) + \pm 1 \cos (γ) \sin (α) & \sin (α) \sin (γ) + \pm 1 \cos (α) \cos (γ) \\ 0 & \cos (α) \cos (γ) + \pm 1 \sin (α) \sin (γ) & - \cos (γ) \sin (α) + \pm 1 \cos (α) \sin (γ) \\ - \pm 1 & 0 & 0 \end{array}) \end{array}

When $\sin (β) = 1$ :

\begin{array}{r} R = (\begin{array}{c} 0 & \cos (γ) \sin (α) - \cos (α) \sin (γ) & \cos (α) \cos (γ) + \sin (α) \sin (γ) \\ 0 & \cos (α) \cos (γ) + \sin (α) \sin (γ) & \cos (α) \sin (γ) - \cos (γ) \sin (α) \\ - 1 & 0 & 0 \end{array}) \end{array}

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

\begin{array}{r} \begin{matrix} \sin (α \pm β) = \sin α \cos β \pm \cos α \sin β \\ \cos (α \pm β) = \cos α \cos β \mp \sin α \sin β \end{matrix} \end{array}

We can rewrite $R$ as:

\begin{array}{r} R = (\begin{array}{c} 0 & V_{1} & V_{2} \\ 0 & V_{2} & - V_{1} \\ - 1 & 0 & 0 \end{array}) \end{array}

where:

\begin{array}{r} \begin{matrix} V_{1} = \cos (γ) \sin (α) - \cos (α) \sin (γ) = \sin (α - γ) \\ V_{2} = \cos (α) \cos (γ) + \sin (α) \sin (γ) = \cos (α - γ) \end{matrix} \end{array}

We immediately see that $α$ and $γ$ 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 $\sin (β) = - 1$ - see http://www.gregslabaugh.name/publications/euler.pdf.