Gimbal lock

See also: http://en.wikipedia.org/wiki/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.

Example

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:

We want to rotate the aircraft to look something like this:

We might start by doing an x rotation, making a slight roll with the left wing tilting down (rotation about $x$) like this:

Let’s say that the x rotation is -0.2 radians.

Then we do a pitch so we are pointing straight up (rotation around $y$ axis). This is a rotation by $-\pi/2$ radians.

To get to our desired position from here, we need to do a turn of something like 0.2 radians of the nose towards the viewer (and the tail away from the viewer). All we have left is our z rotation (rotation around the $z$ axis. Unfortunately, the current result of a rotation around the $z$ axis has now become the same as a previous rotation around the $x$ axis. To see this, 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.

This does not mean that we cannot do the rotations we need, only that we can’t do them by starting with the most obvious x and y rotations. In fact what we will have to do is first rotate around x by $\pi/2 - 0.2$ radians, then do a y rotation of $-\pi/2 + 0.2$ radians, and finally a z rotation of $-\pi/2$ radians. See the code below for the details.

Mathematics of gimbal lock

See transforms3d.derivations.eulerangles.

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:

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

In our case the y rotation $\beta = -\pi / 2, \cos(\beta) = 0, \sin(\beta) = -1$:

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

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:

$$\begin{split}R = \left[\begin{matrix}0 & - W_{1} & - W_{2}\\0 & W_{2} & - W_{1}\\1 & 0 & 0\end{matrix}\right]\end{split}$$

where:

$$\begin{align}\begin{aligned}W_1 = \sin{\left (\alpha \right )} \cos{\left (\gamma \right )} + \sin{\left (\gamma \right )} \cos{\left (\alpha \right )} = \sin(\alpha + \gamma) \,\\W_2 = - \sin{\left (\alpha \right )} \sin{\left (\gamma \right )} + \cos{\left (\alpha \right )} \cos{\left (\gamma \right )} = \cos(\alpha + \gamma)\end{aligned}\end{align}$$

We immediately see that $\alpha$ and $\gamma$ are going to lead to the same transformations - 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 for the case where $\sin(\beta) = 1$; see http://www.gregslabaugh.name/publications/euler.pdf.

The example in code

Here is what our gimbal lock looks like in code:

>>> import numpy as np
>>> np.set_printoptions(precision=3, suppress=True)  # neat printing
>>> from transforms3d.euler import euler2mat, mat2euler
>>> x_angle = -0.2
>>> y_angle = -np.pi / 2
>>> z_angle = -0.2
>>> R = euler2mat(x_angle, y_angle, z_angle, 'sxyz')
>>> R
array([[ 0.   ,  0.389, -0.921],
       [-0.   ,  0.921,  0.389],
       [ 1.   , -0.   ,  0.   ]])

This isn’t the transformation we actually want because of the gimbal lock. The gimbal lock means that x_angle and z_angle result in rotations about the same axis of the object. So, we can add something to the x_angle and subtract the same value from z_angle to get the same result:

>>> R = euler2mat(x_angle + 0.1, y_angle, z_angle - 0.1, 'sxyz')
>>> R
array([[ 0.   ,  0.389, -0.921],
       [-0.   ,  0.921,  0.389],
       [ 1.   , -0.   ,  0.   ]])

In fact, we could omit the z rotation entirely and put all the rotation into the original x axis rotation and still get the same rotation matrix:

>>> R_dash = euler2mat(x_angle + z_angle, y_angle, 0, 'sxyz')
>>> np.allclose(R, R_dash)
True

So, there is no future in doing our transformations starting with this x and y rotation, if we are rotating with this axis order. We can get the transformation we actually want by doing the rotations in the order x, then z then y, like this:

>>> R = euler2mat(x_angle, z_angle, y_angle, 'sxzy')
>>> R
array([[ 0.   ,  0.199, -0.98 ],
       [-0.199,  0.961,  0.195],
       [ 0.98 ,  0.195,  0.039]])

We can get this same transformation using our original x, y, z rotation order, but using different rotation angles:

>>> x_dash, y_dash, z_dash = mat2euler(R, 'sxyz')
>>> np.array((x_dash, y_dash, z_dash))  # np.array for print neatness
array([ 1.371, -1.371, -1.571])
>>> R = euler2mat(x_dash, y_dash, z_dash, 'sxyz')
>>> R
array([[ 0.   ,  0.199, -0.98 ],
       [-0.199,  0.961,  0.195],
       [ 0.98 ,  0.195,  0.039]])