In the process of working on a quaternion quantization post I needed to round-trip between the semi-standard yaw/pitch/roll representation and quaternions. In the process I discovered just how inaccurate my existing conversion code and all those that I could find on the web are. This is note about the duct-tape & super-glue I applied to lower the round-trip errors to an acceptable range for my purposes. I’m not going to carry through beyond that point. I set myself a target peak error of ~0.00015 degrees. I’ll focus on converting from quaternions but will make some nods to matrices.

Aside: there are a number of methods to map a quaternion to three values which are computationally cheaper and introduce only small errors to round-trip.

Test framework can be found: HERE

## Prelim

Informally Euler angles refer to a parameterization of rotations which is a set of three angles about three predefined axes. It’s usually noted that this gives us 12 possible choices $\left(3\times2\times2\right)$, three for the first axis and two choices for the second and third (since we can’t repeat the immediately previous choice). And then that number is doubled for left vs. right handed systems. And we could double that again for fixed frame (global rotation view/space-fixed) or inertial frame (local rotation view/body-fixed). I’d argue that there are really only two distinct versions: All three axes choices are present (Tait-Bryan) and the first and last are repeats (proper Euler angles). Choice of axis/handedness is just sign flips/permutations and local vs. global is the choice of reading the componsition left-to-right vs. right-to-left.

## Tait-Bryan: ZYX

Seemly the most common choice is that of Tait-Bryan ZYX. This wikipedia picture illustrates the parameterization (except negate the y-axis and reverse the direction of $\phi$…use your imagination):

First I’ll walk through the basics from a quaternion perspective. Define a standard basis set:

$\begin{eqnarray*} \mathbf{x} & = & \left(1,0,0\right) \\ \mathbf{y} & = & \left(0,1,0\right) \\ \mathbf{z} & = & \left(0,0,1\right) \end{eqnarray*}$

We take $\mathbf{z}$ to be up, the default “facing” is $\mathbf{x}$ and (of course) right-handed, then the specific parameterization is:

1. rotate about local $\mathbf{z}$ by $\psi$ (yaw)

2. rotate about local $\mathbf{y}$ by $\theta$ (pitch)

3. rotate about local $\mathbf{x}$ by $\phi$ (roll)

Then the quaternion expression of interest is:

$$$\label{eq:tait} Q = w + \left(x,~y,~z\right) = e^{\frac{\psi}{2}\mathbf{z}}e^{\frac{\theta}{2}\mathbf{y}}e^{\frac{\phi}{2}\mathbf{x}}$$$

and recall that:

$e^{\alpha\mathbf{u}} = \cos\left(\alpha\right) + \sin\left(\alpha\right)\mathbf{u}$

## ZYX to quaternion and matrix

To formulate the Euler to quaternion conversion we simply expand the right most expression of $\eqref{eq:tait}$ and set to the middle expression:

$\begin{eqnarray} w &=& \cos\left(\frac{\phi}{2}\right) \cos\left(\frac{\theta}{2}\right) \cos\left(\frac{\psi}{2}\right) + \sin\left(\frac{\phi}{2}\right) \sin\left(\frac{\theta}{2}\right) \sin\left(\frac{\psi}{2}\right) \label{qw} \\ x &=& \sin\left(\frac{\phi}{2}\right) \cos\left(\frac{\theta}{2}\right) \cos\left(\frac{\psi}{2}\right) - \cos\left(\frac{\phi}{2}\right) \sin\left(\frac{\theta}{2}\right) \sin\left(\frac{\psi}{2}\right) \label{qx} \\ y &=& \cos\left(\frac{\phi}{2}\right) \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\psi}{2}\right) + \sin\left(\frac{\phi}{2}\right) \cos\left(\frac{\theta}{2}\right) \sin\left(\frac{\psi}{2}\right) \label{qy} \\ z &=& \cos\left(\frac{\phi}{2}\right) \cos\left(\frac{\theta}{2}\right) \sin\left(\frac{\psi}{2}\right)-\sin\left(\frac{\phi}{2}\right) \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\psi}{2}\right) \label{qz} \end{eqnarray}$

(really the matrix conversion is hidden in the next section.)

## And back again… as if we can compute with Reals

Given $Q$ we can use the similarity transform to find $\left(\mathbf{x}’,~\mathbf{y}’,~\mathbf{z}’\right)$:

$\begin{eqnarray} \mathbf{x}' &=& Q\mathbf{x}Q^* \nonumber \\ &=& \left(w^2+x^2-y^2-z^2, ~2xy+2wz, ~2xz-2wy\right) \label{xpu} \\ &=& \left(1-2\left(y^2+z^2\right), ~2(xy+wz), ~2(xz-wy)\right) \label{xp} \\ &=& \left(\cos(\theta) \cos(\psi), ~\cos(\theta) \sin(\psi),~ -\sin(\theta)\right) \label{xpt} \\ &=& \left(m_{00}, ~m_{01}, ~m_{02} \right) \nonumber \\ \nonumber \\ \mathbf{y}' &=& Q\mathbf{y}Q^* \nonumber \\ &=& \left(2xy-2wz, ~w^2-x^2+y^2-z^2, ~2wx+2yz\right) \label{ypu} \\ &=& \left(2(xy-wz), ~1-2\left(x^2+z^2\right), ~2(wx+yz)\right) \label{yp} \\ &=& \left(\sin(\theta)\cos(\psi)\sin(\phi)-\sin(\psi)\cos(\phi),~\sin(\theta)\sin(\psi)\sin(\phi)+\cos(\psi)\cos(\phi),~\cos(\theta)\sin(\phi)\right) \label{ypt} \\ &=& \left(m_{10}, ~m_{11}, ~m_{12} \right) \nonumber \\ \nonumber \\ \mathbf{z}' &=& Q\mathbf{z}Q^* \nonumber \\ &=& \left(2wy+2xz, ~2yz-2wx, ~w^2-x^2-y^2+z^2\right) \label{zpu} \\ &=& \left(2(wy+xz), ~2(yz-wx), ~1-2\left(x^2+y^2\right)\right) \label{zp} \\ &=& \left(\sin(\theta)\cos(\psi)\cos(\phi)+\sin(\psi)\sin(\phi),~\sin(\theta)\sin(\psi)\cos(\phi)-\cos(\psi)\sin(\phi),~\cos(\theta)\cos(\phi)\right) \label{zpt} \\ &=& \left(m_{20}, ~m_{21}, ~m_{22} \right) \nonumber \\ \end{eqnarray}$

Where the $m_{ij}$ are to show the equivalent matrix $R$.

$R = \left( \begin{matrix} \mathbf{x}' \\ \mathbf{y}' \\ \mathbf{z}' \\ \end{matrix} \right) = \left( \begin{matrix} m_{00} & m_{01} & m_{02} \\ m_{10} & m_{11} & m_{12} \\ m_{20} & m_{21} & m_{22} \\ \end{matrix} \right)$

Ignoring the degenerate region we can easily find $\theta$ and $\psi$ from $\mathbf{x}’$:

$\begin{eqnarray} \theta &=& \text{asin} \left(-m_{02}\right) \label{thetam} \\ &=& \text{asin} \left(-2\left(xz-wy\right)\right) \label{thetaq} \\ \nonumber \\ \psi &=& \text{atan2}\left(m_{01},~m_{00} \right) \label{psim} \\ &=& \text{atan2}\left(xy+wz,~\frac{1}{2}-\left(y^2+z^2\right)\right) \label{psiq2} \end{eqnarray}$

and $\phi$ from the z components of $\mathbf{y}’$ and $\mathbf{z}’$

$\begin{eqnarray} \phi &=& \text{atan2}\left(m_{12},~m_{22}\right) \\ &=& \text{atan2}\left(yz+wx,~\frac{1}{2}-\left(x^2+y^2\right)\right) \label{phiq1} \end{eqnarray}$

The above computations explode into an infinite number of solutions when $\cos\left(\theta\right)=0$ ($\theta=\pm\frac{\pi}{2}$). The matrix form then becomes:

$\begin{eqnarray*} \theta &=& \frac{\pi}{2} &\to& \left( \begin{matrix} 0 & 0 & -1 \\ \sin\left( \phi - \psi \right) & \cos\left( \phi - \psi \right) & 0 \\ \cos\left( \phi - \psi \right) & -\sin\left( \phi - \psi \right) & 0 \\ \end{matrix} \right) \\ \theta &=& -\frac{\pi}{2} &\to& \left( \begin{matrix} 0 & 0 & 1 \\ -\sin\left(\phi + \psi \right) & \cos\left( \phi + \psi \right) & 0 \\ -\cos\left(\phi + \psi \right) & -\sin\left( \phi + \psi \right) & 0 \\ \end{matrix} \right) \end{eqnarray*}$

So for matrices near degenerate $\phi$ and $\psi$ can be expressed as:

$$$\label{eq:msum} \phi-\text{sgn}\left(m_{02}\right)\psi = \text{atan2}\left(m_{10},m_{20}\right)$$$

and the equivalent quaternions respectively are:

$\frac{1}{\sqrt{2}} \left( \cos\left(\frac{\phi-\psi}{2}\right) + \left( \sin\left(\frac{\phi-\psi}{2}\right),~ \cos\left(\frac{\phi-\psi}{2}\right),~ -\sin\left(\frac{\phi-\psi}{2}\right)\right) \right) \\ \frac{1}{\sqrt{2}} \left( \cos\left(\frac{\phi+\psi}{2}\right) + \left( \sin\left(\frac{\phi+\psi}{2}\right),~ -\cos\left(\frac{\phi+\psi}{2}\right),~ \sin\left(\frac{\phi+\psi}{2}\right)\right) \right)$

and $\phi$ and $\psi$ can be expressed as:

$$$\label{qsum} \phi+\text{sgn}\left(xz-wy\right)\psi = 2~\text{atan2}\left(x,w\right)$$$

So my original implementation looked like the following and other versions I could found were pretty much equivalent. Compute yaw by $\eqref{psiq2}$ and pitch by $\eqref{thetaq}$. Choose a degenerate zone threshold and use either $\eqref{phiq1}$ or $\eqref{qsum}$ for roll:

## The duct-tape and super-glue

Plugging the above into my quantization framework I noted something was rotten in Denmark in the form of unexpectly high errors. Ripping out the quantization confirmed (peak error hitting ~5 degrees). Changing the error plotting to $\abs{z}$ of $x’$ pointed the finger at the approaching denegerate case (suprise, suprise). Sadly the error plot also showed that it was above my target peak error on over 15% of the range so the problem was beyond some minor tweak.

Fine. First the denegerate zone test is based on an expression which is approaching one which can be changed to testing an expression approaching zero. We also have some rounding problems such as subtract a small value from one.

For matrices we could compute $\theta$ from:

$\cos\left(\theta\right)^2 = m_{00}^2+m_{01}^2 = m_{12}^2+m_{22}^2$

and for quaternions the middle term expands to $\eqref{cos2_1}$, the right term to $\eqref{cos2_2}$ and they both reduce to $\eqref{cos2_3}$:

$\begin{eqnarray} \cos\left(\theta\right)^2 &=& \left(\left(w^2-z^2\right)+\left(x^2-y^2\right)\right)^2 + ~\left(2xy+2wz\right)^2 \label{cos2_1} \\ &=& ~(2wx+2yz)^2 + \left(\left(w^2-z^2\right)-\left(x^2-y^2\right)\right)^2 \label{cos2_2} \\ &=& ((w+y)^2+(x-z)^2)((w-y)^2+(x+z)^2) \label{cos2_3} \end{eqnarray}$

The infinite solution case is when $\cos\left(\theta\right)$ is zero and approaching zero can be used to detect the degenerate zone. To state the obvious we can compute $\theta$ using the single parameter arc-tangent function:

$\theta = \text{atan}\left(\frac{\sin\left(\theta\right)}{\sqrt{\cos\left(\theta\right)^2}}\right)$

I’m targeting low hanging fruit, using $\eqref{cos2_1}$, carrying through some scale factors, pulling out some common expression and promoting to double computation yields:

Bingo! Well beyond my target. Okay here are the error plots of some examples in the test framework:

• The original version: orig
• Simply promoting the original to double is useless: orig_d. Slightly better where is doesn’t matter and the same error on the problem range.
• Change to doubles and new method: rev_1. Not great accuracy but way beyond my target. Converting back to singles still has exploding error when $\abs{z}$ is approaching one. (not shown)
• Changing to singles for the majority of the range and using doubles for the remainder: rev_2 (5k) is tight to my max peak error and rev_2 (250k) is roughly where the error start to otherwise explode (probability of ~.97 that the computation is in the common case of using singles).

(EDIT: 20210529) The original version of this post was mean and didn’t have a work version in post (required you dig through the toy code and figure it out). Here is the fma based solution including required helper functions:

Extra credit: the reverse transform is also a source of error. My version is a direct translation of the math into code and isn’t even bothering to pull out common sub-expressions that non-fast math optims cannot remove:

if this is modified to promote to double computation then the error drops a reasonable amount. Example Using rev_1 as the forward xform and double ZYX to quaternion gives us the error plot rev1_id.

## Euler angles: ZXZ

Let’s quickly walk through a proper Euler angle parameterization: ZXZ. The quaternion rotation expression:

$\begin{eqnarray} Q &=& e^{\frac{\gamma}{2}\mathbf{z}}e^{\frac{\beta}{2}\mathbf{x}}e^{\frac{\alpha}{2}\mathbf{z}} \nonumber \\ &=& \cos\left(\frac{\gamma+\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) + \left( \cos\left(\frac{\gamma-\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right), ~\sin\left(\frac{\gamma-\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right), ~\sin\left(\frac{\gamma+\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) \right) \label{zxz} \end{eqnarray}$

and using the similarity transform three times (tap, tap, tap) and converting into a matrix:

$\left( \begin{array}{ccc} \cos (\alpha ) \cos (\gamma )-\sin (\alpha ) \cos (\beta ) \sin (\gamma ) & \sin (\alpha ) \cos (\beta ) \cos (\gamma )+\cos (\alpha ) \sin (\gamma ) & \sin (\alpha ) \sin (\beta ) \\ -\sin (\alpha ) \cos (\gamma )-\cos (\alpha ) \cos (\beta ) \sin (\gamma ) & \cos (\alpha ) \cos (\beta ) \cos (\gamma )-\sin (\alpha ) \sin (\gamma ) & \cos (\alpha ) \sin (\beta ) \\ \sin (\beta ) \sin (\gamma ) & -\sin (\beta ) \cos (\gamma ) & \cos (\beta ) \\ \end{array} \right)$

For quaternions we don’t need the matrix, directly from $\eqref{zxz}$:

$\begin{eqnarray*} s &=& \frac{1}{2}\left(\gamma+\alpha\right) = \text{atan}\left(\frac{z}{w}\right) \\ d &=& \frac{1}{2}\left(\gamma-\alpha\right) = \text{atan}\left(\frac{y}{x}\right) \\ \\ \gamma &=& s+d \\ \alpha &=& s-d \\ \\ \beta &=& 2~\text{atan}\left(\sqrt{\frac{x^2+y^2}{w^2+z^2}}\right) \end{eqnarray*}$

Shoving these into a promote to double implementation gives:

and a quick plot of above with single & double ZXZ to quaternion xform: