James Gregson's Website

Nov 02, 2018

Rotation Matrix Euler Angles

This page describes a basic approach to extracting Euler angles from rotation matrices. This follows what OpenCV does, (which is where I got the process) but I also wanted to understand why this approach was used.

Here I'm considering rotation performed first around x-axis (\(\alpha\)), then y-axis (\(\beta\)) and finally z-axis (\(\gamma\)). For this, the rotation matrix is shown below. Other rotation orders are analogous.

\begin{equation*} R(\alpha,\beta,\gamma) = \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 )} & 0\\\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 )} & 0\\- \sin{\left (\beta \right )} & \sin{\left (\alpha \right )} \cos{\left (\beta \right )} & \cos{\left (\alpha \right )} \cos{\left (\beta \right )} & 0\\0 & 0 & 0 & 1\end{matrix}\right] \end{equation*}

To solve for the Euler angles, we can use a little bit of trigonometry. The best way, in my opinion, is to find pairs of entries that are each products of two factors. Using the identity \(\sin^2 + \cos^2 = 1\) can be used to isolate specific angles, e.g. \(R_{0,0}\) and \(R_{0,1}\) can be used to find \(\cos(\beta)\) which unlocks the rest of the angles:

\begin{align*} (-1)^k \cos(\beta + k \pi) = \sqrt{ \cos^2(\beta) ( \sin^2(\gamma) + \cos^2(\gamma)) } = \sqrt{ R_{0,0}^2 + R_{1,0}^2 } \\ \cos(\beta) \approx \sqrt{ R_{0,0}^2 + R_{1,0}^2 } \end{align*}

Pay attention to the first line: For any true value of \(\beta\), this formula will return the same value for \(\beta + k\pi\) where \(k\) is an arbitrary integer. This will in turn determine the angles \(\alpha\) and \(\gamma\) in order to be compatible. The approximate equality will return \(\beta \in [-\pi/2, \pi/2]\).

Once \(\cos(\beta)\) is found we can find the angles directly using atan2:

\begin{align*} \beta = \mbox{atan2}( -R_{2,0}, \cos(\beta) ) \\ \alpha = \mbox{atan2}\left( \frac{R_{2,1}}{\cos(\beta)}, \frac{R_{2,2}}{\cos(\beta)} \right) \\ \gamma = \mbox{atan2}\left( \frac{R_{1,0}}{\cos(\beta)}, \frac{R_{0,0}}{\cos(\beta)} \right) \end{align*}

The other issue occurs when \(|\cos(\beta)| = 0\), which causes division by zero in the equations for \(\alpha,\gamma\) (\(\beta\) is still well defined). In this case there are a number of options. What OpenCV does is to arbitrarily decide that \(\gamma = 0\), which means that \(\sin(\gamma) = 0\) and \(\cos(\gamma) = 1\). The formulas in this case are:

\begin{align*} \beta = \mbox{atan2}( -R_{2,0}, \cos(\beta) ) \\ \alpha = \mbox{atan2}( -R_{1,2}, R_{1,1} ) \\ \gamma = 0 \end{align*}

There are, of course, other options. Python code implementing these operations is below, along with unit testing for the degenerate and non-degerate cases.

import unittest
import numpy

def x_rotation( theta ):
    """Generate a 4x4 homogeneous rotation matrix about x-axis"""
    c = numpy.cos(theta)
    s = numpy.sin(theta)
    return numpy.array([
        [ 1, 0, 0, 0],
        [ 0, c,-s, 0],
        [ 0, s, c, 0],
        [ 0, 0, 0, 1]
    ])

def y_rotation( theta ):
    """Generate a 4x4 homogeneous rotation matrix about y-axis"""
    c = numpy.cos(theta)
    s = numpy.sin(theta)
    return numpy.array([
        [ c, 0, s, 0],
        [ 0, 1, 0, 0],
        [-s, 0, c, 0],
        [ 0, 0, 0, 1]
    ])

def z_rotation( theta ):
    """Generate a 4x4 homogeneous rotation matrix about z-axis"""
    c = numpy.cos(theta)
    s = numpy.sin(theta)
    return numpy.array([
        [ c,-s, 0, 0],
        [ s, c, 0, 0],
        [ 0, 0, 1, 0],
        [ 0, 0, 0, 1]
    ])

def xyz_rotation( angles ):
    """Generate 4x4 homogeneous rotation in x then y then z order"""
    return numpy.dot( z_rotation(angles[2]), numpy.dot( y_rotation(angles[1]), x_rotation(angles[0]) ) )

def xyz_rotation_angles( R, eps=1e-8, check=False ):
    """Back out the Euler angles that would lead to the given matrix"""

    if check and numpy.linalg.norm( numpy.dot( R.transpose(), R ) - numpy.eye(4) ) > eps:
        raise ValueError('Input matrix is not a pure rotation, R\'R != I')

    cos_beta = numpy.sqrt( R[0,0]**2 + R[1,0]**2 )
    if numpy.abs(cos_beta) > eps:
        alpha = numpy.arctan2( R[2,1]/cos_beta, R[2,2]/cos_beta )
        beta  = numpy.arctan2(-R[2,0], cos_beta )
        gamma = numpy.arctan2( R[1,0]/cos_beta, R[0,0]/cos_beta )
        return numpy.array((alpha,beta,gamma))
    else:
        alpha = numpy.arctan2(-R[1,2], R[1,1] )
        beta  = numpy.arctan2(-R[2,0], cos_beta )
        gamma = 0
        return numpy.array((alpha,beta,gamma))


class test_euler_angles(unittest.TestCase):
    def test_angles(self):
        """Do fuzz testing on arbitrary rotations, limiting beta to valid range"""
        for i in range(1000):
            alpha = (numpy.random.rand()-0.5)*numpy.pi*2
            beta  = (numpy.random.rand()-0.5)*numpy.pi*0.9999
            gamma = (numpy.random.rand()-0.5)*numpy.pi*2

            ang = xyz_rotation_angles( xyz_rotation([alpha,beta,gamma]))
            self.assertAlmostEqual( alpha, ang[0] )
            self.assertAlmostEqual( beta,  ang[1] )
            self.assertAlmostEqual( gamma, ang[2] )

    def test_degeneracies(self):
        """Do fuzz testing on the degenerate condition of beta = +- pi/2"""
        for i in range(1000):
            alpha = (numpy.random.rand()-0.5)*numpy.pi*2
            beta  = numpy.sign(numpy.random.randn())*numpy.pi/2
            gamma = (numpy.random.rand()-0.5)*numpy.pi*2
            R     = xyz_rotation((alpha,beta,gamma))
            ang   = xyz_rotation_angles( R )
            R2    = xyz_rotation( ang )
            # check that beta is recovered, gamma is set to zero and
            # the rotation matrices match to high precision
            self.assertAlmostEqual( beta, ang[1] )
            self.assertAlmostEqual( ang[2], 0 )
            for j in range(16):
                self.assertAlmostEqual( R.ravel()[j], R2.ravel()[j] )

if __name__ == '__main__':
    unittest.main()

Nov 02, 2018

Transformation Matrix Jacobians

Starting with the \(4\times4\) homogeneous transform matrix with parameters \(\beta = [ \theta_x, \theta_y, \theta_z, t_x, t_y, t_z ]^T\) where rotations are performed in XYZ order and using the following substitutions:

\begin{align*} c_x = \cos(\theta_x) \\ s_x = \sin(\theta_x) \\ c_y = \cos(\theta_y) \\ s_y = \sin(\theta_y) \\ c_z = \cos(\theta_z) \\ s_z = \sin(\theta_z) \\ \end{align*}

the vector function mapping a point \(p = [p_x, p_y, p_z, 1]^T\) in the body coordinate system to a point in the world coordinate system \(w = [w_x, w_y, w_z, 1]^T\) is:

\begin{equation*} \begin{bmatrix} w_x \\ w_y \\ w_z \\ 1 \end{bmatrix} = F( p, \beta ) = \left[\begin{matrix}c_{y} c_{z} & - c_{x} s_{z} + c_{z} s_{x} s_{y} & c_{x} c_{z} s_{y} + s_{x} s_{z} & t_{x}\\c_{y} s_{z} & c_{x} c_{z} + s_{x} s_{y} s_{z} & c_{x} s_{y} s_{z} - c_{z} s_{x} & t_{y}\\- s_{y} & c_{y} s_{x} & c_{x} c_{y} & t_{z}\\0 & 0 & 0 & 1\end{matrix}\right]\begin{bmatrix} p_x \\ p_y \\ p_z \\ 1 \end{bmatrix} \end{equation*}

and the Jacobian with respect to the parameters \(\beta\) is:

\begin{equation*} J_F = \left[\begin{matrix}p_{y} \left(c_{x} c_{z} s_{y} + s_{x} s_{z}\right) + p_{z} \left(c_{x} s_{z} - c_{z} s_{x} s_{y}\right) & c_{x} c_{y} c_{z} p_{z} + c_{y} c_{z} p_{y} s_{x} - c_{z}p_{x} s_{y} & - c_{y} p_{x} s_{z} + p_{y} \left(- c_{x} c_{z} - s_{x} s_{y} s_{z}\right) + p_{z} \left(- c_{x} s_{y} s_{z} + c_{z} s_{x}\right) & 1 & 0 & 0\\p_{y} \left(c_{x} s_{y} s_{z} - c_{z} s_{x}\right) + p_{z} \left(- c_{x} c_{z} - s_{x} s_{y} s_{z}\right) & c_{x} c_{y} p_{z} s_{z} + c_{y} p_{y} s_{x} s_{z} - p_{x} s_{y} s_{z} & c_{y} c_{z} p_{x} + p_{y} \left(- c_{x} s_{z} + c_{z} s_{x} s_{y}\right) + p_{z} \left(c_{x} c_{z} s_{y} + s_{x} s_{z}\right) & 0 & 1 & 0\\c_{x} c_{y} p_{y} - c_{y} p_{z} s_{x} & - c_{x} p_{z} s_{y} - c_{y} p_{x} -p_{y} s_{x} s_{y} & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right] \end{equation*}

Python code for these respective operations is below:

def make_transform( beta ):
    c_x = numpy.cos(beta[0])
    s_x = numpy.sin(beta[0])
    c_y = numpy.cos(beta[1])
    s_y = numpy.sin(beta[1])
    c_z = numpy.cos(beta[2])
    s_z = numpy.sin(beta[2])
    t_x = beta[3]
    t_y = beta[4]
    t_z = beta[5]
    return numpy.array([
        [c_y*c_z, -c_x*s_z + c_z*s_x*s_y, c_x*c_z*s_y + s_x*s_z, t_x],
        [c_y*s_z, c_x*c_z + s_x*s_y*s_z, c_x*s_y*s_z - c_z*s_x, t_y],
        [-s_y, c_y*s_x, c_x*c_y, t_z],
        [0, 0, 0, 1]
    ])

def make_transform_jacobian( beta, p ):
    c_x = numpy.cos(beta[0])
    s_x = numpy.sin(beta[0])
    c_y = numpy.cos(beta[1])
    s_y = numpy.sin(beta[1])
    c_z = numpy.cos(beta[2])
    s_z = numpy.sin(beta[2])
    t_x = beta[3]
    t_y = beta[4]
    t_z = beta[5]
    p_x = p[0]
    p_y = p[1]
    p_z = p[2]
    return numpy.array([
        [p_y*(c_x*c_z*s_y + s_x*s_z) + p_z*(c_x*s_z - c_z*s_x*s_y), c_x*c_y*c_z*p_z + c_y*c_z*p_y*s_x - c_z*p_x*s_y, -c_y*p_x*s_z + p_y*(-c_x*c_z - s_x*s_y*s_z) + p_z*(-c_x*s_y*s_z + c_z*s_x), 1, 0, 0],
        [p_y*(c_x*s_y*s_z - c_z*s_x) + p_z*(-c_x*c_z - s_x*s_y*s_z), c_x*c_y*p_z*s_z + c_y*p_y*s_x*s_z - p_x*s_y*s_z, c_y*c_z*p_x + p_y*(-c_x*s_z + c_z*s_x*s_y) + p_z*(c_x*c_z*s_y + s_x*s_z), 0, 1, 0],
        [c_x*c_y*p_y - c_y*p_z*s_x, -c_x*p_z*s_y - c_y*p_x - p_y*s_x*s_y, 0, 0, 0, 1],
        [0, 0, 0, 0, 0, 0]
    ])

I generated these using sympy to build the transformations and used a finite-difference Jacobian function to check the output.