Could someone explain why this formula for quantifying distance between Euler angles is giving me an answer higher than expected?

Question

From this paper I'm trying to use a method for comparing the distance between two rotation matrices. I'm using Φ6(R1,R2) = ||$log(R_1R_2^T )$|| (left side, page 159). When I tried this in python the answer this is providing is off by *sqrt(2). Is that because I'm misinterpretting the notation, or doing something wrong in python?

    Python:
    r1 = R.from_euler('xyz', [0,0,0],degrees = True).as_matrix()
    r2 = R.from_euler('xyz', [0,0,90],degrees = True).as_matrix()

    print("r1")
    print(r1)
    print("r2")
    print(r2)

    angle_dif_mat = logm(np.matmul(r1,r2.transpose()))
    print("Angle dif")
    print(angle_dif_mat)

    print("norm")
    print(np.linalg.norm(angle_dif_mat))
    print("this = expected*sqrt(2)")
    print("expected = pi/2")

Output:

r1
[[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
r2
[[ 2.22044605e-16 -1.00000000e+00  0.00000000e+00]
 [ 1.00000000e+00  2.22044605e-16  0.00000000e+00]
 [ 0.00000000e+00  0.00000000e+00  1.00000000e+00]]
Angle dif
[[-2.22044605e-16  1.57079633e+00  0.00000000e+00]
 [-1.57079633e+00 -2.22044605e-16  0.00000000e+00]
 [ 0.00000000e+00  0.00000000e+00  0.00000000e+00]]
norm
2.2214414690791826
this = expected*sqrt(2)
expected = pi/2

Answer 1

I suggest manually coding each of the steps. You should indeed get $\frac{\pi}{2}$. However, perhaps, you are using the Frobenius norm instead of the L2 norm e.g., $|| \log{\mathbf{R}}||_\text{F} = \sqrt{2} |\theta|$.

Essentially, you are considering the distance between rotations (in the same basis) say $\mathbf{P}$ and $\mathbf{Q}$. This means that you can think of the rotations as $\mathbf{P} = {}^B\mathbf{R}_A$ and $\mathbf{Q} = {}^C\mathbf{R}_A$ where ${}^j\mathbf{R}_i$ denotes the rotation from basis $i$ to basis $j$. Keep in mind that $\mathbf{R}^{-1} = \mathbf{R}^T$, so ${}^j\mathbf{R}_i = {}^i\mathbf{R}_j^{-1} = {}^i\mathbf{R}_j^{T}$.

Thus, when you compute $\mathbf{P} \mathbf{Q}^T$, you are computing ${}^B\mathbf{R}_C = {}^B\mathbf{R}_A {}^A\mathbf{R}_C = {}^B\mathbf{R}_A {}^C\mathbf{R}_A^{T}$, which is the rotation from basis $\mathbf{C}$ to basis $\mathbf{B}$.

Now, the formula you provide is the norm of the logarithm of a rotation matrix. This is based on mapping between $\text{SO}(3)$ and $\text{so}(3)$. This is also equivalent to the angle of rotation if the rotation is in axis-angle representation (which is a common distance metric for rotations). You can more simply just compute the angle of rotation directly from $\mathbf{P} \mathbf{Q}^T$ by

$$ \theta = \arccos{\left(\frac{\text{Tr}(\mathbf{P} \mathbf{Q}^T) - 1 }{2} \right)}.$$

If you would manually compute the logarithm, computing $\theta$ is a substep, so you don't need the logarithm for your purpose i.e., the logarithm (which requires $\theta$) is given by

$$ \log{\mathbf{R}} = \frac{\theta}{2\sin{\theta}} (\mathbf{R} - \mathbf{R}^T).$$