Singularity for Rodrigues Parameters at 180 degree

[federkamm]

README.md says that the Rodrigues Parameters would have a singularity at 180 degree. What does that mean? To my understanding, the Rodrigues Parameters are a 3-vector p that parametrize the rotation R = exp([p]_x) where [.]_x is the skew symmetric cross-product matrix and exp is the matrix exponential function. This function is continous differentiable. To me, a singularity would mean that the differential of p -> exp([p]_x) would not have full rank at some point p. So I tried the point p = (pi, 0, 0) and tried to visually picture the exponential map and their derivative and it looks to me that it is of full rank (i.e. rank 3). Obviously, (pi+dx,0,0) turns further around x but (pi,dy,0) and (pi,0,dz) don't, and for symmetry reason the directions where (pi,dy,0) and (pi,0,dz) turn around cannot fall together (except if they would both only turn around x). So, am I wrong or do you mean something different by "singularity"?

[bjack205]

The mapping from quaternions to rodrigues params is v ./ s, where s and and v are the scalar and vector parts of the quaternion, respectively. This mapping will go singular when s -> 0. The scalar part of the quaternion goes to zero at a rotation of 180 degrees. To see this, you can look at the exponential map, which defines the quaternion to be [cos(theta/2); u * sin(theta/2), where theta is the rotation angle and u is the rotation axis. The term cos(theta/2) = 0 when theta = 180.

[federkamm]

Thank you for the explaination. I confused "Rodrigues Parameters" [tan(theta/2) * u] with "Rodrigues Vector" [theta * u]. Now, I agree that the Rodrigues Parameters have a singularity at theta = 180 degree.