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I want to know Exponential coordinates of end effector $\text{S}\theta$(6-vector) from end effector transformation matrix. I calculated the transformation matrix of the end effector from the forward kinematics. And I took the logarithm of Transformation to get ${[S]\theta} ∈se(3)$.

The question is, is ${[S]\theta}$ the same as ${[S\theta]}$? If so, if I solve the skew-symmetric matrix representation of ${[S\theta]}$, is it an exponential coordinate? If not, how do we get the exponential coordinates?

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The question is, is $[S]θ$ the same as $[Sθ]$

Going by your notation of $[S]\theta\in se(3)$, yes. $se(3)$ is the Lie algebra of the group and $[S]$ forming the matrix form for the spatial vector $S$, yes, multiplication by scalar $\theta$ would still keep you within the Lie algebra because it is a vector space.

From a conceptual lens, while orientations themselves have multiple representations, the derivative of an orientation has one canonical form in this Lie algebra space, and what we're doing when we represent an orientation through this $se(3)$ vector, we're representing it like a normalised velocity from the identity oreintstion in that direction of the end orientation, integrated for $theta$ time (making $[S]\theta$), but equivalently it's the same as a non normalised velocity for unit time (or $[S\theta]$). Like how a displacement could be represented as a normalised velocity vector and a time taken, or as a non normalised velocity vector for 1 second.

If not, how do we get the exponential coordinates?

It's not clear to me what exponential coordinates mean in your context. $se(3)$ is the Lie algebra as I described. The concrete representations of the transforms (homogenous matrices, quaternions, etc.) would be group elements (part of $SE(3)$) and you would get them through the exponential map for the group: $\exp(\cdot) : se(3) \rightarrow SE(3)$

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  • $\begingroup$ Thanks for your advice! $\endgroup$
    – HARCO
    Commented Mar 6, 2023 at 9:19

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