3
$\begingroup$

I'm working on a kinematic estimation algorithm, and am trying to come up with the right terminology.

I'm using the Product-of-Exponentials formulation for kinematic chains (and ignoring position for now):

$R = e^{\omega_1 \theta_1}e^{\omega_2 \theta_2}...e^{\omega_n \theta_n}R_0$

where:
$R \in SO(3)$: orientation of end effector
$\theta_{1..n}$: joint coordinates
$\omega_{1..n} \in so(3)$: joint screw axes

Forward Kinematics: Solve for $R$ given $\omega_{1..n}, \theta_{1..n}$
Inverse Kinematics: Solve for $\theta_{1..n}$ given $R, \omega_{1..n}$
?___?: Solve for $\omega_{1..n}$ given $R, \theta_{1..n}$

My question is, what phrase fills that blank? I've been using the PoE formula for so long that I am starting to forget the DH paradigm but I'd imagine there's an equivalent with DH params too.

The literature I'm comparing against calls this "system identification" but I think that's because they're treating it like a dynamical system...I am treating it like a standard robot kinematics problem.

Is there an agreed-upon term for this process?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ \omega is the so(3) matrix and \theta is the joint coordinate? Shouldn't there be an SO(3) matrix on the end, the orientation when all \theta are zero? $\endgroup$ – Peter Corke Apr 27 '19 at 22:14
  • $\begingroup$ @PeterCorke correct, I had omitted for brevity, just edited for clarity ;) $\endgroup$ – Eric Peltola Apr 28 '19 at 2:26
1
$\begingroup$

Try looking for terms like robot calibration, robot kinematic calibration or kinematic calibration. Have a look at Chapter 6 of [1].

[1] Springer Handbook of Robotics, Eds: B.Siciliano, O.Khatib, 2016

Improve this answer
$\endgroup$
1
  • $\begingroup$ Great! That seems like the right direction. I'll look into that chapter when I can get my hands on a copy. And I'm sure you get this all the time but a sincere thank you for all the work you put into that toolbox. Cheers! $\endgroup$ – Eric Peltola Apr 28 '19 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.