Stéphane on Locomotion - Comments: Divergent components of motionhttps://scaron.info/talks/jrl-2019.html/2019-10-29T16:05:00+01:00Posted by: Attendee #52019-10-29T16:05:00+01:002019-10-29T16:05:00+01:00Attendee #5tag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-question-dcm-unicity<p>Is there unicity of the DCMs or exponential dichotomy?</p>Posted by: Stéphane2019-10-29T16:04:00+01:002019-10-29T16:04:00+01:00Stéphanetag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-reply-least-squares-poles<p>In general we could have four gains on the diagonal of the closed-loop
state-transition matrix, but in practice we often use the same gains for
different directions (for instance, the same gain for both sagittal and lateral
DCM feedback in the LIP). I followed this practice, using a single normalized …</p><p>In general we could have four gains on the diagonal of the closed-loop
state-transition matrix, but in practice we often use the same gains for
different directions (for instance, the same gain for both sagittal and lateral
DCM feedback in the LIP). I followed this practice, using a single normalized
gain <span class="math">\(k > 1\)</span> and scaling it on each axis by a factor consistent with the
equations of motion.</p>
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</script>Posted by: Attendee #42019-10-29T16:04:00+01:002019-10-29T16:04:00+01:00Attendee #4tag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-question-least-squares-poles<p>How did you select the poles in the final least-squares formulation?</p>Posted by: Stéphane2019-10-29T16:03:00+01:002019-10-29T16:03:00+01:00Stéphanetag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-reply-proportional-gain<p>On the real robot, it will depend on your control cycle and in particular on
the bandwidth of the force control loop (admittance control on our robots, see
slide 26). DCM and force control gains are coupled when the two are run at
roughly the same frequency, as is the …</p><p>On the real robot, it will depend on your control cycle and in particular on
the bandwidth of the force control loop (admittance control on our robots, see
slide 26). DCM and force control gains are coupled when the two are run at
roughly the same frequency, as is the case here, and we are not modeling this
coupling. For practical advice, check out this note on <a href="https://github.com/stephane-caron/lipm_walking_controller/wiki/Tuning-the-stabilizer">tuning stabilizer
gains</a>.</p>Posted by: Attendee #32019-10-29T16:03:00+01:002019-10-29T16:03:00+01:00Attendee #3tag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-question-proportional-gain<p>How do you choose the remaining proportional gain on slide 12?</p>Posted by: Stéphane2019-10-29T16:02:00+01:002019-10-29T16:02:00+01:00Stéphanetag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-reply-omega-divergent<p>Yes, the point I had doubts on is about the "of motion" part. Previously, when
the DCM was directly computed by linear combination of the CoM position and
velocity, it was clear that "it diverges" and "it is a component of motion"
imply that it is a DCM. But here …</p><p>Yes, the point I had doubts on is about the "of motion" part. Previously, when
the DCM was directly computed by linear combination of the CoM position and
velocity, it was clear that "it diverges" and "it is a component of motion"
imply that it is a DCM. But here, <span class="math">\(\omega\)</span> appears as a technical choice we
make in order to diagonalize the state-transition matrix after changing
variable.</p>
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</script>Posted by: Attendee #22019-10-29T16:02:00+01:002019-10-29T16:02:00+01:00Attendee #2tag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-question-omega-divergent<p>Why did you seem to doubt that <span class="math">\(\omega\)</span> is a DCM, isn't it clearly divergent?</p>
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var mathjaxscript = document.createElement …</script><p>Why did you seem to doubt that <span class="math">\(\omega\)</span> is a DCM, isn't it clearly divergent?</p>
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</script>Posted by: Stéphane2019-10-29T16:01:00+01:002019-10-29T16:01:00+01:00Stéphanetag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-reply-dcm-unicity<p>No! For instance, in a <a href="https://scaron.info/publications/icra-2018.html">previous
work</a> we had used a different
DCM for the VHIP whose formula included the ZMP as well. Multiplying a DCM by a
non-zero scalar also yields a DCM, there may be "classes" of equivalent DCMs
for some equivalence relation, but I wonder what it …</p><p>No! For instance, in a <a href="https://scaron.info/publications/icra-2018.html">previous
work</a> we had used a different
DCM for the VHIP whose formula included the ZMP as well. Multiplying a DCM by a
non-zero scalar also yields a DCM, there may be "classes" of equivalent DCMs
for some equivalence relation, but I wonder what it could be...</p>Posted by: Stéphane2019-10-29T16:01:00+01:002019-10-29T16:01:00+01:00Stéphanetag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-reply-matrices-slide-4<p>This figure corresponds to:</p>
<ul class="simple">
<li><span class="math">\(A = \begin{bmatrix} +2 & 1 \\ 0 & +1 \end{bmatrix}\)</span> for <span class="math">\(\mathrm{eig}(A) = \{2, 1\}\)</span></li>
<li><span class="math">\(A = \begin{bmatrix} -2 & 1 \\ 0 & +1 \end{bmatrix}\)</span> for <span class="math">\(\mathrm{eig}(A) = \{-2, 1\}\)</span></li>
<li><span class="math">\(A = \begin{bmatrix} -2 & 1 \\ 0 & -1 \end{bmatrix}\)</span> for <span class="math">\(\mathrm{eig}(A) = \{-2, -1\}\)</span></li>
</ul>
Posted by: Attendee #12019-10-29T16:01:00+01:002019-10-29T16:01:00+01:00Attendee #1tag:scaron.info,2019-10-29:/talks/jrl-2019.html//comment-question-matrices-slide-4<p>What matrices did you use to generate the figure on slide 4?</p>