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I am developing a robot arm and after calculating the jacobian matrix and then the kinematic isotropy through:

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I get a kinematic isotropy of zero, due to four zero eigenvalues in my jacobian. What does this mean for the smoothnes of my workspace? I have read that the closer the kinematic isotropy is to 1, the more isotropic the manipulability ellipsoid. Does a kinematic isotropy of zero imply no manipulability?

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  • $\begingroup$ What is the morphology of the arm? $\endgroup$ – Ben Apr 28 '20 at 0:45
  • $\begingroup$ @Ben it is a simple 2 degree of freedom arm. One rotation around the x-axis and subsequently a rotation around the z-axis at an offset. $\endgroup$ – Chusikowski Apr 28 '20 at 7:50
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If you look at the formula, the only way for it to equal zero is if the determinant of the Jacobian equals zero. This tells you that the device is in a singular configuration.

With four eigenvalues zero, are you sure the robot is able to span the workspace in any pose?

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  • $\begingroup$ Definitely not, the robot has only 2 rotational dof, so each point in the workspace can only be reached with one joint configuration. Is that the reason I get 0 kinematic isotropy? $\endgroup$ – Chusikowski Apr 28 '20 at 8:10

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