# Kinematic isotropy equal 0

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

• What is the morphology of the arm? – Ben Apr 28 '20 at 0:45
• @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. – Chusikowski Apr 28 '20 at 7:50

## 1 Answer

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?

• 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? – Chusikowski Apr 28 '20 at 8:10