Hello there, hope you are doing great.

Right now I am trying to control youbot to pick and cube and place it to a certain position, on simulation. However, I believe that the reference trajectory is very prone to singularities. Because of that, I am trying to create a reference trajectory that does not contain singularity possibilities.

I am trying on new initial end-effector configurations for the youbot. However I need to extract Tsb matrix (base frame relative to space frame) from the Tse (end-effector frame, relative to space frame) matrix. Because, in order to tune my PI gains, I need to start the robot with 0 error. How can I manage to do that? For example to get Tse from Tsb I do this:

Tse = Tsb * Tb0 (which has a fixed value) * T0e (comes from Forward Kinematics)

enter image description here


1 Answer 1


Do I understand correctly, that the problem is with the matrix equation you have posted and how to calculate $T_{sb}$ from that equation? You can solve the matrix equation by multiplying from the right with the inverse of the last matrix on the rhs, step by step.

$$\require{cancel} $$

$$T_{se} = T_{sb} \times T_{b0} \times T_{0e}$$

$$T_{se} \times T_{0e}^{-1}= T_{sb} \times T_{b0} \times T_{0e} \times T_{0e}^{-1}$$

$$T_{se} \times T_{0e}^{-1}= T_{sb} \times T_{b0} \times \cancel{ T_{0e} \times T_{0e}^{-1}}$$

$$T_{se} \times T_{0e}^{-1}= T_{sb} \times T_{b0}$$

$$T_{se} \times T_{0e}^{-1} \times T_{b0}^{-1}= T_{sb} \times T_{b0} \times T_{b0}^{-1}$$

$$T_{se} \times T_{0e}^{-1} \times T_{b0}^{-1}= T_{sb} \times \cancel{T_{b0} \times T_{b0}^{-1}}$$

$$T_{se} \times T_{0e}^{-1} \times T_{b0}^{-1}= T_{sb}$$

$$T_{sb} = T_{se} \times T_{0e}^{-1} \times T_{b0}^{-1}$$

  • $\begingroup$ Yes, thank you so much! $\endgroup$
    – kucar
    Commented Oct 16, 2020 at 11:31

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