1
$\begingroup$

I have the following question about the Franka Emika Panda dynamic model. In case I perform the solution of the inverse dynamics problem using Robotics Toolbox (Peter Corke) I get different results compared to the solution using the dynamic model published in the following article (for the simplicity I have neglected the friction and gravitational component in both cases): C. Gaz, M. Cognetti, A. Oliva, P. Robuffo Giordano, A. De Luca, 'Dynamic Identification of the Franka Emika Panda Robot With Retrieval of Feasible Parameters Using Penalty-Based Optimization'. IEEE RA-L, 2019. The dynamic model is available here: https://github.com/marcocognetti/FrankaEmikaPandaDynModel/tree/master/matlab/dyn_model_panda
According to the available information, the dynamic model of Panda in Robotics Toolbox is created based on the kinematic and dynamic parameters available in the mentioned paper, so why does the use of Robotics Toolbox show different results, see attached figure? I am attaching a modified Matlab script demonstrating the different behavior. Thank you in advance for the answer.enter image description here enter link description here

enter link description here

$\endgroup$

1 Answer 1

0
$\begingroup$

@Daniel kajzr! Have you solved the problem? I think you may have forgotten to add tau_f to the inverse dynamic equation of the robot. Because the last three joints are mainly used to overcome the friction torque, you have not calculated the friction torque, so the error of the last three joints is relatively large.

$\endgroup$
1
  • $\begingroup$ Bruce Lee, thank you for your reply. Unfortunately adding tau_f is not the solution, the attached matlab script states in the comments that I have neglected the friction and gravity components in both cases (I updated the post) . The resulting traces in the figure are only the tau= Mddq + Cdq. The problem must be in the inertia parameters, because the publication does not explicitly state whether the moment of inertia is related to the local coordinate system or to the CoM. I tried recalculating the given moments of inertia using Steiner's theorem, but I still did not get identical results. $\endgroup$ Commented Jan 18, 2022 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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