I got a question about the effect of the selection vector on the force mode
I understand that the selection vector defines the compliant axis. But if I choose the selection vector to be [1,1,1,0,0,0] i.e. lack of rotation and I apply force in z-direction telling the robot to apply for example 20N at the edge of a surface what happens in the background? Without [0,0,0] there would be some moments applied at the EEF and the robot will tilt but with 0’s it doesn’t.
**Does the robot just apply moments opposite to those it feels at a given time, or maybe the robot somehow turns off the rotation of the EEF? **
I am trying to understand better what exactly happens when the selection vector is chosen.
Thank you in advance
if you have access to a robot, just try out the different settings available through the Program Node Force Template. If you create a Program and inspect the created script code, this could give you a better understanding of how the settings effect the handling of the robot.
To answer your question directly: Yes, the robot only allows a movement in the x/y/z - Direction of the chosen task_frame. It will not tilt or rotate. The Force you then add in the wrench - vector defines the robot driven movement. In your example it would be still possible to push the robot in x/y - direction, even if only a force is applies in z - direction.
Maybe I didn’t explain clearly what my question was:
When I set the complaint axis of the selection vector to 0 how does that prevent the robot from rotating?
Does it block the motors? Or maybe measures the moments and applies moments in the opposite direction to counteract the rotation thus keeping the robot from rotating?
The robot calculates the allowed trajectory depending on your settings and its current position. It also takes account the currently defined payload to either stay in place or apply the correct force. If you would try pushing it outside the pre-defined boundaries the robot and correspondent joints will apply an opposite force.