I trying to understand how manipulators robot works (6 DOF)

I would like to know if the inverse kinematics (IK) solution give only one or multiple solution if we include the rotation (3x3 matrix) of the TCP ? In others words, can we have the same TCP position and orientation for multiple configuraiton angles of the robot ?

I know that forward kinematics (FK) give only 1 solution (which seems logic) but IK is way more difficult to represent.

Thank in advance !

  • $\begingroup$ There are still a probability that position and orientation you desired would have several possible solution or maybe no solution at all $\endgroup$ – Albert H M Mar 25 at 14:42

can we have the same TCP position and orientation for multiple configuraiton angles of the robot


The number of DOF of a manipulator tells us if it's possible or not to reach a pose (position + orientation) in the operational space at hand (usually 6-dimensional).


  • If we work in a 6D operational space, then we are required to have at least a 6-DOF manipulator to reach a full pose (when this pose is within the allowed range).
  • If we deal with a planar manipulator and we are interested in attaining only a target position (i.e. no orientation), then we need a 2-DOF manipulator at least.
  • Same as above but now we include the orientation of the gripper; it follows that we need at least a 3-DOF manipulator.

Notice how I've highlighted the term "at least".

On the other hand, nothing is said about how many joint configurations can resolve the single given target pose.

In general, we have many joint configurations! A clear example of that is provided by a 6-DOF manipulator that can easily move all its joints while keeping the position as well as the orientation of the tip fixed. The latter can be achieved by resorting to the so-called null-space projection method.

  • 1
    $\begingroup$ Thanks a lot Ugo ! $\endgroup$ – hugosc Mar 26 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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