# Is the geometric inverse problem's solution "continuous" for a redundant robot?

Let's say my redundant robot is at an operationnal position $x$. Is the set of all possible joint configuration "continuous", which would mean that it is possible to explore all the possible configurations without moving the end effector. Is there a way to show that it is true or false? I am using a Kuka LBR robot with 7 dof so maybe there is a specific answer for this one.

I have been searching it and did not find any result but I will gladly accept any link or answer that you may have.

I don't know if there is a formal proof to this, but in general, no the set of all possible joint configurations that correspond to a particular end-effector pose is not continuous. I think of the set as islands in joint space. Where each island has some local continuous joint range, but is disconnected from the other islands.

I think there are a few reasons for this: joint limits, and kinematic limits, and physical construction.

As for joint limits, i think a good example is the "standard" robot roll-pitch-roll wrist like the LBR has. if these joints have limits, it will not be continuous rotation, which affect the arm's ability to stay at the end-effector pose while the rest of the arm moves through its redundancy.

As for physical construction, I think a good example is thinking of a planar 3 link arm which is redundant for end-effector 2D position, (with no rotation). For example: For many points, you could find a way to move the joints to keep the end-effector at the same point. But if the point is too close to the base then unless the links can pass through each other, the links will collide and prevent you from getting to the other configurations.

Lastly, as for kinematic constraints. If you fix the end-effector at a point, you can think of the arm as a four-bar linkage. I can't post an animated gif, but the Wikipedia page has this nice example: http://en.wikipedia.org/wiki/File:4_bar_linkage_animated.gif. You should consider the arm to be the 3 links: AD, DC, and CB. The base of the arm is A and the end-effector is B. (Ignore point E and the large triangle). Here the arm is exploring the entire local null space of redundant solutions that it can reach. However, there are is another set of solutions that can be achieved if the arm is flipped horizontally. (if point C is below the line between A and B.) Clearly the arm cannot move their with the end-effector constrained to stay at B. But it is a valid arm configuration.

• Thank you very much for your answer, I agree but now let's assume that we remove joint limits and physical construction and we just consider the simulation side where only joint positions matter, do you still think it is islands in joint space? Actually this is also what I think but I could not give a good explanation why to my departement head. I guess that mathematically one should verify that if we have the $f$ function such that $x=f(q)$, then for a particular $x_0$ we have that the $f^{-1}(x_0)$ set is composed of different islands but I cannot go further...