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I'm trying to rig a robot using this armature with IK constrain. The behavior should be:

  • Base: rotates only on Y (bone) axe.
  • Arm 1: rotates only on X (bone) axe.
  • Arm 2: rotates only on X (bone) axe.
  • Point down: Automatically points down (stays always perpendicular to the floor). And rotates with Arm 2. Basically it has to copy "Base" bone but inverted.

How can I do it??

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The solution with the help of Gordon Brinkmann. Thank you.

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  • $\begingroup$ To me it looks like the "IK bone" is always pointing down. So, maybe a stupid question: why don't you make the "Point down" the IK bone? This question is similar to: Keep bone in IK chain always pointing down. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$ Commented Nov 26 at 12:37
  • $\begingroup$ I've tried what you suggest, but then the point down bone doesn't rotate with the arm 2 bone. In the last image you can see that the IK bone has not the same rotation over Y axe than the base bone. I need both things, pointing down and same rotation as base bone. $\endgroup$ Commented Nov 26 at 12:49

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What you need is an IK controller directly connected to "Arm 2", because you actually don't want the "Point down" bone to be controlled by the IK - it should only be connected to "Arm 2" if I understand correctly:

ik controller setup

Then the "Point down" bone is still connected to "Arm 2" but not part of the IK chain. Now it only needs a Limit Rotation constraint set to limit the X rotation in World Space to Min/Max -90° (Local with Parent works too I think but then you need to set the limits to 0°):

limit rotation

And now the "IK controller" bone controls the movement of the robot arm, the "Point down" moves with "Arm 2" but is always pointing down while having the same Y rotation as the "Base" bone:

robot arm

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  • $\begingroup$ That`s exactly what I need. I finally use a "Copy rotation" constrain and copy the base rotation. I think it works the same. Thank you very much for your help. $\endgroup$ Commented Nov 26 at 15:22

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