1
$\begingroup$

I always have had a problem understanding Chebychev–Grübler–Kutzbach Formula of Degree of Freedom:

$$d = 6\,n - \sum_{i= 1}^{m}\left( 6 - f_i\right)$$

where $n$ is the number of moving links, $m$ is the number of joints and $f_i$ is the degree of freedom of each $i$-th joint.
I read about it in the Wikipedia and there its said that the formula should work for any non-over constrained system. Hence I assume that it should work for parallel manipulators. But when I try to compute it for some parallel manipulators I don't get the degree of freedom that I expect. Maybe I don't understand what parts to consider as the "moving links". I would appreciate if someone explains this through computing it for the cases of "3-RPR" and "Gough-Stewart" parallel platforms.

$\endgroup$
5
$\begingroup$

You can check here how to apply the formula to a parallel robot.

For a 6DOF Stewart platform

$$d = 6\,n - \sum_{i= 1}^{m}\left( 6 - f_i\right)$$

$n = 13$ links, $m = 18$ joints, six of which with mobility 1 (the prismatic joints), 3 of which with mobility 3 (the spherical joints) and three of which with mobility 2 (the universal joints). There is a different variant, where you have 6 spherical joints. That is a bit more complicated since there is a "lost" (or redundant) degree of freedom between the two spherical joints so you have to subtract this. (The pragmatic explanation is, that if the prismatic joint is between two spherical joints it can freely rotate around its axis, but this rotation does not contribute to the motion of the mechanism)

So $d = 6 \cdot 13 - 6 \cdot (6-1) - 6 \cdot (6-2) - 6 \cdot (6-3) = 78 - 30 - 24 - 18 = 6$ dof

For the 3-RPR, since it is planar mechanism the formula is:

$$d = 3\,n - \sum_{i= 1}^{m}\left( 3 - f_i\right)$$

$n = 7$ links, $m = 9$ joints, all with mobility 1. So $d = 3 \cdot 7 - 9 \cdot (3-1) = 21 - 18 = 3$ dof

However, as far as I know, it is not exactly correct to say, that this formula applies to all non-overconstrained types of mechanisms. There is a special, very rarely used case of parallel robots, called the Isoglide robot to which this formula cannot be applied correctly.

$\endgroup$
2
  • $\begingroup$ I noticed that in all of the cases you wrote here the number of links is one more than the one I had computed. For example in the case of the Stewart platform we have six legs and 6 links that form the moving platform (end-effector) which means we have twelve links in total. Would you please explain why in both cases we have one more ? is it because of the base platform (ground)? $\endgroup$ – Arvin Rasoulzadeh Jul 28 '20 at 8:27
  • 1
    $\begingroup$ The mobile platfrom (TCP) also counts as a link $\endgroup$ – 50k4 Jul 28 '20 at 11:40

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.