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I want to compute the DOF of this system if it composed of (n) such links grasping this common object using Grubler's formula obtaining the solution as a function of (n), considering this to be a spatial system then we have: m = 6, The number of rigid bodies or links including the ground N = 4, The number of joints J = 3, The added DOFs from those joints f = 7, so if we used the formula: DOF = m(N - 1 - J) + f we get: 7, and considering the (n) links then the total degrees of freedom of such a system is 7n. I want to make sure that I didn't make any mistake and I am getting it right. Thanks in advance.

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The formula is

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

$$d = 6N - 1 - \sum_{i=1}^j\ (6 - f_i) = 6(N-1 - J) + \sum_{i=1}^J\ f_i $$

In the book Modern Robotics, there is an additional (-1) in the formula, as shown in the second version of the formula, as they recommend counting the ground as linkage. I do not count ground as linkage, I do not use the -1. For me it is just more logical this way, since ground is not a linkage and then I do not need to subtract 1 form the number of linkages. Furthermore the constant 6 in this version of the formula is denoted by m in the book Modern Robotics.

let's denote the number of chains grasping the object with $c$ in order to avoid confusion with the already used $n$ notation.

$N$ is the number of links including the ground link,

$n$ is the number of links not including the ground link,

$J$ is the number of joints and

$fi$ is the number of degrees of freedom of joint $i$.

so fort of all we have 1 linkage, the object being held commonly (assuming the grasp is rigid) counts as a linkage. Per kinematic chain we have 3 linkages, we have to subtract 1 since the last one is grasping, therefore is part of the first linkage. In other words adding another kinematik chain increases the number of linkages by 2 and not by three, since the last linkage will be part of the rigid body, there is no motion between them, therefore it does not count as a new linkage.

$n = 1 + 2 \cdot c$ for the case in the picture, for $c = 4$ linkages $n = 9$

There are 3 joints per kinematic chain. one of the joints with 3 dof, one of the joints with 1 dof and the last joint ... is tricky, it seems to have 3 dof, but in only counts as 2, since the two spherical joints produce a local rotation motion around the axis between their middle points. This a degree of freedom, which does not effect the motion of the whole chain, therefor it needs to be subtracted (if the number of kinematic chains is greater then 1, $c > 1$). So:

$$\sum_{i= 1}^{J}\left( 6 - f_i\right) = c \cdot ((6 - 3) + (6 - 1) + (6 - 2)) = c \cdot (3 + 5 + 4) = c \cdot 12 = 4 \cdot 12 = 48$$

Therfore, for the mechanism in the picture: $d = 6 \cdot 9 - 48 = 54 - 48 = 6$

For the generalized version, for $c > 1$ number of kinematic chains added:

$d = 6 \cdot (1 + 2 \cdot c) - (c \cdot 12) = 6 + 12 \cdot c - 12 \cdot c$

$d = 6$, independent of the number of chains used.

It makes scene, since one chain has 6 degrees of freedom, if you couple the same chain parallel, is should not loose of gain any degrees of freedom.

If we would add 6 linkages, then we would obtain a structure very similar to the Stewart platform. It is commonly known that the Stewart platform has 6 DOF.

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  • $\begingroup$ I really appreciate your help. I'm using the formula from the book: Modern Robotics and the way you represent the problem is very confusing to me. All that I'm asking is to know the total (dof) of the system expressed in the number of robot arms used. I understood the general idea you showed but can't deploy it to mine. Besides, the (dof) of the system in the picture is (10) and I'm sure of that. $\endgroup$ – Ammar Taha Aug 12 at 10:52
  • $\begingroup$ The formula I used is the same formula described in the book, equation 2.4, page 18 of the preprint edition, albeit I use somewhat different notation $\endgroup$ – 50k4 Aug 12 at 11:23
  • $\begingroup$ The DOF of your mechanism is 6, regardless of your n number of attached kinematic chains. I understand why you might think it is 10. It is not. This is covered in my answer in the part "he last joint ... is tricky, it seems to have 3 dof, but in only counts as two, since the two spherical joints produce a local rotation motion around the axis between their middle points. " $\endgroup$ – 50k4 Aug 12 at 11:25
  • $\begingroup$ I updated the answer to better match the notation in the book and add more details about the lost degree of freedom in the two spherical joints $\endgroup$ – 50k4 Aug 12 at 11:34
  • $\begingroup$ Ok, thanks a lot. $\endgroup$ – Ammar Taha Aug 12 at 13:41

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