I was reading Modern Robotics by Kevin Lynch et al. and this robot was presented. They had introduced Grubler's formula which assumes independent joint which gives $dof = 0$ but this is not the case for the robot and they provide an alternative computation but it isn't clear how they got the values for Grubler's formula for this.


I understand that this is Grubler's formula and this is also from the book.

Grubler's formula

The part that is unclear to me is what is highlighted in red. Text with what I don't understand in red


1 Answer 1


Where they say, "...any one of the three parallel links...has no effect on the motion" that pretty much means you could delete any one of the three links. Doing so leaves N=4 and J=4. Plug them into the formula and get the numbers you highlighted in red.

The conclusion "not independent, as required" is sort of a proof by contradiction. More precisely: The thing can move, with a single dof, only in the specific case where the "parallel" links are indeed parallel, and the same length, making one of the three redundant (i.e. not independent). Otherwise (the general case, with weird lengths and angles) the system would be overconstrained and indeed couldn't move (dof=0) as the formula computes.

  • $\begingroup$ Is it correct to say that the top oval link can only move in the horizontal plane and if 2 of the links were removed then it could move in the horizontal plane and also rotate getting 2 degrees of freedom? $\endgroup$ Commented Sep 4, 2021 at 16:05
  • $\begingroup$ @heretoinfinity the oval link moves in an arc ... it does not slide along a horizontal plane $\endgroup$
    – jsotola
    Commented Sep 4, 2021 at 17:11
  • $\begingroup$ Is the arc 1d or 2d since you get varying x and y coordinates? $\endgroup$ Commented Sep 4, 2021 at 17:15
  • $\begingroup$ There is only one degree of freedom. The three of angle, x, and y, are joined together by their circle relationship. Varying any one of them determines the other two, so only one is free. The fact that x and y form (must form) a circular arc in a 2-D plane only confuses things. $\endgroup$
    – r-bryan
    Commented Feb 1, 2022 at 17:57

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