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I have been taught that higher pair joints (e.g. gears, cams, rollers) deduct 1 degree of freedom due to the fact that they still allow two motions

  1. translation along the tangent surface

  2. rotation around the instantaneous Contact point

However, in my reasoning, the higher pair does constrain 2 motions

  1. as it is non-slipping (like gears and rollers) it gives a relationship between the angle and the location (such as $x=\theta \cdot r$)

  2. as it rotates on a surface it gives a relationship between x and y that it household be on the surface (such as $x = y+2$)

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You are right. A non-slip roller or gear has two constraints, one along the contact normal (no interpenetration) and the second one no relative slip velocity. So the total degrees of freedom added with such a joint is 1.

A slipping roller though only has once constraint (no interpenetration) and thus provides a total of two degrees of freedom to the system.

+-----------+------------+------------+
|   JOINT   |  #MOTIONS  | #REACTIONS |
+-----------+------------+------------+
|   FIXED   |     0      |      3     |
|   PIN     |     1      |      2     |
|   SLIDER  |     1      |      2     |
|   GEAR    |     1      |      2     |
|   ROLLER  |     2      |      1     |
|   NO-ROT  |     2      |      1     |
|   FREE    |     3      |      0     |
+-----------+------------+------------+
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  • $\begingroup$ Thank you, but, consider a 2D system of 2 gears pinned by revolute joints to the floor so that it can rotate and mesh each other, this system should have only 1d.o.f. But when i used the equation of 3(link-1) - 2(higher pair) - (lower pair). According to your table, gear should deduct 2. Making the equation 3(3-1) - 2(2(revolute joint)) - 2(gear) which is 0, where it should be 1 $\endgroup$
    – Derpson
    Commented Sep 24, 2016 at 6:21
  • $\begingroup$ A non pinned gear has 1 reaction (along contact normal) and 2 motions. A pinned gear has two reactions as it cannot also slide tangential to the contact. Are you accounting for the pin separately or together with the gear? $\endgroup$
    – John Alexiou
    Commented Sep 24, 2016 at 19:26
  • $\begingroup$ The gear contact above assumes that the gears will not separate and their only allowed is one motion. But by pinning both centers to the ground you are also imposing a no separation constraint so you have redundant constraints. That's why you end up with 0 DOF. Consider that using a table as above is not the best way to describe a system. Consider each case individually and count the total reaction forces. In the case with two gears, two pin reaction forces along the two pins are superfluous. Only one would suffice. $\endgroup$
    – John Alexiou
    Commented Sep 24, 2016 at 20:01

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