anyone can help me
I do not know what wrong
I know this law of degree of freedom
n = 3 (l – 1) – 2 j – h
l=link , j=lower pair , h=high pair
n = 3 (l – 1) – 2 j – h
n = 3 (4 – 1) – 2 × 4– 0 = 1 one degree of freedom
but of program I think this happen
n = 3 (4 – 1) – 2 × 4– 1 = 0
n = 3 (4 – 1) – 2 × 4– 0 = 1 degree of freedom
n = 3 (4 – 1) – 2 × 4– 1 = 0 but this result of program
this example of 5 link i must have two degree but program give 1 degree
Solved! Go to Solution.
https://en.wikipedia.org/wiki/Chebychev%E2%80%93Gr%C3%BCbler%E2%80%93Kutzbach_criterion
The formula for the planar condition in the link above, appears different from the one you show.
I agree, the depicted mechanism looks to have 2-degees of freedom. Check to make sure none of the pin joints has an extra constraint.
I give my formula form book "Theory Of Machine R.S.Khurmi"
and this not problem
problem is all my mechanism degree of freedom have less than one
I check pin from all bar and I think no any extra constrain but same result all time
Have you attached any motors?
Each motor controls a degree of freedom.
can you explain more and can you tell me how you can define degree of freedom correct
thank you very much
For those who are interested in linkages, this is also worth reading: