The degree of freedom

[jwaller]

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

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[dschenken]

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.

[jwaller]

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

[dschenken]

Have you attached any motors?

[jwaller]

yes i put servo motors

Video Link : 5204

[dschenken · Accepted solution]

Each motor controls a degree of freedom.

[jwaller]

can you explain more and can you tell me how you can define degree of freedom correct

[jwaller]

thank you very much

[dschenken]

For those who are interested in linkages, this is also worth reading:

https://en.wikipedia.org/wiki/Overconstrained_mechanism