Solved
The degree of freedom
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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