Constrained Motion of a Kinematic Chain:

Assume there be 'n' links in a kinematic chain & they are linked through 'j' simple hinges. The degree of freedom of a completely constrained kinematic chain is one. Thus,

1 = 3 (n - 1) - 2 j

Or,

 3 n = 2 j + 4                                             . . . (7.6)

From Eq. (7.6), this is evident that whatever be the value of j, the value of 3 n needs to even by according to Eq. (7.6). It means that a mechanism along simple hinges or pairs must have even number of links. Thus, the value of n may not be 5, 7, 9 etc. for a mechanism with simple lower pairs. Already it has been observed that minimum number of binary links is equivalent to 4. The next higher order mechanism begins with n = 6. From Eq. (7.1(a))

                                 6 = n₂  + n₃ + n₄  + ...

Supposing that the mechanism contain only binary & ternary links, the above equation becomes

 n₂  + n₃  = 6                                              . . . (7.7)

2 j = 3 n - 4

2 j = 3 × 6 - 4 = 14

j = 7

From Eq. (7.7) putting for j

2 × 7 = 2n₂ + 3n₃                  . . . (7.8)

Eqs. (7.7) & (7.8) may be solved. It gives n₂ = 4 and n₃ = 2.

Now Let us consider an eight-link chain & considering it contains binary, ternary and quarternary links, from Eq. (7.1) the described expression is attained:

 8 = n₂  + n₃  + n₄                           . . . (7.9)

2 j = 3 × 8 - 4 = 20   or   j = 10                . . . (7.10)

From Eq. (7.3), Eqs. (7.9) and (7.10) will provide several solutions

If n₂ = 4 they give n₃ = 4 & n₄ = 0

If n₂ = 5 they give n₃ = 2 & n₄ = 1

If n₂ = 6 they give n₃ = 0 & n₄ = 2

Likewise, for n = 10, 12, etc. various solutions will be available and it is not possible to obtain unique solution.