The lengths of the links of a 4- bar linkage with revolute pairs only are p,q,r and s units. Given that p < q < r < s and s+p < q+r which of these links should be the fixed one, for obtaining a ‘double crank’ mechanism?

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The lengths of the links of a 4- bar linkage with revolute pairs only are p,q,r and s units. Given that p < q < r < s and s+p < q+r which of these links should be the fixed one, for obtaining a ‘double crank’ mechanism?

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Greg · Answer 1 · 2022-10-19T19:00:28+0000

Right choice is (a) ink of length p

The best I can explain: For Double crank mechanism Shortest link is fixed.

Here shortest link is ‘P’.

The lengths of the links of a 4- bar linkage with revolute pairs only are p,q,r and s units. Given that p < q < r < s and s+p < q+r which of these links should be the fixed one, for obtaining a ‘double crank’ mechanism?

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