I have a project to design a hand exoskeleton for hand rehabilitation. The problem is that i'm thinked I only need the linkage mechanism with multiple 4 bar linkages. But because of human fingers joints and linkage supports on middle of each phalange I realised that there are 5 bar linkages and I really can't figure it out how to solve it or what equations should I use. The mechanism is driven by one actuator. Any help would be appreciated!

Here's an image for reference.

enter image description here

  • $\begingroup$ are the DIP and PIP joints in the correct place in the Fig 3 ? ... where did the diagram come from? $\endgroup$
    – jsotola
    Jan 14 at 3:15
  • $\begingroup$ They are, it's from an project but there are no steps to resolve it otherwise just saying to subdivided the linkage and try individually solving it. : jbbae.info/publications/… $\endgroup$ Jan 14 at 14:42

With 5 bar linkages, you either need two inputs (one per each bar), or you need to couple the movement from one bar to the other (with gears for example), and then have only one input. Input meaning the rotation angle of the bars connected to the grounded link.

In case it might be useful, here are the equations of the inverse kinematics of a 5 bar linkage. Also remember that for closed chain linkages the direct kinematics is far more complicated than the inverse kinematics (unlike for open chain linkages).

In my opinion there is missing information in the paper you took as reference.

  • $\begingroup$ Victor is correct - you need two inputs to control a five-bar mechanism. With finger exoskeletons, a constraint is commonly made that the distal joint moves at one-half the rate of the middle joint. This is based on biodynamic studies (I do not have the references, but studied this many years ago). $\endgroup$
    – SteveO
    Jun 16 at 13:03

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