Dear colleges,
I am working on a parallel robot where the actuated joints are in one leg, i.e. q1 and q2 in the figure.
Then, I have problems with inverse geometry model (IGM) and kinematic model.
In terms of the IGM, I obtain the equations by suppressing the passive angles:
from the left leg: (x-q1)^2+(y-q2)^2=l^2
from the right leg: (x-d)^2+y^2 = l^2.
However, it seems to be impossible to solve q1 and q2. Is there another way to get IGM for this model?
Also, I differentiate the geometry model and get the kinematic model, which has a similar problem.
A[dx; dy]+B[dq1;dq2] = 0
But the B is a matrix [2(q1-x), 2(q2-y); 0 , 0];
I guess the main difficulty is because all the actuated joints are in the same leg.
Dear Zhongmou Li
Yes also i think so that difficulty is because all the actuated joints are in the same leg.
Actually, there is little research on (model and control) the parallel robot with one leg.
However, here are some attached files that interest from near or far the subject.
Best regards
Hi Zhongmou,
By distinguishing active, passive and compliant-active, compliant-passive joints in the parallel robot model, you should solve your problem. For more details, follow:
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.716.6513&rep=rep1&type=pdf
- https://www.eng.yale.edu/grablab/pubs/Borras_ICRA2013.pdf
- http://booksc.org/dl/7980530/f536ba
- http://pf-mh.uvt.rnu.tn/131/1/Quasi-static_modelling_of_compliant_mechanismsapplication_to_a_2-DOF_underactuated_finger.pdf
Best regards
Hi Zhoumou,
You encounter the problem since you select the position of the joint center of the top revolute joint as the output. In this case, the degree-of-freedom of the output is 1. The left leg has one redundant DOF. Therefore, your formulation does not work. To solve the inverse kinematics in this case, you need to use the approaches for redundant serial mechanisms in the literature.
If you select the position of the joint center of the revolute joint on the left as output, which has 2-DOF. There would be no redundant DOF in the left leg. You could solve this problem easily.
Best regards,
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