0
$\begingroup$

I'm struggling to find the DH parameters for this PUMA-type manipulator that yield the same results as the author (1): enter image description here

The way I'm checking if the parameters I have are correct is by comparing the resulting J11, J21 & J22 matrices with the author. These sub-matrices are the constituents of the wrist Jacobian matrix (Jw).

I tried many different combinations of the DH parameters including:

α =[0,90,0,-90,90,-90]
θ =[0,0,0,0,0,0]
a=[0,0,a2,-a3,0,0]
d=[d1,-d2,0,-d4,0,0]

Which result in the same matrices as the author except for some minor differences. The general wrist Jacobian matrix and the sub-matrices obtained by the author are given by:

enter image description here

Whereas the result I got for J11 was:

$$ \left[ \begin{array}{ccc} -d_2 c_1-s_1 (a_2 c_2-a_3 c_{23}+d_4 s_{23}) & c_1 (d_4 c_{23}-a_2 s_2+a_3 s_{23}) & c_1 (d_4 c_{23}+a_3 s_{23}) \\ c_1 (a_2 c_2-a_3 c_{23}+d_4 s_{23})-d_2 s_1 & s_1 (d_4 c_{23}-a_2 s_2+a_3 s_{23}) & s_1 (d_4 c_{23}+a_3 s_{23}) \\ 0 & a_2 c_2-a_3 c_{23}+d_4 s_{23} & d_4 s_{23}-a_3 c_{23} \\ \end{array}\right] $$

And for the J22 matrix I got:

$$ \left[ \begin{array}{ccc} -c_1 s_{23} & c_4 s_1+c_1 c_{23} s_4 & s_1 s_4 s_5-c_1 (c_3 (c_5 s_2+c_2 c_4 s_5)+s_3 (c_2 c_5-c_4 s_2 s_5)) \\ -s_1 s_{23} & c_{23} s_1 s_4-c_1 c_4 & -c_5 s_1 s_{23}-(c_2 c_3 c_4 s_1-c_4 s_2 s_3 s_1+c_1 s_4) s_5 \\ c_{23} & s_{23} s_4 & c_{23} c_5-c_4 s_{23} s_5 \\ \end{array}\right] $$ And the same J12 matrix as the author.

Perhaps the most pronounced difference here is that every Sin [ θ2 + θ3 ] is replaced with Cos [ θ2 + θ3 ] and vice versa, in addition to some sign differences.

Where am I going wrong here?

(1) Wenfu Xu, Bin Liang, Yangsheng Xu, "Practical approaches to handle the singularities of a wrist-partitioned space manipulator".

$\endgroup$
3
  • $\begingroup$ Which book or paper are you referencing? $\endgroup$
    – Ben
    May 21, 2017 at 1:23
  • $\begingroup$ @Ben "Practical approaches to handle the singularities of a wrist-partitioned space manipulator". $\endgroup$ May 23, 2017 at 11:10
  • $\begingroup$ @TarekIbrahim - I edited your matrices to mirror the style of the images you posted. This was mostly replacing entries like $\sin(\theta_2)$ with $s_2$, etc. $\endgroup$
    – Chuck
    May 23, 2017 at 11:56

1 Answer 1

0
$\begingroup$

In looking at your matrices, I would guess that you (or the author!) have made a mistake somewhere in your alpha $\alpha$ terms. Recall the period shift identity that states:

$$ \sin{\left(\theta + \pi/2 \right)} = +\cos{\theta} \\ \cos{\left(\theta + \pi/2 \right)} = -\sin{\theta} \\ $$ So, if you (or again, the author) were off by 90 degrees ($\pi/2$), then that would explain the fact that you appear to have some sines and cosines swapped and some sign errors, too.

I don't have the time at the moment to go through each step by hand and try to evaluate where you've gone wrong, but I would imagine that you should be able to compare individual joint transforms and find the ones that don't match. Try changing the $\alpha$ values for that joint by $\pm \pi/2$ and see if what you have matches what the author has.

Authors can and do make mistakes, and if you're positive the author made a mistake then the professional thing to do would be to contact that author and/or the journal that published the paper and alert them to the mistake. I would just double- and triple-check that the author was incorrect before taking that step, though (run it by your professors, etc.)

$\endgroup$
1
  • $\begingroup$ I am doing all the calculations using Wolfram Mathematica and I have checked the code I'm using multiple times so I think it's unlikely that there's a mistake at my end. I have also tried all possible combinations of the $\alpha$ values but to no avail. The reason I'm hesitant to contact the authors with what I have is the fact that they have published the same results in several other papers of theirs, not just this one. Also, I have no idea about the method by which they do their calculations. $\endgroup$ May 25, 2017 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.