5
$\begingroup$

I'm working on an exercise about DH-parameters for my robotics course in university and I ran into a problem, because my solution is different from the one given by my professor. I calculated the DH-parameters $\theta=\theta_1-90˚, d=20m, a=0, \alpha=-90˚$. So far my solution is identical with the one of my professor. Now the next task is to calculate the transformation matrix $T_{0,1}$ using the calculated DH-parameters. Now I calculated: \begin{equation} T_{0,1}= \left(\begin{matrix} \cos(\theta_1-90) & 0 & -sin(\theta_1-90) & 0 \\ sin(\theta_1-90) & 0 & cos(\theta_1-90) & 0 \\ 0 & -1 & 0 & 20\\ 0 & 0 & 0 & 1 \end{matrix}\right) \end{equation} Now the solution of my professor states: \begin{equation} T_{0,1}= \left(\begin{matrix} \cos(\theta_1-90) & 0 & -sin(\theta_1-90) & 0 \\ sin(\theta_1-90) & 1 & cos(\theta_1-90) & 0 \\ 0 & -1 & 0 & 20\\ 0 & 0 & 0 & 1 \end{matrix}\right) \end{equation} Which is almost what I got, but not exactly since they put a 1 where I put a 0 in position (2,2). After multiplicating the transformation matrixes of the single DH-parameters I had found that the position (2,2) evaluates to $\cos(\theta_i)\cdot\cos(\alpha_i)$ and since since $cos(-90˚)$ evaluates to 0 it should be 0, or not? Where am I wrong?

$\endgroup$
1
  • 1
    $\begingroup$ Your professor's solution seems fishy. I would check if the two matrices are actually rotation matrices by verifying that M*M^T = I (the matrix multiplied by its transpose should give you the identity). $\endgroup$
    – alecive
    Dec 27, 2018 at 21:16

1 Answer 1

8
$\begingroup$

Your professor has made an error, but he or she is only human.

The upper-left 3x3 matrix must be an orthonormal rotation matrix. Every column of that must have a unit norm. The second column $[0, 1, -1]^T$ has a norm of $\sqrt{2}$ which makes the rotation matrix invalid.

$\endgroup$
1
  • $\begingroup$ Sorry for answering that late. Thanks for your help. I didn’t bother checking that, because I assumed my professor did $\endgroup$
    – Max
    Jan 9, 2019 at 21:03

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.