3
$\begingroup$

I currently have a working kinematic chain made by a set of ten links in D-H convention (with usual parameters [ $A_i, D_i, \alpha_i, \theta_i$]).

But my task currently requires the inversion of some of them. Basically, I would have a part of the chain that is read from the end-effector to the origin, using the same links (and thus the same parameters). Is it possible? How to do so?


Please notice that this is not related to the inversion of the kinematic chain. It's more basic: I want to find the dh parameters of the inverted forward kinematic chain.

Let's put it simple: I have dh parameters for a 2 link planar chain from joint 0 to joint 1, so I can compute its direct kinematics. But what if I want to compute the direct kinematics from joint 1 to joint 0?

Given DH parameters [ $A_i, D_i, \alpha_i, \theta_i$], I can retrieve the transform matrix with this formula:

$G = \left[\begin{matrix} cos(\theta) & -sin(\theta)*cos(\alpha) & sin(\theta)*sin(\alpha) & cos(\theta)*a \\ sin(\theta) & cos(\theta)*cos(\alpha) & -cos(\theta)*sin(\alpha) & sin(\theta)*a \\ 0 & sin(\alpha) & cos(\alpha) & d \\ 0 & 0 & 0 & 1 \end{matrix} \right]$

This is the transform matrix from the i-th link to the (i+1)-th link. Thus, I can invert it to obtain the transform matrix from the (i+1)-th link to the i-th link, but the problem is that this is not working. I believe that the reason is related to the fact that the DH convention doesn't work any more as it is.

Any help?

$\endgroup$
3
$\begingroup$

As you correctly say, if you want to compute the transform from link $(n+1)$ to link $n$, you should simply invert the matrix $G$. The inverse of a homogenous transformation matrix

$G = \left[\begin{matrix} R & p \\ 0 & 1 \end{matrix}\right]$

is

$G^{-1} = \left[\begin{matrix} R^T & -R^T p \\ 0 & 1 \end{matrix}\right]$.

If $G$ is already available, this is probably the fastest way of computing $G^{-1}$. Otherwise, if only the DH parameters are known, you can insert the expression for $G$ to find this simplified expression for the translation of $G^{-1}$:

$-R^T p = \left[\begin{matrix} -a\\ -d \sin(\alpha)\\ -d \cos(\alpha) \end{matrix}\right]$

$\endgroup$
  • $\begingroup$ My problem is to find the DH parameters of the inverted matrix. Is it possible to have them? $\endgroup$ – alecive Oct 16 '13 at 17:11
0
$\begingroup$

Eventually, @alecive came up with an elegant solution he published here: http://dx.doi.org/10.1109/ICRA.2014.6907178

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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