Robotic Work Cell

This image shows a single robot work cell.

$T_{camera}^{part}$ represents the pose of the part as seen by the camera. $T_{camera}^{base}$ represents the pose of the robot's base as seen by the camera.

$$\boxed{T_{base}^{part}=T_{base}^{camera}{T}_{camera}^{part}=(T_{camera}^{base}{)}^{-1}{T}_{camera}^{part}}$$

Suppose:

$$T_{camera}^{part}=\left[\begin{array}{cccc}0& -1& 0& 0\\ -1& 0& 0& -5\\ 0& 0& -1& 19\\ 0& 0& 0& 1\end{array}\right]$$ $$T_{camera}^{base}=\left[\begin{array}{cccc}0& -1& 0& 15\\ -1& 0& 0& 25\\ 0& 0& -1& 20\\ 0& 0& 0& 1\end{array}\right]$$

then,

$$T_{base}^{part}=\left[\begin{array}{cccc}1& 0& 0& 30\\ 0& 1& 0& 15\\ 0& 0& 1& 1\\ 0& 0& 0& 1\end{array}\right]$$

where the position of the part relative to the base frame is $p=[30,15,1{]}^T$. Moreover, the 3 axes of the part's frame points to the same directions as the base frame, since the rotation submatrix is $R=I$.

To determine rotation matrix $R=[x^t,y^t,z^t]$ if the robot is supposed to reach down and pick up the part directly from above,

With respect to the base coordinates, this corresponds to $r^3=-i^3$. To grip to object from the front and back, there needs to be a sliding vector of $r^2=\mp i^2$, and to complete the orthogonal coordinate frame, it is a requirement that $r^1=\pm i^1$. Therefore, there are two possible solutions for the rotation matrix $R$, one being:

$$R=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]$$

Now we can put the position and orientation together to develop the tool configuration needed to grasp the part.

$$T_{base}^{grasp}(q)=\left[\begin{array}{cc}R& p\\ 0& 1\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 30\\ 0& -1& 0& 15\\ 0& 0& -1& 1\\ 0& 0& 0& 1\end{array}\right]$$