Inverse Geometric Model (IGM)

The Inverse Geometric Model comes from mapping the cartesian space \(\vec{x}\) to the joint space \(\theta\). But first, let's take a look at the Direct Geometric Model (DGM), which is represented by the transformation matrix and maps the opposite of the IGM.

\[ ^{j-1}T_{j} = \begin{bmatrix} \cos(\theta_{j}) & -\sin(\theta_{j}) & 0 & d_{j}\\ \cos(\alpha_{j})\sin(\theta_{j}) & \cos(\alpha_{j})\cos(\theta_{j}) & -\sin(\alpha_{j}) & -r_{j}\sin(\alpha_{j})\\ \sin(\alpha_{j})\sin(\theta_{j}) & \sin(\alpha_{j})\cos(\theta_{j}) & \cos(\alpha_{j}) & r_{j}\cos(\alpha_{j})\\ 0 & 0 & 0 & 1 \end{bmatrix} \]

This way we can determine the position and orientation vector \(\vec{x} = \{P_{x}, P_{y}, P_{z}, \hat{x}, \hat{y}, \hat{z}\}\) with the following considerations:

\[ ^{j-1}T_{j} = \begin{bmatrix} s_{x} & n_{x} & a_{x} & P_{x} \\ s_{y} & n_{y} & a_{y} & P_{y} \\ s_{z} & n_{z} & a_{z} & P_{z} \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

So,

\[ P_{x} = ^{j-1}T_{j}(P_{x})\\ P_{y} = ^{j-1}T_{j}(P_{y})\\ P_{z} = ^{j-1}T_{j}(P_{z})\\ \langle \hat{x}, \hat{y}, \hat{z} \rangle = \begin{bmatrix} s_{x} & n_{x} & a_{x} \\ s_{y} & n_{y} & a_{y} \\ s_{z} & n_{z} & a_{z} \end{bmatrix} \]

And to map the base of the robot to the tool or any of its joints:

\[ ^{base}T_{tool} = \prod_{j=1}^{n} {}^{j-1}T_{j} = {}^{0}T_{1} {}^{1}T_{2} {}^{2}T_{3} \dots {}^{j-1}T_{j} \]

The article is still in development