# Algorithmic Modified Denavit-Hartenberg Modeling for Robotic Manipulators Using Line Geometry
## 2 Exponential Map Using MDH
### Joint Twist Using MDH  
Since the exponential map (or adjacent transformation) of Equation (1) using MDH is obtained by the matrix product of Equations (3a) and (3b), the joint twist is derived from the left multiplication rule [21; 22] as follows:  
\[\lceil^{i-1}V_{i}\rceil ={}^{i-1}T_{i}^{-1}(\theta_{i},d_{i})^{i-1}\hat{T}_{i}(\theta_{i},d_{i}) \tag{5a}\] \[=\lceil\dot{\lambda}_{z_{i}}\rceil\] (5b) \[\therefore\,{}^{i-1}V_{i} =\dot{\lambda}_{z_{i}}, \tag{5c}\]  
where \({}^{i-1}V_{i}=\dot{\lambda}_{z_{i}}\in\mathbb{R}^{6}\) implies the \(i\)th joint twist relative to the frame \(\{i-1\}\) represented in the frame \(\{i\}\). It is noted that \(i\)th joint twist \({}^{i-1}V_{i}\) is represented in the frame \(\{i\}\). Now, we can define \(i\)th joint matrix \(E_{i}\) by separating the time-derivative of joint variable \(q_{i}\) from the joint twist as follows:  
\[{}^{i-1}V_{i}=\dot{\lambda}_{z_{i}}:=E_{i}q_{i}, \tag{6}\]  
where \(\dot{\lambda}_{z_{i}}=[0,0,\dot{d}_{i},0,0,\dot{\theta}_{i}]^{T}=[0,0,\dot{q}_ {i,d},0,0,\dot{q}_{i,\theta}]^{T}\) because the joint offset distance \(d_{i0}\) and the joint offset angle \(\theta_{i0}\) as MDH parameters are constants. For given four joint types \(\{R,P,H,RP\}\), each joint has its own joint matrix \(E_{i}\) and time-derivative of joint variable \(\dot{q}_{i}=\{\dot{q}_{i,\theta},\dot{q}_{i,d}\}\) as follows:\[R:E_{i}\dot{q}_{i}=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 1\end{array}\right]\dot{q}_{i,\theta} P:E_{i}\dot{q}_{i}=\left[\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right]\dot{q}_{i,d} \tag{7}\] \[H:E_{i}\dot{q}_{i}=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right]\dot{q}_{i,\theta} RP:E_{i}\dot{q}_{i}=\left[\begin{array}{c}0&0\\ 0&0\\ 0&1\\ 0&0\\ 1&0\end{array}\right]\left[\begin{array}{c}\dot{q}_{i,\theta}\\ \dot{q}_{i,d}\end{array}\right].\]  
The joint matrix \(E_{i}\) represented in the frame \(\{i\}\) can be transformed into either the base frame \(\{0\}\) or the tool coordinate frame \(\{tcf\}\); for this purpose, the adjoint transformation is introduced as follows:  
\[{}^{0}E_{i}={}^{0}\text{Ad}_{i}E_{i}\quad\text{and}\quad{}^{tcf}E_{i}={}^{tcf} \text{Ad}_{i}E_{i}, \tag{8}\]  
where \({}^{0}E_{i}\) and \({}^{tcf}E_{i}\) are called spatial screw and body screw in [4], respectively, and the adjoint transformation is defined by  
\[{}^{a}\text{Ad}_{b}=\left[\begin{array}{cc}{}^{a}R_{b}&\left[\begin{array}[] {cc}{}^{a}r_{b}\\ 0_{3}&\left[\begin{array}{cc}{}^{a}R_{b}\\ \end{array}\right]^{a}R_{b}\end{array}\right]\in\mathbb{R}^{6\times 6}, \tag{9}\]  
using the homogeneous transformation \({}^{a}T_{b}({}^{a}R_{b},{}^{a}r_{b})\), in which \({}^{a}R_{b}\in SO(3)\) implies a three-dimensional rotation matrix, \(\left[\begin{array}{cc}{}^{a}r_{b}\\ \end{array}\right]\) represents three-dimensional skew-symmetric matrix obtained from a three-dimensional displacement vector \({}^{a}r_{b}\in\mathbb{R}^{3}\), and \(0_{3}\) implies three-dimensional zero matrix. Till now, we have revealed how the MDH parameters influence the exponential map, the exponential coordinate vector, the joint twist, and the joint matrix. The next section will discuss the relationships between the joint axes using the line geometry, ultimately, to find the MDH parameters.