Solution:

Five-axis Rhino XR-3 robot:

Following is the link coordinate diagram based on the Denvait Hartenberg algorithm. Using the steps from 0 to 7 we obtain the link coordinate diagram for the Rhino XR-3 robot.

Using the D-H algorithm steps 8 to 12 we obtain the following set of kinematic parameters.

d5 = tool length which can vary from robot to robot depending on which tool is installed. Values for the joint distances d and link length a of the Rhino XR-3 robot are as follows:

$\mathrm{d} 1=26.04 \mathrm{~cm}, \mathrm{~d} 5=16.83 \mathrm{~cm}$

$a 2=22.86 \mathrm{~cm}, \mathrm{a} 3=22.86 \mathrm{~cm}, \mathrm{a} 4=0.95 \mathrm{~cm}$

$ \begin{array}{cccccc}\\ \hline \text { Axis } & \theta & d & a & \alpha & \text { Home } \\\\ \hline 1 & q_1 & d_1 & 0 & -\pi / 2 & 0 \\\\ 2 & q_2 & 0 & a_2 & 0 & -\pi / 2 \\\\ 3 & q_3 & 0 & a_3 & 0 & \pi / 2 \\\\ 4 & q_4 & 0 & a_4 & -\pi / 2 & 0 \\\\ 5 & q_5 & d_5 & 0 & 0 & -\pi / 2 \\\\ \hline\\ \end{array}\\ $

The arm matrix equation is divided into 2 parts:

$\mathrm{T}_{\text {base }}^{\text {wrist }} \\$ and $\mathrm{T}_{\text {wrist }}^{\text {tool }} \$

$ \begin{aligned}\\ &T_{\text {base }}^{\text {wist }}=T_0^1 T_1^2 T_2^3\\\\ &=\left[\begin{array}{crcc} \mathrm{C}_1 & 0 & -\mathrm{S}_1 & 0 \\\\ \mathrm{~S}_1 & 0 & \mathrm{C}_1 & 0 \\\\ 0 & -1 & 0 & d_1 \\\\ 0 & 0 & 0 & 1\\ \end{array}\right]\left[\begin{array}{cccc} \mathrm{C}_2 & -\mathrm{S}_2 & 0 & a_2 \mathrm{C}_2 \\\\ \mathrm{~S}_2 & \mathrm{C}_2 & 0 & a_2 \mathrm{~S}_2 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1\\ \end{array}\right]\left[\begin{array}{cccc} \mathrm{C}_3 & -\mathrm{S}_3 & 0 & a_3 \mathrm{C}_3 \\\\ \mathrm{~S}_3 & \mathrm{C}_3 & 0 & a_3 \mathrm{~S}_3 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\ \end{array}\right]\\\\ &=\left[\begin{array}{cccc} \mathrm{C}_1 \mathrm{C}_2 & -\mathrm{C}_1 \mathrm{~S}_2 & -\mathrm{S}_1 & a_2 \mathrm{C}_1 \mathrm{C}_2 \\\\ \mathrm{~S}_1 \mathrm{C}_2 & -\mathrm{S}_1 \mathrm{~S}_2 & \mathrm{C}_1 & a_2 \mathrm{~S}_1 \mathrm{C}_2 \\\\ -\mathrm{S}_2 & -\mathrm{C}_2 & 0 & d_1-a_2 \mathrm{~S}_2 \\\\ 0 & 0 & 0 & 1 \\ \end{array}\right]\left[\begin{array}{cccc} \mathrm{C}_3 & -\mathrm{S}_3 & 0 & a_3 \mathrm{C}_3 \\\\ \mathrm{~S}_3 & \mathrm{C}_3 & 0 & a_3 \mathrm{~S}_3 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\ \end{array}\right]\\\\ &=\left[\begin{array}{cccc} \mathrm{C}_1 \mathrm{C}_{23} & -\mathrm{C}_1 \mathrm{~S}_{23} & -\mathrm{S}_1 & \mathrm{C}_1\left(a_2 \mathrm{C}_2+a_3 \mathrm{C}_{23}\right) \\\\ \mathrm{S}_1 \mathrm{C}_{23} & -\mathrm{S}_1 \mathrm{~S}_{23} & \mathrm{C}_1 & \mathrm{~S}_1\left(a_2 \mathrm{C}_2+a_3 \mathrm{C}_{23}\right) \\\\ -\mathrm{S}_{23} & -\mathrm{C}_{23} & 0 & d_1-a_2 \mathrm{~S}_2-a_3 \mathrm{~S}_{23} \\\\ 0 & 0 & 0 & 1 \\ \end{array}\right]\\ \end{aligned}\\ $

$ \left.T_{\text {base }}^{\text {urist }} \text { home }\right)=\left[\begin{array}{ccc:c} 1 & 0 & 0 & a_3 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & -1 & 0 & d_1+a_2 \\\\ \hdashline 0 & 0 & 0 & 1\\ \end{array}\right]\\ $

$ \begin{aligned}\\ T_{\text {wrist }}^{\text {too }} &=T_3^4 T_4^5 \\\\ &=\left[\begin{array}{cccc} \mathrm{C}_4 & 0 & -\mathrm{S}_4 & a_4 \mathrm{C}_4 \\\\ \mathrm{~S}_4 & 0 & \mathrm{C}_4 & a_4 \mathrm{~S}_4 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & 0 & 1\\ \end{array}\right]\left[\begin{array}{cccc} \mathrm{C}_5 & -\mathrm{S}_5 & 0 & 0 \\\\ \mathrm{~S}_5 & \mathrm{C}_5 & 0 & 0 \\\\ 0 & 0 & 1 & d_5 \\\\ 0 & 0 & 0 & 1\\ \end{array}\right] \\\\ &=\left[\begin{array}{cccc} \mathrm{C}_4 \mathrm{C}_5 & -\mathrm{C}_4 \mathrm{~S}_5 & -\mathrm{S}_4 & a_4 \mathrm{C}_4-d_5 \mathrm{~S}_4 \\\\ \mathrm{~S}_4 \mathrm{C}_5 & -\mathrm{S}_4 \mathrm{~S}_5 & \mathrm{C}_4 & a_4 \mathrm{~S}_4+d_5 \mathrm{C}_4 \\\\ -\mathrm{S}_5 & -\mathrm{C}_5 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\ \end{array}\right]\\ \end{aligned}\\ $

The arm matrix from the base to the tool for the 5-axis articulated–coordinate robot is as follows:

$ \left[\begin{array}{ccc:c} \mathrm{C}_1 \mathrm{C}_{234} \mathrm{C}_5+\mathrm{S}_1 \mathrm{~S}_5 & -\mathrm{C}_1 \mathrm{C}_{234} \mathrm{~S}_5+\mathrm{S}_1 \mathrm{C}_5 & -\mathrm{C}_1 \mathrm{~S}_{234} & \mathrm{C}_1\left(a_2 \mathrm{C}_2+a_3 \mathrm{C}_{23}+a_4 \mathrm{C}_{234}-d_5 \mathrm{~S}_{234}\right) \\\\ \mathrm{S}_1 \mathrm{C}_{234} \mathrm{C}_5-\mathrm{C}_1 \mathrm{~S}_5 & -\mathrm{S}_1 \mathrm{C}_{234} \mathrm{~S}_5-\mathrm{C}_1 \mathrm{C}_5 & -\mathrm{S}_1 \mathrm{~S}_{234} & \mathrm{~S}_1\left(a_2 \mathrm{C}_2+a_3 \mathrm{C}_{23}+a_4 \mathrm{C}_{234}-d_5 \mathrm{~S}_{234}\right) \\\\ -\mathrm{S}_{234} \mathrm{C}_5 & \mathrm{~S}_{234} \mathrm{~S}_5 & -\mathrm{C}_{234} & d_1-a_2 \mathrm{~S}_2-a_3 \mathrm{~S}_{23}-a_4 \mathrm{~S}_{234}-d_5 \mathrm{C}_{234} \\ \hdashline 0 & 0 & 0 & 1 \\ \end{array}\right] \\ $

The final expression depends on all the kinematic parameters and if we evaluate the arm matrix with the soft home position we obtain the following:

$ T_{\text {base }}^{\text {tool }} \text { (home) }=\left[\begin{array}{ccc:c} 0 & 1 & 0 & a_3+a_4 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & -1 & d_1+a_2-d_5 \\\\ \hdashline 0 & 0 & 0 & 1 \\ \end{array}\right] \\ $