Question 4 of 7: Frame assignment, DH parameters and forward kinematics of the RPR arm (20 marks)

Question 4: Frame assignment, DH parameters and forward kinematics of the RPR arm (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The R–P–R arm of Figure 1: a revolute joint $\theta_1$ about the vertical base axis, a prismatic joint $d_2$ sliding parallel to that axis at a horizontal offset $l_1$, and a revolute joint $\theta_3$ whose axis is horizontal and perpendicular to the offset link. The figure states that all link lengths beyond joint 3 are zero, so the wrist centre $e$ coincides with the joint-3 origin.

Find. (a) a frame assignment, (b) the DH parameter table, and (c) the forward-kinematic transform ${}^{B}_{W}T$ from the base frame to the wrist frame.

[Figure not reproduced: Figure 4.1 — the arm as drawn on the paper, redrawn in elevation. The base revolute joint turns about the vertical, the slider carries the wrist up and down, and joint 3 turns about a horizontal axis. See the official exam paper.]

Approach. Attach a $z$ axis to every joint axis, place the $x$ axes along the common normals, read the four Craig (modified) DH parameters off the sketch, and multiply the three link transforms — then add the constant roll that reconciles the DH wrist frame with the triad drawn on the figure.

1. Affix the frames (part a). Following Craig's rules: $\hat{Z}_i$ lies on joint axis $i$; $\hat{X}_{i-1}$ lies along the common normal from $\hat{Z}_{i-1}$ to $\hat{Z}_i$; and the base frame $\{B\} = \{0\}$ is chosen coincident with $\{1\}$ when $\theta_1 = 0$. That gives $\hat{Z}_1$ vertical at the base, $\hat{Z}_2$ vertical at the far end of the offset link (the slider axis is parallel to the base axis), and $\hat{Z}_3$ horizontal along the joint-3 axis. Because $a_2 = 0$ and $d_3 = 0$, frames $\{2\}$ and $\{3\}$ share the wrist origin.
Figure 4.2 — part (a): frames affixed. $\{0\}$ and $\{1\}$ sit at the base; $\{2\}$, $\{3\}$ and the wrist frame $\{W\}$ share the wrist centre $e$. The joint-3 axis (marked with a cross) runs into the page, parallel to $\hat{Y}_0$ at $\theta_1 = 0$.
2. Read the DH parameters (part b). With frames placed, each line of the table is read directly off the sketch: $a_{i-1}$ is the distance along $\hat{X}_{i-1}$ between the two $z$ axes, $\alpha_{i-1}$ the twist about $\hat{X}_{i-1}$, $d_i$ the offset along $\hat{Z}_i$ and $\theta_i$ the joint angle about $\hat{Z}_i$.

Modified (Craig) DH parameters for the RPR arm; the joint variable of each row is shown in bold

$i$	$\alpha_{i-1}$	$a_{i-1}$	$d_i$	$\theta_i$	Joint type
1	$0$	$0$	$0$	$\boldsymbol{\theta_1}$	revolute
2	$0$	$l_1$	$\boldsymbol{d_2}$	$0$	prismatic
3	$-90^\circ$	$0$	$0$	$\boldsymbol{\theta_3}$	revolute

3. Write the three link transforms. Substituting each row into the standard modified-DH form $$ {}^{i-1}_{\ \ i}T = \begin{bmatrix} c\theta_i & -s\theta_i & 0 & a_{i-1} \\ s\theta_i c\alpha_{i-1} & c\theta_i c\alpha_{i-1} & -s\alpha_{i-1} & -s\alpha_{i-1}d_i \\ s\theta_i s\alpha_{i-1} & c\theta_i s\alpha_{i-1} & c\alpha_{i-1} & c\alpha_{i-1}d_i \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ gives $$ {}^{0}_{1}T = \begin{bmatrix} c_1 & -s_1 & 0 & 0 \\ s_1 & c_1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0&0&0&1\end{bmatrix},\quad {}^{1}_{2}T = \begin{bmatrix} 1 & 0 & 0 & l_1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & d_2 \\ 0&0&0&1\end{bmatrix},\quad {}^{2}_{3}T = \begin{bmatrix} c_3 & -s_3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -s_3 & -c_3 & 0 & 0 \\ 0&0&0&1\end{bmatrix}, $$ with the shorthand $c_i = \cos\theta_i$, $s_i = \sin\theta_i$.
4. Reconcile frame $\{3\}$ with the drawn wrist triad. The figure labels the wrist axes with $z$ pointing up and $x$ along the arm, whereas the DH rules force $\hat{Z}_3$ onto the joint axis (horizontal). The two differ by a constant roll of $+90^\circ$ about $\hat{X}_3$, which is a fixed property of the tool mounting and not a joint variable: $$ {}^{3}_{W}T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0&0&0&1 \end{bmatrix} = R_X(90^\circ). $$ No $l_3$ appears, exactly as the question notes.
5. Multiply out the kinematic equations (part c). Chaining the four factors, $$ {}^{B}_{W}T = {}^{0}_{1}T\ {}^{1}_{2}T\ {}^{2}_{3}T\ {}^{3}_{W}T = \boxed{\begin{bmatrix} c_1c_3 & -s_1 & c_1s_3 & l_1c_1 \\ s_1c_3 & c_1 & s_1s_3 & l_1s_1 \\ -s_3 & 0 & c_3 & d_2 \\ 0 & 0 & 0 & 1 \end{bmatrix}}. $$ The wrist centre is therefore $[x_e,\ y_e,\ z_e]^{T} = [\,l_1c_1,\ l_1s_1,\ d_2\,]^{T}$.
6. Sanity-check the structure. Three independent checks all pass. (i) The rotation block factors as $R_Z(\theta_1)R_Y(\theta_3)$, which is what the mechanism does — turn about the vertical, then about a horizontal axis. (ii) The wrist centre always lies at radius $l_1$ from the base axis and at height $d_2$, so the reachable surface is a cylinder of radius $l_1$: the arm cannot change its reach, only its azimuth and height. (iii) At $\theta_1 = \theta_3 = 0$ the transform reduces to a pure translation, i.e. $\{W\}$ is parallel to $\{B\}$, matching the figure.
7. Numerical spot-check. For $l_1 = 0.500$ m, $\theta_1 = 30^\circ$, $d_2 = 0.750$ m and $\theta_3 = -40^\circ$, $$ {}^{B}_{W}T = \begin{bmatrix} 0.6634 & -0.5000 & -0.5567 & 0.4330 \\ 0.3830 & 0.8660 & -0.3214 & 0.2500 \\ 0.6428 & 0 & 0.7660 & 0.7500 \\ 0&0&0&1 \end{bmatrix}, $$ whose position column $[0.4330,\ 0.2500,\ 0.7500]^{T}$ m is exactly $[l_1\cos 30^\circ,\ l_1\sin 30^\circ,\ d_2]^{T}$.