Link frames, DH table and link transforms for a 3-DOF arm: 22-Mec-B12 Robotics, December 2019, Q6

Question 6: Link frames, DH table and link transforms for a 3-DOF 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. A three-revolute arm whose geometry is fixed by the stated axis relationships and the four dimensions marked on Figure 3.

Given data
Quantity	Symbol	Value
Joint 1	$\theta_{1}$	revolute about the vertical base axis
Joint 2	$\theta_{2}$	revolute, axis horizontal and perpendicular to joint 1, intersecting it
Joint 3	$\theta_{3}$	revolute, axis parallel to joint 2
Height of the joint-2 axis above the base	$L_{1}$	as marked in Figure 3
Distance between the joint-2 and joint-3 axes	$L_{2}$	as marked
Joint-3 axis to the wrist, and wrist to the tool point	$L_{3},\,L_{4}$	as marked (tool frame only)

Find. A consistent set of link frames {0}–{3}, the DH table they generate, and the three link transforms.

Part (a): link frames attached to the arm in the zero configuration. $Z_0=Z_1$ runs up the joint-1 axis; $Z_2$ and $Z_3$ point out of the page along the parallel joint-2 and joint-3 axes; $X_0=X_1=X_2$ and $X_3$ run outward along the arm.

Approach. Place $\hat Z_{i}$ on joint axis $i$, take $\hat X_{i}$ along the common normal from $\hat Z_{i}$ to $\hat Z_{i+1}$, and read the four DH parameters of each row straight off the resulting picture. Then substitute them into the standard modified-DH link transform.

(a) Attach $\hat Z$ to every joint axis. $\hat Z_{1}$ points up along the vertical joint-1 axis. $\hat Z_{2}$ lies along the joint-2 axis, which is horizontal and cuts the joint-1 axis at height $L_{1}$; the sense is chosen so that a positive $\theta_{2}$ lifts the arm, matching the sense drawn on Figure 3. $\hat Z_{3}$ is parallel to $\hat Z_{2}$, on the joint-3 axis a distance $L_{2}$ out along the arm.
Attach $\hat X$ along the common normals. $\hat Z_{1}$ and $\hat Z_{2}$ intersect, so the common normal has zero length and $\hat X_{1}$ is simply the direction perpendicular to both, $\hat X_{1}=\hat Z_{1}\times\hat Z_{2}$, which points outward along the arm. $\hat Z_{2}$ and $\hat Z_{3}$ are parallel, so their common normal is the segment joining them and $\hat X_{2}$ also points outward along the arm. $\hat X_{3}$ continues outward toward the tool. In the drawn zero configuration all three $\hat X$ axes are parallel, which is precisely what makes $\theta_{2}=\theta_{3}=0$ there.
Fix the base frame {0}. Frame {0} is placed on the joint-1 axis at the base with $\hat Z_{0}=\hat Z_{1}$ and $\hat X_{0}=\hat X_{1}$ when $\theta_{1}=0$; the only difference from {1} is the height $L_{1}$, which appears as $d_{1}$. (Craig’s narrower convention would slide {0} up to the joint-2 height so that $d_{1}=0$ and carry $L_{1}$ in the station transform $^{S}_{0}T$ instead. Either is acceptable; keeping $d_{1}=L_{1}$ has the advantage that every dimension marked on the figure appears somewhere in the model.)
(b) Read the DH table. With $\alpha_{i-1}$ the twist from $\hat Z_{i-1}$ to $\hat Z_{i}$ about $\hat X_{i-1}$, $a_{i-1}$ the distance along $\hat X_{i-1}$, $d_{i}$ the offset along $\hat Z_{i}$ and $\theta_{i}$ the joint variable, the table is $$\begin{array}{c|cccc} i & \alpha_{i-1} & a_{i-1} & d_{i} & \theta_{i}\\\hline 1 & 0 & 0 & L_{1} & \theta_{1}\\ 2 & 90^\circ & 0 & 0 & \theta_{2}\\ 3 & 0 & L_{2} & 0 & \theta_{3}\end{array}$$ The $90^\circ$ twist in row 2 is the “joints 1 and 2 perpendicular” statement; the zero twist in row 3 is “joints 2 and 3 parallel”; and $a_{1}=0$ records that axes 1 and 2 intersect.
(c) Substitute row 1 into the link transform. With $\alpha_{0}=0$, $a_{0}=0$, $d_{1}=L_{1}$, $$\boxed{^{0}_{1}T=\begin{bmatrix}c_{1} & -s_{1} & 0 & 0\\ s_{1} & c_{1} & 0 & 0\\ 0 & 0 & 1 & L_{1}\\ 0 & 0 & 0 & 1\end{bmatrix}}$$ a pure rotation about the vertical combined with the fixed lift $L_{1}$.
Substitute row 2. With $\alpha_{1}=90^\circ$ ($c\alpha=0$, $s\alpha=1$), $a_{1}=0$ and $d_{2}=0$, $$\boxed{^{1}_{2}T=\begin{bmatrix}c_{2} & -s_{2} & 0 & 0\\ 0 & 0 & -1 & 0\\ s_{2} & c_{2} & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}}$$ The row of zeros with a $-1$ is the tell-tale of the right-angle twist: it maps $\hat Z_{2}$ onto $-\hat Y_{1}$.
Substitute row 3. With $\alpha_{2}=0$, $a_{2}=L_{2}$ and $d_{3}=0$, $$\boxed{^{2}_{3}T=\begin{bmatrix}c_{3} & -s_{3} & 0 & L_{2}\\ s_{3} & c_{3} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}}$$ — the planar form expected for two parallel axes.
Check the frames against the drawing. Multiplying out, $$^{0}P_{3ORG}=\begin{bmatrix}L_{2}c_{1}c_{2}\\ L_{2}s_{1}c_{2}\\ L_{1}+L_{2}s_{2}\end{bmatrix}.$$ At $\Theta=0$ this is $[L_{2},0,L_{1}]^{T}$ — the arm outstretched horizontally at height $L_{1}$, exactly as Figure 3 shows. Setting $\theta_{2}=90^\circ$ raises the joint-3 axis to $[0,0,L_{1}+L_{2}]^{T}$, confirming that the positive sense of $\theta_{2}$ lifts the arm, and $\theta_{1}=90^\circ$ swings it to $[0,L_{2},L_{1}]^{T}$. Note also that $\theta_{3}$ does not appear: the origin of {3} lies on joint axis 3, so joint 3 only reorients frame {3}.
Attach the tool frame (for completeness). Frame {T} sits a further $L_{3}+L_{4}$ along $\hat X_{3}$ with the orientation drawn in Figure 3, so $^{3}_{T}T=\mathrm{Trans}(L_{3}+L_{4},\,0,\,0)$ and $^{0}_{T}T={}^{0}_{1}T\,{}^{1}_{2}T\,{}^{2}_{3}T\,{}^{3}_{T}T$.

Final results
Quantity	Symbol	Value
Row 1 of the DH table	$(\alpha_{0},a_{0},d_{1},\theta_{1})$	$(0,\,0,\,L_{1},\,\theta_{1})$
Row 2 of the DH table	$(\alpha_{1},a_{1},d_{2},\theta_{2})$	$(90^\circ,\,0,\,0,\,\theta_{2})$
Row 3 of the DH table	$(\alpha_{2},a_{2},d_{3},\theta_{3})$	$(0,\,L_{2},\,0,\,\theta_{3})$
First link transform	$^{0}_{1}T$	$\begin{bmatrix}c_{1} & -s_{1} & 0 & 0\\ s_{1} & c_{1} & 0 & 0\\ 0 & 0 & 1 & L_{1}\\ 0 & 0 & 0 & 1\end{bmatrix}$
Second link transform	$^{1}_{2}T$	$\begin{bmatrix}c_{2} & -s_{2} & 0 & 0\\ 0 & 0 & -1 & 0\\ s_{2} & c_{2} & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$
Third link transform	$^{2}_{3}T$	$\begin{bmatrix}c_{3} & -s_{3} & 0 & L_{2}\\ s_{3} & c_{3} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$
Origin of {3} in {0}	$^{0}P_{3ORG}$	$[L_{2}c_{1}c_{2},\;L_{2}s_{1}c_{2},\;L_{1}+L_{2}s_{2}]^{T}$
Tool transform	$^{3}_{T}T$	$\mathrm{Trans}(L_{3}+L_{4},\,0,\,0)$

Concept & Theory

Topic. Link-frame attachment and the Denavit–Hartenberg parameters (Craig’s modified convention).

Key relations. $\hat Z_{i}$ on joint axis $i$; $\hat X_{i}$ along the common normal $\hat Z_{i}\to\hat Z_{i+1}$; $\alpha_{i-1}$, $a_{i-1}$, $d_{i}$, $\theta_{i}$; $^{i-1}_{\;\;i}T=R_{X}(\alpha_{i-1})D_{X}(a_{i-1})R_{Z}(\theta_{i})D_{Z}(d_{i})$.

Why this works. The DH scheme replaces six numbers per joint with four by insisting that $\hat X_{i}$ lie along the common normal, which forces two of the six to be zero. The four survivors then read directly off the mechanism: a twist of $90^\circ$ whenever consecutive axes are perpendicular, zero twist whenever they are parallel, a zero link length whenever they intersect, and an offset wherever the arm is built with a step. Because the zero configuration is drawn with all $\hat X$ axes parallel, every joint variable in the table is zero in the picture, which is the quickest self-check on a frame assignment.

Common pitfall. Assigning frames that do not put the drawn pose at $\Theta=0$, so every subsequent transform carries a hidden constant offset. Also, mixing Craig’s modified parameters with the classical Denavit–Hartenberg ones: in the modified form the twist and length carry the index $i-1$ and the frame sits at the near end of the link, and the two conventions produce different matrices from the same drawing.

Source. Craig, Introduction to Robotics, 4th ed., §3.4–3.7 (link description, frame attachment, Example 3.3 and the three-link arm); Spong et al., 2nd ed., §3.2 for the classical alternative.

Question 6 of 7: Link frames, DH table and link transforms for a 3-DOF arm

Question 6: Link frames, DH table and link transforms for a 3-DOF arm (20 marks)