Question 3 of 7: Axis assignment, DH table and forward kinematics (20 marks)

Question 3: Axis assignment, DH table and forward kinematics (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 arm of Figure 1 read off the source drawing: a prismatic joint sliding along the horizontal arm, a revolute joint turning about that same horizontal axis, and a second prismatic joint sliding perpendicular to it, with the printed link lengths 1, 1 and 2.

Find. (a) a consistent set of frame assignments; (b) the four DH parameters for each of the three links; (c) the forward-kinematic transform $^{0}_{3}T$.

[Figure not reproduced: Figure 3.1 — part (a). The manipulator redrawn in its rest pose with the axis assignments marked. Joints 1 and 2 share the horizontal arm axis; joint 3 slides perpendicular to it and its direction turns with q₂. See the official exam paper.]

Approach. Place a $z$ axis along every joint axis, read the four DH parameters off the resulting geometry, then multiply the three link transforms.

1. Assign the base and end frames (part a). Take the base triad of Figure 1 as $x_{0}$ along the horizontal arm, $z_{0}$ vertically up and $y_{0}$ completing the right-handed set, which makes the drawn end frame $\{e\}$ (with $z_{e}$ up and $x_{e}$ along the arm) parallel to $\{0\}$ in the rest pose. This is the assignment shown in Figure 3.1 and it is the one used for every stated result below.
2. Place a $z$ axis on each joint axis, as the DH convention demands. Joint 1 slides along the arm and joint 2 turns about the arm, so those two joint axes are collinear; joint 3 slides perpendicular to them. The DH $z$ axes are therefore$$z_{0}\parallel z_{1}\parallel \hat{x}_{0}\ \text{(the arm)},\qquad z_{2}\perp z_{1},\ \text{vertical at rest}.$$Note that the DH base frame necessarily has its $z$ along the arm, which is a relabelling of the drawn triad, not a different physical frame — a constant change of basis relates the two, and it is applied once at the end.
3. Read the four parameters for link 1. Because $z_{0}$ and $z_{1}$ are collinear there is no common-normal offset and no twist, so $a_{1}=0$ and $\alpha_{1}=0$. The joint is prismatic, so $\theta_{1}=0$ is constant and the variable is the offset along $z_{0}$, which comprises the fixed 1 unit from the base to joint 2 plus the stroke:$$d_{1}=1+q_{1}.$$
4. Read the four parameters for link 2. The revolute joint turns about $z_{1}$, and the axes $z_{1}$ and $z_{2}$ intersect at joint 3 and are perpendicular, giving $a_{2}=0$ and $\alpha_{2}=+90^\circ$. The 1 unit from joint 2 to joint 3 lies along $z_{1}$, so $d_{2}=1$. Aligning $x_{1}$ with $x_{2}$ in the rest pose costs a constant quarter turn, hence$$\theta_{2}=90^\circ+q_{2}.$$That constant offset is a bookkeeping consequence of the axis assignment, not a physical feature of the arm; a different but equally valid assignment moves it elsewhere.
5. Read the four parameters for link 3. The second prismatic joint slides along $z_{2}$ and carries the 2-unit link out to the end point, so $a_{3}=0$, $\alpha_{3}=0$ and$$d_{3}=2+q_{3},$$with $\theta_{3}=90^\circ$ constant, chosen so that $x_{3}$ lands on the $x_{e}$ drawn in Figure 1. Collecting all twelve entries gives the DH table.

6. Multiply the link transforms and express the result in the drawn base frame (part c). Forming $A_{1}A_{2}A_{3}$ and rewriting in the $(x_{0}\ \text{along the arm},\ z_{0}\ \text{up})$ frame of Figure 3.1 gives the forward kinematics$$\boxed{^{0}_{3}T=\begin{bmatrix}1 & 0 & 0 & 2+q_{1}\\0 & \cos q_{2} & -\sin q_{2} & -(2+q_{3})\sin q_{2}\\0 & \sin q_{2} & \cos q_{2} & (2+q_{3})\cos q_{2}\\0 & 0 & 0 & 1\end{bmatrix}}$$The structure is worth reading aloud: the orientation is a pure rotation $R_{x}(q_{2})$, the $x$ coordinate of the tip is the two fixed unit links plus the first stroke, and the remaining two coordinates are the second link swung round a circle of radius $2+q_{3}$.
7. Check the transform at two poses that can be read straight off the drawing. At the rest pose $q=(0,0,0)$ the matrix reduces to the identity rotation with translation $(2,0,2)$ — two units along the arm and two units up, exactly as Figure 1 shows. Turning only the revolute joint to $q_{2}=90^\circ$ moves the tip to $(2,-2,0)$, i.e. the vertical link has swung down into the horizontal plane, which is the physically correct response to a right-handed rotation about $+x_{0}$. Both checks are geometric rather than algebraic, so they would catch a sign error that re-deriving the same algebra would not.