Question 6 of 7: Jacobian, reachable workspace and dexterous workspace (20 marks)

Question 6: Jacobian, reachable workspace and dexterous workspace (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 forward kinematics of Question 3 and the three stated joint ranges.

Find. (a) the manipulator Jacobian; (b) the three requested workspace projections; (c) a statement about the dexterous workspace.

Approach. Differentiate the tip position from Question 3 to get the linear Jacobian, append the angular rows by inspection of the joint types, then map the joint ranges through the kinematics to obtain the workspace.

1. Write the tip position as a function of the joint variables. From the translation column of $^{0}_{3}T$,$$^{0}P_{e}=\begin{bmatrix}2+q_{1}\\ -(2+q_{3})\sin q_{2}\\ (2+q_{3})\cos q_{2}\end{bmatrix}.$$The Jacobian is nothing more than the derivative of this map, column by column with respect to each joint variable in turn.
2. Differentiate to obtain the linear velocity block. Taking $\partial\,{}^{0}P_{e}/\partial q_{i}$ for $i=1,2,3$ gives$$J_{v}=\begin{bmatrix}1 & 0 & 0\\ 0 & -(2+q_{3})\cos q_{2} & -\sin q_{2}\\ 0 & -(2+q_{3})\sin q_{2} & \cos q_{2}\end{bmatrix}.$$Reading the columns physically: sliding joint 1 moves the tip along $x_{0}$ at unit rate; turning joint 2 carries the tip around a circle at a speed proportional to the current radius; sliding joint 3 drives the tip radially outward.
3. Append the angular velocity block by inspection. Prismatic joints contribute no angular velocity at all, and the only revolute joint turns about $z_{1}=\hat{x}_{0}$, so$$J_{\omega}=\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}.$$Stacking the two blocks gives the full manipulator Jacobian referred to frame $\{0\}$:$$\boxed{^{0}J=\begin{bmatrix}1 & 0 & 0\\0 & -(2+q_{3})\cos q_{2} & -\sin q_{2}\\0 & -(2+q_{3})\sin q_{2} & \cos q_{2}\\0 & 1 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}\quad(6\times 3)}$$It is $6\times 3$ rather than square because the arm has only three joints; the two identically-zero angular rows are the algebraic statement that this arm can never generate angular velocity about $y_{0}$ or $z_{0}$.
4. Confirm the Jacobian never degenerates within the stated limits. The determinant of the linear block is$$\det J_{v}=-(2+q_{3}),$$which for $0\lt q_{3}\lt 1$ lies between $-3$ and $-2$ and so never vanishes. The arm therefore has no interior singularity in position: it can produce a tip velocity in any Cartesian direction from every reachable point. That is a direct consequence of the prismatic joint never being able to retract onto its own axis within the given range.
5. Map the joint ranges into Cartesian space (part b). The three limits translate directly:$$2\lt x_{e}\lt 3,\qquad \rho=\sqrt{y_{e}^{2}+z_{e}^{2}}\in(2,3),\qquad q_{2}\ \text{sweeping a full turn}.$$A full revolution of $q_{2}$ at every radius in $(2,3)$ fills an annulus in the $y_{0}$–$z_{0}$ plane, and sweeping that annulus along $x$ from 2 to 3 gives the reachable workspace as a hollow cylinder (a thick-walled tube of length 1, inner radius 2, outer radius 3) with its axis on $x_{0}$.
6. Project that solid onto the three requested planes. The annulus has area $\pi(3^{2}-2^{2})=15.708$ and the tube encloses the same figure as its volume because its length is unity. The projections are: on $x_{0}$–$y_{0}$, a rectangular band $2\lt x\lt 3$, $|y|\lt 3$; on $y_{0}$–$z_{0}$, the annulus $2\lt\rho\lt 3$; on $x_{0}$–$z_{0}$, the same band $2\lt x\lt 3$, $|z|\lt 3$. Note the two side views show solid rectangles rather than pairs of strips, because points at every $|y|$ from 0 to 3 do exist in the annulus (they simply sit at different $z$), and a projection cannot preserve the hole.

Part (c) — the dexterous workspace. The dexterous workspace is the set of points the end effector can reach with every possible orientation. This arm has only three joints, of which just one is revolute, and Question 3 showed that its orientation is always $^{0}_{3}R=R_{x}(q_{2})$ — a single-parameter family of rotations about one fixed axis. Attaining an arbitrary orientation requires three independent rotational degrees of freedom spanning $SO(3)$, and the two identically-zero rows of $J_{\omega}$ state plainly that two of the three are missing.

The dexterous workspace of this manipulator without the wrist is therefore empty: there is no point at all that the end point can reach in all orientations. This is exactly why the spherical wrist of Question 5 is bolted on — it supplies the missing three rotational degrees of freedom, and because its link lengths are taken as zero it adds orientation freedom without altering the reachable workspace computed above.