Jacobian of a planar 3R arm by velocity propagation: 22-Mec-B12 Robotics, December 2019, Q7

Question 7: Jacobian of a planar 3R arm by velocity propagation (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 planar three-revolute arm of Figure 4, with all three joint axes parallel and normal to the plane of motion.

Given data
Quantity	Symbol	Value
Link 1 length	$l_{1}$	base joint to joint 2
Link 2 length	$l_{2}$	joint 2 to joint 3
Link 3 (hand) length	$l_{3}$	joint 3 to the tip
Joint variables	$\theta_{1},\theta_{2},\theta_{3}$	all revolute, axes parallel to $\hat Z$
Frame {4}	—	origin at the tip, orientation identical to {3}, i.e. $^{3}P_{4ORG}=[l_{3},0,0]^{T}$

Find. The $3\times3$ Jacobian $^{4}J(\Theta)$ that maps the joint rates to the tip twist expressed in frame {4}.

[Figure not reproduced: Figure 4 redrawn. Each joint angle is measured from the extension of the previous link, so the DH table is $(0,0,0,\theta_1),(0,l_1,0,\theta_2),(0,l_2,0,\theta_3)$ and frame {4} is a pure $l_3$ translation from {3}. See the official exam paper.]

Approach. Start from a stationary base and propagate angular and linear velocity outward link by link, rotating each result into the next frame as it goes, then read the Jacobian off the coefficients of the joint rates.

State the propagation rules. For a revolute joint $i+1$, $$^{i+1}\omega_{i+1}={}^{i+1}_{\;\;i}R\,{}^{i}\omega_{i}+\dot\theta_{i+1}{}^{i+1}\hat Z_{i+1},\qquad{}^{i+1}v_{i+1}={}^{i+1}_{\;\;i}R\left({}^{i}v_{i}+{}^{i}\omega_{i}\times{}^{i}P_{i+1}\right).$$ All frames here are planar, so $^{i+1}_{\;\;i}R=R_{Z}(-\theta_{i+1})$ and every $\omega$ is along $\hat Z$.
Link 1. With $^{0}\omega_{0}=0$ and $^{0}v_{0}=0$, $$^{1}\omega_{1}=\begin{bmatrix}0\\0\\ \dot\theta_{1}\end{bmatrix},\qquad{}^{1}v_{1}=0.$$
Link 2. Using $^{1}P_{2}=[l_{1},0,0]^{T}$, $^{1}\omega_{1}\times{}^{1}P_{2}=[0,\,l_{1}\dot\theta_{1},\,0]^{T}$, and rotating into {2} by $R_{Z}(-\theta_{2})$, $$^{2}\omega_{2}=\begin{bmatrix}0\\0\\ \dot\theta_{1}+\dot\theta_{2}\end{bmatrix},\qquad{}^{2}v_{2}=\begin{bmatrix}l_{1}s_{2}\dot\theta_{1}\\ l_{1}c_{2}\dot\theta_{1}\\ 0\end{bmatrix}.$$
Link 3. Using $^{2}P_{3}=[l_{2},0,0]^{T}$ so that $^{2}\omega_{2}\times{}^{2}P_{3}=[0,\,l_{2}(\dot\theta_{1}+\dot\theta_{2}),\,0]^{T}$, and rotating into {3} by $R_{Z}(-\theta_{3})$, $$^{3}v_{3}=\begin{bmatrix}l_{1}s_{23}\dot\theta_{1}+l_{2}s_{3}(\dot\theta_{1}+\dot\theta_{2})\\ l_{1}c_{23}\dot\theta_{1}+l_{2}c_{3}(\dot\theta_{1}+\dot\theta_{2})\\ 0\end{bmatrix},\qquad{}^{3}\omega_{3}=\begin{bmatrix}0\\0\\ \dot\theta_{1}+\dot\theta_{2}+\dot\theta_{3}\end{bmatrix},$$ where $s_{23}=\sin(\theta_{2}+\theta_{3})$ and $c_{23}=\cos(\theta_{2}+\theta_{3})$ arise from the product-to-sum collapse $c_{3}s_{2}+s_{3}c_{2}=s_{23}$.
Step out to the tip. Frame {4} shares the orientation of {3}, so no rotation is involved and only the moment arm changes: with $^{3}P_{4}=[l_{3},0,0]^{T}$, $$^{4}\omega_{4}={}^{3}\omega_{3},\qquad{}^{4}v_{4}={}^{3}v_{3}+{}^{3}\omega_{3}\times{}^{3}P_{4}={}^{3}v_{3}+\begin{bmatrix}0\\ l_{3}(\dot\theta_{1}+\dot\theta_{2}+\dot\theta_{3})\\ 0\end{bmatrix}.$$
Collect the coefficients. Writing $[{}^{4}v_{x},\,{}^{4}v_{y},\,{}^{4}\omega_{z}]^{T}={}^{4}J\,\dot\Theta$, $$\boxed{^{4}J(\Theta)=\begin{bmatrix}l_{1}s_{23}+l_{2}s_{3} & l_{2}s_{3} & 0\\ l_{1}c_{23}+l_{2}c_{3}+l_{3} & l_{2}c_{3}+l_{3} & l_{3}\\ 1 & 1 & 1\end{bmatrix}}$$ The third column, $[0,\,l_{3},\,1]^{T}$, is the pure wrist contribution: joint 3 alone swings the tip through a moment arm $l_{3}$ perpendicular to the hand.
Locate the singularities. Expanding the determinant and simplifying, $$\det{}^{4}J=l_{1}l_{2}\sin\theta_{2},$$ independent of both $\theta_{1}$ and $\theta_{3}$. The arm is therefore singular exactly at $\theta_{2}=0$ (fully outstretched) and $\theta_{2}=180^\circ$ (folded back on itself), each a one-parameter family of configurations rather than an isolated pose. At those poses the tip can no longer move along the line joining joints 1 and 3.
Numerical check. With $l_{1}=1.0$, $l_{2}=0.8$, $l_{3}=0.5$ and $\Theta=(0.30,\,0.85,\,-0.55)\ \text{rad}$, $$^{4}J=\begin{bmatrix}-0.1226 & -0.4182 & 0\\ 2.1374 & 1.1820 & 0.5\\ 1 & 1 & 1\end{bmatrix},\qquad \det{}^{4}J=0.6010=l_{1}l_{2}\sin(0.85).$$ Central-differencing the forward kinematics and rotating the result into {4} reproduces the first two rows to seven decimals, which is the independent confirmation that the propagation was carried out correctly.

Two practical uses follow immediately. Inverting the relation, $\dot\Theta={}^{4}J^{-1}\,{}^{4}\mathcal{V}$, gives the joint rates needed to command a wanted tip twist, and the static force map $\tau={}^{4}J^{T}\,{}^{4}\mathcal{F}$ gives the joint torques that balance a load at the hand. The transpose relation needs no inversion and therefore stays valid even at the singular poses just identified.

Final results
Quantity	Symbol	Value
Jacobian referred to frame {4}	$^{4}J$	$\begin{bmatrix}l_{1}s_{23}+l_{2}s_{3} & l_{2}s_{3} & 0\\ l_{1}c_{23}+l_{2}c_{3}+l_{3} & l_{2}c_{3}+l_{3} & l_{3}\\ 1 & 1 & 1\end{bmatrix}$
Tip linear velocity, $x$ in {4}	$^{4}v_{x}$	$(l_{1}s_{23}+l_{2}s_{3})\dot\theta_{1}+l_{2}s_{3}\dot\theta_{2}$
Tip linear velocity, $y$ in {4}	$^{4}v_{y}$	$(l_{1}c_{23}+l_{2}c_{3}+l_{3})\dot\theta_{1}+(l_{2}c_{3}+l_{3})\dot\theta_{2}+l_{3}\dot\theta_{3}$
Tip angular velocity	$^{4}\omega_{z}$	$\dot\theta_{1}+\dot\theta_{2}+\dot\theta_{3}$
Determinant	$\det{}^{4}J$	$l_{1}l_{2}\sin\theta_{2}$
Singular configurations	$\theta_{2}$	$0$ and $180^\circ$, for any $\theta_{1},\theta_{3}$
Numerical check ($l=1.0,\,0.8,\,0.5$; $\Theta=0.30,\,0.85,\,-0.55$ rad)	$\det{}^{4}J$	$0.6010$

Concept & Theory

Topic. The manipulator Jacobian obtained by outward velocity propagation.

Key relations. $^{i+1}\omega_{i+1}={}^{i+1}_{\;\;i}R\,{}^{i}\omega_{i}+\dot\theta_{i+1}\hat Z_{i+1}$; $^{i+1}v_{i+1}={}^{i+1}_{\;\;i}R({}^{i}v_{i}+{}^{i}\omega_{i}\times{}^{i}P_{i+1})$; $\mathcal{V}=J(\Theta)\dot\Theta$; $\tau=J^{T}\mathcal{F}$; a singularity is $\det J=0$.

Why this works. Velocity propagation is the mechanical analogue of the chain rule: each link adds its own joint rate to the angular velocity it inherits, and adds the cross-product of that inherited angular velocity with its own length to the linear velocity. Rotating into the next frame at every step keeps the entries as short trigonometric sums instead of the long expansions that a differentiate-the-forward-kinematics approach produces. Expressing the answer in a frame at the tip is what makes $\det J$ collapse to $l_{1}l_{2}\sin\theta_{2}$, because the reference frame rotates with the hand and the $\theta_{1}$ dependence drops out.

Common pitfall. Forgetting to rotate the inherited velocity into the new frame, which leaves the columns expressed in a mixture of frames. Also, treating a singularity as an isolated point: here $\det J=0$ fixes only $\theta_{2}$, so the singular set is a two-dimensional surface in joint space and a trajectory can drift into it without ever passing through an obviously extreme pose.

Source. Craig, Introduction to Robotics, 4th ed., §5.3–5.8 and Example 5.3 (velocity propagation for a three-link planar arm; Jacobians in different frames; singularities); Spong et al., 2nd ed., ch. 4.

Question 7 of 7: Jacobian of a planar 3R arm by velocity propagation

Question 7: Jacobian of a planar 3R arm by velocity propagation (20 marks)