Static joint torques and singular configurations of a two-link arm (20 marks): 22-Mec-B12 Robotics, May 2016, Q7

Question 7: Static joint torques and singular configurations of a two-link 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 $2\times2$ Jacobian of a planar two-link arm, written in the shorthand $s_{1}=\sin\theta_{1}$, $c_{12}=\cos(\theta_{1}+\theta_{2})$, and a required static tip force of 20 units directed along $\hat{X}_{0}$.

Find. (a) the joint torques $\tau_{1}$ and $\tau_{2}$ that hold that force; (b) all configurations at which the Jacobian becomes singular.

Figure 7.1 — the two-link planar arm in a general pose and in its two singular configurations. Both singularities occur when the second link lines up with the first.

Approach. Use the force–torque duality $\tau=J^{T}F$ for part (a), and set the determinant of the square Jacobian to zero for part (b).

Invoke the static force–torque relationship. Equating virtual work done at the joints with virtual work done at the tip, $\tau^{T}\delta\Theta=F^{T}\delta X$ with $\delta X=J\,\delta\Theta$, gives the duality$$\tau=J^{T}\,{}^{0}F.$$It is the transpose, not the inverse: no matrix inversion is needed for statics, which is why this relation holds even at singular configurations where $J^{-1}$ does not exist.
Transpose the Jacobian and apply the given force. With $^{0}F=\begin{bmatrix}20 & 0\end{bmatrix}^{T}$,$$\tau=\begin{bmatrix}-l_{1}s_{1}-l_{2}s_{12} & l_{1}c_{1}+l_{2}c_{12}\\ -l_{2}s_{12} & l_{2}c_{12}\end{bmatrix}\begin{bmatrix}20\\ 0\end{bmatrix},$$so only the first column of $J^{T}$ survives:$$\boxed{\tau_{1}=-20\left(l_{1}\sin\theta_{1}+l_{2}\sin(\theta_{1}+\theta_{2})\right),\qquad \tau_{2}=-20\,l_{2}\sin(\theta_{1}+\theta_{2})}$$
Interpret the result as a moment balance. The quantity $l_{1}s_{1}+l_{2}s_{12}$ is the $y$ coordinate of the tip, and $l_{2}s_{12}$ is the $y$ offset of the tip from joint 2. Each torque is therefore the applied force multiplied by the perpendicular distance from its own joint to the line of action of that force — ordinary statics, recovered automatically by the Jacobian transpose. The negative signs say the joints must resist a force pushing in $+x$, which is the expected sense.
Evaluate a numerical illustration. Take $l_{1}=l_{2}=1$ m with $\theta_{1}=30^\circ$ and $\theta_{2}=60^\circ$, so $\theta_{1}+\theta_{2}=90^\circ$:$$\tau_{1}=-20(0.5+1)=-30.0\ \text{N}\cdot\text{m},\qquad \tau_{2}=-20(1)=-20.0\ \text{N}\cdot\text{m}.$$The tip then sits at $y=1.5$ m and joint 2 at $y=0.5$ m, so the moment arms are 1.5 m and 1.0 m about joints 1 and 2 respectively, and $20\times1.5=30$, $20\times1.0=20$ — the moment-arm reading of step 3 is confirmed.
Form the determinant for part (b). Expanding the $2\times2$ determinant and collecting terms,$$\det J=\left(-l_{1}s_{1}-l_{2}s_{12}\right)\left(l_{2}c_{12}\right)-\left(-l_{2}s_{12}\right)\left(l_{1}c_{1}+l_{2}c_{12}\right),$$in which the two $l_{2}^{2}s_{12}c_{12}$ terms cancel identically, leaving$$\det J=l_{1}l_{2}\left(s_{12}c_{1}-s_{1}c_{12}\right)=l_{1}l_{2}\sin\!\left((\theta_{1}+\theta_{2})-\theta_{1}\right),$$$$\boxed{\det J=l_{1}l_{2}\sin\theta_{2}}$$Remarkably, $\theta_{1}$ has vanished entirely: whether the arm is singular depends only on the elbow angle, never on where the whole arm is pointing.
Identify the singular configurations. Since $l_{1}$ and $l_{2}$ are non-zero link lengths, $\det J=0$ if and only if $\sin\theta_{2}=0$, that is$$\boxed{\theta_{2}=0^\circ\quad\text{or}\quad\theta_{2}=180^\circ}$$At $\theta_{2}=0^\circ$ the arm is fully extended and the tip lies at radius $l_{1}+l_{2}$ on the outer boundary of the workspace — a boundary singularity. At $\theta_{2}=180^\circ$ the arm is folded back on itself at radius $|l_{1}-l_{2}|$, which is an interior singularity (it degenerates to the origin when $l_{1}=l_{2}$).
Describe what is lost at a singularity. In both cases the two links are collinear, the two columns of $J$ become parallel and its rank drops from 2 to 1. The tip can then no longer be moved in the radial direction by any combination of joint rates, however large: it has lost a degree of freedom instantaneously. Dually — and this is the practical significance for a design engineer — the arm can resist an arbitrarily large radial force with zero joint torque, because $J^{T}$ maps that force direction into the null vector. Any control scheme based on $\dot\Theta=J^{-1}\dot X$ will command unbounded joint rates near these poses, which is why trajectories are planned to keep $\theta_{2}$ away from $0^\circ$ and $180^\circ$.

Question 7 — results
Part	Quantity	Result
(a)	$\tau_{1}$	$-20\left(l_{1}\sin\theta_{1}+l_{2}\sin(\theta_{1}+\theta_{2})\right)$
(a)	$\tau_{2}$	$-20\,l_{2}\sin(\theta_{1}+\theta_{2})$
(a)	example $l_{1}=l_{2}=1$ m, $30^\circ/60^\circ$	$\tau_{1}=-30.0$, $\tau_{2}=-20.0\ \text{N}\cdot\text{m}$
(b)	$\det J$	$l_{1}l_{2}\sin\theta_{2}$ (independent of $\theta_{1}$)
(b)	singularities	$\theta_{2}=0^\circ$ (boundary, radius $l_{1}+l_{2}$); $\theta_{2}=180^\circ$ (interior, radius $\|l_{1}-l_{2}\|$)

Topic. Force–torque duality and manipulator singularities.
Key relations. $\tau=J^{T}F$; $\det J=l_{1}l_{2}\sin\theta_{2}$ for the planar two-link arm; singular when the links are collinear.
Why this works. The principle of virtual work makes the same Jacobian serve two purposes: it maps joint rates to tip velocity, and its transpose maps tip force to joint torque. The transpose always exists, so statics is defined everywhere, whereas the velocity inverse problem breaks down exactly where the determinant vanishes. The two facts are the same fact viewed from opposite sides.
Common pitfall. Using $J^{-1}$ or $(J^{T})^{-1}$ for part (a). It is not needed, and reaching for it invites the false conclusion that the torques are undefined at a singularity — whereas in truth they are perfectly finite there and it is the velocity problem that fails. A second pitfall is quoting only $\theta_{2}=0$ and missing the folded-back pose.
Source. Craig, Introduction to Robotics, 4th ed., §5.9 (static forces in manipulators) and §5.8 (singularities); Spong et al., 2nd ed., §4.9 and §4.10.

Question 7 of 7: Static joint torques and singular configurations of a two-link arm (20 marks)

Question 7: Static joint torques and singular configurations of a two-link arm (20 marks)