Tool-configuration Jacobian Matrix

The tool-configuration Jacobian matrix gives the instantaneous relation between joint velocity and tool-configuration velocity.

flowchart LR
    DK["Direct kinematics<br/>x = w(q)"] --> D["Differentiate"] --> V["Tool Jacobian<br/>V(q)"]
    QD["Joint velocity<br/>q-dot"] --> V --> XD["Tool velocity<br/>x-dot"]

From the direct kinematics relation:

$$x=w(q)$$

differentiate both sides with respect to time:

$$\boxed{\dot{x}=V(q)\dot{q}}$$

where:

• $q\in {\mathbb{R}}^n$ is the joint-variable vector
• $x\in {\mathbb{R}}^6$ is the tool-configuration vector
• $\dot{q}$ is the joint velocity
• $\dot{x}$ is the tool-configuration velocity
• $V(q)\in {\mathbb{R}}^{6\times n}$ is the tool-configuration Jacobian matrix

Each entry of $V(q)$ is a partial derivative:

$$\boxed{V_{kj}(q)=\frac{\partial w_k(q)}{\partial q_j}},1\le k\le 6,1\le j\le n$$

Select the required Jacobian

The output selected by the problem determines which function must be differentiated.

From the complete tool-configuration function

When the complete tool-configuration function is given directly:

$$x=w(q)$$

differentiate every component of $w(q)$ with respect to every joint variable:

$$\boxed{V(q)=\frac{\partial w(q)}{\partial q}}$$

If $w(q)\in {\mathbb{R}}^6$ and $q\in {\mathbb{R}}^n$, then $V(q)\in {\mathbb{R}}^{6\times n}$. This matrix includes both the position and orientation parts of the tool configuration.

From a homogeneous transformation matrix

A homogeneous transformation from the base frame to the TCP has the form:

$$T_{\mathrm{base}}^{\mathrm{TCP}}(q)=\left[\begin{array}{cc}{R}_{\mathrm{base}}^{\mathrm{TCP}}(q)& p_{\mathrm{base}}^{\mathrm{TCP}}(q)\\ 0& 1\end{array}\right]$$

The first three entries of the fourth column give the TCP position:

$$p(q)=p_{\mathrm{base}}^{\mathrm{TCP}}(q)$$

When the problem uses position only, differentiate this position vector:

$$\boxed{A(q)=\frac{\partial p(q)}{\partial q}},\boxed{\dot{p}=A(q)\dot{q}}$$

If $p(q)\in {\mathbb{R}}^3$ and $q\in {\mathbb{R}}^n$, then $A(q)$ has size $3\times n$. It is the linear block of the Manipulator Jacobian and the first three rows of the complete tool-configuration Jacobian $V(q)$.

Important

Some position-only problems denote $\partial p(q)/\partial q$ by $V(q)$. In the precise lecture notation, this matrix is $A(q)$. Reserve $V(q)$ for the complete $6\times n$ tool-configuration Jacobian $\partial w(q)/\partial q$.

If the problem also requires orientation, the fourth column alone is insufficient. Form the complete $w(q)$ from the position and the required orientation representation, then differentiate the complete $w(q)$.

Important

The joint vector $q$ and position vector $p(q)$ are different. The joint vector is the input, and the position vector is the output of direct kinematics. Differentiate the selected output with respect to the components of $q$.

Planar RR position example

For a two-link planar RR robot:

$$q=\left[\begin{array}{c}{\theta }_1\\ {\theta }_2\end{array}\right]$$

Extract the fourth column of $T_{\mathrm{base}}^{\mathrm{TCP}}(q)$:

$$p(q)=\left[\begin{array}{c}{a}_1\cos {\theta }_1+a_2\cos ({\theta }_1+{\theta }_2)\\ a_1\sin {\theta }_1+a_2\sin ({\theta }_1+{\theta }_2)\\ 0\end{array}\right]$$

Differentiate $p(q)$ with respect to $q$:

$$A(q)=\left[\begin{array}{cc}-a_1\sin {\theta }_1-a_2\sin ({\theta }_1+{\theta }_2)& -a_2\sin ({\theta }_1+{\theta }_2)\\ a_1\cos {\theta }_1+a_2\cos ({\theta }_1+{\theta }_2)& a_2\cos ({\theta }_1+{\theta }_2)\\ 0& 0\end{array}\right]$$

The third row is zero because the robot moves in the $x$-$y$ plane. Use Joint-space Singularities to select and apply the correct rank or determinant test.

For a fixed operating point $q$, $V(q)$ is a linear transformation:

$$\dot{q}\stackrel{V(q)}{\to }\dot{x}$$

Important

The direct kinematics function $w(q)$ maps the joint configuration $q$ to the complete tool configuration $x$. The Jacobian $V(q)$ maps instantaneous joint velocity $\dot{q}$ to instantaneous tool-configuration velocity $\dot{x}$ at the current operating point $q$.

Four-axis SCARA robot

For the four-axis SCARA robot, define:

$$C_1=\cos q_1,S_1=\sin q_1$$ $$C_{1-2}=\cos (q_1-q_2),S_{1-2}=\sin (q_1-q_2)$$

The tool-configuration function is:

$$w(q)=\left[\begin{array}{c}{a}_1{C}_1+a_2{C}_{1-2}\\ a_1{S}_1+a_2{S}_{1-2}\\ d_1-q_3-d_4\\ 0\\ 0\\ -\exp (q_4/\pi )\end{array}\right]$$

The first three components give the tool-tip position. The last three components give the scaled approach vector. The fourth joint variable $q_4$ is the tool roll angle.

Differentiate $w(q)$ with respect to $q$:

$$V(q)=\left[\begin{array}{cccc}-a_1{S}_1-a_2{S}_{1-2}& a_2{S}_{1-2}& 0& 0\\ a_1{C}_1+a_2{C}_{1-2}& -a_2{C}_{1-2}& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& -\frac{\exp (q_4/\pi )}{\pi }\end{array}\right]$$

The zero pattern gives these direct conclusions:

• The tool-tip $x$ and $y$ coordinates depend only on $q_1$ and $q_2$
• The tool-tip $z$ coordinate depends only on $q_3$
• The tool orientation depends only on $q_4$

This sparse matrix shows the simple kinematic structure of the SCARA robot.

Related concepts

• Use Joint-space Singularities to identify configurations where $V(q)$ loses rank
• Use Generalised Inverses when $V(q)$ is rectangular
• Use Resolved-motion Rate Control to calculate $\dot{q}$ from a required $\dot{x}$
• See Inverse Kinematics for the position-level relation $x=w(q)$