Induced Joint Torques and Forces

When a tool contacts the environment, the external tool force and moment must be balanced by internal joint torques and forces.

flowchart LR
    F["Tool wrench<br/>F-tool"] --> JT["Jacobian transpose<br/>J-transpose(q)"] --> TAU["Joint effort<br/>tau"]
    TAU --> FLEX["Joint flexibility<br/>Gamma-inverse"] --> DQ["Joint deflection<br/>delta-q"]
    DQ --> J["Jacobian<br/>J(q)"] --> DU["Tool deflection<br/>delta-u"]

Define the external tool wrench:

$$F^{tool}=\left[\begin{array}{c}{f}^{tool}\\ n^{tool}\end{array}\right]\in {\mathbb{R}}^6$$

where:

• $f^{tool}\in {\mathbb{R}}^3$ is the force at the tool tip
• $n^{tool}\in {\mathbb{R}}^3$ is the moment at the tool tip

Let $\tau \in {\mathbb{R}}^n$ be the joint torque and force vector.

• For a revolute joint, ${\tau }_k$ is the actuator torque about joint axis $k$
• For a prismatic joint, ${\tau }_k$ is the actuator force along joint axis $k$

Equivalent joint torques and forces

For frictionless joints, the Manipulator Jacobian gives:

$$\boxed{\tau =J^T(q)F^{tool}}$$

The Jacobian transpose maps the external tool wrench to the equivalent internal joint torques and forces that maintain static equilibrium.

This relation can also be used in the opposite direction. If joint sensors measure $\tau$, the external tool wrench can be inferred by solving:

$$J^T(q)F^{tool}=\tau$$

This principle supports active compliance. The controller calculates the actuator torques and forces that produce a required contact force and moment during operations such as part insertion and part mating.

Four-axis SCARA example

For the four-axis SCARA robot:

$${\tau }_1=(-a_1{S}_1-a_2{S}_{1-2})f_1^{tool}+(a_1{C}_1+a_2{C}_{1-2})f_2^{tool}+n_3^{tool}$$ $${\tau }_2=a_2{S}_{1-2}{f}_1^{tool}-a_2{C}_{1-2}{f}_2^{tool}-n_3^{tool}$$

Warning

The coefficient of $f_1^{tool}$ in ${\tau }_2$ is $a_2{S}_{1-2}$. This follows from the second column of $J(q)$ in $\tau =J^T(q)F^{tool}$.

$${\tau }_3=-f_3^{tool}$$ $${\tau }_4=-n_3^{tool}$$

The physical interpretation is:

• ${\tau }_3$ is a force because joint 3 is prismatic
• ${\tau }_1$, ${\tau }_2$, and ${\tau }_4$ are torques because their joints are revolute
• Joint 3 alone compensates for the vertical force component
• Joints 1 and 2 together compensate for the horizontal force components
• Joints 1, 2, and 4 compensate for the vertical moment component
• The manipulator structure bears the horizontal moment components

Tool compliance and stiffness

An external tool wrench causes a small tool displacement. The displacement depends on the applied wrench and the stiffness of the arm.

For a small joint displacement $\Delta q_k$, the joint controller produces a restoring torque or force:

$$\Delta {\tau }_k={\gamma }_k\Delta q_k,1\le k\le n$$

where ${\gamma }_k>0$ is the stiffness of joint $k$.

The joint stiffness matrix is:

$$\Gamma =\mathrm{diag}\{{\gamma }_1,{\gamma }_2,\dots ,{\gamma }_n\}$$

For rigid links and an infinitesimal tool deflection $\Delta u$:

$$\boxed{\Delta u=J(q){\Gamma }^{-1}{J}^T(q)F^{tool}}$$

The force-to-deflection sequence is:

$$F^{tool}\stackrel{J^T(q)}{\to }\Delta \tau \stackrel{{\Gamma }^{-1}}{\to }\Delta q\stackrel{J(q)}{\to }\Delta u$$

Define the tool compliance matrix:

$$\boxed{G(q)=J(q){\Gamma }^{-1}{J}^T(q)}$$

Then:

$$\boxed{\Delta u=G(q)F^{tool}}$$

If $G(q)$ is invertible:

$$\boxed{F^{tool}=G^{-1}(q)\Delta u}$$

The matrix $G^{-1}(q)$ is the tool stiffness matrix.

Important

The compliance matrix maps tool wrench to tool deflection. The stiffness matrix maps tool deflection to tool wrench.

Near a joint-space singularity, the determinant of $G(q)$ becomes small, and the components of the tool stiffness matrix become large.

See Manipulator Jacobian of a Four-Axis SCARA Robot (Adept One).