Wrenches

Wrenches

A wrench is the force-side dual of a twist: a $6\times 1$ generalized force that stacks a linear force $f\in {\mathbb{R}}^3$ on top of a moment (torque) $\tau \in {\mathbb{R}}^3$, both acting at the origin of a chosen reference frame:
${\mathcal{F}}_a=\left[\begin{array}{c}{f}_a\\ {\tau }_a\end{array}\right]\in {\mathbb{R}}^6.$
The subscript indicates which frame the quantities are expressed in. A physical force applied at an arbitrary point on a rigid body produces both a force and a torque at the frame origin, so any external interaction with a rigid body is represented as a wrench at some chosen frame.

Wrenches pair with twists to produce power — the instantaneous work done per unit time:
$P={\mathcal{F}}_a^T{V}_a=f_a^T{v}_a+{\tau }_a^T{\omega }_a.$
Power is a scalar and therefore invariant under change of frame, but the individual wrench and twist components transform, so both must be expressed in the same frame for the formula to be correct.

For a robot manipulator, the key relationship between task-space wrenches and joint-space torques is the force-Jacobian duality: if the end-effector experiences a wrench $\mathcal{F}$, the joint torques required to resist (or generate) it are
${\rm Y}=J^T(\theta )\mathcal{F},$
where $J(\theta )$ is the robot Jacobian in the same frame as $\mathcal{F}$. This identity — the transpose of the velocity mapping $V=J\dot{\theta }$ — is the foundation of force control, impedance control, and hand-guided teaching.

Frame Consistency adjoint of $SE(3)$, but the transformation law is the transpose-inverse of the twist law: ${\mathcal{F}}_b={\mathrm{Ad}}_{g_{ba}}^T{\mathcal{F}}_a$. Mixing frames silently — e.g., using a spatial Jacobian with a body-frame wrench — produces wrong torques. Always verify that $J$ and $\mathcal{F}$ are expressed in the same frame.

Wrenches and twists transform between frames via the

Intuition

A force applied to a rigid body has two effects: it pushes the body (linear force) and it twists the body about some axis (moment). A wrench records both together. Viewing a wrench as the dual of a twist is the cleanest way to understand why: twists encode “linear and angular velocity together,” wrenches encode “linear force and torque together,” and the two pair through a scalar (power or work) that does not depend on how the observer labels their axes.

The transpose Jacobian identity ${\rm Y}=J^T\mathcal{F}$ follows from equating power computed in task space with power computed in joint space. A virtual displacement $\delta V=J\delta \theta$ in task space corresponds to virtual work ${\mathcal{F}}^T\delta V={\mathcal{F}}^{T}J\delta \theta$, which must equal the joint-space virtual work ${{\rm Y}}^T\delta \theta$. Matching terms gives ${\rm Y}=J^T\mathcal{F}$. This is the same principle that governs the static equilibrium of any mechanism — forces and torques transform through the kinematic chain via the transpose of the velocity map.

Properties

Structure

(Force-moment pair) A wrench is a linear force stacked with an angular torque, both at a chosen origin.

Writing ${\mathcal{F}}_a=(f_a,{\tau }_a)$ makes explicit that the force acts through the origin of frame $\{A\}$ and the moment acts about the origin of $\{A\}$. Any real-world contact force applied at a point $p\ne o_a$ produces an equivalent wrench at $o_a$ with $f=f_{\text{applied}}$ and $\tau =p\times f_{\text{applied}}+{\tau }_{\text{applied}}$.

(Dual to twists) Wrenches live in the dual space of twists.

Twists $V\in \mathfrak{se}(3)$ describe infinitesimal motion; wrenches $\mathcal{F}\in \mathfrak{se}(3{)}^{\ast }$ describe generalized force. The pairing ${\mathcal{F}}^{T}V$ produces power, matching units of energy per time. This duality is why the transpose (not the inverse) of the Jacobian maps task-space forces to joint-space torques.

Power and Work

(Power) $P={\mathcal{F}}_a^T{V}_a=f_a^T{v}_a+{\tau }_a^T{\omega }_a$, frame-independent in magnitude.

Power is the sum of linear power $f\cdot v$ and rotational power $\tau \cdot \omega$. Both terms must use the same frame; mixing body $\tau$ with spatial $\omega$ produces nonsense. Power is positive when the wrench does positive work on the body (the body absorbs energy) and negative when the body does work on the environment.

(Virtual work) A virtual displacement $\delta V$ produces virtual work $\delta W={\mathcal{F}}^T\delta V$.

Virtual work is the static analogue of power: it treats $\delta V$ as a hypothetical infinitesimal displacement rather than an actual velocity. Setting total virtual work to zero for all admissible $\delta V$ gives the principle of virtual work, a convenient alternative to Newton’s laws for constrained systems.

Frame Transformation

(Wrench transform) ${\mathcal{F}}_b={\mathrm{Ad}}_{g_{ba}}^T{\mathcal{F}}_a$, the transpose-inverse of the twist transform.

Twists transform as $V_b={\mathrm{Ad}}_{g_{ba}}{V}_a$; wrenches transform by the transpose to preserve the power pairing: ${\mathcal{F}}_b^T{V}_b={\mathcal{F}}_a^T({\mathrm{Ad}}_{g_{ba}}^T{\mathrm{Ad}}_{g_{ba}})V_a={\mathcal{F}}_a^T{V}_a$ iff the adjoint on the right is the inverse of the adjoint on the left. This is automatic for elements of $SE(3)$ since ${\mathrm{Ad}}_{g^{-1}}={\mathrm{Ad}}_g^{-1}$.

(Body vs spatial) A body-frame wrench ${\mathcal{F}}_b$ acts “at the tool”; a spatial-frame wrench ${\mathcal{F}}_s$ acts “at the world origin.”

For contact forces at the end-effector, body-frame wrenches are natural (a force sensor mounted at the tool reads body-frame values). For gravity-compensation analyses, spatial wrenches are natural (gravity acts in the world frame). Converting between them uses the adjoint of the current end-effector pose.

Force-Jacobian Duality

(Transpose Jacobian) Joint torques required to produce an end-effector wrench are ${\rm Y}=J^T(\theta )\mathcal{F}$.

This is the static counterpart of the velocity map $V=J\dot{\theta }$. When the manipulator is in static equilibrium (no acceleration) and the end-effector experiences external wrench $\mathcal{F}$, the joint actuators must supply ${\rm Y}=J^T\mathcal{F}$ to hold position. Gravity compensation is a special case: the wrench is gravitational, and ${{\rm Y}}_g=J^T{\mathcal{F}}_g$ gives the torques needed to hold the arm up.

(Force control) Commanding joint torques ${{\rm Y}}_{\text{cmd}}=J^T{\mathcal{F}}_{\text{des}}$ produces the desired end-effector wrench in contact.

This is the basis of direct-force control: instead of servoing position, command the torques that would produce the desired contact force via the transpose Jacobian. Accuracy is limited by friction, Jacobian estimation errors, and unmodeled dynamics, so practical implementations combine force feedback with torque feedforward.

(Singularities) At a kinematic singularity, $J^T$ has nontrivial null space — some end-effector wrenches produce zero joint torque.

When $J$ loses rank (see singularity analysis), its transpose has wrenches in its null space. Physically, external forces applied in those directions are absorbed by the mechanism as constraint forces (the mechanism resists them passively) and produce no torque at the actuators. A fully extended arm, for instance, resists axial pushes through its link structure without needing joint torque.

Examples

Wrench from an Off-Origin Force

A force $f=(0,0,-10)$ N is applied at point $p=(0.3,0,0)$ m in frame $\{A\}$. What is the equivalent wrench at the origin of $\{A\}$?

Solution $f_a=(0,0,-10)$ N (unchanged — translation doesn't affect force).

Force component:

Moment component: ${\tau }_a=p\times f=(0.3,0,0)\times (0,0,-10)=(0,3,0)$ N·m.

Wrench: ${\mathcal{F}}_a=(0,0,-10,0,3,0{)}^T$.

A downward force offset from the origin creates a moment about the $y$-axis at the origin — exactly what happens when you hold a weight in an extended hand and feel the torque at your shoulder.

Joint Torques to Hold a Payload

A 2-link planar manipulator with link lengths ${\ell }_1={\ell }_2=1$ m is in configuration $\theta =(\pi /2,0)$ (arm pointing straight up). A 10 N downward force is applied at the end-effector. What joint torques hold the arm in place?

Solution $(0,2)$ m.

End-effector position:

Planar Jacobian at $\theta$: $J=\left[\begin{array}{cc}-2& -1\\ 0& 0\end{array}\right]$ (the first row is $-{\ell }_1\sin {\theta }_1-{\ell }_2\sin ({\theta }_1+{\theta }_2)$, etc.; at $\theta =(\pi /2,0)$ this evaluates as shown).

Applied wrench (planar 2D force): $\mathcal{F}=(0,-10{)}^T$ N.

Joint torques: ${\rm Y}=J^T\mathcal{F}=\left[\begin{array}{cc}-2& 0\\ -1& 0\end{array}\right]\left[\begin{array}{c}0\\ -10\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$.

Zero torque required — the force is along the arm’s axial direction, in the null space of $J^T$. This is a textbook singularity: at full extension, pushing straight back along the arm axis is supported passively by the mechanism.

See Also

• 5.1. Rigid Body Velocity — the velocity dual of a wrench
• 5.2. Robot Jacobians — the matrix whose transpose maps wrenches to joint torques
• 5.3. Singularities and Manipulability — configurations where $J^T$ has nontrivial null space
• 6.3. Manipulator Equations of Motion — the dynamic equation where ${\rm Y}$ is the applied joint torque