A Mathematical Review

A.5 Spatial velocities and forces

Given two 3D coordinate frames A and B, the spatial velocity, or twist, \hat{\bf v}_{BA} of B with respect to A is given by the 6D composition of the translational velocity {\bf v}_{BA} of the origin of B with respect to A and the angular velocity \boldsymbol{\omega}_{BA}:

\hat{\bf v}_{BA}\equiv\left(\begin{matrix}{\bf v}_{BA}\\
\boldsymbol{\omega}_{BA}\end{matrix}\right). (A.24)

Similarly, the spatial force, or wrench, \hat{\bf f} acting on a frame B is given by the 6D composition of the translational force {\bf f}_{B} acting on the frame’s origin and the moment \boldsymbol{\tau}, or torque, acting through the frame’s origin:

\hat{\bf f}_{B}\equiv\left(\begin{matrix}{\bf f}_{B}\\
\boldsymbol{\tau}_{B}\end{matrix}\right). (A.25)
Figure A.9: Two frames A and B rigidly connected within a rigid body and moving with respect to a third frame C.

If we have two frames A and B rigidly connected within a rigid body (Figure A.9), and we know the spatial velocity \hat{\bf v}_{BC} of B with respect to some third frame C, we may wish to know the spatial velocity \hat{\bf v}_{AC} of A with respect to C. The angular velocity components are the same, but the translational velocity components are coupled by the angular velocity and the offset {\bf p}_{BA} between A and B, so that

{\bf v}_{AC}={\bf v}_{BC}+{\bf p}_{BA}\times\boldsymbol{\omega}_{BC}.

\hat{\bf v}_{AC} is hence related to \hat{\bf v}_{BC} via

\left(\begin{matrix}{\bf v}_{AC}\\
\boldsymbol{\omega}_{AC}\end{matrix}\right)=\left(\begin{matrix}{\bf I}&[{\bf p%
0&{\bf I}\end{matrix}\right)\,\left(\begin{matrix}{\bf v}_{BC}\\

where [{\bf p}_{BA}] is defined by (A.22).

The above equation assumes that all quantities are expressed with respect to the same coordinate frame. If we instead consider \hat{\bf v}_{AC} and \hat{\bf v}_{BC} to be represented in frames A and B, respectively, then we can show that

{}^{A}\hat{\bf v}_{AC}={\bf X}_{BA}\,{}^{B}\hat{\bf v}_{BC}, (A.26)


{\bf X}_{BA}\equiv\left(\begin{matrix}{\bf R}_{BA}&[{\bf p}_{BA}]{\bf R}_{BA}%
0&{\bf R}_{BA}\end{matrix}\right). (A.27)

The transform {\bf X}_{BA} is easily formed from the components of the rigid transform {\bf T}_{BA} relating B to A.

The spatial forces \hat{\bf f}_{A} and \hat{\bf f}_{B} acting on frames A and B within a rigid body are related in a similar way, only with spatial forces, it is the moment that is coupled through the moment arm created by {\bf p}_{BA}, so that

\boldsymbol{\tau}_{A}=\boldsymbol{\tau}_{B}+{\bf p}_{BA}\times{\bf f}_{B}.

If we again assume that \hat{\bf f}_{A} and \hat{\bf f}_{B} are expressed in frames A and B, we can show that

{}^{A}\hat{\bf f}_{A}={\bf X}^{*}_{BA}\,{}^{B}\hat{\bf f}_{B}, (A.28)


{\bf X}^{*}_{BA}\equiv\left(\begin{matrix}{\bf R}_{BA}&0\\
~{}[{\bf p}_{BA}]{\bf R}_{BA}&{\bf R}_{BA}\end{matrix}\right). (A.29)