Relative Velocity of Two Frames

Introduction

Suppose there is an inertial coordinate system $\mathcal{I}$ and two moving coordinate systems, one is $\mathcal{A}$ and the other is $\mathcal{G}$. Suppose we know that the angular velocity and linear velocity of these two moving coordinate systems relative to the inertial coordinate system, that are $^\mathcal{I}v_\mathcal{A}$, $^\mathcal{I}\omega_\mathcal{a}$, $^\mathcal{I}v_\mathcal{g}$ and $^\mathcal{I}\omega_\mathcal{g}$ respectively. And the transformation matrices of the two relative to the inertial coordinate system $\mathcal{I}$ at this moment are known, that are $^\mathcal{I}_\mathcal{A}T$ and $^\mathcal{I}_\mathcal{G}T$. So how to calculate the relative linear velocity and the relative angular velocity of $\mathcal{G}$ relative to the $\mathcal{A}$ coordinate system?

For the linear velocity

Observing the origin of the $\mathcal{G}$ coordinate system in the $\mathcal{I}$ coordinate system, we have:

\[{^\mathcal{I}O_\mathcal{G}} = {^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}P_\mathcal{G}} + {^\mathcal{I}O_\mathcal{A}}\]

Calculate the derivative of both sides simultaneously, we have:

\[\dot{^\mathcal{I}O_\mathcal{G}} = \dot{^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}P_\mathcal{G}} + {^\mathcal{I}_\mathcal{A}R} \dot{^\mathcal{A}P_\mathcal{G}} + \dot{^\mathcal{I}O_\mathcal{A}}\] \[{^\mathcal{I}v_\mathcal{g}} = [^\mathcal{I}\omega_\mathcal{a}] {^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}P_\mathcal{G}} + {^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}v_\mathcal{g}} + {^\mathcal{I}v_\mathcal{A}}\] \[{^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}v_\mathcal{g}} = {^\mathcal{I}v_\mathcal{g}} - {^\mathcal{I}v_\mathcal{A}} - [^\mathcal{I}\omega_\mathcal{a}] {^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}P_\mathcal{G}}\] \[{^\mathcal{A}v_\mathcal{g}} = {^\mathcal{I}_\mathcal{A}R}^T ({^\mathcal{I}v_\mathcal{g}} - {^\mathcal{I}v_\mathcal{A}} - [^\mathcal{I}\omega_\mathcal{a}] {^\mathcal{I}_\mathcal{A}R} {^\mathcal{A}P_\mathcal{G}})\]

For the angular velocity

For the rotation matrix of two coordinate systems, we have:

\[{^\mathcal{I}_\mathcal{G}R} = {^\mathcal{I}_\mathcal{A}R} \cdot {^\mathcal{A}_\mathcal{G}R}\]

Calculate the derivatives of the both sides:

\[\dot{^\mathcal{I}_\mathcal{G}R} = \dot{^\mathcal{I}_\mathcal{A}R} \cdot {^\mathcal{A}_\mathcal{G}R} + {^\mathcal{I}_\mathcal{A}R} \cdot \dot{^\mathcal{A}_\mathcal{G}R}\] \[[^\mathcal{I}\omega_\mathcal{g}] {^\mathcal{I}_\mathcal{G}R} = [^\mathcal{I}\omega_\mathcal{a}] {^\mathcal{I}_\mathcal{A}R} \cdot {^\mathcal{A}_\mathcal{G}R} + {^\mathcal{I}_\mathcal{A}R} [^\mathcal{A}\omega_\mathcal{g}] {^\mathcal{A}_\mathcal{G}R}\] \[\begin{aligned} [^\mathcal{I}\omega_\mathcal{g}] &= [{^\mathcal{I}\omega_\mathcal{a}}] + {^\mathcal{I}_\mathcal{A}R} \cdot [^\mathcal{A}\omega_\mathcal{g}] \cdot (^\mathcal{I}_\mathcal{A}R)^T \\ &= [{^\mathcal{I}\omega_\mathcal{a}}] + [{^\mathcal{I}_\mathcal{A}R}^\mathcal{A}\omega_\mathcal{g}] \end{aligned}\] \[^\mathcal{A}\omega_\mathcal{g} = ({^\mathcal{I}_\mathcal{A}R})^T \cdot({^\mathcal{I}\omega_\mathcal{g}} - {^\mathcal{I}\omega_\mathcal{a}})\]