$$\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}$$

• Rigid Body Kinematics
• Measuring of the Pull-back Acceleration of the material points that are affixed with an accelerometer
• Computing the Pull-back Acceleration Field
• Computing the Acceleration Field
• Skew-symmetric square root of a symmetric tensor
• Computing the Acceleration Field via Integration

Computing the Pull-back Acceleration Field

After operating both the left and right sides of the first expression in the previous section with $\boldsymbol{Q}^{T}$, we get

$$\begin{align} \boldsymbol{Q}^{\rm T}(t) \boldsymbol{a}(\boldsymbol{X}, t)&= \boldsymbol{Q}^{\rm T}(t)\ddot{\boldsymbol{Q}}(t)\boldsymbol{X} + \boldsymbol{Q}^{\rm T}(t)\ddot{\boldsymbol{c}}(t) \label{eq:OperatedAccelerationField} \end{align}$$

Defining $\begin{align} \boldsymbol{\varphi}^{*}(\boldsymbol{a})(\boldsymbol{X},t) &:= \boldsymbol{Q}^{\rm T}(t)\boldsymbol{a}(\boldsymbol{X},t),\tag{Def.1}\label{eq:PullBackAccelerationFieldDef}\\ \boldsymbol{P}(t) &:=\boldsymbol{Q}^{\rm T}(t)\ddot{\boldsymbol{Q}}(t), \tag{Def.2}\label{eq:PDef} \end{align}$

and $\begin{align} \boldsymbol{q}(t) &:= \boldsymbol{Q}^{\rm T}(t)\ddot{\boldsymbol{c}}(t), \tag{Def.3} \label{eq:Def3} \end{align}$ equation $\eqref{eq:OperatedAccelerationField}$ can be written as

$\begin{align} \boldsymbol{\varphi}^{*}(\boldsymbol{a})(\boldsymbol{X},t)&= \boldsymbol{P}(t)\boldsymbol{X}+\boldsymbol{q}(t). \label{eq:PullBackAccelerationField} \end{align}$ The field $\boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)$ is the pull-back of the acceleration field $\boldsymbol{a}$.

Let the accelerometers be glued onto the material points $\mathcal{P}^{0}$, $\mathcal{P}^{1}$, $\mathcal{P}^{2}$, and $\mathcal{P}^{3}$. We denote the position vectors of the material point $\mathcal{P}^{\iota}$ in the reference configurationas $\boldsymbol{X}^{\iota}$, where $\iota=0,1,2,3$. It follows from $\eqref{eq:PullBackAccelerationField}$ that the pull-back acceleration vector at the material points, $\mathcal{P}^{\iota}$ is

$\begin{align} \boldsymbol{\varphi}^{*}(\boldsymbol{a})(\boldsymbol{X}^{\iota},t) &= \boldsymbol{P}(t)\boldsymbol{X}^{\iota}+\boldsymbol{q}(t), \label{eq:m3} \end{align}$ where $\iota=0,1,2,3$,

In order to obtain $\boldsymbol{P}$ in a compact form we define the vectors

$\begin{align} \Delta \boldsymbol{X}^{\mathscr{i}}&=\boldsymbol{X}^{\mathscr{i}}-\boldsymbol{X}^{\mathbb{0}}, \end{align}$ and $\begin{align} \Delta\boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)^{\mathscr{i}}(t)&= \boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)^{\mathscr{i}}(t)-\boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)^{0}(t),\\ \end{align}$ where $\mathscr{i}=1,2,3$, and $\begin{align} \boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)^{\iota}(t)&= \boldsymbol{\varphi}^{*}(\boldsymbol{a})(\boldsymbol{X}^{\iota},t), \end{align}$ where $\iota=0,1,2,3$.

If follows from $\eqref{eq:m3}$ that $\begin{align} \Delta\boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)^{\mathscr{i}}(t) &= \boldsymbol{P}(t) \Delta\boldsymbol{X}^{\mathscr{i}}(t), \label{eq:PhiPX} \end{align}$ where $\mathscr{i}=1,2,3$

Solving the equation $\eqref{eq:PhiPX}$ for $\mathscr{i}=1,2,3$ we get $\begin{align} \boldsymbol{P}(t)= \Delta\boldsymbol{\varphi}^*\left(\boldsymbol{a}\right)^{\mathscr{i}}(t)\cdot \Delta\boldsymbol{X}^{\mathscr{j}}(t) \Delta\boldsymbol{Y}^{\mathscr{i}}\otimes \Delta\boldsymbol{Y}^{\mathscr{j}}, \label{eq:Psol} \end{align}$ where $\Delta \boldsymbol{Y}^{\mathscr{i}}$, $\mathscr{i}=1,2,3$ are the reciprocal vectors corresponding to $\Delta \boldsymbol{X}^{\mathscr{i}}$, $\mathscr{i}=1,2,3$ $\begin{align} \Delta\boldsymbol{Y}^{\mathscr{1}}&= \frac{\Delta\boldsymbol{X}^{\mathscr{2}} \times \Delta\boldsymbol{X}^{\mathscr{3}}}{\Delta\boldsymbol{X}^{\mathscr{1}} \cdot \left(\Delta\boldsymbol{X}^{\mathscr{2}}\times \Delta\boldsymbol{X}^{\mathscr{3}}\right)}\\ \Delta\boldsymbol{Y}^{\mathscr{2}}&= \frac{\Delta\boldsymbol{X}^{\mathscr{3}} \times \Delta\boldsymbol{X}^{\mathscr{1}}}{\Delta\boldsymbol{X}^{\mathscr{1}} \cdot \left(\Delta\boldsymbol{X}^{\mathscr{2}}\times \Delta\boldsymbol{X}^{\mathscr{3}}\right)}\\ \Delta\boldsymbol{Y}^{\mathscr{3}}&= \frac{\Delta\boldsymbol{X}^{\mathscr{1}} \times \Delta\boldsymbol{X}^{\mathscr{2}}}{\Delta\boldsymbol{X}^{\mathscr{1}} \cdot \left(\Delta\boldsymbol{X}^{\mathscr{2}}\times \Delta\boldsymbol{X}^{\mathscr{3}}\right)} \end{align}$

After computing $\boldsymbol{P}(t)$ from $\eqref{eq:Psol}$, the vector $\boldsymbol{q}(t)$, defined in $\eqref{eq:Def3}$, can be computed from $\eqref{eq:m3}$ with, e.g., $\iota=0$.

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