Appendix A: Derivations of Eqs. (6) and (12)
The angular velocity \({\boldsymbol{\omega }}(t)\) and infinitesimal rotation \(\delta {\boldsymbol{\pi }}\) in the body coordinate system for a rotation motion \({\mathbf{A}}(t)\) are calculated by
$$ \widetilde{{\boldsymbol{\omega }}}={\mathbf{A}}^{T}\dot{\mathbf{A}},\quad \widetilde{\delta {\boldsymbol{\pi }}}={\mathbf{A}}^{T}\delta {\mathbf{A}}. $$
The variation of an arbitrary function \(c\left ({\mathbf{A}}\right )\) can be calculated by
$$ \delta {c}=\frac{\partial {c}}{\partial {\mathbf{A}}}\delta {\mathbf{A}}= \frac{\partial {c}}{\partial {\mathbf{A}}}{\mathbf{A}} \widetilde{\delta {\boldsymbol{\pi }}}=\sum _{a=1}^{3}{\sum _{b=1}^{3}{\sum _{c=1}^{3}{ \sum _{d=1}^{3}{\frac{\partial {c}}{\partial {A}_{ab}}{A}_{ac}\epsilon _{bcd} \delta \pi _{d}}}}}, $$
where \(\epsilon _{abc}\) is the alternative tensor such that \(\left ({\boldsymbol{ x}}\times {\boldsymbol{ y}}\right )_{a}=\sum _{b=1}^{3}{ \sum _{c=1}^{3}{\epsilon _{abc}x_{b}y_{c}}}\), and hence
$$ \frac{\partial {c}}{\partial \pi _{d}}=\sum _{a=1}^{3}{\sum _{b=1}^{3}{ \sum _{c=1}^{3}{\epsilon _{bcd}\frac{\partial {c}}{\partial {A}_{ab}}{A}_{ac}}}}, $$
which is equivalent to
$$ \left (\frac{\partial {c}}{\partial {\boldsymbol{\pi }}}\right )^{T}=\sum _{a=1}^{3}{ \frac{\partial {c}}{\partial {\mathbf{A}}_{a:}}\times \left ({ \mathbf{A}}_{a:}\right )^{T}}. $$
(28)
Now consider a special case that \(c={\boldsymbol{ a}}^{T}{\mathbf{A}}^{T}(t){\boldsymbol{ c}}\), where \({\boldsymbol{ a}}\) and \({\boldsymbol{ c}}\) are constant vectors. Then direct calculations yield
$$ \left (\frac{\partial {c}}{\partial {\boldsymbol{\pi }}}\right )^{T}= \widetilde{\boldsymbol{ a}}{\mathbf{A}}^{T}{\boldsymbol{ c}},\quad \frac{\partial ^{2}{c}}{\partial {\boldsymbol{\pi }}\partial {\boldsymbol{\pi }}^{T}}= \widetilde{\boldsymbol{ a}} \widetilde{\left ({\mathbf{A}}^{T}{\boldsymbol{ c}}\right )}. $$
(29)
Moreover, suppose the rotation motion is described by a three-component parameter \({\boldsymbol{\vartheta }}\), and the angular velocity and the infinitesimal rotation can be written as [11]
$$ {\boldsymbol{\omega }}={\boldsymbol{ D}}\left ({\boldsymbol{\vartheta }}\right ) \dot{\boldsymbol{\vartheta }},\quad \delta {\boldsymbol{\pi }}={\boldsymbol{ D}}\left ({ \boldsymbol{\vartheta }}\right )\delta {\boldsymbol{\vartheta }}, $$
(30)
respectively, and by Eqs. (4) and (30),
$$ \frac{\partial {\mathscr {K}}}{\partial \dot{\boldsymbol{\vartheta }}^{T}}=\left ( \frac{\partial {\boldsymbol{\omega }}}{\partial \dot{\boldsymbol{\vartheta }}}\right )^{T} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}={\boldsymbol{ D}}^{T} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}},\quad \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\vartheta }}^{T}}=\left ( \frac{\partial {\boldsymbol{\omega }}}{\partial {\boldsymbol{\vartheta }}}\right )^{T} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}+{\boldsymbol{ D}}^{T} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\pi }}^{T}}. $$
Then direct calculation yields
$$ \frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial \dot{\boldsymbol{\vartheta }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\vartheta }}^{T}} ={ \boldsymbol{ D}}^{T}\left (\frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}\right )+\left ( \dot{\boldsymbol{ D}}^{T}-\left ( \frac{\partial {\boldsymbol{\omega }}}{\partial {\boldsymbol{\vartheta }}}\right )^{T} \right )\frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}-{ \boldsymbol{ D}}^{T}\frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\pi }}^{T}}. $$
(31)
In Appendix C, the following formula will be proved:
$$ \dot{\boldsymbol{ D}}^{T}-\left ( \frac{\partial {\boldsymbol{\omega }}}{\partial {\boldsymbol{\vartheta }}}\right )^{T}={ \boldsymbol{ D}}^{T}\widetilde{\boldsymbol{\omega }}. $$
(32)
Substituting Eq. (32) into Eq. (31) yields
$$ \frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial \dot{\boldsymbol{\vartheta }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\vartheta }}^{T}}={\boldsymbol{ D}}^{T} \left [\frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}+{\boldsymbol{\omega }} \times \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\pi }}^{T}}\right ]. $$
(33)
By Eqs. (3), (4), and (29),
$$ \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}=d^{\alpha }{ \boldsymbol{ q}}_{\alpha }\times \left ({\mathbf{A}}^{T}\dot{\mathbf{r}} \right )+M^{\alpha \beta }{\boldsymbol{ q}}_{\alpha }\times \left ( \dot{\boldsymbol{ q}}_{\beta }+{\boldsymbol{\omega }}\times {\boldsymbol{ q}}_{\beta } \right ),\quad \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\pi }}^{T}}=d^{ \alpha }\left (\dot{\boldsymbol{ q}}_{\alpha }+{\boldsymbol{\omega }}\times { \boldsymbol{ q}}_{\alpha }\right )\times \left ({\mathbf{A}}^{T} \dot{\mathbf{r}}\right ), $$
which can be substituted into Eq. (33), and Eq. (6) can be derived by direct calculations.
Moreover, suppose the virtual work done by the applied force can be written as
$$ \delta {\mathscr {W}}=\delta {\boldsymbol{\pi }}\cdot {\boldsymbol{ t}}+\cdots =\delta { \boldsymbol{\vartheta }}^{T}{\boldsymbol{ D}}^{T}{\boldsymbol{ t}}+\cdots ,$$
and the Lagrange equations with constraints can be written as
$$ \frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial \dot{\boldsymbol{\vartheta }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\vartheta }}^{T}}+\left ( \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\vartheta }}}\right )^{T}{ \boldsymbol{\lambda }}={\boldsymbol{ D}}^{T}{\boldsymbol{ t}}, $$
where \({\mathbf{C}}=0\) indicates the holonomic constraint equations. Direct calculations yield that
$$ \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\vartheta }}}= \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\pi }}} \frac{\partial {\boldsymbol{\pi }}}{\partial {\boldsymbol{\vartheta }}}= \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\pi }}}{\boldsymbol{ D}}. $$
The Lagrange equations with constraints for the three-component parameter \({\boldsymbol{\vartheta }}\) are
$$ \frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial \dot{\boldsymbol{\vartheta }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\vartheta }}^{T}}+\left ( \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\vartheta }}}\right )^{T}{ \boldsymbol{\lambda }}={\boldsymbol{ D}}^{T}\left [\frac{d}{dt} \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}+{\boldsymbol{\omega }} \times \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\pi }}^{T}}+\left ( \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\pi }}}\right )^{T}{ \boldsymbol{\lambda }}\right ]={\boldsymbol{ D}}^{T}{\boldsymbol{ t}}. $$
The choice of the three-component parameter \({\boldsymbol{\vartheta }}\) is arbitrary, and in any specific configuration, it is always possible to find a proper \({\boldsymbol{\vartheta }}\) such that \({\boldsymbol{ D}}\) is invertible. Hence the rotation equations for constrained flexible bodies can be written as
$$ \frac{d}{dt}\frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}+{ \boldsymbol{\omega }}\times \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\omega }}^{T}}- \frac{\partial {\mathscr {K}}}{\partial {\boldsymbol{\pi }}^{T}}+\left ( \frac{\partial {\mathbf{C}}}{\partial {\boldsymbol{\pi }}}\right )^{T}{ \boldsymbol{\lambda }}={\boldsymbol{ t}}, $$
(34)
where the partial derivatives with respect to the infinitesimal rotation \(\delta {\boldsymbol{\pi }}\) can be evaluated by Eq. (28) or Eq. (29). Equation (12) can be derived by substitution of Eqs. (6) and (10) into Eq. (34).
Appendix B: A simplified ANCF/RNCF element for space tethers
Consider the \(k\)th element of the tether with a uniform isotropic material and a uniform cross section, whose undeformed configuration is a straight line, with the arclength coordinate \(x_{k-1}\leq {x}\leq {x}_{k}\). The generalized coordinates of the element are
$$ {\boldsymbol{ q}}_{1}(t)={\mathbf{p}}\left (x_{k-1},\,t\right ),\quad { \boldsymbol{ q}}_{2}(t)={\mathbf{p}}_{x}\left (x_{k-1},\,t\right ),\quad { \boldsymbol{ q}}_{3}(t)={\mathbf{p}}\left (x_{k},\,t\right ),\quad { \boldsymbol{ q}}_{4}(t)={\mathbf{p}}_{x}\left (x_{k},\,t\right ), $$
and Hermite polynomials are adopted to interpolate the position vector, so the corresponding shape functions are
$$ \begin{aligned} &s^{1}\left (x\right )=1-3\xi ^{2}+2\xi ^{3},\qquad s^{2}\left (x \right )=l_{k}(\xi -2\xi ^{2}+\xi ^{3}),\\ &s^{3}\left (x\right )=3 \xi ^{2}-2\xi ^{3},\qquad s^{4}\left (x\right )=l_{k}(\xi ^{3}-\xi ^{2}), \end{aligned} $$
where \(l_{k}=x_{k}-x_{k-1}\) and \(\xi =\left (x-x_{k-1}\right )/{l_{k}}\). The kinetic energy of the element in the RCS can be written as
$$ {\mathscr {K}}_{k}=\frac{1}{2}m_{k}\dot{\mathbf{r}}\cdot \dot{\mathbf{r}}+d_{k}^{i}\dot{\mathbf{r}}\cdot \left ({\mathbf{A}}{ \boldsymbol{ v}}_{i}\right )+\frac{1}{2}M_{k}^{ij}{\boldsymbol{ v}}_{i}\cdot { \boldsymbol{ v}}_{j}, $$
(35)
where \(m_{k}\) is the mass of the element, and
$$ {\boldsymbol{ v}}_{i}=\dot{\boldsymbol{ q}}_{i}+{\boldsymbol{\omega }}\times {\boldsymbol{ q}}_{i}, \quad d_{k}^{i}=\int _{x_{k-1}}^{x_{k}}{\rho s^{i}\left (x\right )dx}, \quad M_{k}^{ij}=\int _{x_{k-1}}^{x_{k}}{\rho s^{i}\left (x\right )s^{j} \left (x\right )dx}. $$
(36)
The reference configuration can be described by
$$ \mathring{\boldsymbol{ q}}_{1}=\left (x_{k-1},\;0,\;0\right )^{T},\quad \mathring{\boldsymbol{ q}}_{2}=\left (1,\;0,\;0\right )^{T},\quad \mathring{\boldsymbol{ q}}_{3}=\left (x_{k},\;0,\;0\right )^{T},\quad \mathring{\boldsymbol{ q}}_{4}=\left (1,\;0,\;0\right )^{T}, $$
and it is easy to check that \(\left \| \mathring{\mathbf{p}}_{x}\left (x\right )\right \| =1\). The axial strain of the cable can be written as
$$ \varepsilon \left (x,\,t\right )=\frac{1}{2}\left ({\boldsymbol{ r}}_{x} \left (x,\,t\right )\cdot {\boldsymbol{ r}}_{x}\left (x,\,t\right )-1 \right ). $$
where \({\boldsymbol{ r}}_{x}\left (x,\,t\right )={\mathbf{A}}\left (t\right ){\mathbf{p}}_{x}\left (x,\,t\right )\) by Eq. (1). The unit tangential vector in the element is
$$ {\boldsymbol{ i}}\left (x,\,t\right )= \frac{{\boldsymbol{ r}}_{x}\left (x,\,t\right )}{\left \| {\boldsymbol{ r}}_{x}\left (x,\,t\right )\right \| }, $$
and the curvature of the tether can be evaluated by
$$ {\boldsymbol{\kappa }}\left (x,\,t\right )= \frac{{\partial {\boldsymbol{ i}}\left (x,\,t\right )}/\partial {x}}{\left \| {\boldsymbol{ r}}_{x}\left (x,\,t\right )\right \| }= \frac{{\boldsymbol{ r}}_{x}\left (x,\,t\right )\times {\boldsymbol{ r}}_{xx}\left (x,\,t\right )}{\left \| {\boldsymbol{ r}}_{x}\left (x,\,t\right )\right \| ^{3}}, $$
The stretch in the tether is expected to be small in practice, such that \(\left \| {\boldsymbol{ r}}_{x}\left (x,\,t\right )\right \| \approx 1\), and
$$ {\boldsymbol{ i}} \approx {\boldsymbol{ r}}_{x},\quad {\boldsymbol{\kappa }}\cdot {\boldsymbol{\kappa }}= \frac{\left ({\boldsymbol{ r}}_{x}\cdot {\boldsymbol{ r}}_{x}\right )\left ({\boldsymbol{ r}}_{xx}\cdot {\boldsymbol{ r}}_{xx}\right )-\left ({\boldsymbol{ r}}_{x}\cdot {\boldsymbol{ r}}_{xx}\right )^{2}}{\left \| {\boldsymbol{ r}}_{x}\right \| ^{6}} \approx \left ({\mathbf{p}}_{x}\cdot {\mathbf{p}}_{x}\right )\left ({ \mathbf{p}}_{xx}\cdot {\mathbf{p}}_{xx}\right )-\left ({\mathbf{p}}_{x} \cdot {\mathbf{p}}_{xx}\right )^{2}. $$
Then the elastic potential energy of the element can be calculated by
$$\begin{aligned} {\mathscr {U}}_{k}^{e} =&\frac{1}{2}\int _{x_{k-1}}^{x_{k}}{EA \varepsilon ^{2}\left (x,\,t\right )dx}+\frac{1}{2}\int _{x_{k-1}}^{x_{k}}{EI \left \| {\boldsymbol{\kappa }}\left (x,\,t\right )\right \| ^{2}dx}\\ =&\frac{1}{2} \kappa _{k}^{\mathit{ijmn}}Q_{ij}(t)Q_{mn}(t)+\frac{1}{2}\upsilon _{k}^{\mathit{ijmn}}P_{ij}(t)P_{mn}(t) \end{aligned}$$
for \(i,j,m,n=1,2,3,4\),
$$ \begin{aligned} &\kappa _{k}^{\mathit{ijmn}}=\int _{x_{k-1}}^{x_{k}}{EAs_{x}^{i}\left (x\right )s_{x}^{j} \left (x\right )s_{x}^{m}\left (x\right )s_{x}^{n}\left (x\right )dx}, \qquad Q_{ij}(t)=\frac{1}{2}\left ({\boldsymbol{ q}}_{i}(t)\cdot { \boldsymbol{ q}}_{j}(t)-\mathring{\boldsymbol{ q}}_{i}\cdot \mathring{\boldsymbol{ q}}_{j}\right ),\\ & P_{ij}(t)=\frac{1}{2}{ \boldsymbol{ q}}_{i}(t)\cdot {\boldsymbol{ q}}_{j}(t), \end{aligned} $$
and
$$\begin{aligned} \upsilon _{k}^{\mathit{ijmn}} =&\int _{x_{k-1}}^{x_{k}}EI\left [2\left (s_{x}^{i}s_{x}^{j}s_{xx}^{m}s_{xx}^{n}+s_{xx}^{i}s_{xx}^{j}s_{x}^{m}s_{x}^{n} \right )\right. \\ &\left.{}-\left (s_{xx}^{i}s_{x}^{j}s_{xx}^{m}s_{x}^{n}+s_{x}^{i}s_{xx}^{j}s_{xx}^{m}s_{x}^{n}+s_{xx}^{i}s_{x}^{j}s_{x}^{m}s_{xx}^{n}+s_{x}^{i}s_{xx}^{j}s_{x}^{m}s_{xx}^{n} \right )\right ]dx. \end{aligned}$$
Then the elastic force is
$$ {\boldsymbol{ F}}_{ke}^{i}=\left (\kappa _{k}^{\mathit{ijmn}}Q_{mn}+\upsilon _{k}^{\mathit{ijmn}}P_{mn} \right ){\boldsymbol{ q}}_{j}. $$
(37)
The virtual work done by the thrust force and the gravitational force on the element is
$$\begin{aligned} \delta {\mathscr {W}}_{k} =&\int _{x_{k-1}}^{x_{k}}{\delta {\boldsymbol{ r}}^{p} \cdot \left \{ c\max \left (0,V_{0}-V_{1}\right )\left [{\boldsymbol{ i}} \times \left ( \frac{r_{\oplus }{\mathbf{r}}}{\left \| {\mathbf{r}}\right \| ^{2}} \times {\boldsymbol{ i}}\right )\right ]-\rho g \frac{r_{\oplus }^{2}\mathbf{r}}{\left \| {\mathbf{r}}\right \| ^{3}} \right \} dx} \\ \approx &\int _{x_{k-1}}^{x_{k}}{\delta {\boldsymbol{ r}}^{p} \cdot \left \{ c\max \left (0,V_{0}-V_{1}\right )\left [{\boldsymbol{ r}}_{x} \times \left ( \frac{r_{\oplus }{\mathbf{r}}}{\left \| {\mathbf{r}}\right \| ^{2}} \times {\boldsymbol{ r}}_{x}\right )\right ]-\rho g \frac{r_{\oplus }^{2}\mathbf{r}}{\left \| {\mathbf{r}}\right \| ^{3}} \right \} dx} \\ =&\delta {\mathbf{r}}\cdot \left ({\boldsymbol{ F}}_{kt}+{ \boldsymbol{ F}}_{kg}\right )+\delta {\boldsymbol{\pi }}\cdot \left [{\boldsymbol{ q}}_{i} \times \left ({\boldsymbol{ Q}}_{kt}^{i}+{\boldsymbol{ Q}}_{kg}^{i}\right ) \right ]+\delta {\boldsymbol{ q}}_{i}\cdot \left ({\boldsymbol{ Q}}_{kt}^{i}+{ \boldsymbol{ Q}}_{kg}^{i}\right ), \end{aligned}$$
where the generalized forces are
$$\begin{aligned} \begin{aligned} &{\boldsymbol{ F}}_{kt}=\zeta _{k}^{ij}\left ({\mathbf{A}}{\boldsymbol{ q}}_{i} \right )\times \left [ \frac{r_{\oplus }{\mathbf{r}}}{\left \| {\mathbf{r}}\right \| ^{2}} \times \left ({\mathbf{A}}{\boldsymbol{ q}}_{j}\right )\right ],\qquad { \boldsymbol{ F}}_{kg}=-m_{k}g \frac{r_{\oplus }^{2}{\mathbf{r}}}{\left \| {\mathbf{r}}\right \| ^{3}}, \\ & {\boldsymbol{ Q}}_{kt}^{i}=\iota _{k}^{\mathit{ijm}}{\boldsymbol{ q}}_{j}\times \left ( \frac{r_{\oplus }{\mathbf{A}}^{T}{\mathbf{r}}}{\left \| {\mathbf{r}}\right \| ^{2}} \times {\boldsymbol{ q}}_{m}\right ),\qquad {\boldsymbol{ Q}}_{kg}^{i}=-gd_{k}^{i} \frac{r_{\oplus }^{2}{\mathbf{A}}^{T}{\mathbf{r}}}{\left \| {\mathbf{r}}\right \| ^{3}}, \end{aligned} \end{aligned}$$
(38)
in which \(\zeta _{k}^{ij}=c\max \left (0,\;V_{0}-V_{1}\right )\hat{\zeta }_{k}^{ij}\) and \(\iota _{k}^{\mathit{ijm}}=c\max \left (0,\;V_{0}-V_{1}\right )\hat{\iota }_{k}^{\mathit{ijm}}\) with
$$ \hat{\zeta }_{k}^{ij}=\int _{x_{k-1}}^{x_{k}}{s_{x}^{i}\left (x \right )s_{x}^{j}\left (x\right )dx},\qquad \hat{\iota }_{k}^{\mathit{ijm}}= \int _{x_{k-1}}^{x_{k}}{s^{i}\left (x\right )s_{x}^{j}\left (x\right )s_{x}^{m} \left (x\right )dx}. $$
The constants \(d_{k}^{i}\), \(M_{k}^{ij}\), and \(\hat{\zeta}_{k}^{ij}\) can be analytically evaluated to get the following matrices
$$ {\boldsymbol{ d}}_{k}=m_{k}\left( \textstyle\begin{array}{c} \frac{1}{2} \\ \frac{l_{k}}{12} \\ \frac{1}{2} \\ -\frac{l_{k}}{12} \end{array}\displaystyle \right),\; {\mathbf{M}}_{k}=m_{k}\left( \textstyle\begin{array}{cccc} \frac{13}{35} & \frac{11l_{k}}{210} & \frac{9}{70} & -\frac{13l_{k}}{420} \\ \frac{11l_{k}}{210} & \frac{l_{k}^{2}}{105} & \frac{13l_{k}}{420} & -\frac{l_{k}^{2}}{140} \\ \frac{9}{70} & \frac{13l_{k}}{420} & \frac{13}{35} & -\frac{11l_{k}}{210} \\ -\frac{13l_{k}}{420} & -\frac{l_{k}^{2}}{140} & -\frac{11l_{k}}{210} & \frac{l_{k}^{2}}{105} \end{array}\displaystyle \right),\; \hat{\boldsymbol{\zeta}}_{k}=\left( \textstyle\begin{array}{cccc} \frac{6}{5l_{k}} & \frac{1}{10} & -\frac{6}{5l_{k}} & \frac{1}{10} \\ \frac{1}{10} & \frac{2l_{k}}{15} & -\frac{1}{10} & -\frac{l_{k}}{30} \\ -\frac{6}{5l_{k}} & -\frac{1}{10} & \frac{6}{5l_{k}} & -\frac{1}{10} \\ \frac{1}{10} & -\frac{l_{k}}{30} & -\frac{1}{10} & \frac{2l_{k}}{15} \end{array}\displaystyle \right) $$
and the constant tensor \(\hat{\iota}_{k}^{ijm}\) can be evaluated to get the following \(\hat{\boldsymbol{\iota}}_{k}^{i}\) matrices with
$$\begin{aligned} \begin{aligned} \hat{\boldsymbol{\iota}}_{k}^{1}=\left( \textstyle\begin{array}{cccc} \frac{3}{5l_{k}} & -\frac{1}{70} & -\frac{3}{5l_{k}} & \frac{4}{35} \\ -\frac{1}{70} & \frac{43l_{k}}{420} & \frac{1}{70} & -\frac{l_{k}}{60} \\ -\frac{3}{5l_{k}} & \frac{1}{70} & \frac{3}{5l_{k}} & -\frac{4}{35} \\ \frac{4}{35} & -\frac{l_{k}}{60} & -\frac{4}{35} & \frac{13l_{k}}{420} \end{array}\displaystyle \right),\qquad \hat{\boldsymbol{\iota}}_{k}^{2}=\left( \textstyle\begin{array}{cccc} \frac{9}{70} & \frac{l_{k}}{140} & -\frac{9}{70} & \frac{3l_{k}}{140} \\ \frac{l_{k}}{140} & \frac{l_{k}^{2}}{120} & -\frac{l_{k}}{140} & -\frac{l_{k}^{2}}{840} \\ -\frac{9}{70} & -\frac{l_{k}}{140} & \frac{9}{70} & -\frac{3l_{k}}{140} \\ \frac{3l_{k}}{140} & -\frac{l_{k}^{2}}{840} & -\frac{3l_{k}}{140} & \frac{l_{k}^{2}}{168} \end{array}\displaystyle \right) \\ \hat{\boldsymbol{\iota}}_{k}^{3}=\left( \textstyle\begin{array}{cccc} \frac{3}{5l_{k}} & \frac{4}{35} & -\frac{3}{5l_{k}} & -\frac{1}{70} \\ \frac{4}{35} & \frac{13l_{k}}{420} & -\frac{4}{35} & -\frac{l_{k}}{60} \\ -\frac{3}{5l_{k}} & -\frac{4}{35} & \frac{3}{5l_{k}} & \frac{1}{70} \\ -\frac{1}{70} & -\frac{l_{k}}{60} & \frac{1}{70} & \frac{43l_{k}}{420} \end{array}\displaystyle \right),\qquad \hat{\boldsymbol{\iota}}_{n}^{4}=\left( \textstyle\begin{array}{cccc} -\frac{9}{70} & -\frac{3l_{k}}{140} & \frac{9}{70} & -\frac{l_{k}}{140} \\ -\frac{3l_{k}}{140} & \frac{-l_{k}^{2}}{168} & \frac{3l_{k}}{140} & \frac{l_{k}^{2}}{840} \\ \frac{9}{70} & \frac{3l_{k}}{140} & -\frac{9}{70} & \frac{l_{k}}{140} \\ -\frac{l_{k}}{140} & \frac{l_{k}^{2}}{840} & \frac{l_{k}}{140} & \frac{-l_{k}^{2}}{120} \end{array}\displaystyle \right) \end{aligned} \end{aligned}$$
Moreover, the stiffness tensors \(\kappa _{k}^{\mathit{ijmn}}\) and \(\upsilon _{k}^{\mathit{ijmn}}\) can be evaluated analytically too, but since they are complicated, it is better to calculate them numerically using Gauss quadratures.
Appendix C: Proof of formula (32)
Suppose the rotational matrix \({\mathbf{A}}=\left ({\boldsymbol{ x}},\,{\boldsymbol{ y}},\,{\boldsymbol{ z}} \right )\) is described by a three-component-parameter vector \({\boldsymbol{\vartheta }}\). The angular velocity \({}\boldsymbol{\omega }\) can be calculated from \(\widetilde{\boldsymbol{\omega }}={\mathbf{A}}^{T}\dot{\mathbf{A}}=\left ({ \boldsymbol{ x}},\,{\boldsymbol{ y}},\,{\boldsymbol{ z}}\right )^{T}\left ( \dot{\boldsymbol{ x}},\,\dot{\boldsymbol{ y}},\,\dot{\boldsymbol{ z}}\right )\). Hence the angular velocity can be explicitly written as
$$ \omega =\left ( \textstyle\begin{array}{c} {\boldsymbol{ z}}\cdot \dot{\boldsymbol{ y}} \\ {\boldsymbol{ x}}\cdot \dot{\boldsymbol{ z}} \\ {\boldsymbol{ y}}\cdot \dot{\boldsymbol{ x}} \end{array}\displaystyle \right )=\sum _{a=1}^{3}{\left ( \textstyle\begin{array}{c} {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a} \\ {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a} \\ {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a} \end{array}\displaystyle \right )\dot{\vartheta }_{a}}={\boldsymbol{ D}}\left ({\boldsymbol{\vartheta }} \right )\dot{\boldsymbol{\vartheta }} ,\quad {\boldsymbol{ D}}=\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c} {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1} & {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2} & {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3} \\ {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1} & {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2} & {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3} \\ {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1} & {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2} & {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3} \end{array}\displaystyle \right ), $$
(39)
where \({\boldsymbol{ x}}_{a}\equiv {\partial {\boldsymbol{ x}}}/{\partial \vartheta _{a}}\), \({\boldsymbol{ y}}_{a}\equiv {\partial {\boldsymbol{ y}}}/{\partial \vartheta _{a}}\), and \({\boldsymbol{ z}}_{a}\equiv {\partial {\boldsymbol{ z}}}/{\partial \vartheta _{a}}\), and \({\boldsymbol{ D}}\) is the same as that in Eq. (30). So the time derivative of \({\boldsymbol{ D}}\) is
$$ \dot{\boldsymbol{ D}}=\sum _{a=1}^{3}{\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c} {\boldsymbol{ z}}_{a}\cdot {\boldsymbol{ y}}_{1} & {\boldsymbol{ z}}_{a}\cdot { \boldsymbol{ y}}_{2} & {\boldsymbol{ z}}_{a}\cdot {\boldsymbol{ y}}_{3} \\ {\boldsymbol{ x}}_{a}\cdot {\boldsymbol{ z}}_{1} & {\boldsymbol{ x}}_{a}\cdot { \boldsymbol{ z}}_{2} & {\boldsymbol{ x}}_{a}\cdot {\boldsymbol{ z}}_{3} \\ {\boldsymbol{ y}}_{a}\cdot {\boldsymbol{ x}}_{1} & {\boldsymbol{ y}}_{a}\cdot { \boldsymbol{ x}}_{2} & {\boldsymbol{ y}}_{a}\cdot {\boldsymbol{ x}}_{3} \end{array}\displaystyle \right )\dot{\vartheta }_{a}}+\sum _{a=1}^{3}{\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c} {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1a} & {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2a} & {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3a} \\ {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1a} & {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2a} & {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3a} \\ {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1a} & {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2a} & {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3a} \end{array}\displaystyle \right )\dot{\vartheta }_{a}}, $$
(40)
and the partial derivative \({\partial {\boldsymbol{\omega }}}/{\partial {\boldsymbol{\vartheta }}}\) can be directly calculated by
$$ \frac{\partial {\boldsymbol{\omega }}}{\partial {\boldsymbol{\vartheta }}}=\sum _{a=1}^{3}{ \left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c} {\boldsymbol{ z}}_{1}\cdot {\boldsymbol{ y}}_{a} & {\boldsymbol{ z}}_{2}\cdot { \boldsymbol{ y}}_{a} & {\boldsymbol{ z}}_{3}\cdot {\boldsymbol{ y}}_{a} \\ {\boldsymbol{ x}}_{1}\cdot {\boldsymbol{ z}}_{a} & {\boldsymbol{ x}}_{2}\cdot { \boldsymbol{ z}}_{a} & {\boldsymbol{ x}}_{3}\cdot {\boldsymbol{ z}}_{a} \\ {\boldsymbol{ y}}_{1}\cdot {\boldsymbol{ x}}_{a} & {\boldsymbol{ y}}_{2}\cdot { \boldsymbol{ x}}_{a} & {\boldsymbol{ y}}_{3}\cdot {\boldsymbol{ x}}_{a} \end{array}\displaystyle \right )\dot{\vartheta }_{a}}+\sum _{a=1}^{3}{\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c} {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1a} & {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2a} & {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3a} \\ {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1a} & {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2a} & {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3a} \\ {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1a} & {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2a} & {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3a} \end{array}\displaystyle \right )\dot{\vartheta }_{a}}. $$
(41)
Then by Eqs. (40) and (41) the left-hand side of Eq. (32) becomes
$$ \dot{\boldsymbol{ D}}^{T}-\left ( \frac{\partial {\boldsymbol{\omega }}}{\partial {\boldsymbol{\vartheta }}}\right )^{T}= \sum _{a=1}^{3}{\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c} {\boldsymbol{ z}}_{a}\cdot {\boldsymbol{ y}}_{1}-{\boldsymbol{ z}}_{1}\cdot { \boldsymbol{ y}}_{a}, & {\boldsymbol{ x}}_{a}\cdot {\boldsymbol{ z}}_{1}-{\boldsymbol{ x}}_{1} \cdot {\boldsymbol{ z}}_{a}, & {\boldsymbol{ y}}_{a}\cdot {\boldsymbol{ x}}_{1}-{ \boldsymbol{ y}}_{1}\cdot {\boldsymbol{ x}}_{a} \\ {\boldsymbol{ z}}_{a}\cdot {\boldsymbol{ y}}_{2}-{\boldsymbol{ z}}_{2}\cdot { \boldsymbol{ y}}_{a}, & {\boldsymbol{ x}}_{a}\cdot {\boldsymbol{ z}}_{2}-{\boldsymbol{ x}}_{2} \cdot {\boldsymbol{ z}}_{a}, & {\boldsymbol{ y}}_{a}\cdot {\boldsymbol{ x}}_{2}-{ \boldsymbol{ y}}_{2}\cdot {\boldsymbol{ x}}_{a} \\ {\boldsymbol{ z}}_{a}\cdot {\boldsymbol{ y}}_{3}-{\boldsymbol{ z}}_{3}\cdot { \boldsymbol{ y}}_{a}, & {\boldsymbol{ x}}_{a}\cdot {\boldsymbol{ z}}_{3}-{\boldsymbol{ x}}_{3} \cdot {\boldsymbol{ z}}_{a}, & {\boldsymbol{ y}}_{a}\cdot {\boldsymbol{ x}}_{3}-{ \boldsymbol{ y}}_{3}\cdot {\boldsymbol{ x}}_{a} \end{array}\displaystyle \right )\dot{\vartheta }_{a}}. $$
(42)
Since \({\boldsymbol{ x}}\), \({\boldsymbol{ y}}\), \({\boldsymbol{ z}}\) are unit vectors, \({\boldsymbol{ x}}\cdot {\boldsymbol{ x}}_{a}={\boldsymbol{ y}}\cdot {\boldsymbol{ y}}_{a}={ \boldsymbol{ z}}\cdot {\boldsymbol{ z}}_{a}=0\), and their partial derivatives can be expressed in the basis \(\left \{ {\boldsymbol{ x}},\,{\boldsymbol{ y}},\,{\boldsymbol{ z}}\right \} \) as
$$ {\boldsymbol{ x}}_{a}=\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right ){ \boldsymbol{ y}}+\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ){ \boldsymbol{ z}},\quad {\boldsymbol{ y}}_{a}=\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right ){\boldsymbol{ z}}+\left ({\boldsymbol{ x}}\cdot { \boldsymbol{ y}}_{a}\right ){\boldsymbol{ x}},\quad {\boldsymbol{ z}}_{a}=\left ({ \boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right ){\boldsymbol{ x}}+\left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ){\boldsymbol{ y}}. $$
Hence their dot products can be calculated by
$$ {\boldsymbol{ x}}_{a}\cdot {\boldsymbol{ y}}_{b}=\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ x}}_{a}\right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{b} \right ),\quad {\boldsymbol{ y}}_{a}\cdot {\boldsymbol{ z}}_{b}=\left ({ \boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right )\left ({\boldsymbol{ x}}\cdot { \boldsymbol{ z}}_{b}\right ),\quad {\boldsymbol{ z}}_{a}\cdot {\boldsymbol{ x}}_{b}= \left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right )\left ({\boldsymbol{ y}} \cdot {\boldsymbol{ x}}_{b}\right ), $$
which can be substituted into Eq. (42) to yield that
$$\begin{aligned} &\dot{\boldsymbol{ D}}^{T}-\left ( \frac{\partial {\boldsymbol{\omega }}}{\partial {\boldsymbol{\vartheta }}}\right )^{T} \\ &\quad\! =\! \sum _{a=1}^{3}\!{\left ( \textstyle\begin{array}{c@{\ }c@{\ }c} \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right )\left ({\boldsymbol{ x}} \cdot {\boldsymbol{ y}}_{1}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right )\left ({\boldsymbol{ y}}\cdot { \boldsymbol{ z}}_{1}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ x}}_{1}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ) \\ \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right )\left ({\boldsymbol{ x}} \cdot {\boldsymbol{ y}}_{2}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right )\left ({\boldsymbol{ y}}\cdot { \boldsymbol{ z}}_{2}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ x}}_{2}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ) \\ \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right )\left ({\boldsymbol{ x}} \cdot {\boldsymbol{ y}}_{3}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right )\left ({\boldsymbol{ y}}\cdot { \boldsymbol{ z}}_{3}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ x}}_{3}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ) \end{array}\displaystyle \! \right )\!\dot{\vartheta }_{a}}. \end{aligned}$$
(43)
On the other hand, the right-hand side of Eq. (32) can be directly calculated by Eq. (39), which yields
$$\begin{aligned} &{\boldsymbol{ D}}^{T}\widetilde{\boldsymbol{\omega }} \\ &\quad \!=\! \sum _{a=1}^{3}\!{\left ( \textstyle\begin{array}{c@{\ }c@{\ }c} \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1}\right )\left ({\boldsymbol{ y}} \cdot {\boldsymbol{ x}}_{a}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1}\right )\left ({\boldsymbol{ x}}\cdot { \boldsymbol{ z}}_{a}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}\right ) \\ \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2}\right )\left ({\boldsymbol{ y}} \cdot {\boldsymbol{ x}}_{a}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2}\right )\left ({\boldsymbol{ x}}\cdot { \boldsymbol{ z}}_{a}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}\right ) \\ \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3}\right )\left ({\boldsymbol{ y}} \cdot {\boldsymbol{ x}}_{a}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3}\right )\left ({\boldsymbol{ x}}\cdot { \boldsymbol{ z}}_{a}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}\right ) \end{array}\displaystyle \!\right )\!\dot{\vartheta }_{a}}. \end{aligned}$$
(44)
Since the vectors \({\boldsymbol{ x}}\), \({\boldsymbol{ y}}\), \({\boldsymbol{ z}}\) are perpendicular to each other, that is, \({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}={\boldsymbol{ y}}\cdot {\boldsymbol{ z}}={ \boldsymbol{ z}}\cdot {\boldsymbol{ x}}=0\), their time derivatives indicate that
$$ {\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{a}=-{\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}, \quad {\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{a}=-{\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}, \quad {\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{a}=-{\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}, $$
which can be substituted to Eq. (44), yielding that
$$\begin{aligned} &{\boldsymbol{ D}}^{T}\widetilde{\boldsymbol{\omega }} \\ &\quad =\!\sum _{a=1}^{3}\!\left ( \textstyle\begin{array}{c@{\ }c@{\ }c} \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{1}\right )\left ({\boldsymbol{ x}} \cdot {\boldsymbol{ z}}_{a}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{1} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{1}\right )\left ({\boldsymbol{ y}}\cdot { \boldsymbol{ x}}_{a}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{1} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{1}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{1} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ) \\ \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{2}\right )\left ({\boldsymbol{ x}} \cdot {\boldsymbol{ z}}_{a}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{2} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{2}\right )\left ({\boldsymbol{ y}}\cdot { \boldsymbol{ x}}_{a}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{2} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{2}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{2} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ) \\ \left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{3}\right )\left ({\boldsymbol{ x}} \cdot {\boldsymbol{ z}}_{a}\right )-\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ z}}_{3} \right )\left ({\boldsymbol{ x}}\cdot {\boldsymbol{ y}}_{a}\right ), & \left ({ \boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{3}\right )\left ({\boldsymbol{ y}}\cdot { \boldsymbol{ x}}_{a}\right )-\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ x}}_{3} \right )\left ({\boldsymbol{ y}}\cdot {\boldsymbol{ z}}_{a}\right ), & \left ({ \boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{3}\right )\left ({\boldsymbol{ z}}\cdot { \boldsymbol{ y}}_{a}\right )-\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ y}}_{3} \right )\left ({\boldsymbol{ z}}\cdot {\boldsymbol{ x}}_{a}\right ) \end{array}\displaystyle \!\right )\!\dot{\vartheta }_{a}. \end{aligned}$$
(45)
Equalities (43) and (45) are obvious, which can be checked term by term. So Eq. (32) is proved. □
Moreover, since the three-component parameter \({\boldsymbol{\vartheta }}\) can be arbitrarily chosen, the proof is independent of any specific \({\boldsymbol{\vartheta }}\).
