Based on the original Rocket Workbench on SourceForge in CVS at:
https://sourceforge.net/projects/rocketworkbench
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291 lines
11 KiB
291 lines
11 KiB
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\documentclass[10pt]{article}
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\usepackage{vmargin}
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\usepackage{amsmath}
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\newcommand{\er}{\vec{e}_r}
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\newcommand{\el}{\vec{e}_\lambda}
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\newcommand{\eu}{\vec{e}_\mu}
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\begin{document}
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\section{Inertial acceleration of the vehicle mass center in body
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reference frame}
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Generally, the absolute acceleration of a particle $A$ in a moving
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reference frame of center B is:
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$$\vec{a}_A = \vec{a}_B + \vec{\alpha}\times\vec{r} +
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\vec{\omega}\times(\vec{\omega}\times\vec{r})
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+2\vec{\omega}\times\vec{v}_{rel} + \vec{a}_{rel}$$
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Choosing an earth centered reference frame with constant angular
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velocity ($\vec{\omega}^E$) and no acceleration yield to the simpler form
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$$\vec{a}_{A_E} = \vec{\omega}^E_E\times(\vec{\omega}^E_E\times\vec{r}^E_E)
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+2\vec{\omega}^E_E\times\vec{v}^{rel}_E + \vec{a}^{rel}_E$$
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where the subscript $E$ indicate that the vector are expressed in the
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earth reference frame.
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Transforming this vectorial notation to matrix notation and noting
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that the angular velocity of $F_B$ relative to $F_E$ is
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$\vec{\omega}^B - \vec{\omega}^E$ yield to:
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$$\vec{a}_{A_E} = \tilde{\omega}^E_E\tilde{\omega}^E_E\vec{r}^E_E
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+2\tilde{\omega}^E_E\vec{v}^{rel}_E + \vec{a}^{rel}_E$$
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We could now expressed this equation in term of component in the body
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fixed reference frame.
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$$\vec{a}_{A_B} = L_{BE}\vec{a}_{A_E} =
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L_{BE}(\tilde{\omega}^E_E\tilde{\omega}^E_E\vec{r}^E_E
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+2\tilde{\omega}^E_E\vec{v}^{rel}_E + \vec{a}^{rel}_E)$$
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$$ = L_{BE}\tilde{\omega}^E_E\tilde{\omega}^E_E\vec{r}^E_E +
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2L_{BE}\tilde{\omega}^E_E\vec{v}^{rel}_E + (\tilde{\omega}^B -
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\tilde{\omega}^E)_B\vec{v}^{rel}_B + \vec{a}^{rel}_B$$
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We could then transform
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$$L_{BE}\tilde{\omega}^E_E\vec{v}^{rel}_E = (L_{BE}\tilde{\omega}^E_E
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L_{EB})(L_{BE}\vec{v}^{rel}_E) = \tilde{\omega}^E_B\vec{v}^{rel}_B$$
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and
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$$L_{BE}\tilde{\omega}^E_E\tilde{\omega}^E_E\vec{r}^E_E =
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(L_{BE}\tilde{\omega}^E_E L_{EB})(L_{BE}\tilde{\omega}^E_E
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L_{EB})(L_{BE}\vec{r}^E_E) =
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\tilde{\omega}^E_B\tilde{\omega}^E_B\vec{r}^E_B$$
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We now have
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$$\vec{a}_{A_B} = \vec{a}^{rel}_B + (\tilde{\omega}^B -
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\tilde{\omega}^E)_B\vec{v}^E_B +
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\tilde{\omega}^E_B\tilde{\omega}^E_B\vec{r}^E_B +
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2\tilde{\omega}^E_B\vec{v}^{rel}_B$$
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which simplify to
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$$\vec{a}_{A_B} = \vec{a}^{rel}_B + (\tilde{\omega}^B +
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\tilde{\omega}^E)_B\vec{v}^E_B +
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\tilde{\omega}^E_B\tilde{\omega}^E_B\vec{r}^E_B$$
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Defining vector component in $F_B$
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$$\vec{v}^E_B = \begin{bmatrix}u\\v\\w\end{bmatrix}$$
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$$\vec{\omega}^B_B = \begin{bmatrix}p\\q\\r\end{bmatrix}$$
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$$\vec{\omega}^E_B = \begin{bmatrix}p^E_B\\q^E_B\\r^E_B\end{bmatrix} =
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L_{BV}\begin{pmatrix}\cos{\lambda}\\0\\-\sin{\lambda}\end{pmatrix}\omega^E$$
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and matrix notation of angular velocity
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$$\tilde{\omega}^B_B =
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\begin{bmatrix}0&-r&q\\r&0&-p\\q&p&0\end{bmatrix}$$
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$$\tilde{\omega}^E_B =
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\begin{bmatrix}0&-r^E_B&q^E_B\\r^E_B&0&-p^E_B\\q^E_B&p^E_B&0\end{bmatrix}$$
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\section{Acceleration of vehicle mass center espressed in a spherical
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coordinate system relative to rotating earth}
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The spherical coordinate system is compose of the three vector
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($\er$, $\el$, $\eu$), where $\er$ point in the direction of the mass
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center of the vehicle, $\el$ point in the north direction and is
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perpendicular to $\er$, $\eu$ point east and is perpendicular to both
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previous vectors. We could also define the direction of $\er$ with two
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angle, $\mu$ is the angle between the equatorial projection of $\er$
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and a reference meridian on the earth. $\lambda$ is the angle between
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$\er$ and the equator. We could then define the relative velocity and
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acceleration of the point C relative to earth by:
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$$\vec{v}_{rel} = \dot{r}\er + r\dot{\mu}\cos{\lambda}\eu + r\dot{\lambda}\el$$
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\begin{align*}
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\vec{a}_{rel} = & (\ddot{r} - r\dot{\lambda}^2 -
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r\dot{\mu}^2\cos^2{\lambda})\er +\\
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& ((2\dot{r}\dot{\mu} + r\ddot{\mu})\cos{\lambda} -
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2r\dot{\mu}\dot{\lambda}\sin{\lambda})\eu +\\
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& (2\dot{r}\dot{\lambda} + r\ddot{\lambda} +
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r\dot{\mu}^2\sin{\lambda}\cos{\lambda})\el
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\end{align*}
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We can now derive the inertial velocity and acceleration of C:
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\begin{align*}
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\vec{v}_C & = \vec{v}_O + \vec{v}_{rel} + \vec{\omega}\times\vec{r}\\
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& = \dot{r}\er + r\dot{\mu}\cos{\lambda}\eu +
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r\dot{\lambda}\el + \omega\vec{k}\times r\er\\
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& = \dot{r}\er + r\cos{\lambda}(\dot{\mu} + \omega)\eu
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+ r\dot{\lambda}\el
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\end{align*}
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$$\vec{a}_C = \vec{a}_O + \vec{\alpha}\times\vec{r} +
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\vec{\omega}\times(\vec{\omega}\times\vec{r})
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+2\vec{\omega}\times\vec{v}_{rel} + \vec{a}_{rel}$$
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Each term will be evaluate separatly:
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\begin{align*}
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\vec{\omega}\times(\vec{\omega}\times\vec{r}) & =
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\omega\vec{k}\times(\omega\vec{k}\times r\er)\\
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& = \omega\vec{k}\times(\omega r\cos{\lambda}\eu)\\
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& = \omega^2r\cos{\lambda}(-\cos{\lambda}\er+\sin{\lambda}\el)\\
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& = -\omega^2r\cos^2{\lambda}\er + \omega^2r\cos{\lambda}\sin{\lambda}\el
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\end{align*}
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\begin{align*}
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\vec{\omega}\times\vec{v}_{rel} & = \omega\vec{k}\times(\dot{r}\er +
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r\dot{\mu}\cos{\lambda}\eu + r\dot{\lambda}\el)\\
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& = \omega\dot{r}\cos{\lambda}\eu + \omega
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r\dot{\mu}\cos{\lambda}(-\cos{\lambda}\er + \sin{\lambda}\el) +
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\omega r\dot{\lambda}(-\cos{\lambda})\eu\\
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& = \omega\cos{\lambda}(\dot{r}-r\dot{\lambda})\eu - \omega
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r\dot{\mu}\cos^2{\lambda}\er + \omega r\dot{\mu}\cos{\lambda}\sin{\lambda}\el
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\end{align*}
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The inertial acceleration of C is then:
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\begin{align*}
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\vec{a}_C = & (\ddot{r} - r\dot{\lambda}^2 -
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r\dot{\mu}^2\cos^2{\lambda} - 2\omega
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r\dot{\mu}\cos^2{\lambda} - \omega^2r\cos^2{\lambda})\er\\
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& + ((2\dot{r}\dot{\mu} + r\ddot{\mu})\cos{\lambda} -
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2r\dot{\mu}\dot{\lambda}\sin{\lambda}
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+ 2\omega\cos{\lambda}(\dot{r}-r\dot{\lambda})\eu\\
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& + (2\dot{r}\dot{\lambda} + r\ddot{\lambda} +
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r\dot{\mu}^2\sin{\lambda}\cos{\lambda} +2\omega
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r\dot{\mu}\cos{\lambda}\sin{\lambda} +
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\omega^2r\cos{\lambda}\sin{\lambda})\el
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\end{align*}
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which simplify to:
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\begin{align*}
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\vec{a}_C = & (\ddot{r} - r\dot{\lambda}^2 -
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r(\omega+\dot{\mu})^2\cos^2{\lambda})\er\\
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& + ((2\dot{r}\dot{\mu} + r\ddot{\mu})\cos{\lambda} -
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2r\dot{\mu}\dot{\lambda}\sin{\lambda}
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+ 2\omega\cos{\lambda}(\dot{r}-r\dot{\lambda})\eu\\
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& + (2\dot{r}\dot{\lambda} + r\ddot{\lambda} +
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r\sin{\lambda}\cos{\lambda}(\omega+\dot{\mu})^2)\el
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\end{align*}
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We must now project the inertial acceleration from this vehicled
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carried frame to the bidy fixed frame with the conversion matrix based
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on the euler angle which define the angular position of $F_B$ relative
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to $F_V$.
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$$L_{BV} = \begin{bmatrix} \cos{\theta}\cos{\psi}&
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\cos{\theta}\sin{\psi}& -\sin{\theta}\\
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\sin{\phi}\sin{\theta}\cos{\psi}-\cos{\phi}\sin{\psi}&
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\sin{\phi}\sin{\theta}\sin{\psi}+\cos{\phi}\cos{\psi}&
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\sin{\phi}\cos{\theta}\\
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\cos{\phi}\sin{\theta}\cos{\psi}+\sin{\phi}\sin{\psi}&
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\cos{\phi}\sin{\theta}\sin{\psi}-\sin{\phi}\cos{\psi}&
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\cos{\phi}\cos{\theta}
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\end{bmatrix}
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$$
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\begin{align*}
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\vec{a}_{C_B} & = L_{BV}\vec{a}_C\\
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& = L_{BV}(\vec{\omega}\times(\vec{\omega}\times\vec{r})
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+ 2\vec{\omega}\times\vec{v}_{rel} + \vec{a}_{rel})\\
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& = L_{BV}\begin{bmatrix}r\cos{\lambda}\sin{\lambda}\\ 0\\
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-r\cos^2{\lambda}\end{bmatrix}\omega^2 +
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2\vec{\omega}^E_B\times\vec{v}_{rel_B} +
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\vec{a}_{rel_B} + (\vec{\omega}^B -
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\vec{\omega}^E)_B\times\vec{v}^E_B \\
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& = L_{BV}\begin{bmatrix}r\cos{\lambda}\sin{\lambda}\\ 0\\
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-r\cos^2{\lambda}\end{bmatrix}\omega^2 +
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\vec{a}_{rel_B} + (\vec{\omega}^B +
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\vec{\omega}^E)_B\times\vec{v}^E_B\\
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& = \begin{bmatrix}\dot{u}\\\dot{v}\\\dot{w}\end{bmatrix} +
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\begin{bmatrix}c_x\\c_y\\c_z\end{bmatrix} +
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\begin{bmatrix}(q + q^E_B)w - (r + r^E_B)v\\
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(r + r^E_B)u - (p + p^E_B)w\\
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(p + p^E_B)v - (q + q^E_B)u\end{bmatrix}
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\end{align*}
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We will also use the moment equation:
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\begin{align*}
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L & = I_x\dot{p} - (I_y - I_z)qr\\
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M & = I_y\dot{q} - (I_z - I_x)rp\\
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N & = I_z\dot{r} - (I_x - I_y)pq
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\end{align*}
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\section{System of equation}
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\subsection{Independant variable}
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\begin{align*}
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(\phi, \theta, \psi)\quad & \text{Euler angle defining orientation of the
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vehicle relative to $F_V$}\\
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(u, v, w)\quad & \text{Velocity of the vehicle relative to earth}\\
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(p, q, r)\quad & \text{Absolute angle rate}\\
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(\lambda, \mu, R)\quad & \text{Position of the rocket relative to earth}\\
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(p^E_B, q^E_B, r^E_B)\quad & \text{Earth angular velocity in body axis}\\
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(P, Q, R) \quad & \text{Relative angular velocity of body relative ro $F_V$
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expressed in $F_B$}
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\end{align*}
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\subsection{List of equations}
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$$X - mg\sin{\theta} = m(\dot{u} + c_x + (q^E_B+q)w - (r^E_B+r)v)$$
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$$Y + mg\cos{\theta}\sin{\phi} = m(\dot{v} + c_y + (r^E_B+r)u - (p^E_B+p)w)$$
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$$Z + mg\cos{\theta}\cos{\phi} = m(\dot{w} + c_z + (p^E_B+pvw -
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(q^E_B+q)u)$$
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\begin{align*}
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L & = I_x\dot{p} - (I_y - I_z)qr\\
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M & = I_y\dot{q} - (I_z - I_x)rp\\
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N & = I_z\dot{r} - (I_x - I_y)pq
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\end{align*}
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$$\begin{bmatrix}\dot{\phi}\\\dot{\theta}\\\dot{\psi}\end{bmatrix} =
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\begin{bmatrix}1&\sin{\phi}\tan{\theta}&\cos{\phi}\tan{\theta}\\
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0&\cos{\phi}&-\sin{\phi}\\
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0&\sin{\phi}\sec{\theta}&\cos{\phi}\sec{\theta}\end{bmatrix}
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\begin{bmatrix}P\\Q\\R\end{bmatrix}$$
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$$\begin{bmatrix}P\\Q\\R\end{bmatrix} =
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\begin{bmatrix}p\\q\\r\end{bmatrix} -
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L_{BV}\begin{bmatrix}(\omega^E -
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\dot{\mu})\cos{\lambda}\\-\dot{\lambda}\\
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-(\omega^E + \dot{\mu})\sin{\lambda}\end{bmatrix}$$
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$$\begin{bmatrix}r\dot{\lambda}\\r\dot{\mu}\cos{\lambda}\\\dot{r}
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\end{bmatrix} = L_{VB}\begin{bmatrix}u\\v\\w\end{bmatrix}$$
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$$\vec{\omega}^E_B = \begin{bmatrix}p^E_B\\q^E_B\\r^E_B\end{bmatrix} =
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L_{BV}\begin{pmatrix}\cos{\lambda}\\0\\-\sin{\lambda}\end{pmatrix}\omega^E$$
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$$L_{VB} = \begin{bmatrix}\cos{\theta}\cos{\psi}&
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\sin{\phi}\sin{\theta}\cos{\psi} - \cos{\phi}\sin{\psi}&
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\cos{\phi}\sin{\theta}\cos{\psi} + \sin{\phi}\sin{\psi}\\
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\cos{\theta}\sin{\psi}&
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\sin{\phi}\sin{\theta}\sin{\psi} + \cos{\phi}\cos{\psi}&
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\cos{\phi}\sin{\theta}\sin{\psi} - \sin{\phi}\cos{\psi}\\
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-\sin{\theta}&
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\sin{\phi}\cos{\theta}&
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\cos{\phi}\cos{\theta}\end{bmatrix}$$
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\end{document}
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