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