Done with all transform_to derivatives
Frank Dellaert
parent
d9c185498e
commit
c20498c688
556
doc/math.lyx
556
doc/math.lyx
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@ -1233,25 +1233,25 @@ Hence, the final derivative of an action in its first argument is
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\begin_layout Standard
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\begin_inset FormulaMacro
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\renewcommand{\Rsix}{\mathfrak{\mathbb{R}^{6}}}
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\newcommand{\Rsix}{\mathfrak{\mathbb{R}^{6}}}
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{\mathfrak{\mathbb{R}^{6}}}
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\end_inset
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\begin_inset FormulaMacro
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\renewcommand{\SEthree}{SE(3)}
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\newcommand{\SEthree}{SE(3)}
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{SE(3)}
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\end_inset
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\begin_inset FormulaMacro
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\renewcommand{\sethree}{\mathfrak{se(3)}}
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\newcommand{\sethree}{\mathfrak{se(3)}}
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{\mathfrak{se(3)}}
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\end_inset
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\begin_inset FormulaMacro
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\renewcommand{\xihat}{\hat{\xi}}
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\newcommand{\xihat}{\hat{\xi}}
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{\hat{\xi}}
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\end_inset
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@ -1327,7 +1327,47 @@ v\end{array}\right]\rightarrow\xihat\define\left[\begin{array}{cc}
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Note we follow Frank Park's convention and reserve the first three components
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for rotation, and the last three for translation.
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Applying the exponential map to a twist
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Hence, with this parameterization, the generators for
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\begin_inset Formula $\SEthree$
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\end_inset
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are
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\begin_inset Formula \[
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G^{1}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 0\\
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0 & 0 & -1 & 0\\
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0 & 1 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{}G^{2}=\left(\begin{array}{cccc}
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0 & 0 & 1 & 0\\
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0 & 0 & 0 & 0\\
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-1 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{ }G^{3}=\left(\begin{array}{cccc}
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0 & -1 & 0 & 0\\
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1 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\]
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\end_inset
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\begin_inset Formula \[
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G^{4}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 1\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{}G^{5}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 1\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{ }G^{6}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 1\\
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0 & 0 & 0 & 0\end{array}\right)\]
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\end_inset
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Applying the exponential map to a twist
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\begin_inset Formula $\xi$
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\end_inset
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@ -1482,85 +1522,55 @@ The action of
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by using homogeneous coordinates
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\begin_inset Formula \[
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q=\left[\begin{array}{cc}
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\hat{q}=\left[\begin{array}{c}
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q\\
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1\end{array}\right]=\left[\begin{array}{cc}
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R & t\\
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0 & 1\end{array}\right]\left[\begin{array}{c}
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p\\
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1\end{array}\right]\]
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1\end{array}\right]=T\hat{p}\]
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\end_inset
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\end_layout
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\begin_layout Standard
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The generators for
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\begin_inset Formula $\SEthree$
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\end_inset
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are
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\begin_inset Formula \[
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G_{1}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 0\\
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0 & 0 & -1 & 0\\
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0 & 1 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{}G_{2}=\left(\begin{array}{cccc}
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0 & 0 & 1 & 0\\
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0 & 0 & 0 & 0\\
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-1 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{ }G_{1}=\left(\begin{array}{cccc}
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0 & -1 & 0 & 0\\
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1 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\]
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\end_inset
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\begin_inset Formula \[
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G_{4}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 1\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{}G_{5}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 1\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\end{array}\right)\mbox{ }G_{6}=\left(\begin{array}{cccc}
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 0\\
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0 & 0 & 0 & 1\\
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0 & 0 & 0 & 0\end{array}\right)\]
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\end_inset
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and hence a twist around the origin applies to homogeneous coordinates
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\begin_inset Formula $\hat{p}\in\mathbb{R}^{4}$
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\end_inset
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as
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\begin_inset Formula \[
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\sum_{i}\xi_{i}G_{i}=\left(\begin{array}{cccc}
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0 & -\omega_{z} & \omega_{y} & v_{x}\\
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\omega_{z} & 0 & -\omega_{x} & v_{y}\\
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-\omega_{y} & \omega_{x} & 0 & v_{z}\\
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0 & 0 & 0 & 0\end{array}\right)=\left[\begin{array}{cc}
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\Skew{\omega} & v\\
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0 & 0\end{array}\right]=\xihat\]
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\end_inset
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Hence, at the origin, an incremental action
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We would now like to know what an incremental rotation parameterized by
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\begin_inset Formula $\xi$
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\end_inset
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on a point
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\begin_inset Formula $p$
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would do:
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\begin_inset Formula \[
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\hat{q}(\xi)=Te^{\xihat}\hat{p}\]
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\end_inset
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induces the velocity
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hence the derivative (following the exposition in Section
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\begin_inset CommandInset ref
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LatexCommand ref
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reference "sec:Derivatives-of-Actions"
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\end_inset
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):
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\begin_inset Formula \[
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\left[\begin{array}{cc}
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\deriv{\hat{q}(\xi)}{\xi}=T\deriv{}{\xi}\left(\xihat\hat{p}\right)=TH_{p}\]
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\end_inset
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where
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\begin_inset Formula $\xihat\hat{p}$
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\end_inset
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corresponds to a velocity in
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\begin_inset Formula $\mathbb{R}^{4}$
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\end_inset
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(in the local
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\begin_inset Formula $T$
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\end_inset
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frame):
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\begin_inset Formula \[
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\xihat\hat{p}=\left[\begin{array}{cc}
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\Skew{\omega} & v\\
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0 & 0\end{array}\right]\left[\begin{array}{c}
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p\\
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@ -1570,16 +1580,23 @@ p\\
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\end_inset
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We can write this as a velocity in
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Notice how velocities are anologous to points at infinity in projective
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geometry: they correspond to free vectors indicating a direction and magnitude
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of change.
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\end_layout
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\begin_layout Standard
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By only taking the top three rows, we can write this as a velocity in
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\begin_inset Formula $\Rthree$
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\end_inset
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as the product of a
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, as the product of a
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\begin_inset Formula $3\times6$
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\end_inset
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matrix
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\begin_inset Formula $H(p)$
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\begin_inset Formula $H_{p}$
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\end_inset
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that acts upon the exponential coordinates
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@ -1591,81 +1608,33 @@ We can write this as a velocity in
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\omega\times p+v=-p\times\omega+v=\left[\begin{array}{cc}
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-\Skew p & I_{3}\end{array}\right]\left[\begin{array}{c}
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\omega\\
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v\end{array}\right]=H(p)\xi\]
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v\end{array}\right]=H_{p}\xi\]
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\end_inset
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Now, if we want to apply this in a frame
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\begin_inset Formula $T$
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\end_inset
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, we can (1) transform the point
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\begin_inset Formula $p$
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\end_inset
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to the origin using
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\begin_inset Formula $T^{-1}$
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\end_inset
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, apply the action
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\begin_inset Formula $\xihat$
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\end_inset
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, and transform back to
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\begin_inset Formula $T$
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\end_inset
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.
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In short
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Hence, the final derivative of the group action is
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\begin_inset Formula \[
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\left[\begin{array}{c}
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q\\
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1\end{array}\right]=Te^{\xihat}T^{-1}\left[\begin{array}{c}
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p\\
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1\end{array}\right]=\exp\left(\Ad T\xihat\right)\left[\begin{array}{c}
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p\\
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1\end{array}\right]\]
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\deriv{\hat{q}(\xi)}{\xi}=T\hat{H}_{p}=\left[\begin{array}{cc}
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R & t\\
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0 & 1\end{array}\right]\left[\begin{array}{cc}
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\Skew{-p} & I_{3}\\
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0 & 0\end{array}\right]\]
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\end_inset
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To get the velocity of the point in frame
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\begin_inset Formula $T$
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in homogenous coordinates.
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In
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\begin_inset Formula $\Rthree$
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\end_inset
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, we have
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this becomes:
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\begin_inset Formula \[
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H(p)\Ad T\xi=\left[\begin{array}{cc}
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-\Skew p & I_{3}\end{array}\right]\left[\begin{array}{cc}
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R & 0\\
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\Skew tR & R\end{array}\right]\xi=R\left[\begin{array}{cc}
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-\Skew{T^{-1}p} & I_{3}\end{array}\right]\xi=H(T^{-1}p)\xi\]
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\deriv{q(\xi)}{\xi}=R\left[\begin{array}{cc}
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-\Skew p & I_{3}\end{array}\right]\]
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\end_inset
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where I made use of
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\begin_inset Formula $\Skew{t-p}R=-RR^{T}\Skew{p-t}R=-R\Skew{R^{T}(p-t)}=-R\Skew{T^{-1}p}$
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\end_inset
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.
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This is intuitive in hindsight: we transform
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\begin_inset Formula $p$
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\end_inset
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back to the orgin by
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\begin_inset Formula $T^{-1}p$
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\end_inset
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, apply
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\begin_inset Formula $H(.)$
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\end_inset
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to get a velocity, and only need the rotation
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\begin_inset Formula $R$
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\end_inset
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to transform it back to the orginal frame (as velocities are not affected
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by translation).
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\end_layout
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\begin_layout Section
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@ -1800,30 +1769,6 @@ which maps the angle
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\end_inset
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Note that
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\begin_inset Formula \begin{equation}
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\skew{\theta}\left[\begin{array}{c}
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x\\
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y\end{array}\right]=\theta R_{\pi/2}\left[\begin{array}{c}
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x\\
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y\end{array}\right]=\theta\left[\begin{array}{c}
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-y\\
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x\end{array}\right]\label{eq:RestrictedCross}\end{equation}
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\end_inset
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which acts like a restricted
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\begin_inset Quotes eld
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\end_inset
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cross product
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\begin_inset Quotes erd
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\end_inset
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in the plane.
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\end_layout
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\begin_layout Standard
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The exponential map can be computed in closed form as
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\begin_inset Formula \[
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R=e^{\skew{\theta}}=\left[\begin{array}{cc}
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@ -1891,6 +1836,92 @@ and between in its second argument,
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\end_inset
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\end_layout
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\begin_layout Subsection
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Derivatives of Actions
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\end_layout
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\begin_layout Standard
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In the case of
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\begin_inset Formula $\SOtwo$
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\end_inset
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the vector space is
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\begin_inset Formula $\Rtwo$
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\end_inset
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, and the group action corresponds to rotating a point
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\begin_inset Formula \[
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q=Rp\]
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\end_inset
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We would now like to know what an incremental rotation parameterized by
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\begin_inset Formula $\theta$
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\end_inset
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would do:
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\begin_inset Formula \[
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q(\text{\theta})=Re^{\skew{\theta}}p\]
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\end_inset
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hence the derivative (following the exposition in Section
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\begin_inset CommandInset ref
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LatexCommand ref
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reference "sec:Derivatives-of-Actions"
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\end_inset
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):
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\begin_inset Formula \[
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\deriv{q(\omega)}{\omega}=R\deriv{}{\omega}\left(e^{\skew{\theta}}p\right)=R\deriv{}{\omega}\left(\skew{\theta}p\right)=RH_{p}\]
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\end_inset
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Note that
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\begin_inset Formula \begin{equation}
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\skew{\theta}\left[\begin{array}{c}
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x\\
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y\end{array}\right]=\theta R_{\pi/2}\left[\begin{array}{c}
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x\\
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y\end{array}\right]=\theta\left[\begin{array}{c}
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-y\\
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x\end{array}\right]\label{eq:RestrictedCross}\end{equation}
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\end_inset
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which acts like a restricted
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\begin_inset Quotes eld
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\end_inset
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cross product
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\begin_inset Quotes erd
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\end_inset
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in the plane.
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Hence
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\begin_inset Formula \[
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\skew{\theta}p=\left[\begin{array}{c}
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-y\\
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x\end{array}\right]\theta=H_{p}\theta\]
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\end_inset
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with
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\begin_inset Formula $H_{p}=R_{pi/2}p$
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\end_inset
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.
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Hence, the final derivative of an action in its first argument is
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\begin_inset Formula \[
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\deriv{q(\theta)}{\theta}=RH_{p}=RR_{pi/2}p=R_{pi/2}Rp=R_{pi/2}q\]
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\end_inset
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\end_layout
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\begin_layout Section
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@ -1978,7 +2009,40 @@ Note we think of robots as having a pose
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and hence I switched the order above, reserving the first two components
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for translation and the last for rotation.
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Applying the exponential map to a twist
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\family roman
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\series medium
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\shape up
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\size normal
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\emph off
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\bar no
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\noun off
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\color none
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The Lie group generators are
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\begin_inset Formula \[
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G^{x}=\left[\begin{array}{ccc}
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0 & 0 & 1\\
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0 & 0 & 0\\
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0 & 0 & 0\end{array}\right]\mbox{ }G^{y}=\left[\begin{array}{ccc}
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0 & 0 & 0\\
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0 & 0 & 1\\
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0 & 0 & 0\end{array}\right]\mbox{ }G^{\theta}=\left[\begin{array}{ccc}
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0 & -1 & 0\\
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1 & 0 & 0\\
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0 & 0 & 0\end{array}\right]\]
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\end_inset
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\family default
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\series default
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\shape default
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\size default
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\emph default
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\bar default
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\noun default
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\color inherit
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Applying the exponential map to a twist
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\begin_inset Formula $\xi$
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\end_inset
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@ -2060,8 +2124,89 @@ We can just define all derivatives in terms of the above adjoint map:
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\end_layout
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\begin_layout Part
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Old Stuff
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\begin_layout Subsection
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The derivatives of Actions
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\end_layout
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\begin_layout Standard
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The action of
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\begin_inset Formula $\SEtwo$
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\end_inset
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on 2D points is done by embedding the points in
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\begin_inset Formula $\mathbb{R}^{3}$
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\end_inset
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by using homogeneous coordinates
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\begin_inset Formula \[
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\hat{q}=\left[\begin{array}{c}
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q\\
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1\end{array}\right]=\left[\begin{array}{cc}
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R & t\\
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0 & 1\end{array}\right]\left[\begin{array}{c}
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p\\
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1\end{array}\right]=T\hat{p}\]
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\end_inset
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|
||||
Analoguous to
|
||||
\begin_inset Formula $\SEthree$
|
||||
\end_inset
|
||||
|
||||
, we can compute a velocity
|
||||
\begin_inset Formula $\xihat\hat{p}$
|
||||
\end_inset
|
||||
|
||||
in the local
|
||||
\begin_inset Formula $T$
|
||||
\end_inset
|
||||
|
||||
frame:
|
||||
\begin_inset Formula \[
|
||||
\xihat\hat{p}=\left[\begin{array}{cc}
|
||||
\skew{\omega} & v\\
|
||||
0 & 0\end{array}\right]\left[\begin{array}{c}
|
||||
p\\
|
||||
1\end{array}\right]=\left[\begin{array}{c}
|
||||
\skew{\omega}p+v\\
|
||||
0\end{array}\right]\]
|
||||
|
||||
\end_inset
|
||||
|
||||
By only taking the top two rows, we can write this as a velocity in
|
||||
\begin_inset Formula $\Rtwo$
|
||||
\end_inset
|
||||
|
||||
, as the product of a
|
||||
\begin_inset Formula $2\times3$
|
||||
\end_inset
|
||||
|
||||
matrix
|
||||
\begin_inset Formula $H_{p}$
|
||||
\end_inset
|
||||
|
||||
that acts upon the exponential coordinates
|
||||
\begin_inset Formula $\xi$
|
||||
\end_inset
|
||||
|
||||
directly:
|
||||
\begin_inset Formula \[
|
||||
\skew{\omega}p+v=v+R_{pi/2}p\omega=\left[\begin{array}{cc}
|
||||
I_{2} & R_{pi/2}p\end{array}\right]\left[\begin{array}{c}
|
||||
v\\
|
||||
\omega\end{array}\right]=H_{p}\xi\]
|
||||
|
||||
\end_inset
|
||||
|
||||
Hence, the final derivative of the group action is
|
||||
\begin_inset Formula \[
|
||||
\deriv{q(\xi)}{\xi}=R\left[\begin{array}{cc}
|
||||
I_{2} & R_{pi/2}p\end{array}\right]=\left[\begin{array}{cc}
|
||||
R & R_{pi/2}q\end{array}\right]\]
|
||||
|
||||
\end_inset
|
||||
|
||||
|
||||
\end_layout
|
||||
|
||||
\begin_layout Section
|
||||
|
|
@ -2091,43 +2236,10 @@ where
|
|||
\end_inset
|
||||
|
||||
is an incremental rotation angle.
|
||||
The derivatives of
|
||||
\series bold
|
||||
\emph on
|
||||
rotate
|
||||
\series default
|
||||
\emph default
|
||||
are then found easily, using
|
||||
\begin_inset Formula \begin{eqnarray*}
|
||||
\frac{\partial x'}{\partial\delta} & = & \frac{\partial(x\cos\theta'-y\sin\theta')}{\partial\delta}\\
|
||||
& = & \frac{\partial(x(\cos\theta\cos\delta-\sin\theta\sin\delta)-y(\sin\theta\cos\delta+\cos\theta\sin\delta))}{\partial\delta}\\
|
||||
& = & x(-\cos\theta\sin\delta-\sin\theta\cos\delta)-y(-\sin\theta\sin\delta+\cos\theta\cos\delta)\\
|
||||
& = & -x\sin\theta-y\cos\theta=-y'\end{eqnarray*}
|
||||
|
||||
\end_inset
|
||||
|
||||
|
||||
\end_layout
|
||||
|
||||
\begin_layout Standard
|
||||
\begin_inset Formula \begin{eqnarray*}
|
||||
\frac{\partial y'}{\partial\delta} & = & \frac{\partial(x\sin\theta'+y\cos\theta')}{\partial\delta}\\
|
||||
& = & \frac{\partial(x(\sin\theta\cos\delta+\cos\theta\sin\delta)+y(\cos\theta\cos\delta-\sin\theta\sin\delta))}{\partial\delta}\\
|
||||
& = & x(-\sin\theta\sin\delta+\cos\theta\cos\delta)+y(-\cos\theta\sin\delta-\sin\theta\cos\delta)\\
|
||||
& = & x\cos\theta-y\sin\theta=x'\end{eqnarray*}
|
||||
|
||||
\end_inset
|
||||
|
||||
|
||||
\end_layout
|
||||
|
||||
\begin_layout Standard
|
||||
\begin_inset Formula \[
|
||||
\frac{\partial p'}{\partial p}=\frac{\partial(Rp)}{\partial p}=R\]
|
||||
|
||||
\end_inset
|
||||
|
||||
Similarly, unrotate
|
||||
Derivatives of unrotate
|
||||
\end_layout
|
||||
|
||||
\begin_layout Standard
|
||||
|
|
@ -2210,56 +2322,6 @@ The derivative of a cross product
|
|||
\end_inset
|
||||
|
||||
|
||||
\end_layout
|
||||
|
||||
\begin_layout Section
|
||||
Rot3
|
||||
\end_layout
|
||||
|
||||
\begin_layout Standard
|
||||
An incremental rotation is applied as (switched to right-multiply Jan 25
|
||||
2010)
|
||||
\begin_inset Formula \[
|
||||
R'=R(I+\Omega)\]
|
||||
|
||||
\end_inset
|
||||
|
||||
where
|
||||
\begin_inset Formula $\Omega=\Skew{\omega}$
|
||||
\end_inset
|
||||
|
||||
is the skew symmetric matrix corresponding to the incremental rotation
|
||||
angles
|
||||
\begin_inset Formula $\omega=(\omega_{x},\omega_{y},\omega_{z})$
|
||||
\end_inset
|
||||
|
||||
.
|
||||
The derivatives of
|
||||
\series bold
|
||||
\emph on
|
||||
rotate
|
||||
\series default
|
||||
\emph default
|
||||
are then found easily, using
|
||||
\begin_inset CommandInset ref
|
||||
LatexCommand eqref
|
||||
reference "eq:Dcross1"
|
||||
|
||||
\end_inset
|
||||
|
||||
:
|
||||
\begin_inset Formula \[
|
||||
\frac{\partial(R(I+\Omega)x)}{\partial\omega}=\frac{\partial(R\Omega x)}{\partial\omega}=\frac{\partial(R\left(\omega\times x\right))}{\partial\omega}=R\frac{\partial\left(\omega\times x\right)}{\partial\omega}=R\Skew{-x}\]
|
||||
|
||||
\end_inset
|
||||
|
||||
|
||||
\begin_inset Formula \[
|
||||
\frac{\partial(Rx)}{\partial x}=R\]
|
||||
|
||||
\end_inset
|
||||
|
||||
|
||||
\end_layout
|
||||
|
||||
\begin_layout Section
|
||||
|
|
|
|||
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doc/math.pdf
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doc/math.pdf
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Reference in New Issue