1068 lines
21 KiB
Plaintext
1068 lines
21 KiB
Plaintext
#LyX 1.6.5 created this file. For more info see http://www.lyx.org/
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\lyxformat 345
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\begin_document
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\begin_header
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\textclass article
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\language english
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\quotes_language english
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\tracking_changes false
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\output_changes false
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\author ""
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\author ""
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\end_header
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\begin_body
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\begin_layout Title
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Geometry Derivatives and Other Hairy Math
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\end_layout
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\begin_layout Author
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Frank Dellaert
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\end_layout
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\begin_layout Standard
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\begin_inset FormulaMacro
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\newcommand{\Skew}[1]{[#1]_{\times}}
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{[#1]_{\times}}
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\end_inset
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\end_layout
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\begin_layout Standard
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This document should be kept up to date and specify how each of the derivatives
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in the geometry modules are computed.
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\end_layout
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\begin_layout Section
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Rot2 (in gtsam)
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\end_layout
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\begin_layout Standard
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A rotation is stored as
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\begin_inset Formula $(\cos\theta,\sin\theta)$
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\end_inset
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.
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An incremental rotation is applied using the trigonometric sum rule:
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\begin_inset Formula \[
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\cos\theta'=\cos\theta\cos\delta-\sin\theta\sin\delta\]
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\end_inset
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\begin_inset Formula \[
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\sin\theta'=\sin\theta\cos\delta+\cos\theta\sin\delta\]
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\end_inset
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where
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\begin_inset Formula $\delta$
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\end_inset
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is an incremental rotation angle.
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The derivatives of
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\series bold
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\emph on
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rotate
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\series default
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\emph default
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are then found easily, using
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\begin_inset Formula \begin{eqnarray*}
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\frac{\partial x'}{\partial\delta} & = & \frac{\partial(x\cos\theta'-y\sin\theta')}{\partial\delta}\\
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& = & \frac{\partial(x(\cos\theta\cos\delta-\sin\theta\sin\delta)-y(\sin\theta\cos\delta+\cos\theta\sin\delta))}{\partial\delta}\\
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& = & x(-\cos\theta\sin\delta-\sin\theta\cos\delta)-y(-\sin\theta\sin\delta+\cos\theta\cos\delta)\\
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& = & -x\sin\theta-y\cos\theta=-y'\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \begin{eqnarray*}
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\frac{\partial y'}{\partial\delta} & = & \frac{\partial(x\sin\theta'+y\cos\theta')}{\partial\delta}\\
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& = & \frac{\partial(x(\sin\theta\cos\delta+\cos\theta\sin\delta)+y(\cos\theta\cos\delta-\sin\theta\sin\delta))}{\partial\delta}\\
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& = & x(-\sin\theta\sin\delta+\cos\theta\cos\delta)+y(-\cos\theta\sin\delta-\sin\theta\cos\delta)\\
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& = & x\cos\theta-y\sin\theta=x'\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial p'}{\partial p}=\frac{\partial(Rp)}{\partial p}=R\]
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\end_inset
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Similarly, unrotate
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \begin{eqnarray*}
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\frac{\partial x'}{\partial\delta} & = & \frac{\partial(x\cos\theta'+y\sin\theta')}{\partial\delta}\\
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& = & \frac{\partial(x(\cos\theta\cos\delta-\sin\theta\sin\delta)+y(\sin\theta\cos\delta+\cos\theta\sin\delta))}{\partial\delta}\\
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& = & x(-\cos\theta\sin\delta-\sin\theta\cos\delta)+y(-\sin\theta\sin\delta+\cos\theta\cos\delta)\\
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& = & -x\sin\theta+y\cos\theta=y'\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \begin{eqnarray*}
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\frac{\partial y'}{\partial\delta} & = & \frac{\partial(-x\sin\theta'+y\cos\theta')}{\partial\delta}\\
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& = & \frac{\partial(-x(\sin\theta\cos\delta+\cos\theta\sin\delta)+y(\cos\theta\cos\delta-\sin\theta\sin\delta))}{\partial\delta}\\
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& = & -x(-\sin\theta\sin\delta+\cos\theta\cos\delta)+y(-\cos\theta\sin\delta-\sin\theta\cos\delta)\\
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& = & -x\cos\theta-y\sin\theta=-x'\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial p'}{\partial p}=\frac{\partial(Rp)}{\partial p}=R\]
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\end_inset
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\end_layout
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\begin_layout Section
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Point3
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\end_layout
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\begin_layout Standard
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A cross product
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\begin_inset Formula $a\times b$
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\end_inset
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can be written as a matrix multiplication
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\begin_inset Formula \[
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a\times b=\Skew ab\]
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\end_inset
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where
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\begin_inset Formula $\Skew a$
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\end_inset
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is a skew-symmetric matrix defined as
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\begin_inset Formula \[
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\Skew{x,y,z}=\left[\begin{array}{ccc}
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0 & -z & y\\
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z & 0 & -x\\
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-y & x & 0\end{array}\right]\]
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\end_inset
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We also have
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\begin_inset Formula \[
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a^{T}\Skew b=-(\Skew ba)^{T}=-(a\times b)^{T}\]
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\end_inset
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The derivative of a cross product
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\begin_inset Formula \begin{equation}
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\frac{\partial(a\times b)}{\partial a}=\Skew{-b}\label{eq:Dcross1}\end{equation}
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\end_inset
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\begin_inset Formula \begin{equation}
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\frac{\partial(a\times b)}{\partial b}=\Skew a\label{eq:Dcross2}\end{equation}
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\end_inset
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\end_layout
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\begin_layout Section
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Rot3
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\end_layout
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\begin_layout Standard
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An incremental rotation is applied as (switched to right-multiply Jan 25
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2010)
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\begin_inset Formula \[
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R'=R(I+\Omega)\]
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\end_inset
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where
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\begin_inset Formula $\Omega=\Skew{\omega}$
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\end_inset
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is the skew symmetric matrix corresponding to the incremental rotation
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angles
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\begin_inset Formula $\omega=(\omega_{x},\omega_{y},\omega_{z})$
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\end_inset
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.
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The derivatives of
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\series bold
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\emph on
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rotate
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\series default
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\emph default
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are then found easily, using
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Dcross1"
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\end_inset
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:
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\begin_inset Formula \[
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\frac{\partial(R(I+\Omega)x)}{\partial\omega}=\frac{\partial(R\Omega x)}{\partial\omega}=\frac{\partial(R\left(\omega\times x\right))}{\partial\omega}=R\frac{\partial\left(\omega\times x\right)}{\partial\omega}=R\Skew{-x}\]
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\end_inset
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\begin_inset Formula \[
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\frac{\partial(Rx)}{\partial x}=R\]
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\end_inset
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\end_layout
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\begin_layout Standard
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For composition and transposing of rotation matrices the situation is a
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bit more complex.
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We want to figure out what incremental rotation
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\begin_inset Formula $\Omega'$
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\end_inset
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on the composed matrix, will yield the same change as
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\begin_inset Formula $\Omega$
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\end_inset
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applied to either the first (
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\begin_inset Formula $A$
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\end_inset
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) or second argument
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\begin_inset Formula $(B$
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\end_inset
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).
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Hence, the derivative with respect to the second argument is now easy:
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\begin_inset Formula \begin{eqnarray*}
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(AB)(I+\Omega') & = & A\left[B(I+\Omega)\right]\\
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AB+AB\Omega' & = & AB+AB\Omega\\
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\Omega' & = & \Omega\\
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\omega' & = & \omega\end{eqnarray*}
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\end_inset
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i.e.
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the derivative is the identity matrix.
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\end_layout
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\begin_layout Standard
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For the first argument of
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\series bold
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\emph on
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compose
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\series default
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\emph default
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, we will make use of useful property of rotation matrices
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\begin_inset Formula $A$
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\end_inset
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:
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\begin_inset Formula \begin{eqnarray}
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A\Omega A^{T} & = & A\Omega\left[\begin{array}{ccc}
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a_{1} & a_{2} & a_{3}\end{array}\right]\nonumber \\
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& = & A\left[\begin{array}{ccc}
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\omega\times a_{1} & \omega\times a_{2} & \omega\times a_{3}\end{array}\right]\nonumber \\
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& = & \left[\begin{array}{ccc}
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a_{1}(\omega\times a_{1}) & a_{1}(\omega\times a_{2}) & a_{1}(\omega\times a_{3})\\
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a_{2}(\omega\times a_{1}) & a_{2}(\omega\times a_{2}) & a_{2}(\omega\times a_{3})\\
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a_{3}(\omega\times a_{1}) & a_{3}(\omega\times a_{2}) & a_{3}(\omega\times a_{3})\end{array}\right]\nonumber \\
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& = & \left[\begin{array}{ccc}
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\omega(a_{1}\times a_{1}) & \omega(a_{2}\times a_{1}) & \omega(a_{3}\times a_{1})\\
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\omega(a_{1}\times a_{2}) & \omega(a_{2}\times a_{2}) & \omega(a_{3}\times a_{2})\\
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\omega(a_{1}\times a_{3}) & \omega(a_{2}\times a_{3}) & \omega(a_{3}\times a_{3})\end{array}\right]\nonumber \\
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& = & \left[\begin{array}{ccc}
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0 & -\omega a_{3} & \omega a_{2}\\
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\omega a_{3} & 0 & -\omega a_{1}\\
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-\omega a_{2} & \omega a_{1} & 0\end{array}\right]\nonumber \\
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& = & \Skew{A\omega}\label{eq:property1}\end{eqnarray}
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\end_inset
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where
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\begin_inset Formula $a_{x}$
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\end_inset
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are the
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\emph on
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rows
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\emph default
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of
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\begin_inset Formula $A$
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\end_inset
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, we made use of the orthogonality of rotation matrices, and the triple
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product rule:
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\begin_inset Formula \[
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a(b\times c)=b(c\times a)=c(a\times b)\]
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\end_inset
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The derivative in the first argument of
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\series bold
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\emph on
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compose
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\series default
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\emph default
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can then be found using
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:property1"
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\end_inset
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:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \begin{eqnarray*}
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(AB)(I+\Omega') & = & A(I+\Omega)B\\
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AB+AB\Omega' & = & AB+A\Omega B\\
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B\Omega' & = & \Omega B\\
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\Omega' & = & B^{T}\Omega B=\Skew{B^{T}\omega}\\
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\omega' & = & B^{T}\omega\end{eqnarray*}
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\end_inset
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The derivative of
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\series bold
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\emph on
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transpose
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\series default
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\emph default
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can be found similarly:
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\begin_inset Formula \begin{eqnarray*}
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A^{T}(I+\Omega') & = & \left[A(I+\Omega)\right]^{T}\\
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A^{T}+A^{T}\Omega' & = & (I-\Omega)A^{T}\\
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A^{T}\Omega' & = & -\Omega A^{T}\\
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\Omega' & = & -A\Omega A^{T}\\
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& = & -\Skew{A\omega}\\
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\omega' & = & -A\omega\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Standard
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Let's find the derivative of
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\begin_inset Formula $between(A,B)=compose(inverse(A),B)$
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\end_inset
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, so
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial c(i(A),B)}{\partial a}=dc1(i(A),B)di(A)=-B^{T}A=-between(B,A)\]
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\end_inset
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\begin_inset Formula \[
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\frac{\partial c(i(A),B)}{\partial b}=dc2(i(A),B)=I_{3}\]
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\end_inset
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Similarly, the derivative of
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\begin_inset Formula $unrotate(R,x)=rotate(inverse(R),x)$
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\end_inset
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, so
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial r(i(R),x)}{\partial\omega}=dr1(i(R),x)di(R)=R^{T}\Skew xR=\Skew{R^{T}x}\]
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\end_inset
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\begin_inset Formula \[
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\frac{\partial r(i(R),x)}{\partial x}=i(R)=R^{T}\]
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\end_inset
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\end_layout
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\begin_layout Section
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Pose3 (gtsam, old-style exmap)
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\end_layout
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\begin_layout Standard
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In the old-style, we have
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\end_layout
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\begin_layout Standard
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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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\begin_inset Formula $R'=(I+\Omega)R$
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula $t'=t+dt$
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\end_inset
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\end_layout
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\begin_layout Standard
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In this case, the derivative of transform_from,
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\begin_inset Formula $Rx+t$
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\end_inset
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:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial((I+\Omega)Rx+t)}{\partial\omega}=\frac{\partial(\Omega Rx)}{\partial\omega}=\frac{\partial(\omega\times Rx)}{\partial\omega}=-\Skew{Rx}\]
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\end_inset
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and with respect to
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\begin_inset Formula $dt$
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\end_inset
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is easy:
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\end_layout
|
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial(Rx+t+dt)}{\partial dt}=I\]
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\end_inset
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The derivative of transform_to,
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\begin_inset Formula $R^{T}(x-t)$
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\end_inset
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, noting that
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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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\begin_inset Formula $R'^{T}=R^{T}(I-\Omega)$
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\end_inset
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, is
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\end_layout
|
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\begin_layout Standard
|
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\begin_inset Formula \[
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\frac{\partial(R'^{T}(x-t))}{\partial\omega}=\frac{\partial(R^{T}(I-\Omega)(x-t))}{\partial\omega}=-\frac{\partial(R^{T}\Omega(x-t))}{\partial\omega}=-\frac{\partial(\Skew{R^{T}\omega}R^{T}(x-t))}{\partial\omega}=\Skew{R^{T}(x-t)}\frac{\partial(R^{T}\omega)}{\partial\omega}=\Skew{R^{T}(x-t)}R^{T}\]
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\end_inset
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and with respect to
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\begin_inset Formula $dt$
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\end_inset
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is easy:
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\end_layout
|
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial(R^{T}(x-t-dt))}{\partial dt}=-R^{T}\]
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\end_inset
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\end_layout
|
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\begin_layout Standard
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The derivative of
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\begin_inset Formula $inverse=R^{T},-R^{T}t=R^{T}(I,-t)$
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\end_inset
|
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, first derivative of rotation in rotation argument:
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|
\begin_inset Formula \begin{eqnarray*}
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(I+\Omega')R^{T} & = & \left((I+\Omega)R\right)^{T}\\
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R^{T}+\Omega'R^{T} & = & R^{T}(I-\Omega)\\
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\Omega'R^{T} & = & -R^{T}\Omega\\
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\Omega' & = & -R^{T}\Omega R=-\Skew{R^{T}\omega}\\
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\omega' & = & -R^{T}\omega\end{eqnarray*}
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\end_inset
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\end_layout
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\begin_layout Section
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Pose3 (gtsam, new-style exmap)
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\end_layout
|
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\begin_layout Standard
|
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In the new-style exponential map, Pose3 is composed with a delta pose as
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follows
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\end_layout
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\begin_layout Standard
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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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\begin_inset Formula $R'=(I+\Omega)R$
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula $t'=(I+\Omega)t+dt$
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\end_inset
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\end_layout
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\begin_layout Standard
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The derivative of transform_from,
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\begin_inset Formula $Rx+t$
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\end_inset
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:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial((I+\Omega)Rx+(I+\Omega)t)}{\partial\omega}=\frac{\partial(\Omega(Rx+t))}{\partial\omega}=\frac{\partial(\omega\times(Rx+t))}{\partial\omega}=-\Skew{Rx+t}\]
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\end_inset
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and with respect to
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\begin_inset Formula $dt$
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\end_inset
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is easy:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial(Rx+t+dt)}{\partial dt}=I\]
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\end_inset
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The derivative of transform_to,
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\begin_inset Formula $R^{T}(x-t)$
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\end_inset
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, eludes me.
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The calculation below is just an attempt:
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\end_layout
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\begin_layout Standard
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Noting that
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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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\begin_inset Formula $R'^{T}=R^{T}(I-\Omega)$
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\end_inset
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, and
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\begin_inset Formula $(I-\Omega)(x-(I+\Omega)t)=(I-\Omega)(x-t-\Omega t)=x-t-dt-\Omega x+\Omega^{2}t$
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial(R'^{T}(x-t'))}{\partial\omega}=\frac{\partial(R^{T}(I-\Omega)(x-(I+\Omega)t))}{\partial\omega}=-\frac{\partial(R^{T}(\Omega(x-\Omega t)))}{\partial\omega}\]
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\end_inset
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\begin_inset Formula \[
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-\frac{\partial(\Skew{R^{T}\omega}R^{T}x)}{\partial\omega}=\Skew{R^{T}x}\frac{\partial(R^{T}\omega)}{\partial\omega}=\Skew{R^{T}x}R^{T}\]
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\end_inset
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\begin_inset Formula \[
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=\frac{\partial(R^{T}\Omega^{2}t)}{\partial\omega}+\Skew{R^{T}x}R^{T}\]
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\end_inset
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and with respect to
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\begin_inset Formula $dt$
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\end_inset
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is easy:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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\frac{\partial(R^{T}(x-t-dt))}{\partial dt}=-R^{T}\]
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\end_inset
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\end_layout
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\begin_layout Section
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Line3vd
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\end_layout
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\begin_layout Standard
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One representation of a line is through 2 vectors
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\begin_inset Formula $(v,d)$
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\end_inset
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, where
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\begin_inset Formula $v$
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\end_inset
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is the direction and the vector
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\begin_inset Formula $d$
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\end_inset
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points from the orgin to the closest point on the line.
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\end_layout
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\begin_layout Standard
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In this representation, transforming a 3D line from a world coordinate frame
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to a camera at
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\begin_inset Formula $(R_{w}^{c},t^{w})$
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\end_inset
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is done by
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\begin_inset Formula \[
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v^{c}=R_{w}^{c}v^{w}\]
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\end_inset
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\begin_inset Formula \[
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d^{c}=R_{w}^{c}\left(d^{w}+(t^{w}v^{w})v^{w}-t^{w}\right)\]
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\end_inset
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\end_layout
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\begin_layout Section
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Line3
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\end_layout
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\begin_layout Standard
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For 3D lines, we use a parameterization due to C.J.
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Taylor, using a rotation matrix
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\begin_inset Formula $R$
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\end_inset
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and 2 scalars
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\begin_inset Formula $a$
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\end_inset
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and
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\begin_inset Formula $b$
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\end_inset
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.
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The line direction
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\begin_inset Formula $v$
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\end_inset
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is simply the Z-axis of the rotated frame, i.e.,
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\begin_inset Formula $v=R_{3}$
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\end_inset
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, while the vector
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\begin_inset Formula $d$
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\end_inset
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is given by
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\begin_inset Formula $d=aR_{1}+bR_{2}$
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\end_inset
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.
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\end_layout
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\begin_layout Standard
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Now, we will
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\emph on
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not
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\emph default
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use the incremental rotation scheme we used for rotations: because the
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matrix R translates from the line coordinate frame to the world frame,
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we need to apply the incremental rotation on the right-side:
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\begin_inset Formula \[
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R'=R(I+\Omega)\]
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\end_inset
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Projecting a line to 2D can be done easily, as both
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\begin_inset Formula $v$
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\end_inset
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and
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\begin_inset Formula $d$
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\end_inset
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are also the 2D homogenous coordinates of two points on the projected line,
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and hence we have
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\begin_inset Formula \begin{eqnarray*}
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l & = & v\times d\\
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& = & R_{3}\times\left(aR_{1}+bR_{2}\right)\\
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& = & a\left(R_{3}\times R_{1}\right)+b\left(R_{3}\times R_{2}\right)\\
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& = & aR_{2}-bR_{1}\end{eqnarray*}
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\end_inset
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This can be written as a rotation of a point,
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\begin_inset Formula \[
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l=R\left(\begin{array}{c}
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-b\\
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a\\
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0\end{array}\right)\]
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\end_inset
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but because the incremental rotation is now done on the right, we need to
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figure out the derivatives again:
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\begin_inset Formula \begin{equation}
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\frac{\partial(R(I+\Omega)x)}{\partial\omega}=\frac{\partial(R\Omega x)}{\partial\omega}=R\frac{\partial(\Omega x)}{\partial\omega}=R\Skew{-x}\label{eq:rotateRight}\end{equation}
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\end_inset
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and hence the derivative of the projection
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\begin_inset Formula $l$
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\end_inset
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with respect to the rotation matrix
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\begin_inset Formula $R$
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\end_inset
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of the 3D line is
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\begin_inset Formula \begin{equation}
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\frac{\partial(l)}{\partial\omega}=R\Skew{\left(\begin{array}{c}
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b\\
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-a\\
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0\end{array}\right)}=\left[\begin{array}{ccc}
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aR_{3} & bR_{3} & -(aR_{1}+bR_{2})\end{array}\right]\end{equation}
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\end_inset
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or the
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\begin_inset Formula $a,b$
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\end_inset
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scalars:
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\begin_inset Formula \[
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\frac{\partial(l)}{\partial a}=R_{2}\]
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\end_inset
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\begin_inset Formula \[
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\frac{\partial(l)}{\partial b}=-R_{1}\]
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\end_inset
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\end_layout
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\begin_layout Standard
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Transforming a 3D line
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\begin_inset Formula $(R,(a,b))$
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\end_inset
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from a world coordinate frame to a camera frame
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\begin_inset Formula $(R_{w}^{c},t^{w})$
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\end_inset
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is done by
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\end_layout
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\begin_layout Standard
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\begin_inset Formula \[
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R'=R_{w}^{c}R\]
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\end_inset
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\begin_inset Formula \[
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a'=a-R_{1}^{T}t^{w}\]
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\end_inset
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\begin_inset Formula \[
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b'=b-R_{2}^{T}t^{w}\]
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\end_inset
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Again, we need to redo the derivatives, as R is incremented from the right.
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The first argument is incremented from the left, but the result is incremented
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on the right:
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\begin_inset Formula \begin{eqnarray*}
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R'(I+\Omega')=(AB)(I+\Omega') & = & (I+\Skew{S\omega})AB\\
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I+\Omega' & = & (AB)^{T}(I+\Skew{S\omega})(AB)\\
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\Omega' & = & R'^{T}\Skew{S\omega}R'\\
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\Omega' & = & \Skew{R'^{T}S\omega}\\
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\omega' & = & R'^{T}S\omega\end{eqnarray*}
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\end_inset
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For the second argument
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\begin_inset Formula $R$
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\end_inset
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we now simply have:
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\begin_inset Formula \begin{eqnarray*}
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AB(I+\Omega') & = & AB(I+\Omega)\\
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\Omega' & = & \Omega\\
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\omega' & = & \omega\end{eqnarray*}
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\end_inset
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The scalar derivatives can be found by realizing that
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\begin_inset Formula \[
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\left(\begin{array}{c}
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a'\\
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b'\\
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...\end{array}\right)=\left(\begin{array}{c}
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a\\
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b\\
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0\end{array}\right)-R^{T}t^{w}\]
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\end_inset
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where we don't care about the third row.
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Hence
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\begin_inset Formula \[
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\frac{\partial(\left(R(I+\Omega_{2})\right)^{T}t^{w})}{\partial\omega}=-\frac{\partial(\Omega_{2}R^{T}t^{w})}{\partial\omega}=-\Skew{R^{T}t^{w}}=\left[\begin{array}{ccc}
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0 & R_{3}^{T}t^{w} & -R_{2}^{T}t^{w}\\
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-R_{3}^{T}t^{w} & 0 & R_{1}^{T}t^{w}\\
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... & ... & 0\end{array}\right]\]
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\end_inset
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\end_layout
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\begin_layout Section
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2D Line Segments
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\end_layout
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\begin_layout Standard
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The error between an infinite line
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\begin_inset Formula $(a,b,c)$
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\end_inset
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and a 2D line segment
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\begin_inset Formula $((x1,y1),(x2,y2))$
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\end_inset
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is defined in Line3.ml.
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\end_layout
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\begin_layout Section
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\series bold
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Recovering Pose
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\end_layout
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\begin_layout Standard
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Below is the explanaition underlying Pose3.align, i.e.
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aligning two point clouds using SVD.
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Inspired but modified from CVOnline...
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\end_layout
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\begin_layout Standard
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\emph on
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Our
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\emph default
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model is
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\begin_inset Formula \[
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p^{c}=R\left(p^{w}-t\right)\]
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\end_inset
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i.e.,
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\begin_inset Formula $R$
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\end_inset
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is from camera to world, and
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\begin_inset Formula $t$
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\end_inset
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is the camera location in world coordinates.
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The objective function is
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\begin_inset Formula \begin{equation}
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\frac{1}{2}\sum\left(p^{c}-R(p^{w}-t)\right)^{2}=\frac{1}{2}\sum\left(p^{c}-Rp^{w}+Rt\right)^{2}=\frac{1}{2}\sum\left(p^{c}-Rp^{w}-t'\right)^{2}\label{eq:J}\end{equation}
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\end_inset
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where
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\begin_inset Formula $t'=-Rt$
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\end_inset
|
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is the location of the origin in the camera frame.
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Taking the derivative with respect to
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\begin_inset Formula $t'$
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\end_inset
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and setting to zero we have
|
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\begin_inset Formula \[
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\sum\left(p^{c}-Rp^{w}-t'\right)=0\]
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\end_inset
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or
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\begin_inset Formula \begin{equation}
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t'=\frac{1}{n}\sum\left(p^{c}-Rp^{w}\right)=\bar{p}^{c}-R\bar{p}^{w}\label{eq:t}\end{equation}
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\end_inset
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here
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\begin_inset Formula $\bar{p}^{c}$
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\end_inset
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and
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\begin_inset Formula $\bar{p}^{w}$
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\end_inset
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are the point cloud centroids.
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Substituting back into
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:J"
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\end_inset
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, we get
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\begin_inset Formula \[
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\frac{1}{2}\sum\left(p^{c}-R(p^{w}-t)\right)^{2}=\frac{1}{2}\sum\left(\left(p^{c}-\bar{p}^{c}\right)-R\left(p^{w}-\bar{p}^{w}\right)\right)^{2}=\frac{1}{2}\sum\left(\hat{p}^{c}-R\hat{p}^{w}\right)^{2}\]
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\end_inset
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Now, to minimize the above it suffices to maximize (see CVOnline)
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\begin_inset Formula \[
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\mathop{trace}\left(R^{T}C\right)\]
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\end_inset
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where
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\begin_inset Formula $C=\sum\hat{p}^{c}\left(\hat{p}^{w}\right)^{T}$
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\end_inset
|
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is the correlation matrix.
|
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Intuitively, the cloud of points is rotated to align with the principal
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axes.
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This can be achieved by SVD decomposition on
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\begin_inset Formula $C$
|
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\end_inset
|
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\begin_inset Formula \[
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C=USV^{T}\]
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\end_inset
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and setting
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\begin_inset Formula \[
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R=UV^{T}\]
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\end_inset
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Clearly, from
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:t"
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\end_inset
|
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we then also recover the optimal
|
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\begin_inset Formula $t$
|
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\end_inset
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as
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\begin_inset Formula \[
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t=\bar{p}^{w}-R^{T}\bar{p}^{c}\]
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\end_inset
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\end_layout
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\end_body
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\end_document
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