above, the phantom dimensions are padded by 1\%.
The other important geometry variables for scanning phantoms are
-the \emph{view diameter}, \emph{scan diameter}, and \emph{focal
-length}. These variables are input into \ctsim\ in terms of
+the \emph{view diameter}, \emph{scan diameter}, \emph{focal
+length}, and \emph{center-detector length}. These variables are input into \ctsim\ in terms of
ratios rather than absolute values.
\subsubsection{Phantom Diameter}\index{Phantom!Diameter}
However, for divergent geometry scanning (equilinear and equiangular),
the \emph{focal length ratio} should be set at \texttt{2} or more
to avoid artifacts. Moreover, a value of less than \texttt{1} is
-physically impossible and it analagous to have having the x-ray
+physically impossible and it analagous to having the x-ray
source inside of the \emph{view diameter}.
+\subsubsection{Center-Detector Length}\index{Center-Detector length}
+The \emph{center-detector length},
+\latexonly{$c$,}\latexignore{\emph{C},}
+is the distance from the center of
+the phantom to the center of the detector array. The center-detector length is set as a ratio,
+\latexonly{$c_r$,}\latexignore{\emph{CR},}
+of the view radius. The center-detector length is
+calculated as
+\latexonly{\begin{equation}f = (v_d / 2) c_r\end{equation}}
+\latexignore{\\\centerline{\emph{F = (Vd / 2) x CR}}}
+
+For parallel geometry scanning, the center-detector length doesn't matter.
+A value of less than \texttt{1} is physically impossible and it analagous to
+having the detector array inside of the \emph{view diameter}.
+
\subsection{Parallel Geometry}\label{geometryparallel}\index{Parallel geometry}\index{Scanner!Parallel}
length} decreases the size of the detector array.
For equiangular geometry, the detectors are equally spaced around a arc
-covering an angular distance of
-\latexonly{$2\,\alpha$.}\latexignore{\emph{2 \alpha}.}
+covering an angular distance of $\alpha$ as viewed from the source. When
+viewed from the center of the scanning, the angular distance is
+\latexonly{$$\pi + \alpha - 2 \, \cos^{-1} \Big( \frac{s_d / 2}{c} \Big)$$}
+\latexignore{\\\emph{pi + \alpha - 2 x acos ((Sd / 2) / C))}\\}
The dotted circle
\latexonly{in figure~\ref{equiangularfig}}
indicates the positions of the detectors in this case.
\latexonly{$\alpha$}\latexignore{\emph{alpha}} and the \emph{focal
length}. This length,
\latexonly{$d_l$,}\latexignore{Dl,} is calculated as
-\latexonly{\begin{equation} d_l = 4\,f \tan (\alpha / 2)\end{equation}}
-\latexignore{\\\centerline{\emph{4 x F x tan(\alpha/2)}}}
+\latexonly{\begin{equation} d_l = 2\,(f + c) \tan (\alpha / 2)\end{equation}}
+\latexignore{\\\centerline{\emph{2 x (F + C) x tan(\alpha/2)}}}
\latexonly{This geometry is shown in figure~\ref{equilinearfig}.}
\begin{figure}
\centerline{\image{10cm;0cm}{equilinear.eps}}