r491: no message

[ctsim.git] / doc / ctsim-concepts.tex

diff --git a/doc/ctsim-concepts.tex b/doc/ctsim-concepts.tex
index bd6bffbb32563ce68df9ec51b23ff895a2286abc..521ee4662547358b6521e9e47271d3c4db204593 100644 (file)
--- a/doc/ctsim-concepts.tex
+++ b/doc/ctsim-concepts.tex
@@ -115,15 +115,18 @@ of this square.
 The two other important variables are the field-of-view-ratio ($f_{vR}$) 
 and the focal-length-ratio ($f_{lR}$).  These are used along with $l_p$ to
 define the focal length and the field of view (not ratios) according to
-\begin{equation}
+\latexonly{\begin{equation}
 f_l = \sqrt{2} (l_p/2)(f_{lR})= (l_p/\sqrt{2}) f_{lR}
 \end{equation}
 \begin{equation}
 f_v = \sqrt{2}l_p f_{vR}
-\end{equation}
+\end{equation}}
 So the field of view ratio is specified in units of the phantom diameter,
 whereas the focal length is specified in units of the phantom radius.  The 
-factor of $\sqrt(2)$ can be understood if one refers to figure 1, where
+factor of 
+\latexonly{$\sqrt(2)$}
+\latexignore{sqrt(2)} 
+can be understood if one refers to figure 1, where
 we consider the case of a first generation parallel beam CT scanner.
 
 \subsubsection{Parallel Geometry}\label{geometryparallel}\index{Concepts,Scanner,Geometries,Parallel}
@@ -168,21 +171,24 @@ See figure 2.
 
 For these geometries, the following logic is executed:  A variable dHalfSquare
 $d_{hs}$ is defined as
-\begin{equation}
+\latexonly{\begin{equation}
 d_{hs} = (f_v)/(2\sqrt{2}) = (l_p/2) f_{vR}
-\end{equation}
+\end{equation}}
 This is then subtracted from the focal length $f_l$ as calculated above, and 
 assigned to a new variable $\mathrm{dFocalPastPhm} = f_l - d_{hs}$.  Since $f_l$ and 
 $d_{hs}$ are derived from the phantom dimension and the input focal length and field of view ratios, one can write, 
+\latexonly{
 \begin{equation}
 \mathrm{dFocalPastPhm} = f_l -d_{hs} 
        = \sqrt{2}(l_p/2) f_{lR} - (l_p/2) f_{vR} = l_p(\sqrt{2}f_{lR} - f_{vR})
 \end{equation}
+}
 If this quantity is less than or equal to zero, then at least for some
 projections  the source is inside the phantom.  Perhaps a figure will help at
 this point. Consider first the case where $f_{vR} = f_{lR} =1 $, figure 3. The
 square in the figure bounds the phantom and has sides $l_p$.  For this case
 then, 
+\latexonly{
 \[ 
 f_l=\sqrt{2}l_p/2 = l_p/\sqrt{2},
 \]
@@ -197,6 +203,7 @@ Then
 \[
 \mathrm{dFocalPastPhm} = ({l_p}/{2}) (\sqrt{2}-1)
 \]
+}
 \begin{figure}
 \includegraphics[height=0.5\textheight]{ctsimfig3.eps}
 \caption{Equilinear and equiangluar geometry when focal length ratio =