\latexonly{$p_l$}\latexignore{\emph{Pl}}
be the width and height of this square. The diameter of this boundary box,
\latexonly{$p_d$,}\latexignore{\emph{Pd},}
-is then
+\rtfsp is then
\latexignore{\\$$\emph{Pl x sqrt(2)}$$\\}
\latexonly{$$p_d = p_l \sqrt{2}$$}
CT scanners actually collect projections around a circle rather than a
square. The diameter of this circle is also the diameter of the boundary
square
-\latexonly{$p_d$.}\latexignore{\emph{Pd}.}
+\latexonly{$p_d$.}\latexignore{\rtfsp\emph{Pd}.}
These relationships are diagrammed in figure 2.1.
\subsubsection{View Diameter}
The \emph{view diameter} is the area that is being processed during scanning of phantoms as
well as during rasterization of phantoms. By default, the \emph{view diameter}
-is set equal to the \emph{phantom diameter}. It may be useful, especially for
+\rtfsp is set equal to the \emph{phantom diameter}. It may be useful, especially for
experimental reasons, to process an area larger (and maybe even smaller) than
the phantom. Thus, during rasterization or during projections, \ctsim\ will
ask for a \emph{view ratio},
-\latexonly{$v_{R}$.}\latexignore{\emph{VR}.}
+\latexonly{$v_r$.}\latexignore{\rtfsp \emph{VR}.}
The \emph{view diameter} is then set as
-\latexonly{$$v_d = p_d v_{R}$$}\latexignore{\\$$\emph{Vd = Pd x VR}$$}
+\latexonly{$$v_d = p_d v_r$$}\latexignore{\\$$\emph{Vd = Pd x VR}$$}
By using a
-\latexonly{$v_{R}$}\latexignore{\emph{VR}}
+\latexonly{$v_r$}\latexignore{\emph{VR}}
less than 1, \ctsim\ will allow
for a \emph{view diameter} less than
\emph{phantom diameter}.
By default, the entire \emph{view diameter} is scanned. For experimental
purposes, it may be desirable to scan an area either larger or smaller than
the \emph{view diameter}. Thus, the concept of \emph{scan ratio}
-\latexonly{$s_{R}$}\latexignore{\emph{SR}}
+\latexonly{$s_r$}\latexignore{\emph{SR}}
is born. The scan diameter
\latexonly{$s_d$}\latexignore{\emph{Sd}}
is the diameter over which x-rays are collected and is defined as
-\latexonly{$$s_d = v_d s_{R}$$}\latexignore{\\$$\emph{Sd = Vd x SR}$$\\}
+\latexonly{$$s_d = v_d s_r$$}\latexignore{\\$$\emph{Sd = Vd x SR}$$\\}
By default and for all ordinary scanning, the \emph{scan ratio} is to
\texttt{1}. If the \emph{scan ratio} is less than \texttt{1},
you can expect significant artifacts.
\latexonly{$f$,}\latexignore{\emph{F},}
is the distance of the X-ray source to the center of
the phantom. The focal length is set as a ratio,
-\latexonly{$f_{R}$,}\latexignore{\emph{FR},}
+\latexonly{$f_r$,}\latexignore{\emph{FR},}
of the view radius. Focal length is
calculated as
-\latexonly{$$f = (v_d / 2) f_R$$}\latexignore{\\$$\emph{F = (Vd / 2) x FR}$$}
+\latexonly{$$f = (v_d / 2) f_r$$}\latexignore{\\$$\emph{F = (Vd / 2) x FR}$$}
For parallel geometry scanning, the focal length doesn't matter. However,
divergent geometry scanning (equilinear and equiangular), the \emph{focal
For these divergent beam geometries, the \emph{fan beam angle} needs
to be calculated. For real-world CT scanners, this is fixed at the
time of manufacture. \ctsim, however, calculates the \emph{fan beam angle},
-\latexonly{$\alpha$,}\latexignore{\emph{alpha},}
-from the diameter of the \emph{scan diameter} and the \emph{focal length}
+$\alpha$ from the diameter of the \emph{scan diameter} and the \emph{focal length}
\latexignore{\\$$\emph{alpha = 2 x asin ( (Sd / 2) / f)}$$\\}
-\latexonly{$$\alpha = 2 \sin^{-1} ((s_d / 2) / f)$$}
+\latexonly{\begin{equation}\label{alphacalc}\alpha = 2 \sin^{-1} ((s_d / 2) / f)\end{equation}}
This is illustrated in figure 2.3.
\begin{figure}
\image{10cm;0cm}{alphacalc.eps}
\emph{fan beam angle} is by varying the \emph{focal length} since the
\emph{scan diameter} by the size of the phantom.
-$$s_d = p_d v_R s_R$$
-If $v_r = 1$ and $s_R = 1$, then $s_d = p_d$. Further, $f = f_R v_R (p_d / 2)$
-Plugging these equations into the above equation,
-$$\alpha = 2\,\sin^{-1} \frac{p_d / 2}{f_R (p_d / 2)}$$
-$$\alpha = 2\,\sin^{-1} (1 / f_R)$$
+To illustrate, the \emph{scan diameter} can be defined as
+\latexonly{$$s_d = v_r s_r p_d$$}\latexignore{\\$$Sd = Vr x Sr x Pd$$\\}
+
+If $v_r = 1$ and $s_r = 1$, then $s_d = p_d$. Further, $f = f_r v_r (p_d / 2)$
+Plugging these equations into
+\latexignore{the above equation,}\latexonly{equation~\ref{alphacalc},}
+We have,
+\latexonly{
+\begin{eqnarray}
+\alpha &= 2\,\sin^{-1} \frac{p_d / 2}{f_r (p_d / 2)} \nonumber \\
+&= 2\,\sin^{-1} (1 / f_r)
+\end{eqnarray}
+}
Thus, $\alpha$ depends only upon the \emph{focal length ratio}.
projects / ctsim.git / commitdiff