-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{$$f_l=\sqrt{2}l_p/2 = l_p/\sqrt{2}$$,
-$$f_v = \sqrt{2}l_p$$,
-and
-$$d_{hs} = {l_p}/{2}$$.
-Then
-$$\mathrm{dFocalPastPhm} = ({l_p}/{2}) (\sqrt{2}-1)$$
-}
-\begin{figure}
-\image{5cm;0cm}{ctsimfig3.eps}
-\caption{Equilinear and equiangluar geometry when focal length ratio =
-field of view ratio = 1.}
-\end{figure}
-The angle $\alpha$ is now defined as shown in figure 3, and the detector
-length is adjusted to subtend the angle $2\alpha$ as shown. Note that the
-size of the detector array may have changed and the field of view is not
-used.
-For a circular array of detectors, the detectors are spaced around a
-circle covering an angular distance of $2\alpha$. The dotted circle in
-figure 3 indicates the positions of the detectors in this case. Note that
-detectors at the ends of the range would not be illuminated by the source.
-
-Now, consider increasing the focal length ratio to two leaving the
-field of view ratio as 1, as in Figure 4. Now the detectors array is
-denser, and the real field of view is closer to that specified, but note
-again that the field of view is not used. Instead, the focal length is
-used to give a distance from the centre of the phantom to the source, and
-the detector array is adjusted to give an angular coverage to include the
-whole phantom.
+
+Empiric testing with \ctsim\ shows that for very large \emph{fan beam angles},
+greater than approximately
+\latexonly{$120^\circ$,}\latexignore{120 degrees,}
+there are significant artifacts. The primary way to manage the
+\emph{fan beam angle} is by varying the \emph{focal length} since the
+\emph{scan diameter} by the size of the phantom.
+
+To illustrate, the \emph{scan diameter} can be defined as
+\latexonly{$$s_d = s_r v_r p_d$$}\latexignore{\\$$Sd = Sr x Vr x Pd$$\\}
+
+Further, $f$ can be defined as \latexonly{$$f = f_r (v_r p_d /
+2)$$}\latexignore{\\$$F = FR x (VR x Pd)$$\\}
+
+Substituting these equations into \latexignore{the above
+equation,}\latexonly{equation~\ref{alphacalc},} We have,
+\latexonly{
+\begin{eqnarray}
+\alpha &= 2\,\sin^{-1} \frac{s_r v_r p_d / 2}{f_r v_r (p_d / 2)} \nonumber \\
+&= 2\,\sin^{-1} (s_r / f_r)
+\end{eqnarray}
+} \latexignore{\\$$\alpha = 2 sin (Sr / Fr$$\\}
+
+Since in normal scanning $s_r$ = 1, $\alpha$ depends only upon the
+\emph{focal length ratio}.
+
+\subsubsection{Detector Array Size}
+In general, you do not need to be concerned with the detector
+array size. It is automatically calculated by \ctsim. For those
+interested, this section explains how the detector array size is
+calculated.
+
+For parallel geometry, the detector length is equal to the scan
+diameter.
+
+For divergent beam geometries, the size of the detector array also
+depends upon the \emph{focal length}. Increasing the \emph{focal
+length} decreases the size of the detector array while increasing
+the \emph{scan diameter} increases the detector array size.
+
+For equiangular geometry, the detectors are spaced around a circle
+covering an angular distance of
+\latexonly{$2\,\alpha$.}\latexignore{\emph{2 \alpha}.} The dotted
+circle in