+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}.
+