  We relate some terms on the boundary of the Newton polygon of the Alexander
polynomial $\Delta(x,y)$ of a rational link to the number and length of
monochromatic twist sites in a particular diagram that we call the standard
form. Normalize $\Delta(x,y)$ so that no $x^{-1}$ or $y^{-1}$ terms appear, but
$x^{-1}\Delta(x,y)$ and $y^{-1}\Delta(x,y)$ have negative exponents, and so
that terms of even total degree are positive and terms with odd total degree
are negative. If the rational link has a reduced alternating diagram with no
self crossings, then $\Delta(-1, 0) = 1$. If the standard form of the rational
link has $m$ monochromatic twist sites, and the $j^{\textrm{th}}$ monochromatic
twist site has $\hat{q}_j$ crossings, then $\Delta(-1, 0) =
\prod_{j=1}^{m}(\hat{q}_j+1)$. Our proof employs Kauffman's clock moves and a
lattice for the terms of $\Delta(x,y)$ in which the $y$-power cannot decrease.
