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<H1><A NAME="SECTION0015250000000000000000"></A>
<A NAME="sec:example_25"></A><A NAME="24692"></A>
<BR>
7.25 Global distribution of antipodes
</H1>
<P>
As promised in Section <A HREF="node144.html#sec:example_23">7.23</A>, we will study antipodes. The antipode of a point at
<!-- MATH
$(\phi, \lambda)$
-->
<IMG
WIDTH="43" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img193.png"
ALT="$(\phi, \lambda)$"> is the point at <!-- MATH
$(-\phi, \lambda + 180)$
-->
<IMG
WIDTH="96" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img194.png"
ALT="$(-\phi, \lambda + 180)$">. We seek an answer
to the question that has plagued so many for so long: Given the distribution of
land and ocean, how often is the antipode of a point on land also on land? And
what about marine antipodes? We use <A NAME="tex2html1484"
HREF="../man/grdlandmask.html"><I><B>grdlandmask</B></I></A><A NAME="25624"></A> and <A NAME="tex2html1485"
HREF="../man/grdmath.html"><I><B>grdmath</B></I></A><A NAME="25629"></A>
to map these distributions and calculate the area of the Earth (in percent)
that goes with each of the three possibilities. To make sense of our <A NAME="tex2html1486"
HREF="../man/grdmath.html"><I><B>grdmath</B></I></A><A NAME="25634"></A>
equations below, note that we first calculate a grid that is +1 when a point and its
antipode is on land, -1 if both are in the ocean, and 0 elsewhere. We then
seek to calculate the area distribution of dry antipodes by only pulling out the nodes
that equal +1. As each point represent an area approximated by <!-- MATH
$\Delta \phi \times \Delta \lambda$
-->
<IMG
WIDTH="61" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img195.png"
ALT="$\Delta \phi \times \Delta \lambda$">
where the <!-- MATH
$\Delta \lambda$
-->
<IMG
WIDTH="25" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img196.png"
ALT="$\Delta \lambda$"> term's actual dimension depends on <IMG
WIDTH="48" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img197.png"
ALT="$\cos (\phi)$">, we need
to allow for that shrinkage, normalize our sum to that of the whole area of the Earth,
and finally convert that ratio to percent. Since the <!-- MATH
$\Delta \lambda$
-->
<IMG
WIDTH="25" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img196.png"
ALT="$\Delta \lambda$">, <IMG
WIDTH="24" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
SRC="img198.png"
ALT="$\Delta \phi$"> terms
appear twice in these expressions they cancel out, leaving the somewhat
intractable expressions below where the sum of <IMG
WIDTH="48" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img197.png"
ALT="$\cos (\phi)$"> for all <IMG
WIDTH="14" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
SRC="img199.png"
ALT="$\phi$"> is known to equal <IMG
WIDTH="46" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
SRC="img200.png"
ALT="$2N_y / \pi$">:
<P>
<BR CLEAR="ALL">
<HR>
<BR>
<PRE>#!/bin/sh
# GMT EXAMPLE 25
#
# Purpose: Display distribution of antipode types
# GMT progs: gmtset, grdlandmask, grdmath, grd2xyz, gmtmath, grdimage, pscoast, pslegend
# Unix progs: cat
#
# Create D minutes global grid with -1 over oceans and +1 over land
ps=example_25.ps
D=30
grdlandmask -Rg -I${D}m -Dc -A500 -N-1/1/1/1/1 -F -Gwetdry.nc
# Manipulate so -1 means ocean/ocean antipode, +1 = land/land, and 0 elsewhere
grdmath wetdry.nc DUP 180 ROTX FLIPUD ADD 2 DIV = key.nc
# Calculate percentage area of each type of antipode match.
grdmath -Rg -I${D}m -F Y COSD 60 $D DIV 360 MUL DUP MUL PI DIV DIV 100 MUL = scale.nc
grdmath key.nc -1 EQ 0 NAN scale.nc MUL = tmp.nc
grd2xyz tmp.nc -S -ZTLf > key.b
ocean=`gmtmath -bi1s -Ca -S key.b SUM UPPER RINT =`
grdmath key.nc 1 EQ 0 NAN scale.nc MUL = tmp.nc
grd2xyz tmp.nc -S -ZTLf > key.b
land=`gmtmath -bi1s -Ca -S key.b SUM UPPER RINT =`
grdmath key.nc 0 EQ 0 NAN scale.nc MUL = tmp.nc
grd2xyz tmp.nc -S -ZTLf > key.b
mixed=`gmtmath -bi1s -Ca -S key.b SUM UPPER RINT =`
# Generate corresponding color table
cat << END > key.cpt
-1.5 blue -0.5 blue
-0.5 gray 0.5 gray
0.5 red 1.5 red
END
# Create the final plot and overlay coastlines
gmtset ANNOT_FONT_SIZE_PRIMARY +10p PLOT_DEGREE_FORMAT dddF
grdimage key.nc -Sn -JKs180/9i -B60/30:."Antipodal comparisons":WsNE -K -Ckey.cpt -Y1.2i \
-U/-0.75i/-0.95i/"Example 25 in Cookbook" > $ps
pscoast -R -J -O -K -Wthinnest -Dc -A500 >> $ps
# Place an explanatory legend below
pslegend -R0/9/0/0.5 -Jx1i/-1i -O -Dx4.5/0/6i/0.3i/TC -Y-0.2i -Fthick >> $ps << END
N 3
S 0.15i s 0.2i red 0.25p 0.3i Terrestrial Antipodes [$land %]
S 0.15i s 0.2i blue 0.25p 0.3i Oceanic Antipodes [$ocean %]
S 0.15i s 0.2i gray 0.25p 0.3i Mixed Antipodes [$mixed %]
END
rm -f *.nc key.* .gmt*
</PRE>
<BR CLEAR="ALL">
<HR>
<P>
In the end we obtain a funny-looking map depicting the antipodal distribution as
well as displaying in legend form the requested percentages (Figure <A HREF="#fig:GMT_example_25">7.25</A>).
Note that the script is set to evaluate a global 30 minute grid for expediency (<IMG
WIDTH="51" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
SRC="img201.png"
ALT="$D = 30$">), hence
several smaller land masses that do have terrestrial antipodes do not show up. If you want
a more accurate map you can set the parameter <IMG
WIDTH="16" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img202.png"
ALT="$D$"> to a smaller increment (try 5 and wait a
few minutes).
<P>
The call to <A NAME="tex2html1487"
HREF="../man/grdimage.html"><I><B>grdimage</B></I></A><A NAME="25640"></A> includes the <B>-Sn</B> to suspend interpolation and only
return the value of the nearest neighbor. This option is particularly practical for plotting
categorical data, like these, that should not be interpolated.
<P>
<DIV ALIGN="CENTER"><A NAME="fig:GMT_example_25"></A><A NAME="25646"></A>
<TABLE>
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 7.25:</STRONG>
Global distribution of antipodes.</CAPTION>
<TR><TD>
<DIV ALIGN="CENTER"><IMG
WIDTH="554" HEIGHT="329" ALIGN="BOTTOM" BORDER="0"
SRC="img203.png"
ALT="\includegraphics[width=\textwidth]{scripts/example_25}"></DIV></TD></TR>
</TABLE>
</DIV>
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Paul Wessel
2010-07-14
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