Anyone sad enough (have enough time on their hands) to work out how long the route would actually take, either in reality or at 768mph?
A lot can be estimated with automation:
2431 legs measured, 153 legs skipped, milage is 464452.142804452
takes 25.1981414281929 days
That's as the crow flies between each point, instant rise to top speed from each stop, stop time not included, and I'm short on latitude / longitude data for around 75 locations which means some legs not measured. The full performance metrics of the stock are not clear to me, and I would add a 60% optimism factor to the above figure before quoting it - so I would estimate 40 days and 40 nights if putting a serious proposal to a minister.
Too right I don't have the time
- but a chance to write some Perl which I ain't done in a while
open FH,"1x01.html";
read FH,$content,-s "1x01.html";
@places = $content =~ /\[([A-Z]{3})\]/g;
print @places;
open FH,"rstats2015.txt";
while ($xs = <FH>) {
($tlc, $lat, $long) = (split "\t",$xs)[1,4,5];
if ($lat ne "NULL" and $long ne "NULL") {
$sinfo{$tlc} = [$lat,$long] ;
}
print ("$tlc $lat $long\n");
}
foreach $goto (@places) {
print ($sinfo{$goto});
if ($sinfo{$comefrom} and $sinfo{$goto}) {
$x1 = $sinfo{$comefrom}[0];
$y1 = $sinfo{$comefrom}[1];
$x2 = $sinfo{$goto}[0];
$y2 = $sinfo{$goto}[1];
$dleg = sqrt((abs($x1-$x2)*50)**2 + (abs($y1-$y2)*75)**2);
print ("Have $comefrom $goto $dleg\n");
$acf += $dleg;
$good++;
} else {
print ("Ooops $comefrom $goto\n");
$bad++;
}
$comefrom = $goto;
}
print ("$good legs measured, $bad legs skipped, milage is $acf\n");
$takes = $acf / 768 / 24;
print ("takes $takes days\n")
File rstats2015.txt can be downloaded from
http://www.wellho.net/data/rstats2015.xyz ; file 1x01.html is a "save as
HTML▸ " of the real time trains web page.
Click on the map to explore geographics
