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November 20th, 2008, 11:42 AM

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Converting Easting/Northing to Latitude/Longitude
Does anybody have a prewritten class that can do this in C#? I need to be able to pass it an UK easting/northing (e.g. 262600, 191900) and return a Lat/Long (e.g. 51.3456, 3.17624).
Your help would be most appreciated.
Elwap.

November 20th, 2008, 02:12 PM

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Why thank you. Your answer was most informative. I'll be sure to seek advice here again.

November 20th, 2008, 02:26 PM


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:)
It was meant to be a funnier way of saying, "I don't know"
I have written the code in PHP, but I don't know where I put it, so it's not that hard (provided you don't try to understand the math).
/ Sam Judson : Wrox Technical Editor /

November 20th, 2008, 04:31 PM

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I certainly don't want to understand the math, as I've seen the formulas. That's why we have encapsulation :)
If you do find the code, I'd really appreciate it. Or if you can point me to a website which has a no frills way of describing how to implement the formula, I'd be most grateful. There must be a converter out there. Here's great online VB.NET to C# converter I use for code snippets all the time:
http://www.developerfusion.com/tools.../vbtocsharp/

November 21st, 2008, 05:20 AM


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Code:
<?php
$lon = 0;
$lat = 0;
function convertXToLong($xCoord, $yCoord)
{
global $lat, $lon;
$E = $xCoord;
$N = $yCoord;
$a = 6377563.396;
$b = 6356256.910; // Airy 1830 major & minor semiaxes
$F0 = 0.9996012717; // NatGrid scale factor on central meridian
$lat0 = 49*pi()/180;
$lon0 = 2*pi()/180; // NatGrid true origin
$N0 = 100000;
$E0 = 400000; // northing & easting of true origin, metres
$e2 = 1  ($b*$b)/($a*$a); // eccentricity squared
$n = ($a$b)/($a+$b);
$n2 = $n*$n;
$n3 = $n*$n*$n;
$lat=$lat0;
$M=0;
do {
$lat = ($N$N0$M)/($a*$F0) + $lat;
$Ma = (1 + $n + (5/4)*$n2 + (5/4)*$n3) * ($lat$lat0);
$Mb = (3*$n + 3*$n*$n + (21/8)*$n3) * sin($lat$lat0) * cos($lat+$lat0);
$Mc = ((15/8)*$n2 + (15/8)*$n3) * sin(2*($lat$lat0)) * cos(2*($lat+$lat0));
$Md = (35/24)*$n3 * sin(3*($lat$lat0)) * cos(3*($lat+$lat0));
$M = $b * $F0 * ($Ma  $Mb + $Mc  $Md); // meridional arc
} while ($N$N0$M >= 0.00001); // ie until < 0.01mm
$cosLat = cos($lat);
$sinLat = sin($lat);
$nu = $a*$F0/sqrt(1$e2*$sinLat*$sinLat); // transverse radius of curvature
$rho = $a*$F0*(1$e2)/pow(1$e2*$sinLat*$sinLat, 1.5); // meridional radius of curvature
$eta2 = $nu/$rho1;
$tanLat = tan($lat);
$tan2lat = $tanLat*$tanLat;
$tan4lat = $tan2lat*$tan2lat;
$tan6lat = $tan4lat*tan2lat;
$secLat = 1/$cosLat;
$nu3 = $nu*$nu*$nu;
$nu5 = $nu3*$nu*$nu;
$nu7 = $nu5*$nu*$nu;
$VII = $tanLat/(2*$rho*$nu);
$VIII = $tanLat/(24*$rho*$nu3)*(5+3*$tan2lat+$eta29*$tan2lat*$eta2);
$IX = $tanLat/(720*$rho*$nu5)*(61+90*$tan2lat+45*$tan4lat);
$X = $secLat/$nu;
$XI = $secLat/(6*$nu3)*($nu/$rho+2*$tan2lat);
$XII = $secLat/(120*$nu5)*(5+28*$tan2lat+24*$tan4lat);
$XIIA = $secLat/(5040*$nu7)*(61+662*$tan2lat+1320*$tan4lat+720*$tan6lat);
$dE = ($E$E0);
$dE2 = $dE*$dE;
$dE3 = $dE2*$dE;
$dE4 = $dE2*$dE2;
$dE5 = $dE3*$dE2;
$dE6 = $dE4*$dE2;
$dE7 = $dE5*$dE2;
$lat = $lat  $VII*$dE2 + $VIII*$dE4  $IX*$dE6;
$lon = $lon0 + $X*$dE  $XI*$dE3 + $XII*$dE5  $XIIA*$dE7;
$lat = $lat*180/pi();
$lon = $lon*180/pi();
}
?>
/ Sam Judson : Wrox Technical Editor /

November 21st, 2008, 06:41 AM

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Thanks, that worked a treat. :)


