OpenStreetMap – Converting Lat/Lon to Pixel-XY on Cropped Map

openstreetmapspherical-geometrywgs84

I have a GPX-Track containing LatLon coordinates based on WGS84. The track should be drawn on a cropped map, and for this, I need the pixel coordinates. The map tiles are from OSM

I'm not quite sure how to do this, but I figured out a way (without implementing it yet, since I don't have the formulars right now).

Convert LatLon to Spherical Mercator which is used by OSM
Linear transformation spherical mercator -> pixel coordinates.
For step 2., 2 identical points are needed. I guess I can use the map's corner points.

Is there a better way to do this?

Best Answer

Solved the Problem. Code here: http://verbose-mode.de/2013/02/cropped-maps-und-wgs84-koordinaten/

Related Solutions

Coordinate System – How to Map Projection from Lat, Lon to Pixel

For files referenced in this answer, see https://gist.github.com/gagern/6636176.

I have some doubts about your input data. Using the data base of postal codes you mentioned, I computed the center of gravity for every polygon to obtain a set of points in Germany. Feeding that data into your algorithm, as posted in the question, I obtained the following image:

Image of postal codes using original code

Modifying the map dimensions, e.g. $mapWidth = 1000 and $mapHeight = 1500, I was able to see all of Germany. So maybe as a first step would be comparing your set of coordinates to mine. To help with this, I included the postal code for every data point in my data file. I hope this helps you compare them to your data.

Nevertheless, getting parameters right in your code seems a bit tricky. I'd suggest an alternative where the chosen bounding box in geographic coordinates will exactly match the chosen size of the map. So here is the code I'd use:

// This map would show Germany:
$south = deg2rad(47.2);
$north = deg2rad(55.2);
$west = deg2rad(5.8);
$east = deg2rad(15.2);

// This also controls the aspect ratio of the projection
$width = 1000;
$height = 1500;

// Formula for mercator projection y coordinate:
function mercY($lat) { return log(tan($lat/2 + M_PI/4)); }

// Some constants to relate chosen area to screen coordinates
$ymin = mercY($south);
$ymax = mercY($north);
$xFactor = $width/($east - $west);
$yFactor = $height/($ymax - $ymin);

function mapProject($lat, $lon) { // both in radians, use deg2rad if neccessary
  global $xFactor, $yFactor, $west, $ymax;
  $x = $lon;
  $y = mercY($lat);
  $x = ($x - $west)*$xFactor;
  $y = ($ymax - $y)*$yFactor; // y points south
  return array($x, $y);
}

[GIS] The ratio Pixel distance to geographical distance

There are two really good article in the Bing Maps documentation around the tiling system and scale/resolution:

http://msdn.microsoft.com/en-us/library/bb259689.aspx

http://msdn.microsoft.com/en-us/library/aa940990.aspx

Assuming you have, or can get the latlong value of the center of the map, or the center of the ship for more accuracy, you can then calculate the ground resolution in meter's/pixel for that level of latitude. At zoom level 19 a single pixel is approximately 30cm of distance, or 1 foot at the equator. The formula for this is:

ground_resolution = (cos(Latitude * pi/180) * 2 * pi * 6378137) / (256 * 2^zoomLevel)

Once you have this you can get approximate distances by converting your pixel distances to meters.

Best Answer

Related Solutions

Coordinate System – How to Map Projection from Lat, Lon to Pixel

[GIS] The ratio Pixel distance to geographical distance

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