PHP Map projections

Question

I have Googled myself to death.. I am attempting to write 2 php functions that will return X and Y from Lat and Long, in both Mercator and flat non-projected (grid) maps. Pr

Answer 1

This presumes your lat/long coords are decimal values, and don't change direction from north/south or east/west within the visible range of the map. If they do, such values should be kept to a single format, and made negativel (eg: 1.0 S would become -1.0 N)

First you'd set the following variables, either finding them with PHP or if you know them already in the script:

$width=//width of image
$height=//height of image

$long_at_left=//Longitude of left-hand coordinates
$long_at_right=//Longitude of right-hand coordinates
$lat_at_left=//Latitude of left-hand coordinates
$lat_at_right=//Latitude of right-hand coordinates

$target_long=//Longitude you want to find
$target_lat=//Latitude you want to find

Then use:

$xtarget=$target_long-$long_at_left;
$ytarget=$target_lat-$lat_at_top;

$xdist=$long_at_left-$long_at_right;
$ydist=$lat_at_top-$lat_at_bottom;

$x=round(($xtarget/$xdist)*$width); //Percentage of distance times width
$y=round(($ytarget/$ydist)*$height); //Percentage of distance times height

Or something of that form should do the trick.

Answer 2

The proportional approach in the previous answer won't work. Mercator projections are quite non-linear.

Here's how I overlay generated images onto a Google or Bing map. In my case, I'm creating a GD image of polygons that will be the overlay. It's much faster to do the polygons in the GD library than the map providers APIs.

First, set up scaling from a standard latitude longitude to a WGS84 projection. Degrees to mercator x-y coordinates in meters.

http://gisgeography.com/wgs84-world-geodetic-system/

// $minlat = minimum image latitude

// $minlon = minimum image longitude

// $maxlat = maximum image latitude

// $maxlon = maximum image longitude

// $latbounds = Image height (in pixels)

// $lonbounds = Image width (in pixels)

$lonrange = abs($maxlon - $minlon);
$WGS84min = log(tan((90.+$minlat)*M_PI/360.))/(M_PI/180.);
$WGS84min = (int) ($WGS84min * 2037598.34/180);
$WGS84max = log(tan((90.+$maxlat)*M_PI/360.))/(M_PI/180.);
$WGS84max = (int) ($WGS84max * 2037598.34/180);
$WGS84diff = $WGS84max - $WGS84min;
$WGS84factor = $latbounds/$WGS84diff;

Then for each latitude/longitude I want to calculate the actual X-Y coordinates on the image.

// $lon1 = the longitude of the point to be converted into image coordinates

// $lat1 = the latitude of the point to be converted into image coordates

X is easy

$x = (int) ((abs($lon1-$minlon)/$lonrange)*$lonbounds);

Y is a bit harder, first calculating to WGS84, and then mapping to the image. Last step, inverting the Y coordinates since the display order is upside down.

$y1 = log(tan((90.+$lat1)*M_PI/360.))/(M_PI/180.);
$y1 = $y1 * 2037598.34/180;
$y1 = (int) (($y1- $WGS84min)*$WGS84factor);
$y  = $latbounds - $y1;

when the image file is complete, use GD to save the image and then use the example in the API library to display your overlay.

https://developers.google.com/maps/documentation/javascript/examples/overlay-simple