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. Problem being every calculation I have ran across assumes your map(s) has the same lat and lon at the corners, and then the result is in meters. ugh
Here is what I have .. maps of different sizes, different lat's, long's at the 4 corners. I downloaded the Proj4 php port, but with zero documentation and more code then I need, I was over whelmed ...
HELP !!
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.
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
来源:https://stackoverflow.com/questions/10887692/php-map-projections