我有一份莫斯科市地图。我们在谷歌地图的图像上添加了一些艺术元素，但GPS坐标和像素之间的相对关系保持不变。
问题：如何将我们拥有的各种数据点的GPS坐标转换为图像中的像素坐标？
理想情况下，我可以用JavaScript来完成，但PHP也可以。
我知道在小尺度（例如城市尺度）上很容易做到这一点（需要了解一个图片角落的地理坐标，然后分别学习在OX和OY轴上图片上一个像素在地理坐标上的“价格”）。
但在大尺度（国家尺度）上，一个像素的“价格”将不是恒定的，并且会变化得足够强烈，因此上述描述的方法无法应用。
如何解决大尺度的问题？
更新：
我没有使用谷歌地图API，我只有：物体的地理坐标（它们来自谷歌地图），我仍然在我的网站上拥有一个简单的*.gif图片，在其中我必须画出对应于地理坐标的点。
现在，从WGS84点到图像像素的过程如下：
你可以这样使用proj4js库：
// include the library
<script src="lib/proj4js-combined.js"></script> //adjust the path for your server
//or else use the compressed version
// creating source and destination Proj4js objects
// once initialized, these may be re-used as often as needed
var source = new Proj4js.Proj('EPSG:4326'); //source coordinates will be in Longitude/Latitude, WGS84
var dest = new Proj4js.Proj('EPSG:3785'); //destination coordinates in meters, global spherical mercators projection, see http://spatialreference.org/ref/epsg/3785/
// transforming point coordinates
var p = new Proj4js.Point(-76.0,45.0); //any object will do as long as it has 'x' and 'y' properties
Proj4js.transform(source, dest, p); //do the transformation. x and y are modified in place
//p.x and p.y are now EPSG:3785 in meters
Proj4js.Proj = OpenLayers.Class({...})这一行报错，提示“OpenLayers未定义”。该如何解决这个问题？ - Kalinin你需要在你的语言中实现Google地图API的投影。我有这个的C#源代码：
public class GoogleMapsAPIProjection
{
private readonly double PixelTileSize = 256d;
private readonly double DegreesToRadiansRatio = 180d / Math.PI;
private readonly double RadiansToDegreesRatio = Math.PI / 180d;
private readonly PointF PixelGlobeCenter;
private readonly double XPixelsToDegreesRatio;
private readonly double YPixelsToRadiansRatio;
public GoogleMapsAPIProjection(double zoomLevel)
{
var pixelGlobeSize = this.PixelTileSize * Math.Pow(2d, zoomLevel);
this.XPixelsToDegreesRatio = pixelGlobeSize / 360d;
this.YPixelsToRadiansRatio = pixelGlobeSize / (2d * Math.PI);
var halfPixelGlobeSize = Convert.ToSingle(pixelGlobeSize / 2d);
this.PixelGlobeCenter = new PointF(
halfPixelGlobeSize, halfPixelGlobeSize);
}
public PointF FromCoordinatesToPixel(PointF coordinates)
{
var x = Math.Round(this.PixelGlobeCenter.X
+ (coordinates.X * this.XPixelsToDegreesRatio));
var f = Math.Min(
Math.Max(
Math.Sin(coordinates.Y * RadiansToDegreesRatio),
-0.9999d),
0.9999d);
var y = Math.Round(this.PixelGlobeCenter.Y + .5d *
Math.Log((1d + f) / (1d - f)) * -this.YPixelsToRadiansRatio);
return new PointF(Convert.ToSingle(x), Convert.ToSingle(y));
}
public PointF FromPixelToCoordinates(PointF pixel)
{
var longitude = (pixel.X - this.PixelGlobeCenter.X) /
this.XPixelsToDegreesRatio;
var latitude = (2 * Math.Atan(Math.Exp(
(pixel.Y - this.PixelGlobeCenter.Y) / -this.YPixelsToRadiansRatio))
- Math.PI / 2) * DegreesToRadiansRatio;
return new PointF(
Convert.ToSingle(latitude),
Convert.ToSingle(longitude));
}
}
来源：
FromPixelToCoordinates 方法难道不应该返回 PointF(Convert.ToSingle(longitude), Convert.ToSingle(latitude)) 吗？[+1] - TLamaGmap2.fromLatLngToContainerPixel(latlng)
例如：
var gmap2 = new GMap2(document.getElementById("map_canvas"));
var geocoder = new GClientGeocoder();
geocoder.getLatLng( "1600 Pennsylvania Avenue NW Washington, D.C. 20500",
function( latlng ) {
var pixel_coords = gmap2.fromLatLngToContainerPixel(latlng);
window.alert( "The White House is at pixel coordinates (" +
pixel_coodrs.x + ", " + pixel_coords.y + ") on the " +
"map image shown on this page." );
}
);
GMap2.fromLatLngToDivPixel(latlng:GLatLng)
如果您想支持多个地图投影（即支持使用不同投影的许多不同地图），那么您绝对需要使用像PROJ.4这样的库，但我不确定您在Javascript或PHP中会找到什么。
// Adapted from : http://blog.cppse.nl/x-y-to-lat-lon-for-google-maps
window.geo = {
glOffset: Math.pow(2,28), //268435456,
glRadius: Math.pow(2,28) / Math.PI,
a: Math.pow(2,28),
b: 85445659.4471,
c: 0.017453292519943,
d: 0.0000006705522537,
e: Math.E, //2.7182818284590452353602875,
p: Math.PI / 180,
lonToX: function(lon) {
return Math.round(this.glOffset + this.glRadius * lon * this.p);
},
XtoLon: function(x) {
return -180 + this.d * x;
},
latToY: function(lat) {
return Math.round(this.glOffset - this.glRadius *
Math.log((1 + Math.sin(lat * this.p)) /
(1 - Math.sin(lat * this.p))) / 2);
},
YtoLat: function(y) {
return Math.asin(Math.pow(this.e,(2*this.a/this.b - 2*y/this.b)) /
(Math.pow(this.e, (2*this.a/this.b - 2*y/this.b))+1) -
1/(Math.pow(this.e, (2*this.a/this.b - 2*y/this.b))+1)
) / this.c;
},
deltaLonPerDeltaX: function(deltaX, zoom) {
// 2^(7+zoom) pixels <---> 180 degrees
return deltaX * 180 / Math.pow(2, 7+zoom);
},
deltaLatPerDeltaY: function(deltaY, zoom, startLat) {
// more complex because of the curvature, we calculte it by difference
var startY = this.latToY(startLat),
endY = startY + deltaY * Math.pow(2, 28-7-zoom),
endLat = this.YtoLat(endY);
return ( endLat - startLat ); // = deltaLat
}
}
http://www.johndcook.com/blog/2009/04/27/converting-miles-to-degrees-longitude-or-latitude/
我的方法不需要使用库，也可以使用裁剪后的地图。这意味着它可以只使用从墨卡托投影图像中提取的部分来工作。也许对某些人有帮助：https://dev59.com/Z3I95IYBdhLWcg3w5SSh#10401734
我遇到了一些困难 - 我同时拥有OpenStreetMap和Google街景地图，并希望投影外部图形图像
var map = new OpenLayers.Map({
div:"map-id",
allOverlays: true
});
var osm = new OpenLayers.Layer.OSM("OpenStreeMao");
var gmap = new OpenLayers.Layer.Google("Google Streets", {visibility: false});
map.addLayers([osm,gmap]);
var vectorLayer = new OpenLayers.Layer.Vector("IconLayer");
var lonlatObject = new OpenLayers.LonLat(24.938622,60.170421).transform(
new OpenLayers.Projection("EPSG:4326"), map.getProjectionObject()
);
console.log(lonlatObject);
var point = new OpenLayers.Geometry.Point(lonlatObject.lon, lonlatObject.lat);
console.log(point);
var point2 = new OpenLayers.Geometry.Point(lonlatObject.x, lonlatObject.y);
console.log(point2);
var feature = new OpenLayers.Feature.Vector(point, null, {
externalGraphic: "http://cdn1.iconfinder.com/data/icons/SUPERVISTA/networking/png/72/antenna.png",
graphicWidth: 72,
graphicHeight: 72,
fillOpacity: 1
});
vectorLayer.addFeatures(feature);
map.addLayer(vectorLayer);
map.setCenter(
new OpenLayers.LonLat(24.938622,60.170421).transform(
new OpenLayers.Projection("EPSG:4326"), map.getProjectionObject()
),
12);
map.addControl(new OpenLayers.Control.LayerSwitcher());