OSM:从球形墨卡托"EPSG:900913"中的投影坐标转换​​关系.到"EPSG:4326"；座标 [英] OSM: convert ties from projected coordinates in spherical mercator "EPSG:900913" to "EPSG:4326" coordinates

问题描述

我正在使用带有图层的地图(来自示例):

I'm using a map with a layer (from the example):

var lonLat = new OpenLayers.LonLat(40.4088576, -86.8576718)
                .transform(
                    new OpenLayers.Projection("EPSG:4326"), // transform from WGS 1984
                    map.getProjectionObject() // to Spherical Mercator Projection
                );

在moveend上，我得到中心坐标:

on moveend i'm get center coordinates:

map.getCenter();
map.getZoom();

和缩放级别:4925535.4503328，-9668990.0134335、12

and zoom level: 4925535.4503328, -9668990.0134335, 12

使用文档中的算法

public PointF TileToWorldPos(double tile_x, double tile_y, int zoom) 
{
    PointF p = new Point();
    double n = Math.PI - ((2.0 * Math.PI * tile_y) / Math.Pow(2.0, zoom));

    p.X = (float)((tile_x / Math.Pow(2.0, zoom) * 360.0) - 180.0);
    p.Y = (float)(180.0 / Math.PI * Math.Atan(Math.Sinh(n)));

    return p;
}

我得到Y〜90和X〜432662

i get Y ~ 90, and X ~ 432662

但是我需要坐标范围:-180..180

but i need coordinates in bounds: -180..180

类似的东西:40.4088576，-86.8576718

something like: 40.4088576, -86.8576718

怎么了?

推荐答案

为什么不让OpenLayers为您重新投影呢?因此，如果您希望将中心点设为WGS84，则只需执行以下操作:

Why not just get OpenLayers to project it back for you? So if you want the centre point as WGS84 then just do:

var center = map.getCenter().transform(map.getProjectionObject(),
                                       new OpenLayers.Projection("EPSG:4326"));

我想您会发现无论如何，它都会做您想要的事情，至少在我理解正确的问题之后……

I think you'll find that will do what you want anyway, at least if I've understood the question right...

顺便说一句，EPSG:4326是您似乎正在寻找的自然WGS84纬度/经度值-大数字是球形墨卡托投影的坐标，即EPSG:900913.

By the way, EPSG:4326 is the natural WGS84 lat/lon values that you seem to be looking for - the large numbers are the projected coordinates in spherical mercator, which is EPSG:900913.

这篇关于OSM:从球形墨卡托"EPSG:900913"中的投影坐标转换​​关系.到"EPSG:4326"；座标的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！