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我有一个使用谷歌静态地图的 JAVA 项目，经过数小时的工作，我无法正常工作，我会解释一切，希望有人能帮助我。

我使用的是静态地图（480 像素 x 480 像素），地图的中心是 lat=47，lon=1.5，缩放级别是 5。

现在我需要的是当我点击这个静态地图上的一个像素时能够得到纬度和经度。经过一番搜索，我发现我应该使用墨卡托投影（对吗？），我还发现每个缩放级别都会使水平和垂直维度的精度翻倍，但我找不到正确的公式来链接像素、缩放级别和纬度/隆...

我的问题只是从像素中获取纬度/经度，了解中心的坐标和像素以及缩放级别......

先感谢您 ！

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4 回答 4

2

使用墨卡托投影。

如果您投影到[0, 256)by的空间[0,256]：

LatLng(47,=1.5) is Point(129.06666666666666, 90.04191318303863)

在缩放级别 5，这些等同于像素坐标：

x = 129.06666666666666 * 2^5 = 4130
y = 90.04191318303863 * 2^5 = 2881

因此，地图的左上角位于：

x = 4130 - 480/2 = 4070
y = 2881 - 480/2 = 2641

4070 / 2^5 = 127.1875
2641 / 2^5 = 82.53125

最后：

Point(127.1875, 82.53125) is LatLng(53.72271667491848, -1.142578125)
于 2012-05-04T20:45:27.733 回答
1

Google-maps 使用地图瓦片来有效地将世界划分为 256^21 像素瓦片的网格。基本上，这个世界是由 4 个最小缩放的瓷砖组成的。当您开始缩放时，您会得到 16 个图块，然后是 64 个图块，然后是 256 个图块。它基本上是一个四叉树。因为这样的 1d 结构只能展平 2d，所以您还需要 Mercantor 投影或转换为 WGS 84。这是一个很好的资源Convert long/lat to pixel x/y on a given picture。谷歌地图中有将经纬度对转换为像素的功能。这是一个链接，但它说瓷砖只有 128x128：http: //michal.guerquin.com/googlemaps.html。

  1. Google Maps V3 - 如何计算给定边界的缩放级别
  2. http://www.physicsforums.com/showthread.php?t=455491
于 2012-05-04T05:47:57.357 回答
1

基于上面 Chris Broadfoot 的答案中的数学以及Stack Overflow 上的 Mercator Projection 的其他一些代码，我得到了这个

public class MercatorProjection implements Projection {

    private static final double DEFAULT_PROJECTION_WIDTH = 256;
    private static final double DEFAULT_PROJECTION_HEIGHT = 256;

    private double centerLatitude;
    private double centerLongitude;
    private int areaWidthPx;
    private int areaHeightPx;
    // the scale that we would need for the a projection to fit the given area into a world view (1 = global, expect it to be > 1)
    private double areaScale;

    private double projectionWidth;
    private double projectionHeight;
    private double pixelsPerLonDegree;
    private double pixelsPerLonRadian;

    private double projectionCenterPx;
    private double projectionCenterPy;

    public MercatorProjection(
            double centerLatitude,
            double centerLongitude,
            int areaWidthPx,
            int areaHeightPx,
            double areaScale
    ) {
        this.centerLatitude = centerLatitude;
        this.centerLongitude = centerLongitude;
        this.areaWidthPx = areaWidthPx;
        this.areaHeightPx = areaHeightPx;
        this.areaScale = areaScale;

        // TODO stretch the projection to match to deformity at the center lat/lon?
        this.projectionWidth = DEFAULT_PROJECTION_WIDTH;
        this.projectionHeight = DEFAULT_PROJECTION_HEIGHT;
        this.pixelsPerLonDegree = this.projectionWidth / 360;
        this.pixelsPerLonRadian = this.projectionWidth / (2 * Math.PI);

        Point centerPoint = projectLocation(this.centerLatitude, this.centerLongitude);
        this.projectionCenterPx = centerPoint.x * this.areaScale;
        this.projectionCenterPy = centerPoint.y * this.areaScale;
    }

    @Override
    public Location getLocation(int px, int py) {
        double x = this.projectionCenterPx + (px - this.areaWidthPx / 2);
        double y = this.projectionCenterPy + (py - this.areaHeightPx / 2);

        return projectPx(x / this.areaScale, y / this.areaScale);
    }

    @Override
    public Point getPoint(double latitude, double longitude) {
        Point point = projectLocation(latitude, longitude);

        double x = (point.x * this.areaScale - this.projectionCenterPx) + this.areaWidthPx / 2;
        double y = (point.y * this.areaScale - this.projectionCenterPy) + this.areaHeightPx / 2;

        return new Point(x, y);
    }

    // from https://stackoverflow.com/questions/12507274/how-to-get-bounds-of-a-google-static-map

    Location projectPx(double px, double py) {
        final double longitude = (px - this.projectionWidth/2) / this.pixelsPerLonDegree;
        final double latitudeRadians = (py - this.projectionHeight/2) / -this.pixelsPerLonRadian;
        final double latitude = rad2deg(2 * Math.atan(Math.exp(latitudeRadians)) - Math.PI / 2);
        return new Location() {
            @Override
            public double getLatitude() {
                return latitude;
            }

            @Override
            public double getLongitude() {
                return longitude;
            }
        };
    }

    Point projectLocation(double latitude, double longitude) {
        double px = this.projectionWidth / 2 + longitude * this.pixelsPerLonDegree;
        double siny = Math.sin(deg2rad(latitude));
        double py = this.projectionHeight / 2 + 0.5 * Math.log((1 + siny) / (1 - siny) ) * -this.pixelsPerLonRadian;
        Point result = new org.opencv.core.Point(px, py);
        return result;
    }

    private double rad2deg(double rad) {
        return (rad * 180) / Math.PI;
    }

    private double deg2rad(double deg) {
        return (deg * Math.PI) / 180;
    }
}

这是原始答案的单元测试

public class MercatorProjectionTest {

    @Test
    public void testExample() {

        // tests against values in https://stackoverflow.com/questions/10442066/getting-lon-lat-from-pixel-coords-in-google-static-map

        double centerLatitude = 47;
        double centerLongitude = 1.5;

        int areaWidth = 480;
        int areaHeight = 480;

        // google (static) maps zoom level
        int zoom = 5;

        MercatorProjection projection = new MercatorProjection(
                centerLatitude,
                centerLongitude,
                areaWidth,
                areaHeight,
                Math.pow(2, zoom)
        );

        Point centerPoint = projection.projectLocation(centerLatitude, centerLongitude);
        Assert.assertEquals(129.06666666666666, centerPoint.x, 0.001);
        Assert.assertEquals(90.04191318303863, centerPoint.y, 0.001);

        Location topLeftByProjection = projection.projectPx(127.1875, 82.53125);
        Assert.assertEquals(53.72271667491848, topLeftByProjection.getLatitude(), 0.001);
        Assert.assertEquals(-1.142578125, topLeftByProjection.getLongitude(), 0.001);

        // NOTE sample has some pretty serious rounding errors
        Location topLeftByPixel = projection.getLocation(0, 0);
        Assert.assertEquals(53.72271667491848, topLeftByPixel.getLatitude(), 0.05);
        // the math for this is wrong in the sample (see comments)
        Assert.assertEquals(-9, topLeftByPixel.getLongitude(), 0.05);

        Point reverseTopLeftBase = projection.projectLocation(topLeftByPixel.getLatitude(), topLeftByPixel.getLongitude());
        Assert.assertEquals(121.5625, reverseTopLeftBase.x, 0.1);
        Assert.assertEquals(82.53125, reverseTopLeftBase.y, 0.1);

        Point reverseTopLeft = projection.getPoint(topLeftByPixel.getLatitude(), topLeftByPixel.getLongitude());
        Assert.assertEquals(0, reverseTopLeft.x, 0.001);
        Assert.assertEquals(0, reverseTopLeft.y, 0.001);

        Location bottomRightLocation = projection.getLocation(areaWidth, areaHeight);
        Point bottomRight = projection.getPoint(bottomRightLocation.getLatitude(), bottomRightLocation.getLongitude());
        Assert.assertEquals(areaWidth, bottomRight.x, 0.001);
        Assert.assertEquals(areaHeight, bottomRight.y, 0.001);
    }

}

如果您（比如说）使用航空摄影，我觉得该算法没有考虑墨卡托投影的拉伸效果，因此如果您感兴趣的区域不是相对靠近赤道，它可能会失去准确性。我想您可以通过将 x 坐标乘以中心的 cos（纬度）来近似它？

于 2017-01-07T23:58:15.533 回答
0

值得一提的是，您实际上可以让谷歌地图 API 为您提供像素坐标的纬度和经度坐标。

虽然它在 V3 中有点令人费解，但这里有一个如何做到这一点的示例。
（注意：这是假设您已经有一张地图和要转换为 lat&lng 坐标的像素顶点）：

let overlay  = new google.maps.OverlayView();
overlay.draw = function() {};
overlay.onAdd = function() {};
overlay.onRemove = function() {};
overlay.setMap(map);

let latlngObj = overlay.fromContainerPixelToLatLng(new google.maps.Point(pixelVertex.x, pixelVertex.y);

overlay.setMap(null); //removes the overlay

希望对某人有所帮助。

更新：我意识到我做了这两种方式，两者仍然使用相同的方式来创建覆盖（所以我不会复制那个代码）。

let point = new google.maps.Point(628.4160703464878, 244.02779437950872);
console.log(point);
let overlayProj = overlay.getProjection();
console.log(overlayProj);
let latLngVar = overlayProj.fromContainerPixelToLatLng(point);
console.log('the latitude is: '+latLngVar.lat()+' the longitude is: '+latLngVar.lng());
于 2017-08-17T05:45:33.193 回答