Question

我有一个使用谷歌静态地图的JAVA项目，经过几个小时的工作，我无法工作，我会解释一切，我希望有人能够帮助我。

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，缩放级别为5。

现在我需要的是当我点击此静态地图上的像素时能够获得lat和lon。经过一些搜索，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在水平和垂直尺寸上的精度都加倍，但我找不到合适的公式来链接像素，缩放级别和lat /经度...

我的问题只是关于从像素获取lat / lon，知道中心的坐标和像素以及缩放级别......

提前谢谢！

Answer 1

使用墨卡托投影。

如果您通过[0, 256)投射到[0,256]的空格：

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

在缩放级别5，这些等于像素坐标：

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)

Answer 2

Google-maps使用地图图块将世界有效划分为256 ^ 21像素图块的网格。基本上世界是由最低变焦的4个瓷砖组成。当你开始变焦时，你得到16个瓷砖，然后是64个瓷砖，然后是256个瓷砖。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。 Google地图中的功能可以从lat-long对转换为像素。这是一个链接，但它说瓷砖只有128x128：http://michal.guerquin.com/googlemaps.html。

Answer 3

根据Chris Broadfoot上面的答案和some other code on Stack Overflow for the Mercator Projection中的数学，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到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接近，它可能会失去准确性赤道。我猜您可以通过将x坐标乘以中心的cos（纬度）来近似它吗？

Answer 4

似乎值得一提的是，您实际上可以使用google maps API为您提供纬度和频率。像素坐标的纵坐标。

虽然在V3中有点令人费解但这里有一个如何做到这一点的例子。
（注意：假设您已经有一个地图并且像素顶点要转换为lat＆amp; lng坐标）：

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

希望能有所帮助。

更新：我意识到我做了两种方式，两种方式仍然使用相同的方式创建叠加层（所以我不会复制该代码）。

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());

从谷歌静态地图中的像素坐标获取lon / lat

4 个答案: