将经度/纬度转换为 X/Y 坐标

Question

我使用 Google Maps API 创建了一张地图，突出显示了明尼苏达州的所有县。基本上，我使用一组经度/纬度坐标创建了县多边形。这是生成的地图的屏幕截图：-

用户要求之一是能够拥有类似的地图作为图像，以便他们可以将其嵌入到他们的 PowerPoint/主题幻灯片中。我找不到任何有用的 Google Maps API 可以让我按原样保存自定义地图（如果您知道一种方法，请告诉我），所以我想我应该用 Java 中的 Graphics2D 来绘制它。

在阅读了将经度/纬度转换为 X/Y 坐标的公式后，我最终得到了以下代码： -

private static final int    EARTH_RADIUS    = 6371;
private static final double FOCAL_LENGTH    = 500;

...

BufferedImage bi = new BufferedImage(WIDTH, HEIGHT, BufferedImage.TYPE_INT_RGB);
Graphics2D g = bi.createGraphics();

for (Coordinate coordinate : coordinates) {
    double latitude = Double.valueOf(coordinate.getLatitude());
    double longitude = Double.valueOf(coordinate.getLongitude());

    latitude = latitude * Math.PI / 180;
    longitude = longitude * Math.PI / 180;

    double x = EARTH_RADIUS * Math.sin(latitude) * Math.cos(longitude);
    double y = EARTH_RADIUS * Math.sin(latitude) * Math.sin(longitude);
    double z = EARTH_RADIUS * Math.cos(latitude);

    double projectedX = x * FOCAL_LENGTH / (FOCAL_LENGTH + z);
    double projectedY = y * FOCAL_LENGTH / (FOCAL_LENGTH + z);

    // scale the map bigger
    int magnifiedX = (int) Math.round(projectedX * 5);
    int magnifiedY = (int) Math.round(projectedY * 5);

    ...
    g.drawPolygon(...);
    ...
}

生成的地图与 Google Maps API 使用同一组经度/纬度生成的地图类似。然而，它看起来有点倾斜，看起来有点不对劲，我不知道如何解决这个问题。

如何使县的形状看起来与上面 Google Maps API 生成的形状一样？

非常感谢。

最终解决方案

感谢@QuantumMechanic 和@Anon，我终于找到了解决方案。

墨卡托投影在这里确实发挥了作用。我正在使用 Java 地图投影库 来执行墨卡托投影的计算。

private static final int    IMAGE_WIDTH     = 1000;
private static final int    IMAGE_HEIGHT    = 1000;
private static final int    IMAGE_PADDING   = 50;

...

private List<Point2D.Double> convertToXY(List<Coordinate> coordinates) {
    List<Point2D.Double> xys = new ArrayList<Point2D.Double>();

    MercatorProjection projection = new MercatorProjection();

    for (Coordinate coordinate : coordinates) {
        double latitude = Double.valueOf(coordinate.getLatitude());
        double longitude = Double.valueOf(coordinate.getLongitude());

        // convert to radian
        latitude = latitude * Math.PI / 180;
        longitude = longitude * Math.PI / 180;

        Point2D.Double d = projection.project(longitude, latitude, new Point2D.Double());

        // shift by 10 to remove negative Xs and Ys
        // scaling by 6000 to make the map bigger
        int magnifiedX = (int) Math.round((10 + d.x) * 6000);
        int magnifiedY = (int) Math.round((10 + d.y) * 6000);

        minX = (minX == -1) ? magnifiedX : Math.min(minX, magnifiedX);
        minY = (minY == -1) ? magnifiedY : Math.min(minY, magnifiedY);

        xys.add(new Point2D.Double(magnifiedX, magnifiedY));
    }

    return xys;
}

...

通过使用生成的 XY 坐标，地图看起来是倒置的，这是因为我相信 Graphics2D 的 0,0 从左上角开始。因此，我需要通过从图像高度中减去该值来反转 Y，如下所示：-

...

Polygon polygon = new Polygon();

for (Point2D.Double point : xys) {
    int adjustedX = (int) (IMAGE_PADDING + (point.getX() - minX));

    // need to invert the Y since 0,0 starts at top left
    int adjustedY = (int) (IMAGE_HEIGHT - IMAGE_PADDING - (point.getY() - minY));

    polygon.addPoint(adjustedX, adjustedY);
}

...

这是生成的地图：-

非常完美！

更新 01-25-2013

以下是根据宽度和高度（以像素为单位）创建图像映射的代码。在本例中，我不依赖 Java Map Project Library，而是提取相关公式并将其嵌入到我的代码中。与上面依赖任意缩放值的代码示例（上面的示例使用 6000）相比，这使您可以更好地控制地图生成。

public class MapService {
    // CHANGE THIS: the output path of the image to be created
    private static final String IMAGE_FILE_PATH = "/some/user/path/map.png";

    // CHANGE THIS: image width in pixel
    private static final int IMAGE_WIDTH_IN_PX = 300;

    // CHANGE THIS: image height in pixel
    private static final int IMAGE_HEIGHT_IN_PX = 500;

    // CHANGE THIS: minimum padding in pixel
    private static final int MINIMUM_IMAGE_PADDING_IN_PX = 50;

    // formula for quarter PI
    private final static double QUARTERPI = Math.PI / 4.0;

    // some service that provides the county boundaries data in longitude and latitude
    private CountyService countyService;

    public void run() throws Exception {
        // configuring the buffered image and graphics to draw the map
        BufferedImage bufferedImage = new BufferedImage(IMAGE_WIDTH_IN_PX,
                                                        IMAGE_HEIGHT_IN_PX,
                                                        BufferedImage.TYPE_INT_RGB);

        Graphics2D g = bufferedImage.createGraphics();
        Map<RenderingHints.Key, Object> map = new HashMap<RenderingHints.Key, Object>();
        map.put(RenderingHints.KEY_INTERPOLATION, RenderingHints.VALUE_INTERPOLATION_BICUBIC);
        map.put(RenderingHints.KEY_RENDERING, RenderingHints.VALUE_RENDER_QUALITY);
        map.put(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
        RenderingHints renderHints = new RenderingHints(map);
        g.setRenderingHints(renderHints);

        // min and max coordinates, used in the computation below
        Point2D.Double minXY = new Point2D.Double(-1, -1);
        Point2D.Double maxXY = new Point2D.Double(-1, -1);

        // a list of counties where each county contains a list of coordinates that form the county boundary
        Collection<Collection<Point2D.Double>> countyBoundaries = new ArrayList<Collection<Point2D.Double>>();

        // for every county, convert the longitude/latitude to X/Y using Mercator projection formula
        for (County county : countyService.getAllCounties()) {
            Collection<Point2D.Double> lonLat = new ArrayList<Point2D.Double>();

            for (CountyBoundary countyBoundary : county.getCountyBoundaries()) {
                // convert to radian
                double longitude = countyBoundary.getLongitude() * Math.PI / 180;
                double latitude = countyBoundary.getLatitude() * Math.PI / 180;

                Point2D.Double xy = new Point2D.Double();
                xy.x = longitude;
                xy.y = Math.log(Math.tan(QUARTERPI + 0.5 * latitude));

                // The reason we need to determine the min X and Y values is because in order to draw the map,
                // we need to offset the position so that there will be no negative X and Y values
                minXY.x = (minXY.x == -1) ? xy.x : Math.min(minXY.x, xy.x);
                minXY.y = (minXY.y == -1) ? xy.y : Math.min(minXY.y, xy.y);

                lonLat.add(xy);
            }

            countyBoundaries.add(lonLat);
        }

        // readjust coordinate to ensure there are no negative values
        for (Collection<Point2D.Double> points : countyBoundaries) {
            for (Point2D.Double point : points) {
                point.x = point.x - minXY.x;
                point.y = point.y - minXY.y;

                // now, we need to keep track the max X and Y values
                maxXY.x = (maxXY.x == -1) ? point.x : Math.max(maxXY.x, point.x);
                maxXY.y = (maxXY.y == -1) ? point.y : Math.max(maxXY.y, point.y);
            }
        }

        int paddingBothSides = MINIMUM_IMAGE_PADDING_IN_PX * 2;

        // the actual drawing space for the map on the image
        int mapWidth = IMAGE_WIDTH_IN_PX - paddingBothSides;
        int mapHeight = IMAGE_HEIGHT_IN_PX - paddingBothSides;

        // determine the width and height ratio because we need to magnify the map to fit into the given image dimension
        double mapWidthRatio = mapWidth / maxXY.x;
        double mapHeightRatio = mapHeight / maxXY.y;

        // using different ratios for width and height will cause the map to be stretched. So, we have to determine
        // the global ratio that will perfectly fit into the given image dimension
        double globalRatio = Math.min(mapWidthRatio, mapHeightRatio);

        // now we need to readjust the padding to ensure the map is always drawn on the center of the given image dimension
        double heightPadding = (IMAGE_HEIGHT_IN_PX - (globalRatio * maxXY.y)) / 2;
        double widthPadding = (IMAGE_WIDTH_IN_PX - (globalRatio * maxXY.x)) / 2;

        // for each country, draw the boundary using polygon
        for (Collection<Point2D.Double> points : countyBoundaries) {
            Polygon polygon = new Polygon();

            for (Point2D.Double point : points) {
                int adjustedX = (int) (widthPadding + (point.getX() * globalRatio));

                // need to invert the Y since 0,0 starts at top left
                int adjustedY = (int) (IMAGE_HEIGHT_IN_PX - heightPadding - (point.getY() * globalRatio));

                polygon.addPoint(adjustedX, adjustedY);
            }

            g.drawPolygon(polygon);
        }

        // create the image file
        ImageIO.write(bufferedImage, "PNG", new File(IMAGE_FILE_PATH));
    }
}

结果：图像宽度 = 600 像素，图像高度 = 600 像素，图像内边距 = 50 像素

结果：图像宽度 = 300 像素，图像高度 = 500 像素，图像内边距 = 50 像素

Answer 1

绘制地图的一个大问题是地球的球形表面无法方便地转换为平面表示。有许多不同的预测试图解决这个问题。

墨卡托是最简单的一种：它假设等纬度的线是平行的水平线，而等纬度的线是平行的。等经度是平行的垂直线。这对于纬度有效（无论您在哪里，1 纬度大约等于 111 公里），但对于经度无效（经度的表面距离与纬度的余弦成正比）。

然而，只要温度低于 45 度（明尼苏达州的大部分地区都是如此），墨卡托投影就可以很好地工作，并创建大多数人会从小学地图中识别的形式。这非常简单：只需将这些点视为绝对坐标，然后缩放到您在其中绘制它们的任何空间。无需触发。

Answer 2

要将 lat/lon/alt（lat 以北度为单位，lon 以东度为单位，alt 以米为单位）转换为以地心固定坐标 (x,y,z)，请执行以下操作：

double Re = 6378137;
double Rp = 6356752.31424518;

double latrad = lat/180.0*Math.PI;
double lonrad = lon/180.0*Math.PI;

double coslat = Math.cos(latrad);
double sinlat = Math.sin(latrad);
double coslon = Math.cos(lonrad);
double sinlon = Math.sin(lonrad);

double term1 = (Re*Re*coslat)/
  Math.sqrt(Re*Re*coslat*coslat + Rp*Rp*sinlat*sinlat);

double term2 = alt*coslat + term1;

double x=coslon*term2;
double y=sinlon*term2;
double z = alt*sinlat + (Rp*Rp*sinlat)/
  Math.sqrt(Re*Re*coslat*coslat + Rp*Rp*sinlat*sinlat);

Answer 3

请记住，地图的外观取决于用于渲染地图的投影。谷歌地图似乎使用墨卡托投影（或与之非常相似的投影）。你的算法相当于什么投影？如果您希望 2D 表示看起来像 Google 的一样，则需要使用相同的投影。