將經度/緯度轉換為X / Y坐標

Question

我使用Google Maps API創建了一張地圖，突出了所有明尼蘇達州的縣。 基本上，我使用一組經度/緯度坐標創建了縣多邊形。 這是生成的地圖的屏幕截圖： -

用戶要求之一是能夠將類似的地圖作為圖像，以便它們可以將其嵌入到PowerPoint /主題幻燈片中。 我找不到任何有用的Google Maps API，它允許我按原樣保存我的自定義地圖（如果你知道一種方式，請告訴我），所以我想我應該用Java中的Graphics2D繪制它。

在閱讀了將經度/緯度轉換為X / Y坐標的公式后，我最終得到以下代碼： -

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(...);
    ...
}

生成的地圖類似於Google Maps API使用相同的經度/緯度生成的地圖。 然而，它似乎有點傾斜，它看起來有點偏，我不知道如何解決這個問題。

如何使縣的形狀看起來就像上面的Google Maps API生成的那樣？

非常感謝。

最終解決方案

我終於通過@QuantumMechanic和@Anon找到了解決方案。

墨卡托投影真的在這里訣竅。 我正在使用Java Map Projection Library來執行墨卡托投影的計算。

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;
}

...

通過使用生成的XY坐標，地圖似乎是反轉的，這是因為我相信graphics2D的0,0從左上角開始。 所以，我需要通過從圖像高度減去值來反轉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);
}

...

這是生成的地圖： -

這十分完美！

更新01-25-2013

這是基於寬度和高度（以像素為單位）創建圖像映射的代碼。 在這種情況下，我不依賴於Java Map項目庫，而是提取出相關的公式並將其嵌入到我的代碼中。 與依賴於任意縮放值的上述代碼示例（上面的示例使用6000）相比，這使您可以更好地控制地圖生成。

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));
    }
}

結果：圖像寬度= 600px，圖像高度= 600px，圖像填充= 50px

結果：圖像寬度= 300像素，圖像高度= 500像素，圖像填充= 50像素

Answer 1

繪制地圖的一個大問題是地球的球面不能方便地轉換成平面表示。 有許多不同的預測試圖解決這個問題。

墨卡托是最簡單的之一：它假設相等緯度的線是平行的水平線，而相等經度的線是平行的垂直線。 這適用於緯度（1度緯度，大約等於111 km，無論您身在何處），但對經度無效（經度的表面距離與latitutude的余弦成正比 ）。

然而，只要你低於大約45度（明尼蘇達州的大部分地區），墨卡托投影效果很好，並創建大多數人將從他們的小學地圖中識別的形式。 它非常簡單：只需將點視為絕對坐標，然后縮放到您繪制它們的任何空間。不需要三角形。

Answer 2

請記住，地圖的外觀是用於渲染地圖的投影的函數。 谷歌地圖似乎使用墨卡托投影（或與其非常相似的東西）。 您的算法等同於什么投影？ 如果您希望2D表示看起來像Google，則需要使用相同的投影。

Answer 3

要將lat / lon / alt（緯度為北緯，lon度數為東，alt為米）轉換為以地球為中心的固定坐標（x，y，z），請執行以下操作：

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);