如何在墨卡托地图(JPEG)上获得x,y的纬度,经度？

Question

我有一个墨卡托投影图作为JPEG,我想知道如何将给定的x,y坐标与其纬度和经度相关联.我看过Gudermannian函数,但老实说我不明白如何使用该函数并应用它.即,它期待什么输入？我发现的实现(JavaScript)似乎在-PI和PI之间取得了一个范围,但是我的y值(以像素为单位)与该范围之间的相关性是什么？

此外,我发现这个功能需要一个纬度并返回谷歌地图的瓷砖,谷歌地图也使用墨卡托.似乎如果我知道如何反转这个功能,我会非常接近我的答案.

/*<summary>Get the vertical tile number from a latitude
using Mercator projection formula</summary>*/

    private int getMercatorLatitude(double lati)
    {
        double maxlat = Math.PI;

        double lat = lati;

        if (lat > 90) lat = lat - 180;
        if (lat < -90) lat = lat + 180;

        // conversion degre=>radians
        double phi = Math.PI * lat / 180;

        double res;
        //double temp = Math.Tan(Math.PI / 4 - phi / 2);
        //res = Math.Log(temp);
        res = 0.5 * Math.Log((1 + Math.Sin(phi)) / (1 - Math.Sin(phi)));
        double maxTileY = Math.Pow(2, zoom);
        int result = (int)(((1 - res / maxlat) / 2) * (maxTileY));

        return (result);
    }

Answer 1

这里有一些代码...如果您需要更多解释,请告诉我.

    /// <summary>
    /// Calculates the Y-value (inverse Gudermannian function) for a latitude. 
    /// <para><see cref="http://en.wikipedia.org/wiki/Gudermannian_function"/></para>
    /// </summary>
    /// <param name="latitude">The latitude in degrees to use for calculating the Y-value.</param>
    /// <returns>The Y-value for the given latitude.</returns>
    public static double GudermannianInv(double latitude)
    {
        double sign = Math.Sign(latitude);
        double sin = Math.Sin(latitude * RADIANS_PER_DEGREE * sign);
        return sign * (Math.Log((1.0 + sin) / (1.0 - sin)) / 2.0);
    }

    /// <summary>
    /// Returns the Latitude in degrees for a given Y.
    /// </summary>
    /// <param name="y">Y is in the range of +PI to -PI.</param>
    /// <returns>Latitude in degrees.</returns>
    public static double Gudermannian(double y)
    {
        return Math.Atan(Math.Sinh(y)) * DEGREES_PER_RADIAN;
    }