java - 将 WGS84 转换为 OSGB36

Question

是否有任何已知的 Java 库允许我将 WGS 84 线转换为 OSGB36，或者是否有我可以使用的好公式？我目前正在使用这个，但它不是很准确，所以我想知道是否有更好的我可以使用。

private static double[] Wgs84ToBNG(double inLat, double inLon) {

    double lat = inLat * Math.PI / 180.0;
    double lon = inLon * Math.PI / 180.0;
    double a = 6377563.396; // Airy 1830 major & minor semi-axes
    double b = 6356256.910;
    double F0 = 0.9996012717; // NatGrid scale factor on central meridian
    double lat0 = 49 * Math.PI / 180.0; // NatGrid true origin
    double lon0 = -2 * Math.PI / 180.0;
    double N0 = -100000; // northing & easting of true origin, metres
    double E0 = 400000;
    double e2 = 1 - (b * b) / (a * a); // eccentricity squared
    double n = (a - b) / (a + b), n2 = n * n, n3 = n * n * n;
    double cosLat = Math.cos(lat), sinLat = Math.sin(lat);
    double nu = a * F0 / Math.sqrt(1 - e2 * sinLat * sinLat); // transverse
                                                                // radius of
                                                                // curvature
    double rho = a * F0 * (1 - e2)
            / Math.pow(1 - e2 * sinLat * sinLat, 1.5); // meridional radius
                                                        // of curvature
    double eta2 = nu / rho - 1;
    double Ma = (1 + n + (5 / 4) * n2 + (5 / 4) * n3) * (lat - lat0);
    double Mb = (3 * n + 3 * n * n + (21 / 8) * n3) * Math.sin(lat - lat0)
            * Math.cos(lat + lat0);
    double Mc = ((15 / 8) * n2 + (15 / 8) * n3)
            * Math.sin(2 * (lat - lat0)) * Math.cos(2 * (lat + lat0));
    double Md = (35 / 24) * n3 * Math.sin(3 * (lat - lat0))
            * Math.cos(3 * (lat + lat0));
    double M = b * F0 * (Ma - Mb + Mc - Md); // meridional arc
    double cos3lat = cosLat * cosLat * cosLat;
    double cos5lat = cos3lat * cosLat * cosLat;
    double tan2lat = Math.tan(lat) * Math.tan(lat);
    double tan4lat = tan2lat * tan2lat;
    double I = M + N0;
    double II = (nu / 2) * sinLat * cosLat;
    double III = (nu / 24) * sinLat * cos3lat * (5 - tan2lat + 9 * eta2);
    double IIIA = (nu / 720) * sinLat * cos5lat
            * (61 - 58 * tan2lat + tan4lat);
    double IV = nu * cosLat;
    double V = (nu / 6) * cos3lat * (nu / rho - tan2lat);
    double VI = (nu / 120)
            * cos5lat
            * (5 - 18 * tan2lat + tan4lat + 14 * eta2 - 58 * tan2lat * eta2);
    double dLon = lon - lon0;
    double dLon2 = dLon * dLon;
    double dLon3 = dLon2 * dLon;
    double dLon4 = dLon3 * dLon;
    double dLon5 = dLon4 * dLon;
    double dLon6 = dLon5 * dLon;
    double N = I + II * dLon2 + III * dLon4 + IIIA * dLon6;
    double E = E0 + IV * dLon + V * dLon3 + VI * dLon5;
    double[] returnValue = { E, N };
    return returnValue;
} 

Answer 1

看看这个page

在“程序/源代码”部分，您会找到标题为“WGS84 Lat/Long <=> OSGB36 Grid References”的小节。在本小节中，有几个指向提供此功能的实用程序的链接。

在此其他post您可以找到此类任务的源代码和解释。