Question

我在C＃项目上工作，我有一个给定的经度和经度点我需要找到第二个点的纬度和经度，该点是x度角，y是离开给定点的距离。

谢谢你，
Agisilaos

Answer 1

请参阅此页面查看Destination point given distance and bearing from start point

Answer 2

这是从WGS-84改编的http://www.movable-type.co.uk/scripts/latlong-vincenty-direct.html（非球形地球）版本：

private const double wgs84_major = 6378.137;
private const double wgs84_minor = 6356.7523142;
private const double wgs84_flattening = 1D / 298.257223563;

public static bool PointFromDistance(double latitude, double longitude, double angleRadians, double distanceMetres, out double newLatitude, out double newLongitude)
{
    double a = wgs84_major * 1000;
    double b = wgs84_minor * 1000;
    double f = wgs84_flattening;

    double s = distanceMetres;
    double sinAlpha1 = Math.Sin(angleRadians), cosAlpha1 = Math.Cos(angleRadians);

    double tanU1 = (1 - f) * Math.Tan(latitude * Math.PI / 180D);
    double cosU1 = 1 / Math.Sqrt((1 + tanU1 * tanU1)), sinU1 = tanU1 * cosU1;
    double sigma1 = Math.Atan2(tanU1, cosAlpha1);
    double sinAlpha = cosU1 * sinAlpha1;
    double cosSqAlpha = 1 - sinAlpha * sinAlpha;
    double uSq = cosSqAlpha * (a * a - b * b) / (b * b);
    double A = 1 + uSq / 16384 * (4096 + uSq * (-768 + uSq * (320 - 175 * uSq)));
    double B = uSq / 1024 * (256 + uSq * (-128 + uSq * (74 - 47 * uSq)));

    double sigma = s / (b * A), sigmaP = 2 * Math.PI;
    double cos2SigmaM = 0;
    double sinSigma = 0;
    double cosSigma = 0;
    double deltaSigma = 0;
    while (Math.Abs(sigma - sigmaP) > 1e-12)
    {
        cos2SigmaM = Math.Cos(2 * sigma1 + sigma);
        sinSigma = Math.Sin(sigma);
        cosSigma = Math.Cos(sigma);
        deltaSigma = B * sinSigma * (cos2SigmaM + B / 4 * (cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM) -
            B / 6 * cos2SigmaM * (-3 + 4 * sinSigma * sinSigma) * (-3 + 4 * cos2SigmaM * cos2SigmaM)));
        sigmaP = sigma;
        sigma = s / (b * A) + deltaSigma;
    }

    double tmp = sinU1 * sinSigma - cosU1 * cosSigma * cosAlpha1;
    double lat2 = Math.Atan2(sinU1 * cosSigma + cosU1 * sinSigma * cosAlpha1,
        (1 - f) * Math.Sqrt(sinAlpha * sinAlpha + tmp * tmp));
    double lambda = Math.Atan2(sinSigma * sinAlpha1, cosU1 * cosSigma - sinU1 * sinSigma * cosAlpha1);
    double C = f / 16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha));
    double L = lambda - (1 - C) * f * sinAlpha *
        (sigma + C * sinSigma * (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM)));

    double revAz = Math.Atan2(sinAlpha, -tmp);  // final bearing

    newLatitude = lat2 * 180D / Math.PI;
    newLongitude = longitude + L * 180D / Math.PI;

    return true;
}

知道一个点的长度，知道给定点的x距离和y度

2 个答案: