WEBGIS开发 已知一个点的经纬度和两点间的方位角 求另外一个点的经纬度

JS实现：

在WEBGIS开发中经常会遇到一些点位坐标的计算问题，恰好今天工作中遇到一个，就是在已知一个点的经纬度和两点间的方位角，求另外一个点的经纬度：

function calc(lng, lat, brng, dist) {
   var a = 6378137;
    var b = 6356752.3142;
    var f = 1 / 298.257223563;

    var lon1 = lng * 1;
    var lat1 = lat * 1;
    var s = dist;
    var alpha1 = brng * (Math.PI / 180)
    var sinAlpha1 = Math.sin(alpha1);
    var cosAlpha1 = Math.cos(alpha1);
    var tanU1 = (1 - f) * Math.tan(lat1 * (Math.PI / 180));
    var cosU1 = 1 / Math.sqrt((1 + tanU1 * tanU1)), sinU1 = tanU1 * cosU1;
    var sigma1 = Math.atan2(tanU1, cosAlpha1);
    var sinAlpha = cosU1 * sinAlpha1;
    var cosSqAlpha = 1 - sinAlpha * sinAlpha;
    var uSq = cosSqAlpha * (a * a - b * b) / (b * b);
    var A = 1 + uSq / 16384 * (4096 + uSq * (-768 + uSq * (320 - 175 * uSq)));
    var B = uSq / 1024 * (256 + uSq * (-128 + uSq * (74 - 47 * uSq)));
    var sigma = s / (b * A), sigmaP = 2 * Math.PI;
    while (Math.abs(sigma - sigmaP) > 1e-12) {
        var cos2SigmaM = Math.cos(2 * sigma1 + sigma);
        var sinSigma = Math.sin(sigma);
        var cosSigma = Math.cos(sigma);
        var 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;
    }

    var tmp = sinU1 * sinSigma - cosU1 * cosSigma * cosAlpha1;
    var lat2 = Math.atan2(sinU1 * cosSigma + cosU1 * sinSigma * cosAlpha1,
        (1 - f) * Math.sqrt(sinAlpha * sinAlpha + tmp * tmp));
    var lambda = Math.atan2(sinSigma * sinAlpha1, cosU1 * cosSigma - sinU1 * sinSigma * cosAlpha1);
    var C = f / 16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha));
    var L = lambda - (1 - C) * f * sinAlpha *
        (sigma + C * sinSigma * (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM)));

    var revAz = Math.atan2(sinAlpha, -tmp); // final bearing

    var lngLatObj = { lng: lon1 + L * (180 / Math.PI), lat: lat2 * (180 / Math.PI) }
    return lngLatObj;
}

其中lng经度, lat纬度, brng方位角, dist两点之间的距离