/**
* 根据一个经纬度及距离角度，算出另外一个经纬度
* @param {*} lng 经度 113.3960698
* @param {*} lat 纬度 22.941386
* @param {*} brng 方位角 45 ---- 正北方：000°或360° 正东方：090° 正南方：180° 正西方：270°
* @param {*} dist 90000距离(米)
*
*/
mapNumberUtil.getLonAndLat = function (lng,lat,brng,dist){
//大地坐标系资料WGS-84 长半径a=6378137 短半径b=6356752.3142 扁率f=1/298.2572236
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 = mapNumberUtil.rad(brng);
var sinAlpha1 = Math.sin(alpha1);
var cosAlpha1 = Math.cos(alpha1);

var tanU1 = (1-f) * Math.tan(mapNumberUtil.rad(lat1));
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+mapNumberUtil.deg(L),lat:mapNumberUtil.deg(lat2)}
return lngLatObj;
}

有任何问题欢迎在评论区留言

或者添加学姐微信
DKA-2018

梦缠绕的时候 发表于 2019-8-5 10:55
或者添加学姐微信
DKA-2018

学姐，mapNumberUtil.rad和mapNumberUtil.deg方法是什么啊