• 根据经纬度与距离计算另一个经纬度


    /**
    * 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)
    *
    * 1、角度转换为弧度公式:弧度=角度×(π ÷度180 )
    * 2、弧度转换为角度公式: 角度=弧度×(180÷π)
    */
    mapHelper.prototype.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 = brng * (Math.PI / 180)//mapNumberUtil.rad(brng);
        var sinAlpha1 = Math.sin(alpha1);
        var cosAlpha1 = Math.cos(alpha1);
    
        //var tanU1 = (1 - f) * Math.tan(mapNumberUtil.rad(lat1));
        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;
    }
    

      抄的https://www.cnblogs.com/object360/p/11301494.html

    只是将弧度与角度直接写在了里面

  • 相关阅读:
    标准化:释放物联网全部潜力的万能钥匙
    Linux系统中如何删除交换文件,来解决突然关闭命令行,导致文件无法打开的问题
    echarts设置图例以及网络格局的相关细节设置
    JSON.parse()与JSON.stringify()的区别,具体例子,一看就明白
    HttpServletRequest
    HttpServletResponse
    JavaWeb核心之Servlet
    Http协议和Tomcat服务器
    dom4j解析
    DTD和Schema约束
  • 原文地址:https://www.cnblogs.com/zhoushangwu/p/12922817.html
Copyright © 2020-2023  润新知