百度地图根据坐标计算两个点的角度和风向

ゝ一纸荒年。 2023-10-04 16:52 14阅读 0赞

话不多说，直接上代码：

/**
     * 计算角度
     *
     * @param lat1
     * @param lng1
     * @param lat2
     * @param lng2
     * @return
     */
    public static double getJiaoDu(double lng1, double lat1, double lng2, double lat2) {
        double x1 = lng1;
        double y1 = lat1;
        double x2 = lng2;
        double y2 = lat2;
        double pi = Math.PI;
        double w1 = y1 / 180 * pi;
        double j1 = x1 / 180 * pi;
        double w2 = y2 / 180 * pi;
        double j2 = x2 / 180 * pi;
        double ret;
        if (j1 == j2) {
            if (w1 > w2) ret = 270; //北半球的情况，南半球忽略
            else if (w1 < w2) ret = 90;
            else ret = -1;//位置完全相同
        } else {
            ret = 4 * Math.pow(Math.sin((w1 - w2) / 2), 2) - Math.pow(Math.sin((j1 - j2) / 2) * (Math.cos(w1) - Math.cos(w2)), 2);
            ret = Math.sqrt(ret);
            double temp = (Math.sin(Math.abs(j1 - j2) / 2) * (Math.cos(w1) + Math.cos(w2)));
            ret = ret / temp;
            ret = Math.atan(ret) / pi * 180;
            if (j1 > j2) {// 1为参考点坐标
                if (w1 > w2) ret += 180;
                else ret = 180 - ret;
            } else if (w1 > w2) ret = 360 - ret;
        }
        double jiaodu = ret;
        System.out.println("度数为ret: " + ret);
        return jiaodu;
    }

/**
     * 根据角度计算方向
     *
     * @param jiaodu
     * @return
     */
    public String StringDirection(double jiaodu) {
        if ((jiaodu <= 10) || (jiaodu > 350)) return "东";
        if ((jiaodu > 10) && (jiaodu <= 80)) return "东北";
        if ((jiaodu > 80) && (jiaodu <= 100)) return "北";
        if ((jiaodu > 100) && (jiaodu <= 170)) return "西北";
        if ((jiaodu > 170) && (jiaodu <= 190)) return "西";
        if ((jiaodu > 190) && (jiaodu <= 260)) return "西南";
        if ((jiaodu > 260) && (jiaodu <= 280)) return "南";
        if ((jiaodu > 280) && (jiaodu <= 350)) return "东南";
        return null;
    }

有需要可单独拿出去使用，根据角度判断方向是根据项目地图需求来做的判断，可根据你们的项目需要来进行修改哦。