java工具类 六 根据经纬度计算距离

Posted 2021-01-30 shaoye007

Java实现根据经纬度计算距离

在项目开发过程中，需要根据两地经纬度坐标计算两地间距离，所用的工具类如下:

Demo1:

 

public static double getDistatce(double lat1, double lat2, double lon1, double lon2) { 
        double R = 6371; 
        double distance = 0.0; 
        double dLat = (lat2 - lat1) * Math.PI / 180; 
        double dLon = (lon2 - lon1) * Math.PI / 180; 
        double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) 
                + Math.cos(lat1 * Math.PI / 180) 
                * Math.cos(lat2 * Math.PI / 180) * Math.sin(dLon / 2) 
                * Math.sin(dLon / 2); 
        distance = (2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a))) * R; 
        return distance; 
    }

Demo2:

 

private static final double EARTH_RADIUS = 6378.137 * 1000; 
    private static double rad(double d) 
    { 
       return d * Math.PI / 180.0; 
    } 
    public static double GetDistance(double lat1, double lng1, double lat2, double lng2) 
    { 
       double radLat1 = rad(lat1); 
       double radLat2 = rad(lat2); 
       double a = radLat1 - radLat2; 
       double b = rad(lng1) - rad(lng2); 
       double s = 2 * Math.asin(Math.sqrt(Math.pow(Math.sin(a/2),2) +  
        Math.cos(radLat1)*Math.cos(radLat2)*Math.pow(Math.sin(b/2),2))); 
       s = s * EARTH_RADIUS ; 
       s = Math.round(s * 10000) / 10000; 
       return s; 
    } 

Demo3:

 

private static final double EARTH_RADIUS = 6378137;//赤道半径(单位m)  
      
    /** 
     * 转化为弧度(rad) 
     * */  
    private static double rad(double d)  
    {  
       return d * Math.PI / 180.0;  
    }  
      
    /** 
     * 基于余弦定理求两经纬度距离 
     * @param lon1 第一点的精度 
     * @param lat1 第一点的纬度 
     * @param lon2 第二点的精度 
     * @param lat3 第二点的纬度 
     * @return 返回的距离，单位km 
     * */  
    public static double LantitudeLongitudeDist(double lon1, double lat1,double lon2, double lat2) {  
        double radLat1 = rad(lat1);  
        double radLat2 = rad(lat2);  
  
        double radLon1 = rad(lon1);  
        double radLon2 = rad(lon2);  
  
        if (radLat1 < 0)  
            radLat1 = Math.PI / 2 + Math.abs(radLat1);// south  
        if (radLat1 > 0)  
            radLat1 = Math.PI / 2 - Math.abs(radLat1);// north  
        if (radLon1 < 0)  
            radLon1 = Math.PI * 2 - Math.abs(radLon1);// west  
        if (radLat2 < 0)  
            radLat2 = Math.PI / 2 + Math.abs(radLat2);// south  
        if (radLat2 > 0)  
            radLat2 = Math.PI / 2 - Math.abs(radLat2);// north  
        if (radLon2 < 0)  
            radLon2 = Math.PI * 2 - Math.abs(radLon2);// west  
        double x1 = EARTH_RADIUS * Math.cos(radLon1) * Math.sin(radLat1);  
        double y1 = EARTH_RADIUS * Math.sin(radLon1) * Math.sin(radLat1);  
        double z1 = EARTH_RADIUS * Math.cos(radLat1);  
  
        double x2 = EARTH_RADIUS * Math.cos(radLon2) * Math.sin(radLat2);  
        double y2 = EARTH_RADIUS * Math.sin(radLon2) * Math.sin(radLat2);  
        double z2 = EARTH_RADIUS * Math.cos(radLat2);  
  
        double d = Math.sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)+ (z1 - z2) * (z1 - z2));  
        //余弦定理求夹角  
        double theta = Math.acos((EARTH_RADIUS * EARTH_RADIUS + EARTH_RADIUS * EARTH_RADIUS - d * d) / (2 * EARTH_RADIUS * EARTH_RADIUS));  
        double dist = theta * EARTH_RADIUS;  
        return dist;  
    }

Demo4:

 

//google map
private static final  double EARTH_RADIUS = 6378137;//赤道半径(单位m)  
      
    /** 
     * 转化为弧度(rad) 
     * */  
    private static double rad(double d)  
    {  
       return d * Math.PI / 180.0;  
    }  
    /** 
     * 基于googleMap中的算法得到两经纬度之间的距离,计算精度与谷歌地图的距离精度差不多，相差范围在0.2米以下 
     * @param lon1 第一点的精度 
     * @param lat1 第一点的纬度 
     * @param lon2 第二点的精度 
     * @param lat3 第二点的纬度 
     * @return 返回的距离，单位km 
     * */  
    public static double GetDistance(double lon1,double lat1,double lon2, double lat2)  
    {  
       double radLat1 = rad(lat1);  
       double radLat2 = rad(lat2);  
       double a = radLat1 - radLat2;  
       double b = rad(lon1) - rad(lon2);  
       double s = 2 * Math.asin(Math.sqrt(Math.pow(Math.sin(a/2),2)+Math.cos(radLat1)*Math.cos(radLat2)*Math.pow(Math.sin(b/2),2)));  
       s = s * EARTH_RADIUS;  
       s = Math.round(s * 10000) / 10000;  
       return s;  
    }

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