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  • 给出两点经纬度算距离

     1 <span style="font-size:14px;font-weight: normal;"> 
     2 private static final  double EARTH_RADIUS = 6378137;//赤道半径(单位m)  
     3       
     4     /** 
     5      * 转化为弧度(rad) 
     6      * */  
     7     private static double rad(double d)  
     8     {  
     9        return d * Math.PI / 180.0;  
    10     }  
    11       
    12     /** 
    13      * 基于余弦定理求两经纬度距离 
    14      * @param lon1 第一点的精度 
    15      * @param lat1 第一点的纬度 
    16      * @param lon2 第二点的精度 
    17      * @param lat3 第二点的纬度 
    18      * @return 返回的距离,单位km 
    19      * */  
    20     public static double LantitudeLongitudeDist(double lon1, double lat1,double lon2, double lat2) {  
    21         double radLat1 = rad(lat1);  
    22         double radLat2 = rad(lat2);  
    23   
    24         double radLon1 = rad(lon1);  
    25         double radLon2 = rad(lon2);  
    26   
    27         if (radLat1 < 0)  
    28             radLat1 = Math.PI / 2 + Math.abs(radLat1);// south  
    29         if (radLat1 > 0)  
    30             radLat1 = Math.PI / 2 - Math.abs(radLat1);// north  
    31         if (radLon1 < 0)  
    32             radLon1 = Math.PI * 2 - Math.abs(radLon1);// west  
    33         if (radLat2 < 0)  
    34             radLat2 = Math.PI / 2 + Math.abs(radLat2);// south  
    35         if (radLat2 > 0)  
    36             radLat2 = Math.PI / 2 - Math.abs(radLat2);// north  
    37         if (radLon2 < 0)  
    38             radLon2 = Math.PI * 2 - Math.abs(radLon2);// west  
    39         double x1 = EARTH_RADIUS * Math.cos(radLon1) * Math.sin(radLat1);  
    40         double y1 = EARTH_RADIUS * Math.sin(radLon1) * Math.sin(radLat1);  
    41         double z1 = EARTH_RADIUS * Math.cos(radLat1);  
    42   
    43         double x2 = EARTH_RADIUS * Math.cos(radLon2) * Math.sin(radLat2);  
    44         double y2 = EARTH_RADIUS * Math.sin(radLon2) * Math.sin(radLat2);  
    45         double z2 = EARTH_RADIUS * Math.cos(radLat2);  
    46   
    47         double d = Math.sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)+ (z1 - z2) * (z1 - z2));  
    48         //余弦定理求夹角  
    49         double theta = Math.acos((EARTH_RADIUS * EARTH_RADIUS + EARTH_RADIUS * EARTH_RADIUS - d * d) / (2 * EARTH_RADIUS * EARTH_RADIUS));  
    50         double dist = theta * EARTH_RADIUS;  
    51         return dist;  
    52 }
    53 </span> 
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  • 原文地址:https://www.cnblogs.com/qiangshu/p/4596705.html
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