根据经纬度,计算距离

复制代码
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;  
    }
复制代码

 

posted @   IWing  阅读(716)  评论(0编辑  收藏  举报
编辑推荐:
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
· AI与.NET技术实操系列:向量存储与相似性搜索在 .NET 中的实现
· 基于Microsoft.Extensions.AI核心库实现RAG应用
· Linux系列:如何用heaptrack跟踪.NET程序的非托管内存泄露
· 开发者必知的日志记录最佳实践
阅读排行:
· TypeScript + Deepseek 打造卜卦网站:技术与玄学的结合
· Manus的开源复刻OpenManus初探
· 写一个简单的SQL生成工具
· AI 智能体引爆开源社区「GitHub 热点速览」
· C#/.NET/.NET Core技术前沿周刊 | 第 29 期(2025年3.1-3.9)
点击右上角即可分享
微信分享提示