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; }