java 百度地图判断两点距离2

package baiduApi;

public class BaiDuMap {
static double DEF_PI = 3.14159265359; // PI
static double DEF_2PI= 6.28318530712; // 2*PI
static double DEF_PI180= 0.01745329252; // PI/180.0
static double DEF_R =6370693.5; // radius of earth
//适用于近距离
public static double GetShortDistance(double lon1, double lat1, double lon2, double lat2)
{
double ew1, ns1, ew2, ns2;
double dx, dy, dew;
double distance;
// 角度转换为弧度
ew1 = lon1 * DEF_PI180;
ns1 = lat1 * DEF_PI180;
ew2 = lon2 * DEF_PI180;
ns2 = lat2 * DEF_PI180;
// 经度差
dew = ew1 - ew2;
// 若跨东经和西经180 度,进行调整
if (dew > DEF_PI)
dew = DEF_2PI - dew;
else if (dew < -DEF_PI)
dew = DEF_2PI + dew;
dx = DEF_R * Math.cos(ns1) * dew; // 东西方向长度(在纬度圈上的投影长度)
dy = DEF_R * (ns1 - ns2); // 南北方向长度(在经度圈上的投影长度)
// 勾股定理求斜边长
distance = Math.sqrt(dx * dx + dy * dy);
return distance;
}
//适用于远距离
public static double GetLongDistance(double lon1, double lat1, double lon2, double lat2)
{
double ew1, ns1, ew2, ns2;
double distance;
// 角度转换为弧度
ew1 = lon1 * DEF_PI180;
ns1 = lat1 * DEF_PI180;
ew2 = lon2 * DEF_PI180;
ns2 = lat2 * DEF_PI180;
// 求大圆劣弧与球心所夹的角(弧度)
distance = Math.sin(ns1) * Math.sin(ns2) + Math.cos(ns1) * Math.cos(ns2) * Math.cos(ew1 - ew2);
// 调整到[-1..1]范围内,避免溢出
if (distance > 1.0)
distance = 1.0;
else if (distance < -1.0)
distance = -1.0;
// 求大圆劣弧长度
distance = DEF_R * Math.acos(distance);
return distance;
}
public static void main(String[] args) {
double mLat1 = 23.5539530; // point1纬度
double mLon1 = 114.8903920; // point1经度
double mLat2 = 23.5554550;// point2纬度
double mLon2 = 114.8868890;// point2经度
double distance = BaiDuMap.GetShortDistance(mLon1, mLat1, mLon2, mLat2);
System.out.println(distance);
}

}

原文地址:https://www.cnblogs.com/zhangzhiqin/p/9774272.html