工作中遇到了点到直线的距离，给出一个点的经纬度，求解这个点到

一条道路的垂直距离。道理表示使用起止点，起止点同样也是经纬度，

PS:好久没有用到高数了，真心觉得自己全部忘记了，公式推导了好久，终于搞定了垂足问题。

 

 

//实现的功能：给出某一个点的经纬度和某一条道路的 // 起止经纬度，计算出点到道路上的垂直距离， // 如果垂足不在道路上，则返回点与道路起止点的最小距离 /**  * Created by yyk .  */ public class PointToLine {
      static double DEF_PI = 3.14159265359;     public static void main(String[] args) {
       //点到直线的垂足         double[] Point = new double[]{2,2};         int R = 6371;         double[] StrLonLat = new double[]{1,1};//StrLonLat EndLonLat表示直线的起止经纬度         double[] EndLonLat = new double[]{1,3};         double[] Pedal = pointtoline(Point, StrLonLat, EndLonLat);//Pedal表示垂足的经纬         //SimpleDateParam dfs=new SimpleDateParam("yyyy-MM-dd HH:mm:ss.SSS");         //long start=dfs.timestamp();         double PtoP = EDistance(Point, Pedal, R);//计算考虑了地球的弧度问题         double PtoStr = EDistance(Point, StrLonLat, R);         double PtoEnd = EDistance(Point, EndLonLat, R);         double min = MIN(PtoP, PtoStr, PtoEnd);//要求的值         System.out.println("min"+min);       //  long end=dfs.timestamp();        // long time=start-end;       //  System.out.println("Time:"+time);     }

public static double EDistance(double[] LonLat1, double[] LonLat2, int R) {
              //将两点经纬度转换为三维直角坐标             double x1 = Math.cos(LonLat1[0] / 180 * DEF_PI);             double y1 = Math.sin(LonLat1[0] / 180 * DEF_PI);             double z1 = Math.tan(LonLat1[1] / 180 * DEF_PI);             double x2 = Math.cos(LonLat2[0] / 180 * DEF_PI);             double y2 = Math.sin(LonLat2[0] / 180 * DEF_PI);             double z2 = Math.tan(LonLat2[1] / 180 * DEF_PI);             //根据直角坐标求两点间的直线距离（即弦长）             double L = Math.sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) + (z1 - z2) * (z1 - z2));             //根据弦长求两点间的距离（即弧长）             double Eudis = R * DEF_PI * 2 * (Math.asin(0.5 * L / R)) / 180;             return Eudis;         }         public static double[] pointtoline(double[] point, double[] strlonlat, double[] endlonlat) {
              double A = endlonlat[1] - strlonlat[1];             double B = strlonlat[0] - endlonlat[0];             double C = strlonlat[1] * endlonlat[0] - strlonlat[0] * endlonlat[1];             double[] Pedal = new double[2];             Pedal[0] = (point[0] * B * B - point[1] * A * B-A*C) / (A * A + B * B);             Pedal[1] = (point[1]*A*A - A * B * point[0] - B * C) / (A * A + B * B);             System.out.println("垂足经度:"+Pedal[0]+"垂足纬度:"+Pedal[1]);             return Pedal;         }         public static double MIN(double pp, double ps, double pe) {
              double mindis;             if (pp < ps && pp < pe)                 mindis = pp;             else if (ps < pe)                 mindis = ps;             else                 mindis = pe;             if(mindis==pp)                 System.out.println("pedal in lineAB");             else                 System.out.println("pedal is not in lineAB");             return mindis;         }     }