點到直線的距離

工作中遇到了點到直線的距離，給出一個點的經緯度，求解這個點到

一條道路的垂直距離。道理表示使用起止點，起止點同樣也是經緯度，

PS:好久沒有用到高數了，真心覺得自己全部忘記了，公式推導了好久，終於搞定了垂足問題。

 

 

//實現的功能：給出某一個點的經緯度和某一條道路的
// 起止經緯度，計算出點到道路上的垂直距離，
// 如果垂足不在道路上，則返回點與道路起止點的最小距離

/**
 * Created by yyk .
 */
public class PointToLine {
    static double DEF_PI = 3.14159265359;
    public static void main(String[] args) {

     //點到直線的垂足
        double[] Point = new double[]{2,2};
        int R = 6371;
        double[] StrLonLat = new double[]{1,1};//StrLonLat EndLonLat表示直線的起止經緯度
        double[] EndLonLat = new double[]{1,3};
        double[] Pedal = pointtoline(Point, StrLonLat, EndLonLat);//Pedal表示垂足的經緯
        //SimpleDateParam dfs=new SimpleDateParam("yyyy-MM-dd HH:mm:ss.SSS");
        //long start=dfs.timestamp();
        double PtoP = EDistance(Point, Pedal, R);//計算考慮了地球的弧度問題
        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) {
            //將兩點經緯度轉換為三維直角坐標
            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);
            //根據直角坐標求兩點間的直線距離（即弦長）
            double L = Math.sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) + (z1 - z2) * (z1 - z2));
            //根據弦長求兩點間的距離（即弧長）
            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("垂足經度:"+Pedal[0]+"垂足緯度:"+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;
        }

    }