点至直线的距离和垂足点计算

心已赠人 2021-11-29 21:12 769阅读 0赞

点至直线的距离和垂足点计算

//点到直线的垂足点  
public static Coordinate getFootPoint(Coordinate point, Coordinate pnt1, Coordinate pnt2)  
\{  
double A=pnt2.y-pnt1.y; //y2-y1  
double B=pnt1.x-pnt2.x; //x1-x2;  
double C=pnt2.x\*pnt1.y-pnt1.x\*pnt2.y; //x2\*y1-x1\*y2  
if (A \* A + B \* B < 1e-13) \{  
return pnt1; //pnt1与pnt2重叠  
\}  
else if (Math.abs(A \* point.x + B \* point.y + C) < 1e-13) \{  
return point; //point在直线上(pnt1\_pnt2)  
\}  
else \{  
double x = (B \* B \* point.x - A \* B \* point.y - A \* C) / (A \* A + B \* B);  
double y = (-A \* B \* point.x + A \* A \* point.y - B \* C) / (A \* A + B \* B);  
Coordinate fpoint = new Coordinate();  
fpoint.x = x;  
fpoint.y = y;  
return fpoint;  
\}  
\}  
public static double getDistancePoineToLine\_planeCoord(Coordinate point, Coordinate pnt1, Coordinate pnt2)  
\{ //平面坐标中  
double A=pnt2.y-pnt1.y; //y2-y1  
double B=pnt1.x-pnt2.x; //x1-x2;  
double C=pnt2.x\*pnt1.y-pnt1.x\*pnt2.y; //x2\*y1-x1\*y2  
if (A \* A + B \* B < 1e-13) \{ //pnt1与pnt2重叠  
double dx = point.x - pnt1.x;  
double dy = point.y - pnt1.y;  
return Math.sqrt(dx \* dx + dy \* dy);  
\}  
else if (Math.abs(A \* point.x + B \* point.y + C) < 1e-13) \{  
return 0; //point在直线上(pnt1\_pnt2)  
\}  
else \{  
double distance = Math.abs(A \* point.x + B \* point.y + C) / Math.sqrt(A \* A + B \* B);  
return distance;  
\}  
\}  
public static double getDistance\_wgs84(Coordinate p1, Coordinate p2)  
\{  
return getDistance\_wgs84(p1.x,p1.y,p2.x,p2.y);  
\}  
//计算两经纬度点的距离，单位米  
public static double getDistance\_wgs84(double p1\_x,double p1\_y, double p2\_x,double p2\_y)  
\{  
double lat1=p1\_x;  
double lon1=p1\_y;  
double lat2=p2\_x;  
double lon2=p2\_y;  
//  
double earch\_radius=6371008.8; // 地球半径 平均值 米  
//用haversine公式计算球面两点间的距离。  
//经纬度转换成弧度  
double h\_lat1=lat1\*Math.PI/180.0;  
double h\_lon1=lon1\*Math.PI/180.0;  
double h\_lat2=lat2\*Math.PI/180.0;  
double h\_lon2=lon2\*Math.PI/180.0;  
//差值  
double vlon=Math.abs(h\_lon1-h\_lon2);  
double vlat=Math.abs(h\_lat1-h\_lat2);  
//  
double h=HaverSin(vlat)+Math.cos(h\_lat1)\*Math.cos(h\_lat2)\*HaverSin(vlon);  
//  
double d=2\*earch\_radius\*Math.asin(Math.sqrt(h));  
//  
return d;  
\}  
public static double HaverSin(double theta)  
\{  
double v=Math.sin(theta/2);  
return v\*v;  
\};  
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//点到直线的距离 （http://www.my516.com）  
public static double getDistanceByPointToLine\_笛卡尔坐标(Coordinate point, Coordinate pnt1, Coordinate pnt2)  
\{  
double dis = 0;  
if (pnt1.x == pnt2.x)  
\{  
if (pnt1.y == pnt2.y)  
\{  
double dx = point.x - pnt1.x;  
double dy = point.y - pnt1.y;  
dis = Math.sqrt(dx \* dx + dy \* dy);  
\}  
else  
dis = Math.abs(point.x - pnt1.x);  
\}  
else  
\{  
double lineK = (pnt2.y - pnt1.y) / (pnt2.x - pnt1.x);  
double lineC = (pnt2.x \* pnt1.y - pnt1.x \* pnt2.y) / (pnt2.x - pnt1.x);  
dis = Math.abs(lineK \* point.x - point.y + lineC) / (Math.sqrt(lineK \* lineK + 1));  
\}  
//  
return dis;  
\}  
\---------------------

转载于:https://www.cnblogs.com/ly570/p/11044520.html