C#实现计算一个点围绕另一个点旋转指定弧度后坐标值的方法

本文实例讲述了C#实现计算一个点围绕另一个点旋转指定弧度后坐标值的方法。

分享给大家供大家参考。

具体如下:

1.示例图

P(x1,y1)以点A(a,b)为圆心,旋转弧度为θ,求旋转后点Q(x2,y2)的坐标

2.实现方法

先将坐标平移,计算点(x1-a,y1-b)围绕原点旋转后的坐标,再将坐标轴平移到原状态


/// 
/// 结构:表示一个点
/// 
struct Point
{
 //横、纵坐标
 public double x, y;
 //构造函数
 public Point(double x, double y)
 {
  this.x = x;
  this.y = y;
 }
 //该点到指定点pTarget的距离
 public double DistanceTo(Point p)
 {
  return Math.Sqrt((p.x - x) * (p.x - x) + (p.y - y) * (p.y - y));
 }
 //重写ToString方法
 public override string ToString()
 {
  return string.Concat("Point (",
   this.x.ToString("#0.000"), ',',
   this.y.ToString("#0.000"), ')');
 }
}
/// 
/// 计算点P(x,y)与X轴正方向的夹角
/// 
/// 横坐标
/// 纵坐标
/// 夹角弧度
private static double radPOX(double x,double y)
{
 //P在(0,0)的情况
 if (x == 0 && y == 0) return 0;
 //P在四个坐标轴上的情况:x正、x负、y正、y负
 if (y == 0 && x > 0) return 0;
 if (y == 0 && x < 0) return Math.PI;
 if (x == 0 && y > 0) return Math.PI / 2;
 if (x == 0 && y < 0) return Math.PI / 2 * 3;
 //点在第一、二、三、四象限时的情况
 if (x > 0 && y > 0) return Math.Atan(y / x);
 if (x < 0 && y > 0) return Math.PI - Math.Atan(y / -x);
 if (x < 0 && y < 0) return Math.PI + Math.Atan(-y / -x);
 if (x > 0 && y < 0) return Math.PI * 2 - Math.Atan(-y / x);
 return 0;
}
/// 
/// 返回点P围绕点A旋转弧度rad后的坐标
/// 
/// 待旋转点坐标
/// 旋转中心坐标
/// 旋转弧度
/// true:顺时针/false:逆时针
/// 旋转后坐标
private static Point RotatePoint(Point P, Point A, 
 double rad, bool isClockwise = true)
{
 //点Temp1
 Point Temp1 = new Point(P.x - A.x, P.y - A.y);
 //点Temp1到原点的长度
 double lenO2Temp1 = Temp1.DistanceTo(new Point(0, 0));
 //∠T1OX弧度
 double angT1OX = radPOX(Temp1.x, Temp1.y);
 //∠T2OX弧度(T2为T1以O为圆心旋转弧度rad)
 double angT2OX = angT1OX - (isClockwise ? 1 : -1) * rad;
 //点Temp2
 Point Temp2 = new Point(
  lenO2Temp1 * Math.Cos(angT2OX),
  lenO2Temp1 * Math.Sin(angT2OX));
 //点Q
 return new Point(Temp2.x + A.x, Temp2.y + A.y);
}


3.Main函数调用


static void Main(string[] args)
{
 //求两点间长度
 Point A = new Point(0, 0);
 Point B = new Point(3, 4);
 Console.WriteLine("Length of AB: " + A.DistanceTo(B));
 Point P = new Point(5, -5);
 Console.WriteLine(P.ToString() + 'n');
 //绕原点(0,0)逆时针旋转
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 9, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 10, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 11, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 12, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 13, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 14, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 15, false));
 Console.WriteLine(RotatePoint(P, new Point(0, 0), Math.PI / 4 * 16, false));
 Console.WriteLine();
 //绕点(2.5,2.5)顺时针旋转
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 1));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 2));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 3));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 4));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 5));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 6));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 7));
 Console.WriteLine(RotatePoint(P, new Point(2.5, 2.5), Math.PI / 4 * 8));
 Console.ReadLine();
}


4.运行结果:

希望本文所述对大家的C#程序设计有所帮助。

0.235068s