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
/// <summary>
/// 结构：表示一个点
/// </summary>
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" ), ')' );
}
}
/// <summary>
/// 计算点P(x,y)与X轴正方向的夹角
/// </summary>
/// <param name="x">横坐标</param>
/// <param name="y">纵坐标</param>
/// <returns>夹角弧度</returns>
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;
}
/// <summary>
/// 返回点P围绕点A旋转弧度rad后的坐标
/// </summary>
/// <param name="P">待旋转点坐标</param>
/// <param name="A">旋转中心坐标</param>
/// <param name="rad">旋转弧度</param>
/// <param name="isClockwise">true:顺时针/false:逆时针</param>
/// <returns>旋转后坐标</returns>
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。
dy("nrwz");
查看更多关于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的详细内容...