点矩阵旋转

对于棋盘格角点标定时，有些情况是棋盘格有一定角度旋转，角点也随之有一个旋转角，在写入矩阵就会有很多问题，需要对原角点按旋转角逆转回原来标准情况。

对棋盘格所有点以中心点为圆心，按顺时针旋转alpha°。

数学原理
如下图， 在2维坐标上，有一点p(x, y) , 直线opの长度为r, 直线op和x轴的正向的夹角为a。 直线op围绕原点做逆时针方向b度的旋转，到达p’ (s,t)

 

s = r cos(a + b) = r cos(a)cos(b) – r sin(a)sin(b)   (1.1)
t = r sin(a + b) = r sin(a)cos(b) + r cos(a) sin(b)  (1.2)
其中 x = r cos(a)  , y = r sin(a)
代入(1.1), (1.2) ,
s = x cos(b) – y sin(b)    (1.3)
t = x sin(b) + y cos(b)    (1.4)

void rotatecorners(HObject &image, vector<cv::Point2f> &angleBDpoints,double angle)
{
	double radian = angle * PI / 180;    //注意是弧度
	Mat cvb;                             //读取棋盘格
	int indexx = 0;
	int min_ = 3000;
	for (int i = 0; i < angleBDpoints.size() - 1; i++)
	{
		if ((abs(angleBDpoints[i].x - cvb.cols / 2) + abs(angleBDpoints[i].y - cvb.rows / 2)) < min_)
		{
			min_ = abs(angleBDpoints[i].x - cvb.cols / 2) + abs(angleBDpoints[i].y - cvb.rows / 2);
			indexx = i;
		}
	}
	Point2f pc;
	pc.x = angleBDpoints[indexx].x;
	pc.y = angleBDpoints[indexx].y;
	//==========顺时针
	for (int i = 0; i < angleBDpoints.size(); ++i)
	{
		Point2f xy;
		xy.x = (angleBDpoints[i].x-pc.x)*cos(radian) + (angleBDpoints[i].y-pc.y)*sin(radian)+pc.x;
		xy.y = (angleBDpoints[i].y-pc.y)*cos(radian) - (angleBDpoints[i].x-pc.x)*sin(radian)+pc.y;
		angleBDpoints[i] = xy;
	}
}