平面坐标系,直线绕一固定点旋转,C++实现(利用opencv显示)

参考博客:https://blog.csdn.net/faithmy509/article/details/80235631
推导公式图解

平面中，一个点(x,y)绕任意点(x0,y0)逆时针旋转a度后的坐标(r_x,r_y)

r_x= (x - x0)*cos(a) - (y - y0)*sin(a) + x0 ;
r_y= (x - x0)*sin(a) + (y - y0)*cos(a) +y0 ;

平面中，一个点(x,y)绕任意点(dx,dy)顺时针旋转a度后的坐标(r_x,r_y)

r_x= (x - x0)*cos(a) - (y - y0)*sin(-a) + x0 ;
r_y= (x - x0)*sin(-a) + (y - y0)*cos(a) +y0 ;

C++实现

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define PI                      3.141592654
using namespace std;
using namespace cv;

struct point {
    double x;
    double y;
};

void rotatePoint(double angle, point &rotate_pt, point origin_pt, point center_pt)
{
    double x0 = center_pt.x;
    double y0 = center_pt.y;

    double x = origin_pt.x;
    double y = origin_pt.y;

    rotate_pt.x = (x - x0) * cos(angle* PI / 180) - (y - y0) * sin(angle* PI / 180) + x0;
    rotate_pt.y = (x - x0) * sin(angle* PI / 180) + (y - y0) * cos(angle* PI / 180) + y0;
}

int main() {
    Mat m_image = Mat(500, 500, CV_8UC3, Scalar(0, 0, 0));
    vector<point> origin_line;
    double angle = -25;
    for(int i = 0; i <= 500 ; i+=10)
    {
        point temp_pt;
        temp_pt.x = i;
        temp_pt.y = 2 *i + 4;
        origin_line.push_back(temp_pt);

        circle(m_image, cvPoint(temp_pt.x, temp_pt.y), 3, Scalar(255, 255, 255), -1);//白色点
    }

    for (int i = 0; i < origin_line.size(); i++)
    {
        point origin_pt,center_pt, rotate_pt;
        origin_pt.x = origin_line[i].x;
        origin_pt.y = origin_line[i].y;
        center_pt.x = origin_line[0].x;
        center_pt.y = origin_line[0].y;

        rotatePoint(angle,rotate_pt, origin_pt,center_pt);

        circle(m_image, cvPoint(rotate_pt.x, rotate_pt.y), 2, Scalar(0, 0, 255), -1);//红色点
    }

    imshow("rotate",m_image);
    waitKey();

    return 0;
}

运行后可得到下列图像:

白色是原始线段,红色是旋转后的线段,由于opencv的坐标系的Y轴和我们正常的坐标系相反,所以角度输入为负时看起来逆时针旋转,实际上在正常的坐标系下是上述推导和公式是正确的.