Rotate points by an angle
Hello, i am trying to rotate a set of points in a vector<points> by an user-defined angle and found a solution at SO. In the following code the dimension of the output image (rotated by 45 degree) is correct but the position of the points seem to be shifted. Can someone give me a tip, what the problem is?
cv::Point rotate2d(const cv::Point& inPoint, const double& angRad)
{
cv::Point outPoint;
//CW rotation
outPoint.x = std::cos(angRad)*inPoint.x - std::sin(angRad)*inPoint.y;
outPoint.y = std::sin(angRad)*inPoint.x + std::cos(angRad)*inPoint.y;
return outPoint;
}
cv::Point rotatePoint(const cv::Point& inPoint, const cv::Point& center, const double& angRad)
{
return rotate2d(inPoint - center, angRad) + center;
}
int main( int, char** argv )
{
// Create an dark Image with a gray line in the middle
Mat img = Mat(83, 500, CV_8U);
img = Scalar(0);
vector<Point> pointsModel;
for ( int i = 0; i<500; i++)
{
pointsModel.push_back(Point(i , 41));
}
for ( int i=0; i<pointsModel.size(); i++)
{
circle(img, pointsModel[i], 1, Scalar(120,120,120), 1, LINE_8, 0);
}
imshow("Points", img);
// Rotate Points
vector<Point> rotatedPoints;
Point tmpPoint;
cv::Point pt( img.cols/2.0, img.rows/2.0 );
for ( int i=0; i<pointsModel.size(); i++)
{
tmpPoint = rotatePoint(pointsModel[i] , pt , 0.7854);
rotatedPoints.push_back(tmpPoint);
}
Rect bb = boundingRect(rotatedPoints);
cout << bb;
Mat rotatedImg = Mat(bb.height, bb.width, img.type());
rotatedImg = Scalar(0);
for (int i=0; i<rotatedPoints.size(); i++ )
{
circle(rotatedImg, rotatedPoints[i], 1, Scalar(120,120,120), 1, LINE_8, 0);
}
imshow("Points Rotated", rotatedImg);
waitKey();
return 0;
}
Why don't you use getRotationMatrix2D and pDst = (rot2d*Mat(p[i])).t()
Thank you LBerger, I thought about asking the question directly at stackoverflow, because the solution was from there but you were faster with your comment than I was with deleting the question. If the duplication is troublesome, I can remove it here.
To your comment: With Mat(p[i]) you mean the input point, don't you?