emgucv angle detection for rectangle in csharp

Question

I am using following thread to perform angle detection for a rectangle image. Detect centre and angle of rectangles in an image using Opencv

I am stuck at following piece of code.

cv::Point2f edge1 = cv::Vec2f(rect_points[1].x, rect_points[1].y) - cv::Vec2f(rect_points[0].x, rect_points[0].y);
cv::Point2f edge2 = cv::Vec2f(rect_points[2].x, rect_points[2].y) - cv::Vec2f(rect_points[1].x, rect_points[1].y);

cv::Point2f usedEdge = edge1;
if(cv::norm(edge2) > cv::norm(edge1)) usedEdge = edge2;

cv::Point2f reference = cv::Vec2f(1,0); // horizontal edge

angle = 180.0f/CV_PI * acos((reference.x*usedEdge.x + reference.y*usedEdge.y) / (cv::norm(reference) *cv::norm(usedEdge)));

I am not able to figure out following few lines which i required to convert in emgu csharp.

cv::Point2f edge1 = cv::Vec2f(rect_points[1].x, rect_points[1].y) - cv::Vec2f(rect_points[0].x, rect_points[0].y);
cv::Point2f edge2 = cv::Vec2f(rect_points[2].x, rect_points[2].y) - cv::Vec2f(rect_points[1].x, rect_points[1].y);

angle = 180.0f/CV_PI * acos((reference.x*usedEdge.x + reference.y*usedEdge.y) / (cv::norm(reference) *cv::norm(usedEdge)));

if(cv::norm(edge2) > cv::norm(edge1)) usedEdge = edge2;
cv::Point2f reference = cv::Vec2f(1,0);

Can anyone help me how to resolve the same? Any help or suggestion will be highly appreciated?

George Kerwood · Accepted Answer · 2020-06-19T17:52:32.043

2

The Point2f here are simply points, having float precision properties of X and Y, being used to store 2D vectors of I and J. Their method if declaration is setting the edges to be the vector between two points, i.e. the delta between those two points. In C#, I would write this as:

float deltaX = rect_points[1].X - rect_points[0].X;
float deltaY = rect_points[1].Y - rect_points[0].Y;
PointF edge1 = new PointF(deltaX, deltaY);

OR of course...

PointF edge1 = new PointF(rect_points[1].X - rect_points[0].X, rect_points[1].Y - rect_points[0].Y);
PointF edge2 = new PointF(rect_points[2].X - rect_points[1].X, rect_points[2].Y - rect_points[1].Y);

These PointF are now the two vectors, or edges, that join at rect_points[1]. Next, norm is performed in order to compare the magnitude of the two. This is simply Pythagoras if we perform the same manually:

edge1Magnitude = Math.Sqrt(Math.Pow(edge1.X, 2) + Math.Pow(edge1.Y, 2));
edge2Magnitude = Math.Sqrt(Math.Pow(edge2.X, 2) + Math.Pow(edge2.Y, 2));

The longer of the edges, that with the greatest magnitude, is considered the "primary", or longer edge the rectangle:

PointF primaryEdge = edge1Magnitude > edge2Magnitude ? edge1 : edge2;
double primaryMagnitude = edge1Magnitude > edge2Magnitude ? edge1Magnitude : edge2Magnitude;

Finally, to find the angle between the primaryEdge, and a horizontal vector, reference. This is the acos, of the "Dot Product", of the two, or:

PointF reference = new PointF(1,0);
double refMagnitude = 1;
double thetaRads = Math.Acos(((primaryEdge.X * reference.X) + (primaryEdge.Y * reference.Y)) / (primaryMagnitude * refMagnitude));
double thetaDeg = thetaRads * 180 / Math.PI;

Now, thetaDeg is the angle between edge1 and the horizontal, in degrees.

edited Jun 19 '20 at 17:52

answered Jun 19 '20 at 07:33

George Kerwood

1,248
8
19

Please help for below lines. if(cv::norm(edge2) > cv::norm(edge1)) usedEdge = edge2; cv::Point2f reference = cv::Vec2f(1,0); – manthan Jun 19 '20 at 11:34
Please help for cv::Point2f reference = cv::Vec2f(1,0); – manthan Jun 19 '20 at 12:11
@mBangali That should of been clear from the rest with a little effort from you, but I've now included it. – George Kerwood Jun 19 '20 at 12:18
Thank you for your help. I already tried a lot before coming here. – manthan Jun 19 '20 at 12:28
No problem, I've also answered your other question :) – George Kerwood Jun 19 '20 at 12:30
I am getting NaN for angle. I have posted code in question. – manthan Jun 19 '20 at 17:08
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/216296/discussion-between-george-kerwood-and-mbangali). – George Kerwood Jun 19 '20 at 17:15
Few images are of 180 degrees but detected as 90 degrees and vice versa happening. – manthan Jun 23 '20 at 17:28
The algorithm returns the angle between the longest edge of the rectangle and the x-axis. It's the same algorithm as the one you asked to copy. – George Kerwood Jun 23 '20 at 17:51
I wanted to know is there anyway to upgrade algorithm to resolve issue or I have to find something else. – manthan Jun 23 '20 at 18:03
What is “the issue”? Why is it wrong that they are at 90 degrees? Maybe ask a new question with some example image and an explanation of what result you want or were expecting. – George Kerwood Jun 23 '20 at 18:35
New question added https://stackoverflow.com/questions/62550517/emgucv-pan-card-improper-skew-detection-in-c-sharp – manthan Jun 24 '20 at 08:33

emgucv angle detection for rectangle in csharp

1 Answers1