How draw axis of ellipse

Question

I am using fitellipse of Opencv and C++, and I'm getting these values:

/// Find the rotated rectangles and ellipses for each contour
vector<RotatedRect> minRect( contours.size() );
vector<RotatedRect> minEllipse( contours.size() );

for( int i = 0; i < contours.size(); i++ )
{
   minRect[i] = minAreaRect( Mat(contours[i]) );

   if( contours[i].size() > 5 )
      minEllipse[i] = fitEllipse( Mat(contours[i]) );

   // ...
}

float xc    = minEllipse[element].center.x;
float yc    = minEllipse[element].center.y;
float a     = minEllipse[element].size.width  / 2;   
float b     = minEllipse[element].size.height / 2;
float theta = minEllipse[element].angle;

But with these values how can I draw the axis of an ellipse, for example of the following ellipse?

NOTE: Element is an ellipse stored in minEllipse.

Are you asking how to draw a line segment given its endpoint, length, and slope angle? — n. m. could be an AI, Oct 30 '15 at 09:32
I want draw rotated axis of ellipse with point center of ellipse, length of radius, and angle, because the ellipse in image are rotated so I want draw the new axis inside ellipse. — user3704922, Oct 30 '15 at 09:40
It looks like you need help with geometry rather than programming. — n. m. could be an AI, Oct 30 '15 at 09:44
I think that axis are posible to draw with only these values, but I search and I don't found, and all geometry methods that I used don't give me good results. — user3704922, Oct 30 '15 at 09:50
If you cannot draw it on paper, then you are at a wrong site. — n. m. could be an AI, Oct 30 '15 at 10:02

Jan Rüegg · Accepted Answer · 2015-10-30T10:58:43.663

You can use minEllipse[element].points to get the four corners of the rotated bounding rectangle, like described here.

Then you only need to calculate the average of the two points on each side of the rectangle to get the endpoints for the axes...

Point2f vertices[4];
minEllipse[element].points(vertices);
line(image, (vertices[0] + vertices[1])/2, (vertices[2] + vertices[3])/2, Scalar(0,255,0));
line(image, (vertices[1] + vertices[2])/2, (vertices[3] + vertices[0])/2, Scalar(0,255,0));

Bob__ · Answer 2 · 2015-10-30T11:46:50.253

3

You are probably looking for those formulas:

ct = cos(theta)
st = sin(theta)

LongAxix0.x = xc - a*ct
LongAxis0.y = yc - a*st
LongAxis1.x = xc + a*ct
LongAxix1.y = yc + a*st

ShortAxix0.x = xc - b*st
ShortAxix0.y = yc + b*ct
ShortAxis1.x = xc + b*st
ShortAxix2.y = yc - b*ct

edited Oct 30 '15 at 11:46

answered Oct 30 '15 at 11:02

Bob__

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score 0 · Answer 3 · edited May 23 '17 at 12:09

But with these values how can I draw the axis of an ellipse?

The axis of the ellipse are passing through its centre:

float xc = minEllipse[element].center.x;
float yc = minEllipse[element].center.y;

the start and end points of the axis could be at an offset from the centre defined by the ellipse's width and height, i.e.:

// horizontal axis start/ end point

// coordinates
int HxStart = xc - size.width  / 2;
int HyStart = yc;

int HxEnd = xc + size.width  / 2;
int HyEnd = yc;

// points
Point Hstart(HxStart, HyStart);
Point Hend(HxEnd, HyEnd);

// horizontal axis
Line horizontalAxis(Hstart, Hend);

// vertical axis start/ end point
int VxStart = xc;
int VyStart = yc - size.height  / 2;

int VxEnd = xc;
int VyEnd = yc + size.height  / 2;

// ----//----

Now, you can rotate the axis (the above for points) by the provided angle theta, around the centre of the ellipse.

Having the above and knowing how to construct a line you can build the two axis at any given angle theta.

How draw axis of ellipse

3 Answers3