Boundary detection-What function will be the best for this?

edit savecancel

You can follow the below step to find the minimum and maximum boundary.

1. Load image.
2. Blur to remove noise.
3. Convert to HSV colour space for segment the grass.
4. Use OpenCV inRange() to filter green colour.

5. Morphological operation to fill-out the black region. Here you will get binary image with boundary of 0 value.

6. Scan each column until black pixel to find minimum and maximum boundary.

Here is the code

#include <iostream>
#include <stdio.h>
#include <opencv2/opencv.hpp>

using namespace cv;
using namespace std;

int main( int argc, char** argv )
{
  Mat edge,dst,threshold;
  Mat HSV;
  int dilation_size =4;

  Mat src = imread( "src.jpg",1);
  blur( src, src, Size(2,2) );
  cvtColor(src,HSV,CV_BGR2HSV);
  inRange(HSV,Scalar(44,90,30),Scalar(76,255,255),threshold);
  Mat element = getStructuringElement( MORPH_RECT,Size( 2*dilation_size + 1, 2*dilation_size+1 ), Point( dilation_size, dilation_size ) );
  morphologyEx( threshold, dst, MORPH_CLOSE, element );


  int minBound=dst.rows;
  int maxBound=0;
  int tmp=0;

  for(int i = 0; i < dst.cols; i++){
     for(int j = 0; j < dst.rows; j++){
        if(dst.at<uchar>(j, i)==0){
        tmp=j;
        break;
        }
     }
    if(tmp<minBound) minBound=tmp;
    if(tmp>maxBound) maxBound=tmp;
   }

  line(dst, Point(0,minBound), Point(dst.cols,minBound),  Scalar(0), 1,8, 0); //minmum boudary
  line(dst, Point(0,maxBound), Point(dst.cols,maxBound),  Scalar(255), 1,8, 0); // maximum boundary
  line(src, Point(0,minBound), Point(dst.cols,minBound),  Scalar(0,0,255), 1,8, 0); //minmum boudary
  line(src, Point(0,maxBound), Point(dst.cols,maxBound),  Scalar(0,0,255), 1,8, 0); // maximum boundary

  imshow( "source", src );
  imshow( "threshold", threshold );
  imshow( "dst", dst );
  waitKey(0);
  return 0;
}

Result :-

Threshold Image:-

Boundary:-

You can chose any value between minimum and maximum boundary, like the row which have maximum transition from 255 to 0, and that may be depend on your source image.

Hope these helpful....