Calculate distance (stereo vision) [closed]
For my project I am using parts of the next code:
To track objects of a specific color I implemented this method:
My question is: How can I calculate the distance to the tracked colored objects?
Thank you in advance!
*The application calls the method for the left and right frame. This is not efficient...
**I need to calculate detectedObject.Zcor
DetectedObject Detect(IplImage *frame)
{
//Track object (left frame and right frame)
//Calculate average position
//Show X,Y,Z coordinate and detected color
color_image = frame;
imgThreshold = cvCreateImage(cvSize(color_image->width,color_image->height), IPL_DEPTH_8U, 1);
cvInitFont(&font, CV_FONT_HERSHEY_PLAIN, 1, 1, 0, 1.4f, CV_AA);
imgdraw = cvCreateImage(cvGetSize(color_image),8,3);
cvSetZero(imgdraw);
cvFlip(color_image, color_image, 1);
cvSmooth(color_image, color_image, CV_GAUSSIAN, 3, 0);
threshold = getThreshold(color_image);
cvErode(threshold, threshold, NULL, 3);
cvDilate(threshold, threshold, NULL, 10);
imgThreshold = cvCloneImage(threshold);
storage = cvCreateMemStorage(0);
contours = cvCreateSeq(0, sizeof(CvSeq), sizeof(CvPoint), storage);
cvFindContours(threshold, storage, &contours, sizeof(CvContour), CV_RETR_CCOMP, CV_CHAIN_APPROX_NONE, cvPoint(0,0));
final = cvCreateImage(cvGetSize(color_image),8,3);
for(; contours!=0; contours = contours->h_next)
{
CvRect rect = cvBoundingRect(contours, 0);
cvRectangle(color_image,
cvPoint(rect.x, rect.y),
cvPoint(rect.x+rect.width, rect.y+rect.height),
cvScalar(0,0,255,0),
2,8,0);
string s = to_string(rect.x) + "," + to_string(rect.y);
char const* pchar = s.c_str();
cvPutText(frame, pchar, cvPoint(rect.x, rect.y), &font, cvScalar(0,0,255,0));
detectedObject.Xcor = rect.x;
detectedObject.Ycor = rect.y;
}
cvShowImage("Threshold", imgThreshold);
cvAdd(final,imgdraw,final);
detectedObject.Zcor = 0;
return detectedObject;
First of all this is a project that is not included inside the OpenCV library. For questions, adress the author, not this Q&A forum. Look at the FAQ. Topic closed.