OpenCV Q&A Forum - RSS feedhttp://answers.opencv.org/questions/OpenCV answersenCopyright <a href="http://www.opencv.org">OpenCV foundation</a>, 2012-2018.Mon, 04 Dec 2017 22:01:04 -0600face recognition by Euclidean space distance seeking for helphttp://answers.opencv.org/question/179732/face-recognition-by-euclidean-space-distance-seeking-for-help/In the n dimensional Euclidean space,
If
The first point is **P1 [x1,x2,...,xn]**
The second point is **P2 [y1,y2,...,yn]**
so ，the Euclidean space distance from P1 to P2 is
**d(P1,P2) = sqrt ( (x1-y1)^2 + (x2-y2)^2 + ... + (xn-yn)^2 )**
so , in the face recognition field , for example , in many open source ，
every point have x coordinate value and y coordinate value ,
I think it is ,
The first face is **P1 [(x1,y1),(x2,y2),...,(xn,yn)]**
The second face is **P2 [(a1,b1),(a2,b2),...,(an,bn)]**
so ，the Euclidean space distance from P1 to P2 is ________________________________ ?
the result is too complex , even is unsolvable .
I don't know if I have a problem with my understanding ?strivingMon, 04 Dec 2017 22:01:04 -0600http://answers.opencv.org/question/179732/Algorithm behind findHomography() and the kind of homography returned by ithttp://answers.opencv.org/question/170601/algorithm-behind-findhomography-and-the-kind-of-homography-returned-by-it/Hi folks!
I am dealing with the estimation and decomposition of homographies for a couple of days now and the very moment I think I got it I stumble upon new and even more mind-boggling questions. What I would like to understand is how a homography can be decomposed into parameters R and t how to estimate the camera path with the aforementioned parameters. I am well aware of the plethora of different algorithms to do that but I have a few questions that I could not answer after hours of googling and literature research.
As far as I know homographies are transformations which transform one set of points from one image plane to another and that they are the most general kind of transformation containing rotation, translation and non-linear scaling parameters. Today I read that there are two kinds of homographies: projective and euclidean homographies. From what I've read thus far, Euclidean homographies directly express the reconstruction of the relative motion between the views and the normal of the plane.
- Is that correct?
- Is it save to assume that a Euclidean homography can be thought of as a "real world" homography which describes my rotation and translation from one camera position to the next?
- Are there any other important differences between projective and Euclidean homographies?
- Which algorithm is used in findHomography() in OpenCV 2.4.13?
- Does findHomography() in OpenCV 2.4.13 return a projective or euclidean homography?
- When do I need a projective or a euclidean homography?
Your help is deeply appreciated!!!
Thank you
Conundraah
ConundraahWed, 02 Aug 2017 09:23:45 -0500http://answers.opencv.org/question/170601/OpenCV Distance Metricshttp://answers.opencv.org/question/4601/opencv-distance-metrics/Hello everyone, I'm doing a benchmark test on keypoint detectors, descriptors and matching of those for a research project. I'm aware that there are great benchmark tests out there but this will be done for a specific experimental environment. Descriptors will be Orb and Freak and detectors are ORB, SURF and maybe FAST. I will use BruteForceMathcer.
So, in order to make the test fair, I decided I will use approximately same amount of keypoints and the same distances. But I can't seem to find a detailed explanation of the distance metrics used in opencv.
For instance; when I use "BruteForceMatcher<L2<float> > matcher;" when both keypoint detector and descriptor is orb, it gives me a correct match between two points with coordinates point1(130,339) and point2(130,340). Obviously the distance between those two is 1, but when I look at the matches vector, the distance value is 272.71964 which is very confusing for me.
My question is, is there any documentation that explains why this is the case? I googled it, but haven't found a decent explaination. If not, I would really appreciate if you could explain this.
Thank youaliciadominicaTue, 27 Nov 2012 02:47:13 -0600http://answers.opencv.org/question/4601/