2

The 3D reconstruction literature all mentions the following relationship:

Given a point m1 in the first image, the homogeneous coordinates of the epipolar line l2 in the second image corresponding to m1 are l2 = Fm1.

However, I can't get my head around what the coordinates of l2 represent in old-fashioned cartesian coordinates (i.e., the image coordinates). It's a single point, so how would I, say, physically draw the line on the image?

  • Is it a vector direction from the epipole?
  • Do l2 and the epipole specify two points on the epipolar line?
  • Is it something else that I'm not grasping?

I hope it's the third choice, because the other two don't seem to work out in my case.

Jabberwock
  • 173
  • 1
  • 9
  • Someone docked me a point, but didn't bother to tell me why. The relationship is so ubiquitous in the literature that I can't imagine it's because the question isn't clear. If it's because I apparently didn't do enough research, I did. I've been trying to find an explanation for about a month now on my own. The closest thing I could find is in Ch 9 of the (also ubiquitous) Hartley & Zisserman book (http://www.robots.ox.ac.uk/~vgg/hzbook/). There they talk about it being the cross product of m1 and the epipole (relative to the camera) - but that's a vector in 3D space, not on the image. – Jabberwock Jan 16 '17 at 16:21
  • I'm going to go out on a limb and say the downvote was probably from someone who thought this question was off-topic, as its more of a math than a programming question. – code11 Jan 16 '17 at 20:30
  • I guess that's possible. But this is where the questions for epipolar geometry and fundamental matrices seem to show up. If there's a better place to ask it, I'd gladly ask it there. And the implementation is definitely programming (specifically OpenCV/Python/Blender). Would the Signal Processing group be better? – Jabberwock Jan 16 '17 at 23:07
  • I actually did some work on this in my masters, but I've apparently forgotten it all. Try this link, I think its what I used to learn about it: http://docs.opencv.org/trunk/da/de9/tutorial_py_epipolar_geometry.html – code11 Jan 17 '17 at 00:05
  • @code11, thanks for taking the time to reply. I had already read that page, but I searched some more and found the answer. – Jabberwock Jan 17 '17 at 02:42

1 Answers1

1

I found the answer here: http://www.cs.cmu.edu/~16385/lectures/Lecture18.pdf.
The coordinates of the line l = Fm1 represent the line equation:

lxx + lyy + lz = 0

So all I need to draw the line in image 2 are l, the epipole e2, and another (arbitrary) x value.

It's also somehow related to the cross product, since l also equals e2 x m2. I don't really understand that part, though.

Jabberwock
  • 173
  • 1
  • 9