1
$\begingroup$

According to Single Axis Geometry by Fitting Conics by Jiang et al., one can compute the image of the circular points in a picture from conics which are the images of circles. Fit two conics to tracked points, then compute the intersection points of the two conics. There are always 4 intersection points including complex and infinite points according to Bezout’s theorem, of which 0-4 are real points. If there is a unique pair of complex conjugates, the image of circular points $\mathbf i$ and $\mathbf j$ are exactly this pair of complex conjugates. When there is no real intersection points we obtain two pairs of complex conjugate points and two possible solutions. The paper says this double solution ambiguity can be removed by using any additional conic.

My question is how can we do that?

Let's say we were able to obtain two conics $C_1$, $C_2$. As you can see here these conics have no real intersection points therefore we have two pairs of complex conjugate points:

\begin{align*} x_1 &= 1265.62 - 136.414i & y_1 &= 790.132 + 38.3403i \\ x_2 &= 1265.62 + 136.414i & y_2 &= 790.132 - 38.3403i \\ x_3 &= 334.378 + 136.414i & y_3 &= 790.132 + 38.3403i \\ x_4 &= 334.378 - 136.414i & y_4 &= 790.132 - 38.3403i \end{align*}

The paper says we can use any additional conic to resolve this problem. Let's add another conic $C_3$. The intersection points of $C_1$ and $C_3$ are:

\begin{align*} x_1 &= 800 - 9360.25i & y_1 &= 1696.11 - 0.00008i \\ x_2 &= 800 + 9360.25i & y_2 &= 1696.11 + 0.00008i \\ x_3 &= 800 - 2347.41i & y_3 &= 541.402 - 0.00002i \\ x_4 &= 800 + 2347.41i & y_4 &= 541.402 + 0.00002i \end{align*}

The intersection points of $C_2$ and $C_3$ are:

\begin{align*} x_1 &= 800 - 1767.11i & y_1 &= 446.355 - 0.0001i \\ x_2 &= 800 + 1767.11i & y_2 &= 446.355 + 0.0001i \\ x_3 &= 800.238 - 555946i & y_3 &= 91806.1 + 0.039i \\ x_4 &= 800.238 + 555946i & y_4 &= 91806.1 - 0.039i \end{align*}

Here is the picture of all three conics: Here is the picture of all three

So how can we say what intersection points are the image of circular points?

$\endgroup$
2
  • $\begingroup$ Paul, I removed the second question about Laguerre's formula since asking two questions in one post is too confusing. If you care about that, please ask that separately. I also edited your question to make the math look nicer. Have a look at the source code to see how you can achieve this yourself in the future. Also see this tutorial on math formatting. It might have been useful to include the equations of the conics under consideration, not just their points of intersection. $\endgroup$
    – MvG
    Apr 9, 2015 at 14:35
  • $\begingroup$ The equations of the conics are under the hyperlinks. Thank you for help. $\endgroup$
    – Paul
    Apr 9, 2015 at 15:20

1 Answer 1

Reset to default
0
$\begingroup$

If your three conics are the images of three circles in a common plane or in parallel planes, then these three conics must have a conjugate complex pair of points in common, which will be the images of $\mathbf i$ and $\mathbf j$. In your setup, there is no such common point, therefore at least one of the conics either isn't the image of a circle at all, or is the image of a circle in some other plane not parallel to the others.

I'm not sure if this is what you did, but simply picking any conic $C_3$ will be of no use. It really has to be a conic fitted to points of a tracked circular movement.

This whole approach of identifying the images of the circle points reminds me of this other beautiful question.

$\endgroup$
1
  • $\begingroup$ Yes, all three conics are fitted to points of a tracked circular movement. But there could be inaccuracy in tracking points or fitting conics. Is there any way to use this technic in situation when data is scattered so it could turn out that conics do not lie on parallel planes? How can I be sure they do lie on parallel planes? $\endgroup$
    – Paul
    Apr 9, 2015 at 16:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .