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Represent a point in 2-d projective space $\mathbb P^2$ with homogeneous coordinate $(x_1,x_2,x_3)^T$. Suppose this point is located on a conic, my question why it can be parameterized as $$\begin{bmatrix}x_1 \\ x_2 \\x_3\end{bmatrix} =A\begin{bmatrix}1 \\ \theta \\\theta^2\end{bmatrix} =\begin{bmatrix}a_{11}+a_{12}\theta+a_{13}\theta^2 \\ a_{21}+a_{22}\theta+a_{23}\theta^2 \\ a_{31}+a_{32}\theta+a_{33}\theta^2 \\\end{bmatrix}$$ where A is a non-singular 3$\times$3 matrix?

To my understanding, a circle which is a conic case in $\mathbb P^2$ can be parameterized as $$(x_1-x_3a)^2+(x_2-x_3b)^2=x_3r^2$$ It doesn't coincide with above parameterization,who can help? Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 76.

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2 Answers 2

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Look at the rational parametrization of the unit circle using the tangent half-angle formulas:

$$x = \frac{1-t^2}{1+t^2} \qquad y = \frac{2t}{1+t^2}$$

which can describe any point on the circle except for $(-1,0)$ which corresponds to $t=\infty$. Now this certainly matches the pattern you have:

$$\begin{bmatrix}1-t^2\\2t\\1+t^2\end{bmatrix}= \begin{bmatrix}1&0&-1\\0&2&0\\1&0&1\end{bmatrix} \begin{bmatrix}1\\t\\t^2\end{bmatrix}$$

I'm using $t$ not $\theta$ to make it very clear that this parameter is not the angle, although it is the tangens of half the angle.

Now any other non-degenerate real conic can be obtained from the unit circle via a projective transformation (in a non-unique way). So combining the above parametrization of the unit circle with a projective transformation will give you the desired result.

When mapping one conic to another, you essentially get to choose the images of three points on the conic, and then everything else follows. So you will likely need to designate three points on the conic in order to find a unique map to the unit circle. In this case there is a way to avoid the detour to the unit circle and go from these three points, as I'll explain in a different answer. I think this answer here is very useful as a short intuitive explanation of why this works, but in practice I'd rather follow the other approach to determine a specific matrix.

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  • $\begingroup$ Good job! From the shallower to the deeper. While contents of this post is my desired answer, I will take time to digest materials of your another post. $\endgroup$
    – Finley
    Nov 5, 2017 at 4:43
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You can use three points on a conic to define a projective basis for points on that conic, and with respect to that parametrization the property you look for holds true.

A conic can be defined as the set of points which see four given points at a certain cross-ratio, i.e. the lines joining a generic point on the conic to the designated four points will have the same cross ratio for all points on the conic. In this sense a conic defined by five points $P_1$ through $P_5$ can be defined as

$$\{P\mid(P_1,P_2;P_3,P_4)_P=(P_1,P_2;P_3,P_4)_{P_5}\}$$

or in words, the point $P$ sees $P_1$ through $P_4$ under the same cross ratio as $P_5$ does. In a way, $P_1$ through $P_3$ establish a projective basis on the conic. The position of $P_4$ with respect to that basis is a constant, which can be used as the parameter to describe $P_4$ with respect to the basis and the conic. In this sense a parameter $t$ describes a point $Q_t$ on the conic like this:

$$(P_1,P_2;P_3,Q_t)_{P_4}=t=(P_1,P_2;P_3,Q_t)_{P_5}$$

In case you are not familiar with this cross-ratio notation, here is the same expressed in terms of $3\times3$ determinants written as $[\ldots]$ brackets:

$$\frac{[P_1,P_3,P_4][P_2,Q_t,P_4]}{[P_2,P_3,P_4][P_1,Q_t,P_4]}=t= \frac{[P_1,P_3,P_5][P_2,Q_t,P_5]}{[P_2,P_3,P_5][P_1,Q_t,P_5]}$$

You can also rewrite the left side as

$$[P_1,P_3,P_4]\langle P_2\times P_4, Q_t\rangle =t[P_2,P_3,P_4]\langle P_1\times P_4, Q_t\rangle$$

and then

$$\langle([P_1,P_3,P_4]P_2-t[P_2,P_3,P_4]P_1)\times P_4, Q_t\rangle=0$$

which is the condition for $Q_t$ to lie on a certain line, the coordinates of which are linear in $t$. So if you have two such equations, one for the left half and one for the right half of the conditions above, then you are in fact intersecting two lines, both of them linear in $t$. The intersection of two lines can be computed using the cross product, so the result will be quadratic in $t$. And if you do all of the computations with $t$ as a symbolic parameter, you will end up with polynomials in $t$ which you can write as a matrix.

Note that $P_1$ itself corresponds to $t=\infty$. From a topological point of view, you cannot have a bijective map between the real line you use for your parameters and conics which are topological circles. You have to cut the conic at some point, or use homogeneous parametrization, i.e. something of the form

$$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= A\begin{bmatrix}u^2\\tu\\t^2\end{bmatrix}\qquad (t,u)\neq(0,0)$$

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