7
$\begingroup$

In this paper, page 92, the so called fundamental matrix in computer vision is derived.

Some notation:

$M = (x,y,z)^T$ is a 3d point and $ \left[ \begin{array}{cc} M\\ 1 \end{array} \right] $ represents the homogeneous coordinate $(x,y,z,1)^T$

There are two pinhole cameras defined as:

$P_i$ a 3x4 projection matrix for camera $i$:

$$m_i = P_i \left[ \begin{array}{cc} M\\ 1 \end{array} \right] = s_i \left[ \begin{array}{cc} u_i\\ v_i\\ 1 \end{array} \right]$$

The projection can be decomposed into $A_i\left[R_i\ t_i\right]$, where $A_i$ are the intrinsic parameters of the camera (an upper 3x3 triangular matrix), $R_i$ is a 3x3 rotation matrix (rotates the camera relative to the coordinate systems axises), and $t_i$ is a 3x1 translation vector. Note: $\left[R_i\ t_i\right]$ is a 3x4 matrix ($t$ is the last column).

$A_i$ defines the camera intrinsic parameters (focal length, scale factors, etc) I don't think these parameters are relevant to my question ,but the matrix looks like this:

$$ A_i = \begin{bmatrix}a&b&c\\0&d&e\\0&0&1\end{bmatrix} $$

The first camera ($i=1$) is positioned at origo and its axises aligns with the coordinate axises:

$$P_1 = A_1 \left[I\ \Bbb{0} \right]$$

The second camera is translated by $t$ and rotated by $R$ and has its own intrinsic parameters $A_2$:

$$P_2 = A_2\left[R\ t \right]$$

With these notations, we have the following two equations to project the 3d point $M$ to the plane of each camera:

$$ s_1m_1 = A_1[I\ 0] \left[ \begin{array}{cc} M\\ 1 \end{array} \right] \tag{1} $$ $$ s_2m_2 = A_2[R\ t] \left[ \begin{array}{cc} M\\ 1 \end{array} \right] \tag{2} $$

Epipolar geometry

Given two images from two different cameras of the same scene, a ray from the camera center to a point $M$ will project as a line in each camera plane. $M$ is projected to $m_1$. There is a line $l_{m_1}$ in the other camera plane in which $m_2$ must be on. This line is called the epipolar line.

An image explains it better:

enter image description here

So if we know $m_1$, and we need to find $m_2$ (the corresponding point to $m_1$), then we could limit the search to the epipolar line $l_2$ (which goes through $e_2$ and $m_2$). In this way, we can search in one dimension instead of two for the corresponding points in the other camera image. Of course we cannot find $l_2$ via $m_2$ since we are looking for $m_2$.

The fundamental matrix $F$ is defined such that: $l_2 = F\ m_1$. And the constraint that $m_2$ will be on this line is: $m_2^T l_2 = 0$.

So the constraint to find $F$ is: $m_2^T F m_1 = 0$.

Unfortunately I'm not able to see (neither geometrically nor algebraically) how $F$ is derived / deduced.

Question

From these two equations, in the paper (and I've seen it in other papers as well), the following is deduced (which proofs the existence of $F$):

$$ m_2^{T}A_2^{-T}TRA_1^{-1}m_1= 0 $$

with the note: "by eliminating $M$, $s1$ and $s2$", and: $T$ is an antisymmetric matrix defined by $t$ such that, where $\times$ is the cross product, $Tx = t \times x$ for all 3D vectors.

How exactly can these be eliminated given (1) and (2)?

I thought I could eliminate $M$ this way:

$ A_2^{-1}s_2m_2 = RM+t $

$RA_1^{-1}s_1m_1 + t = RM+t$

$RA_1^{-1}s_1m_1 + t = A_2^{-1}s_2m_2$

At this point, I suppose T is used since $Tt$ would be $t$ cross $t$ which is zero.

$TRA_1^{-1}s_1m_1 = TA_2^{-1}s_2m_2$

Thankful for a hint or two.

$\endgroup$
1
$\begingroup$

You were almost there. From equations (1) and (2), we have $s_1A_1^{-1}\tilde{\mathbf m}_1 = M$ and $s_2A_2^{-1}\tilde{\mathbf m}_2 = RM + \mathbf t = s_1RA_1^{-1}\tilde{\mathbf m}_1 + \mathbf t$. The cross product of two vectors is orthogonal to them both, so we have $$\begin{align} 0 = (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge s_2A_2^{-1}\tilde{\mathbf m}_2) &= (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge(s_1RA_1^{-1}\tilde{\mathbf m}_1 + \mathbf t)) \\ &= (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge s_1RA_1^{-1}\tilde{\mathbf m}_1 + \mathbf t\wedge\mathbf t) \\ &= (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge s_1RA_1^{-1}\tilde{\mathbf m}_1) \\ &= s_1s_2(A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge RA_1^{-1}\tilde{\mathbf m}_1) \\ &= s_1s_2\tilde{\mathbf m}_2^T(A_2^{-1})^T TRA_1^{-1}\tilde{\mathbf m}_1.\end{align}$$ We now drop the irrelevant scale factors and are left with the desired equation.

$\endgroup$
4
  • $\begingroup$ Thanks! It is interesting to note that there is a degenerate case where RM is in the same direction as t. Which could happen if the two cameras are somewhat facing each other and M lies between the two cameras. $\endgroup$
    – j-a
    May 23 '17 at 7:57
  • $\begingroup$ @j-a When $M$, $C_1$ and $C_2$ are colinear, you don’t have a unique plane. The resulting fundamental matrix will be rank-deficient, I believe. $\endgroup$
    – amd
    May 23 '17 at 8:45
  • $\begingroup$ Indeed. Thanks! $\endgroup$
    – j-a
    May 23 '17 at 11:42
  • $\begingroup$ I can’t help feeling that this derivation is a bit contrived. Seems like there should be a way to derive this same equation from the condition that $C_1$, $C_2$, $m_1$ and $m_2$ are coplanar. $\endgroup$
    – amd
    May 23 '17 at 16:12
0
$\begingroup$

It seems helpful to note that for any pair of vectors $x,y$ on $3$ components, we have $$ x^T[m_2^{T}A_2^{-T}TRA_1^{-1}m_1]y = \\ (A_2^{-1}m_2 x)^T(TRA^{-1}m_1y) = \\ (A_2^{-1}m_2 x)^T(t \times [RA^{-1}m_1y]) $$ It would be sufficient to prove that the image of $A_2^{-1}m_2$ lies in the plane spanned by $t$ and the image of $RA^{-1} m_1$, assuming that I've interpreted the data types here correctly.

$\endgroup$
5
  • $\begingroup$ I see you are starting from the result, to prove that it is true. That will also give me insights thanks. A^-1m_1 = M, so RA^-1m_1 is RM. But what would RMy be? Geometrically it is true that these must lie in the same plane. (RM+t, t, RM), somewhat confused by (RM+t, t, RMy) though? Why is x and y required to make the proof? (Would also appreciate any hint on how to go from the first two equations to end up with the result). $\endgroup$
    – j-a
    May 18 '17 at 21:49
  • $\begingroup$ It is often convenient to show that a matrix $M$ is zero by showing that $x^TMy$ is zero for every $x$ and $y$. See my latest edit; that should be the "the image of $RA^{-1}m_1$". $\endgroup$ May 18 '17 at 22:13
  • $\begingroup$ Honestly, it's really hard for me to follow how you've defined all of the matrices here. It is not clear to me, for example, that we should have $A_2^{-1}m_2 = M$. Given that I can't understand what the pieces are, I can't tell you how to go from the equations through the proof. $\endgroup$ May 18 '17 at 22:16
  • $\begingroup$ Thanks for answering I will update the question and try clarify everything to the best of my understanding $\endgroup$
    – j-a
    May 18 '17 at 22:44
  • $\begingroup$ I updated the question, is it a bit better now? $\endgroup$
    – j-a
    May 18 '17 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.