Ytukyg

I previously asked a question with regards to what the matrix $mathrm{H}_{3 times 3}$ is/represents in the following textbook excerpt:

In applying projective geometry to the imaging process, it is customary to model the world as a $3$D projective space, equal to $mathbb{R}^3$ along with points at infinity. Similarly the model for the image is the $2$D projective plane $mathbb{P}^2$. Central projection is simply a map from $mathbb{P}^3$ to $mathbb{P}^2$. If we consider points in $mathbb{P}^3$ written in terms of homogeneous coordinates $(mathrm{X}, mathrm{Y}, mathrm{Z}, mathrm{T})^T$ and let the centre of projection be the origin $(0, 0, 0, 1)^T$, then we see that the set of all points $(mathrm{X}, mathrm{Y}, mathrm{Z}, mathrm{T})^T$ for fixed $mathrm{X}$, $mathrm{Y}$, and $mathrm{Z}$, but varying $mathrm{T}$ form a single ray passing through the point centre of projection, and hence all mapping to the same point. Thus, the final coordinates of $(mathrm{X}, mathrm{Y}, mathrm{Z}, mathrm{T})$ is irrelevant to where the point is imaged. In fact, the image point is the point in $mathbb{P}^2$ with homogeneous coordinates $(mathrm{X}, mathrm{Y}, mathrm{Z})^T$. Thus, the mapping may be represented by a mapping of $3$D homogeneous coordinates, represented by a $3 times 4$ matrix $mathrm{P}$ with the block structure $P = [I_{3 times 3} | mathbf{0}_3]$, where $I_{3 times 3}$ is the $3 times 3$ identity matrix and $mathbf{0}_3$ a zero 3-vector. Making allowance for a different centre of projection, and a different projective coordinate frame in the image, it turns out that the most general imaging projection is represented by an arbitrary $3 times 4$ matrix of rank $3$, acting on the homogeneous coordinates of the point in $mathbb{P}^3$ mapping it to the imaged point in $mathbb{P}^2$. This matrix $mathrm{P}$ is known as the camera matrix.

In summary, the action of a projective camera on a point in space may be expressed in terms of a linear mapping of homogeneous coordinates as

$$begin{bmatrix}
x \
y \
w end{bmatrix} = mathrm{P}_{3 times 4}
begin{bmatrix}
mathrm{X} \
mathrm{Y} \
mathrm{Z} \
mathrm{T} \ end{bmatrix}$$

Furthermore, if all the points lie on a plane (we may choose this as the plane $mathrm{Z} = 0$) then the linear mapping reduces to

$$begin{bmatrix}
x \
y \
w end{bmatrix} = mathrm{H}_{3 times 3}
begin{bmatrix}
mathrm{X} \
mathrm{Y} \
mathrm{T} \ end{bmatrix}$$

which is a projective transformation.

It is now clear to me that I didn't understand this section properly. Specifically, it is not clear to me why choosing the plane $mathrm{Z} = 0$ means that the linear mapping

$$begin{bmatrix}
x \
y \
w end{bmatrix} = mathrm{P}_{3 times 4}
begin{bmatrix}
mathrm{X} \
mathrm{Y} \
mathrm{Z} \
mathrm{T} \ end{bmatrix}$$

reduces to

$$begin{bmatrix}
x \
y \
w end{bmatrix} = mathrm{H}_{3 times 3}
begin{bmatrix}
mathrm{X} \
mathrm{Y} \
mathrm{T} \ end{bmatrix}$$

More specifically, I don't understand why setting $mathrm{Z} = 0$ necessitates getting rid of the entirety of the 3rd column of $mathrm{P}_{3 times 4}$, which gets us $mathrm{H}_{3 times 3}$ (as per bounceback's answer in the aforementioned question)?

And I'm also wondering whether choosing the plane $mathrm{Z} = 5$ (or any other plane) instead of $mathrm{Z} = 0$ would still reduce the transformation from

$$begin{bmatrix} x \ y \ w end{bmatrix} = mathrm{P}_{3 times 4} begin{bmatrix} mathrm{X} \ mathrm{Y} \ 0 \ mathrm{T} \ end{bmatrix}$$

to

$$begin{bmatrix} x \ y \ w end{bmatrix} = mathrm{H}_{3 times 3} begin{bmatrix} mathrm{X} \ mathrm{Y} \ mathrm{T} \ end{bmatrix},$$

where

$$mathrm{H}_{3 times 3} = begin{bmatrix}1&0&0\0&1&0\0&0&0end{bmatrix}$$

?

I would greatly appreciate it if people could please take the time to clarify this.

Lets write $P_{3times4}=begin{bmatrix}r_{11}&r_{12}&r_{13}&tau_1\ r_{21}&r_{22}&r_{23}&tau_2\r_{31}&r_{32}&r_{33}&tau_3\end{bmatrix}$
and call the $r$ terms $R_{3times3}$ and the $tau$ terms $tau_{3times1}$
thus $P_{3times 4}=[R,tau]$ (I dropped the dimension indices for brevity, but $tau$ is a column vector). Now $P_{3times 4} U_4$ becomes $R begin{bmatrix} X\Y\Z end{bmatrix} +Ttau=R U_3 +Ttau$. where I defined $U_3 =begin{bmatrix} X\Y\Z end{bmatrix}$.

Before dealing with the $Z=0$ case, lets look at the case where the plane does not pass through the origin: all points $U_3$ are on a plane, hence they obey $N^T U_3 =d$ where $N$ is a unit vector normal to the plane and $-dneq0$ is the distance of the origin from the plane. For this case we just have $1= {1over d} N^T U_3$ and therefor $$R U_3 +Ttau= R U_3 +(Ttau){1over d}(N^T U_3)$$ so that finally $P_{3times 4} U_4$ can be written as $[R +{Tover d}(tau N^T)]U_3$ and $H_{3times3}=[R+ {Tover d}(tau N^T)]$ is called a "Homography".

The degenerate case is where $d=0$ and which also contains the case $Z=0$ by setting $N^T=[0,0,1]$. In this case instead of pushing $1= {1over d} N^T U_3$ we push $0= N^T U_3$ . So $$R U_3 +Ttau = (R (I-N N^T) +tau N^T )(U_3+T N)$$ , In the specific case $N^T=[0,0,1]$ we have $left[R begin{bmatrix} 1&0&0\0&1&0\0&0&0end{bmatrix} +begin{bmatrix} 0&0&tau_x\0&0&tau_y\0&0&tau_zend{bmatrix}right]begin{bmatrix} X\Y\Tend{bmatrix}$

If 2D is initial affine space we compose 1P - projective space by
$
left(
begin{matrix}
tilde{x} \
tilde{y}
end{matrix}
right)=
left(
begin{matrix}
a & b \
c& d
end{matrix}
right)
left(
begin{matrix}
x \
y
end{matrix}
right)
$.

Left hand are homogeneous coordinates of 1P - point. Right hand coordinates represent a vector in 2D.

In a similar way we compose 2P from 3D:
$
left(
begin{matrix}
tilde{x} \
tilde{y} \
tilde{z}
end{matrix}
right) =
left(
begin{matrix}
a_{11}& a_{12} & a_{13} \
a_{21}& a_{22}& a_{23} \
a_{31}& a_{32}& a_{33} \
end{matrix}
right)
left(
begin{matrix}
x \
y \
z \
end{matrix}
right)
$.

$
frac{x}{omega}=frac{p_{11}X+p_{12}Y+p_{13}Z+p_{14}T}{p_{31}X+p_{32}Y+p_{33}Z+p_{34}T} \
frac{y}{omega}=frac{p_{21}X+p_{22}Y+p_{23}Z+p_{24}T}{p_{31}X+p_{32}Y+p_{33}Z+p_{34}T}
$.

If Z=0 then
$
frac{x}{omega}=frac{p_{11}X+p_{12}Y+p_{14}T}{p_{31}X+p_{32}Y+p_{34}T} \
frac{y}{omega}=frac{p_{21}X+p_{22}Y+p_{24}T}{p_{31}X+p_{32}Y+p_{34}T}
$.

If Z=5 then
$
frac{x}{omega}=frac{p_{11}X+p_{12}Y+p_{13}.5T+p_{14}T}{p_{31}X+p_{32}Y+p_{33}.5T+p_{34}T} \
frac{y}{omega}=frac{p_{21}X+p_{22}Y+p_{23}.5T+p_{24}T}{p_{31}X+p_{32}Y+p_{33}.5T+p_{34}T}
$.