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I'm having some trouble understanding this passage from Multiple View Geometry in Computer Vision (Second Edition)

enter image description here

(I know screenshotting text is frowned upon but it was the easiest way for me to ask this question)

1) I understand that the line between $(0, 0, 0, 1)^T$ and $(X, Y, Z, T)^T$ with fixed $X$, $Y$, $Z$, and varying $T$ creates a ray passing through the point $(X, Y, Z, T)$ and $(0, 0, 0, 1)$ in $\mathbb{P}^3$. When they say "all mapping to the same point" what are they referring to exactly? And what is the "final coordinate" they are referring to? I understand that the all the points on the ray are homogenous coordinates of $(X, Y, Z, 1)^T$ but that's about it.

2) Why does the image point in $\mathbb{P}^2$ have homogenous coordinates $(X, Y, Z)^T$? Is it the same $Z$ in $(X, Y, Z, 1)^T$?

3) I see that the $P$ they describe simply gets rid of the 4th component of the homogenous coordinate, $T$, and we are left with $(X, Y, Z)^T$. Where did $(x, y, w)^T$ come from?

4) In the second example where they show a projective transformation, the vector $(X, Y, T)^T$ does not exist in $\mathbb{P}^3$ right? So why does it even come up? Can't we just use $(X, Y, 0, T)^T$ ?

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I understand that the line between (0,0,0,1)T and (X,Y,Z,T)T with fixed X, Y, Z, and varying T creates a ray passing through the point (X,Y,Z,T) and (0,0,0,1) in ℙ3. When they say "all mapping to the same point" what are they referring to exactly? And what is the "final coordinate" they are referring to? I understand that the all the points on the ray are homogenous coordinates of (X,Y,Z,1)T but that's about it.

We start by imagining a central projection in 3D. The set-up is as follows: we choose a plane $P$ to project onto (this corresponds to our photographic paper). We choose a point $c$ called the "center" to project through (this corresponds to the center of the lens). It is important that $c$ is not on $P$. Now we project as follows: given any other point $a \neq c$, we consider the unique line $\overline{ac}$. The result of the projection is the intersection of the line with the plane $P$ (if the line is parallel to $P$ the result is the point at infinity).

[Here are some questions: 1. We said it is important that $c$ not be on $P$; can you see why? 2. Suppose that $a$ happen to be on the plane $P$; what is the result of the projection?]

Now, one interesting property of this process is that if we choose any other point on the line $\overline{ac}$, say $d \neq a, c$, then $d$ is projected through $a$ onto $P$ to exactly the same point as $c$ (because the line $\overline{ac}$ and the line $\overline{ad}$ are the same line; and will have the same intersection with $P$).

In the quoted paragraph, all points of the ray (except the center point which we are not allowed to choose; it's like a blind-spot) are projected to the same point in $\mathbb{P}^2$, just like $c$ and $d$ above. We can think of the ray as being generated by the process of varying $T$, so that when the process stops a point is chosen; that is what they call the ``final coordinat'' (so what they are really saying is that $T$ doesn't matter).

Why does the image point in ℙ2 have homogenous coordinates (X,Y,Z)T? Is it the same Z in (X,Y,Z,1)T?

It is the same $Z$. This is because of the specific choice of the center of projection; and because they are thinking of $\mathbb{P}^2 \subseteq \mathbb{P}3$ corresponding to setting the last coordinate to be $0$ (so all points of the form $(A, B, C, 0)$). [Given any point $(X,Y,Z,T) \neq (0, 0, 0, 1)$ what is the equation of the unique line passing through $(X, Y, Z, T)$ and $(0, 0, 0, 1)$? What is the unique point of intersection of this line with the plane $T=0$? What happens to the point if we use homogeneous coordinates?]

I see that the P they describe simply gets rid of the 4th component of the homogenous coordinate, T, and we are left with (X,Y,Z)T. Where did (x,y,w)T come from?

As we have just mentioned, the coordinates may change if our center is not $(0, 0, 0, 1)$ and/or if we think of $\mathbb{P}^2$ as embedded in $\mathbb{P}^3$ in some other manner (i.e., instead of all of the points with last coordinate $0$; maybe all of the points with second coordinate $1$ or something like that.) This is what they are saying in the sentence starting with "Making allowance..." I'm assuming they will go into the relevant formulae in detail later on, so you will see how the center and frame affect the result.

In the second example where they show a projective transformation, the vector (X,Y,T)T does not exist in ℙ3 right? So why does it even come up? Can't we just use (X,Y,0,T)T ?

You are right that we can use $(X, Y, 0, T)$. The reason this comes up is that square matrices are a lot nicer to handle than non-square matrices (we have better tools to analyze them). If $Z = 0$, then a $3\times 4$ matrix acting on $(X, Y, 0, T)^T$ will give the same result as an appropriate $3\times 3$ matrix acting on $(X, Y, T)$ [Question: how do you get the $3\times 3$ matrix given the $3\times 4$ matrix?]

Good luck! =)

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  • $\begingroup$ Thank you so much for the thorough answer. I have a few follow ups before I can answer your inline questions. Under the second quote, I still don't quite understand why the $Z$ iOS the same. I also don't know how to find the equation of the line $\endgroup$ – Carpetfizz Jan 1 '18 at 13:46
  • $\begingroup$ Okay, so I've found the equation of the line as $\mathbf{r} = (0, 0, 0, 1)^T + t(X, Y, Z, T-1)^T$. I've also found that this line intersects the plane $T=0$ at $(0, 0, 0, 0)$ although I'm not sure how any of this relates to the problem. Why did we choose $T=0$ as the image plane? I thought that was the plane at infinity $\endgroup$ – Carpetfizz Jan 1 '18 at 14:58
  • $\begingroup$ @Carpetfizz Careful with the intersection. The last coordinate of the line is the parametric equation t(T-1)+1. If you set $t=0$ it becomes $1$, rather then $0$. $\endgroup$ – GLG Jan 1 '18 at 18:07
  • $\begingroup$ Oh right thanks. Do you mind explaining the significance of this? $\endgroup$ – Carpetfizz Jan 1 '18 at 18:22
  • $\begingroup$ Also, I'm not sure why $t=0$. If I plug in $T$ into the equation of the plane, I get $t(T-1)+1 = 0$ and $t = \frac{-1}{T-1}$ $\endgroup$ – Carpetfizz Jan 1 '18 at 18:46

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