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Consider the following example of perspective projection: I have one railway track with two parallel lines meeting at the point $X$ at infinity which is the theoretical approach. This X is called the vanishing point. But practically, this $X$ doesn't exist.

My first question is: How can we say it doesn't exist, but when we see a real image of a railway track, the tracks intersect at $X$?

My second question is: If it exists, how does this $X$ project onto the projection plane during projection? Is it projecting like normal point projection of $(x, y, z)$ onto the view plane $(x_p, y_p, z_p)$?

N. B - I don't want that detailed answer. I just want an intuitive one that is brief and easy to understand.

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  • $\begingroup$ The vanishing point that you are talking about exists on the film (i.e. in 2D) and it is defined by the intersection of the projection of said parallel lines. But the vanishing point isn't unique. There are in fact infinitely many vanishing points. You could for example pick slightly rotated boxes (i.e. different angled parallel lines) and each of those would have their own vanishing point(s). $\endgroup$
    – lightxbulb
    Commented Oct 31, 2021 at 17:09
  • $\begingroup$ @joojaa The projections do intersect. It's what artists use when practicing perspective/construction to check that the angles are correct - they continue the projected "parallel" lines, of e.g. their drawn boxes, and check that they intersect at the same vanishing point. The only way the projection of parallel lines do not intersect is if those are parallel to the projection plane/film. $\endgroup$
    – lightxbulb
    Commented Oct 31, 2021 at 19:23
  • $\begingroup$ @lightxbulb l have asked that question, do you have any straightforward answer? So please give the answer without confusing me. $\endgroup$
    – user17337
    Commented Oct 31, 2021 at 19:55
  • $\begingroup$ What part of my answer did you find unsatisfactory? $\endgroup$
    – lightxbulb
    Commented Oct 31, 2021 at 20:04
  • $\begingroup$ @lightbulb, I asked regarding X, what you about X? $\endgroup$
    – user17337
    Commented Oct 31, 2021 at 20:17

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My first question is how we can say it doesn't exist, but when we see real image of railway track they intersects at X?

Because human vision only sees a perspective on reality. It may appear to cause some intersection at a far distant point, but if you change your perspective (by moving towards said far distant point), no such intersection ever actually happens.

Just because you see a thing doesn't make it real.

My second question is if it exists then during projection how this X is projects in projection plane?

A 2D space cannot capture the entirety of a 3D space. There's too much of it.

As such, any 2D view of a 3D space is going to only present a subsection of it. A perspective projection works by essentially squeezing a triangular prism view of a 3D world into a 2D area. This creates depth distortion, as more distant areas take up less space than areas closer to you.

A "vanishing point" exists after a perspective projection by the nature of the perspective projection.

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  • $\begingroup$ Is it projecting like normal point projection (x,y,z) to in view plane (xp,yp,zp)? $\endgroup$
    – user17337
    Commented Nov 1, 2021 at 5:51
  • $\begingroup$ I don't know what you mean by "normal point projection". Do you mean a perspective projection? And as has been discussed several times now, it's not a projection if it transforms between spaces that have the same dimensions. $\endgroup$ Commented Nov 1, 2021 at 6:19
  • $\begingroup$ question is could I represents vanishing point (x,y,z) transforms to (xp,yp,zp)? $\endgroup$
    – user17337
    Commented Nov 1, 2021 at 6:54
  • $\begingroup$ I think there may be a confusion as to what a computer graphics pipeline does versus what projection means $\endgroup$
    – joojaa
    Commented Nov 1, 2021 at 6:54
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    $\begingroup$ @User4567: Appearance and reality are not the same thing. A mirror does not create a reversed world; it merely reflects light. A lens in front of your retina that bends the path of light does not cause parallel lines to intersect, even if it makes it appear as if they do. Objects don't grow in size if they get closer to you, even if the lens over your retinas makes it appear that they do. $\endgroup$ Commented Nov 4, 2021 at 20:21
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In CG the vanishing point can be found computationally by figuring where every pixel on the view plane merges into a single pixel. Imagine a line from every corner of the screen extending out until they touch. That is the vanishing point. It is very literal in computer graphics because no mater what you render beyond that point it will all merge to a single pixel. Now you can move the camera all over the place and that single pixel size block can be anywhere on the screen, it is not unique. But it is still at the vanishing point.

That is the ideal. In practice the vanishing point is much closer because objects become so small that attempting to render them is more or less pointless.

How does it exist in the view plane? If you start with the camera looking at a block covering the entire view plane then start moving backward, that block will become smaller and smaller until it is the size of a single pixel. The camera can keep moving back but what's the point?

Vanishing points in the real world are much more ephemeral as we can get great big telescopes and move them back, we are about to move vanishing points back in the real world even more with the James Web telescope. But in computer graphics the vanishing point is real and can be computed.

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  • $\begingroup$ it's not my question answer.. I have asked regarding X? I have demanded about X, it's right? $\endgroup$
    – user17337
    Commented Oct 31, 2021 at 20:23
  • $\begingroup$ My answer is literally taking about the X get a magnifying glass and look at that dot in the middle of the X that is what I am talking about. $\endgroup$
    – pmw1234
    Commented Oct 31, 2021 at 20:26
  • $\begingroup$ I have demanded projection about X , (X, y, z) to (xp, yp, zp) it's right? $\endgroup$
    – user17337
    Commented Oct 31, 2021 at 20:29
  • $\begingroup$ And that value can be computed, it can be computed in world space, it can be computed in perspective space, and it can be computed in screen space. You can also compute Y and even Z is you want. $\endgroup$
    – pmw1234
    Commented Oct 31, 2021 at 20:36
  • $\begingroup$ but joojaa is saying totally different things, he is saying vanishing point doesn't exist, and it's projection not possible... $\endgroup$
    – user17337
    Commented Oct 31, 2021 at 20:39
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Vanishing points are a simplified concept that artists use to help them draw perspective images of idealized pinhole cameras. Like so much in art history it's only one possible way to draw 3D imagery. You can use other methods but they are more convoluted and not widely taught to people. Artists have a person in the loop so that person can use judgement so it's the quick and dirty way is good enough.

There is no need to know or even draw a line to a vanishing point to draw a correct perspective image. You can also use advanced methods to draw fisheye perspectives where straight lines are no longer straight.

  • A vanishing point does not exist in 3D space ($R^3$).
  • A vanishing point exists in a projection $R^2$.
  • But it does exist in a homogenious projective space. But needs to be modelled separately from normal geometry as you can not rely on taking it with you form 3d space. Like normal points

Therefore you can not transform a vanishing point from 3D to 2D which is what you originally asked. But you can create a model where its true.

Mathematically such a point exists at the limit of an infinite line, but there's no set where they overlap. Intersects at infinity is OK. But in practice, there's no difference between a mathematical point and an apparent point. They transform the same way in spaces they exist in.

Now, it is important that we do not regard a 2D image consisting of pixels. We might be rendering vector graphics and thus computing points coordinates not pixels. Yet for practical reasons, an imaging system has to have some minimal width for a line for it to be visible.

Again, as I said earlier, any point sufficiently far away from the camera in 3D is at the mathematically approximate location of a vanishing point in 2D. For all intents and purposes, this is a vanishing point. This point behaves like any other point. But is this in any way useful for your line of query? Probably not since this definition would mean that there are lots of vanishing points in any image, horizon line being all the vanishing points for all lines of a plane.* Your line questioning is sufficiently close to a XY problem to never have a good answer. Consider asking what you want to achieve instead.

Is a vanishing point real? This is a really deep metaphysical question**. In some sense, it's not, if you give me the ability to zoom infinitely then you will never reach it. But you can still identify it and see it. It both does and does not exist at the same time, it's a concept. There is no contradiction here just different models that have different results. The concept exists, after all, it has a Wikipedia page, but so does spider man. Real-world is like that, with lots and lots of conflicting models.***

* If you haven't noticed it, is not very fruitful to ask: "Where is the vanishing point of this image, this camera, or this model?". Because there are probably too many of them. It is however much more fruitful to ask: "Where is the vanishing point of this line?"

** ask a philosophy.se?

*** this reminds me of the fractal nature of the world. If you ask how long is the coastline of a country is you get different answers depending on how accurately you measure. The length of this may be infinite. Which admittedly is not very useful and probably not accepted as real in any pub quiz.

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  • $\begingroup$ There's nothing metaphysical about a vanishing point. It's the intersection point of the projection of two lines that are parallel in 3D (whose projection however is not parallel). In homogeneous coordinates the vanishing point in extended 3D (i.e. including infinity) can be represented as $(x_p,y_p,z_p, 0)$. All lines parallel to this direction intersect at this point at infinity. $\endgroup$
    – lightxbulb
    Commented Nov 1, 2021 at 9:49
  • $\begingroup$ @lightxbulb it is not intesect but there is a limit point yes. But even so its a concept that has a solution in some models but not in others. $\endgroup$
    – joojaa
    Commented Nov 1, 2021 at 12:45
  • $\begingroup$ Yes, the formal meaning of intersect at infinity (this is valid terminology from projective geometry) implies that they agree in the limit. So all lines parallel to a specific vector $(x,y,z)$ intersect at $(x,y,z,0)$ (one typically makes the vector unit length in order to get a unique representation). $\endgroup$
    – lightxbulb
    Commented Nov 1, 2021 at 12:48
  • $\begingroup$ @lightxbulb yes but like i said this only applies to the homogenious model, it does not follow that there is a point at infinity in normal 3d cartesian space. $\endgroup$
    – joojaa
    Commented Nov 1, 2021 at 12:53
  • $\begingroup$ @lightxbulb im not saying that it does not exist in a 2d projection as such but rather i is not a concept that exists everywere. Even here it relies on being able to model infinity somehow in your model $\endgroup$
    – joojaa
    Commented Nov 1, 2021 at 12:56
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The vanishing point does not exist in the 3d model, only in the projected image.

If it exists, how does this X project onto the projection plane during projection?

It's the limit of the projection of the straight rail tracks as distance goes to inifinity,

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  • $\begingroup$ could you elaborate more your this statement "It's the limit of the projection of the straight rail tracks as distance goes to inifinity" $\endgroup$
    – user17337
    Commented Nov 1, 2021 at 14:27
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My first question is: How can we say it doesn't exist, but when we see a real image of a railway track, the tracks intersect at X?

You do not see the tracks intersect at X or anywhere else because the tracks do not intersect. However, when they get far enough away and the angles line up, you can no longer distinguish the tracks as two separate entities, and that is what we call the vanishing point. It might be easier to understand vanishing points in the real world before dealing with projections. when you look at a road or a track as it goes around a bend there is a point where the inside of the curve obscures the far side so what is around the bend is no longer visible, but as you approach the corner, you can see further and further around it. That is also known as the vanishing point and it moves as your view point moves.

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