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I synchronised with @David k's first and second answer regarding perspective and parallel projection. From the first question I understand that image on the screen is typically described in horizontal and vertical coordinates, that is 2D points. But from second question I understand that image on the screen are realistic which uses perspective projection has coordinates 3D.

My confusion is that how we show the image on the screen with 2D points by perspective projection?

The Point (x,y,z) is projected to position (xp,yp,zvp) on the view plane.Since the view plane is placed at position zvp along the zv axis. So when zvp=0 Projectors (projection vectors) are not converged to projection reference point in parallel projection. But in perspective projection when zvp=0, it is reduced to parallel projection,but projectors are converged to centre of projection. Am I correct above these concepts?

The vanishing point in perspective projection which is 3D coordinates could be represents in projective plane as (x, y, z, 0). Am I correct?

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  • Hi. Welcome to GDSE. This question doesn't seem to be about graphic design, and I think it would be better suited to Computer Graphics Stack Exchange. Sorry, but I'm voting to close it.
    – Billy Kerr
    Oct 27 '21 at 11:04
  • Sorry, but most of use really do not have used to the notation and coordinate system that you use. I guess it's based on some book. That makes your writings to look nonsense even in case they happen to be logical and well formulated ones. Your recent question was solvable because you included the relevant book pages and the case was an elementary misreading what the book said. I must admit I haven't now a slightest idea what you mean with term (x, y, z, 0). BTW. Parallel projection can be seen as the limit case where projection center is moved to infinity to projection direction.
    – user287001
    Oct 27 '21 at 11:40
  • @user287001 I linked two question which may be helpful to give the answer. Only you have an caliber to solve the question. Please help me.
    – Alok Maity
    Oct 27 '21 at 11:49
  • Some of your problems seem to be confusion caused by the difference between coordinate systems here drive.google.com/file/d/1DO2pwP8u6GPs6pjnOg0q19JMr6k8dIhr/… and in the book excerpt in this question math.stackexchange.com/questions/4288470/… The coordinate systems use partially same letters, but they are very different. The 2 linked questions do not tell what YOU mean with (x, y, z, 0). I have seen combination (x, y, z, 0) numerous times in different contexts, for ex. in physics.
    – user287001
    Oct 27 '21 at 12:12
  • @user287001 Any point goes to infinity, we just write in 2D (inifiny,infinity),we can't represents in 2D. That's why we use projective plane to represent infinite point which is of the form (x, y, 0) which represents inifiny because (x/0,y/0). In projective plane P^2 (x, y, 0) is called infinite point in 2D . But in perspective projection has vanishing point 3D(infinite point) and this point could be represented by homogenous coordinates by (x, y,z, 0)
    – Alok Maity
    Oct 27 '21 at 12:30
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Seemingly there's no flood of answers from competent mathematicians in this site, but here's one from the rest of us.

At first about presenting a perspective image with 2D coordinates.

To be fully defined a plane in Euclidean 3D space needs 3 points which are not in the same line:

enter image description here

I have here a grey polygon which is a part my plane which is defined by my green, orange and blue point.

The green one is my plane base point. Vector P points to it from my origin.

Orange and blue are my orientation points. Vector A points from the base point to the orange orientation point. Vector B points from the base point to the blue one.

For an arbitary point (purple) in my plane there exist always such numbers u and v that vector P + uA + vB points from the origin to the arbitary point.

When u and v vary through all real number value combinations the end tip of P + uA + vB walks through every point of my plane. Thus u and v can be considered as the internal coordinates of my plane.

Just in my drawing obviously u is about +2 and v is about +0.5

Every plane have an infinite number of three point sets (green, orange, blue) which define the same plane. For practical simplicity we often have perpendicular A and B which are both one unit long.

In even simpler cases we use in whole 3D space a coordinate system where A has the same direction as the X-axis and B has the same direction as the Y axis. But that's not a must. You should see that every point in a plane is possible to tell with 2 internal plane coordinate numbers (u and v) if the base point and the orientation points are known or equivalently vectors P, A and B are known.

Perspective images of 3D objects are drawn by the projector lines to the imaging plane. The image is a part of a plane, so every image point must be possible to present with 2 internal coordinate numbers of the known (=vectors P, A and B are also known) plane.

In your recent linked book example we can simply drop out the non-changing Zvp which was assumed to be a constant and defined from the start. The rest remaining coordinates Xp and Yp of the image point on the imaging plane have the role of my u and v.

Reduction from perspective projection to parallel one: That does not happen by setting Zvp=0 because the observer or actually the projection center in computer graphics talk doesn't move to infinity - its place doesn't depend on the place of the imaging plane (or actually the view plane in your talk)

My knowledge of projective coordinate systems for extended 3D space is too thin for proper answering to the (x,y,z,0) -question.

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  • Let the moderator clean the board. I'm afraid questions this elementary will not get friendly treatment there and my even more elementary answer would get there good laughs (at best).
    – user287001
    Oct 27 '21 at 21:37
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    User crossposted this on computergraphics so if i were to answer i would answer there. Please dont answer questions that clearly dont belong on graphicdesign. Anyway, compuer graphics program dont set the z value, theres no need to i can just ignore it. As for clip space coordinates not my forte but technically all kinds of undefined behaviours happen outside the clipspace frustum so not really a good ide to contemplate on these too much.
    – joojaa
    Oct 27 '21 at 21:38
  • no he got answers to all his questions save one i think this may be puhing things a bit far.
    – joojaa
    Oct 27 '21 at 21:39
  • Anyway, its not so much that i think they dont like to answer as the chain of questioning being a bit unproductive. So none of the questions really make any meaningful impact to OPs knowledge as theres no self reflection on the root cause of the unexpected problems. Therefore its very likely to be a sort of hydra that keeps getting more heads to kill. What we really need is fire to burn some of the assumptions.
    – joojaa
    Oct 27 '21 at 21:58
  • @user how u is about +2 and v is about +0.5-- I seem u is +1?
    – Alok Maity
    Oct 28 '21 at 12:31

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