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If I do this (excuse the transpose, I'm used to looking at matrices the other way round):

import bpy
from  mathutils import Matrix
ob =  bpy.context.object

m = Matrix( ([1, 0, 0, 0],
             [0, 1, 0, 0],
             [0, 0, 1, 0],
             [0, 0, 1, 0] )  )

m.transpose()

for vert in ob.data.vertices:
    vert.co = m @ vert.co

I get a translation of the mesh by 1 in object-Z. Good. So somewhere, vert.co must be understood to be (x,y,z,w=1), otherwise the translation wouldn't be picked up.

However, if I do this:

import bpy
from  mathutils import Matrix
ob =  bpy.context.object

m = Matrix( ([1, 0, 0, 0],
             [0, 1, 0, 0],
             [0, 0, 1, 1],
             [0, 0, 0, 0] )  )

m.transpose()

for vert in ob.data.vertices:
    vert.co = m @ vert.co

.. I'm hoping for a basic perspective transform, in which the w of every coordinate is set to its z, and then, in normalization, the whole coordinate is divided through by its w before using its x,y,z. But, OK, nothing happens.

Is there any way of getting at the implicit 'w' of the homogeneous vector, or do I have to divide it by hand?

Is there a way using matrix multiplication like this to make projections in the API?

It could be my Blender, it could be my math. I don't know....

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  • $\begingroup$ @Robert Thanks for taking a look. The 0 at the end of the diagonal, i thought, was justified by w -> 0*x + 0*y + 1*z + 0*w I don't want to add 1 to w. $\endgroup$
    – Robin Betts
    Dec 6, 2019 at 9:28
  • 1
    $\begingroup$ Sorry for the confusion w=1 was only meant for the translation, not for the projection. Otherwise you'd be setting w to zero, which would place the vertex at infinity (and perform a division by zero when attempting to convert to 3D). $\endgroup$
    – Robert Gützkow
    Dec 6, 2019 at 12:08

1 Answer 1

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Make the vert.co 4d

From Perspective projection of transformation matrix wiki

The simplest perspective projection uses the origin as the center of projection, and the plane at ${\displaystyle z=1}$ as the image plane. The functional form of this transformation is then ${\displaystyle x'=x/z}$ ; ${\displaystyle y'=y/z}$. We can express this in homogeneous coordinates as:

$\begin{pmatrix}x_c\\ y_c \\ z_c \\ w_c\end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0\end{pmatrix} \begin{pmatrix}x\\ y \\ z \\ w=1\end{pmatrix}$

$\begin{pmatrix}x'\\ y' \\ z' \\ 1\end{pmatrix} = 1 / w_c \begin{pmatrix}x_c\\ y_c \\ z_c \\ w_c\end{pmatrix}$

To emulate above would map the vert co to 4d using w of the result

Is there any way of getting at the implicit 'w' of the homogeneous vector, or do I have to divide it by hand?

yeah suppose I've done that.

import bpy
from  mathutils import Matrix
from math import radians
ob =  bpy.context.object

m = Matrix( ([1, 0, 0, 0],
             [0, 1, 0, 0],
             [0, 0, 1, 1],
             [0, 0, 0, 0] )  )

m.transpose()

R = Matrix.Rotation(radians(30), 4, 'X')
#m = R @ m
for vert in ob.data.vertices:
    v = m @ vert.co.to_4d() # sets vert.co.w to 1

    '''
    # vert.co.w is set to vert.co.z
    v =  vert.co.to_4d()
    v.w = v.z
    v = m @ v
    '''
    vert.co = 1 / v.w * v.to_3d()

If run on default cube, flattens it to the plane z=1. However others will fail with zero divide error.

The question is how to handle $\ w_c = 0$

One property of homogeneous coordinates is that they allow you to have points at infinity (infinite length vectors), which is not possible with 3D coordinates. Points at infinity occur when W=0. If you try and convert a W=0 homogeneous coordinate into a normal W=1 coordinate, it results in a bunch of divide-by-zero operations:

This means that homogeneous coordinates with W=0 can not be converted into 3D coordinates. What use does this have? Well, directional lights can be though of as point lights that are infinitely far away. When a point light is infinitely far away the rays of light become parallel, and all of the light travels in a single direction. This is basically the definition of a directional light.

Playing around with meshes using above Not sure how useful the answer is. In first example in question above, worth noting m.translation = (0, 0, 1) whereas in second it is zero. Converting 3d vector to 4d sets w to 1. Which I assume is done for us since it doesn't spit the dummy multiplying a 4x4 with a 3x1.

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  • $\begingroup$ 'spit the dummy' ;D .. to_4d() was what I was missing. After playing around, conclude there is no implicit division available for homogeneous coordinates.. it is helpful to be reassured of that. I was hoping just to combine matrices, but it does look as if there has to be a manual step. Division by 0 normally handled by near clipping plane.. have to watch out for it. $\endgroup$
    – Robin Betts
    Dec 6, 2019 at 10:52

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