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I have a mesh stored in .obj format with a texture image in .png format. For any "defined" pixel in the texture image, I'd like to compute the corresponding 3D coordinates when the texture image is mapped to the 3D mesh. For "defined" pixels, I mean the texture pixels that correspond to points on the surface of the mesh, not the texture pixels between the texture patches.

The .obj file format is as follows:

v 0.123 0.234 0.345 1.0
v ...
...
vt 0.500 1
vt ...
...
vn 0.707 0.000 0.707
vn ...
...
f 1/1 2/2 3/3
f ...
...
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  • \$\begingroup\$ What do you need this for? There's not necessarily a defined mapping between texture pixels (texels) and 3D coordinates. A single texel may map to zero, one, or an arbitrary number of points in 3D space. \$\endgroup\$ Aug 3 '21 at 20:05
  • \$\begingroup\$ @JonathanS. I'm trying to do remeshing by mapping texels to points in 3D. Is there reference about this kind of mapping given mesh files and texture images? Thanks! \$\endgroup\$
    – chaohuang
    Aug 3 '21 at 20:45
  • \$\begingroup\$ I sadly don't know of any good references. You might be able to do this using the texture coordinate derivatives. Build two base vectors for the triangle's plane, one in increasing U direction, one in increasing V direction. You can then use these vectors to build a 2D grid that you can step through and get the coordinates of every texel. \$\endgroup\$ Aug 3 '21 at 21:07
  • \$\begingroup\$ I think we can do affine texture mapping with little effort. For perspective texture mapping I have no idea (well, if you had quads, you could solve a perspective projection - not exactly what I link, but same approach). Also, I remind you there are cases where the UV coordinates of a triangle in space don't make a triangle. For the simpler example, if the three vertex of a triangle in space have the same UV coordinates, then that point on the texture maps to the whole triangle. In other cases a point on the texture maps to a line. \$\endgroup\$
    – Theraot
    Aug 4 '21 at 9:41
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This answer only covers Affine Texture Mapping. Consider this a partial solution.


I'll be assuming you can process your obj to a list of triangles. Each triangle having three vertex. And each vertex having their 3D space coordinates (x, y, z) and their texture coordinates (u, v).


For each triangle, sort the points by one texture coordinate - let us say -u. Once sorted, let us call these points A, B, and C.

You are going to define - at most - two ranges: The range from A.u to B.u, and the range from B.u to C.u. And I say at most, because perhaps some of these coordinates are equal. If they are, you don't need that range.

The start and end of each range define two lines, which constraint the texture space on the u axis. On The v axis, you will constraint the texture space by two line equations of the form v = mu + n.

For the first range, the lines equations correspond to the lines l_AB and l_AC. For the second range, the lines are l_AC, l_BC.

You can figure out the line equations with a little algebra. The slope is the good old formula:

m = (B.v - A.v)/(B.u - A.u)

And for the intercept, we can replace one of the points:

A.v = m(A.u) + n

=>

n = A.v - m(A.u)

Remember that we are doing the lines l_AB and l_AC for the first range, which cross at the point A. And the lines l_AC and l_BC for the second, which cross at the point C. Thus, we can use the v coordinates of the other points to determine which line has lower values in each range.

Now, you can iterate over the texture. For example, for the first range:

if (B.v < C.v)
{
    l_1 = l_AB
    l_2 = l_AC
}
else
{
    l_1 = l_AC
    l_2 = l_AB
}

for (int u = (int)floor(A.u); u <= (int)ceil(B.u); u++)
{
    for (int v = (int)floor(l_1(u)); v <= (int)ceil(l_2(u)); v++)
    {
        // Something with u and v
    }
}

I don't know how you want to handle texels that are partially on a triangle. However, I'm suggesting you include them. Which is the reason behind the rounding (floor and ceil) above.


We need to convert these coordinates to 3D, of course. For that we will have to express these u and v coordinates as a linear combination of the lines of the triangle.

We can use a single conversion for the whole triangle - being the points of a triangle always coplanar. Let us base it on the point A. So we will work assuming A = 0.

We will define vectors in texture space: dB = B - A, dC = C - A and d = uv - A. So we could say:

p * dB + q * dC = d

Which is actually two equations:

p * dB.u + q * dC.u = d.u
p * dB.v + q * dC.v = d.v

Since we have two linear equations and two unknown (p, q) we can solve the system. Let us do elimination, Gauss-Jordan style:

   p     q
┌                  ┐
│ dB.u  dC.u │ d.u │
│            │     │
│ dB.v  dC.v │ d.v │
└                  ┘

Divide the first equation by dB.u =>

     p      q
┌                         ┐
│  dB.u    dC.u  │  d.u   │
│ ------  ------ │ -----  │
│  dB.u    dB.u  │  dB.u  │
│                │        │
│  dB.v    dC.v  │  d.v   │
└                         ┘

=>

   p      q
┌                       ┐
│        dC.u  │  d.u   │
│  1    ------ │ ------ │
│        dB.u  │  dB.u  │
│              │        │
│ dB.v   dC.v  │  d.v   │
└                       ┘

Subtract from the second equation the first equation dB.v times =>

     p                    q
┌                                                       ┐
│                        dC.u       │        d.u        │
│    1                  ------      │       ------      │
│                        dB.u       │        dB.u       │
│                                   │                   │
│                             dC.u  │             d.u   │
│ dB.v -(dB.v)1  dC.v -(dB.v)------ │ d.v -(dB.v)------ │
│                             dB.u  │             dB.u  │
└                                                       ┘

=>

  p              q
┌                                              ┐
│              dC.u       │        d.u         │
│ 1           ------      │       ------       │
│              dB.u       │        dB.u        │
│                         │                    │
│            dB.v * dC.u  │        dB.v * d.u  │
│ 0  dC.v - ------------- │ d.v - ------------ │
│               dB.u      │          dB.u      │
└                                              ┘

Divide the second equation by dC.v - (dB.v * dC.u)/dB.u =>

  p               q
┌                                                   ┐
│               dC.u        │         d.u           │
│ 1            ------       │        ------         │
│               dB.u        │         dB.u          │
│                           │                       │
│             dB.v * dC.u   │         dB.v * d.u    │
│     dC.v - -------------  │  d.v - ------------   │
│                dB.u       │           dB.u        │
│ 0  ---------------------- │ --------------------  │
│             dB.v * dC.u   │          dB.v * dC.u  │
│     dC.v - -------------  │  dC.v - ------------- │
│                dB.u       │             dB.u      │
└                                                   ┘

=>

  p    q
┌                                   ┐
│     dC.u  │         d.u           │
│ 1  ------ │        ------         │
│     dB.u  │         dB.u          │
│           │                       │
│           │         dB.v * d.u    │
│           │  d.v - ------------   │
│           │           dB.u        │
│ 0    1    │ --------------------  │
│           │          dB.v * dC.u  │
│           │  dC.v - ------------- │
│           │             dB.u      │
└                                   ┘

=>

  p    q
┌                                             ┐
│     dC.u  │               d.u               │
│ 1  ------ │              ------             │
│     dB.u  │               dB.u              │
│           │                                 │
│           │    dB.u * d.v     dB.v * d.u    │
│           │   ------------ - ------------   │
│           │        dB.u          dB.u       │
│ 0    1    │ ------------------------------- │
│           │   dB.u * dC.v     dB.v * dC.u   │
│           │  ------------- - -------------  │
│           │        dB.u          dB.u       │
└                                             ┘

=>

  p    q
┌                                           ┐
│     dC.u  │              d.u              │
│ 1  ------ │             ------            │
│     dB.u  │              dB.u             │
│           │                               │
│           │    dB.u * d.v - dB.v * d.u    │
│           │   -------------------------   │
│           │              dB.u             │
│ 0    1    │ ----------------------------- │
│           │   dB.u * dC.v - dB.v * dC.u   │
│           │  ---------------------------  │
│           │              dB.u             │
└                                           ┘

=>

  p    q
┌                                         ┐
│     dC.u  │             d.u             │
│ 1  ------ │            ------           │
│     dB.u  │             dB.u            │
│           │                             │
│           │   dB.u * d.v - dB.v * d.u   │
│ 0    1    │ --------------------------- │
│           │  dB.u * dC.v - dB.v * dC.u  │
└                                         ┘

Subtract from the first equation the second equation dC.u/dB.u times =>

  p         q
┌                                                                    ┐
│     dC.u     dC.u  │  d.u       dB.u * d.v - dB.v * d.u      dC.u  │
│ 1  ------ - ------ │ ------ - --------------------------- * ------ │
│     dB.u     dB.u  │  dB.u     dB.u * dC.v - dB.v * dC.u     dB.u  │
│                    │                                               │
│                    │            dB.u * d.v - dB.v * d.u            │
│ 0         1        │          ---------------------------          │
│                    │           dB.u * dC.v - dB.v * dC.u           │
└                                                                    ┘

=>

  p  q
┌                                                      ┐
│      │  d.u       dB.u * d.v - dB.v * d.u      dC.u  │
│ 1  0 │ ------ - --------------------------- * ------ │
│      │  dB.u     dB.u * dC.v - dB.v * dC.u     dB.u  │
│      │                                               │
│      │            dB.u * d.v - dB.v * d.u            │
│ 0  1 │          ---------------------------          │
│      │           dB.u * dC.v - dB.v * dC.u           │
└                                                      ┘

=>

  p  q
┌                                    ┐
│      │   dC.v * d.u - dC.u * d.v   │
│ 1  0 │ --------------------------- │
│      │  dB.u * dC.v - dB.v * dC.u  │
│      │                             │
│      │   dB.u * d.v - dB.v * d.u   │
│ 0  1 │ --------------------------- │
│      │  dB.u * dC.v - dB.v * dC.u  │
└                                    ┘

This last simplification was computer assisted. Thanks Wolfram|Alpha.

So we have:

k = dB.u * dC.v - dB.v * dC.u
p = (dC.v * d.u - dC.u * d.v) / k
q = (dB.u * d.v - dB.v * d.u) / k

We can incorporate that to our code, and use those components to find the 3D coordinates.

VectorUV dB = B.uv - A.uv
VectorUV dC = C.uv - A.uv

if (B.v < C.v)
{
    l_1 = l_AB
    l_2 = l_AC
}
else
{
    l_1 = l_AC
    l_2 = l_AB
}

k = dB.u * dC.v - dB.v * dC.u
for (int u = (int)floor(A.u); u <= (int)ceil(B.u); u++)
{
    for (int v = (int)floor(l_1(u)); v <= (int)ceil(l_2(u)); v++)
    {
        VectorUV d = new VectorUV(u - A.u, v - A.v)
        p = (dC.v * d.u - dC.u * d.v) / k
        q = (dB.u * d.v - dB.v * d.u) / k
        coords = A.xyz + (B.xyz - A.xyz) * p + (C.xyz - A.xyz) * q
        // …
    }
}

I remind you that the code above only handles the first range of a triangle. For the other range you need another for loop. And, of course, you would be doing that for all triangles.


You would have an array where you can store those coordinates. Make it an array of sets, because you may find that the same texel is mapped to one position in 3D, multiple, or none. And you might find a texel has a position multiple times, in particular on the edges.


By the way. We have a division by zero if k = dB.u * dC.v - dB.v * dC.u = 0. I believe this happens if the shape in texture space does not make a triangle. I suggest you consider these cases as pathological.

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