draw 3d faces as 2d - math

Question

I have 3d mesh and I would like to draw each face a 2d shape.
What I have in mind is this:
for each face
1. access the face normal
2. get a rotation matrix from the normal vector
3. multiply each vertex to the rotation matrix to get the vertices in a '2d like ' plane
4. get 2 coordinates from the transformed vertices
I don't know if this is the best way to do this, so any suggestion is welcome.
At the moment I'm trying to get a rotation matrix from the normal vector,
how would I do this ?
UPDATE:
Here is a visual explanation of what I need:
At the moment I have quads, but there's no problem
converting them into triangles.
I want to rotate the vertices of a face, so that
one of the dimensions gets flattened.
I also need to store the original 3d rotation of the face.
I imagine that would be inverse rotation of the face
normal.
I think I'm a bit lost in space :)
Here's a basic prototype I did using Processing:
void setup(){
size(400,400,P3D);
background(255);
stroke(0,0,120);
smooth();
fill(0,120,0);
PVector x = new PVector(1,0,0);
PVector y = new PVector(0,1,0);
PVector z = new PVector(0,0,1);
PVector n = new PVector(0.378521084785,0.925412774086,0.0180059205741);//normal
PVector p0 = new PVector(0.372828125954,-0.178844243288,1.35241031647);
PVector p1 = new PVector(-1.25476706028,0.505195975304,0.412718296051);
PVector p2 = new PVector(-0.372828245163,0.178844287992,-1.35241031647);
PVector p3 = new PVector(1.2547672987,-0.505196034908,-0.412717700005);
PVector[] face = {p0,p1,p2,p3};
PVector[] face2d = new PVector[4];
PVector nr = PVector.add(n,new PVector());//clone normal
float rx = degrees(acos(n.dot(x)));//angle between normal and x axis
float ry = degrees(acos(n.dot(y)));//angle between normal and y axis
float rz = degrees(acos(n.dot(z)));//angle between normal and z axis
PMatrix3D r = new PMatrix3D();
//is this ok, or should I drop the builtin function, and add
//the rotations manually
r.rotateX(rx);
r.rotateY(ry);
r.rotateZ(rz);
print("original: ");println(face);
for(int i = 0 ; i < 4; i++){
PVector rv = new PVector();
PVector rn = new PVector();
r.mult(face[i],rv);
r.mult(nr,rn);
face2d[i] = PVector.add(face[i],rv);
}
print("rotated: ");println(face2d);
//draw
float scale = 100.0;
translate(width * .5,height * .5);//move to centre, Processing has 0,0 = Top,Lef
beginShape(QUADS);
for(int i = 0 ; i < 4; i++){
vertex(face2d[i].x * scale,face2d[i].y * scale,face2d[i].z * scale);
}
endShape();
line(0,0,0,nr.x*scale,nr.y*scale,nr.z*scale);
//what do I do with this ?
float c = cos(0), s = sin(0);
float x2 = n.x*n.x,y2 = n.y*n.y,z2 = n.z*n.z;
PMatrix3D m = new PMatrix3D(x2+(1-x2)*c, n.x*n.y*(1-c)-n.z*s, n.x*n.z*(1-c)+n.y*s, 0,
n.x*n.y*(1-c)+n.z*s,y2+(1-y2)*c,n.y*n.z*(1-c)-n.x*s,0,
n.x*n.y*(1-c)-n.y*s,n.x*n.z*(1-c)+n.x*s,z2-(1-z2)*c,0,
0,0,0,1);
}
Update
Sorry if I'm getting annoying, but I don't seem to get it.
Here's a bit of python using Blender's API:
import Blender
from Blender import *
import math
from math import sin,cos,radians,degrees
def getRotMatrix(n):
c = cos(0)
s = sin(0)
x2 = n.x*n.x
y2 = n.y*n.y
z2 = n.z*n.z
l1 = x2+(1-x2)*c, n.x*n.y*(1-c)+n.z*s, n.x*n.y*(1-c)-n.y*s
l2 = n.x*n.y*(1-c)-n.z*s,y2+(1-y2)*c,n.x*n.z*(1-c)+n.x*s
l3 = n.x*n.z*(1-c)+n.y*s,n.y*n.z*(1-c)-n.x*s,z2-(1-z2)*c
m = Mathutils.Matrix(l1,l2,l3)
return m
scn = Scene.GetCurrent()
ob = scn.objects.active.getData(mesh=True)#access mesh
out = ob.name+'\n'
#face0
f = ob.faces[0]
n = f.v[0].no
out += 'face: ' + str(f)+'\n'
out += 'normal: ' + str(n)+'\n'
m = getRotMatrix(n)
m.invert()
rvs = []
for v in range(0,len(f.v)):
out += 'original vertex'+str(v)+': ' + str(f.v[v].co) + '\n'
rvs.append(m*f.v[v].co)
out += '\n'
for v in range(0,len(rvs)):
out += 'original vertex'+str(v)+': ' + str(rvs[v]) + '\n'
f = open('out.txt','w')
f.write(out)
f.close
All I do is get the current object, access the first face, get the normal, get the vertices, calculate the rotation matrix, invert it, then multiply it by each vertex.
Finally I write a simple output.
Here's the output for a default plane for which I rotated all the vertices manually by 30 degrees:
Plane.008
face: [MFace (0 3 2 1) 0]
normal: [0.000000, -0.499985, 0.866024](vector)
original vertex0: [1.000000, 0.866025, 0.500000](vector)
original vertex1: [-1.000000, 0.866026, 0.500000](vector)
original vertex2: [-1.000000, -0.866025, -0.500000](vector)
original vertex3: [1.000000, -0.866025, -0.500000](vector)
rotated vertex0: [1.000000, 0.866025, 1.000011](vector)
rotated vertex1: [-1.000000, 0.866026, 1.000012](vector)
rotated vertex2: [-1.000000, -0.866025, -1.000012](vector)
rotated vertex3: [1.000000, -0.866025, -1.000012](vector)
Here's the first face of the famous Suzanne mesh:
Suzanne.001
face: [MFace (46 0 2 44) 0]
normal: [0.987976, -0.010102, 0.154088](vector)
original vertex0: [0.468750, 0.242188, 0.757813](vector)
original vertex1: [0.437500, 0.164063, 0.765625](vector)
original vertex2: [0.500000, 0.093750, 0.687500](vector)
original vertex3: [0.562500, 0.242188, 0.671875](vector)
rotated vertex0: [0.468750, 0.242188, -0.795592](vector)
rotated vertex1: [0.437500, 0.164063, -0.803794](vector)
rotated vertex2: [0.500000, 0.093750, -0.721774](vector)
rotated vertex3: [0.562500, 0.242188, -0.705370](vector)
The vertices from the Plane.008 mesh are altered, the ones from Suzanne.001's mesh
aren't. Shouldn't they ? Should I expect to get zeroes on one axis ?
Once I got the rotation matrix from the normal vector, what is the rotation on x,y,z ?
Note: 1. Blender's Matrix supports the * operator 2.In Blender's coordinate system Z point's up. It looks like a right handed system, rotated 90 degrees on X.
Thanks

Answer 1

That looks reasonable to me. Here's how to get a rotation matrix from normal vector. The normal is the vector. The angle is 0. You probably want the inverse rotation.
Is your mesh triangulated? I'm assuming it is. If so, you can do this, without rotation matrices. Let the points of the face be A,B,C. Take any two vertices of the face, say A and B. Define the x axis along vector AB. A is at 0,0. B is at 0,|AB|. C can be determined from trigonometry using the angle between AC and AB (which you get by using the dot product) and the length |AC|.

Answer 2

You created the m matrix correctly. This is the rotation that corresponds to your normal vector. You can use the inverse of this matrix to "unrotate" your points. The normal of face2d will be x, i.e. point along the x-axis. So extract your 2d coordinates accordingly. (This assumes your quad is approximately planar.)
I don't know the library you are using (Processing), so I'm just assuming there are methods for m.invert() and an operator for applying a rotation matrix to a point. They may of course be called something else. Luckily the inverse of a pure rotation matrix is its transpose, and multiplying a matrix and a vector are straightforward to do manually if you need to.
void setup(){
size(400,400,P3D);
background(255);
stroke(0,0,120);
smooth();
fill(0,120,0);
PVector x = new PVector(1,0,0);
PVector y = new PVector(0,1,0);
PVector z = new PVector(0,0,1);
PVector n = new PVector(0.378521084785,0.925412774086,0.0180059205741);//normal
PVector p0 = new PVector(0.372828125954,-0.178844243288,1.35241031647);
PVector p1 = new PVector(-1.25476706028,0.505195975304,0.412718296051);
PVector p2 = new PVector(-0.372828245163,0.178844287992,-1.35241031647);
PVector p3 = new PVector(1.2547672987,-0.505196034908,-0.412717700005);
PVector[] face = {p0,p1,p2,p3};
PVector[] face2d = new PVector[4];
//what do I do with this ?
float c = cos(0), s = sin(0);
float x2 = n.x*n.x,y2 = n.y*n.y,z2 = n.z*n.z;
PMatrix3D m_inverse =
new PMatrix3D(x2+(1-x2)*c, n.x*n.y*(1-c)+n.z*s, n.x*n.y*(1-c)-n.y*s, 0,
n.x*n.y*(1-c)-n.z*s,y2+(1-y2)*c,n.x*n.z*(1-c)+n.x*s, 0,
n.x*n.z*(1-c)+n.y*s,n.y*n.z*(1-c)-n.x*s,z2-(1-z2)*c, 0,
0,0,0,1);
face2d[0] = m_inverse * p0; // Assuming there's an appropriate operator*().
face2d[1] = m_inverse * p1;
face2d[2] = m_inverse * p2;
face2d[3] = m_inverse * p3;
// print & draw as you did before...
}

Answer 3

For face v0-v1-v3-v2 vectors v3-v0, v3-v2 and a face normal already form rotation matrix that would transform 2d face into 3d face.
Matrix represents coordinate system. Each row (or column, depending on notation) corresponds to axis coordinate system within new coordinate system. 3d rotation/translation matrix can be represented as:
vx.x vx.y vx.z 0
vy.x vy.y vy.z 0
vz.x vz.y vz.z 0
vp.x vp.y vp.z 1
where vx is an x axis of a coordinate system, vy - y axis, vz - z axis, and vp - origin of new system.
Assume that v3-v0 is an y axis (2nd row), v3-v2 - x axis (1st row), and normal - z axis (3rd row). Build a matrix from them. Then invert matrix. You'll get a matrix that will rotate a 3d face into 2d face.
I have 3d mesh and I would like to draw each face a 2d shape.
I suspect that UV unwrapping algorithms are closer to what you want to achieve than trying to get rotation matrix from 3d face.

Answer 4

That's very easy to achieve: (Note: By "face" I mean "triangle")
Create a view matrix that represents a camera looking at a face.
Determine the center of the face with bi-linear interpolation.
Determine the normal of the face.
Position the camera some units in opposite normal direction.
Let the camera look at the center of the face.
Set the cameras up vector point in the direction of the middle of any vertex of the face.
Set the aspect ratio to 1.
Compute the view matrix using this data.
Create a orthogonal projection matrix.
Set the width and height of the view frustum large enough to contain the whole face (e.g. the length of the longest site of a face).
Compute the projection matrix.
For every vertex v of the face, multiply it by both matrices: v * view * projection.
The result is a projection of Your 3d faces into 2d space as if You were looking at them exactly orthogonal without any perspective disturbances. The final coordinates will be in normalized screen coordinates where (-1, -1) is the bottom left corner, (0, 0) is the center and (1, 1) is the top right corner.