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I am implementing perspective from scratch for an academic project. I am using "Computer Graphics: principles and practices", by Foley, van Dam, Feiner and Hughes (second edition in C).

I just followed the book by implementing all the matrices transformations needed to traslate, rotate, shear, scale, project, transform from perspective to parallel canonical view volumes and for clipping. The book apparently uses a right-handed coordinate system. However, I ended up with primitives appearing in a left-handed coordinate system and I cannot explain why.

Here are the matrices that I used:

Translation:
1, 0, 0, dx
0, 1, 0, dy
0, 0, 1, dz
0, 0, 0, 1

Rotation (to align a coordinate system (rx, ry, rz) to XYZ):
rx1, rx2, rx3, 0
ry1, ry2, ry3, 0
rz1, rz2, rz3, 0
0  , 0  , 0  , 1

Scale:
sx, 0 , 0 , 0
0 , sy, 0 , 0
0 , 0 , sz, 0
0 , 0 , 0 , 1

Shear XY:
1, 0, shx, 0
0, 1, shy, 0
0, 0, 1  , 0
0, 0, 0  , 1

Projecting onto a plane at z = d, with PRP at origin, looking in the positive z direction:
1, 0, 0  , 0
0, 1, 0  , 0
0, 0, 1  , 0
0, 0, 1/d, 0

Then given VRP, VPN, PRP, VUP, f and b (and the direction of projection dop), reduce the space to the canonical viewing volume for perspective using P:

rz = VPN / |VPN|
rx = (VUP x rz) / |VUP x rz|
ry = rz x rx
P = ScaleUniform(-1 / (vrp1Z + b)) * 
    Scale(-2 * vrp1Z / deltaU, -2 * vrp1Z / deltaV, 1) * 
    Shear(-dopX / dopZ, -dopY / dopZ) * 
    T(PRP) * 
    R(rx, ry, rz) * 
    T(-VRP)

Where vrp1 is ShearXY * T(-PRP) * (0, 0, 0, 1), deltaU and deltaV the width and height of the viewing window. dop is computed as CW - PRP, where CW is the center of the viewing window.

Then Projection(d) * P gives me the projection matrix.

I projected simple lines representing the unit vectors on x, y and z, but the representation that I obtained drawn on the screen was clearly a left handed coordinate system... Now I need to work in a right handed coordinate system, so is there a way to know where I did wrong?

Here is the code I used: As you can see, the Z component of the scale matrix is of opposite sign, since clipping wasn't working properly because something was right-handed and something left-handed, but I couldn't discern what exactly, so I swapped the sign of the scale because it wasn't needed in a left-hand system.

Vector rz = vpn.toUnitVector();
Vector rx = vup.cross(rz).toUnitVector();
Vector ry = rz.cross(rx).toUnitVector(); 

Vector cw  = viewWindow.getCenter();
Vector dop = cw - prp;

Matrix t1 = Matrix::traslation(-vrp[X], -vrp[Y], -vrp[Z]);
Matrix r  = Matrix::rotation(rx, ry, rz);   
Matrix t2 = Matrix::traslation(-prp[X], -prp[Y], -prp[Z]);   
Matrix partial = t2 * r * t1;

Matrix shear  = Matrix::shearXY(-dop[X] / dop[Z], -dop[Y] / dop[Z]);
Matrix inverseShear = Matrix::shearXY(dop[X] / dop[Z], dop[Y] / dop[Z]);

Vector vrp1 = shear * t2 * Vector(0, 0, 0, 1);

Matrix scale = Matrix::scale(
    2 * vrp1[Z] / ((viewWindow.xMax - viewWindow.xMin) * (vrp1[Z] + b)),
    2 * vrp1[Z] / ((viewWindow.yMax - viewWindow.yMin) * (vrp1[Z] + b)),
    1 / (vrp1[Z] + b)); // HERE <--- WAS NEGATIVE
Matrix inverseScale = Matrix::scale(
    ((viewWindow.xMax - viewWindow.xMin) * (vrp1[Z] + b)) / (2 * vrp1[Z]),
    ((viewWindow.yMax - viewWindow.yMin) * (vrp1[Z] + b)) / (2 * vrp1[Z]),
    (vrp1[Z] + b));

float zMin = -(vrp1[Z] + f) / (vrp1[Z] + b);

Matrix parallel = Perspective::toParallelCvv(zMin);
Matrix inverseParallel = Perspective::inverseToParallelCvv(zMin);
Matrix perspective = Perspective::copAtOrigin(-vrp1[Z]);

projection = perspective * shear * partial;
canonicalView = parallel * scale * shear * partial;
canonicalToProjection = perspective * inverseScale * inverseParallel;
Totem
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  • Does this help narrow it down? http://stackoverflow.com/questions/1263072/changing-a-matrix-from-right-handed-to-left-handed-coordinate-system/1263107#1263107 – Charles Clark Oct 23 '13 at 04:50
  • I know the convention of angles' signs, but in the matrix rotation I don't have directly sines and cosines. The rotation matrix's components are the new unit vectors after all the rotation, and it doesn't really matter if the coordinate system is left or right handed. I mean, if the initial system is right-handed, also the rotated system should be. Only that there is nothing telling me how is the initial system: I only have a bunch of points. So I really don't know how it ends up being left handed. – Totem Oct 23 '13 at 16:55

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