I am trying to find the parallel projection matrix to project a 3D arbitrary point to an plane of equation ax+by+cz+d=0. I would like to project my 3D points to the plane, parallel to the normal of the plane. I know how to do it in a classical way but I heard about homogeneous matrix. I found a lot of information on perspective projection matrix but I am a little bit lost, and don't know how to apply that to my problem..
Any help ?
thanks
Although you'll be starting with the plane equation in the form Ax + By + Cz + D = 0, you'll first want to define a local coordinate system that allows you to use a parametric equation of the plane.
The parametric equation of the plane we're interested in requires you to write it in the form of some origin Q (any point on the plane) plus the coordinate u times some vector U in the plane, plus the coordinate v times some vector V also in the plane, but not colinear with U plus the coordinate n times the normal vector N = (A,B,C).
Q + uU + vV + nN
If you had these u,v,n coordinates, you would perform your operation of projecting all points in space onto the plane by simply setting their n coordinate to zero and keeping their u and v coordinates as is.
So we will start with a matrix that can transform from the local coordinate system (local to the plane) in u,v,n to the global coordinate system in x,y,z coordinates. This is really straightforward to write as a 4x4 homogeneous matrix.
[ Ux Vx Nx Qx ]
G = [ Uy Vy Ny Qy ]
[ Uz Vz Nz Qz ]
[ 0 0 0 1 ]
(Reminder that Nx = A Ny = B and Nz = C.)
Call this matrix G (for "global"). The inverse G⁻¹ would transform from global back to local coordinates.
Our matrix F (for "flatten") that squashes the n coordinate to zero is simply:
[ 1 0 0 0 ]
F = [ 0 1 0 0 ]
[ 0 0 0 0 ]
[ 0 0 0 1 ]
So now we can construct your matrix M that begins by transforming from global to local coordinates with G⁻¹, then flattening it to the plane with F, then transforming from local to global coordinates with G. (Remember the order is reversed for matrices because the vector goes on the right.)
M = GFG⁻¹
That's a 4x4 matrix formed by G multiplied by F multiplied by G⁻¹.
There were two steps I skipped.
Q in the plane.U and VWe can choose any origin for the parametric definition of the plane because we won't be using the u or v coordinates.
To find a point Q in the plane to use as its origin, we can find where the plane intersects either the X, Y or Z axis. (It will hit one of those). Simply use any non-zero coordinate of the normal vector. This is like substituting the point (0,0,z) into the plane equation Ax + By + Cz + D = 0 and solving for z. There are three potentially appropriate points depending on which coordinate was non-zero.
(0, 0, -D/C)(0, -D/B, 0)(-D/A, 0, 0)To choose vectors for U and V (they don't actually have to be unit vectors) you just have to choose vectors that are in the plane that are not colinear.
To choose a U vector, we can take any vector and take the cross product of that vector and the normal. The result will be a vector in the plane. We just have to make sure this cross product does not result in a zero-length vector. In other words, the "any vector" we choose must not be colinear with the normal. We can select one of:
X̂ = (1, 0, 0)Ŷ = (0, 1, 0)Ẑ = (0, 0, 1)e.g.: U = N × X̂
One of the above choices will be appropriately not colinear with the normal vector. Check that the cross product is not zero-length before carrying on.
Then, to get the V vector, we simply take V = N × U (cross product). Again, none of these vectors have to be unit vectors because we don't care about the u and v coordinates and we don't care about the magnitude of the n coordinate because we're just going to squash it to zero anyway.
So it's a few steps to construct the matrix, but hopefully this outlines all the work you have to do to get it. Then you can just multiply any point P by the matrix to project it onto the plane.
eg.: P' = MP
Hopefully you're comfortable enough using a 4x4 matrix, and of course, supplying your vectors that you want to transform in homogenous form, which is a 4-vector with coordinates (x,y,z,1).