6

Currently, I'm facing a problem and I think I have found a solution to it. However, I'd be interested to hear your opinion or hints on this.

I have written geometric modeling software and I have users who want to define a number of different bodies (e.g., cylinders, spheres, ellipsoids, etc) by providing the coefficients of the general quadric surface equation:

$$ax^2 + 2bxy + 2cxz + 2dx + \dots=0$$

which, in general, can be written also using a coefficient matrix $Q$ and a row vector $v$, $ v Q v^t $. My solid modeler already provides various different objects like cylinders, ellipsoids, etc but they are defined in a canonical system (e.g., cylinder parallel to the $\,z\,$ axis) and carry a transformation matrix to place them at the right position for rendering. In order to treat bodies defined by general quadrics, I have to do two things:

  1. Determine which kind of surface has been defined

  2. Obtain the transformation matrix which will transform the respective surface to the position/orientation which is implicitly included in the quadrics coefficients.

After doing a bit of algebra I came up with the following idea for 2. I thought to tackle the issue via principal axis transformation. So I would first try to determine the eigenvalues and the eigenvectors for the coefficient matrix $Q.$ The matrix which is represented by the eigenvectors in its columns should transform my coefficient matrix into the canonical system.

If I take the inverse of this eigenmatrix then this should be my transformation matrix, or am I wrong here? The eigenvalues would eventually determine the magnitude of parameters like the radius for spheres & cylinders, the major or minor halfaxes for ellipsoids etc.

I'd appreciate to hear your thoughts on this or if somebody has a simpler solution I'd be happy to discuss this as well.

Rodrigo de Azevedo
  • 23,223
user68158
  • 61
  • This isn't a complete answer, but this has many of the elements that you are looking for, with several links to the relevant literature. – Damien Apr 25 '13 at 01:06

2 Answers2

1

Okay, let's try something like that: $$ C' = MCM^T $$ with $M$ being the sought pose and $(C,C')$ the quadrics in correspondence.

As $C$ is a real & symmetric matrix, if we apply an Eigen decomposition, we can write: $$ V'D'(V')^T = MVD(V)^TM^T $$ Let's re-arrange: $$ V'D'(V')^T = MV(D)(MV)^T $$ Because a rigid transformation would not change the Eigenvalues, I expect that $$ V' = MV $$ and from here: $$ M = V'V^{-1} $$

I have not implemented this though. So let me know if you think there is any mistake in this thought.

Tolga Birdal
  • 1,082
  • 9
  • 18
0

This supplementary material (Chapter 11) of a computer vision course describes the matrix form of quadratic surfaces (Q), how to raytrace them, and how to apply homogenous transformation matrices (M) to them:

$$ Q'=M^{-1}Q(M^{-1})^t $$ with $$ X=\left[\begin{array}{cccc}x\\y\\z\\1\end{array}\right], ~ Q = \left[ \begin{array}{cccc} a & b & c & d \\ b & e & f & g\\ c & f & h & i \\ d & g & i & j \end{array} \right], ~ X^tQX=0, $$

and $X$ being your $v^t$.

user119417
  • 11