From spencer@spline.eecs.umich.edu Wed Oct  5 09:24:27 1988
Path: leah!bingvaxu!sunybcs!boulder!ncar!mailrus!umich!spline.eecs.umich.edu!spencer
From: spencer@spline.eecs.umich.edu (Spencer W. Thomas)
Newsgroups: comp.graphics
Subject: Re: Transforming Quadric Surfaces ( HELP!!!! )
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Date: 5 Oct 88 13:24:27 GMT
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In article <511@uvicctr.UUCP> bcorrie@uvicctr.UUCP (Brian Corrie) writes:
>I have the following problem : Given a quadric of the form
>        Ax^2 + 2Bxy + 2Cxz + 2Dx + Ey^2 + 2Fyz + 2Gy + Hz^2 + 2Iz + J = 0
>
>  I store the coefficients in a 4x4 matrix. I believe this is a kind of
>  standard since I have seen it in several journal articles. The matrix and
>  the form of the equation is given below.
>
>               | A B C D |   | x |
>   [x,y,z,1] * | B E F G | * | y | = 0
>               | C F H I |   | z |
>               | D G I J |   | 1 |
>
>  My problem is that I need to translate and rotate this quadric by a standard
>  3D translation. ( eg. Translate its center to origin ). It is my
>  understanding that this can be done by using a standard 4x4 3D homogenous
>  transformation matrix and its transpose or its inverse.

Yup.  Thats right.

Suppose your transformation is T.  I.e.

[x,y,z,1] * T = [x',y',z',w']

Letting T(-1) is the inverse of T, you have
[x,y,z,1] = [x',y',z',w'] * T(-1)

Let Tt be the transpose of T, then

Tt  | x | = | x' |
    | y |   | y' |
    | z |   | z' |
    | 1 |   | w' |

and
    | x | = Tt(-1) * | x' |
    | y |            | y' |
    | z |            | z' |
    | 1 |            | w' |

Thus,
               | A B C D |   | x |
   [x,y,z,1] * | B E F G | * | y | = 0
               | C F H I |   | z |
               | D G I J |   | 1 |

is the same as

                           | A B C D |            | x |
   [x',y',z',w'] * T(-1) * | B E F G | * Tt(-1) * | y | = 0
                           | C F H I |            | z |
                           | D G I J |            | 1 |

So your new quadric matrix must be T(-1) * M * Tt(-1).

=Spencer (spencer@crim.eecs.umich.edu)