Page 171 - Geometric Modeling and Algebraic Geometry
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9 Intersecting Biquadratic Patches  173
                           9.6.1 Resultant-based method

                           In the parameter domain [0, 1] , the self–intersection curve of the first patch forms
                                                    4
                           the set
                              $                                                          %
                               (u 1 ,v 1 ,u 2 ,v 2 ) ∈ [0, 1] | (u 1 ,v 1 )  =(u 2 ,v 2 ) and x(u 1 ,v 1 )= x(u 2 ,v 2 ) .
                                                  4
                                                                                         (9.26)
                           This locus is the real trace of a complex curve. We assume that it is either empty or of
                           dimension 0 or 1. We do not consider degenerate cases, such as a plane which is cov-
                           ered twice. In the examples presented below (see Section 9.7), the self–intersection
                           locus is a curve in R .
                                           4
                              We use the following change of coordinates to discard the unwanted trivial com-
                           ponent (u 1 ,v 1 )=(u 2 ,v 2 ). Let (u 2 ,v 1 ) be a pair of parameters in [0, 1] , (l, k) ∈ R 2
                                                                                   2
                           and let u 1 = u 2 + l, v 2 = v 1 + lk. If we suppose that we have (u 1 ,v 1 )  =(u 2 ,v 2 ),
                           then l  =0. Hence x(u 1 ,v 1 )= x(u 2 ,v 2 ) if and only if x(u 2 +l, v 1 )= x(u 2 ,v 1 +lk).
                           We suppose now that (u 2 ,v 1 ,l,k) verifies this last relation.
                                  ˜
                              Let T(u 2 ,v 1 ,l,k) be the polynomial  1  [x(u 2 + l, v 1 ) − x(u 2 ,v 2 + lk)], its de-
                                                             l
                           gree in (u 2 ,v 1 ,l,k) is (2, 2, 1, 2) and the monomial support with respect to (l, k)
                           contains only k l, k, l and 1. We can decrease the degree by introducing
                                       2
                                                                       1
                                                               ˜
                                              T(u 2 ,v 1 ,m,k)= mT(u 2 ,v 1 ,  ,k).      (9.27)
                                                                      m
                           Then in T(u 2 ,v 1 ,m,k), the monomial support in (m, k) consists only of 1,m,k 2
                           and km. So, we can write T in a matrix form:

                                                                                    ⎛     ⎞
                                             ⎛                                    ⎞    1
                                               a 1 (u 2 ,v 1 ) b 1 (u 2 ,v 1 ) c 1 (u 2 ,v 1 ) d 1 (u 2 ,v 1 )
                                                                                    ⎜  m  ⎟
                              T(u 2 ,v 1 ,m,k)=  ⎝ a 2 (u 2 ,v 1 ) b 2 (u 2 ,v 1 ) c 2 (u 2 ,v 1 ) d 2 (u 2 ,v 1 ) ⎠ ⎜  ⎟
                                                                                    ⎝ k ⎠
                                                                                        2
                                               a 3 (u 2 ,v 1 ) b 3 (u 2 ,v 1 ) c 3 (u 2 ,v 1 ) d 3 (u 2 ,v 1 )
                                                                                      km
                                                                                         (9.28)
                           By Cramer’s rule, we get
                                            m =  D 2  ,  k =  D 3 ,  and  km =  D 4      (9.29)
                                                       2
                                                           D 1
                                                D 1
                                                                           D 1
                           with


                                    b 1 c 1 d 1        −a 1 c 1 d 1        b 1 −a 1 d 1        b 1 c 1 −a 1
                           D 1 = b 2 c 2 d 2 ,D 2 = −a 2 c 2 d 2 ,D 3 = b 2 −a 2 d 2 ,D 4 = b 2 c 2 −a 2 .









                                    b 3 c 3 d 3        −a 3 c 3 d 3        b 3 −a 3 d 3        b 3 c 3 −a 3
                           Let Q(u 2 ,v 1 ) be the polynomial Q = D D 1 − D D 3 .
                                                            2
                                                                   2
                                                           4       2
                                                            $                           %
                           Lemma 3. The implicitly defined curve (u 2 ,v 1 ) ∈ [0, 1] | Q(u 2 ,v 1 )=0 is the
                                                                          2
                           projection of the self–intersection locus (given by the set (9.26) but in C ) into the
                                                                                      4
                           parameters domain (u 2 ,v 1 ) ∈ [0, 1] .
                                                        2
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