Page 213 - Basic Structured Grid Generation
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202  Basic Structured Grid Generation

                          This means that since for the angle θ in Fig. 8.14

                                                                      2

                                                    d 1   f N     f N
                                             tan θ =   =     +          − 1,
                                                    p 1   p 1     p 1
                        provided that the value of f is essentially the same at M and N (and equal to f N ), so
                                      √
                        that p 1 <f N <  2p 1 ,wehave
                                                              √
                                                    1 < tan θ<  2 + 1,
                        which implies that
                                                       ◦
                                                                  ◦
                                                     45 <θ < 67.5 .
                          If AY is taken as the next active edge, we will have
                                                  2       2      2      2
                                               AY = (2p 2 ) = (p 1 ) + (d 1 ) ,
                        and, using eqn (8.8), we obtain
                                                  1
                                              2          2          2      2
                                          (p 2 ) =   (f N ) + f N (f N ) − (p 1 )  .        (8.9)
                                                  2
                                                            √
                          Sinceweare assuming that p 1 <f N <  2p 1 , it follows that

                                                   1    p 2         1
                                                  √ <      <   1 + √ .
                                                    2   p 1          2
                          If this procedure is repeated iteratively in the case where the value of f is assumed
                        to be effectively constant, the general step involves
                                                   1
                                               2          2           2      2
                                          (p n+1 ) =  (f N ) + f N (f N ) − (p n )         (8.10)
                                                   2
                        and

                                                                       2
                                                                2
                                                d n = f N +  (f N ) − (p n ) .             (8.11)
                          The values of p n then converge to a value p satisfying, according to eqn (8.10),
                                                                        

                                                   2
                                                                        2
                                               p       1             p
                                                    =  2  1 +  1 −         .               (8.12)
                                               f N                 f N  
                          It is straightforward to show that the solution of this equation is
                                                           √
                                                             3
                                                       p =    f N ,                        (8.13)
                                                            2
                        and, furthermore, the corresponding value of d, according to eqn (8.11), is
                                                            3
                                                        d = f N .                          (8.14)
                                                            2
                          These are characteristic measurements for an equilateral triangle of circumradius f N .
                        Thus this algorithm tends to produce approximately equilateral triangles of the target
                        circumradius after a few iterations.
                          Examples of triangulations in domains of various shapes using Voronoi point-
                        insertion are shown in Figs 8.15–8.18.
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