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                                                                       NUMERICAL GRID GENERATION
                            Notethat s 0 isthephysicaldistancebetweenboundarynode1anditsneighbouring
                            interior node 2 in the ξ 2 direction. Similar definitions are introduced at the north
                            boundary ξ 2 = ξ 2max .
                               Now, let θ 0 be the angle of intersection between ξ 1 and ξ 2 grid lines at ξ 2 = 0.
                            Then,
                                                                 1
                                                                      2
                                                  ∇ ξ 1 ·∇ ξ 2 =|  a ||  a | cos θ 0 .         (8.31)
                            Using Equation 8.19, however, it follows that

                                           ∂ξ 1 ∂ξ 2  ∂ξ 1 ∂ξ 2
                             ∇ ξ 1 ·∇ ξ 2 =       +
                                           ∂x 1 ∂x 1  ∂x 2 ∂x 2
                                                             ξ 2 =0
                                         ⎡                                                    ⎤
                                                                                      0.5
                                                   2         2            2         2

                                                               0.5
                                              ∂ξ 1      ∂ξ 1         ∂ξ 2      ∂ξ 2
                                       =  ⎣         +                      +            cos θ 0  ⎦  ,
                                              ∂x 1      ∂x 2         ∂x 1      ∂x 2
                                                                                               ξ 2 =0
                                                                                               (8.32)
                                                      j
                            but, from the definitions of β introduced in Chapter 6,
                                                      i
                                              ∂ξ 1   1 ∂x 2        ∂ξ 2    1 ∂x 2
                                                  =        ,           =−        ,
                                              ∂x 1   J ∂ξ 2        ∂x 1     J ∂ξ 1
                                              ∂ξ 1     1 ∂x 1      ∂ξ 2   1 ∂x 1
                                                  =−        ,          =       .               (8.33)
                                              ∂x 2     J ∂ξ 2      ∂x 2   J ∂ξ 1
                            Substituting these definitions and using Equation 8.30, we can write Equation 8.32
                            as
                                                           ⎡                                 ⎤
                                                                                      0.5
                                                                         2         2

                                  ∂x 2 ∂x 2  ∂x 1 ∂x 1      ds 0    ∂x 2      ∂x 1
                              −          +              =  ⎣              +             cos θ 0  ⎦  .
                                  ∂ξ 1 ∂ξ 2  ∂ξ 1 ∂ξ 2      d ξ 2   ∂ξ 1      ∂ξ 1
                                                    ξ 2 =0
                                                                                              ξ 2 =0
                                                                                               (8.34)
                            Evaluation of ∂x 1 /∂ξ and ∂x 2 /∂ξ 2
                                                2
                            To make further progress, we must evaluate ∂x 1 /∂ξ 2 and ∂x 2 /∂ξ 2 at ξ 2 = 0. This
                            can be done using Equation 8.34. Thus,
                                                              ⎡                           ⎤ −1
                                                                                        0.5
                                                                            2         2


                                           ∂x 2 ∂x 2  ∂x 1 ∂x 1  ds 0  ∂x 2      ∂x 1
                               cos θ 0 =−         +           ⎣              +            ⎦     .
                                           ∂ξ 1 ∂ξ 2  ∂ξ 1 ∂ξ 2  d ξ 2  ∂ξ 1     ∂ξ 1
                                                                                            ξ 2 =0
                                                                                               (8.35)
                                                   $       2
                            Therefore, since sin θ 0 =  1 − cos θ 0 ,
                                                            ⎡                            ⎤ −1
                                                                                       0.5

                                                                           2         2

                                         ∂x 2 ∂x 1  ∂x 1 ∂x 2  ds 0   ∂x 2      ∂x 1
                                sin θ 0 =        −          ⎣               +            ⎦    .
                                         ∂ξ 2 ∂ξ 1  ∂ξ 2 ∂ξ 1  d ξ 2  ∂ξ 1      ∂ξ 1
                                                                                           ξ 2 =0
                                                                                               (8.36)
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