Page 154 - Wind Energy Handbook
P. 154

128                        AERODYNAMICS OF HORIZONTAL-AXIS WIND TURBINES

                                    ð
                                     1
                                       P n (v)P k dv ¼ 0  n 6¼ k              (3:152)
                                      1
          For n ¼ 0 the Legendre polynomial of the second kind is


                                              1   1 þ v
                                      Q 0 (v) ¼ ln                            (3:153)
                                              2   1   v
          For n . 0 the Legendre polynomials of the second kind Q n (v) can be obtained from
          the polynomials of the first kind. For n ¼ 1to4

                                Q 1 (v) ¼ (P 1 (v)Q 0 (v)Þ  1

                                                    3
                                Q 2 (v) ¼ (P 2 (v)Q 0 (v)Þ  v                 (3:154)
                                                    2
                                                    5  2  2
                                Q 3 (v) ¼ (P 3 (v)Q 0 (v)Þ  v þ
                                                    2     3
                                                    35  3  55
                                Q 4 (v) ¼ (P 4 (v)Q 0 (v)Þ   v þ  v
                                                     8     24

          The solutions for Equation (3.149b), with m ¼ 0, are the same as for (3.149a) but
          with imaginary arguments, i.e.,

                                         Ö 2 (ç) ¼ P n (iç)                  (3:155a)

          and

                                         Ö 2 (ç) ¼ Q n (iç)                  (3:155b)

          where

                                               1
                                  Q 0 (iç) ¼ i tan (ç)  for ç , 1

          and

                                          ð
                                Q 0 (iç) ¼ i    tan  1  ç  for ç . 1
                                          2

          For non-zero values of the integer m the solutions of Equation (3.149a) become

                                                    d m
                                    m
                                               2 m=2
                                  P (v) ¼ (1   v )      P n (v)               (3:156)
                                    n                 m
                                                    dv
                        m
          If m . n then P (v) ¼ 0.
                        n
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