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We have Ac/4 = 36.3704 deg and A = 2.6893. Here, xa is the horizontal distance
between the wing aerodynamic center ofthe exposed wing and the center of gravity
in ter.ms of mean aerodynar/uic chord and xa > 0 if the center of gravity is aft of
the aerodynamic center. From the calculations done for Example 3.2 at M - 0.7,
we have
                       3367 M3 _ 0.4810 M2+0.2214 M +0.4918       (0 < M < 0.80)
   -,,,) w = 0.33,
For M = 0.7, we get
                         -                             (-,,,) w = 0.5266
The center of gravity is located 15,9334 m from the leading edge of the fuselage.
From Fig. 3.58, we note that the leading edge of the exposed wing is located
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   14.1275 m from the leading edge of the fuselage. We have Cre - 8.94 m and c -
  6.8072 m. Then,
                                  xcg - 15.9334 -14.1275
                              Crt            8 94
                                    - 0.2020
                    '     x-, = (-., ) - (-,,,.),;
           Cre
                                            -. 0.2020 - 0.5266           '
                             - -0.3246
             xa
                                        -.. = (-,,.) (-: )
                       c'
                                - -0.3:
      _ }.3246(688~~;2)
                             - -0.4263
                                          ~
 Substitution of all these values gives
                                (Cnp)A.W _ 0.0608]rad
                           C7- :
                                          = O.OOllldeg
 The combined wing contribution because of dihedral and sweep is then given by
                       (CnP)wing = (-O.OOOICL +O.OO11C2)/deg
 The fuselage contribution is given by
                        (Cnp)B,W, = [-KNKR, (SS  ) (b)] 7deg
　
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