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We note that
     1,
r,g = ~- 8mrb
 mrcg = ~~: 8mrh
(4.255)
(4.256)
Then
        F = m[(2e x (2e X (R, + Ro) + (ao)e +2C2e x  V;j + (dd,   )b X i:cg
+(D~b X (c7)l.b x lcb,)]
(4.257)
      Specialcases.    When the center of gravity of the vehicle coincides with the
 origin of the body-fxed coordinate system, rcg = O, then Eq. (4.257) assumes the
form
                     F = m[CZe x ?2e X (Re + Ro) + (ao)e + 2C2e x V:]         (4.258)
     For the low-altitude flight dynamic problems, IRol << IRel, then
                          F =m[C2e X ?2e x Re + (ao)e + 2?2e X Ve]              (4.259)
    Let us expand the vector products in Eq. (4.259) and derive the equations of
motion in the usual Cartesian form. Let A, denote the latitude of the ongin of the
 OxeYeZe axes system. Then,
                                                S2e = 2e<2c COS;t- - F-CeS2e sin A                               (4.260)
                                                                Re - -ke Re                                             (4.261)
                                                 1V : = le Vor + Je Voy + ke Voz                                (4.262)
                                             - ZXe + JYe + kze
                 -                      (4.263)
                                             (ao)e = IXe + ,lYe + kZe                              (4.264)
Using the vector cross product rule,
                                       a x b x c = b(a . c) - c(b  a)                         (4.265)
We have
                 g2e X ?2e X Re = S2e(?2e . Re) - Re(?2e . S2e)                   (4.266)
                                                                =  (leg2e cos A  - rCe C2e siri } )(SZeRe siri A)
　
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