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0.5. Then, with Ae = 3.22 and assuming k -~ 0.8, we find (CU,)e - 2.898/rad.
With these calculations, we get (CIUl)e = 3.17099kad.
     Now let us evaluate the body contribution (CU}r)B. We have
                                (Cl4t)B = 2(Cta)B(l - -f )
                                                      (cLr)B  = (Cl.ot)B ( SV,,.)
                                     (CU.)B - 2(k2 - k,)(SV:,, )
so that
                    .     (C2rr)B = 2(k2 - ki) -.2*0.95 - 1.9/rad
With Xm = Xc8 = 5.0 m and lf = 10 m, we get (CLq)B = 1.9/rad. With these calcu-
lations, we obtain (Clq)TW = 3.2462/rad.
EQUATIONS OF MOTION AND ESTIMATION OF STABILITY DERIVATIVES 431
   We have
                         (Cmq)WB = [KWcB) + KB,W)] SS ( : )2(Cmq)e
       +(Cmq)BSS (l:)2/rad
The conrribution of the exposed                   e iS given by
                                           (Cmq):e.  [_qje/' . q,e M =02
We have
         tanAc/4=tanA,E-( 2b )
    ~N-
                                         B = \                S2 Ac/4
                                           Cl = A3 tan2 Ac/4
                                      3
                                       C2 = B
                                                  C3 = AB + 6 cos Ac/4
                                                         C4 = A + 6 cos Ac/4
                                                C5 = A +2cos Ac/4
Substituting, we get Ac/4 = 36.8698 deg, B N 1.0, ci  - 18.7798, C2 - 3, C3  -
 8.0097, 04 - 8.020, and 05 - 4.822. Furthermore,
(Cmq)e,M =0.2 = -0.7Cia cos Ac/4
A(0.5€ + 2g2)
+(24 )+g]
We assume Cia-.27rk-27r * 0.8-.5.02655/rad. With these values we get
(Cmq)e,M =O,Z = -1.2376/rad and then (Cmq)e = -1.2406/rad.
     The body contribution is given by
                                           )7 - VB1(
　
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