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x = (Xac)e - Xcg,le
The exposed taper ratio A,t  - 0.6]5.6  -. 0.1071. The data on Xac givenin Fig.   3.19
is for A, - O and 0.2. Using these two sources of data, we find that the mean value
of (Xac/Cr)e = 0.575 for Ae -. 0.1071. Then, we get (Xoc)e - 0.575 *5.6 -- 3.22.
We have x  = (Xac)e - (Xc8)lc = 3.22 - 2.0 -  1.22. The mean aerodynamic chord
is given by   .
                          -= (23 )(:A~+A )
Substituting Ae  = 0.1071 and Cre = 5.6 m, we get Ce  - 4.1058 m and g  : 0.2971.
Similarly, with theoretical root chord Cr - 6,025 and theoretical taper ratio A, =
0.09958, we get the theoretical mean aerodynamic chord c - 4.0529 m.
       The subso:ruc lift-curve slope of a straight tapered wing is given by the following
expression:
CLat -
27r A
2+
where 8  = ^~~ and the midchord sweep angle Ad2 is given by the following
expression:
tfulAc/2=tariA.E-( b )
With ALE  - 45 deg, Cre = 5.6 m,cr  - 0.6 m forthe exposed wing, we get tan Ac/2  =
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
　
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