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     Assuming thatit is a midwing configuration, determine Ctp, Cnp, CLr, and Cnr
using strip theory for ct = 5 deg and M = 0.20. Compare your'results with those
obtained using Datcom methods.
    So,ution.   The strip theory gives
                                 (Ctp)W = -g(ao + CDJ)
We have ao - 0.1 * 57.3 -. 5.73/rad and
                             CD,I = CDO,I + CDa,ICL
                                                          -. 0.022 + 0.001* 5 - 0.027
Substituting, we get the strip theory value of (Clp)W = -0.9595/rad.
      Now let us determine the valr                                    ie Datcom   method. We have
                   (qp,W = (  - -)::_:p(,jS(,,,,:,:,)/rad
We a\ssumek = l.Oand p = Jj;lj~ = 0.9797.FromFig.  4.25,(pcip/k)cL =O
= -0.43 for ALE - 0, A,--l.0, and A -. 6.0. Because F = O, we find [(qP)r]
(Clp)r=01 = 1.0. With these values, we get (Clp)W = -0.430/rad. Therefore,
the strip theory result differs from the Datcom value considerably, by as much as
lOOgro.
Now let us estimate (Cnp)W. The strip theory gives
                             (Cnp)W = -g(CL - CDa)/rad
Substituting the required values, we get (Cnp)W - -0.0831-Urad.
   Datcoml gives
  (Cnp)W=Clptana,(K-l)+K(CC c~= CLlrad
                                                                                                                                    L = O,M
                                       A +4 cos r                     +0.5(AB + cos Ac/4)ta
     = AB+4 AAt,.,.t,)[A2:[ _ ~At,,]
 ( CC,  ),, = O,M                                                                                       + 0.5(A  +  cos Ac/4)tanl
     x (C~ c =o/rad
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PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
(CC,)c,=o=_[A+6  cA+osAct4)(:-, +   ]
                                                                  A +4 cos Ac/4
               (CLct)e       (CLcr)e
　
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