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K=~:} a"==n_Ae aw2=~A
    1.1CIA
           e= R u     .+ 1R)7rA
R = ai}~ + a217j + a3AI + a4
ai - 0.0004      a2 ~ -0.0080     a3 - 0.0501      a4 - 0.8642
We have
i.    AA
Ai -
  'I      cos ALE
     ao
CLr,e = ~+ _
Substituting ao - 0.1 *57.3 - 5.73/rad and A - 6, we get CUr,e = 0.0767/deg
or 4.3943/rad. With this, CL -. CIAr.ea - 0.3834. Further using Ac/4 = 0 and
A = 1.0, we get R - 0.9632, e - 0.9812, and K  - 0.9942.
   We have B =                     ~- = 1- Substituting, we get (CnplCL)C =O =
(Crip/CL)CL~.M = -0.6 and- (Cnp)W~ -0.2285. Therefore, the stnip theory
prediction of -0.08317/rad is in error by as much as 63%.
     Now let us calculate (Clr)W and (Cnr)W. The strip theory gives
   CL 0.1*-5~0.16
(Clr)W= 3 = 3    1667lrad
    ~CDJ 0.0011*5+0.022  (
(Cr)W= 3 =-      -0.009/rad
                            3
Nowlet us estimate (Clr)W and (Cnr)W using Datcod methods.
(Clr)W = c,(C/,).,_ .M + (A2  )~7rad
                             Num (
(g  C~=O,M = D<. (g  ),.=o.v=o
EQUATlONS OF MOTION AND ESTIMATION OF STABILITY DERiVATIVES 435
           B2)   'AB+2cosAc/4 :an2Ac/4)
 Num=1+2B~AABl2B /,)+(7~  :,,)(~ ')
                :cosAc/4)  ,AB+4~osA 4    8  )
             A+2cosAc/,  tan2Ac/4'
    Den:l+  -.,,,)(- -)
               A+4cosA.c/,    8  /
We have r : O, ALE - Ac]4 = O, A - 6, and M -. 0.15. Substituting and
simplifying, we get (Clr)W - 0.1051/rad, Thus, the str:ip theory prediction of
0.1667 differs by as much as 50o/o compared to the Datconl result.
   We have
                         (Cnr)W = (C/: )cl- + (g  )CDO
Assuaung g=0 from Fig. 4.29, we get (Cnr/CZ) = -0.02 and (Cnr/CD) =
-0.20. We have CL = 0.3834 and CDO -. 0.022. Substituting, we get (Cnr)W =
-0.0073/rad. In this case, the strip theory result of -0.009/rad differs from the
Datcom  result by about 25%.
                                         Example 4.14
   Estimate the low-speed vertical tail contributions (Clp)V, (Cnp)V, (Clr)V, and
(Cnr)V at an angle of attack of 5 deg using the following data: lift-curve slope
au  - 0.07/deg, leading-edge sweep ALE  - 0, vertical tail length /u =' 9.0 m,
wing mean aerodynamic chord c - 3.0 m, wing span b =  15 m, zy = 0.9 m, ratio
of vertical tail area to wing area Sy]S - 0.20, wing aspect ratio A = 6, and wing
taper ratio A, = 0.5.
Solution.   We have
   (l+?dp T7v=0.724+~06S3S~+2f4 +0.009A
                           Cyp.v = -kay  1 + ?dp)-7,SS
                                                      z N zU COS CL - /v sin a
       (qp)V={2(b)( b )ICyp,v
   (Cnp)V=-(;)(l,,osa+zusm,)( b )Cyp.v/rad
               (C,r)V  =  b2(lv cos a + zl sinu)(ZU cos a - ly sin ty)Cyp.v
                          (Cnr)V = b2 (lv cos a + zu SiIICL,)?Cyp.V
　
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