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   Solutron.   We have forsubst       ;peeds
                        (Cip,W,B, = :/s['(-::)/xcp KMAK f + (-,,)A]
                                  +F [CFp KM,+ Aj~p ] +(ACW)z,
We have A = 2.8396, Ac/2 - 25.3070, and A, - 0.170. Using the data given in
Fig. 3.96 for A:0 and A -.0.5 andinterpolating for A = 0.170, we get (CtplCL) Xty2
= -0.00214. For A]cos Ac/2 = 3.1311, from Fig. 3.97, we get (approximately)
K MA = 1.0.  For l}/b = (lf  - 0.5cr)/b = 22.4792/17.3228 - 1.2976.  From  Fig.
3.98, we obtain Kf =1.0, and, from Fig. 3.99, (Cip/CL)A = - 0.0010. From
310          PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Fig, 3.100, we obtain (approximately) C,p/ F =  - 0.00015 using interpolation for
the given value of}.. KMr as given by Fig. 3.101 depends on the Mach number. For
M  - 0.7, we obtain KMr = 1.07. With d - 3.048 m, A - 2.8396, zw :1.27 m,
and b - 17.3228 m from Eqs. (3.369) and (3.370) wc obtain
ACip = -0.0000261]deg2
~=-
                                   (ACtp)z. = 0.0009105
Substituting, we get
       (Ctp)W(B) = CL(-0.00214 - 0.0010) + F(-0.00015 * 1.07 - 0.0000261)
                 +0.0009105 -  -0.00314CL - 0.000186r + 0.0009105
With F : 3.5 deg, we get
                  '    (CLp)WtB) = -0.00314CL+0.000258
The vertical tail contribution is given by
(Ctp)v = -ka, (1+ ?ap) ,,, (SS )    zu cosa ~IU sina
                                        b
From the solution of Example 3.8, we have k -. 0.76,a   - 0.0491]deg, and
(i+ Lau/aPDov  -  1.2012. With zv - 3.8290 m,lu  : 7.7561 m, b -.  17.3228 m,
and Sy/S = 0.1909 and assuming a to be small, we get
      (Cip)v - -0.76*0.0491*1.20120*0.1909  3.8290-7.75
                    17.322~56 )
                               -  -0.0019 + 0.000068cr
where a is in degrees. Summing the wing and tail contributions, we have
                         Cip = cr(0.000068 - 0.00314CLu) - 0.0016
where ct is in degrees and CUr (per degree) was calculated as shown in Fig. 3.60.
     At supersoruc speeds. we have
(Clp)WcB) = -0.061CN (g73) [l+;t.(l+A,E)] (1+ A2  ) (    ~-)
 X[-, +(-, )4'3]+r(Crp+A~p)+(ACtp)z.
where CNa iS per radian, ALE is in radians, and f' is in degrees. We have ALE  -
45 deg  = 0.7853 rad and A  -  0.1705. From the solution of Example 3.2, we have,
for the the wing at M - 2.0, CNa -. 1.81/rad.
    Using Eq. (3.372), we obtain Cip/r = 5.4043 * 10-4 Ctp.
  Using Datcoml for the wing at 1.4 < M < 4.0 (see Chapter 4 for more
information on CiP), wc get
CIP = A(-0.0025 Mz +0.0283 M - 0.1154)
STATIC STABILITY AND CONTROL
311
With A = Ac - 2.8396 and M = 2.0, we get Ctp = -0.1954. With chis, we get
Cip] F = 1.0560 :k l0-4.
    From calculations performed earlier for M - 0.7, we have
                                AC'p, = -0.0000261/deg2
                   :r
                                   (ACtp)zw - 0.0009105
　
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