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To e\raluate ay, we need to evaluate the effective aspect ratio.
        Au(r
       Av.etr= A A,[l+KH(AA  -1)]
From the data given in Example 3.2, we have Au = 1.4869, ,\u = 0.2951, bv  -
5.4864 m, and 2ri  - 3.048 m so that by]2ri = 1.8. Here, 2ri is the aver-
age fuselage depth in the region of the vertical tail. Using these data, we obtain
AvcB)/Av = 1.63 from Fig. 3.77. Because this vehicle does not have a horizontal
tail, Av.HB/Av.B = 1-0 so that the effective aspect ratio Au.etr - 1.63 * 1.4869 -
2.4236. With this, we obtain k  - 0.76 from Fig. 3.75. Using Eq. (3.19), we obtain
Ac/2 = 26.252 deg. Using Eq. (3.16), we obtain ay = 2.8113/rad or 0.0491/deg.
    Then, we need to evaluate the combined sidewash and dynamic pressure ratio
term given by
(l+?dp),7,=0.724+3.06 ,+S~SA~+0.4~f +0.0009A
We have Ac/4 = 36.3704 deg, zr., =  1.27 m, d f.max  - 3.048 m, A  - 2.8306, and
SU/S = 0.1909. Substitution gives
                                  (1+ ;p) 7, = 1.2012
We have lu - 7.7561 m. Using this value, we obtain  V2 - 0.08549. Then, substi-
tuting all the required values, we get
(Cnp.V)nx = 0.0038/deg
The total value of directional stability parameter at M - 0.7 and 8500 m altitude
is given by
                       (C,,p)r-ix = (Cnp)w + (Crrp)B(W) + (Cnp.-V)fix
                                          = -O.OOOICL + O.OOllCZ - 0.0025 + 0.0038
                                          - 0.0013 - O-OOOICL + O.OOllCZldeg
Thus, at M - 0.7, this aircraft has a positi\re directional stability.
    Now let us evaluate Cnp for M - 2.0 at 18,000 m altitude.ln the absence of a
reliable method to estimate the wing contribution at su )ersonic s )eeds and further
noting that wing contribution is generally small, we ignore the wing contribution
to directional stability at supersonic speeds.
      To evaluate the fuselage contribution (C,p)B(W), we need to evaluate K RI, which
depends on the Reynolds number. All the other parameters in that expression for
fuselage contribution remain the same as calculated for M - 0.7.At M  - 2.0 and
 altitude of18,000 m, we find that the Reynolds numberis equal to 11.20x 10J. From
STATIC STABILITY AND CONTROL
291
Fig. 3.74, we get K Rt: 1.97. Substituting in the above expression for fuselage
contribution, we get (Cnp)B(W) - -0.00242]deg.
   Now let us approximately evaluate vertical tail contribution. We have Av.eff =
2.4236, Au - 0.2951, and ALE., = 45 deg. From Fig. 3.14, we get t3 (CNcr)theory  =
3.85/rad, where [3 : -/_      f = 1.732. -fhis gives (CNcr)theory = 2.2228/rad_
Note that p here is not sideslip angle.
   Now we need to apply the sonic leading correction to the above value of CNa
using the data given in Fig. 3.15a. We have Ay = 2.5 so that AYi = Ay/cos A =
3.5360. With this, we obtain from Fig. 3.15a, CNa/(CNcr)theory = 0.825 so that
CNa  -  1.8338/rad or 0.0320/deg, and ay  = 0.032/deg. All the other values in the
expression for vertical tail contribution remain unchanged. Substituting, we obtain
                                 (C,zp.v)rix = 0.00249ldeg
Then,
                                 Cnp = (Cnp)B(W) + (Cnp.V)fix
                                                             - -0.00242 + 0.00249
                                  = 0.00007/deg
　
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