曝光臺 注意防騙 網曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

where KN and KRI are to be obtained from data given in Figs. 3.73 and 3.74.
Using data given in Example 3,2, we get hj : 2.7838 m and h2 - 3.048 m
so that J~ ff/ = 0.9556. Furthermore, we have h - d f.max = 3.048 m so that
h/b f.max -- 3.048l3.2715 = 0.9317, xm~lf  = 15.9334l23.2410 - 0.6856, and
ClSB.S = 23.24102l60.75 -. 8.8913. With these values, we get KN - 0.0016.
        To find K Rt, we need.to find the Reynolds number based on the fuselagelength lf ,
 wluch is 23.2410 m. At an altitude of 8500 m, the kinematic viscosit)r is 3.05 x lO-s
m2/s, and the speed of sound is 305.0 nVs (see Appendix). This gives us a flight
velocity of 213.5 m/s and a Reynolds number (Vlf /v) of 16.27 x l07. Then we
obtain K RI  = 2.06 from Fig. 3.74. We have SB.S  = 60.75 I112, S =  106.0114 1112,
and b - 17.3228 m. Substituting these values in the above equation, we get
                            (Cnp)B(W) = -0.0025]deg
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290           PERFORMANCE, STABILI-fY, DYNAMICS, AND CONTROL
    Now we will evaluate the vertical tail contribution, which is given by
                                                 (Cnp.V)nx = kap (1 + ~p )  7 , V2
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
　
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