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

                N
                                              K2  = 0 - 0.003(0 - 0 + 0) - 0.006(-2,0)
                           - 0.0120
                                ~
     l91.6378
         8uim=-( O~003)  31~02 4 44 +0.012)
                     - -6.4460deg
   The stick force gradient is given by
dFs = 2K'V(Kz + Ch8.,8r)
dV
                                    = 2(-3.1002) * 44.44(0.0120 - 0.003[-6.4460l)
                 - -8.6351Ns/m
The stick force gradient is negative because the aircraft is unstable stick-free.
STATIC STABILITY AND CONTROL
Example 3.5
247
      Estimate the elevator effectiveness for a horizontal tail fitted with an elevator and
having the following characteristics: aspect ratio - 3.78, Ac/2 = 45.5 deg, Ce]C =
0.224,A - 0.586, r7t - 0.141, and rlo - 0.61. The airfoil section is a NACA
65A006. Assume that the Reynolds number Ri - 6.1 x l06.
    Solution.   We have to determine the elevator effectiveness parameter r.
 For NACA 65A006 airfoil section, we obtain tan(~rEl2)=0.0616 using
Fig. 3.13c.
   From Figs. 3.13a and 3.13b, we obtain ao/(ao)ttieory = 0.887 and (ao)cheory =
6.58 so that ao - 5.8365/rad. Using Fig. 3.35, we obtain (a8)a/(a8)C, = 1.083.
Using the data of Fig. 3.36, we obtain Kb - 0.55. From Fig. 3.37b, we obtain
Cl8/(Cl8)trieory = 0.817 and, from Fig. 3.37a, (CLa)aleory = 3.77]rad so that C/8 =
3.08lrad. Then, subsrituting in Eq. (3.98), we obtain r - 0.3143.
                                             Example 3.6
      Estimate the hinge-moment coeffcients for a horizontal tail fitted with an eleva-
tor and having the following data: A = 4.0, Ac/4 = 45 deg, AHL - 45 deg, ct,/c f  =
0.09, cf /c = 0.16, Reynolds number Ri  - 6.1 x l06, airfoil section NACA
65A006, rk - 0.25, r7o = 0.65, and tcl2cf  - 0.08.
So/ution.    The first step is to calculate the balance ratio as given by
BR -.
-. 0.0412
The next step is to evaluate various parameters appearing in Eq. (3.121).
  From Fig. 3.46, we note that Chabal/C;ra, = 1. Using Fig. 3.13c, we obtain
4TE -. 7 deg for the given 65A series airfoil with 6% thickness ratio so that
tan 4-rE]2 = 0.0616. Then, from Fig. 3.13a, we obtain ao/(ao)meofy - 0.887 and,
from Fig. 3.13b, (ao)theory = 6.58/rad so that ao = 5.8365/rad. With this and
using cflc = 0.16 from Fig. 3.47a, we obtain, C;a/(Chtr)theory = 0.65. Using
Fig. 3.47b for t]c -. 0.06, we get (Cha)theory = -0.39 so that cLa  = -0.2535lrad
and Chaw - -0.2535]rad.
     We have c,lc f = 0.09 (streamwise) or cZ]c; = 0.09]cos 45 deg = 0.1272 (nor-
mal to quarter chordline). Similarly, with c f ]c = 0.16, we get c}-lc' = 0.226.
    We have i7i = 0.15 and r7o = 0.65. Using these values and Fig. 3.48b, we get
 (Kct)",  =  1.45 and (Ka)0o  = 2.4. Then, using Eq. (3.125), we obtain Ka  - 0.6188.
Furthermore, using Fig. 3.48c, we get B2 - 0.92.
    From Fig. 3.48a, for A = 4.0, we get ACha,/aoB2Ka cos Ac/4 ~ 0.012. With
all these values, we get ACha - 0.0296/rad. Substitution in Eq. (3.121) gives
Chtr = -0.1028/rad or -0.0018/deg.
    Now let us calculate Ch8.e. From F,g. 3.49, we firicl (Ch8.e)bal/CL8.e = 1. Ffom
 Fig. 3.50a, using c f /c = 0.16 and aol(ao)rriemy = 0.887, we find cL8.e/cch8.e)theory
=0.90. Using t/c = 0.06 and cf /c = 0.16 from Fig. 3.50b, we find (Cha.e)a}eory
= -0.83 so that C28.e = -0.747lrad and Ch8.ew -. -0.747.
　
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