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                                 ~
            &~ )-=  :g ),x- C.. C,. (1-: )
                                     = 0.05+(   OO~27) (~~) (1 _ 0.5)
                             - 0.1175
                                                                          160 *1000
                       Vrrim:   3 00
                                                                    = 44.44 m/s
                    ~
                        Ku = -(-5.0616) (~0;2~)1450.0.1175
                                          -. 191.6378
            ~
                                    K' = (1/2)*1.225(-5.0616)   .
                             - -3.1002
                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
　
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