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Nm.cum =  ac.w - (:C/- f +-
                                                                                              a ,,
[(1-: )+2p-, (1+:,)]  (3.233)
d8e      CL(Xc8 - Nm.rum)
d7z == --      Cm8
(3.234)
    \/2
 R = g~~_j
 ,~7
sin ~:  v
           n
sl : g/~-/
         V
JV~-
-6{-
           I
256           PERFORMANCE, STABILiTY, DYNAMICS, AND CONTROL
N:.tum = xac,w - (:C/  )f +
        x (1 - ,CC,,,.)
(d/.),um=G,(Vg),7tSeCeCc.,&~.),,._ (3.236)
                                          Example 3.7
   A trainer aircraft is initially in a steady level flight at an altitude of 5000 m
(cr = 0.6) at a forward speed'9f 100-0 m/s. The aircraft then climbs to 6000 m at
which it enters into a dive and recovers approximately in a semicircular path so
that, at the bottom of the pull- out, it is once again at 5000 m altitude.
      Determine  1) the elevator setting above that required for initiallevel fligh~/an2d
2) the stick force per g, based on the following data: wing loading = isoo r
xac.w = 0.24, xc8 = 0.25, Cm f  = 0.15CL, aw = 0.1, ar -0.08, t - 0-5,lr - 8 m,
VI :0.6, e - 0.35a, Ctia -  - 0.003]deg,  Ch8.e -  - 0.006/deg,  r7r = 0.9, Gi -
1.0 rad/m, Se - 2.0 m2, and Ce = 0.6 m.
So/ution.    From the information grven, we find that R - 1000 m. Then,
           2W
                       CL = p\/-2S
                                                                    2 *1500
                                              = ~225~ 0.6*1002
      - 0.4
    W
ALi - pgSr
  1500
=  ~225~0 6~9.81* 8
                                     -. 25.4842
                    Cm8 = -a,Virlrr
                                                 -. -0.08 * 0.6 * 0.9 * 0.5
                        - -0.0216
Nv =  ac.w - (-,.,)f -+ V 7  [(1- :  )+-,,,]
      -. 0.24 - 0.15 +   0.08 *0.6*0.9
                                                                    Ot)9)  [1- 0.35+ 20~25 54~8432 ]
    - 0.4193
　
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