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where S2 :  V/R and R is the radius oftum. The component C2y is equivalent to a
pitch rate q about the body y axis. This apparent pitching motion gives rise to an
induced velocity qlr at the horizontal tail (see Fig. 3.65) resulting in an increase
in the horizontal tail angle of attack.
                                                                                            q Lt
                            Act/.r - -V                          (3.225)
I
      S2 sin ~lr
= -V                         (3.226)
il
Fig. 3.64    Airplane in a coordinated turn.
D
STATIC STABILITY AND CONTROL
Aa
255
Fig. 3.65     Induced angle of attack at horizontal tail due to pitch rate in a coordinated
turn.
From Chapter 2, we have
Then,
                     t, ,
Aar'.r = tV-g (. - 1)
(3.227)
(3.228)
(3.229)
(3.230)
Proceeding in a similar fashion as we did before for the case of a pull-up in a
vertical plane, we obtain
darr.r
dC  = 2/, (1+')
(3.231)
&g ), = a+&~:)f -  .V.  [(l-~)+2p, (1+')]  (3.232)
                                                ar Vi o,
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
　
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