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时间:2010-06-01 00:51来源:蓝天飞行翻译 作者:admin
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at - (aw - iw + /r - €) + Acr/.r                             (3.192)
d~Yt   ]
d~ ='  (1-:1 )+2;_
Stick-fixed maneuverpoint.   We have
Cm - CLXa + Cmac*rn + Cmf - a,a;Vi 77t
(3.193)
(3.194)
              = CLXa + Cmac.rn + Cm f  - at[aw - iw + it - € + Aat.t] Vi r7,        (3.195)
Differentiating with respect to CL, we get
                                                                                    ar fVi r/,
~C~),=a+(tlC~_)f--..V,,7,(1-:1: : 7/ (3.196)
                 2)ul
The stick-fixed maneuver point Nm is that position of the center of gravity where
(dCm]dCL)m - O and is given by
Nm =  a.- (:C/  ),+V 7  [(1- ~) + 2p~,]            (3.197)
                                            Hm = Nm - Xc8
For a stable airplane, Hm is positive.
(3.198)
 L' -. L + AL
n=1+ \CC,
252           PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
      Elevator required per g.     Because ofincreased stability level during the ma-
neu'ver, the magnitude of the elevator deflection required for trim also increases.
This increase in elevator defiection is usually expressed by the parameter (d8e/dn),
which is the elevator deflection per unit increase in load factor above unity.
From Eq. (3.101), we have
                                                                                  d8e                         ( dC, )m
                  ~ig ), =-  .                 (3.199)
With
      xcg - Nm
  Cm8
(3.200)
 ddC~), = dd8 ddr;                    (3.201)
and
                           dn .  1
           (3.202)
 .                            d~L = CL
we obtain
                                                      d8c         . CL(Xcg - Nm)
                               -            (3.203)
                              d7z -- --    CmS
Usually, for a stable airplane, Xc8  <  Nm and Cm8  < 0 so that we have d8e]dn  < 0-
 
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