曝光臺 注意防騙 網(wǎng)曝天貓店富美金盛家居專營店坑蒙拐騙欺詐消費者

                                 Cf ='( o9~o R~258
        0.455
                N
                                                  = ( o9~o~0 66 x~05~2 58
                                           -. 0.004418
                N
                      Df = 2lp V2SCf
                                              = ~: *1.225 * 502 * 45 * 0.004418
                                  - 304.4829 N
Example 1.5
    For a double wedge airfoil of 2 m chord and 4% thickness ratio, determine the
lift and wave-drag coefficients per unit span at a Mach number of 2.0 and a -
5 deg.
Cf = :k328
  1.328
    = .~~j R~_T~

:!
 $
 i
 li
a4
 . I:g
 ~
 li
 .~
:a
 ~
64                 PERFORMANCE, STABILtTY, DYNAMICS, AND CONTROL
             5
a = 5 deg = 575 3  = 0.08726 rad
      4a
           Cl = ./ff/-l
                            4 * 0.08726
    = ~--I-
            - 0.2015
          4
Cdw = j~--f [-2+(-)2]
     - 0.02128
                                                        Example 1.6
  For a two-dimensional rectangular wing having a NACA 23012 airfoil, the
critical Mach number is 0.672 and the lift-curve slope is 0.104/deg. What should
be the leading-edge sweep angle to have a critical Mach number of 1.5? What is
the lift-curve slope of this swept wing?
So/ution. Wehave
M,r,s =  M A
cosA -- _Mcr
  =M
 0.672
      = ~5
        - 0.448
A - 63.3845 deg
The lift-curve slope is. given by
                                                    aos - a cos A
                                                                              - 0.104 * 0.448
                                       - 0.0466/deg
1:13 Summarlv
      In ttus chapter, we reviewed the basic principles of aerodynamics applicable to
　
中國航空網(wǎng) m.k6050.com
航空翻譯 www.aviation.cn
本文鏈接地址：動力機械和機身手冊1(39)