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At subso:tuc speeds, the wing contribution is evaluated using Eq. (4.594). The
vertical tail contribution is evaluated using Eq. (4.598). At supersonic speeds,
Datcom/ data are used. The details of this method are not given in the text.
     The calculated values are presented in Fig. 4.35c.
426              PERFORMAhICE, STABILrfY, DYNAMICS, AND CONTROL
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Fig.436    C,r,CU,and C~ for the aircraft ofExample 4.10.
      For supersonic speeds, no general method of calculation is available. However,
a crude e~timate of the vertical tailis obtained using Eq. (4.618) with a,, evaluated
at supersonic speeds.
     The calculated values of Cyr are presented in Fig. 4.36b.
Cnr: Wehave
                                       C,zr = (Cnr)W + (Cnr)V
The wing contribution was estimated using Datcom data7 with CDO -. 0.0025.
The vertical tail contribution at subsonic speeds is estimated using the Eq. (4.620).
 No general method is available for supersonic speeds. However, using Eq.  (4.620),
 crude estimates for vertical tailcontribution arc obtained at supersonic speeds. The
 calculated values are presented in Fig. 4.36c.
   Lateral-directional acceleration derivatives.  The derivatives11C,~8;te~zBalauned
  CnB_were obtzuned using Eqs. (4.632), (4.634), and (4.635). The calc
 are plotted in Fig. 4.37.
EQUATIONS OF MOTION AND ESTIMATION OF STABILITY DERIVATIVES 427
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Fig. 4.37    C p, CtB, and C,,p for the aircraft ofExample 4.10.
                                         Example 4.11
    An aircraft with a wing loading of 2000 N/m2 is in a steady level fiight with a
velocity ofl00 m/s,The drag polar of the aircraftis given by CD = 0.025 + 0.05CZ.
Assuming p = 1*25 kg/m3, CL = 0.08a deg, /r = 3c, Sr/S = 0.35, e = 0.35tx deg,
ar = O.O(xyr deg, and r7r = 0,9, determine the stability derivatives C;ca, Cxu, Cza,
Czu, Czd, Cmq, and Cm Z.
Solution. Wehave
Then
                                    2 * 2000
C~ = p(.U). = ~25~00
-- 0.35 * 3.0 -. 1.05
        CD = CDO +kC~. = 0.025 + 0.05 * 0.322 = 0.03012
CDa = 2kCLCLtr = 2 * 0.05 * 0.32(0.08 * 57.3) = 0.1467/rad
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428              PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
      Cxu = -2CD - CDu - -2*0.03012 - -0.06024
           Cxcr - CL - CDcr = 0.32 - 0.1467 : 0.1733/rad
                  Cz" - -2CL - CLu - -2 *0.32 - -0.64
C. za -- -CLa - CD = -0.08 *57.3 - 0.03012 - -4.614/rad
           Cz" - -CLcr
      =-2at,,,.(-,.) .
               N
                                 -. -2(0.06 * 57.3)1.05 * 0.9 * 0.3
               ~
               - -1.9494/rad
               ~
　
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