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E
1.5     2     2.5     3     3.5     4
      Mach Number
Fig.3.61   Variation ofx"e and C,na with Mach number.
243
  From Figs. 3.60 and 3.61 we observe that present calculations are in quite
good agreement with APAS results except around M - 1, where both methods are
approximate.
                                     Example 3.3
    An aircraft has the following data: xcg  - 0.3, Xac - 0.24, CL.w  = O.lO((Xn +
2.5), CL.max = 1.2, C   c.w = 0.06, e = 0.3a, Cnf = 0.05 + O.lCL, ar = 0.08/deg,
th = 0.9, Vi  = 0.6, rw = O,/r = 2deg, Cha  = -0.002]deg,Ch&  = -0.003/deg,
and r - 0.20.
     Determine l) the angle of attack in steady level flight if the elevator is locked in
neutral position, 2) the permissible most forward position of the center of gravity
if the maximum up elevator defiection is limited to 30 deg, and 3) stick-free neutral
point and stick-free margin.
where
Cm = Cmo + (:C~ ),,x C~
Cmo = Cm*ac + Cmjo - ai V l r7t(cL*n.oL - iw + /t)
- 0.06 + 0.05 - 0.08 * 0.6 * 0.9 * (-2.5 - 0+ 2)
- 0.1316
244            PERFORMANCE, STABIUTY, DYNAMICS, AND CONTROL
and ,
             &C/ ),x =  c8 -  a,+ &C/ )f - -.. (i- L) v,,i,
                                    = 0.30 - 0.24 + 0.1 - (0.08l0.10)(1 - 0.3)0.6 * 0.9
　
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