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STATIC STABILITY AND CONTROL
239
Fig. 3.59'   Variation offuselage cross-sectional area of generic tailless airplane.
     The variations of fuselage cross-sectional area SB and its first derivative dSBldx
are shown in Fig. 3.59. The leading dimensional characteristics of this vehicle are
as follows.
    Wing: Span b = 17.3228 m, leading sweep ALE = 45 deg, dihedral angle F =
3.5 deg, theoretical area S = 106.0114 m2, exposed area Sexp = 73.6282 m2, theo-
retical aspect ratio A = 2.8306, exposed aspec7ratio Ae - 2.6893, theoretical taper
ratio A = 0.1427, exposed taper ratio At = 0.1705,exposed root chord Cre = 8.94 m,
theoretical root chord Cr - T0:6766 m, Cr =  1.5236 m, sectional (two-dimensional)
lift-curve slope ao = 0.0877/deg, mean aerodynamic chord c - 6.8072 m, vertical
distance between the center of gravity and the quarter chordline of the wing root
chord zu, - 1.27 m, and the airfoil section geometry parameter Ay = 2.5.
      Fusektge: Overalllengthlf  = 23.2410 m,length of the nose In = 8.7122 m, dis-
tance between the fuselage leading edge and the leading edge of the exposed wing
root chord IN - 14.1275 m, fuselage height at a distance of lf l4 from the lead-
ing edge is 2.7838 m, fuselage height at a distance of 31f /4 from leading edge is
3.048 m, maximum width b f,   x = 3.2715 m, maximum height d f.max = 3.048 m,
maximum cross-sectional area SB,max = 8.3193 m2, and projected side area SB.S =
60,75 II12.
   Itertical Tadr: Leading-edge sweep ALE.,=45 deg, theoretical area Sy -
 20.2426 m2, root chord (exposed) = 4.5826 m,root chord (theoretical) -. 5.6977 m,
 span (theoretical) bv - 5.4864 m, tip chord c, = 1.6815 m, taper ratio (theoretical)
240          PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
= 0.2951, theoretical aspect ratio Av = 1.4869, horizontal distance between the
center ofgravity and the vertical tail aerodynamic center/y : 7.7561 m, vertical dis-
tance between the fuselage centerline and the aerodynamic center zu - 3.8290 m,
and the vertical tail airfoil section geometry parameter Ay = 2.50.
    For a center of gravity location at.15.9334 m from the fuselage leading edge,
calculate the wing~body lift-curve slope and pitching-moment-curve slope for
subsonic/supersonic speeds.
   So/ution.   The given configuration has alow-aspect ratio wing and the wing-
span-to-body-diameter/width ratio b/b f.max is 5.2951, which is small. Therefore,
it is preferable to use the combined wing-body approach.
   From the given data, we obtain the following: Ac/4 =36.3704 deg, Ac/2=
25.3106 deg, Ae = 2.6893, k = 0.8, Se,p]S = 0.6945, k2 - ki  - 0.89 (see Fig. 3.6),
and Ayi - 3.5355 [Eq. (3.22)l.
   CalcrilotioFi of CLa,WB. To begin with, let us calculate the lift-curve slope of
the fuselage nose. Using Eq. (3.26): we obtain CLa.N = 0.0025/deg for subsonic
Mach numbers. For supersonic Mach numbers, we use the data given in Fig. 3.lOa
for ogive-cylinders. The calculated values of CLa.N (referenced to wing area S)
were curve fitted to obtain the following expression:
CLa.N  = 0.0033 + 0.00035 Mldeg
   The lift-curve slope of the exposed wing was calculated using Eq. (3.16) for
subsonic speeds based on exposed aspect ratio Ae. This equation gives an analytical
expression for CLa,e in terms of Mach number.
   The lift-curve slope of the exposed wing at supersonic speeds was calculated
using the data of Fig. 3.14 and applying the correction obtained from Fig. 3.15a
using AYi = Ay/cos ALE = 3.536. The calculated values of Ct.cr.e were curve
fitted to obtain the following expression for 1.2 < M < 5.0,
CLa,e = 0.0038 M2 _ 0.03088 M + 0.0791ldeg
Knowing CLa.N and CLce.e at subsonic and supersonic speeds, we can now find
KN using Eq. (3.25).
   Using Eq. (3.27), we obtain KWcB) = 1.17, which is applicable for both sub-
sonic and supersonic Mach numbers. We get KBcW) = 0.285 using Eq. (3.28) for
subsonic speeds. For supersonic speeds, we have to use the data of Fig. 3.18b for
KBtW) because pAe(l + A,t)[(tan A LE/p) +  1]  >  4. These values of KB(W) were
curve fitted to obtain the following expression applicable for 1.2 S M < 4.0:
KBcW) = 0.0063 M2 _ 0.0645 M + 0.2362
With these values, we are now in a position to calculate the wing-body lift coeffi-
cient using the following equation:
          Sexp
CLa.WB = [KN + KW(B) + KBcW)lC      S
　
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