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60                 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
S(x)
CDW
x
a)
ffi
      . b)
Fig.1.61 Area-ruleconcept
interceptor aircraft.20,21 Wind-tunnel tests on the original design indicated that
the transonic drag was so high that the aircraft would not be able to f(y past Mach
1. This prediction was foun9'to be true when the prototype aircraft fiew in 1953.
The concept of area rule was then applied, and the fuselage of the F-102 was
redesigned. In the late 1954, the modified F-102 flew past Mach 1 successfully.
After this fiight validation, several aircraft were redesigned on the basis of the area
rule concept. These include the Convair F-106, the Convair'B-58, and the Vought
F8V. Some recent examples are the Air Force B-l and the Boeing 747 aircraft.
     It should be noted that with the application of the area rule concept, an aircraft
configuration can be derived to give a minimum wave drag only at one flight Mach
number. At any other flight Mach number, the benefits of area rule may be lost and
the wave drag may bc substantially higher.
Example 1-1
    A fiying wing with an area of 27.75 m2 has a NACA 2412 airfoil section. The
weight of the flying wing is 2270.663 kg and the aspect ratio is 6. For level fiight
at an altitude of 1500 m (a = 0.864) and a velocity of 160 km/h, determine the
angle of attack, induced-drag coefficient,lift-to-dra~ ratio, and the drag. Assume
e - 0.95.
REVIEW OF BASIC AERODYNAMIC PRINCIPLES                  61
    Solution.   For the NACA 2412 airfo11,3 we have ao - O.lOtUdeg and Ccto -
0,0060. We have V = 160 km/h:44.44 m/s.
     From Eq. (1.52), we have
a = ~+
Here, we have to multiply the value of ao by 57.3 in the denominator because we
are given ao/deg.
     Substituting ao - 0.104, A - 6, and e -.0.95, we get a = 0.079/deg.
Further,
C~= p-: Vft,S
                                                                        2270.663 * 9.81
                             = ~  225~0 864*44.442*27.75
                          - 0.768
Then,
              CL
                                 ct = -
                              a
and and
0.768
= O 079
= 9.7203 deg
CDr = kC~.
   =( ~ )CZ
 0.7682
                                                                 7t6 * 0.95
                                        - 0.03294
                                      CD = CDO + CDi
                                                                         -. 0.006 + 0.03294
　
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