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 he = const curve is tangential to two equal-valued Ps curves. At this point, the air-
 craft follows a small segment BC of a constant energy height untilit finds another
 curve of equal value of Px, which it does at point C.In this process of tracking the
 segment~C, the aircraft is essentially in a dive and is accelerating to supersonic
speeds. On reaching point C, the aircraft resumes its climb in supersonic region
 along the path CD. At point D, another change occurs and the aircraft follows the
path DE to its final destination, represented by point E, during which it is in a
 decelerating climb. Even though such a flight path is an unusual one to rfly, it is still
 the optimal compared to any other possible flight path connecting energy states A
 and 9:

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104            PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Table 2.1    Climb performance ofsome :urcraft
Aircraft                 Type         Sea level, R]Cmu.    Minimum time to climb
Mirage III
Mirage 2000
F 111
F-4
F-14
F-15
F-16
Interceptor
Interceptor
Fighter
Fighter
Interceptor
Fighter
Fighter
  5.0 km/min
14.94 km/min
10.94 km/min
  8.534 knVmin
  9.14 km/min
15.24 km/min
15.24 km/nun
3.0 mt to 11 km
2.4 mt to l4.94 km
N/A
N/A
N/A
N/A
N/A  .
   For a modern (supersonic) combat aircraft with sufficiently high excess-thrust
capability, transonic discontinuities and closed contours in supersonic regions dis-
appear, and the optimal climb schedule will be quite similar to that of the subsonic
aircraft as shown in Fig. 2.19.
   The (R]C)  x and the minimum time to climb to a given altitude are impor-
tant performance metrics for interceptor/fighter aircraft. Table 2.1 gives these two
parameters for some modern combat aircraft.
Example 2.5
       For the propeller airplane of Example 2.3, determine the maximum climb angle,
maximum rate of climb, and corresponding speed and lift coefficients.
      So/ution.     Using the power-available and power-required curves of Fig. 2.14,
the rate of climb R/C is plotted against velocity V in Fig. 2.21. Also plotted in
Fig. 2.21 is the climb angle y obtained by dividing RlC by V.
       From this graph, we find that the maximum rate of climb is  10.5 m/s and occurs
at 50 m/s. The corresponding lift coe:fficient is found to be 1.084.
    The steepest climb angle is -21.8 deg and occurs at 19.0 m/s but needs a lift
coefficient of 7.54, which is in excess of the maximum lift coefficient of 1.75.
Therefore, the permissible steepest climb angle for this aircraft occurs at Vstdl of
39.4323 m/s and is equal to 14.5 deg.
                                           Example 2.6
   An aircraft weighing 156,960 N is powered by a jet engine whose thrust is
 independent offlight speed. The maximum rate ofclimb at sealevel occurs at 152.5
m/s. The wing area is 46 m2. The drag polar is given by CD = 0.016 + 0.045C2
and CL,max = 1.5. Determine 1) the sea le'vel thrust To developed by the engme,
2) lift coefficient, 3) maximum and minimum speeds in level flight at sealevel, and
4) the absolute ceiling assunung that the thrust varies with altitude as T = Toa0.6.
    Solution,   We are given  W =156,960.0 N, S-46 I112, CDO -.0.016, k -.
　
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