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         Vmax - Umax VR
                          - 271.5436 m/s
Umin = ~
 =     ==:i
        -. 0.1137
          Vmin -. Unun VR
                    - 11.4993 m/s
Vstall -
- 64.3357 m/s
94                  PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Because  VJmn <  Vsum,level flightis not possible at Vnvn. Therefore, the nunimum
level fiight speed at 8000 m altitude is VsraU - 64.3357 m/s.
    At absolute ceiling, z -.1- Therefore,
T = .p~r
             1.0 * 21.6777
 45,000
 - 2075.866 N
   T
o : To
2075.866
 20,000
      - 0.1038
p - ooa
     ~  1.225 * 0.1038
  = 0.1271kgtm3
From atmospheric tables (see Appendix), the corresponding altitude, which in this
case is the absolute ceiling of the airplane, is found to be approximately equal to
17.7 km.
2.5 ClimbingFlight
         For all altitudes below the absolute ceiling, we have two solutions for steady level
fiight, alow-speed solution Vnun and a high-speed solution V   x. What happens if
we try to fiy at any speed V, Vnun < V < Vmax usirig fullthrust or power? Now we
have excess thrust or excess power, and steady level flightis notpossible.What may
happen is that either 1) the ;rlplane will accelerate to Vmax if the altitude is forced
to remain constant or 2) the airplane- will climb. In this section, we will discuss
climbing flight. To begin with,let us consider the steady or constant velocity climb
and then the more general case of accelerated climb. For simplicity,let us assume
that the weight of the airplane remains constant during the climb.
  Consider an air)lane in a steady climbing flight The fiight path of such an
airplane is a straight line in vertical plane. From Eqs. (2.7) and (2.8), we have
                               L- Wcosy -0
                          T-D- Wsiny-0
and the kineinatic equations are
                                                  x - V cos y
                                       h: V siny
(2.99)
(2,100)
(2.101)
(2.102)
AIRCRAFT PERFORMANCE
95
From Eq. (2.100), we obtain
                 .    T-D
           siny- W
                                              Excess thrust
           = We~g~t       (2.103)
We observe that the climb angle y is directly proportional to the excess thrust per
unit weight. Therefore, steepest climb Ymax occurs when the excess thrust per unit
weight is maximum.
     Let R/C denote the rate of climb. Then
R/C =h
        = V sin y
          V(T - D)
 -W
         Ptl - PR
 =W
     - Ps                            (2.104)
where Ps is the excess power per unit weight. Usually, Ps is called specific excess
　
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