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a)
Velocit)r m/s
  c)
tf
m
-0
f:
a
o
o
c
J
b)
Velodty m/s
  d)
Fig. 2.30     Turning performance of propeller aircraft of Example 2.9.
the aircraft is given by CD  = 0.025 + 0.05CZ, The reciprocating engine produces
840 kW, and the propulsive efficiency of the engme-propeller combination is 0.85.
Determine the lift coefficient, velocity, load factor, turn rate, and radius of turn for
1) fastest sustained turn rate, 2) sharpest sustained tum, and 3) maximum load
factor.
      So/ution.    Because we cannot obtain an analytical solution to the tunung per-
 formance of the propeller aircraft for fastest and sharpest sustained tums, we have
 to use a graphical method. The calculated values of load factor n, lift coe:fficient
 CL, rate of turn co, and radius of turn are plotted against the velocity as shown in
Fig. 2.30.
     From Fig. 2.30b, we observe that for V < 60 m/s, the required lift coefficient
 exceeds CL,max of1.75. Hence, the aircraft cannot achieve the predicted load factor,
 rate of turn, and radius of tum for velocities lower than 60 m/s. Therefore, we have
 eomax  - 0.344 rad/s or 19.7112 deg/s, Rmn  -  174.5 m, wluch occur at  V  : 60 m/s.
        The turning performance for maximumload factor has analytical solution. Using
Eq. (2.240), w9Qfind nmax = 2.3968. The corresponding velocity Vn,max as given
 by Eq. (2.239) is equal to 72.969 m/s. The corresponding rate of tum and radius of
turn were found to be equal to 16.7798 deg/s and 249.1757 m. These values agree
with the graphical solution given in Fig. 2.30.
AIRCRAFT PERFORMANCE
141
                                         Example 2.10
      Ajet aircraft weighs 78,480 N and has a wing area of 30 m2, CL,max of 1.2, sea
level thrust of 39,240 N, drag polar CD  = 0.012+0.12CZ,  and alimit load factor
of 8.0. For turning tlight at sea level, determine the turn rate, radius of turn, load
factor,lift coe:fficient, andlift-to-drag ratio when fiying for the following cases: 1)
MSTR, 2) SST, 3) maximum load factor, and 4) turning at corner velocity. Cases
1-3 correspond to coordinated turns in a horizontal plane.
     So/ution.    Let us first consider cases 1-3 of constant-velocity tums in a hori-
zontal plane.
Case l,MSTR:   We have
                1
Em - 2Jk~
                   1
     =: N
 2W6 fi~_
    - 13.1762
VR -
- .      .4F6:~-
\1  .22s~3ol7 o 0~2
     -  116.2160 m/s
  V - VR
                -  116.2160 m/s
z = TT43
          39,240 *13.1762
  = ~8 480
  - 6.5881
n=    -
 =.  .  =Tr_
   - 3.4894
p, = ,os.' (3 4~894)
     = 73.3466 deg

   .i
    t

-[~
~
~
o
p
_o
142                PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
CL - TI
-. 3.4894 *
~ 1.1034
(D - tOmBX
g.J~-=2
~-
    VR
              9.8 h~6~T172
-N
- 1-62~60
   - 0.2822racVs
     - 16.1693 deg/s
 27r
t -. --
   co
    - 22.2650 s
    V
R ~ --
   to
116.216
 = 02822
- 411.8214 m
The lift-to-drag ratio is gi'ven by [refer to Exercise 2.19(b)]
          E=( + )J~-o
                   3.4894       1
         = ~+34984g4)./61i_i f
　
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