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r12=~W~S~ k   2k      lcbo t4kV)  . (2.234)
l22 : A /t - BIr4
A-
B -,
k'?7p P'(kW)p
   2k(Wl S)
p2CDO
4k~W~S -
(2.235)
(2.236)
(2.237)
For turning in horizontal plane with maximum load factor,
                                               2nddV  = A - 4B V3 = o                                (2.238)
Because n # 0, the velocity that maximizes the load factor is given by
                       ~D(W/s) g
           Vn,max- k- P2~kYF
             :DO
With this, the maximum load factor in turn is given by
                      nmax -0.6874 '[k'r7pP'(kW)]2pEn  ~
                    ,   k(W/S)
(2.239)
(2.240)
130             PERFORMANCE, STABfLITY, DYNAMICS, AND CONTROL
Then,
co = g~7
 R = -, //;2max
gV~=l
                                 v
C~ = p2S~W
               X
Time for one complete turn of 27r radians is given by
h7r = X-j:
(2.241)
(2.242)
(2.243)
(2.244)
    Turning performance of jet aircraft.   For jet aircraft whose thrust is inde-
pendent of fiight velocity, analytical solution can be obtained as follows.
   We have
  TEm
  z= W
D=2~V (.2+-,)
Then, Eq. (2.204) can be expressed in nondimensional form as
or
zW  V
E -2Vg (.2+~)=0
u4 _ 2zu2 +nL = 0
n=~
(2.245)
(2.246)
(2.247)
(2.248)
(2.249)
     Maximum }sustained turn rate.    As said before, the maximum turn rate gen-
erated while holding a constant altitude is called the MSTR. Using Eq. (2.249),
Eq. (2.211) can be written as
  g^     7
 CD =           f R
(  ,V )2 = 2z - r12 - ~
(2.250)
(2.251)
AIRCRAFT PERFORMANCE
where VR iS the reference velocity given by
For MSTR or COmax,
because co + O, we get
VR ~
u -. 1
131
(2.252)
(2.253)
(2.254)
(2.255)
thus, the maximum turn rate occurs when V - VR. Substituting u - 1 in Eqs.
(2.249) and (2.250), we obtain
                n=\  IC            (2.256)
            gj~-2-       (2.2s7)
                                 COmax = -VR
The lift coefficient CL can be obtained using the nondimensional expression for
drag in Eq. (2.246) to express the ratio ofinduced drag to zero-lift drag as follows:
                                 kCl_ = _, = n2 = 2z - 1                      (2.258)
               CDO  U4
so that
CL -
                                   = j~'fcZ ~                         (2.259)
　
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