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usually performs better in close combat  Some of the modern high-performance
combat aircraft such as the F-15 or the F-16 are capable of producing turn rates as
high as 20 deg/s.
   For given values of engine tbrust or power, wing loading, and aircraft drag
polar, the problem of determining optimal values of turn rate, the radius of turn
subject to the restrictions imposed by the structural limit load factor nimi, and
aerodynamic limitation CL,max is a nonlinear programming problem and is be-
yond the scope of this text. Interested readers may refer elsewhere.l  Instead, we
will present a simple analysis that gives a fair estimation of optimum tuming
performance.
     Amongst various possible types of turning fiights in a horizontal plane, we will
 study three cases that are of specialinterest:  1) turning tlight with maximum turn
rate, 2) sharpest turn or nunimum radius tum, and 3) turning flight with maximum
possible load factor.
    Turning flight of prope/ler aircraft.   Consider the constant-velocity, coordi-
nated turning flight of a propeller aircraft in a horizontal plane. In general, for
propeller airplanes graphical or numerical methods have to be used to determine
the turning performance. However, for the propeller air 71ane whose power devel-
oped by the engine and the propulsive efficiency are independent of flight velocity,
the problem of turning performance can be formulated in analytical form, but the
resulting equations often need numerical or graphical solutions.
AIRCRAFT PERFORMANCE
127
   For the propeller aircraft, power available and power required are the basic
quantities. In view of this, multiply Eq. (2.204) by velocity to obtain
or
TV -DV -O
(2.213)
k'r7p P(kW) - gpSCDO V3 _ (2kp S3   ) = 0              (2.214)
where T V = k'rlpP(kW) is the power available Pa in Nm/s, P(kW) is the power
developed by the reciprocating piston engine in kW, k' = 1000 is the constant
to convert the kW into Nm/s, and r7p is the propulsive efficiency. As said before,
we assume that for a given altitude, both P(kW) and r7p are independent of fiight
velocity. Solving for load factor n, we get
   1 rk'r7pP'(kW)p(W/S)V
n = W~S [_-    2k
     p2CDO \/4n0.5
  4k j    (2.215)
where P'(kW) is the power per unit weight or the specific power in kW/N.
    Maximum sustained turn rate.   Here we consider the case when the aircraft
attempts to make a constant-velocity turn in a horizontal plane with maximum
rate of turn or angular velocity in turn. The maximum turn rate so generated while
holding a constant altitude is called the maximum sustained turn rate (MSTR).
    From Eq. (2.211), we have
For the MSTR,
where
  g'  7
co - -,                                 (2.216)
        V
dw   - (-11))
dV=g VV~, -
       -. O
\/ dt12 _ (F1,2 _ 1) = 0
2 dV
(2.217)
(2.218)
Cll12          ,pP/(kW)(WlS)  p2CDO 3kV )     (2.219)
d-V =~W~S (k 7r Jp 2k
Substitution in Eq. (2.218) gives an expression of the form
aV4+bV +c=0
(2.220)

128              PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
A solution ofthis equation gives the velocity that maximizes the turn rate.However,
this equation has no analytical solution. Therefore, either a graphical or numerical
solution has to be obtained. Let us assume that  V f r is such a solution. Then,
   1  r7ppP'(kW)(WlS)Vfr p2CD0~4r 0.5
n=.,(- 2k   4k
            g-  I
      COmax=  It '
                   R = g_},-f
                                        1
                                                   COS /1, -  -.
　
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