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AIRCRAFT PERFORMANCE
71
usual in performance analyses to assume that the thrust is aligned with the flight
path (e = 0).
For the special case of an airplane whose flight path is a straight line in the verti-
cal plane and whose fiight'velocit)r is constant, acceleration terms on the right-hand
sides of Eqs. (2.5) and (2.6) vanish. The performance of an airplane based on this
assumption is called static performance. Examples of static performance are steady
level fiight, steady climb, range, and endurance in constant- velocity cruise. Per-
formance problems that involve acceleration terms are accelerated climbs, takeoff,
and landing.
Assuming that the thrust vector is aligned with the fiight path, equations for
static performance are given by
T-D- Wsiny -0
L- Wcosy -O
(2.7)
(2.8)
In addition to these equations of motion, we also need the following kinematic
relations to calculate distances with respect to the ground:
x -. V cos y
h = V siny
(2.9)
(2.10)
where x and h are the horizontal and 'vertical distances measured with respect to the
origin of a suitable coordinate system fixed on the ground. The "." over a symbol
denotes differentiation with respect to time. :fhus, x -. dx/dt is the horizontal
velocity with respect to the ground, and h : dhldt is the rate of increase of
altitude or the rate of climb. In calm air, the velocity with respect to the ground
and that with respect to air are equal. In the presence of the wind, x = V + VLV,
where vw is the wind velocity and "-" applies to the headwind and "+" for the
tailwind.
Let us introduce some nondimensional parameters so that we can express the
above equations in a more compact form. Let
E = g-
L
n - -.
W
TEm
z= W
V
li -. -
VR
VR -
(2.12)
(2.13)
(2.14)
(2.15)
Here, E is known as the lift-to-drag ratio and the maximum value of E is denoted
by Em. The parameter n, which is the ratio of lift to weight, is called the load
factor. The parameter z is the nondimensional thrust, and u is the nondimcnsional
:!
\
/,
i}
;.
:
72 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Then,
which gives
CD CDO+kCZ
C~ -- - CL
d CD -CDO+kC2
d~ C)=~
-. 0
CL -
(g )-..= 2~
Inverting Eq. (2.19) we obtain
Em = (g ),ax
1
= 2~
Let C2 denote the value oflift coefficient when E = Em. Then,
Cz =
(2.16)
(2.17)
(2.1 8)
(2.19)
(2.20)
(2.21)
Schematic variations of CL, CD, and E with angle of attack are shown in Fig. 2.3.
The drag of the aircraft is given by .
D = ;:p V2S(CDO + kCZ) (2.22)
With L - n W, we have
, 2n W
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