Transcript AE 301 Aerodynamics I

Range and Endurance
• When designing or comparing aircraft, two
parameters usually come to mind:
– Range: the horizontal distance an airplane can travel on a
single fueling. The cruise portion of a flight is associated
with flying for range.
– Endurance: the amount of time an airplane can remain
aloft on a single fueling. The loiter phase of a mission is
associated with flying for endurance.
• We will calculate the range and endurance for pistonpropeller and turbojet aircraft separately.
• All our calculations will assume still air conditions I.e. no head or tail winds.
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R and E - Propeller Aircraft
• Both range and endurance of an aircraft will depend
upon the rate at which fuel is burned.
• The common parameter used to define this rate is
called the Specific Fuel Consumption, SFC.
Weight of fuel
lb of fuel
SFC 

EnginePower T ime (bhp) hour
• The SFC is considered a constant for an engine type varying vary little with throttle setting or flight
conditions.
• Typical range of values: 0.4 - 0.7 lb/hp/hr
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R and E - Propeller Aircraft (continued)
• From the fact that the SFC is a constant, we can
deduce some simple range and endurance relations.
• For a long endurance, we would like to burn the
minimum fuel per hour. From the SFC then:
lb of fuel
SFC  PR
 SFC  (bhp) 
hour

• Thus, for long endurance we want to fly with a high
propeller efficiency and at minimum power required!
• From our previous result for power required, it
follows that for max endurance, we want to fly at the
velocity such that CL3/2/CD is maximum.
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R and E - Propeller Aircraft (continued)
• For a long range, we would like to burn the minimum
fuel per mile traveled. Dividing our previous result by
velocity (miles/hour) gives:
lb of fuel SFC  (bhp) SFC  TR


mile
V

• Thus, for long range we want to fly with a high
propeller efficiency and at minimum thrust required!
• From our previous result for thrust required, it follows
that for max range, we want to fly at the velocity
such that CL/CD is maximum.
(Remember, these results are for propeller driven aircraft - jets
will be different!)
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R and E - Propeller Aircraft (continued)
• Now let’s work at some quantitative relations.
• First, since SFC does not have consistent units, a new
fuel burn rate is introduced:
lb of fuel
1

N of fuel 1 
c

 c 
 
(ft  lb/sec) sec ft
(W ) sec m 

• Note that in the British system, c and SFC are related
via the conversion factors:
SFC
SFC
c

ftlb
sec
550sechp 3600hour  1980000hourftlbhp 
• Is other references, if these conversion factors are
part of the equations, they are using SFC not c!
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R and E - Propeller Aircraft (continued)
• The weight of an aircraft is usually broken down into
various components of which only the fuel portion
varies during normal flight.
W  WOEW  WPayload  Wf
• The rate at which fuel is burned depends upon the
SFC and the power setting:
dW  dWf  cPdt
• Rearrange gives dt  dW  cP which can be integrated
to get the endurance:
t1
W0
W1
dW
E   dt  
 cP
t0
W0
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dW
E 
cP
W1
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R and E - Propeller Aircraft (continued)
• Similarly, the range is found by multiplying time by
velocity and integrating:
V dW
dx  V dt 
 cP
W0
x1
W1
V dW
V dW
R   dx  
R 
 cP
cP
x0
W0
W1
• These integral relations are useful for calculating the
R and E for a given mission where V and P may vary
through the flight.
• However, they are not very useful for rapid R and E
estimations for a given airplane. For that purpose,
we use the Breguet relations!
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Breguet Propeller Equations
• The Breguet relations are approximate expressions
for R and E obtained by making assumptions about a
typical flight profile.
• For range, assume steady, level flight so that L=W
and PR=VTR=VD.
0
0
V dW
VWdW
L dW
R 


cP
cV  DW
cD W
W1
W1
W1
W0
W
W
• Now assume that c and  are constant, and that the
aircraft is flown at a velocity such that L/D = CL/CD
remains a constant.
• (Note that V and  are not necessarily constant!)
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Breguet Propeller Eqns (continued)
• With these assumptions, the integral can be
evaluated to get:
 Wo 

R
ln
c C D  W1 
 CL
• This relations tells us that to maximize range, we
want:
– The largest possible propeller efficiency
– The lowest possible specific fuel consumption
– The largest possible weight fraction of fuel, Wf/W
– A large (L/D)max and to fly at the velocity where this is
achieved.
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Breguet Propeller Eqns (continued)
• Perform a similar procedure with the endurance
equation.
W
W
W
dW
WdW
L dW
E
0
0
 cP   cV
W1
W1
 DW

0
 cV
W1

D W
• To go further, we need to relate velocity to lift via:
V  2W / SCL 
E
W0
C L   SC L dW
 cC
W1
D
2W
W

W0

W1
C 3 / 2   S dW
L
cC D
2 W 3/ 2
• Now assume that c,  and  are constant, and that
the aircraft is flown at a velocity such that CL3/2/CD
remains a constant
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Breguet Propeller Eqns (continued)
• With these assumption, integration yields:
E
C 3 / 2
L
cCD
 1
1 
2  S  1/ 2  1/ 2 
W0 
 W1
• This relations tells us that to maximize endurance, we
want:
– The largest possible propeller efficiency
– The lowest possible specific fuel consumption
– The largest possible weight fraction of fuel, Wf/W
– A large (CL3/2/CD)max and to fly at the velocity where this is
achieved.
– Flying at the highest density possible, I.e. sea level!
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R and E - Jet Aircraft
• Jet aircraft differ from propeller aircraft primarily in
the fact that a jet engine produces thrust directly
while a piston engine produces power.
• The fuel consumption for jet aircraft is thus based
upon T and called the Thrust Specific Fuel
Consumption, TSFC.
Weight of fuel
lb of fuel
TSFC 

EngineT hrusthour lb of thrusthour
• As with piston engines, TSFC is a nearly a constant
for a given powerplant.
• Typical range of values: 0.5 - 1.0 lb/lb/hr
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R and E - Jet Aircraft (continued)
• From the fact that the TSFC is a constant, we can
deduce some simple range and endurance relations.
• For a long endurance, we would like to burn the
minimum fuel per hour. From the TSFC then:
lb of fuel
 TSFC  (lb of thrust )
hour
• Thus, for long endurance on a jet, we want to fly
with a minimum thrust required!
• From our previous result for thrust required, it follows
that for max endurance, we want to fly at the
velocity such that CL/CD is maximum.
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R and E - Jet Aircraft (continued)
• For a long range, we would like to burn the minimum
fuel per mile traveled. Dividing our previous result by
velocity (miles/hour) gives:
lb of fuel TSFC  (lb of thrust) TSFC  TR


mile
V
V
• To reach a conclusion from this relation, lets assume
TR=D and substitute velocity from our CL definition
TR D 1
2W
1

 2  V SCD  2  
SCD 
V V
 SCL
1
2
CD
 SW 1/ 2
CL
• Thus, among other things, for max range we want to
fly at a velocity such that CL1/2/CD is a maximum!
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R and E - Jet Aircraft (continued)
• Now let’s work at the quantitative relations.
• First, since TSFC does not have consistent units, a
new fuel burn rate is introduced:

N of fuel 1 
 ct 


( N ) sec sec 

lb of fuel
1
ct 

(lb of thrust)sec sec
• Note that ct is the same in either unit system.
Converting to TSFC results in:
ct 
TSFC
sec
3600hour

• As before, always check units and conversion factors
when using other references!
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R and E - Jet Aircraft (continued)
• The rate at which jets burn fuel is given by:
dW  dWf  ctTdt
• Rearranging and integrating for endurance gives:
t1
W1
dW
E   dt  
 ctT
t0
W0
W0
dW
E 
cT
W1 t
• Similarly, range will be calculated by:
x1
R   dx 
x0
W1
V dW
W  ctT
0
R
W0
V dW
W ctT
1
• Note how similar these equations are to those for
propellers - only the denominator is different.
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Breguet Jet Equations
• Now make assumptions similar to those for propellers
previously to get the Breguet Jet equations from
these integral relations.
• For endurance, assume steady, level flight so that
L=W and T=TR=D.
E
W0
W0
W0
dW
WdW
L dW


W ctT W ct DW W ct D W
1
1
1
• Now with ct a constant and assuming the aircraft is
flown at a velocity such that L/D = CL/CD remains a
constant.
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Breguet Jet Eqns (continued)
• Pulling out theses constants and integrating yields:
1 CL  Wo 
E
ln 
ct CD  W1 
• This relations tells us that to maximize endurance, we
want:
– The lowest possible thrust specific fuel consumption
– The largest possible weight fraction of fuel, Wf/W
– A large (L/D)max and to fly at the velocity where this is
achieved.
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Breguet Jet Eqns (continued)
• Perform a similar procedure with the jet range
equation.
W0
W
W
0
V dW 0 VWdW
V L dW
R 


cT T
ct DW
cD W
W1
W1
W1 t
• To go further, relate velocity to lift via:
V  2W / SCL 
W0
CL
R 
cC
W1 t D
W
1/ 2
0
CL
2W dW

  SC L W W1 ct C D
2
dW
  S W 1/ 2
• Now assume that ct and  are constant, and that
the aircraft is flown at a velocity such that CL1/2/CD
remains a constant
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Breguet Jet Eqns (continued)
• With these assumptions, integration yields:
1/ 2
C
2 L
R
ct CD
2


W
S
1/ 2
0
 W11/ 2

• This relations tells us that to maximize range, we
want:
– The lowest possible specific fuel consumption
– The largest possible weight fraction of fuel, Wf/W
– A large (CL1/2/CD)max and to fly at the velocity where this is
achieved.
– Flying at the lowest density possible, I.e. high altitude!
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Breguet Eqns - Summary
• Here is a summary of the Breguet equations:
• For piston-propellers:
 Wo 

R
ln
c C D  W1 
 CL
E
C 3 / 2
L
cCD
 1
1 
2  S  1/ 2  1/ 2 
W0 
 W1
• For turbojets:
1/ 2
2 CL
R
ct CD
2


W
S
AE 301 Aerodynamics I
1/ 2
0
W
1/ 2
1

1 CL  Wo 
E
ln 
ct CD  W1 
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CL and CD relations - Summary
• For (CL3/2/CD)max,
CD,i=3CD,0
C L3 / 2 3eARC D , 0 

CD
4C D , 0
3/ 4
C L  3eARC D , 0
• For (CL/CD)max,
C L  eARC D , 0
• For (CL1/2/CD)max,
CL 
1
3
eARC D ,0
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CD,i=CD,0
C L 1 eAR

C D 2 C D ,0
CD,i=1/3 CD,0
C L1/ 2

CD
203

1
3
eARC D , 0 
1/ 4
4
3
CD ,0
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Accelerated Flight
• Thus far, all the performance parameters we have
considered have been for an aircraft in unaccelerated
flight - so called static performance.
• Now lets begin to consider the performance of an
aircraft experiencing accelerations - either along or
perpendicular to the flight path.
• For the first case - accelerations along the flight path
- we will consider the two most extreme situations:
takeoff and landing.
• For the second - accelerations normal to the flight
path - we will consider turning flight.
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Takeoff Performance
• An aircraft under acceleration must obey Newton’s
second law:
F  ma
or
aF
m
• If we further assume that the force and thus the
acceleration is constant, then by successive
integrations over time we can get:
V  at or
t  V  Vm
a
F
2
2
2
1
1 F
Vm
V
s  2 at or
s2
 m
m
F
2F
• In these relations, t and s, are the time and distance
required to achieve a target velocity, V.
 
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
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Takeoff Performance (continued)
• Now let’s concentrate on our particular situation as
illustrated below:
L
T
D
R
W
• In addition to the familiar forces of L, W, T, and D, we
also have a ground resistance force, R.
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Takeoff Performance (continued)
• This resistance force is due to the rolling friction
between the tires and the ground.
• Assume this force is proportional to the normal force
of contact:
R  r W  L
• The friction coefficient used here, r, will also depend
upon the type of runway surface. Typical values are:
r = 0.02 for smooth, hard runway (asphalt or concrete)
r = 0.1 for grass runway (unmowed and uncompressed)
• With this new force, a summary of the forces in the
flight path direction gives:
F  T  D  R  T  D  r W  L
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Takeoff Performance (continued)
• One special note concerning drag applies to both
takeoff and landing.
– Due to the interaction of the ground, the wing tip vortices
are weakened.
– The result is both an enhanced lift, and a decreased induced
drag - this is called the ground effect.
– To account for this drag decrease we will use:

CL2 

D  q S  CD,0  
eAR 

where  depends upon the height of the wing above the
ground, h, and the wing span, b:
2

16h / b

2
1  16h / b
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0  1
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Takeoff Performance (continued)
• If the airplane forces are plotted schematically for a
jet aircraft (fighter), we would see something like:
• Note that all our forces
vary as the aircraft
gains speed.
Forces
W
T
• The net axial force also
varies, but not as much.
T-D-(W-L)
• Thus, is is reasonable to
replace this net force
with an average value.
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L
D+(W-L)
D
(W-L)
sLO
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s
Takeoff Performance (continued)
• Thus, assuming that thrust is nearly constant, we can
define an effective axial force.
Feff  T  D  r W  Lavg  const
• To determine an average value for the resistance
forces, D and R, it is suggested that the value at 0.7
the lift off velocity is used:
D  r W  Lavg  D  r W  L0.7V
LO
• And, in turn, the lift off velocity is typically 1.2 times
the stall speed at takeoff:
VLO  1.2Vstall ,TO  1.2
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2W
  SC L ,max,TO
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Takeoff Performance (continued)
• Putting all these assumptions together finally gives us
the take off ground roll distance:
sLO
2
2
W / g 
VLO
m
VLO


2 Feff 2 T  D   r (W  L)avg
sLO
1.44W 2

g SCL ,max T  D   r (W  L)avg
or




• One final assumption Anderson notes is when the
thrust is much greater than the resisting forces. For
this case:
1.44W 2
sLO 
g SCL,maxT
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Takeoff Performance (continued)
• To summarize our results for take off performance:
sLO
1.44W 2

g SCL ,max T  D   r (W  L)avg


– Takeoff distance increases with the square of aircraft weight!
– If thrust variations with altitude are included, T  , then
takeoff distance increases inversely with the square of
density, s  1/2!
– Takeoff distance decrease with higher wing area, S, or
higher takeoff CL,max.
– Takeoff distance decreases with either higher thrust or
reduced ground resistance.
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Takeoff Performance (continued)
• One final note: the FAA regulations for aircraft do
not specify takeoff distance by the ground roll, but by
the distance needed to clear an obstacle:
sLO
h = 50ft (FAR 23)
or 35ft (FAR 25)
sTO or sTOFL
• For small propeller airplanes (FAR 23 certified), a
good approximation is to assume sTO=1.7sLO.
• For large transport aircraft (FAR 25 certified) both the
regulations and aircraft performance is more
complicated - see other references for details!
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Landing Performance
• Calculations for landing performance are very similar
to those for takeoff performance.
• The primary differences
are:
Forces
W
– Most airplanes land with
the engines at idle - T ~ 0.
L
– The aircraft starts with an
initial touchdown velocity,
VT, and decelerates to rest.
– The resistance forces, drag
and ground friction are
intentionally large.
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D+(W-L)
D
s
r(W-L)
sL
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Landing Performance (continued)
• The balance of forces now gives:
0
dV
F  T  D   r W  L   m
dt
• And, as with takeoff, we will assume a constant
average axial force can be defined by:
Feff  D  r W  Lavg  D  r W  L0.7VT
• The distance relation is the same except for a
negative sign to account for starting with velocity VT
and decelerating to rest:
VT2 m
sL  
2 Feff
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Landing Performance (continued)
• Putting all these together gives the relation:
VT2 m
VT2 W / g 
sL 

2 Feff 2D   r (W  L)avg
• Anderson assumes that touch-down will occur at 1.3
times the stall speed at landing, such that:
VT  1.3Vstall , L  1.3
2W
  SC L ,max,L
– or when
CL , L 
CL ,max,L
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1.32

CL ,max,L
1.69
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Landing Performance (continued)
• With this, our final landing distance relations is:
1.69W 2
sL 
g SCL,max,L D   r (W  L)avg
• To increase the resisting forces and thus shorten
landing distances, a number of force enhancing
methods are used:
– First, brakes are normally applied. This has the effect of
increasing the rolling friction. For paved, dry runways, a
value of r=0.4 is typical.
– Also the profile drag associated with landing flaps is
intentionally high - thus increasing Cd,0.
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Landing Performance (continued)
– Spoilers are used on many aircraft. These upper surface,
split flap like devices act to abruptly decrease the wing lift
once on the ground which helps to improve the rolling
friction. Spoilers also increase the profile drag.
– Drag chutes use to be common on high performance
fighters. And of course landing arresting systems such as on
carrier ships can be used.
– Finally, many airplanes can produce reversed thrust either
through a mechanical system on jet engines or reversed
prop pitch on piston/turboprop engines. To include reversed
thrust, simply add in this term with the other resisting
forces:
sL 
g SCL ,max,L
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1.69W 2
TR  D   r (W  L)avg

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
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Landing Performance (continued)
• Finally as with takeoff, the FAA regulations for aircraft
do not specify landing distance by the ground roll
alone, but include the distance needed to clear an
obstacle before landing:
sLG
h = 50ft (FAR 23)
or 50ft (FAR 25)
sL
• In addition, for FAR 25 certified aircraft, the FAA
defines a landing field length, SFL, as SL/0.6. This
provides an additional safety factor to account for
pilot technique.
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Turning Flight
• Now consider situations where the acceleration is
normal to the flight path. The most important of
these is banked turn as illustrated below.

L
Fr
W
R
• A side force, Fr, is generated by banking the aircraft
and tilting the lift vector.
• This side force acts as to produce centripetal
acceleration that pulls the aircraft around in an arc of
radius R
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Turning Flight (continued)
• Note that since the lift vector is tilted by the bank
angle, , to maintain level flight, the lift force must
exceed the weight according to:
L
 W
L cos  W
Fr
• If we define a load factor, n, as the ratio of lift to
weight, then by vector math:
n
L
W
Fr  L2  W 2  W n2  1
• Thus, the centripetal force is related to the load
factor.
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Turning Flight (continued)
• From physics, we know that the
centripetal force is calculated by:
V2
Fr  m
R
• Thus, the turn radius, R, is a function of
the aircraft velocity and load factor:

R
V2
V2
Rm

Fr g n 2  1
• Similarly for the turning rate:
d V g n 2  1



dt R
V
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Pullup Maneuver
• In a pullup maneuver the aircraft
suddenly experiences an increase
in lift above that necessary to
support the weight.

• This is normally achieved by
rotating the craft up to a higher
angle-of-attack through use of
the elevator control surface.
R
L
• As a result of the imbalance of
forces in the vertical direction, the
flight path begins to curve
upwards as shown in the sketch.
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Pullup Maneuver (continued)
• The centripetal force for this maneuver is now in the
vertical plane:
Fr  L  W  W (n 1)
• The radius of curvature of the flight path and the
pitch-up rate are similarly given by:
V2
V2
Rm

Fr g (n  1)

V g (n  1)

R
V
• Note that these relations are equally valid for a
pushover maneuver - a “pullup” with n < 1.0.
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Pulldown Maneuver
• The pulldown maneuver is the
inverted form of the pullup
maneuver.
• In this case, both lift and weight
are contributing to the pitch rate
- as a result, simply rolling an
airplane inverted will initiate this
maneuver.
W
L
R

• By convention, we will still
consider lift and thus load factor
positive as sketched.
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Pulldown Maneuver (continued)
• Since the lift and weight are now in the same
direction, the downward acting centripetal forces is:
Fr  L  W  W (n  1)
• And the flight path radius of curvature and pitch
down rate are given by:
V2
V2
Rm

Fr g (n  1)
V g (n  1)


R
V
• Note that as long as n > -1, R and  are positive for n=-1 the plane is in level flight; for n < -1 the
plane begins climbing in a “pushup” maneuver.
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Maneuver Summary
• A highly maneuverable aircraft is one for which the
maneuver radius, R, is small and the turn/pitch rate,
, is large.
• To achieve this, all the equations derived so far
indicate that a low velocity and high load factor are
desirable!
• High load factors, of course, require strong structures
- but the maximum load factor also depends upon
the CL,max of the aircraft and it’s flight velocity.
• To see how load factor and velocity are inter-related,
we use a so-called V-n diagram.
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