Aircraft Flight Dynamics Robert F. Stengel Lecture7 Gliding, Climbing, and Turning Performance

Gliding, Climbing, and Turning
Flight Performance

Robert Stengel, Aircraft Flight Dynamics,
MAE 331, 2012
!
Copyright 2012 by Robert Stengel. All rights reserved. For educational use only.!
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•  Flight envelope"
•  Minimum glide angle/rate"
•  Maximum climb angle/rate"
•  V-n diagram"
•  Energy climb"
•  Corner velocity turn"
•  Herbst maneuver"
The Flight Envelope
Flight Envelope Determined by
Available Thrust"
•  Flight ceiling defined by
available climb rate"
–  Absolute: 0 ft/min"
–  Service: 100 ft/min"
–  Performance: 200 ft/min"
•  Excess thrust provides the
ability to accelerate or climb"
•  Flight Envelope: Encompasses all altitudes
and airspeeds at which an aircraft can fly "
–  in steady, level flight "
–  at fixed weight"
Additional Factors Define the
Flight Envelope"

•  Maximum Mach number"
•  Maximum allowable
aerodynamic heating"
•  Maximum thrust"
•  Maximum dynamic
pressure"
•  Performance ceiling"
•  Wing stall"
•  Flow-separation buffet"
–  Angle of attack"
–  Local shock waves"
Piper Dakota Stall Buffet"
/>!
Boeing 787 Flight
Envelope
(HW #5, 2008)"
Best
Cruise
Region"
Gliding Flight

D = C
D
1
2
ρ
V
2
S = −W sin
γ

C
L
1
2
ρ
V
2
S = W cos
γ

h = V sin
γ

r = V cos
γ
Equilibrium Gliding Flight"
Gliding Flight"
•  Thrust = 0"
•  Flight path angle < 0 in gliding flight"
•  Altitude is decreasing"
•  Airspeed ~ constant"
•  Air density ~ constant "

tan
γ
= −
D
L
= −
C

D
C
L
=

h

r
=
dh
dr
;
γ
= −tan
−1
D
L
#
$
%
&
'
(
= −cot
−1
L
D
#
$
%

&
'
(
•  Gliding flight path angle "
•  Corresponding airspeed "
V
glide
=
2W
ρ
S C
D
2
+ C
L
2
Maximum Steady Gliding Range"
•  Glide range is maximum when
γ
is least negative, i.e.,
most positive"
•  This occurs at (L/D)
max
"
Maximum Steady
Gliding Range"
•  Glide range is maximum when
γ
is least negative,
i.e., most positive"

•  This occurs at (L/D)
max
"

tan
γ
=

h

r
= negative constant =
h − h
o
( )
r − r
o
( )
Δr =
Δh
tan
γ
=
−Δh
−tan
γ
= maximum when
L
D
= maximum

γ
max
= −tan
−1
D
L
#
$
%
&
'
(
min
= −cot
−1
L
D
#
$
%
&
'
(
max
Sink Rate "
•  Lift and drag define
γ
and V in gliding equilibrium"
D = C
D

1
2
ρ
V
2
S = −W sin
γ
sin
γ
= −
D
W
L = C
L
1
2
ρ
V
2
S = W cos
γ
V =
2W cos
γ
C
L
ρ
S



h = V sin
γ
= −
2W cos
γ
C
L
ρ
S
D
W
$
%
&
'
(
)
= −
2W cos
γ
C
L
ρ
S
L
W
$
%
&
'

(
)
D
L
$
%
&
'
(
)
= −
2W cos
γ
C
L
ρ
S
cos
γ
1
L D
$
%
&
'
(
)
•  Sink rate = altitude rate, dh/dt (negative)"
•  Minimum sink rate provides maximum endurance"
•  Minimize sink rate by setting ∂(dh/dt)/dC

L
= 0 (cos
γ
~1)"
Conditions for Minimum
Steady Sink Rate"


h = −
2W cos
γ
C
L
ρ
S
cos
γ
C
D
C
L
$
%
&
'
(
)
= −
2W cos
3

γ
ρ
S
C
D
C
L
3/2
$
%
&
'
(
)
≈ −
2
ρ
W
S
$
%
&
'
(
)
C
D
C
L
3/2

$
%
&
'
(
)
C
L
ME
=
3C
D
o
ε
and C
D
ME
= 4C
D
o
L/D and V
ME
for Minimum Sink Rate"
V
ME
=
2W
ρ
S C
D

ME
2
+ C
L
ME
2
≈
2 W S
( )
ρ
ε
3C
D
o
≈ 0.76V
L D
max
L
D
( )
ME
=
1
4
3
ε
C
D
o
=

3
2
L
D
( )
max
≈ 0.86
L
D
( )
max
L/D for Minimum Sink Rate"
•  For L/D < L/D
max
, there are two solutions"
•  Which one produces minimum sink rate?"
L
D
( )
ME
≈ 0.86
L
D
( )
max
V
ME
≈ 0.76V
L D
max

Gliding Flight of the
P-51 Mustang"

Loaded Weight = 9,200 lb (3,465 kg)
L / D
( )
max
=
1
2
ε
C
D
o
=16.31
γ
MR
= −cot
−1
L
D
$
%
&
'
(
)
max
= −cot
−1

(16.31) = −3.51°
C
D
( )
L/D
max
= 2C
D
o
= 0.0326
C
L
( )
L/D
max
=
C
D
o
ε
= 0.531
V
L/D
max
=
76.49
ρ
m / s

h

L/D
max
= V sin
γ
= −
4.68
ρ
m / s
R
h
o
=10km
= 16.31
( )
10
( )
=163.1 km
Maximum Range Glide"

Loaded Weight = 9,200 lb (3,465 kg)
S = 21.83 m
2
C
D
ME
= 4C
D
o
= 4 0.0163
( )

= 0.0652
C
L
ME
=
3C
D
o
ε
=
3 0.0163
( )
0.0576
= 0.921
L D
( )
ME
=14.13

h
ME
= −
2
ρ
W
S
$
%
&
'

(
)
C
D
ME
C
L
ME
3/2
$
%
&
&
'
(
)
)
= −
4.11
ρ
m / s
γ
ME
= −4.05°
V
ME
=
58.12
ρ
m / s

Maximum Endurance Glide"
Climbing Flight
•  Rate of climb, dh/dt = Specific Excess Power "
Climbing Flight"


V = 0 =
T − D −W sin
γ
( )
m
sin
γ
=
T − D
( )
W
;
γ
= sin
−1
T − D
( )
W


γ
= 0 =
L −W cos
γ

( )
mV
L = W cos
γ


h = V sin
γ
= V
T − D
( )
W
=
P
thrust
− P
drag
( )
W
Specific Excess Power (SEP) =
Excess Power
Unit Weight
≡
P
thrust
− P
drag
( )
W
•  Flight path angle " •  Required lift"

•  Note significance of thrust-to-weight ratio and wing loading"
Steady Rate of Climb"


h = V sin
γ
= V
T
W
"
#
$
%
&
'
−
C
D
o
+
ε
C
L
2
( )
q
W S
( )
*
+

,
,
-
.
/
/
€
L = C
L
q S = W cos
γ
C
L
=
W
S
#
$
%
&
'
(
cos
γ
q
V = 2
W
S
#
$

%
&
'
(
cos
γ
C
L
ρ


h = V
T
W
!
"
#
$
%
&
−
C
D
o
q
W S
( )
−
ε
W S

( )
cos
2
γ
q
*
+
,
-
.
/
= V
T h
( )
W
!
"
#
$
%
&
−
C
D
o
ρ
h
( )
V
3

2 W S
( )
−
2
ε
W S
( )
cos
2
γ
ρ
h
( )
V
•  Climb rate "
•  Necessary condition for a maximum with
respect to airspeed"
Condition for Maximum
Steady Rate of Climb"


h = V
T
W
!
"
#
$
%
&

−
C
D
o
ρ
V
3
2 W S
( )
−
2
ε
W S
( )
cos
2
γ
ρ
V

∂

h
∂
V
= 0 =
T
W
"
#

$
%
&
'
+V
∂
T /
∂
V
W
"
#
$
%
&
'
(
)
*
+
,
-
−
3C
D
o
ρ
V
2
2 W S

( )
+
2
ε
W S
( )
cos
2
γ
ρ
V
2
Maximum Steady "
Rate of Climb: "
Propeller-Driven Aircraft"
∂
P
thrust
∂
V
= 0 =
T
W
"
#
$
%
&
'
+V

∂
T /
∂
V
W
"
#
$
%
&
'
(
)
*
+
,
-
•  At constant power"

∂

h
∂
V
= 0 = −
3C
D
o
ρ
V

2
2 W S
( )
+
2
ε
W S
( )
ρ
V
2
•  With cos
2
γ
~ 1, optimality condition reduces to"
•  Airspeed for maximum rate of climb at maximum power, P
max
"
V
4
=
4
3
!
"
#
$
%
&
ε

W S
( )
2
C
D
o
ρ
2
; V = 2
W S
( )
ρ
ε
3C
D
o
= V
ME
Maximum Steady Rate
of Climb: "
Jet-Driven Aircraft"
•  Condition for a maximum at constant thrust and cos
2
γ
~ 1"
•  Airspeed for maximum rate of climb at maximum thrust, T
max
"

∂


h
∂
V
= 0
0 = ax
2
+ bx + c and V = + x
= −
3C
D
o
ρ
2 W S
( )
V
4
+
T
W
#
$
%
&
'
(
V
2
+
2

ε
W S
( )
ρ
= −
3C
D
o
ρ
2 W S
( )
V
2
( )
2
+
T
W
#
$
%
&
'
(
V
2
( )
+
2
ε

W S
( )
ρ
Optimal Climbing Flight
What is the Fastest Way to Climb from
One Flight Condition to Another?"
•  Specific Energy "
•  = (Potential + Kinetic Energy) per Unit Weight"
•  = Energy Height"
Energy Height"
•  Could trade altitude with airspeed with no change in energy
height if thrust and drag were zero"
Total Energy
Unit Weight
≡ Specific Energy =
mgh + mV
2
2
mg
= h +
V
2
2g
≡ Energy Height, E
h
, ft or m
Specific Excess Power"
dE
h
dt

=
d
dt
h +
V
2
2g
!
"
#
$
%
&
=
dh
dt
+
V
g
!
"
#
$
%
&
dV
dt
•  Rate of change of Specific Energy "
= V sin
γ

+
V
g
"
#
$
%
&
'
T − D −mgsin
γ
m
"
#
$
%
&
'
= V
T − D
( )
W
= V
C
T
− C
D
( )
1
2

ρ
(h)V
2
S
W
= Specific Excess Power (SEP) =
Excess Power
Unit Weight
≡
P
thrust
− P
drag
( )
W
Contours of Constant
Specific Excess Power"
•  Specific Excess Power is a function of altitude and airspeed"
•  SEP is maximized at each altitude, h, when"
d SEP(h)
[ ]
dV
= 0
Subsonic Energy Climb"
•  Objective: Minimize time or fuel to climb to desired altitude
and airspeed"
Supersonic Energy Climb"
•  Objective: Minimize time or fuel to climb to desired altitude
and airspeed"
The Maneuvering Envelope

•  Maneuvering envelope: limits
on normal load factor and
allowable equivalent airspeed"
–  Structural factors"
–  Maximum and minimum
achievable lift coefficients"
–  Maximum and minimum
airspeeds"
–  Protection against
overstressing due to gusts"
–  Corner Velocity: Intersection
of maximum lift coefficient
and maximum load factor"
Typical Maneuvering Envelope:
V-n Diagram"
•  Typical positive load factor limits"
–  Transport: > 2.5"
–  Utility: > 4.4"
–  Aerobatic: > 6.3"
–  Fighter: > 9"
•  Typical negative load factor limits"
–  Transport: < –1"
–  Others: < –1 to –3"
C-130 exceeds maneuvering envelope"
/>!
Maneuvering Envelopes (V-n Diagrams)
for Three Fighters of the Korean War Era"
Republic F-84"
North American F-86"
Lockheed F-94"

Turning Flight
•  Vertical force equilibrium"
Level Turning Flight"
L cos
µ
= W
•  Load factor"
n =
L
W
=
L
mg
= sec
µ
,"g"s
•  Thrust required to maintain level flight"
T
req
= C
D
o
+
ε
C
L
2
( )
1
2

ρ
V
2
S = D
o
+
2
ε
ρ
V
2
S
W
cos
µ
#
$
%
&
'
(
2
= D
o
+
2
ε
ρ
V
2

S
nW
( )
2
µ
: Bank Angle
•  Level flight = constant altitude"
•  Sideslip angle = 0"
•  Bank angle"
Maximum Bank Angle in
Level Flight"
cos
µ
=
W
C
L
qS
=
1
n
= W
2
ε
T
req
− D
o
( )
ρ

V
2
S
µ
= cos
−1
W
C
L
qS
$
%
&
'
(
)
= cos
−1
1
n
$
%
&
'
(
)
= cos
−1
W
2

ε
T
req
− D
o
( )
ρ
V
2
S
*
+
,
,
-
.
/
/
•  Bank angle is limited by "
€
µ
: Bank Angle
C
L
max
or T
max
or n
max
•  Turning rate"

Turning Rate and Radius in Level Flight"


ξ
=
C
L
qS sin
µ
mV
=
W tan
µ
mV
=
g tan
µ
V
=
L
2
− W
2
mV
=
W n
2
− 1
mV
=

T
req
− D
o
( )
ρ
V
2
S 2
ε
− W
2
mV
•  Turning rate is limited by "
C
L
max
or T
max
or n
max
•  Turning radius "

R
turn
=
V

ξ
=

V
2
g n
2
− 1
Maximum Turn Rates"
“Wind-up
turns”"
•  Corner velocity"
Corner Velocity Turn"
•  Turning radius "
R
turn
=
V
2
cos
2
γ
g n
max
2
− cos
2
γ
V
corner
=
2n
max

W
C
L
mas
ρ
S
•  For steady climbing or diving flight"
sin
γ
=
T
max
− D
W
Corner Velocity Turn"
•  Time to complete a full circle "
t
2
π
=
V cos
γ
g n
max
2
− cos
2
γ
•  Altitude gain/loss "
Δh

2
π
= t
2
π
V sin
γ
•  Turning rate "


ξ
=
g n
max
2
− cos
2
γ
( )
V cos
γ
Not a turning rate comparison"
/>!
Herbst Maneuver"
•  Minimum-time reversal of direction"
•  Kinetic-/potential-energy exchange"
•  Yaw maneuver at low airspeed"
•  X-31 performing the maneuver
"
Next Time:

Aircraft Equations of Motion

Reading
Flight Dynamics, 155-161
Virtual Textbook, Parts 8,9