Gliding, Climbing, and Turning Flight Performance · 3 Gliding Flight • Thrust = 0 • Flight...
Transcript of Gliding, Climbing, and Turning Flight Performance · 3 Gliding Flight • Thrust = 0 • Flight...
1
Gliding, Climbing, and Turning Flight Performance
Robert Stengel, Aircraft Flight Dynamics, MAE 331, 2018
Copyright 2018 by Robert Stengel. All rights reserved. For educational use only.http://www.princeton.edu/~stengel/MAE331.html
http://www.princeton.edu/~stengel/FlightDynamics.html
• Conditions for gliding flight• Parameters for maximizing climb angle and rate• Review the V-n diagram• Energy height and specific excess power• Alternative expressions for steady turning flight• The Herbst maneuver
Learning Objectives
Reading:Flight Dynamics
Aerodynamic Coefficients, 130-141
1
Review Questions§ How does air density decrease with altitude?§ What are the different definitions of airspeed?§ What is a “lift-drag polar”?§ Power and thrust: How do they vary with
altitude?§ What factors define the “flight envelope”?§ What were some features of the first
commercial transport aircraft?§ What are the important parameters of the
“Breguet Range Equation”?§ What is a “step climb”, and why is it important?
2
2
Gliding Flight
3
Equilibrium Gliding Flight
4
CD12ρV 2S = −W sinγ
CL12ρV 2S =W cosγ
!h =V sinγ!r =V cosγ
3
Gliding Flight• Thrust = 0• Flight path angle < 0 in gliding flight• Altitude is decreasing• Airspeed ~ constant• Air density ~ constant
tanγ = −D
L= −
CD
CL
=hr=dhdr; γ = − tan−1 D
L#
$%
&
'(= −cot−1
LD#
$%
&
'(
Gliding flight path angle
Corresponding airspeed
Vglide =2W
ρS CD2 + CL
25
Maximum Steady Gliding Range
• Glide range is maximum when γ is least negative, i.e., most positive
• This occurs at (L/D)max 6
4
Maximum Steady Gliding Range
• Glide range is maximum when γ is least negative, i.e., most positive
• This occurs at (L/D)max
γmax = − tan−1 D
L#
$%
&
'(min
= −cot−1 LD#
$%
&
'(max
7
tanγ =!h!r= negative constant =
h − ho( )r − ro( )
Δr = Δhtanγ
= −Δh− tanγ
= maximum when LD
= maximum
Sink Rate, m/s Lift and drag define γ and V in gliding equilibrium
h =V sinγ
= −2W cosγCLρS
DW$
%&
'
()= −
2W cosγCLρS
LW$
%&
'
()DL
$
%&
'
()
= −2W cosγCLρS
cosγ 1L D$
%&
'
()
Sink rate = altitude rate, dh/dt (negative)
8
D = CD12ρV 2S = −W sinγ
sinγ = − DW
L = CL12ρV 2S =W cosγ
V = 2W cosγCLρS
5
• Minimum sink rate provides maximum endurance• Minimize sink rate by setting ∂(dh/dt)/∂CL = 0 (cos γ ~1)
Conditions for Minimum Steady Sink Rate
h = − 2W cosγCLρS
cosγ CD
CL
$
%&
'
()
= −2W cos3γ
ρSCD
CL3/2
$
%&
'
() ≈ −
2ρWS
$
%&
'
()CD
CL3/2
$
%&
'
()
CLME=
3CDo
εand CDME
= 4CDo9
L/D and VME for Minimum Sink Rate
VME =2W
ρS CDME
2 + CLME2≈
2 W S( )ρ
ε3CDo
≈ 0.76VL Dmax
LD( )
ME=14
3εCDo
=32
LD( )
max≈ 0.86 L
D( )max
10
6
L/D for Minimum Sink Rate• For L/D < L/Dmax, there are
two solutions• Which one produces smaller
sink rate?
LD( )
ME≈ 0.86 L
D( )max
VME ≈ 0.76VL Dmax
11
HL-10
M2-F1
M2-F2
M2-F3
X-24A
X-24B
Historical FactoidsLifting-Body Reentry Vehicles
12
7
Climbing Flight
13
Rate of climb, dh/dt = Specific Excess Power
Climbing Flight
!V = 0 =T − D −W sinγ( )
m
sinγ =T − D( )W
; γ = sin−1 T − D( )W
!γ = 0 =L −W cosγ( )
mVL =W cosγ
!h =V sinγ =VT − D( )W
=Pthrust − Pdrag( )
W
Specific Excess Power (SEP) = Excess PowerUnit Weight
≡Pthrust − Pdrag( )
W
• Flight path angle • Required lift
14
8
Note significance of thrust-to-weight ratio and wing loading
Steady Rate of Climb
h =V sinγ =V TW"
#$
%
&'−
CDo+εCL
2( )qW S( )
*
+,,
-
.//
L = CLqS =W cosγ
CL =WS
⎛⎝⎜
⎞⎠⎟cosγq
V = 2 WS
⎛⎝⎜
⎞⎠⎟cosγCLρ
!h =V TW
⎛⎝⎜
⎞⎠⎟ −
CDoq
W S( ) −ε W S( )cos2 γ
q⎡
⎣⎢
⎤
⎦⎥
=VT h( )W
⎛⎝⎜
⎞⎠⎟−CDo
ρ h( )V 3
2 W S( ) −2ε W S( )cos2 γ
ρ h( )V
Climb rate
15
Necessary condition for a maximum with respect to airspeed
Condition for Maximum Steady Rate of Climb
h =V TW!
"#
$
%&−
CDoρV 3
2 W S( )−2ε W S( )cos2 γ
ρV
∂ h∂V
= 0 = TW"
#$
%
&'+V
∂T /∂VW
"
#$
%
&'
(
)*
+
,-−3CDo
ρV 2
2 W S( )+2ε W S( )cos2 γ
ρV 2
16
9
Maximum Steady Rate of Climb:
Propeller-Driven Aircraft
∂Pthrust∂V
= 0 = TW"
#$
%
&'+V
∂T /∂VW
"
#$
%
&'
(
)*
+
,-• At constant power
∂ h∂V
= 0 = −3CDo
ρV 2
2 W S( )+2ε W S( )ρV 2
• With cos2γ ~ 1, optimality condition reduces to
• Airspeed for maximum rate of climb at maximum power, Pmax
V 4 =43!
"#$
%&ε W S( )2
CDoρ2
; V = 2W S( )ρ
ε3CDo
=VME
17
Maximum Steady Rate of Climb:
Jet-Driven AircraftCondition for a maximum at constant thrust and cos2γ ~ 1
Airspeed for maximum rate of climb at maximum thrust, Tmax
∂ h∂V
= 0
0 = ax2 + bx + c and V = + x
−3CDo
ρ2 W S( )V
4 + TW
⎛⎝⎜
⎞⎠⎟V
2 +2ε W S( )
ρ= 0
18
−3CDo
ρ2 W S( ) V
2( )2 + TW
⎛⎝⎜
⎞⎠⎟ V
2( ) + 2ε W S( )ρ
= 0
Quadratic in V2
10
Optimal Climbing Flight
19
What is the Fastest Way to Climb from One Flight Condition to
Another?
20
11
• Specific Energy • = (Potential + Kinetic
Energy) per Unit Weight• = Energy Height
Energy Height
Can trade altitude for airspeed with no change in energy height if thrust and drag are zero 21
≡ Energy Height, Eh , ft or m
Specific Energy ≡ Total EnergyUnit Weight
= mgh +mV2 2
mg= h + V
2
2g
Specific Excess Power
dEh
dt=ddt
h+V2
2g!
"#
$
%&=
dhdt+Vg!
"#
$
%&dVdt
Rate of change of Specific Energy
=V sinγ + Vg
⎛⎝⎜
⎞⎠⎟T − D −mgsinγ
m⎛⎝⎜
⎞⎠⎟ =V
T − D( )W
= Specific Excess Power (SEP)
= Excess PowerUnit Weight
≡Pthrust − Pdrag( )
W
22
=VCT −CD( ) 1
2ρ(h)V 2S
W
12
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
23
maxwrt V
SEP(h)[ ]
Subsonic Minimum-Time Energy ClimbObjective: Minimize time to climb to desired
altitude and airspeed
24
• Minimum-Time Strategy:• Zoom climb/dive to intercept SEPmax(h) contour• Climb at SEPmax(h)• Zoom climb/dive to intercept target SEPmax(h) contour
Bryson, Desai, Hoffman, 1969
13
Subsonic Minimum-Fuel Energy ClimbObjective: Minimize fuel to climb to desired
altitude and airspeed
25
• Minimum-Fuel Strategy:• Zoom climb/dive to intercept [SEP (h)/(dm/dt)] max contour• Climb at [SEP (h)/(dm/dt)] max
• Zoom climb/dive to intercept target[SEP (h)/(dm/dt)] max contour
Bryson, Desai, Hoffman, 1969
Supersonic Minimum-Time Energy Climb
Objective: Minimize time to climb to desired altitude and airspeed
26
• Minimum-Time Strategy:• Intercept subsonic SEPmax(h) contour• Climb at SEPmax(h) to intercept matching zoom climb/dive contour• Zoom climb/dive to intercept supersonic SEPmax(h) contour• Climb at SEPmax(h) to intercept target SEPmax(h) contour• Zoom climb/dive to intercept target SEPmax(h) contour
Bryson, Desai, Hoffman, 1969
14
Checklistq Energy height?q Contours?q Subsonic minimum-time climb?q Supersonic minimum-time climb?q Minimum-fuel climb?
27
dEh
dmfuel
= dEh
dtdt
dmfuel
= 1!mfuel
dhdt
+ Vg
⎛⎝⎜
⎞⎠⎟dVdt
⎡
⎣⎢
⎤
⎦⎥
SpaceShipOneAnsari X Prize, December 17, 2003
Brian Binnie, *78Pilot, Astronaut
28
15
SpaceShipOne Altitude vs. RangeMAE 331 Assignment #4, 2010
29
SpaceShipOne State Histories
30
16
SpaceShipOne Dynamic Pressure and Mach Number Histories
31
The Maneuvering Envelope
32
17
• 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
33
Maneuvering Envelopes (V-n Diagrams)for Three Fighters of the Korean War Era
Republic F-84
North American F-86
Lockheed F-94
34
18
Turning Flight
35
• Vertical force equilibrium
Level Turning Flight
L cosµ =W• Load factor
n = LW = L mg = secµ,"g"s
• Thrust required to maintain level flight
µ :Bank Angle
• Level flight = constant altitude• Sideslip angle = 0
36
Treq = CDo+ εCL
2( ) 12 ρV2S = Do +
2ερV 2S
Wcosµ
⎛⎝⎜
⎞⎠⎟
2
= Do +2ε
ρV 2SnW( )2
19
Bank angle
Maximum Bank Angle in Steady Level Flight
Bank angle is limited by
€
µ :Bank Angle
CLmaxor Tmax or nmax
37
cosµ = WCLqS
= 1n
=W 2εTreq − Do( )ρV 2S
µ = cos−1 WCLqS
⎛⎝⎜
⎞⎠⎟
= cos−1 1n
⎛⎝⎜
⎞⎠⎟
= cos−1 W 2εTreq − Do( )ρV 2S
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Turning rate
Turning Rate and Radius in Level Flight
Turning rate is limited by
CLmaxor Tmax or nmax
Turning radius
Rturn =
Vξ=
V 2
g n2 −138
!ξ = CLqS sinµmV
= W tanµmV
= g tanµV
= L2 −W 2
mV
= W n2 −1mV
=Treq − Do( )ρV 2S 2ε −W 2
mV
20
Maximum Turn Rates
“Wind-up turns”
39
• Corner velocity
Corner Velocity Turn
• Turning radius Rturn =V 2 cos2 γ
g nmax2 − cos2 γ
Vcorner =2nmaxWCLmas
ρS
• For steady climbing or diving flight sinγ = Tmax − D
W
40
21
Corner Velocity Turn
• Time to complete a full circle
t2π =V cosγ
g nmax2 − cos2 γ
• Altitude gain/loss Δh2π = t2πV sinγ
• Turning rate
ξ =g nmax
2 − cos2 γ( )V cosγ
41
Checklistq V-n diagram?q Maneuvering envelope?q Level turning flight?q Limiting factors?q Wind-up turn?q Corner velocity?
42
22
Herbst Maneuver• Minimum-time reversal of direction• Kinetic-/potential-energy exchange• Yaw maneuver at low airspeed• X-31 performing the maneuver
43
Next Time:Aircraft Equations of Motion
Reading:Flight Dynamics,
Section 3.1, 3.2, pp. 155-161
44
What use are the equations of motion?How is the angular orientation of the airplane described?
What is a cross-product-equivalent matrix?What is angular momentum?
How are the inertial properties of the airplane described?How is the rate of change of angular momentum calculated?
Learning Objectives
23
Supplemental Material
45
Gliding Flight of the P-51 Mustang
Loaded Weight = 9,200 lb (3,465 kg)
L /D( )max =1
2 εCDo
= 16.31
γ MR = −cot−1 LD
⎛⎝⎜
⎞⎠⎟ max
= −cot−1(16.3) = −3.5°
CD( )L/Dmax = 2CDo= 0.0326
CL( )L/Dmax =CDo
ε= 0.531
VL/Dmax =76.49
ρm/s
!hL/Dmax =V sinγ = − 4.68ρm/s
Rho=10km = 16.31( ) 10( ) = 163.1 km
Maximum Range Glide
Loaded Weight = 9,200 lb (3,465 kg)S = 21.83m2
CDME= 4CDo
= 4 0.0163( ) = 0.0652
CLME=
3CDo
ε=
3 0.0163( )0.0576
= 0.921
L D( )ME = 14.13
!hME = − 2ρ
WS
⎛⎝⎜
⎞⎠⎟
CDME
CLME3/2
⎛
⎝⎜⎞
⎠⎟= − 4.11
ρm/s
γ ME = −4.05°
VME =58.12
ρm/s
Maximum Endurance Glide
46