Flight and Orbital Mechanics
Transcript of Flight and Orbital Mechanics
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1Challenge the future
Flight and Orbital Mechanics
Lecture slides
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Semester 1 - 2012
Challenge the future
DelftUniversity ofTechnology
Flight and Orbital MechanicsLecture hours 1,2 – Unsteady Climb
Mark Voskuijl
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Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash Boeing 727 (1974)
• Analytical solution
• Example exam question
• Summary
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IntroductionQuestion
What is the most efficient way (minimum time) to go
from take-off at sea-level to Mach 1.5 at 15,000 m?
A. Climb at airspeed for max . At 15,000 m accelerate to Mach 2
B. Climb at airspeed for max RC. At 15,000 m accelerate to Mach 2
C. Climb at airspeed for max RC to 11,000 m. Descent and accelerate to Mach 1.2 at 9,000 m, climb and accelerate to Mach 2 at 11,000 m. Decelerated climb to 15,000 m, Mach 1.5
D. Accelerate at sea-level to Mach 1.5, climb to 15,000 m at airspeed for max RC
A. Climb at airspeed for max . At 15,000 m accelerate to Mach 2
B. Climb at airspeed for max RC. At 15,000 m accelerate to Mach 2
C. Climb at airspeed for max RC to 11,000 m. Descent and accelerate to Mach 1.2 at 9,000 m, climb and accelerate to Mach 2 at 11,000 m. Decelerated climb to 15,000 m, Mach 1.5
D. Accelerate at sea-level to Mach 1.5, climb to 15,000 m at airspeed for max RC
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15000
11000
9000
0
Altitude [m]
Subsonic
M = 1.2(Supersonic)
M = 2
M = 1.5
IntroductionSolution
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Introduction
Typical problem AE1102
• What is the maximum rate of climb of Aircraft X at a given
altitude?
Typical problem AE2104
• What is the minimum time to climb from altitude A to altitude B
for Aircraft X?
Difference with AE1102 – Flight mechanics
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IntroductionDifference with AE1102 – Flight mechanics
V
Point performance versus Path performance
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Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash B727
• Analytical solution
• Example exam question
• Summary
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How does a pilot perform a climb?
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How does a pilot perform a climb?
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How does a pilot perform a climb?
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How does a pilot perform a climb?Airspeed indicator
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How does a pilot perform a climb?
212t sp p V
2102t s Ep p V
2 2 01 102 2 E EV V V V
Airspeed indicator
E IV V
For low Mach numbers:
One equation, two unknowns… assume sea level conditions:
Relationship true airspeed – equivalent airspeed:
The indicated airspeed is almost the same as the equivalent airspeed (instrument errors)
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Large difference!!!
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How does a pilot perform a climb
• Climb at constant indicated airspeed and constant power
setting
• True airspeed is therefore increasing
• The climb is unsteady
Summary
0dV
dt
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Typical climb profile
Altitude
Distance
3 km
19 km
Conclusion: very faint curvature; the climb is almost a straight line
So, the climb is quasi-rectilinear
0d
dt
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Summary
• A typical climb is performed at constant indicated airspeed and at
a constant power setting. Therefore, the true airspeed is actually
increasing. Since airspeed is not constant, it is an unsteady
climbing flight
• The climb is almost a straight line. It is therefore a quasi-rectilinear
flight
0dV
dt 0
d
dt
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Is this a problem?
• Pilot flies at minimum airspeed
max
min
2 1
L
WV
S C
max
,min min
0 0
2 1e
L
WV V
S C
• Minimum airspeed depends on altitude
• Equation for minimum equivalent airspeed
• Minimum airspeed seen by pilot is independent of altitude!
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Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash B727, 1974
• Analytical solution
• Example exam question
• Summary
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Equations of motion
Free Body Diagram Kinetic Diagram
T
Zb
Ze
Xe
V
Zb
Ze
Xe
dmV
dt
dVm
dt
Xb Xb
D
V
L
W
TReminder: Angle of attack; angle between aircraft body axis and airspeed Flight path angle; angle between airspeed vector and horizon Pitch attitude; angle between horizon and aircraft body axisLift is by definition perpendicular to airspeedDrag is parallel to the airspeedThrust is fixed to the aircraft and therefore has an angle of attack T with respect to the airspeed
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Equations of motion
/ /: cos sin
: cos sin
TV
TV
F ma
dVF T D W m
dt
dF L W T mV
dt
0
Unsteady quasi-rectilinear climbing flight
Extra assumption: T = 0,Approximation: cos 1, sin 0
= 0= 1
= 1
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Equations of motion
sindV
T D W mdt
L W
sindV
TV DV WV mVdt
Unsteady climb
a r
W dV dHP P V W
g dt dt
Rewrite to power equation by multiplying with airspeed
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22AE2104 Flight and Orbital Mechanics |
Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash B727, 1974
• Analytical solution
• Example exam question
• Summary
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Story – Crash Boeing 727, 1974
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Story – Crash Boeing 727, 1974
• B727 Northwest Orient
• 1 Dec 1974
• Flight J.F. Kennedy Airport – Buffalo
• Checklist: Pitot heaters off
• 16000 ft IAS = 305 kts RC = 2500 ft/min
• >16000 ft IAS > 340 kts RC = 5000 ft/min
• Comment pilot “We ‘re light”
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Story – Crash Boeing 727, 1974
• 23000 ft IAS = 405 kts RC = 6500 ft/min
• Overspeed horn: “Pull back and let her climb”
• Stick shaker: “Mach Buffet”
“Guess we’ll have to pull her up further”
• 24800 ft stall @ = 30°
• Still pulling Deep stall
• Horizontal stabiliser damaged
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Story – Crash Boeing 727, 1974
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Story – Crash Boeing 727, 1974
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Story – Crash Boeing 727, 1974
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Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash B727, 1974
• Analytical solution
• Example exam question
• Summary
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Analytical solution
What is the difference between the climb performance in steady flight and in unsteady flight?
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Analytical solution
Steady Climb
Equation of motion // V
0 sinT D W sinW dV
T D Wg dt
0 sina rP P WV
sina r
W dVV P P WV
g dt
V dV V dV dh V dVRC
g dt g dh dt g dh
sina rP P V dVV
W g dt
Multiply with airspeed
Intermezzo
Multiply with airspeed
sina rst
P PV RC
W
a rst
P PRC
W
st
V dVRC RC
g dt
Unsteady climb
Equation of motion // V
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Analytical solution
st
V dVRC RC RC
g dh
Unsteady climb continued
1
1st
RC
V dVRC
g dh
result
What does this actually mean physically?
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Analytical solution
• The excess power is partially used to accelerate (kinetic energy)
and partially to climb (potential energy)
• The rate of climb in an unsteady climb is therefore smaller than in a
steady climb.
1; 0
1st
RC dV
V dVRC dh
g dh
This is called the kinetic energy correction factor
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34AE2104 Flight and Orbital Mechanics |
Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash B727, 1974
• Analytical solution
• Example exam question
• Summary
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Example exam question
An aircraft carries out a quasi-rectilinear symmetrical climbing flight at constant EAS in the
troposphere of the ISA
Calculate at flight altitude H = 10 km the ratio between the actual rate of climb in the unsteady
climbing flight and the rate of climb in the steady climbing flight at an instantaneous Mach
number of M = 0.8 (attention M is not constant)
Carefully derive the “kinetic energy correction factor” for this flight using the equations of
motion in unsteady flight.
Given for the troposphere (ISA):
0 1
0 0
0
1 ; ;
0.0065 [K/m]
g
RHdp g dH p RT
T
dT
dH
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Solution (1/2)
0
1
1st
RC
V dVRC
g dH
Constant equivalent airspeed
0
EV V
In an exam you will be most likely be asked to derive this ratio first by deriving the equations of motion
So the derivative in the kinetic energy correction factor depends on atmospheric properties
Gas law:
p RT
0 0RT
d dp
dH dH
0
Hydrostatic equation
dpg
dH
0
0 0 0
dR gdT
dH p dH p
2
0 0
11 1
2
V dV dV
g dH g dH
2 0
0
11
2
EdV
g dH
02
0
12
E
dV
g dH
0 0
2
dT dppR RT
dH dHp
0 0
2
R RTdT dp
p dH dHp
0 points
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Solution (2/2)
2 2
0 0E E0
0
So the correction factor becomes
V V1+ 1+ 1
2g 2
gR dT R dT
p dH p p g dH
2 2
0
Instantaneous Mach number is given; so rewrite:
EV V
2 20E
0 0
V1+ 1 1 1
2 2
R dT M R dT
p g dH g dH
2 2
Fill in the given values:
0.8 1.4 287.051 1 1 0.0065 1 1.36
2 2 9.81
M R dT
g dH
1
0.731.36st
RC
RC
So, the actual rate of climb in an unsteady climb is actually only 73% of the maximum
achievable rate of climb in steady flight (for this flight condition)
2 2
0 0
1M RT M p
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Content
• Introduction
• How does a pilot perform a climb?
• Equations of motion
• Story – Crash B727, 1974
• Analytical solution
• Example exam question
• Summary
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Summary
• A typical climb is performed at constant indicated airspeed and at
a constant power setting. Therefore, the true airspeed is actually
increasing. Since airspeed is not constant, it is an unsteady
climbing flight
• The climb is almost a straight line. It is therefore a quasi-rectilinear
flight
• Corresponding equations of motion:sin
dVT D W m
dt
L W
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Summary
• The rate of climb in an unsteady climb is smaller than in a steady climb
because part of the excess energy is used to accelerate
• You must be able to derive this ratio from the equations of motion
• You must be able to calculate this ratio (see example exam question)
• For more background information: read Ruijgrok – Elements of
airplane performance section 14.2
1; 0
1st
RC dV
V dVRC dh
g dh
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Questions?