minimum-time and -fuel optimization6. minimum-time and fuel problems mae 546 2018.pptx created date:...

Minimum-Time and -Fuel Optimization !

Robert Stengel! Optimal Control and Estimation MAE 546 !

Princeton University, 2018

Copyright 2018 by Robert Stengel. All rights reserved. For educational use only.http://www.princeton.edu/~stengel/MAE546.html

http://www.princeton.edu/~stengel/OptConEst.html

!! Climbing flight of an airplane!! Energy and power!! Choice of control variable!! Numerical optimization!! Minimum time-to-climb problem!! Minimum fuel-to-climb problem!! Stability and control (supplement)

1

Flight Envelope Determined by Available Thrust

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

2

Energy Height and Specific Excess Power!

3

•! Specific Energy •! = (Potential + Kinetic Energy) per Unit Weight•! = Energy Height

Energy Height

Could trade altitude for airspeed with no change in energy height if thrust and drag were zero

Total EnergyUnit Weight

! Specific Energy =mgh + mV 2 2

mg= h +

V 2

2g! Energy Height, Eh , ft or m

4

Specific Excess Power

dEh

dt= ddt

h + V2

2g!"#

$%&= dhdt

+ Vg

!"#

$%&dVdt

dEh

dt= Specific Excess Power (SEP) = Excess Power

Unit Weight!

Pthrust " Pdrag( )W

dEh

dt=V sin! + V

g"#$

%&'

T ( D( )m

( gsin!)*+

,-.

dEh

dt=V T ! D( )

W=V

CT !CD( ) 12"(h)V 2S

W

•! Specific Power

!V = T ! D( ) m ! gsin"!h =V sin"

5

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

6

Subsonic Energy ClimbObjective: Minimize time or fuel to climb to

desired altitude and airspeed

Approximate Optimal Trajectory produced by a Switching Curve 7

Angle of Attack Control“Point-Mass” Model

!V = Tmax ! CDo+ " CL#

#$% &'2( ) 12 (V 2S$

%)&'*m ! gsin+

!+ = 1V

CL## 12(V 2S,

-./01 m ! gcos+$

%)&'*

!h =V sin!!r =V cos!

!mfuel = " SFC( ) Tmax( )

8

Minimum Time-to-Climb Problem

Vo = 100 m/s; ! o = 0 rad; ho = 0 m (sea level); ro = 0 mInitial and final conditions

!! End time, tf, is open, thrust takes maximum value, and control variable is angle of attack,

!! Fuel expended, but simulated vehicle mass held constant

Vf = 200 m/s; ! f = open; hf = 10,000 m; rf = open

! t( )

9

1,000-sec Trajectory,!Simple Angle of Attack Control

Angle of Attack History

!! No optimization, open-loop control of point-mass model!! Interchange of kinetic and potential energy!! Lightly damped long-period oscillation 10

Altitude vs. Velocity, !Simple Angle of Attack Control,

tf = 1,000 sec

!! Lightly damped, long-period oscillation!! Kinetic/potential energy interchange

11

Alternative: Pitch Angle Control

!! = 1

VCL"

# $!( ) 12%V 2S&

'()*+ m $ gcos!,

-./01

! = " #$

!! = 1

VCL"

12#V 2S$ %CL"

12#V 2S!&

'()*+ m % gcos!,

-./01

Controlling pitch angle introduces flight path angle damping

(see Supplemental Material for details)

! = Angle of Attack, rad" = Pitch Angle, rad# = Flight Path Angle, rad

12

Angle of Attack or Pitch Angle Control?

!!

! =" + #

13

!h =V sin!

1,000-sec Trajectory,!Simple Pitch Angle Control

Note difference in pitch-angle and angle-of-attack profiles

! =" + #

14

Pitch Angle History

!! No optimization, open-loop control of point-mass model!! Inherent damping!! Long-period oscillation does not occur

Altitude vs. Velocity!Simple Pitch Angle Control, !

tf = 1,000 sec!! Increased damping eliminates oscillation!! Pitch angle chosen to produce kinetic energy increase

followed by potential energy increase

15

Minimum Time-to-Climb Optimization Problem!

16

Minimum-Time Cost Function

Terminal cost provides trajectory

objective

J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }

+ 12

1+ xT t( )Qx t( ) + uT t( )Ru t( )"# $%{ }t0

t f

& dt

17

•! Integrand•! 1 is the integrand for minimizing time•! Small quadratic term

•! provides non-singular trajectory control with ad hoc damping and regulation [good], but

•! penalizes non-zero values of state and control [not so good]

V0 = 100 m/s! 0 = 0 radh0 = 0 m (sea level)r0 = 0 m

xdes t f( ) =Vf = 200 m/s

! f = open

hf = 10,000 m

rf = open

"

#

$$$

%

$$$

Minimum-Time Cost Function with Augmented Trajectory Damping

J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }

+ 12

1+ xT t( )Qx t( ) + !xT t( )Q !x !x t( ) + uT t( )Ru t( )"# $%{ }t0

t f

& dt

•! State rate weighting •! is unbiased by non-zero state or control•! provides damping

!x t( ) = f x t( ),u t( )!" #$

J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }

+ 12

1+ xT t( )Qx t( ) + f T x t( ),u t( )"# $%Q !xf x t( ),u t( )"# $% + uT t( )Ru t( )"# $%{ }

t0

t f

& dt18

Weights Used in Optimization

P =

100 0 0 00 20 0 00 0 0.4 00 0 0 0

!

"

####

$

%

&&&&

Q =

10!2 0 0 00 1 0 00 0 0 00 0 0 0

"

#

$$$$

%

&

''''

Q !x =

1 0 0 00 1 0 00 0 0 00 0 0 0

!

"

####

$

%

&&&&

R = 1

Terminal Penalty State Penalty

State-Rate Penalty

Control Penalty

x1 =V : Velocity, m/sx2 = ! : Flight path angle, radx3 = h : Height, mx4 = r : Range, mu = " : Pitch angle, rad

19

•! Optimization algorithm is still minimizing at iteration cutoff

Fixed End-Time Cost History, tf = 575 sec, 36 Iterations

•! Adaptive steepest-descent optimization algorithm

20

•! Ad hoc variation of final time (TBD)

Why is tf = 575 sec Approximately the Minimum-Time Solution?

tf = 600 secJ* = ~750

tf = 550 secJ* = ~3480tf = 575 sec

J* = ~935

Desired final state not achievable with shorter time

21

~Minimum-Time Altitude vs. Velocity, tf = 575 sec, 36 Iterations

Recall Energy-Height Approximation

Kinetic energy increaseTrade for potential energy increase

Velocity Increase and Shallow Dive to satisfy terminal condition

22

Power loss with altitude greater in the simulation than in E-H approximation

~Minimum-Time Trajectory, tf = 575 sec, 36 Iterations

Velocity Flight Path Angle

Height Range

23

Pitch Angle

Angle of Attack

Minimum Fuel-to-Climb Optimization Problem!

24

Minimum-Fuel Cost Function with Augmented Trajectory Damping

J t f( ) = 12 x t f( )! xdes t f( )"# $%TPf x t f( )! xdes t f( )"# $%{ }

+ 12

!m + xT t( )Qx t( ) + !xT t( )Q !x !x t( ) + uT t( )Ru t( )"# $%{ }t0

t f

& dt

•! Same cost-function weights•! Different Integrand

–! Fuel-flow rate in the integrand for minimizing total fuel use

25

Minimum-Fuel Cost History, tf = 575 sec, 36 Iterations

Adaptive optimization algorithm takes two bad iterations

… but regains convergence, with small oscillation in terminal cost

26

Minimum-Fuel Altitude vs. Velocity, tf = 575 sec, 36 Iterations

27

~Minimum-Fuel Trajectory, tf = 575 sec, 36 Iterations

Virtually identical to minimum-time solution

… but less fuel is used. Minimum time: 849 kg. Minimum fuel: 831 kg

Velocity Flight Path Angle

Height Range

Pitch AngleAngle of Attack

28

Minimum-Fuel Adjoint Vector History, "(t), tf = 575 sec, 36 Iterations

!! t f( ) = "#"x

t f( )$%&

'()

T

= *xT t f( )PfCost sensitivity to state perturbations over time

Velocity Adjoint, "1

Flight Path Angle Adjoint, "2

Height Adjoint, "3

Range Adjoint, "4

29

Next Time:!Neighboring-Optimal Control via Linear-Quadratic Feedback!

!

Reading!OCE: Section 3.7 !

30

SSuupppplleemmeennttaall MMaatteerriiaall!!

31

Modal Properties of the System !

32

Linearized Equations for Velocity and Flight Path Angle Perturbations, Using

Angle of Attack as the Control

! !V! !"

#

$%%

&

'((=

TV ) DV( ) )g

LVVo

0

#

$

%%%

&

'

(((

!V!"

#

$%%

&

'((+

~ 0L*Vo

#

$

%%%

&

'

(((!*

TV !! Thrust m( )

!V= Sensitivity of acceleration-due-to-thrust to velocity variation

= 0 for this problem because thrust = maximum thrust" #

DV !! Drag m( )

!V= Sensitivity of acceleration-due-to-drag to velocity variation > 0

State : !x = !V!"

#

$%%

&

'((

Control : !u = !)

LV !! Lift m( )

!V= Sensitivity of acceleration-due-to-lift to velocity variation > 0

L" !! Lift m( )

!"= Sensitivity of acceleration-due-to-lift to angle-of-attack variation > 0

33

Characteristic Equation and Stability, Using Angle of Attack as

the Control Variable

sI! F = sI!!DV !gLVVo

0

"

#

$$$

%

&

'''=

s + DV( ) g

! LV Vos

= s s + DV( ) + g LV Vo= s2 + DVs + g

LVVo

= s2 + 2() ns +) n2 = 0

0 = s2 + 2!" ns +" n2

" n = g LV Vo! 2 g

V0!

13.9Vo m / s( ) ; Period ! 0.453Vo, sec

! = DV

2 g LV Vo

!22

CD

CL

#$%

&'(

Characteristic Equation

Natural Frequency and Damping Ratio

34

"Total Damping"= 2!" n = DV

Linearized Equations for Velocity and Flight Path Angle Perturbations, Using Pitch Angle as the Control

Replace angle of attack by pitch angle for control

State : !x = !V!"

#

$%%

&

'((

Control : !u = !) = !* + !"

! !V! !"

#

$%%

&

'((=

)DV )gLVVo

0

#

$

%%%

&

'

(((

!V!"

#

$%%

&

'((+

~ 0L*Vo

#

$

%%%

&

'

(((!*

! !V! !"

#

$%%

&

'((=

)DV )gLVVo

0

#

$

%%%

&

'

(((

!V!"

#

$%%

&

'((+

~ 0L*Vo

#

$

%%%

&

'

(((!+ ) !"( )

=)DV )gLVVo

) L*Vo

#

$

%%%

&

'

(((

!V!"

#

$%%

&

'((+

~ 0L*Vo

#

$

%%%

&

'

(((!+

35

Characteristic Equation and Stability, Using Pitch Angle as the

Control Variable

sI! F = sI!!DV !gLVVo

! L"Vo

#

$

%%%

&

'

(((=

s + DV( ) g

! LV Vos + L"

Vo( )= s + L"

Vo( ) s + DV( ) + g LV Vo= s2 + L"

Vo+ DV( )s + g LV Vo

+ DVL"Vo( )

= s2 + 2)* ns +* n2 = 0

0 = s2 + 2!" ns +" n2

" n = g LV Vo+ DV

L#Vo( )

! =

L#Vo

+ DV( )2 g LV Vo

+ DVL#Vo( )!! Natural frequency is increased

!! Damping is increased

36"Total Damping"= 2!" n =

L#Vo

+ DV( )

Definitions and Numerical Values for Variables and Constants

V =Velocity, m/s! = Flight path angle, radh = Altitude (or height), mr = Range, mm = Mass, kg" = Angle of attack, rad; "max = 10°# = Air density = #SLe

$%h = 1.225e$h/9,042kg/m3

g = Gravitational acceleration = 9.801 m/s2

Angle of attack has linear effect on lift and quadratic effect on drag

T = TSL!!SL

"#$

%&'(T = TSL e)*h( )(T = Thrust, N;

(T = Throttle setting, %, = 100% for minimum-time/fuel problemSFC = Specific Fuel Consumption = 10 g/kN-sCL = CL+

+ = Lift coefficient = 5.7+

CD = CDo

+ ,CL2( )

1) V Vsound( )2= Drag coefficient

= 0.025 + 0.072CL2( ) 1) V Vsound( )2

S = Reference area = 21.5 m2

mE = Vehicle mass = 4,550 kg ~ constant

37

Supersonic Energy ClimbObjective: Minimize time or fuel to climb to

desired altitude and airspeed

Approximate Optimal Trajectory produced by a Switching Curve 38

Energy State Profile, tf = 570 sec

Energy State : E =1mg

mV 2

2+ mgh

!"#

$%&=V 2

2g+ h, meters

Specific Excess Power :

SEP = Vmg

Thrust ! Drag( )

!! Optimized Energy State is monotonic and always increasing

!! Rate of change decreases with altitude

39