Flight and Orbital Mechanics - TU Delft OCW Flight and Orbital Mechanics 2 | Content •Introduction •Summary previous lecture •Performance diagram •Optimal climb •Solution

1Challenge the future

Flight and Orbital Mechanics

Lecture slides

Semester 1 - 2012

Challenge the future

DelftUniversity ofTechnology

Flight and Orbital MechanicsLecture hours 3, 4 – Minimum time to climb

Mark Voskuijl

2AE2104 Flight and Orbital Mechanics |

Content

• Introduction

• Summary previous lecture

• Performance diagram

• Optimal climb

• Solution low speed aircraft

• Energy height

• Solution high speed aircraft (M>1)

• Example exam question

• Summary


Content

• Introduction



• Optimal climb


• Energy height



• Summary


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


15000

11000

9000

0

Altitude [m]

Subsonic

M = 1.2(Supersonic)

M = 2

M = 1.5

IntroductionSolution


Content

• Introduction



• Optimal climb


• Energy height



• Summary


Summary previous lecture

• A typical climb (civil subsonic aircraft) 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

W dVT D W

g dt

L W


Summary previous lecture

• 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


Content

• Introduction



• Optimal climb

• Energy height




• Summary


Performance diagram

P

V

Pa

a rsteady

P PRC

W

Pr


Aerodynamic forces (year 1) Lift – Drag polar

0

2

LD D

CC C

Ae


Aerodynamic forces (year 1) Lift

0

2

LD D

CC C

AeL W

Drag

So, one part of the drag decreases (!) with airspeed (1/V2) and one part increases with airspeed (V2)

212DD C V S

0

22 21 1

2 2L

D

CD C V S V S

Ae

0

22 21 1

2 22 2 4

4 1 1D

WD C V S V S

S V Ae

0

221

2 212

D

WD C V S

Ae V S

212LC V S W

2

2 1L

WC

S V


Aerodynamic forces (year 1) Drag as a function of airspeed

Aircraft are quite unique in the sense that drag increases when airspeed decreases!

0

0

210 2

2

212

i

D

i

D D D

D C V S

WD

Ae V S



P

V

Pa

Pr

V1 V2

V3

a rsteady

P PRC

W

V3

V1 V2

V4

RCsteady

V

Determine magnitude of difference and divide by aircraft weight (W)


Performance diagramHow does it change with altitude?

Note: this sheet is from the first year lecture


P

V

Pa

Pr

a rsteady

P PRC

W

V3 V4

RCsteady

V

H2

H1


Performance diagramRate of climb as a function of airspeed and altitude


Performance diagram

H

V

RC = 6 m/s

RC = 2 m/s

Rate of climb as a function of airspeed and altitude


Content

• Introduction



• Optimal climb

• Energy height




• Summary


Optimal climb

• During climb at constant indicated airspeed, the aircraft is

accelerating. V(H) is fixed

• Easy flight technique for the pilot

• But is this optimal???

• What is optimal?


Optimal climb

• Minimum time to climb (time)

• Minimum fuel consumed during climb (fuel)

• Distance covered during climb (distance)

• Noise during climb (noise)

What is optimal?


Optimal climb

• Minimum time to climb (time)

• Minimum fuel consumed during climb (fuel)

• Distance covered during climb (distance)

• Noise during climb (noise)

We will focus on the first option, but the other three are optimal as

well

What is optimal?


Optimal climb

• Objective: to find the true airspeed as a function of altitude that

yields the minimum time to climb

dH dHRC dt

dt RC

2

1

H

H

dHt dt

RC

0

1

1steady

RC

V dVRC

g dH

2

1

0

1H

steadyH

V dV

g dHt dH

RC

• Time t is not minimal if the integrand is

minimal at every altitude H because the

term dV/dH is in the integrand

• Variational calculus is necessary


Variational calculus?

Choosing maximum RCsteady at sea level seems like a good starting point

2

1

0

1H

steadyH

V dV

g dHt dH

RC

Minimize integrand:• RCsteady maximum• dV/dH <0

RCsteady max

dV/dH<0

• At H = 0, RCst is maximal at V = 100

• The integrand can be minimized more by choosing (dV/dH)1 < 0

• At altitude H > 0 no optimum V can be chosen

• Conclusion: No local optimum but global optimum (consider complete flight path)


Optimal climb

Solution

• Simplify (low speed aircraft)

• Energy method (high subsonic / supersonic aircraft)

2

1

0

1H

steadyH

V dV

g dHt dH

RC


Content

• Introduction



• Optimal climb


• Energy height



• Summary


Low speed aircraft


Low speed aircraft

0

0 0

1H H

st

V dV

g dHdHt dH

RC RC

0

H

st

dHt

RC

Assumption: Low speed, low altitude dV/dH is small

Choose maximum steady rate of climb at each altitude


Low speed aircraft

H1H2

V

P RCsteady H

V

sin

a r

a rsteady

W dVT D W

g dt

P P W dVRC V

W g dt

P PRC

W


Low speed aircraft

a r

st

P PRC

W

can be assumed constant as function of airspeed

for propeller aircraft

aP

0,min ,max

3st r L opt DRC P C C Ae

Solution – propeller aircraft

Pa

Pr

V

Pr,min

,

2 1 1opt

L opt

WV

S C

0

Corresponding constant

constant because

e

i i e

V V

V V V

Note: this optimal CL was determined in the first year lectures!


Low speed aircraft

• Conclusion: if the pilot selects the airspeed for the maximum

steady rate of climb at the ground and if he / she keeps the

indicated airspeed constant, then the climb will be optimal

• Optimal in this case means minimum time to climb (assuming

dV/dH is small)

Summary


Low speed aircraftStory – World War I


Low speed aircraft

• Allied airfoils (British / French) versus German airfoils

Story – World War I

• Thin airfoils have low CL,max

• Therefore, German aircraft

were able to fly slower

British / French) versus

German airfoils


Low speed aircraftStory – World War I

Vmin

(thin profiles)Vmin

(thick profiles)

0

,

0 ,

3

0.035

6

0.8

1.26 (large !)

2 1

oL D

D

L opt

E ground

L opt

C C Ae

C

A

e

C

WV V

S C

Pa

Pr

Numerical example


Content

• Introduction



• Optimal climb


• Energy height



• Summary



Energy height

• Altitude and flight speed are rapidly interchangeable (exchange

of kinetic energy and potential energy)

• Increasing the total energy is much slower because it depends

on the excess power

2 21 12 2

WEnergy mV mgH V WH

g

V

H

Concept to solve the minimum time to climb problem

Line of constant energy


Energy height

2

2

e

e

EnergyH

W

VH H

g

A

C

D

B


Energy height

Large energy

Small energy

Optimal combination of V and H to increase the energy

V

H

So: “Minimum time to climb” problem approximated by another problem: find Vopt (H) at which minimum time to total energy


Energy height

210 2

E mg H mV

0

sinW dV

T D Wg dt

e a rsteady

dH P PRC

dt W

Equation of MotionEnergy height

212

0

WE WH V

g

2

02e

E VH H

W g

2

0

1

2

edH dH dV

dt dt g dt

0

1sin

dV T D

g dt W

0

sin a rP PV dVV

g dt W

2

0

1

2

a rP PdV dH

g dt dt W

Energy method


Minimum time to climb

es

dHRC

dt

e

st

dHdt

RC

2

1

(time to energy height)

e

e

H

e

stH

dHt

RC


Optimum path subsonic aircraft

(for each constant energy height line, find the maximum steady rate of climb)


Content

• Introduction



• Optimal climb


• Energy height



• Summary


Solution high speed aircraft

Altitude h

Exc

ess

pow

er,

ft/s

ec

h1<h2<h3

h1

h3

h2

Mach


Supersonic aircraft

• Dip in excess power is caused

by transonic drag rise:

supersonic aircraft

High subsonic aircraft

0

2

D D LC C M k M C





• Mig-29: Climb from sea level to 6000 m in less than 1 minute

Example performance


Content

• Introduction



• Optimal climb


• Energy height



• Summary


Example exam question


Example exam question


Content

• Introduction



• Optimal climb


• Energy height



• Summary


Summary


Questions?

Flight and Orbital Mechanics - TU Delft OCW Flight and Orbital Mechanics 2 | Content •Introduction •Summary previous lecture •Performance diagram •Optimal climb •Solution

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