Geometry of Optimal Coverage for Space-based Targets for Space - - PowerPoint PPT Presentation

Geometry of Optimal Coverage for Space-based Targets for Space based Targets with Visibility Constraints

• Dr. Belinda G. Marchand

Assistant Professor The University of Texas at Austin; The Aerospace Corporation Member of the Technical Staff Senior Member of the Technical Staff Senior & Chris Kobel The Aerospace Corporation Project Leader Senior Project Leader Senior

1

Problem Statement

• Define algorithm for maximizing coverage of

space based targets within the bounds of a pre-specified altitude band.

• Assumptions:

– No visibility below given altitude (tangent height) No visibility below given altitude (tangent height) – Focus only on space based targets – Sensor range pre-defined – Sensor range pre-defined

2

BTH Coverage Problem g

S R

θ

MAX BTH COVERAGE = MAX COVERAGE ANGLE

3

Optimal Satellite Height f M i ATH C for Maximum ATH Coverage

( ) ( )

2 2

2 cos h R h R R R h R φ = + + +

1

sin

e t

R h φ

− ⎛

⎞ + = ⎜ ⎟

( ) ( )

2 cos

s e low e low e

h R h R R R h R φ = + + − + − sin

e t e low

R h φ = ⎜ ⎟ + ⎝ ⎠

4

Maximizing Visibility ithi B d d Altit d R within a Bounded Altitude Range

• Goal: To maximize the area of intersection

between the following curves (2D) or surfaces (3D)

– UTAS Upper Target Altitude Shell – LTAS Lower Target Altitude Shell g – RS Range Shell – TL Tangent Line TL Tangent Line

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Factors Influencing Area Calculation g

• Satellite Altitude
• Separation of UTAS and LTAS
• Size of RS and where it intersects the TL
• Size of RS and where it intersects the TL
• Intersections of TL with UTAS and LTAS

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Step 1: Formulate Conditions & Eqns For Curve Intersections

S

& Eqns. For Curve Intersections

U2B R U1B S U2B L2B L1B U1B B2 B1 A2 T2 T1 A1 ˆ x L2A rt rl L1A U2A ru U1A

A

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Step 2: Computing the Coverage Area as a function of Satellite Altitude as a function of Satellite Altitude

• Initially, if coverage exists at all, the area of coverage can be

h h f R2 ( id UTAS) ( b l thought of as πR2 – (area outside UTAS) – (area below LTAS) – (area below TL)

• Each of these three terms depend on the size of the RS and

h i b h UTAS d h LTAS the separation between the UTAS and the LTAS

• There is no single equation that generally describes the

coverage area. Thus, all special cases must be identified a priori

• The area may be represented as a piecewise continuous

function, but it is a highly nonlinear function.

• Identifying the optimal height is best accomplished by

understanding the geometrical structure of the problem and through adequate numerical analysis.

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Step 3: Identify Special Cases Depending on Location of Critical Intersections Depending on Location of Critical Intersections

S

2B

U

1B

U

2B

L

1B

L

1B 2

T

1

T

2

B

1

B

2A

L

1A

L

2A

L

1A

L

2A

U

1A

U

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Step 4: Identify Simplest Form of Area Equation for each Possible Case for each Possible Case

• There are multiple ways of formulating the same

i diffi l h h area equation, some more difficult than others.

• Divide area calculation into basic shapes

Triangles – Triangles – Arc segments – Circular Sectors

• Computation depends only on Cartesian

coordinates of Primary and Secondary Intersections

• Composite area equation depends only on
• Composite area equation depends only on

elementary components

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Constrained Search Space p

• Satellite MUST be located:

– Above the THS – Below no-coverage altitude:

1.5 2 x 10

4

R

g

( )

3

2 2 2 2 s u t t

r R r r r = + − +

0.5 1

3

s

r

u

r

t

r

• 1
• 0.5
• 2
• 1.5
• 1
• 0.5

0.5 1 1.5 2 x 10

4

• 2
• 1.5

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Intersections of RS with the U/LTAS /

( )

2 2 2

( )

2 2 2 u s

r r R + −

( )

1 1

2 2 2 B B s

x y y R + − =

1 1

2 2 2 B B u

x y r + =

( )

1 2

2

B B s

y y r = =

1 2 2

2 2 B B u B

x x r y = − = −

( )

1 1

2 2 2 A A s

x y y R + − =

2 2 2

x y r + =

( )

1 2

2 2 2

2

l s A A s

r r R y y r + − = =

2 2

1 1

A A l

x y r + =

1 2 1

2 2 A A l A

x x r y = − = −

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Intersection of the TL with the LTAS

The equation for the TL that connects the satellite to the THS is given by,

y y −

( )

s s

y m x x y = − +

Where m denotes the slope of the line,

t s t s

y y m x x − = − r ⎛ ⎞

and

1

sin , cos , and cos .

t t t t t t t t s

r x r y r r θ θ θ

− ⎛

⎞ = = = ⎜ ⎟ ⎝ ⎠

The intersection of the TL with the LTAS is identified from the solution to the f ll i f i

1 / 1 /

2 2 2

A B A B

L L l

x y r + =

( )( ) ( )

1 / 2 /

2 2 2 2 2 2

2 4 4 1 2 1

A B A B

s s s l L L

mr m r m r r x x m − ± − + − = − = +

following system of equations:

( )

m

1 / 1 / A B A B

L L s

y mx r = +

1 / 2 / 1 / A B A B A B

L L L s

y y mx r = = +

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Intersection of the TL with the UTAS

The intersections of the TL with the UTAS are similarly identified through the solution to the following system of equations

1 / 1 /

2 2 2

A B A B

U U u

x y r + =

to the following system of equations,

1 / 1 / A B A B

U U s

y mx r = +

The solution is subsequently identified as,

( )( ) ( )

1 / 2 /

2 2 2 2 2 2

2 4 4 1 2 1

A B A B

s s s u U U

mr m r m r r x x m − ± − + − = − = +

q y ,

( )

2 1 m +

1 / 2 / 1 / A B A B A B

U U U s

y y mx r = = +

1 / 2 / 1 / A B A B A B

U U U s 14

Intersection of the TL with the RS

The intersection of the TL with the RS is identified from the solution to the following system of equations

( )

1 1

2 2 2, T T s

x y y R + − =

system of equations,

1 1

.

T T s

y mx y = +

The solution to the above system is given by,

1 2

2 ,

1

T T

R x x m = − = + y y mx r = = +

1 2 1

.

T T T s

y y mx r = = +

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Sample Area Calculation p

16

Geometrical Components p

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Triangle Area and Semiperimeter g p

• ∆’s are a large component of the coverage area geometry.
• Define the area of a ∆ as a function of the semiperimeter,

“s”, and the sides of the ∆; “a”, “b”, and “c”:

( )

• “s” easily computed from available shell intersections

( ) ,

2 a b c s + + =

• s easily computed from available shell intersections
• Subsequently, the area of a triangular section is given by:

( ) ( )( )( ) ( ) ( )( )( )

, , a b c s s a s b s c = − − −

• Α

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Arc Segments g

• B

B

• B2

B1 cT 3

( )

3 3

2

1 , cos 2 2 2

T u u T u SECTOR TRIANGLE

c r r c r φ φ

Σ

= −

• Α

ru ru

3

1

2sin 2

T u

c r φ

− ⎛

⎞ = ⎜ ⎟ ⎝ ⎠ ru ru φ

19

Area of Intersection B t T Ci l Between Two Circles

Th l t th l ft f th The example to the left focuses on the Intersection of the RS with the LTAS. Note, in each case, the area of intersection is given by the sum of g y the area of two arc segments. However, that equation can vary by a constant factor depending on the geometry

• f the intersection (4 types)
• f the intersection (4 types)

20

Area of Intersection f RS ith L/UTAS

• f RS with L/UTAS

21

Composite Triangles: Type 1 p g yp

L2A T2 A2

( ) ( ) ( ) ( )

R B T B L T L B T B L T L T L R B T + Α Α Α Α Α Α

( ) ( ) ( ) ( )

1

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

, , , , , , , ,

l l

r R B T B L T L B T B L T L r T L R B T

Λ Σ Σ

= − +

• Α

Α Α Α Α Α

22

Composite Triangle: Type 2 p g yp

S L2B L2A

B

( ) ( ) ( )

2

2 1 2 1 2 1 2 1 1 2

, , , , , ,

l l

r L S L L L S L S L L L S r L L

Λ Σ

= −

• Α

Α Α Α Α

( ) ( ) ( )

23

“Teardrop” Sections p

( )

( ) ( ) ( ) ( )

2

2 2 1 2 1 2 1 2 2 2

, , , ; , ,

s

R PP R r PP R r r r R PP R PP R PP R

π Σ

⎧ + < + ⎪ = ⎨ ≥ ⎪ A A A A A

• (

) ( )

2 2 1 2 1 2

, , , ;

s

R PP R r PP R r r

Σ

+ ≥ + ⎪ ⎩A A

• 24

Summary of Special Cases y p

• Primary cases:
• Subcases due to RS Size

– Rt ≤ Rs < Rl – Rl ≤ Rs < Ru

2 2b

T S U S <

– Ru ≤ Rs < Rs3

2 2 2 2 2 2 b b b a

U S T S L S L S T S L S ≤ < ≤ <

• Subcases due to

existence of intersections

2 2 2 2 2 2 b a a a

L S T S U S L S T S ≤ < ≤

existence of intersections

( ) ( )

1 2

0 entry/exit t / it A A B B =

2 2 a

L S T S ≤

( )

1 2

0 entry/exit B B =

25

Area Geometry: S t llit B l UTAS Satellite Below UTAS

26

Area Geometry: S t llit Ab UTAS Satellite Above UTAS

27

28

29

30

Coverage Area Analysis Tool g y

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R = 5000 km, hl = 1000 km, hu = 5000 k d h 100 k km, and ht = 100 km.

4 x 10

7

2)

2 3 4

X: 1353 Y: 4.015e+007

f Coverage (km

2

1000 2000 3000 4000 5000 6000 7000 8000 9000 1 Altitude (km) Area of 40 60 80 (deg) 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 20 40

θout

Altitude (km)

32

hu = 5600 km, hl = 600 km (5000x5000 grid)

u l

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Optimal Altitude Space p p

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Conclusions

• Optimal satellite altitude non-intuitive
• Graphical tools helpful in design process
• Ongoing work

– Multi-objective optimization for constellation design with applications to constrained ATH coverage problem.

• The results of the current investigation represent

useful startup solutions for numerical

• ptimization process
• ptimization process.
• Results also provide physical insight into the

expected trends. p

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