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 for Maximum ATH Coverage f M i ATH C − ⎛ ⎞ R + h 2 ( ( ) ) 2 ( ( ) ) 1 ⎜ ⎜ e e t t ⎟ ⎟ = = + + + + − + + φ φ − φ φ = = h h R R h h R R 2 2 R R R R h h cos cos R R sin sin s e low e low e ⎝ + ⎠ R h e low 4

Maximizing Visibility within a Bounded Altitude Range ithi B d d Altit d R • Goal: To maximize the area of intersection between the following curves (2D) or surfaces (3D) – UTAS � U pper T arget A ltitude S hell – LTAS � L ower T arget A ltitude S hell g – RS � R ange S hell – TL � T angent L ine TL � T angent L ine 5

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 6

Step 1: Formulate Conditions & Eqns For Curve Intersections & Eqns. For Curve Intersections S S R U 2 B U 2 B U 1 B U 1 B B 2 B 1 L 2 B L 1 B A 2 A 1 T 1 T 2 x ˆ r t r l L 2 A L 1 A A r u 7 U 2 A U 1 A

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 thought of as π R 2 – (area outside UTAS) – (area below R 2 h h f ( id UTAS) ( b l LTAS) – (area below TL) • Each of these three terms depend on the size of the RS and the separation between the UTAS and the LTAS h i b h UTAS d h 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. 8

Step 3: Identify Special Cases Depending on Location of Critical Intersections Depending on Location of Critical Intersections S U U 2 B 1 B 1 B B T B T 2 2 1 1 L L 2 B 1 B L L L L 2 A 2 A 1 A 1 A U 9 U 2 A 1 A

Step 4: Identify Simplest Form of Area Equation for each Possible Case for each Possible Case • There are multiple ways of formulating the same area equation, some more difficult than others. i diffi l h h • 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 10

Constrained Search Space p • Satellite MUST be located: – Above the THS – Below no-coverage altitude: g 4 x 10 2 2 ( ) 2 2 2 r = R + r − r + r s u t t 3 1.5 R 1 r s 3 0.5 r u r 0 t -0.5 -1 -1.5 11 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 4 x 10

Intersections of RS with the U/LTAS / ( ( ) ) 2 2 2 r + r − R 2 2 u s ( ( ) ) y = y = 2 2 2 2 x + y − y = R B B B B s 2 r 1 2 1 1 s 2 2 2 x + y = r 2 2 x = − x = r − y B B u 1 1 B B u B 1 2 2 ( ) 2 2 2 r + r − R 2 l s ( ) y = y = 2 2 x + y − y = R A A A A s 2 r 1 2 1 1 s 2 2 2 x x + + y y = = r r 2 2 2 2 x = − x = r − y A A l 1 1 A A l A 1 2 1 12

Intersection of the TL with the LTAS The equation for the TL that connects the satellite to the THS is given by, ( ) = − + y m x x y s s Where m denotes the slope of the line, y y − − y y t s m = x − x t s and − ⎛ ⎛ ⎞ ⎞ r r 1 ⎜ t ⎟ x = r sin θ , y = r cos θ , and θ = cos . t t t t t t t ⎝ ⎠ r s The intersection of the TL with the LTAS is identified from the solution to the following system of equations: f ll i f i ( )( ) 2 2 2 2 2 − 2 mr ± 4 m r − 4 1 + m r − r s s s l 2 2 2 x + y = r = − = x x L L l L L ( ( ) ) 2 1 A B / 1 A B / 1 / A B 2 A B / 2 1 + m m = + = = + y mx r y y mx r L L s L L L s 1 A B / 1 A B / 1 / A B 2 A B / 1 / A B 13

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 to the following system of equations, 2 2 2 x + y = r U U u 1 / A B 1 / A B y = mx + r U U s 1 / A B 1 / A B The solution is subsequently identified as, q y , ( )( ) 2 2 2 2 2 − 2 mr ± 4 m r − 4 1 + m r − r s s s u x = − x = U U ( ( ) ) 2 1 / A B 2 A B / 2 1 2 1 + + m m y = y = mx + r U U U U U U s s 1 / 1 / A B A B 2 2 A B A B / / 1 / 1 / A B A B 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 system of equations, 2 ( ) 2 2 , x + y − y = R T T s 1 1 y = mx + y . T T s 1 1 The solution to the above system is given by, R x = − x = 2 , T T 1 2 + 1 m y y = = y y = = mx mx + + r r . T T T s 1 2 1 15

Sample Area Calculation p 16

Geometrical Components p 17

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”: ( ( ) ) , a + b + c s = 2 • “s” easily computed from available shell intersections s easily computed from available shell intersections • • Subsequently, the area of a triangular section is given by: ( ( ) ) ( ( )( )( )( )( ) ) Α a b c , , = s s − a s − b s − c � 18

Arc Segments g �������� �������� �������� �������� ���������� ���������� �������� �������� B 2 B B B 1 c r 1 φ ���������� ���������� ( ) � T u 2 c T 3 Α r c , = φ r − 3 cos Σ u T u 2 2 2 3 � � �� � SECTOR TRIANGLE − ⎛ ⎞ c T 1 ⎜ ⎟ φ = 2sin 3 ⎝ ⎠ 2 r u r u r u r u r u φ 19

Area of Intersection B t Between Two Circles T Ci l Th The example to the left focuses on the l t th l ft f th 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 of the intersection (4 types) of the intersection (4 types) 20

Area of Intersection of RS with L/UTAS f RS ith L/UTAS 21

Composite Triangles: Type 1 p g yp L 2 A T 2 A 2 ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) Α Α Α Α r R B T , , R B T , B L B L , T L T L = Α Α Α Α B T B T , B L B L , T L T L − Α Α r T L , T L + + Α Α R B T R B T , � Λ l 2 2 2 2 2 2 2 2 2 2 2 2 Σ l 2 2 Σ 2 2 1 22

Composite Triangle: Type 2 p g yp S L 2 A L 2 B B ( ( ) ) ( ( ) ) ( ( ) ) Α Α r L S , , L L , L S = Α Α L S , L L , L S − Α r L L , � Λ l 2 1 2 1 2 1 2 1 Σ l 1 2 2 23

“Teardrop” Sections p ⎧ ( ) ( ) 2 2 A + A < + R PP , , R r PP , ; R r r ⎪ � 1 2 Σ 1 2 s ( ) = ⎨ A r R PP , , π 1 2 � ( ( ) ) ( ( ) ) 2 ⎪ ⎪ 2 2 2 2 ⎩ A A + A A ≥ ≥ + R PP R PP , , R R r PP , PP ; R R r r � 1 2 Σ 1 2 s 24

Summary of Special Cases y p • Primary cases: • Subcases due to RS Size – R t ≤ R s < R l – R l ≤ R s < R u T S < U S 2 2 b – R u ≤ R s < R s3 U S ≤ T S < L S 2 b 2 2 b • Subcases due to L S ≤ T S < L S existence of intersections existence of intersections 2 2 b b 2 2 2 2 a a ( ) A A = 0 entry/exit L S ≤ T S < U S 1 2 2 a 2 2 a ( ( ) ) B B B B = 0 entry/exit 0 t / it L S L S ≤ ≤ T S T S 1 2 2 a 2 25

Area Geometry: S t llit B l Satellite Below UTAS UTAS 26

Area Geometry: S t llit Ab Satellite Above UTAS UTAS 27

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