Metric Distances 28 Great Circle Distances North Pole (90N lat) - - PowerPoint PPT Presentation

Metric Distances

28

Great Circle Distances

W

0º Equator (0º lat) Greenwich (Prime) Meridian (0º lon)

N S E

North Pole (90ºN lat) South Pole (90ºS lat) International Dateline (180º lon) Latitude (y) Longitude (x) (lon, lat) = (x, y) = (140ºW, 24ºN) = (–140º, 24º)

(Meridian) (Parallel)

13.35 mi

R

(x2,y2) (x1,y1) B C A a c b North Pole Prime

l

• n

2

= x

2

l

• n

1

= x

1

lat2 = y2 l a t

1

= y

1

Equator Meridian

29

Circuity Factor

30

road road 1 2 1 2

: , where usually 1.15 1.5 ( , ), estimated road distance from to

i i

GC GC

d Circuity Factor g g d d g d P P P P     



2-D Euclidean Distance

             

                                          

2 2 1 1,1 2 1,2 1 2 2 2 1 2,1 2 2,2 3 2 2 1 3,1 2 3,2

1 1 2 3 , 7 1 4 5 x p x p d d x p x p d x p x p x P d

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Minisum Distance Location

   

2 2 1 ,1 2 ,2 3 1 * * *

1 1 7 1 4 5 ( ) ( ) ( ) x arg min ( ) ( )

i i i i i

d x p x p TD d TD TD TD



               



x

P x x x x x

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120°

Fermat’s Problem (1629):

Given three points, find fourth (Steiner point) such that sum to others is minimized (Solution: Optimal location corresponds to all angles = 120°)

Minisum Weighted-Distance Location

• Solution for 2-D+ and

non-rectangular distances:

– Majority Theorem: Locate NF at EFj if – Mechanical (Varigon frame) – 2-D rectangular approximation – Numerical: nonlinear unconstrained optimization

• Analytical/estimated derivative

(quasi-Newton, fminunc)

• Direct, derivative-free (Nelder-

Mead, fminsearch)

1 * * *

number of EFs ( ) ( ) arg min ( ) ( )

m i i i

m TC w d TC TC TC



   



x

x x x x x

Varignon Frame

1

, where 2

m j i i

W w W w



 

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Convex vs Nonconvex Optimization

34