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

Metric Distances 28

Great Circle Distances North Pole (90ºN lat) North Pole C Prime (Meridian) Meridian b a International Dateline (180º lon) Latitude ( y ) Longitude ( x ) (Parallel) ( lon , lat ) = ( x , y ) A ( x 1 , y 1 ) = (140ºW, 24ºN) N = ( – 140º, 24º) B c W E ( x 2 , y 2 ) l a t lat 2 = y 2 0 º Equator (0º lat) 1 S = y 1 x 1 = Equator n Greenwich (Prime) 1 o Meridian (0º lon) l x n = l o 2 2 South Pole (90ºS lat) 13.35 R mi 29

Circuity Factor d     road i Circuity Factor : g , where usually 1.15 g 1.5 d GC i   d g d ( P P , ), estimated road distance from to P P road GC 1 2 1 2 30

2-D Euclidean Distance   1 1        x 2 3 , P 7 1  4 5         2 2    x p x p   1 1,1 2 1,2   d 1         2 2      d d x p x p   2 1 2,1 2 2,2     d       3 2 2    x p x p   1 3,1 2 3,2   31

Minisum Distance Location   1 1     P 7 1    4 5       2   2 d ( ) x x p x p i 1 i ,1 2 i ,2 3   TD ( ) d ( ) x x i  i 1  * x arg min TD ( ) x x 120°  * * TD TD ( ) x 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°) 32

Minisum Weighted-Distance Location • Solution for 2-D+ and non-rectangular distances:  m number of EFs – Majority Theorem: Locate NF at m   TC ( ) x w d ( ) x m W   i i  EF j if , where w W w  i 1 j i 2  i 1  * x arg min TC ( ) x – Mechanical (Varigon frame) x – 2-D rectangular approximation  * * TC TC ( x ) – Numerical: nonlinear unconstrained optimization • Analytical/estimated derivative (quasi-Newton, fminunc ) • Direct, derivative-free (Nelder- Mead, fminsearch ) Varignon Frame 33

Convex vs Nonconvex Optimization 34