Complexity of the Geological Framework and Use of Mathematics - Geomathematics: Theoretical Foundations, Applications and Future Developments - page 34

Fig. 1.20 Artificial
example of polygon
bounding region with events
at points C i ( i ¼ 1,2,3)
(Source: Agterberg 1994 ,
Fig. 3)

A 3

A 4

A 6

A 2

θ 34

C 3

A 5

C 1

θ 11

A 7

C 2 θ 27

A 8

A 1

and Diggle ( 1991 , 1993 ). In it the study region is approximated by a concave
or self-intersecting polygon with m sides. Each circle through two points
( cf . Fig. 11.19 ) intersects this polygon in 2 k 2 m sides. Once these points of

intersection have been determined, it is relatively easy to calculate the value of w ij
for each of the n ·( n -1) circles. This result also can be obtained by successively
determining for each side whether it has 0, 1 or 2 points of intersection with one of
the circles.
An artificial example with relatively few events is given in Fig. 1.20 for
clarification. It is assumed that the n events occur at points labeled C i ( i ¼ 1,

2, | C j C i |. The vertices

of the polygon are ordered moving in the clockwise direction. Each point C i can be
related to each side A k A k+1 by a triangle. The angle opposite the side will be written
as

... , n ). The distance between two events satisfies r ij ¼ | C i C j | ¼

A k C i A k+1 . This angle is defined to be positive if the entire triangle or a
portion of it bounded by the side belongs to the study area; negative otherwise. The
sign of the angle will be written as s (

ʸ ik ¼ ∠

ʸ ik ) ¼ ʸ ik /| ʸ ik |. Two positive angles ( ʸ 11 , ʸ 27 )

and one negative angle ( ʸ 24 ) are shown in Fig. 1.20 for example. In practice, the

angle A k C i A k+1 can be determined as the difference between two angles

measured in the clockwise direction from a given direction. Note that

ʸ ik ¼ ∠

ʸ ik < ˀ ( i ¼ 1,

2, ... , n ; k ¼ 1, 2, ... , m ) because events occur inside (and not on) the boundary, and

Σ k ʸ ik ¼ , n ) because negative angles, if they occur, are cancelled out

by a surplus of positive angles.
It is convenient to rewrite A k +1 as B k for k

2 ˀ ( i ¼ 1, 2, ...

¼ 1, 2, ... , m 1 setting B m ¼ A 1 . Each

triangle then has three sides that can be written as a ik ¼ | C i B k |, b ik ¼ | C i A k |, and as

c ik ¼

| A i B k |. For individual triangles, the double subscripts can be deleted without
creating confusion because each triangle is determined fully by its three corner