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Appendix 5: Determination of the Mean Length of a Ley

Let P₁=(x₁,y₁), P₂=(x₂,y₂) be two points chosen at random inside a k×1 rectangle (k≥1), i.e.

x1,x2 are random numbers between 0 and k y1,y2 are random numbers between 0 and 1

Let

m= y2−y1 , c=y1−mx1, Q=√(1+1/m2). x2−x1

If 0≤c≤1 then

if  km+c≤0 calculate d=cQ if  0≤km+c≤1 calculate d=k |m| Q if  km+c≥1 calculate d=(1−c)Q

If c≤0 then

if  0≤km+c≤1 calculate d=(km+c)Q if  km+c≥1 calculate d=Q

If c≥1 then

if  0≤km+c≤1 calculate d=(1−km−c)Q if  km+c≤0 calculate d=Q

For any pair of selected points P₁, P₂, one of the above conditions is satisfied, thus leading to a value of d=ley length.  (Referring to Fig. 1 of the Introduction, if P₁ is P and P₂ is Q, then d=length AB.)

The experiment is repeated for 1000 pairs of points P₁, P₂, and the results classified into class intervals of length.  The results are interesting:

For k=1, a square 1 unit by 1 unit:

d (in units):	0–0.2	0.2–0.4	0.4–0.6	0.6–0.8	0.8–1.0	1.0–1.2	1.2–1.4	1.4–
Frequency:	0	2	11	42	89	726	129	1

For k=1.5, a rectangle 1 unit by 1.5 units:

d (in units):	0–0.2	0.2–0.4	0.4–0.6	0.6–0.8	0.8–1.0	1.0–1.2	1.2–1.4	1.4–1.6	1.6–1.8	1.8–
Frequency:	0	3	7	14	36	281	139	409	111	0

Notice the two peaks for frequency in the second case.  This happens whenever k>1, one peak being about d=1 unit and the other about d=k units. 

The mean ley length, L, can be computed either from a class distribution, such as given above, or, more exactly, by summing the 1000 d-values as they are calculated, and dividing by 1000 at the end. 

2014: For a theoretical discussion, see the update to this appendix