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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 (k1), i.e.

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

Let

m= y2y1 ,c=y1mx1,Q=(1+1/m2).
x2x1

If 0c1 then

if  km+c0calculate d=cQ
if  0km+c1calculate d=k |m| Q
if  km+c1calculate d=(1c)Q

If c0 then

if  0km+c1calculate d=(km+c)Q
if  km+c1calculate d=Q

If c1 then

if  0km+c1calculate d=(1kmc)Q
if  km+c0calculate 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): 0–0.20.2–0.40.4–0.60.6–0.80.8–1.0 1.0–1.21.2–1.41.4–
Frequency: 02114289 7261291

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

d (in units): 0–0.20.2–0.40.4–0.60.6–0.80.8–1.0 1.0–1.21.2–1.41.4–1.61.6–1.81.8–
Frequency: 0371436 2811394091110

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