Talk:Ley line/Archive 1

Here's a handwaving order-of-magnitude argument for the probability of ley lines.

Consider a set on n points in an area with approximate diameter d. Consider a valid line to be one where every point is within distance w/2 of the line (that is, lies on a track of width w.

Consider all the unordered sets of k points from the n points, of which there are

$$ \frac {n!} {(n-k)!k!} $$

What is the probability that any given set of points is co-linear in this way? Let's very roughly consider the line between the "leftmost" and "rightmost" two points of the k selected points (for some arbitary left/right axis: we can choose top and bottom for the exceptional vertical case). For each of the remaining points, the probability that the point is "near enough" to the line is roughly w/d. Write &epsilon; = w/d

So, the expected number of k-point ley lines is very roughly

$$ \frac {n!} {(n-k)!k!} \epsilon^{k-2}$$

So, set w = 50m, d = 30 km, (pencil line on an OS map) thus &epsilon; = 0.001666...

Then, we can tabulate the expected number of k-point lines found in a single map as follows:

w = 50m, d = 30 km

k      n      expected points

3 from 10:    0.20 4 from 10:    0.00 5 from 10:    0.00 6 from 10:    0.00 7 from 10:    0.00

3 from 20:    1.90

4 from 20:    0.01 5 from 20:    0.00 6 from 20:    0.00 7 from 20:    0.00

3 from 50:   32.67 4 from 50:    0.64 5 from 50:    0.01 6 from 50:    0.00 7 from 50:    0.00

3 from 60:   57.03 4 from 60:    1.35 5 from 60:    0.03 6 from 60:    0.00 7 from 60:    0.00

3 from 70:   91.23 4 from 70:    2.55 5 from 70:    0.06 6 from 70:    0.00 7 from 70:    0.00

3 from 100: 269.50 4 from 100:  10.89 5 from 100:   0.35 6 from 100:   0.01 7 from 100:   0.00

3 from 200: 2189.00 4 from 200: 179.68 5 from 200:  11.74 6 from 200:   0.64 7 from 200:   0.03

Notice the rapid increase of the number of expected lines with n.

Ley line proponents generally view five point ley lines as definitive proof that a line is real. Note that as there are roughly 400 OS maps covering the UK, if there were 50 points in each, there would be an expected 256 4-point leys and 4 five-point leys found nationally.

Increasing the value of w dramatically increases the number of expected lines on a single map, with effects that increase with the value of k.

Increasing the size of a map will increase d, thus reducing &epsilon;, but will increase n: for example, if a 30 km map has 50 points, a 60 km map with the equivalent density of significant points will have 200, although it will have a value of &epsilon; that is half of that for the 30 km map.

w = 50m, d = 60 km

3 from 100: 134.75 4 from 100:   2.72 5 from 100:   0.04 6 from 100:   0.00 7 from 100:   0.00

3 from 150: 459.42 4 from 150:  14.07 5 from 150:   0.34 6 from 150:   0.01 7 from 150:   0.00

3 from 200: 1094.50 4 from 200:  44.92 5 from 200:   1.47 6 from 200:   0.04 7 from 200:   0.00

3 from 250: 2144.17 4 from 250: 110.34 5 from 250:   4.52 6 from 250:   0.15 7 from 250:   0.00

3 from 300: 3712.58 4 from 300: 229.72 5 from 300:  11.33 6 from 300:   0.46 7 from 300:   0.02

Note how the probability of finding a 5-point ley on the 60km map with n=200 is much more than four times that of finding a 5-point ley on the 30km map with n=50. With 100 such maps nationally, the expected national number of 4-point leys would be 4492, with 147 5-point leys.

Combining all these 100 maps into a single national map 600 km in "diameter" containing 20,000 points gives the following expected number of leys:

3 from 20000: 111094445.00 4 from 20000: 46282408.68 5 from 20000: 15424384.07 6 from 20000: 4283479.99 7 from 20000: 1019570.23 8 from 20000: 212336.12 9 from 20000: 39305.78 10 from 20000: 6548.01 11 from 20000: 991.63 12 from 20000: 137.65

Conclusion
Finding leys gets enormously easier, the bigger the area you consider, for a given density of points. Thus, finding leys with many points over long distances is disproportionately easy, not disproportionately hard as intuition might suggest.

Code
Python code for the above: def factorial(x): prod = 1L for y in range(1, x+1): prod = prod * y   return prod

def combin(k, n): return factorial(n) / (factorial(n-k) * factorial(k))

w = 50.0 d = (60000*10) eps = w/d for n in [20000]: print for k in [3, 4, 5, 6, 7, 8, 9]: print " %d from %3d: %7.2f" % (k, n, combin(k, n)*(eps**(k-2)))

Hm. The computation on the main ley line page should be computed using the exact data, not the approximation. User:The Anome

Is all this mathematical content original research by a Wikipedian? If it is, I don't think it should be included. (See What Wikipedia is not, under "What Wikipedia entries are not", point no. 10.) If it isn't, you need to attribute it to its source. We're supposed to summarise what other people have reported about things, you see, rather than come up with novel content ourselves. -- Oliver P. 11:26 8 Jul 2003 (UTC)

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Feel free to take it out, but please transfer it to the talk page if you do: I'm sure it can be made citable if needed. I'm not sure it's "novel" work: it's just a back-of-envelope computation that I could not track down an online source for. The results tie in with the extlinked computer simulation and the observed real-world results, which is a pretty good sanity check. -- The Anome 11:53 8 Jul 2003 (UTC)