Having reviewed my numbers from last night, I see that I made an
error on the vertical, and "re-used" some space for
neighboring vertical squares (still good on the horizontal though.
Therefore, I have to drop a single row of 10 and the number becomes
115.
I also worked up my numbers per the stated criteria of a 0.05
buffer on all edges. For this style, I get 106, as
follows:
row num
cumulative side length
1
10
0.000
2
10 528.000
3
10 1056.000
4
9 1513.261
5
10 1970.523
6
9 2427.784
7
10 2885.046
8
9 3342.307
9
10 3799.568
10
9 4256.830
11
10 4714.091
106
With the new max side length of 4,752 feet, approx 38 feet
remains, not enough to convert another row.
Bill... I think all the other kids left for the playground (AKA a
cache).
Tim
Your answer is more accurate than mine, but does not
technically meet the stated criteria of a ".05 mile buffer zone."
Of course, as you point out, a "floating" buffer zone, while not
meeting the stated goal, does allow for more
points.
Same problem we were solving before, just with a
square that is 528' shorter on a side, so it is 4752 on a side.
That means you get 10 in the long rows and 9 in the short rows
(triangle pattern for max density). Rows are still 457.261'
apart, except you can put a couple the full 528 apart and get extra
rows of 10. I think 96 is right.
Bill in Willcox
From:
az-geocaching-bounces@listserv.azgeocaching.com
[mailto:az-geocaching-bounces@listserv.azgeocaching.com] On Behalf
Of Tim Giron
Sent: Monday, March 14, 2005 9:02 PM
To: listserv@azgeocaching.com
Subject: Re: [Az-Geocaching] Maximum square mile cache
density
With this
new constraint (which I will call interlocking horizontal neighbors),
I will offer 2 answers to be debated. The first is 120, since
now the rows are all 10 "dots" wide and they just shift back
and forth, forming triangles. However, this leaves approx 250
feet wasted at the top of the square (since there is no longer a
benefit to "squaring up" the sides. So, I will throw
out another, softer number of 125 which is the average for two squares
stacked vertically (the 250 feet from each add up to enough to make
another row which takes 457 feet, and they split the number in the
row).
Tim
Team
AZFastFeet
Okay, maybe I opened a can of worms here...
I should be more specific...I was trying to figure out
how many caches can fit into a square mile, leaving enough buffer
zone (.05 mile) around the edges, so each square mile around the
area in question can also have the same amount of caches?
Any math geniuses out there? Anyone?
Scott and I were discussing this today, and also
called it a "Power Grid"...
Maybe on Terracaching.com....thinking,
thinking.....