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.....
>>