Re: not long to go


Mike Flowers
 

I know, Steve - I was just playing for time while gathering statistics. So
here goes my complete SWAG at the problem:



The population density of the planet (including all land area) is about 105
people per square mile. If Antarctica is eliminated (since it has zero
population density - not exactly correct, but let's assume this.), the world
population density rises only to 115 people per square mile.



Each sub-square, such as EL29hk is equal to 5 nautical miles by 2.5 nautical
miles.



One international nautical mile converts to 1.150779 miles (statute).



So a Maidenhead sub-square is 5.753895 miles by 2.8769475 miles or
16.5536538355125 square miles.



Doing the math we get: an average of 115 people per square mile X
16.5536538355125 square miles equals 1903.670191083938 - say an average 1904
persons per Maidenhead sub-square.



The average percentage of the population that are Amateur Radio operators
can be roughly expratolated from the following jumbled table:



Japan 1,296,059 1.012 1999

United States 738,497 0.239 2012

Thailand 176,278 0.275 2006

South Korea 141,000 0.288 2000

Germany 75,262 0.092 2007

Taiwan 68,692 0.296 1999

Canada 69,183 0.201 2011

Spain 58,700 0.127 1999

United Kingdom 58,426 0.094 2000

Russia 38,000 0.026 1993

Brazil 32,053 0.016 1997

Italy 30,000 0.049 1993

Indonesia 27,815 0.011 1997

People's Rep China 20,000 0.001 2008

France 18,500 0.028 1997

Ukraine 17,265 0.037 2000

Argentina 16,889 0.042 1999

Poland 16,000 0.041 2000

Australia 15,328 0.067 2000

India 15,679 0.001 2000

Denmark 8,668 0.156 2012

Slovenia 6,500 0.317 2000

South Africa 6,000 0.012 1994

Norway 5,302 0.106 2000



0.14725 Average percentage of population
that are Amateur Radio operators = %.0014735





Another visit to the calculator yields: an average 1904 persons per
Maidenhead sub-square X percentage of the population licensed Amateurs
.0014735 equals 2.805544 - 2, perhaps 3 licensed Amateurs per Maidenhead
sub-square on average.



A generous estimate is that the average probability of working any one
Maidenhead sub-square is 635:1 - give or take a point or two.







- 73 de Mike, <http://www.qrz.com/db/K6MKF> K6MKF,
<http://www.qrz.com/db/W6NAG> W6NAG, VP <http://www.ncdxc.org/> NCDXC,
<http://www.scdr.org/> Dachshund Rescue, <http://www.waiohuli.com/> Maui



From: 070@yahoogroups.com [mailto:070@yahoogroups.com] On Behalf Of melachri
Sent: Friday, June 29, 2012 5:43 PM
To: 070@yahoogroups.com
Subject: [070] Re: not long to go





I meant the QSO probability problem you proposed.

As someone who DOES chase grid squares, I thought it would be interesting.

Steve

--- In 070@yahoogroups.com <mailto:070%40yahoogroups.com> , "Mike Flowers"
<mike.flowers@...> wrote:

Something NEW & BIG starts in:1Days



Just one days, apparently . ;>)



- 73 de Mike, <http://www.qrz.com/db/K6MKF> K6MKF,
<http://www.qrz.com/db/W6NAG> W6NAG, VP <http://www.ncdxc.org/> NCDXC,
<http://www.scdr.org/> Dachshund Rescue, <http://www.waiohuli.com/> Maui



From: 070@yahoogroups.com <mailto:070%40yahoogroups.com>
[mailto:070@yahoogroups.com <mailto:070%40yahoogroups.com> ] On Behalf Of
melachri
Sent: Friday, June 29, 2012 5:23 PM
To: 070@yahoogroups.com <mailto:070%40yahoogroups.com>
Subject: [070] Re: not long to go





I just hope that it doesn't take 358 years for someone to solve THIS
problem!

Steve

--- In 070@yahoogroups.com <mailto:070%40yahoogroups.com>
<mailto:070%40yahoogroups.com> , "Mike Flowers"
<mike.flowers@> wrote:

Alas, Steve, I must demur .



Est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc
marginis
exiguitas non caperet. - I have discovered a truly marvelous proof of
this,
which this margin (email) is too narrow to contain. - Pierre de Fermat






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