Re: decimal longitude and latitude problem
- From: "jack" <jack1981@xxxxxxxxx>
- Date: 16 Nov 2006 18:25:51 -0800
very detail answer... thanks a lot.
Dave Seaman wrote:
On 16 Nov 2006 05:32:38 -0800, steve wrote:
Please help...I need to know the formula for calculating a particular
distance from a specific decimal longitude and latitude. For exampl,
for a house located at -77.209528 ,38.716728, what is the latitude and
longitude range for all houses withing .25 miles or .5 miles
It's a problem in spherical trigonometry. Set up a spherical triangle
having two of its vertices at the specified points. The third vertex
should be either the north pole or the south pole, whichever seems
convenient. Then apply the law of cosines for sphereical triangles:
cos c = cos a cos b + sin a sin b cos C
where
c is the distance between the two points,
a is the polar distance (co-latitude) of the first point,
b is the polar distance of the second point,
C is the difference of the longitudes.
If you solve for c in decimal degrees, then multiply by 60 to convert to
nautical miles. In other words, one minute of arc = one nautical mile =
1.15 statute miles, approximately.
For the second part of your question, you need to know that one degree of
latitude is 60 nautical miles, while one degree of longitude is 60*sin L
nautical miles, where L is the latitude.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
.
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