[OSM-dev] Question about Calculating Radius from GPS cord

[Fire Girl]

 I think I am following you! :)

Here is the math I have so far....

Calculation: 5 miles
of: 45.5 latitude
and: 25.5 longitude

-------------------------------

so, for the latitude, (i.e. add and subtract 2.5/69ths of a degree of
latitude)

With 5 miles assumed, and using that forumla, the distance from 45.5
would be:

45.505797 (on the plus side)
45.494203 (on the minus side)

Right? :)


but on the Longitude side, (i.e. 2.5/f) .... f = 69 * sin(latitude)

.... i am not so clear what Latitude number I should be using there and
what that might look like. 

How do I look at least so far in my calculations? :) 


  ----- Original Message -----
  From: "David Earl"
  To: "Fire Girl"
  Subject: Re: [OSM-dev] Question about Calculating Radius from GPS
  cord
  Date: Fri, 08 Aug 2008 16:02:10 +0100


  On 08/08/2008 14:30, Fire Girl wrote:
  > I am working with OSM data, and would like to be able to spec out
  > 5 mile bounding boxes from certain GPS points.
  >
  > After research into this problem, I am to understand that each
  > degree of latitude is approximately 69 miles (111 kilometers)
  > apart with a slight variance (68.703 - 69.407 miles) between the
  > equator and the poles, and that each degree of longitude is
  > widest at the equator @ 69.172 miles (111.321 kilometers) and
  > gradually shrinks to zero at the poles. : ) :)
  >
  > So what does this mean? If I want to take a input point, like lets
  say,
  >
  > 167.9 lat
  > -29.1 lon
  >
  > or
  >
  > -63.1
  > 18.1
  >
  > Can someone say with authority, what the 'calculus' would be to
  > definitivly construct a NSWE bounding box with a 5 mile radius
  > around those points?.... that would be basically close enough or
  > accurate? :)


  If you want accuracy, then you are asking the wrong question,
  because the "bounding box" will be a curved section of a surface of
  the earth (or, rather, the approximation to it defined by the
  ellipsoid for the datum you're using for the lat/lon coordinates -
  unless you want to start taking altitude and terrain into account),
  not a flat straight-edged box. So if you want to talk in terms of
  flat bounding boxes, you have to start asking "in what projection?"
  etc.

  If all you want is the lat/lon of the corners an approximate box
  parallel to lines of latitude and longitude approx 5 miles across,
  then
  you could indeed just add and subtract 2.5/69ths of a degree of
  latitude and 2.5/f of a degree of longitude, where f is the approx
  length of one degree at that latitude, i.e. f = 69 * sin(latitude)
  miles (with the reasonable approximation that the earth is a
  sphere, so the trigonometry is trivial).

  David

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