If you're dealing with a relatively range of latitude (within a few
degrees), then I think it may solve the problem. For example, if you're
working generally near latitude 40 and you want to work in x,y in miles,
then couldn't you just:
gen x=69*latitude
gen y=53*longitude
and then euclidian distances would be fairly close to true distances...
Michael Blasnik
[email protected]
----- Original Message -----
From: "Glen Waddell" <[email protected]>
To: <[email protected]>
Sent: Monday, September 29, 2003 4:05 PM
Subject: st: RE: Re: Latitude/longitude in spatwmat
> Michael Blasnik writes
>
> > Well, latitude to distance is fairly simple -- it's just the
> circumference of
> > the earth divided by 360, which equals about 69 miles. The longitude
> conversion
> > will vary with latitude. ... You can use some trig if you need to
> get more
> > precise.
>
> Agreed. The distance from {lat1,lon1} to {lat2,lon2} is simple. It is
> equal to
>
> radius*(acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(lon2-lon1)))
>
> where radius is of Earth in miles and lat/lon are in radians. However,
> this does not solve the problem of projecting latitude and longitude
> onto an {x,y} plane.
>
>
> Glen
>
>
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]] On Behalf Of Michael
> Blasnik
> Sent: Monday, September 29, 2003 12:13 PM
> To: [email protected]
> Subject: st: Re: Latitude/longitude in spatwmat
>
>
> Well, latitude to distance is fairly simple -- it's just the
> circumference of the earth divided by 360, which equals about 69 miles.
> The longitude conversion will vary with latitude. At the equator (lat=0
> degrees), one degree longitude equals one degree latitude. The value
> drops as move away from the equator of course. I think that at
> latitudes of 30, 40, and 50 degrees it's about 60, 53, and 44 miles
> respectively. You can use some trig if you need to get more precise.
>
> Michael Blasnik
> [email protected]
>
>
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