[postgis-users] Re: st_dwithin in SRID 4269

[MarkW]

Okay, you said to correct you!
A minute of latitude is the unit that is a nautical mile. If each second
where a mile, there would be 3600 nautical miles in a degree - rather than
the actual 60.

Mark

On Sat, Apr 25, 2009 at 2:05 PM, <nicklas.aven at jordogskog.no> wrote:

>  Hallo
>
> Someone have to correct me if I'm wrong but as I remamber it from my
> nautical experience from quite along ago:
> A nautical mile is 1852 meters yes. But not a degree. the distance of a
> degree is depending on if you are talking about a longitude or latitude
> degree. A latitude degree can be converted to meters. I think one degree is
> divided in 60 minuites. And one miniuite is divided in 60 seconds and if I
> remember right on second is the same as one nautical mile; 1852 meter. But
> if we are talking about longitude degrees it depends on how close you are to
> the poles. If you are just one meter from the pole you just have to take a
> walk of some meters to pass all 360 degrees but if you are walking at the
> equator you will have to walk and swim about 40000 km.
>
> This is why you have to transform your geometry to a meter-based srid
> before calculating distances right.
>
> What you first have to do is to tell the system what srid your data
> originally has if that isn't already done when loaded. That you do with
> st_setsrid:
> http://www.postgis.org/documentation/manual-svn/ST_SetSRID.html
>
> then when the system knows what you have (that is stored together with your
> geometry), then you can transform it to desired meterbased srid with
> st_transform:
> http://www.postgis.org/documentation/manual-svn/ST_Transform.html
>
> Hope things get clearer (and hope I'm right)
>
> /Nicklas
>
> 2009-04-25 Sachin Srivastava wrote:
>
>
> >
> >>
> 2009/4/25 Sean Fulton <seanasy at gmail.com>
> > > On 2009-04-25 09:50:55 -0400, Sachin Srivastava <
> sachin.srivastava1984 at gmail.com> said:
> >
> > I am a newbie, could somebody explain what distance transformation
>  should i
> > use to eliminate discrepancy in results for the following 2 queries
> >
> > 1) SELECT * from table1 where st_dwithin(geom1, geom2, distance);
> > *(Note, geom1 and geom2 are in SRID 4269)
> > *
> > 2) SELECT * from table1 where st_dwithin(transform(geom1, 2163),
> > transform(geom2,2163), distance);
> > *
> > Note:Here distance is in meters
> >
> > The things that i understood so far are, distance has to be in the same
> SRID
> > as the two geoms, I knew distance in meters, so the second query will
> give
> > me correct results however what transformation should i do in the first
> > query distance value to get the same result as of query 2.
> >
> >
> Sachin,
> >
> > I believe the answer is that you use query 2 that you provided.  Query 1
> will always give you distance in degrees which is going to be meaningless.
> You will never get meters out of Query 1 without transforming the
> coordinates as you did in Query 2.  You could try to convert the result of
> Query 1 from degrees to meters but that's really doing it backwards and I
> wouldn't trust the results.  Query 2 is the correct approach.
> >
> >  Sean
> >
> >
> > _______________________________________________
> > postgis-users mailing list
> >postgis-users at postgis.refractions.net
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> >
>
> >Sean,
> >
> > Thanks a lot for your answer.
> >
> > 1 degree is 1852 meters, so if i do distance/1852, in the Query1 i should
> get the correct results, but that is not happening.
> > I guess that conversion has to do a lot with what SRID i am using.
> >
> > So what you suggest is the best way to find whether two geoms are at
> 'dist' distance apart.  Whether the following query will do in all cases.
> >
> > SELECT * FROM table where st_dwithin(transform(geom1, 2163),
> transform(geom2, 2163), dist);
> >
> > Or It will again depend on which SRID i am using?
> >
> >
> >
> >
> >--
> >An Expert is the one who has made all the mistakes that are possible
> within a narrow field.
> >       ---------- Sachin
> >
>
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>
>
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