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May 23, 2017 at 12:35 history edited CommunityBot
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Oct 27, 2013 at 20:32 history edited user484 CC BY-SA 3.0
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Oct 27, 2013 at 20:03 comment added DavidC @RahulNarain. Thanks. That was due to sloppiness as well as misunderstanding the full meaning of the conversion formula. I did test the current formula using your criterion and it did produce a sphere, so I'm fairly certain that it now works.
Oct 27, 2013 at 19:59 history edited DavidC CC BY-SA 3.0
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Oct 27, 2013 at 19:52 history edited DavidC CC BY-SA 3.0
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Oct 27, 2013 at 17:47 comment added user484 This is a nice approach, but your conversion formulas are incorrect, as you can see by the fact that ParametricPlot3D[{r Cos[#2 Degree], r Cos[# Degree], r Sin[#]} & @@ {lon, lat}, {lon, -180, 180}, {lat, -90, 90}] is not a sphere. The formulas should be {r Cos[#1 Degree] Cos[#2 Degree], r Cos[#1 Degree] Sin[#2 Degree], r Sin[#1 Degree]} & instead. Also you might want to use BoxRatios -> Automatic in the plot so that distances are shown without distortion.
Oct 27, 2013 at 13:51 comment added DavidC @Rahul. Yes, good advice. (Although I used different conversion formulas from decimal degrees to XYZ). Fortunately, Nearest can also handle distances in 3-space.
Oct 27, 2013 at 0:06 history edited DavidC CC BY-SA 3.0
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Oct 27, 2013 at 0:02 history undeleted DavidC
Oct 26, 2013 at 23:58 history edited DavidC CC BY-SA 3.0
complete rewrite
Oct 25, 2013 at 21:31 history deleted DavidC via Vote
Oct 25, 2013 at 20:04 comment added DavidC Thanks rm-rf. I'll look into this more closely
Oct 25, 2013 at 17:14 comment added user484 Since the points represent "some places in a city" and so are likely to be quite close together relative to the size of the Earth, it might be enough to scale the longitudes by $\cos\phi_m$, where $\phi_m$ is the mean latitude, and then continue to use Euclidean distances, no? en.wikipedia.org/wiki/…
Oct 25, 2013 at 15:39 history edited DavidC CC BY-SA 3.0
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Oct 25, 2013 at 15:37 comment added rm -rf @DavidCarraher Because 1 minute E from (0,0) is not the same distance as 1 minute N from (0,0). This difference only gets worse the closer you get to the poles. GeoDistance will tell you that, whereas EuclideanDistance will consider them both equally distant.
Oct 25, 2013 at 15:31 history edited DavidC CC BY-SA 3.0
voronoi diagram
Oct 25, 2013 at 12:30 comment added Dr. belisarius We are talking about 10^9 points here. I don't think the most obvious approaches will be useful here ...
Oct 25, 2013 at 11:34 comment added DavidC Presuming the long/lat, minutes, and seconds are converted to a single decimal, why won't euclidean work? Besides, isn't this a one-off computation?
Oct 25, 2013 at 10:44 comment added rm -rf You'll have to use Nearest with GeoDistance as the distance metric, not Euclidean. This will slow things down tremendously!
Oct 25, 2013 at 10:04 history edited DavidC CC BY-SA 3.0
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Oct 25, 2013 at 9:57 history edited DavidC CC BY-SA 3.0
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Oct 25, 2013 at 9:49 comment added Kuba It will definitelly be faster if you precalculate NearestFunction: near = Nearest[landmarks]; ({#, near[#]} & /@ citypoints). Belisarius showed this lately
Oct 25, 2013 at 9:45 history answered DavidC CC BY-SA 3.0