# How can I find out how to convert this point to lat lon?

I have received a shapefile with measurement points to be analyzed, but no matter how I try, I can't find how to convert to lat lon. I suppose it would be good to know the projection but when I look into the file (with R and summary()) it says "Is projected: NA" and "proj4string : [NA]".

One measurement-point I know is close to the border Peru-Chile has x,y = 354521, 7997417.8.

How can I find out how to convert this point (and many others) to lat lon? I tried to follow the instructions of similar questions here, but to no avail.

Thanks,

/Chris

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It looks like you are using R, so try the proj4 package. Using @mkennedy's guess of UTM zone 19S, you can query a proj4string, and use the package:

library(proj4)
proj4string <- "+proj=utm +zone=19 +south +ellps=WGS84 +datum=WGS84 +units=m +no_defs "

# Source data
xy <- data.frame(x=354521, y=7997417.8)

# Transformed data
pj <- project(xy, proj4string, inverse=TRUE)
latlon <- data.frame(lat=pj\$y, lon=pj\$x)
print(latlon)

Produces:

lat       lon
1 -18.10714 -70.37495

Also, if you need more precision in the output, set options("digits"=12), and you will see more digits used to display the coordinates:

lat            lon
1 -18.107144101 -70.3749500491
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Note: to project sp objects in R, see this excellent answer – Mike T Aug 20 '12 at 6:09

Based on the coordinate values, it's probably UTM zone 19S (south). The question is what geographic coordinate reference system (datum) is being used. Try PSAD56.

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+1. Not too many coordinate systems have values near 8,000,000 (meters): that's 1/5 the way around the world from the origin! – whuber Jan 17 '12 at 23:14

A northing of over 800,000 in Peru would be reasonable. IN the southern hemisphere, 0 northing is at the south pole and it goes up to just under 10,000,000 at the equator, where it starts over with 0 in the northern hemisphere. If you can locate the correct eastings and northings of a dozen or so points, you could warp the whole thing to known coordinates (could get them perhaps from GoogleEarth - depends on the precision you want).

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