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12

It depends on what you mean by 'Universal Coordinate System'. If you wonder whether most Professionals understand Latitude and longitude, well in that case it is pretty much universally understood. But if you ask, whether it is used by everyone, then the answer is a resounding, No. There are many reasons why people use projected coordinate systems instead ...


10

I've answered this question in passing while answering another of your question. The Greenwich Observatory was defined as a prime meridian, based on the observations by the astronomer Sir George Airy in 1851. London was selected as the official prime meridian for international maps by the International Meridian Conference in 1884. When you use a GPS, by ...


8

The principal radius of the WGS84 spheroid is a = 6378137 meters and its inverse flattening is f = 298.257223563, whence the squared eccentricity is e2 = (2 - 1/f)/f = 0.0066943799901413165. The meridional radius of curvature at latitude phi is M = a(1 - e2) / (1 - e2 sin(phi)^2)^(3/2) and the radius of curvature along the parallel is N = a / (1 - e2 ...


6

Here's a Python function that creates a memory layer containing a line at the specified latitude. You call it using createLatitudeLayer(latitude=-23), for example. You can specify which CRS the layer should use by specifying targetCrsEPSG=<EPSG code>. You can also specify how many points to use for creating the line, by setting numpoints=<number ...


5

Simply create a text file with this content: id;wkt 1;LINESTRING(-180 -23, 180 -23) and use Layer -> Add delimited Text Menu entry with semicolon as delimiter and EPSG:4326 as CRS. For meridians, it is better to end the line at 89° when using EPSG:3857: id;wkt 1;LINESTRING(7 -89, 7 89)


5

Looks like ogr2ogr has support for exactly this problem. The below is copied directly from that page: How do I flip coordinates when they are not in the expected order The EPSG has a recommanded order for geographic SRS where the coordinates tuples of a geometry must appear in the (latitude, longitude) order, whereas most GIS will properly display such ...


5

Unless I'm misunderstanding the question, the shape's .extent property is all you need. with open('out.txt', 'wb') as out_text_file, arcpy.da.SearchCursor('path_to_data', ('msa', 'SHAPE@')) as cur: print >>out_text_file, "msa min_lon max_lon min_lat max_lat" for row in cur: msa, ext = row[0], row[1].extent print ...


4

If you're working in Ubuntu, you could use this in a shell script: To get longitude like 81°5.8' (degree and minutes from code in d,m,s): tmp0=`echo 081054800 | cut -c2,3` tmp1=`echo 081054800 | cut -c5` tmp2=`echo 081054800 | cut -c6,7` tmp3=`echo "scale=2; $tmp2/60" | bc -l` echo $tmp0 $tmp1 $tmp3 | awk '{ print $1"°"$2$3"'\''" }' which gives: ...


4

There is nothing in your Question to suggest that you need to perform a conversion. Instead, I would try: point = osgeo.ogr.Geometry(osgeo.ogr.wkbPoint) point.SetPoint(0, 112.73091137, -7.2377761) The original values look like they are from a projected coordinate system whereas you appear to be receiving Twitter data in a geographic coordinate system but ...


4

The short answer is: there was a historical shift between the first global datum and the Greenwich meridian; and it continues to move (slowly) because of continental drift. You can find more detailed information on this site In the late 1950s (under the auspices of the US Navy), the Applied Physics Laboratory (APL) of the Johns Hopkins University began ...


3

The QuickWKT plugin is very useful for putting temporary lines, points or polygons on the map; a trifle buggy but works. Install it and you'll get a new toolbar button 'WKT' - click it and you are given a dialogue with space to paste a WKT string, and a drop-down menu to load examples of WKT strings. I am having trouble with adding a LINESTRING when it ...


3

Your coordinates are in UTM (Universal Transverse Mercator) Eastings and Northings. You can convert your UTM coordinates to lat/long online using many different sites. Here in one. You can also perform this calculation in excel or inside of a database if you can work out the formula. Alternatively you can perform the conversion inside of a GIS such as QGIS ...


3

It's clear that it's not necessary to shift the projection, but the data. Using GDAL >= 1.10.0 compiled with SQLite and SpatiaLite: ogr2ogr russia_shifted.shp world.shp -dialect sqlite -sql "SELECT ShiftCoords(geometry,360,0) FROM world WHERE CNTRY_NAME='Russia'" where shiftX = 360 (degrees) and shiftY = 0. UPDATE: here's the whole workflow... Clip ...


3

There is no meaningful way to link "time the GPS is on" with a particular precision level. The accuracy is driven by many factors ranging from the characteristics of the hardware (especially the antenna) through to atmospheric conditions and GPS satellite constellation configuration. A reasonable approach (using information you can get from most receivers) ...


3

Check NOAA's GSHHG (Global Self-consistent, Hierarchical, High-resolution Geography Database). The database is constantly being updated and maintained (latest update since Im answering this Q: July 1, 2013) GSHHG is: a high-resolution geography data set amalgamated from two data bases in the public domain: World Vector Shorelines (WVS) and CIA World ...


3

If I understand well, you have your data in a Coordinate System (from your description, probably geographic one), and you want to see\capture coordinates in a different system. In QGIS you can do the following: Option 1 -Set you Project to the desired coordinate system Make sure you set the correct CRS for your layers using right-click over the layer ...


3

Welcome Lucy -- your question is a little bit unusual for this site, which mainly caters to people who use geographic information systems professionally or in research. I found it surprisingly hard to google up a site that has a concise history and comparison of navigation tools over time. However, I think the Smithsonian Museum's Time and Navigation site ...


3

After extensive discussion on the question, I think we're finally there. The input data is in EPSG 26918 (UTM zone 18N). So if you want to query against that, you need to use that spatial reference system. If you want to ask in longitude / latitude, you need to transform that into 26918, which is easy to do with ST_Transform. The conversion looks like: ...


2

There is no realistic way that you can provide information to your supervisor (superior has other connotations) on how Google implemented the service behind their geocoding API. Google, in general, is not very forthcoming on technical implementation details for their services, because it represents competitive advantage. Note: this isn't to say that there ...


2

I've made a lot of simplification. My need is to convert from Lambert93 (IGN) to WGS84. For the region of Bordeaux (south/west of France), the result is exactly the same as Geofree. import static java.lang.Math.abs; import static java.lang.Math.atan; import static java.lang.Math.exp; import static java.lang.Math.log; import static java.lang.Math.pow; import ...


2

If you can offer an azimuth, you could generate lines from the points and azimuth. then intersect to find corners. Once corners are generated, build the bounding box from these points to include points 1-4 Ex: point 1 azimuth generates line A point 2 azimuth generates line B point 3 azimuth generates line C point 4 azimuth generates line D intersect lines ...


2

There is no one formula to solve this problem on a spheroid (since it's a partial differential equation which can only be solved by iterative means), but the solution is a trivial implementation of the "direct" problem (given a point, a bearing, and a distance, locate a point). The US National Geodetic Survey has a web site with FORTRAN source (I converted ...


2

Try this one: http://cs2cs.mygeodata.eu/ or use "proj" (command line below assuming UTM zone 17 south and wgs84): # geographical coordinates (ie. longitude,latitude) to utm proj -f "%.4f" +proj=utm +zone=17 +south +ellps=WGS84 < myfile # utm to geographical coordinates (ie. longitude,latitude) proj -f "%.4f" +proj=utm +zone=17 +south +ellps=WGS84 ...


2

I agree with the essence of Devdatta Tengshe's answer, except that I believe the main reason for the common use of plane (projected) coordinates over geographic (lat-lon) coordinates is one of computational convenience. It was and still is very difficult for people or surveyors to do basic CoGo (coordinate geometry) calculations on the sphere (using ...


2

The bunch of different lat/lon datums we have is a relict of the first surveying done on national basis. These had their limits on the borders (mainland or overseas colonies), and noone thought of putting a universal CRS at that time. As a consequence, getting lat/lon coordinates always needs a question about the datum they are in. While local projected ...


2

You can use functions from the rgeos package to extract such regions (e.g. gIntersection, gDifference). I use gDifference in this example, because gIntersection returns a SpatialCollections object here: # define rectangular region y_lim <- c(-1, 1)*23.5 rect_lim <- cbind(c(rep(bbox(ao)["x", ], each=2), bbox(ao)["x", 1]), c(y_lim, ...


2

OpenLayers uses the EPSG:3857 coordinate system, in meters, and not the WGS84 system, in degrees, look at OpenStreetMap Wiki: EPSG:3857 But why use subprocess and ogr2ogr? 1) you can use directly the PostGIS ST_AsGeoJSON function: import psycopg2 conn = psycopg2.connect("dbname='osm' host='localhost' user='me'") cur = conn.cursor() # srid of the layer ...


1

I would suggest taking a look at your dataset visually (you could connect to your database using QGIS for example). That way you can check if a point would actually be in the polygon, and as @BradHards suggests, check that you have the correct lon/lat when constructing the geometry. Your query looks fine - it's not necessary to have true at the end. You can ...


1

If the unknown points are on the specified circles, and if you assume a simple circular reference surface (a sphere, not an ellipsoid), then it is easy to get the angle at A from AC to the unknown points: angle = ctd / scd, where ctd is your so-called cross-track distance (what I'd call arc distance), and scd is your small-circle distance away from A ...


1

What you're looking for is the initial bearing (or forward azimuth), which if followed in a straight line along a great-circle arc will take you from the start point to the end point. Here is some simple JavaScript from this link: var y = Math.sin(dLon) * Math.cos(lat2); var x = Math.cos(lat1)*Math.sin(lat2) - ...



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