# Tag Info

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Accuracy is the tendency of your measurements to agree with the true values. Precision is the degree to which your measurements pin down an actual value. The question is about an interplay of accuracy and precision. As a general principle, you don't need much more precision in recording your measurements than there is accuracy built into them. Using too ...

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This Wikipedia page (Decimal Degrees) has a table on "Accuracy versus decimal places". Also the accuracy of your coordinates depends on the instrument used to collect the coordinates - A-GPS used in cell phones, DGPS etc. decimal places degrees distance 0 1 111  km 1 0.1 11.1  km 2 ...

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The terms are medieval: latitude (n.) late 14c., "breadth," from Old French latitude (13c.) and directly from Latin latitudo "breadth, width, extent, size," from latus "wide," ... . Geographical sense also is from late 14c., literally "breadth" of a map of the known world... In the fourteenth century, most known-world maps made in the Christian and ...

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I'm not an expert in this area, but I have done some reading on the subject, particularly on its history. I think the reason is: Accurate measurement of latitude came first as it was based on astronomical measurements. Longitude was not accurately measurable until a highly accurate time measuring device was developed.

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This question assumes an ellipsoidal model of the earth. Its reference surface is obtained by rotating an ellipse around its minor axis (plotted vertically by convention). Such an ellipse is just a circle that has been stretched horizontally by a factor of a and vertically by a factor of b. Using the standard parameterization of the unit circle, t --> ...

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Although geodesics do look a little like sine waves in some projections, the formula is incorrect. Here is one geodesic in an Equirectangular projection. Clearly it is not a sine wave: (The background image is taken from http://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Equirectangular-projection.jpg/800px-Equirectangular-projection.jpg.) ...

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Same, equivalent terms. lat/long is probably the most common, but of course "long" is a reserved word / data type in many programming languages, so you'll find other ways to shorten longitude. In some contexts, GIS people will use long/lat (i.e. reverse order), since this is the normal X/Y coordinate reference system for a Cartesian plane.

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Manually reversing the rotation should do the trick; there should be a formula for rotating spherical coordinate systems somewhere, but since I can't find it, here's the derivation ( ' marks the rotated coordinate system; normal geographic coordinates use plain symbols): First convert the data in the second dataset from spherical (lon', lat') to (x',y',z') ...

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For a simple mean, then no, you do not want to average the longitude and latitude ordinates. This might work pretty well at lower latitudes, but at higher latitudes it will beging to give poor results, and completely break down near the poles. The method I've used for this type of thing, is to convert the longitude,latitude coordinates to 3d cartesian ...

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Your coordinates are Ordnance Survey GB (OSGB36 Airy 1830 ellipsoid) 431833,336684 (base on a false origin with the Transverse Mercator projection. http://en.wikipedia.org/wiki/Ordnance_Survey_National_Grid To convert them correctly to Lat/Lng Use this site:http://www.nearby.org.uk example: http://www.nearby.org.uk/coord.cgi?p=431833%2C336684 look for the ...

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For an automated ArcGIS solution, try the following in model builder: Add a new field "Latitude". Calculate the "Latitude" field using Calculate Field: !SHAPE.FirstPoint.Y! (see attached field calculator screenshot). This calculation is based on the centroid of each polygon. Create a new shapefile using Sort. Note that the new shapefile will have a new ...

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This isn't so old-fashioned: I remember having to solve exactly this problem back in the 80's when we didn't have scanners readily available and had to lift coordinates and elevations off large-format printed maps for geostatistical analysis. In effect you can already read the longitude accurately along any line of longitude on the map. You want to ...

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Right, a bit of trig, some simple algebra, and a ruler should get you there... assuming it is a conic projection with the north pole at the centre. First you need to determine the location of the north pole. To do that, you need to measure the distance along the bottom of your map of two points, A and B. To keep things positive, you can add a horizontal ...

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It is used in marine navigation to make the maths simpler, as well as instrumentation like sextants. Take the example of a navigator who wants to find the angle of a celestial body with the horizon, he'll first make coarse adjustments on the sextant to get to the nearest degree, then turn the vernier to make fine adjustments. The vernier is marked off in ...

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You can use Python shapefile library pyshp >>> import shapefile >>> sf = shapefile.Reader("shapefiles/blockgroups") >>> shapes = sf.shapes() >>> # Read the bounding box from the 4th shape >>> shapes[3].bbox [-122.485792, 37.786931000000003, -122.446285, 37.811019000000002] >>># Read the 8th point in the ...

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Just like any other function, you will get out the same unit of measurement you put in. In this example you are using lat/long and so will therefore get the distance between your two points (the hypotenuse) in degrees. The number of kilometers/miles/inches/whatever that the output corresponds to depends entirely on where on the surface of the earth the ...

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When someone says that data points are tiny compared to other data layers, the usual problem is that the new points have been assigned a projected coordinate system--probably the one that the other layers are using--but the new points are latitude-longitude and should be assigned a geographic coordinate system. The Add XY tool will default to the data ...

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You can experiment with different transformations on a Danish Ministry of the Environment site. You may have to make multiple guesses (narrowed down perhaps by this experimentation), map out the points, overlay them on related layers, and check for reasonableness. The second set is likely in meters in UTM 32N. "Lat" and "lon" are mixed up: they correspond ...

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Assuming your data has a projected coordinate system: Set data frame to a geographic coordinate system (WGS84). Add your data. (Set datum transformation if nessesary.) Add a new field (ycoordinate). Right click field ycoordinate and calculate y-coordinate of centroid (use coordinate system of data frame for calculation.) Export data to Excel In Excel: sort ...

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I don't think there is any way to georeference a ski resort map - which are often hand-drawn - in a meaningful way. It's an artistic interpretation of the landscape with high degrees of freedom and a focus on being easy to read. Maybe if you describe your final goal - e.g. what do you want to do with the georeferenced image? - we can propose a different ...

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In general, you cannot determine the zone from the coordinates: that's why the zone is needed! In your case, though, it sounds like you are looking within a small area that straddles neighboring zones. In that case, locations in zone 39 will have large positive x-values and locations in zone 40 will have negative or small positive x-values: that's how you ...

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The referenced article is thoughtful. However, I believe there is a "simple and elegant" solution: for geographic datasets, there are two kinds of bounding boxes. Those that do not straddle the +-180 meridian can be stored and searched as always. Those that do straddle the +-180 meridian can be stored in a semi-complementary form: namely, store the range ...

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'Yes' is the short answer. The way you do it depends on what you want to use the data for. The simplest way is to re-project your data into a Lat/Lon coordinate system using ST_Transform() (I think it is just 'Transform()' in SpatiaLite and something weird in Oracle). Alternatively, follow one of these recipes, perhaps using Proj4. Once you have ...

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Unproject the DEM to determine the geographic region it covers. Create the latitude coordinate grid for this region. (In Arc* products prior to version 10, this is done with "\$\$YMap": http://forums.esri.com/Thread.asp?c=93&f=1740&t=182407.) Projecting this grid produces the latitude raster. For the projection, use some form of interpolation ...

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You need to populate your spatial_ref_sys table, e.g.: psql -d your_db -f spatial_ref_sys.sql This will give you SRID 4326, but not 94283, which doesn't appear to be any publicly recognized SRID. To add 900913 (i.e. leetspeek for GOOGLE), see the PostGIS instructions at http://trac.osgeo.org/openlayers/wiki/SphericalMercator

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The coordinate system, USA Contiguous Albers equal area conic, by default is in meters, so any calculations you do should be in meters or square meters. Depending on how large the polygons are, the values could be very large. Did you want to have the values reported in kilometers instead? The Calculate Geometry tool lets you choose a different unit value. ...

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You have to distinguish between the project CRS and the layer CRS. Your data is in lat/lon degrees, so the layer CRS should be EPSG:4326, WGS84. Your project CRS can be anything else. In most cases a projected CRS, where units are metres or feet. On-the-fly projection should always be activated. Just to be sure: the coordinates should have decimal points ...

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If the image is in the least realistic--and it appears to be--then it represents a projection from a portion of Euclidean three-space onto a portion of two-space. We cannot expect this projection to have a nice mathematical formula, because it is likely the artist has introduced irrregular small distortions, but we can expect its formula to somewhat ...

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