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1

I think you are going to have to break (as a node) the road at the polygon edges to do this in vector space. 1) Use Identity Tool in ArcGIS and this should do the trick. Just option that is something like Just ID Remember to recalculate lengths in the attribute table after the identity (just to be sure) It will add the polygon details (in this case thee ...


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You may have multi-part polygons in your file, and the area column would thus be reflecting the total area of all 'parts' of a polygon. You can use the Vector > Geometry Tools > Multiparts to singleparts tool to create single-part polygons, and calculate the $area on those to get the answer you are expecting.


2

@DarrenCope made a very good point, your layer contained inconsistencies. Please note: I only posted this as an answer as to include a screenshot of the results, please accept Darren's answer when he posts one as he came up with the correct solution. I used the Multiparts to singleparts function (via Vector > Geometry Tools > Multiparts to singleparts) ...


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If you have a Spatial Analyst license, then you can Reclassify the slope raster into five classes (0-3, 3-15, etc.). Then, the raster's attribute table will show how many cells are in each class and the area can be easily calculated. If you don't have the license or don't want to make a new raster, then Calavin's suggested method is ideal.


2

One way to get an area by doing a little manual work is to open up your DEM properties window on the symbology tab. Click on Classify... and click on your class break values in the window on the right. There will be an "Elements in Class" number on the bottom right of the window which is the number of pixels in that class. You can then multiply that number ...


3

Disabling the "on-the-fly" projection setting should yield you the correct results when performing analytics via the Field Calculator: Project > Project properties > CRS This bug has been known in early versions of QGIS but has mostly been resolved since QGIS 2.1.


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This might be what you are looking for. http://www.arcgis.com/home/item.html?id=8d2012a2016e484dafaac0451f9aea24


2

If you look at the source for getGeodesicArea, which is actually in the LinearRing class, you will will see from the comments (and the code) that this calculates an approximate area based on a sphere. If you look at GeographicLib, which includes a Javascript port of the C++ library, you will see that they use ellipsoidal calculations, which will be more ...


2

The geodesic area for your example is 19518154994956.3 m² (using GeographicLib). E.g.: var points = [ {lat: 0, lon: 50}, {lat: -5, lon: -60}, {lat: -30, lon: -30} ]; var p = GeographicLib.PolygonArea; var result = p.Area(GeographicLib.Geodesic.WGS84, points, false); var area = result.area; // 19518154994956.285 Here are multiple ...


1

I can confirm some strange results. Let's take a simple area without islands and coasts, like Va1. The measurement tool says it is 111x155km large, so 17224 km² should be ok. It is calculated to 17099 km² by the identify tool and the field calculator. Va2 is rather complicated, because the hole of Iceland has lots of vertices. With laea projection, I get ...



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