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There is a relatively simple exact formula for the area of any spherical quadrangle bounded by parallels (lines of latitude) and meridians (lines of longitude). It can be derived straightforwardly using basic properties of the ellipse (of major axis a and minor axis b) that is rotated around its minor axis to produce the ellipsoid. (The derivation makes a ...


Use spTransform to transform the coordinates to WGS84: library("rgdal") library("rgeos") map <- readOGR(".", "kommuner1983") map_wgs84 <- spTransform(map, CRS("+proj=longlat +datum=WGS84")) plot(map_wgs84, axes=TRUE) gCentroid(map_wgs84) # SpatialPoints: # x y # 1 10.05 55.96 # Coordinate Reference System (CRS) arguments: +proj=longlat ...


The answer to radouxju's question depends on the shape of the pixel when projected onto the ellipsoid. If the coordinate system of the raster is longitude and latitude, then the pixel is a rhumb line rectangle and whuber's answer can be used, or, more generally, you can use the formula for a polygon whose edges are rhumb lines. If the coordinate system is ...


I just entered the coordinates into Google maps. If you change them to 35.205357, -111.59330 it puts you in the Flagstaff area. So The coordinates may have just had their decimal places in the wrong places or missing.


When you change the projection via the properties dialog, I don't believe you are re-projecting the data, you are asserting that the data is actually in the projection you are setting it to. That's why it's all screwed up when you try to use it. You need to use the "Project" tool in the Arc Toolbox. It's all explained here. Even though the article refers to ...


The latitude value could be 35 20 53.57 while the longitude value could be -111 59 33. Or they could both be decimal degrees, with the appropriate insertion or change of position.


I have run into the same problem using QGIS today. Whilst I appreciate that it's late to answer I thought this might help. My KML (WGS84) layers, which were exported from an OSGB (ESPG:27700) Shapefile were all offset to WSW by 3 - 5 Metres. I managed to work around this by exporting to GeoJSON (WGS84) and then exporting the GeoJSON layer to KML format. ...

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