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I figured out the solution (not an exact estimate as it assumes the earth is a perfect sphere - not the case) and thought I should post it to help others that come across this thread. This is the raster calculator formula to use: 85.7476 * Cos($$YMap * 0.0174532925) The 85.7476 km2 comes from ~9.24 km squared, which is the cell area at the equator. This applies to 5 arc minutes and will change depending on your original cellsize. The 0.0174532925 is a conversion factor to convert degrees (from YMap) to radians, in order to satisfy the cosine argument.
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01-11-2012
11:03 AM
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Hi Dan, Thanks for answering. In decimal degrees, yes, the cells are all the same size (0.0833x0.0833) however I would like to have those measurements in hectares. I was under the impression that latitude would have an effect of how many hectares are in each grid cell, once projected. Is this incorrect? I've measured out the areas using the measuring tool, once I reproject the data, and the cells no longer have equal area: there are higher values near the equator and lower values with increasing latitudes. Is there some kind of way I can use the raster calculator to imprint the area of each cell as it's value, in projected units? (sort of like [shape].ReturnArea, but for pixels?) Thanks again!
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01-11-2012
08:40 AM
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Hi, I have a dataset that was originally in a decimal degrees projection. It is a raster of the globe, in 5 arc minute resolution. The final product I need is the same raster, with the each individual cell's value being its area, in hectares. These would essentially be highest around the equator (at around 8,000), and smooth out gradually to ~5,000 near the poles. After doing a lot of research on the forums, I've reprojected the dataset into an equal area projection (cylindrical equal area) as some have recommended. I am however stumped on how to use the raster calculator (or any other tool) to now calculate the area of each cell and assign this value to it in a new raster. Any ideas? Happy to elaborate more if any part of the above is unclear. Thank you!
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01-11-2012
07:49 AM
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