I have a raster layer of which I would need to find the area (m2) per cell. Any suggestions for a simple approach to achieve this?
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1Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.– JonCommented Apr 1, 2019 at 17:17
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1@Jon that's going to change the grid basis. Probably not a good idea. If the raster is already in a cartesian system then multiply the height by the width. If the grid is lat-long then there's some calculations to be done...– SpacedmanCommented Apr 1, 2019 at 17:34
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1I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.– JonCommented Apr 1, 2019 at 17:53
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1Okay, thanks. My pixel size is 0.000833333, -0.000833333. As I understand, this is not in the unit of m2? I have a raster containing values of energy demand per hectare per cell. I would like to convert this to energy demand per cell and thus need to identify the area of the cell.– A.NCommented Apr 1, 2019 at 18:41
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1Then the unit is degrees and area of a pixel in hectares depends on where on the Earth the pixel is located. The answer by @Jon is relevant. If your raster is rather small you can calculate the length of one degree latitude and longitude around your location with some online tool like csgnetwork.com/degreelenllavcalc.html and use the result as an estimate.– user30184Commented Apr 1, 2019 at 19:44
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2 Answers
Your question is not well constrained. If you are wondering how to find the area of a cell from a raster that is unprojected and aligned to WGS84 grid (e.g. epsg:4326), then the following Python code should help.
import numpy as np
import rasterio
fp = 'your_tiff_file.tif'
with rasterio.open(fp) as testif:
rast = testif.read(1)
gt = testif.transform
pix_width = gt[0]
ulX = gt[2]
ulY = gt[5]
rows = testif.height
cols = testif.width
lrX = ulX + gt[0] * cols
lrY = ulY + gt[4] * rows
lats = np.linspace(ulY,lrY,rows+1)
a = 6378137
b = 6356752.3142
# Degrees to radians
lats = lats * np.pi/180
# Intermediate vars
e = np.sqrt(1-(b/a)**2)
sinlats = np.sin(lats)
zm = 1 - e * sinlats
zp = 1 + e * sinlats
# Distance between meridians
# q = np.diff(longs)/360
q = pix_width/360
# Compute areas for each latitude in square km
areas_to_equator = np.pi * b**2 * ((2*np.arctanh(e*sinlats) / (2*e) + sinlats / (zp*zm))) / 10**6
areas_between_lats = np.diff(areas_to_equator)
areas_cells = np.abs(areas_between_lats) * q
areagrid = np.transpose(np.matlib.repmat(areas_cells,cols,1))
This snippet returns the array areagrid that contains the area in square kilometers of each pixel in your 4326 raster. There are a number of considerations "under the hood" that I am not detailing here, but I will note that I have compared this method with reprojection methods and the results have always been very similar.
You asked for a QGIS solution, so you would have to modify this a bit to get it to run in the Python console. Maybe overkill for what you want.
It is possible using the SAGA raster calculator per Getting latitude / longitude in the QGIS Raster Calculator and Creating direction raster in QGIS?
Since you know your cell size is 0.000833333, -0.000833333 degrees (per comment), (or 3 arc-seconds/cell) then, you can use the xpos() and ypos() to convert the degrees into an area.
Assuming a minute of latitude is a standard nautical mile of 1852 meters on a spherical earth:
area_m2 = 0.000833333*60*1852 * 0.000833333*60*1852* cos(ypos())`
or
area_m2 = (0.000833333*60*1852)^2*cos(ypos()*pi()/180)
In the Saga raster calculator, (under Processing/Saga/Raster Calculus/Raster Calculator), choose your layer and use a formula like:
(0.000833333*60*1852)^2*cos(ypos()*pi()/180)
This should produce a new raster layer conformant with your input layer, with each cell holding the square meters of a 3-arc-second cell at that latitude (about 6830m2 in my area, decreasing northward)
