Determine area of cell in raster (QGIS)SUM Area irregular polygons per grid cellProjecting a Raster from WGS84 to planar Equal Area to Calculate Area: How to Select the Cell Size?Geographic vs. projected raster cell size areaCalculating area of polygon within square mile in ArcGIS?Sum area of specific poly attributes within raster grid cellHow to get unique values and cell counts of a raster in QGIS 2.6?Calculate area of raster grid cells in an unprojected raster using ArcGISChanging cell size of raster in QGIS during extraction (clip)QGIS 2D Raster Area CalculationCalculating the area of raster within polygon
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Determine area of cell in raster (QGIS)
SUM Area irregular polygons per grid cellProjecting a Raster from WGS84 to planar Equal Area to Calculate Area: How to Select the Cell Size?Geographic vs. projected raster cell size areaCalculating area of polygon within square mile in ArcGIS?Sum area of specific poly attributes within raster grid cellHow to get unique values and cell counts of a raster in QGIS 2.6?Calculate area of raster grid cells in an unprojected raster using ArcGISChanging cell size of raster in QGIS during extraction (clip)QGIS 2D Raster Area CalculationCalculating the area of raster within polygon
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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?
qgis raster cell-size
asked Apr 1 at 17:15
A.NA.N
261
|
show 3 more comments
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?
qgis raster cell-size
asked Apr 1 at 17:15
A.NA.N
261
1
Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.
– Jon
Apr 1 at 17:17
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...
– Spacedman
Apr 1 at 17:34
1
I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.
– Jon
Apr 1 at 17:53
Thanks. To clarify - with height and width, are you referring to the cell size?
– A.N
Apr 1 at 17:53
1
Then 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.
– user30184
Apr 1 at 19:44
|
show 3 more comments
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?
qgis raster cell-size
asked Apr 1 at 17:15
A.NA.N
261
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?
qgis raster cell-size
qgis raster cell-size
asked Apr 1 at 17:15
A.NA.N
261
asked Apr 1 at 17:15
A.NA.N
261
asked Apr 1 at 17:15
A.NA.N
261
asked Apr 1 at 17:15
A.NA.N
261
asked Apr 1 at 17:15
A.NA.N
261
261
1
Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.
– Jon
Apr 1 at 17:17
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...
– Spacedman
Apr 1 at 17:34
1
I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.
– Jon
Apr 1 at 17:53
Thanks. To clarify - with height and width, are you referring to the cell size?
– A.N
Apr 1 at 17:53
1
Then 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.
– user30184
Apr 1 at 19:44
|
show 3 more comments
1
Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.
– Jon
Apr 1 at 17:17
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...
– Spacedman
Apr 1 at 17:34
1
I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.
– Jon
Apr 1 at 17:53
Thanks. To clarify - with height and width, are you referring to the cell size?
– A.N
Apr 1 at 17:53
1
Then 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.
– user30184
Apr 1 at 19:44
1
1
Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.
– Jon
Apr 1 at 17:17
Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.
– Jon
Apr 1 at 17:17
1
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...
– Spacedman
Apr 1 at 17:34
@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...
– Spacedman
Apr 1 at 17:34
1
1
I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.
– Jon
Apr 1 at 17:53
I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.
– Jon
Apr 1 at 17:53
Thanks. To clarify - with height and width, are you referring to the cell size?
– A.N
Apr 1 at 17:53
Thanks. To clarify - with height and width, are you referring to the cell size?
– A.N
Apr 1 at 17:53
1
1
Then 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.
– user30184
Apr 1 at 19:44
Then 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.
– user30184
Apr 1 at 19:44
|
show 3 more comments
1 Answer
1
active
oldest
votes
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.
answered Apr 1 at 18:02
JonJon
1,4041421
add a comment |
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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.
answered Apr 1 at 18:02
JonJon
1,4041421
add a comment |
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.
answered Apr 1 at 18:02
JonJon
1,4041421
add a comment |
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.
answered Apr 1 at 18:02
JonJon
1,4041421
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.
answered Apr 1 at 18:02
JonJon
1,4041421
answered Apr 1 at 18:02
JonJon
1,4041421
answered Apr 1 at 18:02
JonJon
1,4041421
answered Apr 1 at 18:02
JonJon
1,4041421
1,4041421
add a comment |
add a comment |
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Project your raster layer into an appropriate coordinate reference system (CRS) that uses meters. E.g. find the correct UTM zone and use that CRS.
– Jon
Apr 1 at 17:17
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...
– Spacedman
Apr 1 at 17:34
1
I was assuming the data were in lat/long. Otherwise I thought it was obvious to just look at the pixel resolution, but maybe not.
– Jon
Apr 1 at 17:53
Thanks. To clarify - with height and width, are you referring to the cell size?
– A.N
Apr 1 at 17:53
1
Then 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.
– user30184
Apr 1 at 19:44