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How big is a MODIS 250m pixel in reality?

This appears to be a question that contains its own answer, but I have several reasons to doubt that the pixel size is actually 250m x 250m:

  1. I downloaded MODIS 250m NDVI data from the USGS site, transformed it from Sin to Geographic (SRID 4326) using the MODIS Reprojection Tool (which has since been replaced, but at the time was the official tool).
  2. Using gdalinfo on the resulting file gives the pixel size as (0.002884053983564 X 0.002884053983564). I assume that this measurement is in decimal degrees.
  3. This online calculator gives the length of one degree of latitude at the latitude of my study area as 111,092.7 m, and of longitude as 81,540.9 m. That would make the pixel dimensions 320m in the N-S direction and 235m in the E-W direction.
  4. I've overlaid the NDVI image onto other data, and the two correspond; for example you can clearly see greener vegetation along rivers. This wouldn't happen if they were misaligned or mis-projected.
  5. I can also measure the size of a pixel using QGIS's measurement tool, and the length comes out to about 320 x 235 m, i.e., it agrees with the file metadata.

I have read the documentation and found a few suspicious references in published papers to things like "nominal" pixel size of 250 m, or pixel size of 250 m "at nadir", and this paper makes it look to me as if the physical camera/mirror apparatus would result in pixels that aren't 250m x 250m square, but the authors don't describe in detail the projection and/or transformation used to create the data products so I'm not sure.

  • Note that you are using MODIS L3 data which is resampled onto the MODIS sinusoidal grid. That is the relation between the size of the MODIS L3 pixel and the size of one or more corresponding pixels within a swath is just a decision taken at the stage of processing algorithm design. Could have been resampled onto a different grid altogether. – Dmitri Chubarov Mar 17 at 4:37
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Per https://lpdaac.usgs.gov/dataset_discovery/modis, the viewing swath width of MODIS is 2,330 km, thus a large portion of the image is off-nadir in some way. https://modis.gsfc.nasa.gov/about/specifications.php

The following forum post gives an explanation of how to calculate pixel size based on viewing position. (Note: still an estimate due to factors outlined in the post) https://oceancolor.gsfc.nasa.gov/forum/oceancolor/topic_show.pl?tid=2018

Compute the scan angle, S (in radians), given pixel number:

S = (I-hp)/H

where:

 I is the zero-based pixel index
 hp is 1/2 the total number of pixels (zero-based) (for MODIS each scan is 1354 "1km" pixels, 1353 zero-based, so hp = 676.5)
 H is the sensor altitude divided by the pixel size (for MODIS altitude is approximately 700km, so for "1km" pixels, H = 700/1)

 For 500m pixels, hp = 1353, H = 1400 (700/0.5)
 For 250m pixels, hp = 2706, H = 2800 (700/0.25)

Compute the zenith angle:

Z = asin(1.111*sin(S))  

where Z is the zenith angle.

Compute the Along-track pixel size:

Pt = Pn*9*sin(Z-S)/sin(S)

where Pn is the nadir pixel size (e.g. 1km, 0.5km, 0.25km)

Compute the Along-scan pixel size:

Ps = Pt/cos(Z)

Thus, area is ~ Pt * Ps

Additional information can be found at https://oceancolor.gsfc.nasa.gov/forum/oceancolor/topic_show.pl?tid=277

  • Excellent answer; these equations are very useful. – John Mar 15 at 21:12
  • The equations show that the pixel size varies with pixel position relative to the center of the satellite path, but they do not show size varying with latitude. Am I right to assume that these equations hold everywhere, e.g., even over the poles? If I'm envisioning it correctly, there must be much more overlap between successive passes at the poles than at the equator, meaning that a given location on earth might have different pixel widths in successive passes (i.e., up near the pole it might be close to the center of the path on one pass, further on another). – John Mar 15 at 21:27
  • I'm not familiar enough with the data / couldn't find a specific reference about the differences by latitude. If anyone finds, please add an answer. – smiller Mar 18 at 13:15

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