How big is a MODIS 250m pixel in reality?How to mosaic multiple MODIS images of Multiple months using MRT tool?How to reproject MODIS L2 data with lat/lonHow to remove cloudy pixel from MODIS NDVI (MOD13Q1)How can I create a polygon(filled polygon) from a single pixel stored in a hdf file(MODIS)?Issue with mapping MODIS SIN grid to EASE 2.0 with gdalwarpHow to mosaic multiple modis tiles of several dates using mrt tool?MODIS Landsat overlappingMODIS R Connection errorChange pixel size of MODIS in GEEMODIS 04_L2 projection transformation

A Cautionary Suggestion

Gantt Chart like rectangles with log scale

What's the meaning of “spike” in the context of “adrenaline spike”?

Employee lack of ownership

If curse and magic is two sides of the same coin, why the former is forbidden?

Are there verbs that are neither telic, or atelic?

Can a druid choose the size of its wild shape beast?

Min function accepting varying number of arguments in C++17

How to write cleanly even if my character uses expletive language?

Have researchers managed to "reverse time"? If so, what does that mean for physics?

How could a scammer know the apps on my phone / iTunes account?

Most cost effective thermostat setting: consistent temperature vs. lowest temperature possible

How to create the Curved texte?

Is a party consisting of only a bard, a cleric, and a warlock functional long-term?

My adviser wants to be the first author

Why did it take so long to abandon sail after steamships were demonstrated?

How to simplify this time periods definition interface?

Brexit - No Deal Rejection

How to use deus ex machina safely?

Credit cards used everywhere in Singapore or Malaysia?

PTIJ: Who should I vote for? (21st Knesset Edition)

Bach's Toccata and Fugue in D minor breaks the "no parallel octaves" rule?

Do I need to be arrogant to get ahead?

Does Mathematica reuse previous computations?



How big is a MODIS 250m pixel in reality?


How to mosaic multiple MODIS images of Multiple months using MRT tool?How to reproject MODIS L2 data with lat/lonHow to remove cloudy pixel from MODIS NDVI (MOD13Q1)How can I create a polygon(filled polygon) from a single pixel stored in a hdf file(MODIS)?Issue with mapping MODIS SIN grid to EASE 2.0 with gdalwarpHow to mosaic multiple modis tiles of several dates using mrt tool?MODIS Landsat overlappingMODIS R Connection errorChange pixel size of MODIS in GEEMODIS 04_L2 projection transformation













1















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.










share|improve this question




























    1















    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.










    share|improve this question


























      1












      1








      1


      2






      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.










      share|improve this question
















      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.







      remote-sensing modis pixel modis-reprojection-tool cell-size






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 4 hours ago









      Aaron

      38.2k20109254




      38.2k20109254










      asked 8 hours ago









      JohnJohn

      887




      887




















          1 Answer
          1






          active

          oldest

          votes


















          4














          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






          share|improve this answer

























          • Excellent answer; these equations are very useful.

            – John
            7 hours ago











          • 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
            6 hours ago










          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "79"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fgis.stackexchange.com%2fquestions%2f315691%2fhow-big-is-a-modis-250m-pixel-in-reality%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4














          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






          share|improve this answer

























          • Excellent answer; these equations are very useful.

            – John
            7 hours ago











          • 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
            6 hours ago















          4














          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






          share|improve this answer

























          • Excellent answer; these equations are very useful.

            – John
            7 hours ago











          • 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
            6 hours ago













          4












          4








          4







          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






          share|improve this answer















          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







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          smillersmiller

          2,134217




          2,134217












          • Excellent answer; these equations are very useful.

            – John
            7 hours ago











          • 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
            6 hours ago

















          • Excellent answer; these equations are very useful.

            – John
            7 hours ago











          • 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
            6 hours ago
















          Excellent answer; these equations are very useful.

          – John
          7 hours ago





          Excellent answer; these equations are very useful.

          – John
          7 hours ago













          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
          6 hours ago





          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
          6 hours ago

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Geographic Information Systems Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fgis.stackexchange.com%2fquestions%2f315691%2fhow-big-is-a-modis-250m-pixel-in-reality%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          名間水力發電廠 目录 沿革 設施 鄰近設施 註釋 外部連結 导航菜单23°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.7113923°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.71139計畫概要原始内容臺灣第一座BOT 模式開發的水力發電廠-名間水力電廠名間水力發電廠 水利署首件BOT案原始内容《小檔案》名間電廠 首座BOT水力發電廠原始内容名間電廠BOT - 經濟部水利署中區水資源局

          Prove that NP is closed under karp reduction?Space(n) not closed under Karp reductions - what about NTime(n)?Class P is closed under rotation?Prove or disprove that $NL$ is closed under polynomial many-one reductions$mathbfNC_2$ is closed under log-space reductionOn Karp reductionwhen can I know if a class (complexity) is closed under reduction (cook/karp)Check if class $PSPACE$ is closed under polyonomially space reductionIs NPSPACE also closed under polynomial-time reduction and under log-space reduction?Prove PSPACE is closed under complement?Prove PSPACE is closed under union?

          Is my guitar’s action too high? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strings too stiff on a recently purchased acoustic guitar | Cort AD880CEIs the action of my guitar really high?Μy little finger is too weak to play guitarWith guitar, how long should I give my fingers to strengthen / callous?When playing a fret the guitar sounds mutedPlaying (Barre) chords up the guitar neckI think my guitar strings are wound too tight and I can't play barre chordsF barre chord on an SG guitarHow to find to the right strings of a barre chord by feel?High action on higher fret on my steel acoustic guitar