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I have a world map which I assumed used a Mercator projection but when I tried to plot points on it using a formula for the Spherical Pseudo-Mercator projection (from here: http://wiki.openstreetmap.org/wiki/Mercator#Python) but the Y coordinates seem off. How can Identify which projection is it using? Can someone provide pseudocode/Python for transforming lat,long pairs to x,y pixels for it? If it may help, this image comes from a KML file.

  <GroundOverlay>
    <name>1986-2010</name>
    <altitude>0</altitude>
    <altitudeMode>clampToGround</altitudeMode>
    <Icon>
      <href>KG_1986-2010.png</href>
    </Icon>
    <LatLonBox>
      <north>90</north>
      <south>-90</south>
      <east>180</east>
      <west>-180</west>
    </LatLonBox>
  </GroundOverlay>

Climate projection

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  • Note that KML by specification uses only a single projection, EPSG:4326
    – nickves
    Apr 25, 2017 at 0:43
  • I could not find that in the documentation, thanks. Do you have a link? Apr 26, 2017 at 15:37

2 Answers 2

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GCS WGS 1984 works for me.

I added to ArcGIS, added Esri's world map data (which is stored in GCS_WGS_1984) and did a quick two point (top-left, bottom-right) reference and it seems to fit pretty well.

enter image description here

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The projection system in question is the simple cylindrical (or equi-rectangular or platte carée or cylindrical equidistant) projection.

The distinguishing feature is that each 15° x 15° graticule section (or any two equal lat-lon differences) maps as a square. Another feature is that all meridians (N-S lines) are true to scale.

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  • Thanks, with the projection name I was able to find a formula for me to plot to. If someone needs it: x = ((MAP_WIDTH/360.0) * (180 + lon)); y = ((MAP_HEIGHT/180.0) * (90 - lat)) Apr 24, 2017 at 21:36

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