I've been struggling with this for a while and wondered if someone could lend a hand. I have a map which already has sampling points from centroids in a 35x35km grid that was created in north pole stereographic. I need the grid to represent 5x5km, essentially dividing the grid by 7; however, I can't for the life of me figure it out. I've tried to buffer, I've tried to duplicate and move the affline (large area, so it gets distorted), etc. Densifying the lines creates equal spacing along the edges of the polygons but I can't figure out how to get them to join in lines, because of the offset of the lines due to the projection warp. Attached is a picture of what it looks like - any ideas?enter image description here

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    Why don't you work with this grid in its original coordinate system? That is the only way it will appear as a collection of congruent squares. – whuber Nov 8 '13 at 4:11
  • The grid was originally created in this projection, to provide equal area grid cells that line up for the entirety of north America. It wasn't created using the usual methods for creating grids (fishnets, vector grid, etc), but I believe it was built using hand-converted coordinate points. Sorry if I am misunderstanding something. – fork_in_toaster Nov 8 '13 at 16:07
  • Do you have any specific information on how the grid actually was created, then? (Despite what you write, I strongly suspect it was built in the usual way using a different projection and then reprojected.) – whuber Nov 8 '13 at 16:09
  • I have a readme file: The grid is a polar stereographic projection with a standard longitude of 100 W. These are the formulas used to convert between lat/lon and ix,iy (grid co-ordinates): lat,lon -> ix,iy: ix = k * sin(lon + 100) + 88; iy = - k * cos(lon + 100) + 212; Where k = cos(lat)/(1+sin(lat)) * ( R/35 * (1+SIN(60)) ); and R is the radius of the earth ( 6370.997 km) ix,iy -> lat,lon: lat = 90 - 2*ARCTAN( SQRT(xx + yy) /( R/35 * (1+SIN(60))) ); lon = -100 - ARCTAN(x/y) ; Where x = ix - 88; y = iy - 212; DATE periods: 1994-1998 – fork_in_toaster Nov 8 '13 at 16:23
  • It's safe to assume the polygons were originally drawn as rectangles, implying that wherever four of them meet at a corner, all angles are equal (to 90 degrees). That obviously is not the case in this image, proving it was not made with any conformal projection, much less a stereographic (which is conformal). Therefore the resolution to your problem likely lies in finding out what projection it really uses. – whuber Jan 21 '15 at 16:37

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