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View Full Version : free SW for Transverse Mercator projections?



jumpjack
02-05-2009, 07:45 AM
Does it exist?

I also need Lambert Azimutal Equal Area projections.

RobA
02-05-2009, 10:02 AM
Take a look at Flex Projector (http://www.flexprojector.com/) and G.Projector (http://www.giss.nasa.gov/tools/gprojector/)

One of them may have the projections you require.

-Rob A>

jumpjack
02-05-2009, 01:24 PM
Take a look at Flex Projector (http://www.flexprojector.com/) and G.Projector (http://www.giss.nasa.gov/tools/gprojector/)

One of them may have the projections you require.

-Rob A>

I didn't notice Flex Projctor also supports raster images, I'll try it again.
I also already tried G.Projector, but it lacks Transverse Mercator projections.

I found a dozen of programs: DGALWARP, fGIS, GRASS, PROJ.4, GMT & WinGMT, fTools... but I always miss HOW to use them! They're really complex programs! I just need to know how to get a projection from an image, I can't study tons of pages about cartography! :-(

DGALWARP (http://www.forestpal.com/image_reprojection.html) appears to be quite good to be explained in a forum: does anybody have an idea of the needed command line to put one of these images (http://www.evl.uic.edu/pape/data/Earth/)into Transverse Mercator and Lambert Azimuthal Equal Area?

RobA
02-05-2009, 01:51 PM
Based on your other post, I think what you want is to go from an image like this (equirectangular):

9997

to this:

9998

I did this using Gimp, with the mathmap plugin, using the following mathmap script I wrote:


unit filter sinusoidal (unit stretched image in)
divs = 9;
py = 2;
counter = 0;
while (counter <= divs) do
px = if (x>1-(2*counter/divs)-1/divs && x<=1-(2*counter/divs)+1/divs)
then (x-(1-(2*counter/divs)))/cos(y*pi)+1-(2*counter/divs)
end;
py = if (x>1-(2*counter/divs)-1/divs && x<=1-(2*counter/divs)+1/divs &&
px<=1-(2*counter/divs)+1/divs && px>=1-(2*counter/divs)-1/divs)
then y
else py
end;
counter = counter + 1;
end;
in(xy:[px,py])
end


Change the divs=9 to however many "petals" you want.

edit: Another thing tha might make this easier to assemble is what i did in the next one. I offset bottom half of the image by 1/2 the petal width (in this case 60 px), ran the filter, then offset the bottom half back (-60px), with wrap on, of course.:

9999

-Rob A>

jumpjack
02-06-2009, 02:42 AM
Based on your other post, I think what you want is to go from an image like this (equirectangular):

9997

to this:

9998

I did this using Gimp, with the mathmap plugin, using the following mathmap script I wrote:
that's a very cool news: I'm a developer, so maybe I could easily implement other projections I need... if just knew the formulas/algorithms I need! :D

I already tried "gores" projections, but it's only good up to 70-75 latitude: it's really hard to properly stitch gores at poles, that's why I also need the Lambert azimutal projection centered on poles.
Do you know the formula for it too?

waldronate
02-06-2009, 03:36 AM
See the other thread for a description of how to use Wilbur to do the projection you're asking about. It should be possible to do this with an FT projection definition.

RobA
02-06-2009, 04:14 PM
that's a very cool news: I'm a developer, so maybe I could easily implement other projections I need... if just knew the formulas/algorithms I need! :D

I already tried "gores" projections, but it's only good up to 70-75 latitude: it's really hard to properly stitch gores at poles, that's why I also need the Lambert azimutal projection centered on poles.
Do you know the formula for it too?

Almost every protection formula is at wikipedia:

http://en.wikipedia.org/wiki/Lambert_azimuthal_equal-area_projection

-Rob A>

jumpjack
02-08-2009, 09:15 AM
Based on your other post, I think what you want is to go from an image like this (equirectangular):

9997

to this:

9998

I did this using Gimp, with the mathmap plugin, using the following mathmap script I wrote:


unit filter sinusoidal (unit stretched image in)
divs = 9;
py = 2;
counter = 0;
while (counter <= divs) do
px = if (x>1-(2*counter/divs)-1/divs && x<=1-(2*counter/divs)+1/divs)
then (x-(1-(2*counter/divs)))/cos(y*pi)+1-(2*counter/divs)
end;
py = if (x>1-(2*counter/divs)-1/divs && x<=1-(2*counter/divs)+1/divs &&
px<=1-(2*counter/divs)+1/divs && px>=1-(2*counter/divs)-1/divs)
then y
else py
end;
counter = counter + 1;
end;
in(xy:[px,py])
end


Change the divs=9 to however many "petals" you want.

edit: Another thing tha might make this easier to assemble is what i did in the next one. I offset bottom half of the image by 1/2 the petal width (in this case 60 px), ran the filter, then offset the bottom half back (-60px), with wrap on, of course.:

9999

-Rob A>
I just discovered that GIMP plugins can also be written in Python:
http://www.jamesh.id.au/software/pygimp/

Are you able to port your short source to python? I already know it, so I won't have to learn a new language.

RobA
02-08-2009, 04:16 PM
I just discovered that GIMP plugins can also be written in Python:
http://www.jamesh.id.au/software/pygimp/

Are you able to port your short source to python? I already know it, so I won't have to learn a new language.

If you know python YOU should be able to convert it ;). The math for that mapping is trivial (I got it from wikipedia)

If you are willing to use Gimp then the mathmap script will work fine, as mathmap is available for every OS.

It is much faster than scripting in scheme and much simpler than scripting in python (when performing manipulation of image data.)

If you want a stand-alone python script, I'll leave that to you...

-Rob A>