Take the 2-minute tour ×
Geographic Information Systems Stack Exchange is a question and answer site for cartographers, geographers and GIS professionals. It's 100% free, no registration required.

I need to re-project a PNG from a given projection to the Google maps projection.
I have a grid of the source projection, looks like a Mercator projection. What is the exact name of this projection in the GDAL classification?

I need to re-project this exact region of the map, so if it is difficult to tell exactly, an any projection which would be a good approximation for this region would do.

Sorry for the dumb question, I'm a newbie.

projection grid

Thanks!

share|improve this question
add comment

2 Answers

up vote 8 down vote accepted

To analyze this, I captured the image, magnified it approximately five fold, and digitized 52 locations where coordinate lines cross. For the record, the data are appended below. (Assuming I located the crossings correctly to the nearest pixel, we should expect the errors to be on the order of 1/5 = 0.20 pixel: there's no getting around this basic discretization error.)

Because the meridians are equally spaced, parallel, and vertical, this must be the standard aspect of some cylindrical projection. The measurements indicate the y-spacing is inversely proportional to the cosine of the latitude; that is, y should be the integral of the secant function: this identifies it as some variant of a Mercator projection. As a check, regression of the y-coordinate on the Mercator formula (ln(sec(lat) + tan(lat))) works extremely well, with a typical residual error of 0.18 of a pixel on the screen. (Residuals appear random, are uncorrelated with latitude, longitude, or their squares, and exhibit no heteroscedasticity.) Comparing 0.18 to the expected error of (approximately) 0.20 pixel shows that's as well as one can possibly do with this image.

As a double-check, I created a small shapefile reproducing the graticule shown in the image, projected it with a Mercator projection (based on the sphere, with a standard parallel of 0 degrees), and adjusted the world file of the image to position and rescale it to match the lines of latitude in the projected shapefile. The match is visually perfect:

enter image description here

However, as is readily seen, 12 degrees of longitude in the projected (red) graticule span only 11 degrees of longitude in the original image. This indicates the Mercator projection has been expanded horizontally by a factor of 12/11 (approximately). Such an expansion still gives a valid projection, but the result is no longer conformal (which is one of the exemplary properties of the Mercator).

Note that this distorted Mercator is not necessarily identical to the original projection: especially over small areas of the globe, two different projections can have excellent visual matches. The analysis has established that a distorted Mercator will work well within the accuracy of the original image itself, and (most likely) only for features within the extent of that image. It also successfully rules out many other likely possibilities, including Transverse Mercator (meridians would not be equally spaced and would have to curve towards each other at the top and bottom) and Stereographic (which is conformal, whereas this projection clearly is not).

For those who would care to improve on this analysis, here are the measurements.

X   Y   lon lat
44.248  590.275 13  56
91.999  513.548 14  55
139.427 438.279 15  54
186.855 365.113 16  53
234.283 293.567 17  52
281.873 223.963 18  51
329.301 155.330 19  50
376.567 88.153  20  49
423.995 22.758  21  48
471.423 88.477  22  49
519.013 155.491 23  50
566.603 223.963 24  51
613.707 294.052 25  52
91.789  590.335 14  56
138.889 590.335 15  56
186.930 590.021 16  56
234.187 590.335 17  56
281.601 590.649 18  56
329.015 590.492 19  56
376.900 590.021 20  56
423.842 590.335 21  56
471.256 590.492 22  56
518.984 590.335 23  56
566.712 590.335 24  56
613.340 590.492 25  56
613.654 513.720 25  55
613.340 438.674 25  54
613.340 365.512 25  53
613.026 294.078 25  52
613.497 224.056 25  51
613.183 155.447 25  50
613.497 88.566  25  49
566.398 22.940  24  48
519.455 22.783  23  48
471.256 23.097  22  48
423.999 22.940  21  48
376.429 22.940  20  48
329.329 23.097  19  48
281.287 22.940  18  48
234.030 23.097  17  48
186.930 23.097  16  48
139.517 22.940  15  48
91.632  23.097  14  48
44.061  22.783  13  48
43.904  88.880  13  49
44.061  155.604 13  50
44.532  224.056 13  51
44.218  294.235 13  52
44.375  365.198 13  53
44.061  438.517 13  54
44.218  513.249 13  55
44.375  590.178 13  56
share|improve this answer
2  
wow! that's an answer! –  axk Aug 1 '11 at 17:30
add comment

You can't get the projection just by looking at an image. But we can try to guess.

The mark in the lower right is for IM GW which appears to be a Polish weather service. Searching their site for projections eventually yields multiple references to GUGiK 1992/19.

This may be the same as GUKiK 1980 or System 1992/19 referenced in this file.

Try one of those and see if it lines up.

Update: The scale factor listed in this document [Warning: Word file] for GUKiK 1992/19 is the same as the one listed for System 1992/19. EPSG:2180 might be close enough (or the same), too.

share|improve this answer
    
Thanks! I'll try to re-project with this source projection type. –  axk Jul 30 '11 at 8:55
1  
(+1) Although the meridians are equally spaced, careful measurement shows the parallels are not. Thus the projection is cylindrical. The measurements indicate the y-spacing is inversely proportional to the cosine of the latitude, showing it is likely an equal area (cylindrical) projection. That rules out Transverse Mercator, Equirectangular, or any form of the Stereographic projection, which would eliminate both GUKiK 1992/19 and 1980 from consideration. –  whuber Jul 30 '11 at 17:06
    
@whuber, thanks! I'll try to figure out how to reproject from a cylyndrical projection to Meractor now. –  axk Jul 31 '11 at 9:50
    
Looks like the spacing between the parallels is decreasing towards the equator, and with an equal area cylindrical projections they should increase. –  axk Jul 31 '11 at 21:15
1  
@axk You're correct. I got it reversed. So let's pull out the big guns and drop the data into a regression package. It tells me that, to the accuracy contained in a megapixel version of your image, that y is directly proportional to log(sec(lat) + tan(lat)): in other words, this is a Mercator projection (not transverse!). –  whuber Aug 1 '11 at 14:28
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.