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Is it possible to clip a polygon from a raster layer without being georeferenced? When I try I get this error message "ERROR 1: Unable to compute a transformation between pixel/line and georeferenced coordinates" and am not really sure what it means. I'm speculating that it wants me to georeference the raster layer but I can't as it is a photo, not a map (not a photo of a map) and if I just georeference arbitrarily to try and placate it then the pictures warps.

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  • If its not geospatial data, why not use some other application to cut out a part of the picture?
    – BradHards
    Aug 20 '16 at 11:43
  • @BradHards - I'm doing a load of stats and other stuff using QGIS with my pictures and have ~1000 to do. I want to come up with a method that avoids jumping back and forth between applications.
    – Monica
    Aug 21 '16 at 10:08
  • got the same error message trying to clip a global land survey raster May 11 '19 at 21:21
  • This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
    – Dan C
    May 11 '19 at 22:02
  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review May 11 '19 at 23:03
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While the raster may be left un-georeferenced, a vector layer can not. You need coordinates in a defined CRS for the vertices of the polygon.

What you can do:

Choose an arbitrary projected CRS (like Web mercator EPSG:3857), and set the extent of the raster with gdal_translate -a_ullr (see http://www.gdal.org/gdal_translate.html) exactly to the size of the raster. This prevents the raster from getting stretched. Make sure that the layer and project CRS are EPSG:3857 if it does not look as expected.

Then you can create a polygon layer in the same CRS, and use the clip function.


UPDATE

Your issue seems to be a bit tricky. Gdalwarp expects to do some kind of transforming, but fails on an identical reprojection.

This way it works for me:

gdal_translate -of Gtiff -a_srs EPSG:3857 -a_ullr 0 3456 4608 0 Aarhus.jpg AarhusNew.tif
gdalwarp -q -cutline clippoly.shp -tr 1.0 1.0 -of GTiff -dstnodata 0 AarhusNew.tif -overwrite  clipped.tif

You have to re-create the clip layer, because all coordinates are now positive (whereas negative Y values fail in your example, beeing identical to the internal pixel/line reference).

looking like this:

enter image description here

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  • When I imported/ created the layers previously I have already been setting the CRS as 3857 and getting error message ERROR 1. I'm not really sure what the gdal translate is meant to do but when I run it I get the message "Input file size is 4608, 3456 Computed -srcwin 0 0 4608 -3455 from projected window. Error: Computed -srcwin 0 0 4608 -3455 has negative width and/or height"
    – Monica
    Aug 21 '16 at 12:34
  • I still get exactly the same error message
    – Monica
    Aug 22 '16 at 12:33
  • Can you upload a sample file via dropbox or similar?
    – AndreJ
    Aug 22 '16 at 15:35
  • Dunno if I've done this correctly but here is a link to the file I'm playing with: dropbox.com/s/qbsh4lsj29uqrke/practice.qgs?dl=0 For bonus points, this one has my standard error message on it: dropbox.com/s/wv6lmqbd9a4cy2p/…
    – Monica
    Aug 23 '16 at 18:43
  • The .qgs file does not reveal everything. Adding the tif (or png) and shapefile (with shx/dbf/prj, maybe zipped) might help to reproduce the error.
    – AndreJ
    Aug 24 '16 at 11:09

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