[QGIS-Developer] Coordinates of a pixel

Sat Sep 5 03:14:46 PDT 2020

Glad to hear you found a solution. In general, I usually favour the
approach of developing a solution which works in a local coordinate system
and projecting after the fact. It's usually a lot simpler and allows me to
leave the "rocket science" to the rocket scientists...

One caveat you should be aware of though is that calculating positions on
systems that are not cylindrical (e.g. a conical projection like albers
ea), working at very high latitudes (where cylindrical projections probably
aren't suitable anyway) or using images that have some sort of rotation
applied etc may make things considerably more complex and even totally
break your localised implementation (depending on how intelligent the
solution is). I usually try to build things that are dynamic by somehow
binding units/ CRS to the current map frame or similar logic, but in an
instance like this it's probably better to have a hard-coded
"all-or-nothing" solution.

There may also be other situations where things get a bit fuzzy, but I
think as long as you're using UTM and images aren't rotated you should be
OK (just be aware of these limitations if you plan on implementing this for
users in other regions).

Regards

On Sat, 5 Sep 2020 at 02:58, Miguel-Ángel Fas-Millán <fas.millan at gmail.com>
wrote:

> Hi again,
> Thanks for your replies. Charles was right, the deviation between WGS and
> ETRS was not the problem. Finally I'm using UTM while moving through the
> dataset and only convert to geographic coordinates when I arrive at the
> point of interest. The previous way of doing it was quite more complex and
> it seems it added inaccuracies here and there.
>
> Thanks again for your time,
>
> Miguel
>
> El mar., 1 sept. 2020 a las 19:59, Charles Dixon-Paver (<
> charles at kartoza.com>) escribió:
>
>> I would wager that those deviations between WGS and ETRS are unlikely to
>> be the cause of this discrepancy, unless your implementation somehow
>> compounds the inaccuracies between them. The way it's "calculated" in a GIS
>> is by referencing the image to a coordinate system and then returning the
>> coordinates at a specified location (a process which is totally abstract
>> from image pixels).
>>
>> I haven't tested the code you're using, but from what I can see the most
>> likely culprit is some abstract issue with the implementation which is
>> difficult for us to resolve here (although a bug is certainly possible).
>> For example, if you are using a DSM that is in geographic coordinates, the
>> pixel sizes/ cell resolution are geographic units. That means that each
>> cell is not a uniform distance (in meters) in size, yet the function you
>> link to seems to require an input distance parameter in meters.
>> Transforming units between coordinate systems is not trivial and likely
>> requires expert advice. This may not be the case though, as I may be
>> misunderstanding your requirements entirely.
>>
>> I believe Tims response was merely an example of how to get the details
>> of an image which has been imported into QGIS in a similar manner as what I
>> suggested with a world file. Depending on the usage and accuracy required,
>> you can quite easily use the pixel size and the raster extents/ origin to
>> calculate a pixel position (although not very robust,  I find it is often
>> useful).
>>
>> I hope that helps somewhat rather than confusing the issue further. As I
>> stated before, perhaps someone has an idea of where to point you in the
>> codebase (in either QGIS or upstream projects like proj and GDAL), but I
>> don't think you're entirely on the right path here...
>>
>> On Tue, 1 Sep 2020 at 17:59, Miguel-Ángel Fas-Millán <
>> fas.millan at gmail.com> wrote:
>>
>>> Hi again,
>>>
>>> Charles > I checked that, but no. The deviation is not so high. I
>>> suspect that the way it's calculated in QGIS is taking something into
>>> account that the function I'm using (and the web, both return same result)
>>> is not.  Maybe this? My DSM is in ETRS89: "*Note that the World
>>> Geodetic System WGS84 and the European Reference System ETRS89 are
>>> virtually identical and that coordinate transformation between the two
>>> systems in practical navigation is unnecessary. However, for high-precision
>>> surveying work - be aware that the two systems deviates more than half a
>>> meter.*"
>>> But I don't know how to take that deviation into account in the function
>>> or why it seems to affect only  the longitude.
>>>
>>> Code of the calculateEndingGlobalCoordinates that I'm using:
>>> https://github.com/mgavaghan/geodesy/blob/master/src/main/java/org/gavaghan/geodesy/GeodeticCalculator.java
>>>
>>> Tim > Good to know. Not sure how could I leverage those data. I called
>>> pixels  to the DSM cells but really I'm not interested in the graphical
>>> part, just in how is calculating the coordinates of those cells from the
>>> xllcenter and yllcenter correctly.
>>>
>>>
>>>
>>>
>>> El mar., 1 sept. 2020 a las 15:32, Tim Sutton (<tim at kartoza.com>)
>>> escribió:
>>>
>>>> Hi
>>>>
>>>> Also you can get the extents of the layer in the layer properties ->
>>>> Information tab e.g.:
>>>>
>>>>
>>>> Name DEM_10m
>>>> Path /home/timlinux/gisdata/Rwanda/DEM_10m.tif
>>>> CRS EPSG:4326 - WGS 84 - Geographic
>>>> Extent 28.9592194665249139,-2.8404830825979692 :
>>>> 30.8994723851809887,-1.0469841014800894
>>>> Unit degrees
>>>> Width 21524
>>>> Height 19896
>>>> Data type Int16 - Sixteen bit signed integer
>>>> GDAL Driver Description GTiff
>>>> GDAL Driver Metadata GeoTIFF
>>>> Dataset Description /home/timlinux/gisdata/Rwanda/DEM_10m.tif
>>>> Compression
>>>> Band 1
>>>> RepresentationType=THEMATIC
>>>> STATISTICS_APPROXIMATE=YES
>>>> STATISTICS_MAXIMUM=3821
>>>> STATISTICS_MEAN=1697.9520539082
>>>> STATISTICS_MINIMUM=15
>>>> STATISTICS_STDDEV=341.19933119234
>>>> STATISTICS_VALID_PERCENT=56.38
>>>> More information
>>>> AREA_OR_POINT=Area
>>>> DataType=Generic
>>>> Dimensions X: 21524 Y: 19896 Bands: 1
>>>> Origin 28.9592,-1.04698
>>>> Pixel Size 9.01436962765319623e-05,-9.01436962765319623e-05
>>>>
>>>> Regards
>>>>
>>>> Tim
>>>>
>>>>
>>>> On Tue, Sep 1, 2020 at 9:51 AM Charles Dixon-Paver <charles at kartoza.com>
>>>> wrote:
>>>>
>>>>> I took a cursory glance at the calculator you used and the description
>>>>> you've given, and I think there's a possibility you're simply using an
>>>>> easting value in the calculator. You could try multiplying your input
>>>>> longitude by -1 for a quick fix (although I haven't tested this will work
>>>>> at all).
>>>>>
>>>>> On Tue, 1 Sep 2020 at 03:31, Miguel-Ángel Fas-Millán <
>>>>> fas.millan at gmail.com> wrote:
>>>>>
>>>>>> Thanks for your reply. Well, let me provide more details to explain
>>>>>> why I was asking that.
>>>>>> I have a DSM, with its xllcenter/yllcenter coordinates and I need to
>>>>>> know the coordinates of the (in this case) center of any of the
>>>>>> squares/cells (which, maybe wrongly, I called pixels) represented by this
>>>>>> dataset. To that I've been using a function equivalent to the provided by
>>>>>> this calculator:
>>>>>> https://geodesy.noaa.gov/cgi-bin/Inv_Fwd/forward2.prl , which takes
>>>>>> as parameters the starting point, azimuth, distance, and returns the ending
>>>>>> coordinates.
>>>>>>
>>>>>> For instance, I want to calculate the coordinates of the first cell
>>>>>> of the dataset (top left of the matrix). So I do:
>>>>>>
>>>>>> calculateEndingGlobalCoordinates(Ellipsoid.*WGS84*, start_coords, 0.0,
>>>>>> nrows*cellsize) //0.0 for north
>>>>>>
>>>>>> The problem is that it returns a value with a latitude that seems ok,
>>>>>> but a wrong longitude. When I place that coordinate in QGIS, it is out of
>>>>>> the image, quite at left of the top left corner.
>>>>>>
>>>>>> Unsurprisingly, if I place the coordinates obtained with this method
>>>>>> (or with the online calculator mentioned) in google maps, to check if it
>>>>>> makes sense looking at what's there in the satellite images, they
>>>>>> make no sense at all.
>>>>>>
>>>>>> However, if I take the coordinates appearing in QGIS when I hover one
>>>>>> of the cells, and place them in google maps, it makes perfect sense. With a
>>>>>> few meters of difference, but well, at least it is on the right track. (I
>>>>>> am that sure because I took as reference a ATC tower, which is the highest
>>>>>> element in a wide area).
>>>>>>
>>>>>> That's why I wanted to check what's the difference between the
>>>>>> mentioned method I was using and whatever is made in the code to return
>>>>>> those quite correct coordinates.
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> El mar., 1 sept. 2020 a las 0:01, Charles Dixon-Paver (<
>>>>>> charles at kartoza.com>) escribió:
>>>>>>
>>>>>>> Someone can correct me if I'm wrong, but I think there's a
>>>>>>> misunderstanding here of how this raster data is handled in a GIS. >From my
>>>>>>> understanding of the question, I don't know that what you're asking for is
>>>>>>> available "in the code" the way you expect.
>>>>>>>
>>>>>>> The coordinates are showing the position of the cursor relative to
>>>>>>> the origin of the assigned coordinate reference system. The raster data is
>>>>>>> "projected" onto that reference system, which assigns positions to some
>>>>>>> points on the image and stretches, rotates or distorts the image in
>>>>>>> accordance with the images affine parameters that ensure all the different
>>>>>>> parts of the image remain spatially correct. How the GIS knows where to get
>>>>>>> these parameters varies between data types and file formats.
>>>>>>>
>>>>>>> Playing around with the georeferencer tool in QGIS should give you a
>>>>>>> pretty clear understanding of how this "projection" works.
>>>>>>>
>>>>>>> There are ways to get the coordinates of a position or pixel within
>>>>>>> an image programmatically, the easiest of which that I can think of is
>>>>>>> using values from a world file [1] with an xy position (in pixels) of the
>>>>>>> pixel of interest. The code required for achieving this, however, is
>>>>>>> probably going to be dependant on a wide variety of factors (not
>>>>>>> least of all the CRS units and pixel size). Alternatively, you could likely
>>>>>>> grab the coordinate position of a pixel from within a QGIS project, but
>>>>>>> that doesn't seem to be what you're after. Perhaps a developer familiar
>>>>>>> with the GDAL or QGIS code bases can point you in the direction of some
>>>>>>> wizardry that will achieve what it is you are looking for without a clearer
>>>>>>> understanding of your use case though.
>>>>>>>
>>>>>>> [1] https://en.wikipedia.org/wiki/World_file
>>>>>>>
>>>>>>> On Mon, 31 Aug 2020 at 19:57, Miguel-Ángel Fas-Millán <
>>>>>>> fas.millan at gmail.com> wrote:
>>>>>>>
>>>>>>>> Hi all,
>>>>>>>> New around here and as a QGIS user. I need to find something in the
>>>>>>>> code, I hope you can help me.
>>>>>>>> When you open a DSM and place the mouse on any pixel, you can see
>>>>>>>> the coordinates of that point. Could someone tell me where in the code is
>>>>>>>> made the calculation of those coordinates?
>>>>>>>>
>>>>>>>> Regards,
>>>>>>>>
>>>>>>>> Angel
>>>>>>>>
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>>>>
>>>>
>>>> --
>>>>
>>>> ------------------------------------------------------------------------------------------
>>>> 
>>>>
>>>> Tim Sutton
>>>> Visit http://kartoza.com to find out about open source:
>>>>  * Desktop GIS programming services
>>>>  * Geospatial web development
>>>> * GIS Training
>>>> * Consulting Services
>>>> Skype: timlinux Irc: timlinux on #qgis at freenode.net
>>>> Tim is a member of the QGIS Project Steering Committee
>>>>
>>>> -------------------------------------------------------------------------------------------
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>>>>
>>>
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