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4

In the attribute view create your new columns as decimal (floats). Then enable editing, and use expressions to update the columns. To get the latitude (x-coordinate) of the start of a line feature, use the expression x_at(0). To get the end of a line feature use x_at(-1). Repeat for your longitudes (y-coordinates). You select the column, type in the ...


4

You have your four columns: X-start; Y-start; X-end; Y-end. Using the field calculator for each column update using: $x_at(0) $x_at(-1) $y_at(0) $y_at(-1) This asks for the coordinate at the start of the line (the zero) and the end of the line (the negative one). This will provide you with a decimal latitude and longitude, so ensure the columns are ...


2

Looks like you're using a CRS in metres but trying the style it using a degrees based format. You'll either need to set the grid projection to a geographic one (ie WGS84), or use a different coordinate format.


2

Have a look at How to georeference a vector layer with control points?, in particular the answer concerning the Vector Bender plugin for QGIS. Watch the video and try the plugin, report back if you encounter any issues.


1

Use label expression: def FindLabel ( [OID] ): mxd = arcpy.mapping.MapDocument("CURRENT") lr=arcpy.mapping.ListLayers(mxd)[0] with arcpy.da.UpdateCursor(lr, ('Shape@','POINT_X','POINT_Y'),r'"OID"='+str( [OID] )) as cursor: for row in cursor: row[1]=row[0].firstPoint.X row[2]=row[0].firstPoint.Y cursor.updateRow(row) ...


1

I think you have to carry out three transformations, wich are all affine transformations: scale, rotate and translate. I suggest reading about affine transformation on wikipedia: https://en.wikipedia.org/wiki/Affine_transformation If you basically can carry out an affine transformation, then this might help you to determine the correct parameters: http://...


1

The plugin in the accepted answer is not available anymore. The Affine Transformations is available and is very useful. With this plugin you can create a formula. In the screenshot I shifted all cells with 17.396 (meters) to the north.


1

The points look like this, when drawn in a cartesian coordinate system:


1

Alternatively to your solution (which seems to work in ARCGIS), you can set up an oblique mercaor projection. This would work in QGIS and GDAL-based software as well: PROJ.4 : +proj=omerc +lat_0=55.102476 +lonc=1.42624 +alpha=-43.22 +k=0 +x_0=0 +y_0=0 +gamma=0 +datum=WGS84 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs OGC WKT : PROJCS["unnamed", GEOGCS["WGS ...


1

As a first guess, you can create a custom transverse mercator projection on one of you coresponding points. Preferably the one closest to the DXF origin: +proj=tmerc +lat_0=41.045452 +lon_0=28.895929 +k=1 +x_0=1.03521 +y_0=-6.34918 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs lon_0 and lat_0 are the degree coordinates in WGS84, and x_0 and y_0 ...



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