Create TIN out of Raster with berrier polyline

Nicola — Wed, 05 May 2021 11:17:17 GMT

Hello everyone,

At the moment I'm creating a Toolbox for the visualization of subsurface 3D models. For that I would like to work with TINs. When I do the workflow manually it works perfectly fine. But with my code I have a problem and I don't understand it.

Here is the code for going through my shapefiles, create rasters out of them with the use of the geological map outlines. The output for the raster is correct, it's perfectly in the shape of the geological map. But the TIN is created as a square. As mentioned before, when I do the steps one after another in ArcPro the TIN is also created in the shape of the geological map.

shpFileList = arcpy.ListFiles("*.shp")
for shpFile in shpFileList:
   out_location = arcpy.GetParameterAsText(6)
   arcpy.FeatureClassToGeodatabase_conversion(shpFile,out_location)
   zField = "Extrusion"
   shpFileName = os.path.splitext(shpFile)[0]
   outRaster = shpFileName + "out.img"
   arcpy.Idw_3d(shpFile,zField, outRaster,in_barrier_polyline_features=geo_line_out)
   arcpy.RasterToGeodatabase_conversion(outRaster,out_location)
   out_TIN = shpFileName + "tin"
   arcpy.RasterTin_3d(outRaster, out_TIN)

Maybe anybody can see the mistake?

Best Regards!

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Create TIN out of Raster with berrier polyline