I am creating a new script in graphical modeler. I have an input which is « vector field » as point.

But when I use « TIN interpolation », I’m not able to select an « input of the model ». I can only select layers that are already loaded in QGIS, which is not very useful in graphical modeler!

1 Answer 1


I'm also facing this problem in my model's development, currently, I'm using the Thin Plate Spline (TIN) algorithm as a replacement for QGIS' native TIN algorithm or the Multilevel B-Spline depending on my needs.

I think you should post a bug report as that's not the intended behavior.

See https://github.com/qgis/QGIS/issues/30029, as of 3.22.11 LTR this bug hasn't been resolved, the workaround is using SAGA TIN for a drop-in replacement (albeit with different result compared to the native algorithm).

Edit: I'm currently using Triangulation [saga:triangulation] from SAGA (https://saga-gis.sourceforge.io/saga_tool_doc/7.8.2/grid_gridding_5.html), which need a point layer. So, as an intermediate, I'm passing the vector layer to Vector geometry -> Points along geometry [native:pointsalonglines] with 2m spacing. Both works in model builder and result in an otherwise identical result (to the naked eye, and in my case, after converting back to contour line) to qgis:tininterpolation.

  • As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
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    Commented Oct 3, 2022 at 16:55
  • @katagena adding additional information, you can use saga:triangulation as a drop-in replacement to qgis:tininterpolation. If your vector layer is a structure or break line, you will need to pass the layer to native:pointsalonglines. Both works in model builder. Commented Oct 14, 2022 at 15:51

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