I started to study TIN in QGIS and created my first TIN from a point layer with elevation. There are two interpolation methods called Linear and Clough-Tocher that give very different results. Linear looks like a regular Raster and with Clough-Tocher the result cannot be interpreted at first glance. Further the values of both methods vary a lot, as can be seen in the picture.

enter image description here

I cannot find information in the QGIS Docs what the difference is between the two methods.

What is the difference and why are the results so different?

I am using QGIS 3.4.13.

  • 4
    Really a good question. I can't answer it propperly. Just to comment that extreme values are inside the triangles of the triangulation and the surface satisfaces a continuous slope at each point. The algorithm creates "valleys and mountains" within triangles in order to satisfy that condition. Jan 31, 2020 at 22:06
  • 2
    For Clough-Tocher see: xmswiki.com/wiki/GMS:Clough-Tocher
    – Babel
    Aug 31, 2021 at 16:22
  • I tried out both interpolation methods on a small test dataset, and the results are comparable... Aug 31, 2021 at 16:23

1 Answer 1


It seems that Linear and Clough-Tocher are quite coherent in respect to each other. Out of curiosity, I tried interpolating a small, random dataset with both algorithms, also opting for the creation of the underlying triangulation networks:

enter image description here

As expected, there is only a very limited difference:

enter image description here


With a larger dataset of 10.000 random points, differences surface (pun intended). In particular:

  • the underpinning TIN is identical, while
  • the single faces are interpolated differently

Clough-Tocher interpolation: enter image description here

Linear interpolation: enter image description here

Zoom in:

enter image description here

Also, I find the histograms interesting:

Clough-Tocher interpolation histogram: enter image description here

Linear interpolation histogram: enter image description here

However, if we create a difference layer between the two interpolated rasters, and sample it with the original 10k vector points, we see that on average the two interpolation methods do not differ at vertexes/original points, which shows both behave "well":

{'COUNT': 9984,
'CV': 183.1713180828161,
'EMPTY': 16,
'FILLED': 9984,
'FIRSTQUARTILE': -0.14226750284433365,
'IQR': 0.28140850365161896,
'MAJORITY': -0.20819500088691711,
'MAX': 39.16905975341797,
'MEAN': 0.003276590218837291,
'MEDIAN': -0.001677999971434474,
'MIN': -15.795705795288086,
'MINORITY': -15.795705795288086,
'RANGE': 54.964765548706055,
'STD_DEV': 0.6001773492016894,
'SUM': 32.713476744871514,
'THIRDQUARTILE': 0.1391410008072853,
'UNIQUE': 9924}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.