I am a complete beginner in GIS and taking a class right now, and we are doing some assignments in ArcGIS - ArcMap.

In one of them we were given a shapefile with heightpoints. These had been manually collected using gps, and they had been chosen to best represent the variations of the landscape.

Next we were to build a TIN-model and a Grid (with 3D Analyst toolbox -Interpolation / Natural Neighbor, Z vaule field = SPOT, Output Cell Size = 10 meter) from these points.

And finally to interpolate contour lines with Surface Contour - Contour interval = 10 meter.

Now here comes the question: Which of these methods give the most accurate contour lines, and why?

I have a hard time getting my head aroud this. I have written in my report that the Grid-based contours are more accurate because the Grid has alot more interpolated points to use when drawing the contours. Or is this just making the lines more smooth, not more accurate.

In this example file we are using, the grid cell size set to 10 meters, gives a smaller grid than the TIN-triangles, even when the original data points are close together.

But the grid based contours are interpolated twice? First from the data points to all the centers of all the cells, and then to the contour lines.

And the TIN-based contours are interpolated straight from the data points to new points on the lines that form the edges of the triangles. Are they maybe more accurate but less smooth??

Any help to clear my confusion would be appreciated.

here is a picture of the points, models and countours from ArcMap: http://i.imgur.com/jfzdZEE.jpg

  • There is a comparison with explanations here - Raster vs. TIN. – Barbarossa Mar 21 '14 at 13:53

Firstly you need to understand that contours are not painted on the ground, they are a model. Secondly a TIN and a DEM are also models. With that in mind debating the relative inaccuracies of models of a model seems a little pointless.

When creating a contour from a TIN the process must internally convert to a raster or the only vertices created will be on the facets (lines joining nodes in the triangulation), the cell size used by this process is arbitrary, which is then contoured in the same manner as if you converted it to a raster yourself. In truth the answer lies in the cell size that is arbitrarily chosen by the process and the TIN to raster method, which you are unlikely to know.

If you want to know what the difference is then convert both contours to TIN, TIN to raster and then subtract one from the other, view this against the points you started with and you will find that the differences between both contours are least where there is a TIN node and most where there is significant distances between nodes, this is where interpolation occurs and whatever is happening here is subjective!

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