# SRTM/ 90m DEM - slope calculations

I have a question relating to a comment in a closed discussion (Does a value in srtm dem file refer to the center or corner of a cell?).

The comment was following:

"When talking about any DEM, it's resolution is important. SRTM has a resolution of 90 m for non US areas. Theoretically speaking, you can't really say anything about the height variation within that pixel. So just as PostGIS rasters (or rather All Rasters), it is the value for the entire area of 90m*90m, and not for just some specific point within that square. However, many interpolating algorithms, assume the height is for the center of the pixel, and will interpolate the heights for points in between the pixel centers."

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Does this mean that it is not possible to estimate correctly the height/slope of the landscape from 90m DEM? E.g. if I have calculated slope based on 90m DEM. What is meant by interpolating algorithms, should one interpolate 90m SRTM to get correct height?

• as of 2015, SRTM-1 is globally available outside the US, having a resolution of 1 arcsec in both dimensions (note: this is approximately 30m at the equator, and less towards the poles!). the slope is calculated across neighboring pixels (e.g. between two height values), and since one pixel (value) represents the height of an area of roughly 30m*30m, any derivates (slope, aspect, etc.) are mere approximations. – ThingumaBob Apr 29 '18 at 14:19
• Thank you! But I had to use 90m DEM for my calculations. Could someone comment on SRTM-3 please? – mikbet Apr 29 '18 at 15:04
• ah, the same applies to the lower resultion data; approximation is obviously worse, though... – ThingumaBob Apr 29 '18 at 15:27
• You are confusing array spacing with origin. No, the cell values are not representing the lower left of the cell, they represent the cell center. And, yes at a given scale the derived slope is correct. – Jeffrey Evans Apr 29 '18 at 17:08