Is there a known methodology for removing the trend from a DEM surface?

For example, if I had a DEM of a river, that river would flow from high elevation (upstream) to lower elevations (downstream). Even if the distance from the top of the water surface to the bottom of the river bed were the same across the whole river, the downstream portion of the raster would still have lower elevation values in the DEM. Is it possible to even out these values so that areas of equal depth also possess equal elevations?

When removing this trend it would be ideal to preserve the overall characteristics of the DEM, such as dips in the riverbed.enter image description here

3 Answers 3


@jul is correct that "you need to compute a trend surface, and then subtract it from your initial DEM to obtain the 'detrended' one," but it sounds like simpler procedures are needed in this case to "preserve the dips." If the "trend surface" too closely follows the original DEM, then the residuals will not retain the local characteristics of the surface. Thus, among the techniques to avoid are all the local ones (splines, filters, and--especially--kriging) and the ones to favor are global.

A simple, robust, direct approach is to fit a plane to the DEM in the vicinity of the river. This doesn't take any fancy technology or heavy computations, because (according to Euclid) a plane is determined by three (non-collinear) points in space. Accordingly, select one point (x1',y1',z1') = (coordinates, elevation) at the head of the river, another point (x2',y2',z2') at the downstream end, and a third point at (x0',y0',z0') of your choosing away from the line segment connecting the first two points. (These coordinates are indicated with primes because we will soon change them.) This last point does not have to correspond to a point on or even near the ground surface! In fact, a good initial choice might be to set its elevation to the average of the upstream and downstream elevations, z0' = (z1'+z2')/2.

The calculations are eased by adopting the point (x0',y0',z0') as the origin of a local coordinate system. In these coordinates the other two points are at

(x1,y1,z1) = (x1'-x0',y1'-y0',z1'-z0') 
(x2,y2,z2) = (x2'-x0',y2'-y0',z2'-z0'). 

Any arbitrary location, at (x',y') in the original coordinate system, has coordinates (x,y) = (x'-x0',y'-y0') in this new system. Because any plane passing through the origin (0,0,0) must have an equation of the form z = a*x + b*y, this reduces the problem to the following:

Find an equation in the form z = a*x + b*y for the plane passing through the points (0,0,0), (x1,y1,z1), and (x2,y2,z2).

The unique solution is to compute

u = z1 y2 - z2 y1
v = x1 z2 - x2 z1
w = x1 y2 - x2 y1

in which terms

a = u/w, b = v/w.

Having found these two numbers a and b, and recalling the two original coordinates x0' and y0', a raster calculation of the form

[DEM] - a * ([X'] - x0') - b * ([Y'] - y0')

removes the "tilt" from the DEM. In this expression [X'] refers to the the x-coordinate grid in the original coordinates and [Y'] refers to the original y-coordinate grid. The resulting DEM is guaranteed to have the same elevation (namely, z0') at each of the three points you originally chose; what it does elsewhere depends on the DEM itself!

(I hope some readers appreciate how this approach avoids all references to trigonometry or least squares machinery. :-)

  • Don't forget to add your reference to tilting a raster at forums.arcgis.com/threads/22904-Trouble-with-Hillshading
    – user681
    Jun 24, 2011 at 16:12
  • @Dan Good point: the previous post at gis.stackexchange.com/questions/10959/… looks closely related to this one.
    – whuber
    Jun 24, 2011 at 16:14
  • The previous question was also mine - I felt that I should have refined it in order to more succinctly ask the questions (e.g. need to preserve dips). I've looked around for code that will carry this type of analysis out, but I've been unsuccessful. Are you aware of anything that makes use of ArcObjects that would be useful, or should I be starting from scratch? Thanks for the help - it has been invaluable.
    – Radar
    Jun 24, 2011 at 16:41
  • @Radar First try this analysis by hand. It requires some judgment (you have to pick those three points carefully) and, often, some iteration to do right. That experience will give you a better sense of how the calculations work. You might also compare this approach with the linear trend surface fitting offered within ArcGIS.
    – whuber
    Jun 24, 2011 at 16:51
  • Fair enough - I have already automated the derivation of the upstream and downstream points. I imagine the trickiest part will be getting that 3rd point. Thanks again!
    – Radar
    Jun 24, 2011 at 17:03

I've found de-trending with trend analysis unsatisfactory for rivers. It may work in regions like the midwest but in the west a single linear or polynomial surface never seems to approximate the river gradient very well because of the complex topography. What I've used instead is weighted averaging where I compare the elevation of an upland pixel to a weighted average of the river using a kernel density (see article in Winter 2010 ArcUser). This produces a more "localized" result. It is still possible to get some negative values, but I've found that those values seem to be confined to the river channel and are most common in steep gradient areas. More recently I've experimented with pathdistance approaches to mapping "flood height" but this doesn't won't highlight those degressions that you are interested in.

  • Thank you for your remarks Tom. Welcome to our site--it's good to see you here!
    – whuber
    Aug 25, 2011 at 21:54

You need to compute a trend surface, and then subtract it from your initial DEM to obtain the "detrended" one.

To compute a trend surface, any surface simplification function may be used (filtering, spline smoothing, kriging, etc.). This choice depends on how you want this trend surface to look like.

Here is an example on how to use kriging in arcGIS 10.0 for that.

  • 1
    For my example, is there a preferred method for obtaining the trend surface (e.g. type of surface simplification)?
    – Radar
    Jun 24, 2011 at 15:44
  • That is THE question! It depends on which terrain variations you want to be considered as "trend" and removed, and which ones you want to keep. The stronger the simplification, the less you keep local variations of your terrain. As said by whuber, a trend surface composed of local planes may even be satisfying.
    – julien
    Jun 24, 2011 at 16:17

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