how can i find cliffs with 90o and export them in points?

[KONPETROV]

hello i am trying to find a solution at a problem which have to do with points in vertical cliffs. How can i calculate them?

[DanPatterson_Retired · Accepted solution]

Can I assume that you have a DEM from which you have derived a slope layer in degrees? If so, then you can query that layer to find the areas where the slope is greater than some percentage then convert those area to points.  Given the caveat that you are going to get many points perhaps, to denote one cliff face and that cliffs are rarely exactly 90° (so pick a reasonable value given the lithology or geology of your area)

[XanderBakker]

If this question is related to your other question: how can i find points with same x,y data and different z value  then I would suggest to use LAS Point Statistics As Raster (Data Management) with the Z_RANGE option for this purpose.

You may want to open the other thread up, so more people can make suggestions.

[DanPatterson_Retired]

why is it closed Xander? what was suggested there before any more suggestions are made here?

[KONPETROV]

thank you for your reply, it was closed for 3 people who had answered and by mistake i deleted my post, also the post was for code. I posted something with a more general title to avoid discussions about my code, and be more understanding if someone want to be helped by that at his/her problem. Thanks a lot Mr. Patterson D, yes i have a DEM and i have also derived a slope layer in degrees. One solution is what you suggested i can select area>85o. Mr. Bakker X. I don't have lidar dataset if that is what you mean. The  output i want to have is at the form of x,y,z. Theoretically why i cannot have cliff with 90o in my Dem, i cannot undestand that.  

[XanderBakker]

In a DEM which is a raster file each grid cell can only have a single value. If you have a cliff of 90 degrees, it should have two values at the same grid cell. In a point file you can have points that overlap and each point has its own value. For the points you could have a slope of 90°.

[KONPETROV]

i exported points from contours with interval 3meters, but i cannot find points with x y and different z, i thought that if i keep only the integer of each x y then i can find some, and i did but then there are 2 problems. I have and error of 0.10000 - 0.90000 and i cannot find so many. For example i have points with different height of 10m, 15m, 24m etc, and i am sure that in that area i have cliff with at least 100-150 meter vertical.

[DanPatterson_Retired]

As Xander points out, a vertical cliff can only be denoted by an elevation difference between cells since a single cell cannot contain more than one value.  One other way of determining such features would be to run a filter over the DEM looking for large ranges in elevation differences.  This can be accomplished using focal statistics using a simple 3x3 neighbourhood, or should you wish to examine directionality, you could use a 2x1 or 1x2 filter...in either case you could use the range option.  You would want to have a grid cell spacing that is appropriate to your contour spacing as a minimum which isn't the same as the contour interval.  So if your area has 'cliffs' that are abrupt in their horizontal distance, then a cell size of 10 m or so would work quite well (I am speculating here) but you could experiment with your DEM'...  in any event, see the Focal Statistic help topic and its associated tools

[KONPETROV]

Thank you for your suggestion i will try it and i will post my result.

[KONPETROV]

Sorry i didn't post it earlier, of course this suggestion was right. I got back a range from 15-328 meters for a 3x3 meter analysis. I made a calculation in raster calculator with some other factors also, and now i have classified the region that sutisfies my criterias. Thank you very much.