[GRASS-dev] convert 3D polygons or Tins to a DEM?

[Helena Mitasova]

On Jan 21, 2008, at 11:23 AM, Markus Neteler wrote:

> On Jan 21, 2008 4:22 PM, Martin Landa <landa.martin at gmail.com> wrote:
>> Hi,
>>
>> 2008/1/21, Martin Landa <landa.martin at gmail.com>:
>>> 2008/1/21, Markus Neteler <neteler at osgeo.org>:
>>>> can we convert 3D polygons or tins to a DEM? Possibly a job for
>>>> v.to.rast...
>>>>
>>>> v.to.rast help
>>>> ...
>>>>           use   Source of raster values:
>>>>                 z    - use z coordinate (points or contours only)
>>>>
>>>> Would it be a big mess to extend that?
>>>
>>> I guess wouldn't be so complicated to add support also for faces  
>>> (and
>>> 3d lines in general) to v.to.rast... I can try take a look.
>>
>> just idea,
>>
>> tin ->
>>
>> v.to.points -n
>> v.clean tool=rmdupl
>> v.surf.* -> dem
>
> well, this would only give an approximation AFAIK.

that is right - if you want to preserve the triangle facets in the TIN
you need to do linear interpolation using the 3 points of each  
triangle facet
and you get something like this
(this is TIN converted to raster in some old version of Arc)

http://skagit.meas.ncsu.edu/~helena/gmslab/interp/F1b.gif

If you convert to points and run interpolation or approximation
you can get various surfaces e.g. this

http://skagit.meas.ncsu.edu/~helena/gmslab/interp/F1c.gif
or this
http://skagit.meas.ncsu.edu/~helena/gmslab/interp/F1f.gif

although for some applications the interpolated surface is preferable
to get rid of various artifacts created by TIN and create a  
continuous surface,
there are applications that include designed features and structures  
where
TIN was used in the design and you need to preserve the breaklines
defined by the TIN.

I am not sure whether this should be part of v.to.rast - maybe a  
separate
module (v.surf.linear) that would also compute the slope and aspect
map for each facet would be more appropriate,

Helena

>
> I realize that it is easy with (inclined) triangles but less so
> with whatever-polygons (are that always faces or could it be
> a more complex form?).
>
> Markus
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