[R-sig-Geo] bottom area of a lake calculation

Thomas Adams te@3rd @ending from gm@il@com
Wed Aug 8 16:49:47 CEST 2018


Hello Sophie,

Yes, what you show was my understanding. I'm sure what you want to do can
be done in R, but I'm not sure what packages to use. For doing such a
thing, I use GRASS GIS. Regardless, I think you should re-project your data
from latitude-longitude to Albers Equal Area or Lambert Conic Conformal,
etc., that is area preserving. Eventually, you need to do this to get an
area anyhow.

An additional step would be to reproject, then spatially interpolate the
point locations you have in your spreadsheet into a grid to Albers Equal
Area or Lambert Conic Conformal, etc.

I could take a stab at this and show you...

Regards,
Tom

On Wed, Aug 8, 2018 at 10:39 AM, <sophie.leblanc using neuf.fr> wrote:

> Dear Tom,
>
> thanks for your reply.
> Just to be sure I have explain correctly.
> With marmap I can get the green area but I also want to calculate the red
> one.
> I have attached the kind of data I have.
>
>> I do not have any GIS software.
> I wanted to convert my spherical coordinate to cartesian ones with
>
> x = radius * cos(latitude) * cos(longitude)
> y = radius * cos(latitude) * sin(longitude)
> z = radius * sin(latitude)
> radius= earth_radius-depth : not fully sure...
> Do you think it is correct?
>
> The slope is not the same everywhere around the lake.
> Is it possible to calculate area with your first explanations?
> what R package should I use?
>
> thanks,
> Sophie Leblanc
>
>
>
>
>
>
>
> *http://sophieleblanc6.wix.com/photographie
> <http://sophieleblanc6.wix.com/photographie>*
> De : "Thomas Adams"
> A : sophie.leblanc using neuf.fr,"r-sig-Geo using r-project.org"
> Envoyé: mercredi 8 août 2018 16:07
> Objet : Re: [R-sig-Geo] bottom area of a lake calculation
>
>
> Sophie,
>
> My apologies, I was thinking in terms of using GRASS GIS... but the idea
> is the same
>
> Tom
>
> On Wed, Aug 8, 2018 at 10:06 AM, Thomas Adams <tea3rd using gmail.com> wrote:
>>
>> Sophie,
>>
>> I think a reasonable approximation might be obtained:
>>
>> (1) if you were to calculate the slope at each grid point using
>> r.slope.aspect
>> (2) use r.mapcalc to calculate A/cos(slope), where A is the pixel
>> projected area (e.g., 100mx100m = 10000 m^2) -- lat-long coordinates will
>> not work.
>>
>> Consequently, if the slope were approximately 0, cos(slope) approaches 1
>> and A/cos(slope) = A; as slope approaches 90 deg, cos(slope) becomes very
>> small, so A becomes very large. This works where the slope direction is
>> orthogonal to a pixel edge, so a correction to account for other slope
>> directions may be needed. But, with smaller grids, the error may be
>> negligible for you...
>>
>> Tom
>>
>>
>> On Wed, Aug 8, 2018 at 4:55 AM, <sophie.leblanc using neuf.fr> wrote:
>>>
>>> <span style="font-family:arial,helvetica,sans-serif;
>>> font-size:12px">‌</span>Dear all,<br>
>>> <br>
>>> I would like to calculate the bottom area of a lake. I have already used
>>> marmap to calculate projected area, i.e. surface area.<br>
>>> But i have not found anything to calculate a non projected area. I have
>>> the coordinates of points (longitude, latitude and depth) I can convert
>>> them into xyz. But then I don't know how to calculate the surface I have
>>> found polygonal area calculation but I have a 3D object, a polyheron and I
>>> think I can not use polygonal area.<br>
>>> Do you have any idea? do you know an R package that could help?<br>
>>> <br>
>>> thanks,<br>
>>> Sophie Leblanc<br>
>>> <br>
>>> <!-- +Signature --><font size="2"><em><a href="http://sophieleblanc6.wi
>>> x.com/photographie">http://sophieleblanc6.wix.com/photographie</a></em></font>
>>> <!-- -Signature -->
>>>         [[alternative HTML version deleted]]
>>>
>>> _______________________________________________
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>>
>>
>>
>>
>>
>>
>>
>
>
>
>
>
>



-- 
Thomas E Adams, III
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