Dear all;
Thank you for your reply.
David has explained an interesting method.
David I have DEM file of the region and I have extracted the xyz data from
that.
Also I can extract bathymetry data as xyz file.
I have calculated the storage (volume) of reservoir at the current
elevation.
But the method I have used to calculate the volume is different from your
method. I have crop DEM by the reservoir boundary and then calculate the
volume.
I would be more than happy if you please please explain more or write codes
for me how to get volume at different elevation.
And also about the following function, especially f(Storage).

lm(Elevation~f(Storage)
Sincerely

On Tue, 9 Apr 2024, 21:26 David Stevens via R-help, <r-help@r-project.org>
wrote:

> Water engineer here. The standard approach is to 1) get the storage vs.
> elevation data from the designers of the reservoir or, barring that, 2)
> get the bathymetry data from USBR or state DWR, or, if available, get
> the DEM data from USGS if the survey was done before the reservoir was
> built or 3) get a boat+sonar with GPS  +lots of time and survey the
> bottom elevation yourself. Put the xyz data into ArcGIS and have it
> create the bottom surface, then, with several elevations, integrate the
> xyz data from Z to the bottom to find the storage. Plot the storage at
> each water surface to get an idea of the shape and then use
> lm(Elevation~f(Storage) where f(Storage) may be a cubic or quartic
> polynomial. Then double the Storage and calculate Elevation. This type
> of thing is done everyday by hydrologists.
>
> Good luck
>
> David K Stevens, PhD, PE, Professor
> Civil and Environmental Engineering
> Utah Water Research Laboratory
> Utah State University
> 8200 Old Main Hill
> Logan, UT 84322-8200
> david.stev...@usu.edu
> (435) 797-3229 (office)
>
> On 4/9/2024 8:01 AM, peter dalgaard wrote:
> > So, you know how to get volume for given water level.
> >
> > For the reverse problem, you get in trouble because of the nonlinearity
> inherent in the dependence of surface area on the level.
> >
> > I don't think there is a simple solution to this, save for mapping out
> the volume as a function of water level and solving equations for the water
> level using (say) uniroot(). Which may actually suffice for practical
> purposes.
> >
> > For small changes, finding the derivative of the relation is easy:
> d(volume) = Area * d(level) and this can be used as an approximate relation
> as long as the Area remains nearly constant.
> >
> > However generic questions like doubling the volume are impossible to
> answer without knowledge of the reservoir shape. E.g. in a cylindrical
> reservoir halving the water level also halves the volume, but in a conical
> reservoir, halving the level leaves only 1/8 of the volume.
> >
> > -pd
> >
> >
> >
> >> On 8 Apr 2024, at 05:55 , javad bayat <j.bayat...@gmail.com> wrote:
> >>
> >> Dear all;
> >> Many thanks for your replies. This was not homework. I apologize.
> >> Let me explain more.
> >> There is a dam constructed in a valley with the highest elevation of
> 1255
> >> m. The area of its reservoir can be calculated by drawing a polygon
> around
> >> the water and it is known.
> >> I have the Digital Elevation Model (DEM) of the region (reservoir and
> its
> >> surrounding area). I have calculated the volume of the current reservoir
> >> (7e6 m3) using the following codes.
> >> library(raster)
> >> library(terra)
> >> library(exactextractr)
> >> library(dplyr)
> >> library(sf)
> >> # Calculate volume for polygon
> >> # Read the DEM raster file
> >> r <- rast("E:/...DEM.tif")
> >> # Read the polygon shapefile
> >> p <- st_read("E:/...Dam.shp")
> >>
> >> r <- crop(r, extent(p))
> >> r <- mask(r, p)
> >>
> >> # Extract the cells in each polygon and calculate the area of each cell
> >> x <- exact_extract(r, p, coverage_area = TRUE)
> >> # Extract polygon values as a dataframe
> >> x1 = as.data.frame(x[1])
> >> head(x1)
> >> x1 = na.omit(x1)
> >> # Calculate the height above the minimum elevation in the polygon
> >> x1$Height = max(x1[,1]) - x1[,1]
> >> # Calculate the volume of each cell
> >> x1$Vol = x1[,2] * x1[,3]
> >> sum(x1$Vol)
> >> x2 = x1[,c(1,2,4)]
> >> x2 = sort(x2,'value')
> >> head(x2)
> >> x3 <- aggregate(Vol ~ value, data = x2, FUN = sum)
> >> x4 <- aggregate(coverage_area ~ value, data = x2, FUN = sum)
> >> x5 = cbind(x3, Area = x4[,2])
> >> library(dplyr)
> >> x6 <- x5 %>%
> >>   mutate(V_sum = cumsum(Vol)) %>%
> >>   mutate(A_sum = cumsum(Area))
> >> plot(x6$value~x6$V_sum)
> >>
> >> And I thought that it is possible to get the elevation for a specific
> >> volume by linear model between elevation and volume, as follow:
> >>
> >> # Get a linear model between elevation and the volume
> >> lm1 <- lm(value ~ V_sum, data = x6)
> >> d <- data.frame(V_sum = 14e6)  #
> >> predict(lm1, newdata = d)
> >>
> >> But it is not possible through the LM.
> >> Now I want to know what would be the water level in the reservoir if the
> >> reservoir volume doubled or we adding a known volume to it?
> >> Also what would be the volume if the water level increases to 1250 m?
> >>
> >> I would be more than happy if you help me to do this.
> >> Sincerely
> >>
> >> On Mon, Apr 8, 2024 at 12:23 AM <avi.e.gr...@gmail.com> wrote:
> >>
> >>> John,
> >>>
> >>> Your reaction was what my original reaction was until I realized I had
> to
> >>> find out what a DEM file was and that contains enough of the kind of
> >>> depth-dimension data you describe albeit what may be a very irregular
> cross
> >>> section to calculate for areas and thence volumes.
> >>>
> >>> If I read it correctly, this can be a very real-world problem worthy
> of a
> >>> solution, such as in places like California where they had a tad more
> rain
> >>> than usual and some reservoirs may overflow. Someone else provided what
> >>> sounds like a mathematical algorithm but my guess is what is needed
> here is
> >>> perhaps less analytic since there may be no trivial way to create
> formulas
> >>> and take integrals and so on, but simply an approximate way to
> calculate
> >>> incremental volumes for each horizontal "slice" and keep adding or
> >>> subtracting them till you reach a target and then read off another
> variable
> >>> at that point such as depth.
> >>>
> >>> Some care must be taken as water level has to be relative to something
> and
> >>> many natural reservoirs have no unique bottom level. Some water may
> also be
> >>> stored underground and to the side and pour in if the level lowers or
> can
> >>> be
> >>> used to escape if the level rises.
> >>>
> >>>
> >>> -----Original Message-----
> >>> From: R-help <r-help-boun...@r-project.org> On Behalf Of Sorkin, John
> >>> Sent: Sunday, April 7, 2024 3:08 PM
> >>> To: Rui Barradas <ruipbarra...@sapo.pt>; javad bayat <
> j.bayat...@gmail.com
> >>>> ;
> >>> R-help <R-help@r-project.org>
> >>> Subject: Re: [R] Question regarding reservoir volume and water level
> >>>
> >>> Aside from the fact that the original question might well be a class
> >>> exercise (or homework), the question is unanswerable given the data
> given
> >>> by
> >>> the original poster. One needs to know the dimensions of the reservoir,
> >>> above and below the current waterline. Are the sides, above and below
> the
> >>> waterline smooth? Is the region currently above the waterline that can
> >>> store
> >>> water a mirror image of the region below the waterline? Is the region
> above
> >>> the reservoir include a flood plane? Will the additional water go into
> the
> >>> flood plane?
> >>>
> >>> The lack of required detail in the question posed by the original
> poster
> >>> suggests that there are strong assumptions, assumptions that typically
> >>> would
> >>> be made in a class-room example or exercise.
> >>>
> >>> John
> >>>
> >>> John David Sorkin M.D., Ph.D.
> >>> Professor of Medicine, University of Maryland School of Medicine;
> >>> Associate Director for Biostatistics and Informatics, Baltimore VA
> Medical
> >>> Center Geriatrics Research, Education, and Clinical Center;
> >>> PI Biostatistics and Informatics Core, University of Maryland School of
> >>> Medicine Claude D. Pepper Older Americans Independence Center;
> >>> Senior Statistician University of Maryland Center for Vascular
> Research;
> >>>
> >>> Division of Gerontology and Paliative Care,
> >>> 10 North Greene Street
> >>> GRECC (BT/18/GR)
> >>> Baltimore, MD 21201-1524
> >>> Cell phone 443-418-5382
> >>>
> >>>
> >>>
> >>>
> >>> ________________________________________
> >>> From: R-help <r-help-boun...@r-project.org> on behalf of Rui Barradas
> >>> <ruipbarra...@sapo.pt>
> >>> Sent: Sunday, April 7, 2024 10:53 AM
> >>> To: javad bayat; R-help
> >>> Subject: Re: [R] Question regarding reservoir volume and water level
> >>>
> >>> Às 13:27 de 07/04/2024, javad bayat escreveu:
> >>>> Dear all;
> >>>> I have a question about the water level of a reservoir, when the
> volume
> >>>> changed or doubled.
> >>>> There is a DEM file with the highest elevation 1267 m. The lowest
> >>> elevation
> >>>> is 1230 m. The current volume of the reservoir is 7,000,000 m3 at
> 1240 m.
> >>>> Now I want to know what would be the water level if the volume rises
> to
> >>>> 1250 m? or what would be the water level if the volume doubled
> >>> (14,000,000
> >>>> m3)?
> >>>>
> >>>> Is there any way to write codes to do this in R?
> >>>> I would be more than happy if anyone could help me.
> >>>> Sincerely
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>> Hello,
> >>>
> >>> This is a simple rule of three.
> >>> If you know the level l the argument doesn't need to be named but if
> you
> >>> know the volume v then it must be named.
> >>>
> >>>
> >>> water_level <- function(l, v, level = 1240, volume = 7e6) {
> >>>    if(missing(v)) {
> >>>      volume * l / level
> >>>    } else level * v / volume
> >>> }
> >>>
> >>> lev <- 1250
> >>> vol <- 14e6
> >>>
> >>> water_level(l = lev)
> >>> #> [1] 7056452
> >>> water_level(v = vol)
> >>> #> [1] 2480
> >>>
> >>>
> >>> Hope this helps,
> >>>
> >>> Rui Barradas
> >>>
> >>>
> >>> --
> >>> Este e-mail foi analisado pelo software antivírus AVG para verificar a
> >>> presença de vírus.
> >>> http://www.avg.com/
> >>>
> >>> ______________________________________________
> >>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>> PLEASE do read the posting guide
> >>> http://www.r-project.org/posting-guide.html
> >>> and provide commented, minimal, self-contained, reproducible code.
> >>>
> >>> ______________________________________________
> >>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>> PLEASE do read the posting guide
> >>> http://www.R-project.org/posting-guide.html
> >>> and provide commented, minimal, self-contained, reproducible code.
> >>>
> >>>
> >> --
> >> Best Regards
> >> Javad Bayat
> >> M.Sc. Environment Engineering
> >> Alternative Mail: bayat...@yahoo.com
> >>
> >>      [[alternative HTML version deleted]]
> >>
> >> ______________________________________________
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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