Hello
I am really glad of your answer, but for me it does not work as I would like
So, my data are called cc (is a subset of a bigger spatialdataframe) and the variogram would be
varcc=variogram(pf1_3atm~1,cc, width=500,cutoff=10000) 
 
Then I fix the sill and the range of the variogram
inisill=var(cc@data$pf1_3atm,na.rm=T) 
iniran=var$dist[varcc$gamma>0.9*inisill][1]
varcc.fit=fit.variogram(varcc,vgm(psill=inisill, "Exp", iniran)) # I can not fix the nugget because the output gives errors ("Singular model")
plot(varcc,varcc.fit)
> varcc[1,"gamma"]  this would be the nugget (where the model cuts the axis)
[1] 11.97
The problem in the indicator you suggest is:
varcc.fit$psill[1] / sum(varcc.fit$psill)= 1, because  varcc.fit$psill[1] = sum(varcc.fit$psill)
as the output of varcc.fit gives:
varcc.fit
  model    psill    range
1   Exp 41.46455 1297.040

I have tried (don't know if it is correct):
>  varcc[1,"gamma"]/sum(varcc.fit$psill)
[1] 0.2886803
(and this would mean that the spatial correlation is weak because we have a relative big nugget effect)
Can this be done? 
Thanks a lot again for your attention.

> Date: Thu, 30 Sep 2010 18:39:22 +0200
> From: edzer.pebesma@uni-muenster.de
> To: r-sig-geo@stat.math.ethz.ch
> Subject: Re: [R-sig-Geo] nugget in variogram
> 
> 
> 
> On 09/30/2010 04:13 PM, Arantzazu Blanco Bernardeau wrote:
> > Hello everybody
> > I am new on geostatistics so maybe my question is trivial or is
> > already answered (I can not find the solution in the thread list). I
> > have a variogram where I did not fix the nugget, because doing it
> > was giving "singular models". I would like to know the nugget ratio
> > of my variograms but I don't know how to get then the nugget value.
> > So I did look in the output of variogram and did take the minimum
> > value of gamma. Is this correct?
> 
> Well, it's not the usual way, where you would fit a model:
> 
> library(gstat)
> data(meuse)
> coordinates(meuse) = ~x+y
> v = variogram(log(zinc)~1, meuse)
> v.m = fit.variogram(v, vgm(1, "Sph", 500, 1))
> v[1, "gamma"] # your estimator:
> plot(v, v.m)
> v.m$psill[1] / sum(v.m$psill) # the usual estimator:
> 
> Only when you have plenty of measurements at very short distances
> apart, I can imagine you would use the sample variogram (v) value.
> 
> > Thanks very much for your attention. 
> >  		 	   		  
> > 	[[alternative HTML version deleted]]
> > 
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> > R-sig-Geo@stat.math.ethz.ch
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> 
> -- 
> Edzer Pebesma
> Institute for Geoinformatics (ifgi), University of Münster
> Weseler Straße 253, 48151 Münster, Germany. Phone: +49 251
> 8333081, Fax: +49 251 8339763  http://ifgi.uni-muenster.de
> http://www.52north.org/geostatistics      e.pebesma@wwu.de
> 
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