Re: About isostatic computation and -N+l (gravfft)

Gabriel,

Yes I say your 'avoid demeaning' question but don't understand it. Why 
would you want it? Anyway we can't because it's inherent to  FFT to 
remove the so called DC component (the mean) before applying the 
transform. Sorry if I'm not really answering your question but I'm 
afraid I didn't quite understand what you are expecting to obtain.

> Yes Joaquim, I'm getting exactly that, but I would like to set -N+l. I 
> don't know what happened to the second part of my first e-mail, so I 
> will quote it: "Is there a way to avoid the demeaning? Setting -N+l 
> does nothing, so I don't know whether this is a bug or if -N+a is 
> really needed."
>
> Thank you
> Gabriel
>
>
> Em Quinta-feira, 14 de Novembro de 2013 16:46, J. Luis 
> <[email protected]> escreveu:
> On 14-11-2013 18:26, Gabriel Dragone wrote:
>> Hello,
>>
>> I have a doubt concerning the isostatic calculation done by gravfft 
>> (-T). Using, for example, Te=0 and a compensation depth (-Z) of 35000 
>> m, I would expect that all topography had its root bellow 35000. 
>> Nevertheless, since -N+a is set, the response is a fluctuation around 
>> it.
>
> The FFT method (Parker's), after detrending,  implies that the 
> solution is 'waves' about zero but there is one term in the equation 
> that represents the average compensation depth (that's what -Z does). 
> So naturally your solution should be about the 35 km depth (parts 
> above and others bellow). Isn't that what you get?
>
>
>
>> Is there a way to avoid the demeaning? Setting -N+l does nothing, so 
>> I don't know whether this is a bug or if -N+a is really needed.
>>
>
> The man says "-T implicitly sets -N+a". This is for better convergence 
> of the Parker's method'
>
> Joaquim
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>
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