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• ## Re: May be of interest to those studying peak oil

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• According to my calculations, roughly 1120 years: = ln(1+7000000*ln(1.01))/ln(1.01) (surely this simplifies further but I m not that good at maths). So all the
Message 1 of 4 , May 2, 2007
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According to my calculations, roughly 1120 years:

= ln(1+7000000*ln(1.01))/ln(1.01)

(surely this simplifies further but I'm not that good at maths).

So all the known uranium in the sea could be good for one millenium
of 1% growth, and then it's "lights out" (that's the scary part -
imagine what sort of hi-tech society would exist after 1000 years of
growth, and then plunge them into energy poverty!!). And to think of
the mind-boggling technological advances and infrastructure that
would need to be made to make even a small fraction of that uranium

Why did Lightfoot oppose renewables again?

--- In ASPO_Oz_YoungProf@yahoogroups.com, "Dr Richard Muhlack"
<richard.muhlack@...> wrote:
>
> --- In ASPO_Oz_YoungProf@yahoogroups.com, "James Ward"
> <james.ward@> wrote:
> > Now, suppose global electricity consumption grows at 1% per
year,
> to
> > support global economic growth. Who wants to hazard a guess as
to
> > how long our 7,000,000 year seawater uranium reserve would last
if
> > consumption increased at 1% per year?
>
> Okay,
>
> Let
> Supply[1] = 7e6
> RateofUse[1]=1/Supply[1]
>
> Supply[2]=Supply[1]-RateofUse[1]
> RateofUse[2]=RateofUse[1]*1.01
>
> thus,
>
> Supply[n]=Supply[n-1]-RateofUse[n-1]
> RateofUse[n]=RateofUse[n-1]*1.01
>
> using this crude "brute force" method, Supply < 0 when n = 2706
years
>
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