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Re: May be of interest to those studying peak oil

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  • James Ward
    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
      readily available...

      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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