Re: OT: coins and currency (was: [Theory] Types of numerals)
| From: | tomhchappell <tomhchappell@...> | 
|---|
| Date: | Sunday, January 8, 2006, 22:08 | 
|---|
--- In conlang@yahoogroups.com, Jefferson Wilson <jeffwilson63@F...>
wrote:
>
> Paul Bennett wrote:
> > On Fri, 06 Jan 2006 08:11:15 -0500, Mark J. Reed
<markjreed@M...>
> > wrote:
> >
> >> In US currency, for instance, there are essentially 4 sub-dollar
> >> denominations (1, 5, 10, 25), since half dolalrs are very rare.
As a
> >> result, some values require up to 9 coins (e.g. 94¢ and 99¢).
> >> Reintroduction of a commonly-circulated half-dollar would cut
that
> >> down by one coin; a two-cent piece would reduce it by two more.
That
> >> would yield six denominations and a maximum minimum (:)) of six
coins
> >> per value.
> >
> >
> > I'm sure you're aware of the British system, which is partitioned
1, 2,
> > 5,  10, 20, 50, 100, 200, 500, etc. I have a gut feeling that
it's more
> > optimal than the US system of (essentially) 1, 5, 10, 25, 100,
500,
> > 1000,  2000, which strikes me as more organic but less wieldy.
> >
> > Of course, it shouldn't take much math to prove that the most
optimal
> > system would have units of 1, 2, 4, 8, 16, 32, etc., provided of
course
> > that the general populace could be made sufficiently familiar
with the
> > concept.
>
> Depends on whether you want the lowest number of _coins_ or the
> lowest number of _types_.  Binary is good for the former, but for
> the latter you get the series: 1, 3, 6, 12, 24, etc.  (Something
> to keep in mind for those of us with duodecimal numbering systems
> I think.)  Hmmm, take this series up to 96, round each value to
> the nearest number divisible by 5, and you have the American
> coinage system.
>
> --
> Jefferson
> 
http://www.picotech.net/~
>
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