OT: Golden Ratio (WAS: Re: A3)

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----- Original Message -----
From: Kennedy McEwen <rkm@kennedym.demon.co.uk>
To: <epson-inkjet@leben.com>
Sent: Saturday, April 27, 2002 11:13 AM


> In article
> <20020427042348.JMIB9131.mta04.mail.mel.aone.net.au@[63.34.205.98]>,
> artdepartment <artdepartment@ozemail.com.au> writes

<snip>

> >The basic thinking behind the "A" system is that the 'basic' unit -
A0 is
> >1 square metre in area with proportions that are based on the Golden
> >Mean, that is 1:1.414 (root2). A1 is A0 folded in half across its
long

<snip>

> All very true Tony except that the ratio of sides on the A-Series (and
> the B-Series, for that matter) is not based on the "Golden Mean", also
> known as the Golden Section.

> The Golden Mean is derived from the convergence of the ratio of
> successive terms in the Fibonacci Series, and never approaches the
> square root of two, the difference between the two is very
significant.

> The Fibonacci Series is: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89...
etc.
> each term being the sum of the previous two.  The ratio of successive
> terms is 1, 2, 1.5, 1.6, 1.625, 1.61538... which eventually converges
on
> the Golden Mean of 1.61803398874989, which is significantly different
> from the square root of two.

Kennedy,

Apologies if this post seems picky, but as I know you're a stickler for
accuracy I felt the need to put in my two pen'orth here ;-)

While it's true that the Golden Mean (or Golden Section, or Golden
Ratio) crops up as the limit of the ratio of successive terms in the
Fibonacci Series, to state that it 'is derived from the ... Fibonacci
Series ...' seems to imply that this is it's original definition, which
I'm pretty sure it isn't.

>From what I recall from studying the subject many years ago, the Ancient
Greeks were familiar with this ratio. One of the simpler ways to derive
it is to divide a straight line into two unequal parts such that the
ratio of the length of the whole line to the length of the larger part
is the same as the ratio of lengths of the larger part to the smaller.
As an 'ASCII diagram' this looks like

A----------------------B--------------C
                  x                             1

If the length AB is x and BC is 1 then you get the equation

(x + 1)/x = x/1

Solving for x and taking the positive root gives (sqrt(5) + 1)/2 which
has a non-terminating decimal expansion starting as you describe above.

The Golden Ratio crops up all over the place in geometry. Supposedly
it's an aesthetically pleasing ratio, which is why it's sometimes used
in architecture and painting. Maybe our distinguished Fine Arts
representative Dave Tobie can comment on this :-)

Returning to the elegant way in which the 'A series' of paper sizes is
derived based on sqrt(2), then FWIW you can derive other series of sizes
based on dividing the original rectangular piece of paper into n equally
sized sub-rectangles if the ratio of the sides of the original is 1 to
sqrt(n).

Regards,

Alan Rew

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