Date: May 7, 2013 12:13 AM
Author: JT
Subject: Re: Calculate circumreference and area of circle without Pi is easy

On 7 Maj, 04:43, JT <jonas.thornv...@gmail.com> wrote:
> On 7 Maj, 04:23, JT <jonas.thornv...@gmail.com> wrote:
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> > On 7 Maj, 04:18, donstockba...@hotmail.com wrote:
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> > > On Monday, May 6, 2013 9:03:27 PM UTC-5, JT wrote:
> > > > Knowing that a hexagon built by isocles we can split them into right
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> > > > triangles and have hy^2-0.5hy^2=b^2.
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> > > > And from there we create a new dodecagon knowing the difference
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> > > > between base and hypotenuse we can calculate required heights of new
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> > > > vertices, we plug it in and get hypotenuse of triangle forming new
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> > > > vertice.
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> > > > We repeat all this splitting hexagon, dodecagon, 24, 48, 96, 192...
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> > > > recursively until sought smoothness of the circle is aquired while
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> > > > adding vertices and triangle areas.
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> > > Only one calculation is needed if you just know pi.
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> > Yes but Pi is approximation, my answer would be a fraction and
> > therefor exact for the sought and aquired smoothness.

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> Well a vertice dependent fraction relative the original hypotenuse of
> hexagon. So if we set hypotenuse to 1 for any radius we get a general
> solution fraction for the required smoothness, that need just be
> multiplicated with the the split hexagon hypotenuse length to get the
> circumreference.


Split hexagon hypotenuse length sounds a bit confusing, multiplicate
with vertice length of hexagon.