```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:>>>>>>>>>> > On 7 Maj, 04:18, donstockba...@hotmail.com wrote:>> > > 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>> > > > triangles and have hy^2-0.5hy^2=b^2.>> > > > And from there we create a new dodecagon knowing the difference>> > > > between base and hypotenuse we can calculate required heights of new>> > > > vertices, we plug it in and get hypotenuse of triangle forming new>> > > > vertice.>> > > > We repeat all this splitting hexagon, dodecagon, 24, 48, 96, 192...>> > > > recursively until sought smoothness of the circle is aquired while>> > > > adding vertices and triangle areas.>> > > Only one calculation is needed if you just know pi.>> > Yes but Pi is approximation, my answer would be a fraction and> > therefor exact for the sought and aquired smoothness.>> 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, multiplicatewith vertice length of hexagon.
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