JT
Posts:
1,434
Registered:
4/7/12


Re: Calculate circumreference and area of circle without Pi is easy
Posted:
May 7, 2013 12:13 AM


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 UTC5, JT wrote: > > > > Knowing that a hexagon built by isocles we can split them into right > > > > > triangles and have hy^20.5hy^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, multiplicate with vertice length of hexagon.

