- Posted November 20, 2012 by
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- CIRCLE CIRCUMFERENCE RISE TO THE SECOND POWER 1988 - BOOK 1997, by PANAGIOTIS STEFANIDES
- SEASONS GREETINGS WITH CALENDAR 2013 BY PANAGIOTIS STEFANIDES
- PROPOSED GEOMETRY OF THE PLATONIC TIMAEUS
- VIDEO CLIP , Paper Presented to Conference :Symmetry Festival 2006, Budapest
- VIDEO CLIP SPIRALOGARITHM By Panagiotis Stefanides.SYMMETRY FESTIVAL 2006 BUDAPEST HUNGARY
THE QUADRATURE TRIANGLE IMPLICATIONS
THE QUADRATURE TRIANGLE IMPLICATIONS *
[16/x]^2 -[x]^2- 16=0 PANAGIOTIS STEFANIDES for x = π
http://www.wolframalpha.com/input/?i=%5B16%2Fx%5D%5E2+-%5Bx%5D%5E2-+16%3D0
http://www.wolframalpha.com/input/?i=2+sqrt%282+%28-1%2Bsqrt%285%29%29%29
For x = π =3.14460551.. = 4/ sqrt [ Φ ] = 4/T=
= 4/ {[ sqrt (5) +1 ]/2} = 2 sqrt(2 (-1+sqrt(5)))
FOR THIS VALUE OF π =3.14460551.. :
FOR THE VALUE OF DIAMETER = [16/ π ] = 5.0880786..
CIRCUMFERENCE = 16 = A SQUARE WITH SIDE [4]=QUARTER CIRCLE.
THIS QUARTER SIDE [4] MULTIPLIED BY THE DIAMETER [16/ π ]
EQUALS [4]*[16/π] = [ 64/ 3.14460551..] = 20.3523144..
ROOT OF THIS IS A SQUARE OF SIDE = 4.511353943..
AREA OF THIS SQUARE =20.3523144..
AREA OF CIRCLE OF DIAMETER [16/π] = [π/4]*[16/π]^2 =
= [ 0.786151378..]*[5.0880786..]^2 = [ 0.786151378..]*[25.88854384] =
= 20.35231441..
THIS AREA OF CIRCLE IS EQUAL TO THE SQUARE OF SIDE 4.511353943..
Ref:
http://www.stefanides.gr/pdf/2012_Oct/PHOTO_12.pdf
[ http://www.stefanides.gr/pdf/2012_Oct/PHOTO_11.pdf ]
http://www.stefanides.gr/pdf/BOOK%20_GRSOGF.pdf
http://www.stefanides.gr/pdf/PROPOSED_GEOMETRY_OF_THE_PLATONIC_TIMAEUS_GREEK.pdf.pdf
*Quadrature Triangle Panagiotis Stefanides Coinned Terminology.
Refers to the Plato’s Timaeus Most Beautiful Triangle , Proposed
Interpretation by Panagiotis Stefanides which is Similar to the Kepler[Magirus] Triangle BUT NOT THE SAME and not as BEAUTIFUL to this:
http://www.stefanides.gr/Html/platostriangle.htm
http://en.wikipedia.org/wiki/Kepler_triangle
© Copyright 1986 - 20012 Panasgiotis Stefanides.
http://www.stefanides.gr
ABSTRACT
Proposed Interpretation of the Timaeic Plato
This work, proposes a different interpretation of Plato’s Most Beautiful Triangle.
In section 53/54, of PLATO'S "TIMAEUS", PLATO speaks about the triangular shapes of the Four Elemental Bodies, of their kinds and their combinations:
These Bodies are the Fire (Tetrahedron) the Earth (Cube), the Water (Icosahedron), and the Air (Octahedron). These are bodies and have depth. The depth necessarily, contains the flat surface and the perpendicular to this surface is a side of a triangle and all the triangles are generated by two kinds of orthogonal triangles: the "ISOSCELES" Orthogonal and the "SCALΕΝΕ" Orthogonal. From the two kinds of triangles the "Isosceles" Orthogonal has one nature. (i.e. one rectangular angle and two acute angles of 45 degrees), whereas the "scalene" has infinite (i.e. it has one rectangular angle and two acute angles of variable values having, these two acute angles, the sum of 90 degrees). From these infinite natures we choose one triangle "THE MOST BEAUTIFUL".
Thus, from the many triangles, we accept that there is one of them "THE MOST BEAUTIFUL.
Let us choose then, two triangles, which are the basis of constructing the Fire and the other Bodies : "Το μεν ισοσκελές, το δε τριπλήν κατά δύναμιν έχον της ελάττονος την μείζω πλευράν αεί."
Proposed New Interpretation:
One of these two is the "ISOSCELES" orthogonal triangle, the other is the "SCALENE" orthogonal triangle, its hypotenuse having a value equal to the "CUBE" of the value of its horizontal smaller side and having its vertical bigger side the value of the "SQUARE" of its smaller horizontal side. The value of the smaller horizontal side is equal to the square root of the Golden Number, the ratio of the sides is equal, again, to the Square Root of the Golden Number (geometrical ratio) and the Tangent of the angle between the hypotenuse and the smaller horizontal side is also equal to the Square Root of the Golden Number (Θ =51 49-38-15-9-17-19-54-37-26-24-0 degrees). The product of the smaller horizontal side and that of the hypotenuse is equal to the "SQUARE" of the bigger vertical side, and the following equation holds:
T^4-T^2-1=0 , T = SQRT [(SQRT.(5) + 1)/2].
© Copyright 1986 - 2012 Panasgiotis Stefanides.
http://www.stefanides.gr
- TAGS:
- triangle,
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- quadrature
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