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General: NUMERO DE ORO RELACIONADO CON EL CIRCULO
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Respuesta  Mensaje 1 de 2 en el tema 
De: BARILOCHENSE6999  (Mensaje original) Enviado: 22/05/2017 17:26
 
Resultado de imagen para triangle equilater PHI GOLDEN NUMBER
 
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Respuesta  Mensaje 2 de 2 en el tema 
De: BARILOCHENSE6999 Enviado: 22/05/2017 17:27

kayaked out to some islands, had a hike around, caught some spectacular orange lighting on the evergreen mountains through a sunset cloud gap. Refreshed and noting…

π = 3.14159265358979
4 = 4
4/π = 1.27323954473516
(4/π)^2 = 1.6211389382774
φ = 1.61803398874989
((4/π)^2)-φ = 0.00310494952750973
(((4/π)^2)-φ)/φ = 0.00191896434135394 = 0.191896434135394%

doesn’t work as well that way around, but fuller awareness is good.

Now I’ve got a piece of work on my hands sorting out concise summary of the √5 & φ power ladders and their Schwabe intersection (given paid work). As you can guess, it will end up taking weeks to reach the standard for which I aim, but I’ll keep up the brief updates along the way.

Eventually it will all be done to minimally tolerable standard. Only then will open minds be able to make overall sense of it.

I think at this stage anyone following along should at least be aware of the inner system φ fractal and it’s connection to Schwabe, but I would be surprised if anyone has more than a vague sense of how it hooks to the √5 frame. Even though I have r^2 =1 for the latest √5 scaling (inverse square), I haven’t yet even given any thought to how to illustrate it.

…and to me there’s (effectively) nothing if there’s no illustration. Words are just to convey thoughts on the trail to a picture. Before there’s a clear concise picture, there’s only unfinished thinking.

…but in this particularly simple case, once there’s a picture it’s going to look so trivial as to again land back at nothing …and it is from the nothing within that the wheel derives its infinite utility.

What’s the picture going to look like? See if you can guess before seeing.

 

https://tallbloke.wordpress.com/2015/12/21/why-phi-an-orbital-parameters-test/

 
 
Reply  Message 3 of 4 on the subject 
From: BARILOCHENSE6999 Sent: 21/12/2016 13:48
 

trouble choosing between models — r^2 set:
0.999991869, 0.999987017, 0.999987844, 0.999989971, 0.999992955

so I’ve been looking at physical parameters to help discern

something I noticed:

mass
J / S = 189.9 / 56.846 = 3.34060444

1/(radius^2)
J / S = 0.036962337 / 0.010996863 = 3.361171098

and of course there’s the obvious one we’ve seen pointed out countless times forever:

period:
S / J = 29.447498 / 11.862615 = 2.482378295 ~= 5/2

Given the √5 coming up, notice that
(2 * 2.482378295)^(3/2) = 11.06233845

Of the models based on √5, φ, & 11.06233845 none are uniformly better. Each performs better in a certain segment of the solar system. With r^2 so high, the general pattern is nailed, but even though the residuals are small, they’re sufficiently systematic to demand more scrutinizing attention.

I’m not one to get carried away with details. I like exploring first-order aggregate structure. So I may just report all models side-by-side and leave it there for others to go explore the roots of the subtle but systematic structural nuances. I want to work out and illustrate the geometry before reporting just tables, which are by orders of magnitude inferior to illustrations on the information assimilability scale.

I’m going to rewrite this from above since it’s probably more intuitive reorganized this way:

(φ^5)^(2/3)) ~= 5

2π*5/(2√2) = 2π*5/(2^(3/2)) = 11.10720735
~= φ^5 = 11.09016994

(11.10720735)*(11.09016994) / ( (11.10720735 + 11.09016994) / 2 ) = 11.09868211
2*(11.09868211) = 22.19736421

(1/(J+S)) / (1/(V+E)) = 8.456145629 / 0.380883098 = 22.20141997

5 = 5
2 = 2
φ = 1.61803398874989
1/φ = Φ = 0.618033988749895
2φ = 3.23606797749979
1/2φ = 0.309016994374947
arccos(1/2φ) = 1.25663706143592
(5/2)*arccos(1/2φ) = 3.14159265358979
π = 3.14159265358979
2π = 6.28318530717959
2π/5 = 1.25663706143592
cos(2π/5) = 0.309016994374947
1/cos(2π/5) = 3.23606797749979
2cos(2π/5) = 0.618033988749895
1/(2cos(2π/5)) = 1.61803398874989
1/(2cos(2π/5)) = φ
(5/2)*arccos(1/2φ) = π

 
 
 
Reply  Message 4 of 4 on the subject 
From: BARILOCHENSE6999 Sent: 21/12/2016 13:57
Resultado de imagen para triangle equilater PHI GOLDEN NUMBER


 
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