This thread culminated into the derivation of a more precise calculation of π as shown here:

and elaborated on page 3.Beginning with a circle whose diameter is √5,

place two unit squares inside of the circle side-by-side

(either horizontally or vertically, the latter is shown above)

and find √5 as any diagonal between two opposite corners (AB as shown)

(this diagonal thus beingequalto the diameter of the circle).

Add 1 unit distance to this diagonal (√5 + 1) and find the mid-point (/2).

Note: this midpoint coincides with the circle whose diameter is 2r = 1

by way of rotating the √5 diagonal about the origin of the circle we began with,

and thus"kisses"the square four timesequidistantly, thus a more precise π

can be begotten directly from Φexpressed as a ratio of Φ.

.This ratio is 4/√Φ = π and/or 16/Φ = π²

Concerning:

https://en.wikipedia.org/wiki/Kepler_triangle

I am unsure as to the motive for the above, but the construction of the geometry

is conceptually unsound: concerning the same Kepler triangle, a circle that circumscribes it

completely ignores the imperative need for the diameter of the circle to be √5, the same to which

the Kepler triangle must apply. If the circle is not √5, there is no valid coupling and π remains 'transcendental'

however if coupled to the √5 diameter circle (as shown), the coupled relation permits a geometry

whose co-operation between Φ and π are respectively reflected in the following:

Φ solves for

f(x) = x² - x - 1

π as 4/√Φ solves for

f(x)=x

^{4}+ 16x

^{2}- 256

thus

**Φπ² = 16**

graphed with an overlay of the same geometry:

__________________________________________________________________________

This is important because it confirms the bi-rotation model (mandated by the geometry as intrinsic)

and implies a bi-

*orientation compliment*, the properties of which can be derived inductively -

the implications of which amounts to the capacity to calculate universal roots using universal geometries

that satisfy cubed proportionality (as time and space reciprocally do).