Paradise Trinity Day

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Re: Paradise Trinity Day
Re: Spiraling Squares design
"In pursuit of the Big Bang of Quadrature"
As Paradise has a precise location in the Grand Universe,
Pi has a precise location in this Cartesian Neighborhood:
Location of Pi between D=2 and D=4
(calculate up and down from sqrt(Pi)):
= 4.0
x 2.2567583341910251477923178062431.. 2(sqrt(4/Pi))
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.1283791670955125738961589031215.. sqrt(4/Pi)
= 2.0
1.1283791670955125738961589031215..
= 2/sqrt(Pi), sqrt(Pi)/(Pi/2), 2(sqrt(1/Pi))
Rod
"In pursuit of the Big Bang of Quadrature"
As Paradise has a precise location in the Grand Universe,
Pi has a precise location in this Cartesian Neighborhood:
Location of Pi between D=2 and D=4
(calculate up and down from sqrt(Pi)):
= 4.0
x 2.2567583341910251477923178062431.. 2(sqrt(4/Pi))
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.1283791670955125738961589031215.. sqrt(4/Pi)
= 2.0
1.1283791670955125738961589031215..
= 2/sqrt(Pi), sqrt(Pi)/(Pi/2), 2(sqrt(1/Pi))
Rod

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Re: Paradise Trinity Day
Re: EIO Squares design (in progress)
About EIO Squares ... (Enclosing, Inscribed, 'O' = circle)
Since "Old MacDonald had a farm ..."
"moo moo, baa baa, oink oink, quack quack ... EIO"
(re: https://www.youtube.com/watch?v=EGymNLc87M )
Calculate up and down from perimeter of circle's square ...
Perimeter of Enclosing square for D = 4.0
= 16.0
x 1.1283791670955125738961589031216.. sqrt(4/Pi)
<> 14.179630807244128218385339866729..
Perimeter of circle's square; SoCS = 2(sqrt(Pi))
/ 1.2533141373155002512078826424055.. ((sqrt(Pi)(sqrt(2))/2
= 11.313708498984760390413509793678.. 4(2(sqrt(2)))
Perimeter of Inscribed square for D = 4.0
Confirmation that "Old MacDonald had a farm" ...
Relation of Enclosing/Inscribed perimeters:
16.0 / 11.313708498984760390413509793678..
= 1.4142135623730950488016887242097.. sqrt(2)
1.1283791670955125738961589031216.. sqrt(4/Pi)
x 1.2533141373155002512078826424055.. ((sqrt(Pi)(sqrt(2))/2
= 1.4142135623730950488016887242097.. sqrt(2)
Rod ... ...
"moo moo, baa baa, oink oink, quack quack ... EIO"
Did you know ...
Like "impossible" Quadrature, it's impossible to say fast
"moo moo, baa baa, oink oink, quack quack ... EIO"
but possible with lotsa practice (video tape this ...
sounds like a barnyard with a group of quackers)
About EIO Squares ... (Enclosing, Inscribed, 'O' = circle)
Since "Old MacDonald had a farm ..."
"moo moo, baa baa, oink oink, quack quack ... EIO"
(re: https://www.youtube.com/watch?v=EGymNLc87M )
Calculate up and down from perimeter of circle's square ...
Perimeter of Enclosing square for D = 4.0
= 16.0
x 1.1283791670955125738961589031216.. sqrt(4/Pi)
<> 14.179630807244128218385339866729..
Perimeter of circle's square; SoCS = 2(sqrt(Pi))
/ 1.2533141373155002512078826424055.. ((sqrt(Pi)(sqrt(2))/2
= 11.313708498984760390413509793678.. 4(2(sqrt(2)))
Perimeter of Inscribed square for D = 4.0
Confirmation that "Old MacDonald had a farm" ...
Relation of Enclosing/Inscribed perimeters:
16.0 / 11.313708498984760390413509793678..
= 1.4142135623730950488016887242097.. sqrt(2)
1.1283791670955125738961589031216.. sqrt(4/Pi)
x 1.2533141373155002512078826424055.. ((sqrt(Pi)(sqrt(2))/2
= 1.4142135623730950488016887242097.. sqrt(2)
Rod ... ...
"moo moo, baa baa, oink oink, quack quack ... EIO"
Did you know ...
Like "impossible" Quadrature, it's impossible to say fast
"moo moo, baa baa, oink oink, quack quack ... EIO"
but possible with lotsa practice (video tape this ...
sounds like a barnyard with a group of quackers)

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Re: Paradise Trinity Day
Re: EIO Squares design*
“moo moo, baa baa, oink, oink, quack quack … EIO”
* EOP: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod
“moo moo, baa baa, oink, oink, quack quack … EIO”
* EOP: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod

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Re: Paradise Trinity Day
Re: EIO Squares design
When two triangles adjoin on the squared circles farm,
the hypotenuses transform into a "hypotenusii".
“moo moo, baa baa, oink, oink, quack quack
and a loud yeller hypotenusii ... EIO”
Rod ... ...
When two triangles adjoin on the squared circles farm,
the hypotenuses transform into a "hypotenusii".
“moo moo, baa baa, oink, oink, quack quack
and a loud yeller hypotenusii ... EIO”
Rod ... ...

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
"Convincing concept of Quadrature
with ad infinitum of adjoined triangles
via sqrt(2) in a Cartesian Neighborhood."
Acknowledges triangles created by the 8 points
of a circle upon which rests the circle's square.
Rod
"Convincing concept of Quadrature
with ad infinitum of adjoined triangles
via sqrt(2) in a Cartesian Neighborhood."
Acknowledges triangles created by the 8 points
of a circle upon which rests the circle's square.
Rod

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
see a boxa Pi in this Cartesian Neighborhood.
Rod ... ...
Who knew?! Believers (of Quadraturial persuasion)Convincing concept of Quadrature
see a boxa Pi in this Cartesian Neighborhood.
Rod ... ...

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
is that it was drawn last ... using available Cartesian points.
But such is the essence of "impossible" Quadrature.
Some lines were then redrawn to "push" box into background.
But such is the "impossible" patience of a curious geometer.
Rod
What's remarkable  even mysterious  about the boxa PiConvincing concept of Quadrature
is that it was drawn last ... using available Cartesian points.
But such is the essence of "impossible" Quadrature.
Some lines were then redrawn to "push" box into background.
But such is the "impossible" patience of a curious geometer.
Rod

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
Wedge at bottom left, containing 'i' (for 'impossible' and 'information'),
identifies the three lines representing Pi/2, sqrt(Pi), and 2.0, since ...
2.0/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 1.1283791670955125738961589031215..,
the circlesquaring constant nicknamed "iPi" for "impossible Pi".
Rod
Geometers' secret:Convincing concept of Quadrature
Wedge at bottom left, containing 'i' (for 'impossible' and 'information'),
identifies the three lines representing Pi/2, sqrt(Pi), and 2.0, since ...
2.0/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 1.1283791670955125738961589031215..,
the circlesquaring constant nicknamed "iPi" for "impossible Pi".
Rod

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
Quadrature is best comprehended by visualing movement of triangles on an axis:
Golden vertical line, representing sqrt(Pi), is both long side of adjoined right triangle
on the right and hypotenuse of adjoined right triangle on the left.
Either triangle can pivot in the opposite direction, complementing this Cartesian composition
that clearly shows the geometric relationship of Pi/2, sqrt(Pi), and 2.0.
Rod
Geometers' secret (another):Convincing concept of Quadrature
Quadrature is best comprehended by visualing movement of triangles on an axis:
Golden vertical line, representing sqrt(Pi), is both long side of adjoined right triangle
on the right and hypotenuse of adjoined right triangle on the left.
Either triangle can pivot in the opposite direction, complementing this Cartesian composition
that clearly shows the geometric relationship of Pi/2, sqrt(Pi), and 2.0.
Rod

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
giving evidence of 2/sqrt(Pi) propagation of circlesquaring right triangles.
The distance between three of these triangles* is (2(sqrt(Pi))^2,
hinting that 2/sqrt(Pi) is a Constant Of Interest (COI, pronounced "coy")
for development of a more precise Pi value, relative to Circumference.
* and relation of their hypotenuse to long side
Rod ... ... (cruisin' in my Conveyance Of Interest)
After more "visualization", a new moon appears (on the left of the design),best comprehended by visualizing movement of triangles on an axis
giving evidence of 2/sqrt(Pi) propagation of circlesquaring right triangles.
The distance between three of these triangles* is (2(sqrt(Pi))^2,
hinting that 2/sqrt(Pi) is a Constant Of Interest (COI, pronounced "coy")
for development of a more precise Pi value, relative to Circumference.
* and relation of their hypotenuse to long side
Rod ... ... (cruisin' in my Conveyance Of Interest)

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Re: Paradise Trinity Day
Re: Rights of Pythagoras design
About sqrt(2) spiral with tandem Pi spiral ...
Ratio of hypotenuse to long side of circlesquaring
right triangle maintains at 2/sqrt(Pi) during growth of spirals
(SoCS = Side of Circle's Square):
When D = 1
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 0.88622692545275801364908374167057.. sqrt(Pi)/2, SoCS
When D = sqrt(2)
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2)))/2, SoCS
When D = 2.0
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.7724538509055160272981674833411.. sqrt(Pi), SoCS
Evidence of tandem Pi spiral ...
1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2)))/2, SoCS
/ 0.88622692545275801364908374167057.. sqrt(Pi)/2, SoCS
= 1.4142135623730950488016887242097.. sqrt(2)
1.7724538509055160272981674833411.. sqrt(Pi), SoCS
/ 1.2533141373155002512078826424055.. sqrt(Pi)(sqrt(2)), SoCS
= 1.4142135623730950488016887242097.. sqrt(2)
Analysis/conjecture:
Because Diameter is a whole number every second
sqrt(2) increment, and because both sqrt(2) and Pi spirals
grow in tandem, then the Pi value must be influenced
by sqrt(2) "wholeness" every second increment.
"Either sqrt(2) is transcendental or Pi is not"
(that is, the Pi of geometry  not of advanced math)
Rod
About sqrt(2) spiral with tandem Pi spiral ...
Ratio of hypotenuse to long side of circlesquaring
right triangle maintains at 2/sqrt(Pi) during growth of spirals
(SoCS = Side of Circle's Square):
When D = 1
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 0.88622692545275801364908374167057.. sqrt(Pi)/2, SoCS
When D = sqrt(2)
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2)))/2, SoCS
When D = 2.0
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.7724538509055160272981674833411.. sqrt(Pi), SoCS
Evidence of tandem Pi spiral ...
1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2)))/2, SoCS
/ 0.88622692545275801364908374167057.. sqrt(Pi)/2, SoCS
= 1.4142135623730950488016887242097.. sqrt(2)
1.7724538509055160272981674833411.. sqrt(Pi), SoCS
/ 1.2533141373155002512078826424055.. sqrt(Pi)(sqrt(2)), SoCS
= 1.4142135623730950488016887242097.. sqrt(2)
Analysis/conjecture:
Because Diameter is a whole number every second
sqrt(2) increment, and because both sqrt(2) and Pi spirals
grow in tandem, then the Pi value must be influenced
by sqrt(2) "wholeness" every second increment.
"Either sqrt(2) is transcendental or Pi is not"
(that is, the Pi of geometry  not of advanced math)
Rod

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Re: Paradise Trinity Day
Re: Spi_rally design
(rally of spiraling circlesquaring objects)
While trying to give isosceles right triangles more visibility,
I noticed a missing spiral: the circlesquaring scalene triangle
Now, circlesquaring right triangles are most visible, followed by
scalene triangles, with isosceles right triangles lurking nearby.
Obviously, 2/sqrt(Pi) is the common spiral growth factor.
Rod ... ... (cruisin' back to the rally)
(rally of spiraling circlesquaring objects)
While trying to give isosceles right triangles more visibility,
I noticed a missing spiral: the circlesquaring scalene triangle
Now, circlesquaring right triangles are most visible, followed by
scalene triangles, with isosceles right triangles lurking nearby.
Obviously, 2/sqrt(Pi) is the common spiral growth factor.
Rod ... ... (cruisin' back to the rally)

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Re: Paradise Trinity Day
Re: Spi_rally design
(rally of spiraling circlesquaring objects)
With 2/sqrt(Pi) relationship, both Diameter and SoCS (hypotenuse and long side)
maintain that relationship no matter how many times they're divided by 2 ...
and computers will run out of known Pi digits long before division by 2
gets lost in subquantum values.
Then what happens
Rod
(rally of spiraling circlesquaring objects)
New "out of the box" insight ...2/sqrt(Pi) is the common spiral growth factor.
With 2/sqrt(Pi) relationship, both Diameter and SoCS (hypotenuse and long side)
maintain that relationship no matter how many times they're divided by 2 ...
and computers will run out of known Pi digits long before division by 2
gets lost in subquantum values.
Then what happens
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
All lines of the two similar scalene and two similar right triangles
have sqrt(2) relationship, making this Cartesian Neighborhood
the convincing essence of "impossible" Quadrature.
Rod
All lines of the two similar scalene and two similar right triangles
have sqrt(2) relationship, making this Cartesian Neighborhood
the convincing essence of "impossible" Quadrature.
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
Redrawn to show that every line of the circlesquaring objects in both circles
(the smaller circle a sqrt(2) sibling of the larger) has geometricallyprovable
sqrt(2) relationship with a similar line. HCIT
Rod ... ... (off for similar respite)
Redrawn to show that every line of the circlesquaring objects in both circles
(the smaller circle a sqrt(2) sibling of the larger) has geometricallyprovable
sqrt(2) relationship with a similar line. HCIT
Rod ... ... (off for similar respite)

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
(added to: http://aitnaru.org/images/Khristos_Voskrese.pdf )
Geometers' secret ...
The dark blue (circlesquaring) scalene triangle can be inscribed
in a circle having the same diameter as the small light blue circle
... and both circles host identical scalene triangles.
(a conjunction of circles known as Vesica Piscis)
Rod
(added to: http://aitnaru.org/images/Khristos_Voskrese.pdf )
Geometers' secret ...
The dark blue (circlesquaring) scalene triangle can be inscribed
in a circle having the same diameter as the small light blue circle
... and both circles host identical scalene triangles.
(a conjunction of circles known as Vesica Piscis)
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
relative to Quadrature. More exploration of Spi_rally Similar did not reveal
any significant geometric contribution by Vesica to this Neighborhood.
The geometric objects, angles, and lines of "impossible" Quadrature
are unique  recently "Out of the Box" and "New Millennium"!
Maybe, a circlesquaring scalene triangle is this millennium's V.P.
and incorporating 2/sqrt(Pi), 45degree angle, et al.
Rod
Who knew that the Vesica Piscis was so "ancient history",a conjunction of circles known as Vesica Piscis
relative to Quadrature. More exploration of Spi_rally Similar did not reveal
any significant geometric contribution by Vesica to this Neighborhood.
The geometric objects, angles, and lines of "impossible" Quadrature
are unique  recently "Out of the Box" and "New Millennium"!
Maybe, a circlesquaring scalene triangle is this millennium's V.P.
and incorporating 2/sqrt(Pi), 45degree angle, et al.
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
Vesica is inherent in the sqrt(2) CSC* foundation of this Quadrature,
a foundation constructed again and again for several years!
* Circle inscribed in Square inscribed in Circle
Rod
Of course notdid not reveal any significant geometric contribution by Vesica
Vesica is inherent in the sqrt(2) CSC* foundation of this Quadrature,
a foundation constructed again and again for several years!
* Circle inscribed in Square inscribed in Circle
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
Now, a bit more esoteric but ...
convincing geometric proof that Pi is divisible by 2.
Who knew (and note the thetalike symbol)
Rod ... ...
Now, a bit more esoteric but ...
convincing geometric proof that Pi is divisible by 2.
Who knew (and note the thetalike symbol)
Rod ... ...

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
(that scalene triangle inscribed in a circle).
The short line of the scalene is one side of an inscribed square
(not displayed to limit busyness of "impossible" esotericism).
Rod
Skip the 'theta"  Quadrature prefers its own symbolnote the thetalike symbol
(that scalene triangle inscribed in a circle).
The short line of the scalene is one side of an inscribed square
(not displayed to limit busyness of "impossible" esotericism).
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
composed of 3 overlapping scalene triangles. HCIT
Rod
... especially since the scalene unifies two overlapping trapezoidsQuadrature prefers its own symbol
composed of 3 overlapping scalene triangles. HCIT
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
Who knew that Pi was always divisible by 2 ...
since a circle's area square could have been any square from
an inscribed square to a square enclosing the circle, all of which
are divisible by two in a CSC* Cartesian Neighborhood.
* Circle inscribed in Square inscribed in Circle
Geometers' master secret:
The 8 points (only) of a circle upon which rests the square
was the beginning of this research years ago since area square
must have sides less than the enclosing square and sides
greater than the inscribed square A revelation!
This 8point "revelation" is effective Quadraturial rebuttal
to those who insist (still) that Pi must account for the infinite
number of points on a circle. "On a plate, there's 8 ... only."
(grammarians prefer "there are")
Rod
Who knew that Pi was always divisible by 2 ...
since a circle's area square could have been any square from
an inscribed square to a square enclosing the circle, all of which
are divisible by two in a CSC* Cartesian Neighborhood.
* Circle inscribed in Square inscribed in Circle
Geometers' master secret:
The 8 points (only) of a circle upon which rests the square
was the beginning of this research years ago since area square
must have sides less than the enclosing square and sides
greater than the inscribed square A revelation!
This 8point "revelation" is effective Quadraturial rebuttal
to those who insist (still) that Pi must account for the infinite
number of points on a circle. "On a plate, there's 8 ... only."
(grammarians prefer "there are")
Rod

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Re: Paradise Trinity Day
Re: Spi_rally Similar design
The dark blue lines show that the diagonal of the smaller trapezoid
has same length as midpoint to midpoint of the larger trapezoid
... when the circles geometrically associate by sqrt(2).
Rod ("Pi with two forks, please")
Geometers' secret:Pi was always divisible by 2
The dark blue lines show that the diagonal of the smaller trapezoid
has same length as midpoint to midpoint of the larger trapezoid
... when the circles geometrically associate by sqrt(2).
Rod ("Pi with two forks, please")

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Re: Paradise Trinity Day
Re: Two Similar design*
(further exploration of Spi_rally Similar)
Two Similar contrasts the circlesquaring objects of the largest and smallest circles,
a contrast that includes the Happy Family of Quadrature's objects: trapezoid,
scalene, right, and isosceles right triangles.
This juxtaposition creates visual confirmation that the largest circle
(and its objects) have twice the dimensions of the smallest circle.
* added to: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod
(further exploration of Spi_rally Similar)
Whereas Spi_rally Similar focused on the associated sqrt(2) circle of the largest circle,Pi was always divisible by 2
Two Similar contrasts the circlesquaring objects of the largest and smallest circles,
a contrast that includes the Happy Family of Quadrature's objects: trapezoid,
scalene, right, and isosceles right triangles.
This juxtaposition creates visual confirmation that the largest circle
(and its objects) have twice the dimensions of the smallest circle.
* added to: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod

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Re: Paradise Trinity Day
Re: Two Similar design
must end in an even number ... according to this CSC
precision (re: sqrt(2)^2 = 2). What is every odd number?
An interim calculation?
If every Pi division (or multiplication) by sqrt(2) must produce
an even ending digit (maintains CSC precision), what sequences
of Pi's digits remain as "authentic" Pi?
Maybe sqrt(2)'s ending digit is the chain's weak link.
Rod
Obviously, every Pi calculation to collect more digitsPi was always divisible by 2
must end in an even number ... according to this CSC
precision (re: sqrt(2)^2 = 2). What is every odd number?
An interim calculation?
If every Pi division (or multiplication) by sqrt(2) must produce
an even ending digit (maintains CSC precision), what sequences
of Pi's digits remain as "authentic" Pi?
Maybe sqrt(2)'s ending digit is the chain's weak link.
Rod