Paradise Trinity Day
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Re: Paradise Trinity Day
Re: Quinale! 2:1(16 design
(aka "The Art of Pi")
Simplified (even the PDF):
http://aitnaru.org/images/Quinale!.pdf
Rod
(aka "The Art of Pi")
Simplified (even the PDF):
http://aitnaru.org/images/Quinale!.pdf
Rod
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Re: Paradise Trinity Day
Re: Quinale! 2:1(16 design
(aka "The Art of Pi")
Simplified (even PDF): http://aitnaru.org/images/Quinale!.pdf
but still here: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf
Rod ... ...
(aka "The Art of Pi")
Simplified (even PDF): http://aitnaru.org/images/Quinale!.pdf
but still here: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf
Rod ... ...
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Re: Paradise Trinity Day
Re: Quinale! 2:1(16 design
(aka "The Art of Pi")
a^2 + b^2 = c^2 thus Pi + (4 - Pi) = 4
Who knew
Rod (now that's "transcendental")
(aka "The Art of Pi")
aka "Pythagorean Pi" since ...If c = 2, a = sqrt(Pi), b = sqrt(4-Pi)
a^2 + b^2 = c^2 thus Pi + (4 - Pi) = 4
Who knew
Rod (now that's "transcendental")
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Re: Paradise Trinity Day
Re: Scanale! design
(more "Art of Pi")
Two overlapping scalene triangles, associated by sqrt(2).
(of course, circle-squaring in this Neighborhood)
Rod ... ...
(more "Art of Pi")
Two overlapping scalene triangles, associated by sqrt(2).
(of course, circle-squaring in this Neighborhood)
Rod ... ...
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Re: Paradise Trinity Day
Re: Scanale! design
(more "Art of Pi")
Updated to show 3 overlapping, circle-squaring scalene triangles,
with largest and smallest an increment (or decrement) of sqrt(2)^2.
Rod
(more "Art of Pi")
Updated to show 3 overlapping, circle-squaring scalene triangles,
with largest and smallest an increment (or decrement) of sqrt(2)^2.
Rod
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Re: Paradise Trinity Day
Re: Scanale! design
(more "Art of Pi")
Geometers' secret ...
All 3 overlapping, circle-squaring scalene triangles are similar; each line
of the middle triangle has sqrt(2) relationship to the respective line
of both the smaller and larger triangles ... what we would expect
of similar geometry in sqrt(2)-nested circles.
Of course, this proves that the largest and smallest triangles
differ by a factor of 2 (precisely), proving Pi/2 is "transcendental".
Rod ... ...
(more "Art of Pi")
Geometers' secret ...
All 3 overlapping, circle-squaring scalene triangles are similar; each line
of the middle triangle has sqrt(2) relationship to the respective line
of both the smaller and larger triangles ... what we would expect
of similar geometry in sqrt(2)-nested circles.
Of course, this proves that the largest and smallest triangles
differ by a factor of 2 (precisely), proving Pi/2 is "transcendental".
Rod ... ...
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Re: Paradise Trinity Day
Re: Scanale! design
(more "Art of Pi")
Geometers' secret ...
The three similar vertices of the two scalene triangles
in "adjacent" circles are equidistant ... precisely.
Rod
(more "Art of Pi")
Geometers' secret ...
The three similar vertices of the two scalene triangles
in "adjacent" circles are equidistant ... precisely.
Rod
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Re: Paradise Trinity Day
Re: Scanale! design
(more "Art of Pi")
Who knew?!
Pi was always precisely divisible by 2:
Where D = 2, C = Pi(D),
2/sqrt(2) = side of inscribed square
and C/4 = Pi/2.
Rod
(more "Art of Pi")
Who knew?!
Pi was always precisely divisible by 2:
Where D = 2, C = Pi(D),
2/sqrt(2) = side of inscribed square
and C/4 = Pi/2.
Rod
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Re: Paradise Trinity Day
Re: Scanale! design
(more "Art of Pi")
Simple message of the geometry:
Obviously, Pi cannot run off into infinity
without sqrt(2) close behind (or in front).
"Either sqrt(2) is transcendental or Pi is not"
Rod ... ...
(more "Art of Pi")
Simple message of the geometry:
Obviously, Pi cannot run off into infinity
without sqrt(2) close behind (or in front).
"Either sqrt(2) is transcendental or Pi is not"
Rod ... ...
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Re: Paradise Trinity Day
Re: Scanale! design
(the "Art of Perfect Pi")
All astute geometers know ...
a circle of Diameter = 2/sqrt(Pi)
and Circumference = 2(sqrt(Pi))
has an Area = 1 "A Perfect Pi"
D = 2/sqrt(Pi) = 1.1283791670955125738961589031215..
C = Pi(D) = 3.1415926535897932384626433832795..
x 1.1283791670955125738961589031215..
= 3.5449077018110320545963349666823.. = 2(sqrt(Pi))
A = Pi(r^2)
= 3.1415926535897932384626433832795..
x ((1.1283791670955125738961589031215..)/2)^2
= 3.1415926535897932384626433832795..
x 0.31830988618379067153776752674503..
= 1
Rod
(the "Art of Perfect Pi")
All astute geometers know ...
a circle of Diameter = 2/sqrt(Pi)
and Circumference = 2(sqrt(Pi))
has an Area = 1 "A Perfect Pi"
D = 2/sqrt(Pi) = 1.1283791670955125738961589031215..
C = Pi(D) = 3.1415926535897932384626433832795..
x 1.1283791670955125738961589031215..
= 3.5449077018110320545963349666823.. = 2(sqrt(Pi))
A = Pi(r^2)
= 3.1415926535897932384626433832795..
x ((1.1283791670955125738961589031215..)/2)^2
= 3.1415926535897932384626433832795..
x 0.31830988618379067153776752674503..
= 1
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design*
"A quantum of Pi for D=2 and A = 1"
* added to: http://aitnaru.org/images/Quinale!.pdf
That's sum Pi!
Rod
"A quantum of Pi for D=2 and A = 1"
* added to: http://aitnaru.org/images/Quinale!.pdf
That's sum Pi!
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design
"A quantum of Pi for D=2 and A = 1"
Geometers' quantum secret of Entangled Pi (a riddle):
The circle is to the square and the square to the circle
as the circle to the circle and the square to the square
bestow the square root of Pi geometrically entangled.
Rod
"A quantum of Pi for D=2 and A = 1"
Geometers' quantum secret of Entangled Pi (a riddle):
The circle is to the square and the square to the circle
as the circle to the circle and the square to the square
bestow the square root of Pi geometrically entangled.
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design
"A quantum of Pi for D=2 and A = 1"
Geometers' quantum secret of Entangled Pi
(a previous riddle ... or conundrum):
"To square the circle one must circle the square."
Rod
"A quantum of Pi for D=2 and A = 1"
Geometers' quantum secret of Entangled Pi
(a previous riddle ... or conundrum):
"To square the circle one must circle the square."
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design
"Entangled Pi" refers to 2/sqrt(Pi), the constant
that defines the circle-squaring right triangle
(new Pythagorian perspective).
The '~Pi' symbol has good potential.
2/sqrt(Pi) = 1.1283791670955125738961589031215..
/2 = 0.56418958354775628694807945156077..
^2 = 1.2732395447351626861510701069801..
Rod ... ...
Geometers' secret ...To square the circle one must circle the square
"Entangled Pi" refers to 2/sqrt(Pi), the constant
that defines the circle-squaring right triangle
(new Pythagorian perspective).
The '~Pi' symbol has good potential.
2/sqrt(Pi) = 1.1283791670955125738961589031215..
/2 = 0.56418958354775628694807945156077..
^2 = 1.2732395447351626861510701069801..
Rod ... ...
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Re: Paradise Trinity Day
Re: Entangled Pi design
"Pi Fork" identified in inner circle to authenticate "impossible" Quadrature
and featuring line length ratios of sqrt(Pi) and 2/sqrt(Pi).
Rod
Geometers' secret ...To square the circle one must circle the square
"Pi Fork" identified in inner circle to authenticate "impossible" Quadrature
and featuring line length ratios of sqrt(Pi) and 2/sqrt(Pi).
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design*
Geometers' secret (about "Pi Fork" ratios) ...
Given: circle-squaring right triangle (red) with light blue "Pi Fork"
Line length ratios = 2/sqrt(Pi)
- hypotenuse to long side
- two long crossed blue lines
Line length ratios = sqrt(Pi)
- top of right triangle to short blue line
- hypotenuse to adjoined blue line
- long side to adjoined blue line
Rod ... ...
* updated in: http://aitnaru.org/images/Quinale!.pdfTo square the circle one must circle the square
Geometers' secret (about "Pi Fork" ratios) ...
Given: circle-squaring right triangle (red) with light blue "Pi Fork"
Line length ratios = 2/sqrt(Pi)
- hypotenuse to long side
- two long crossed blue lines
Line length ratios = sqrt(Pi)
- top of right triangle to short blue line
- hypotenuse to adjoined blue line
- long side to adjoined blue line
Rod ... ...
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Re: Paradise Trinity Day
Re: Entangled Pi design
Looks esoteric but it's just extrapolation of lines and objects
from the foundational 2/sqrt(Pi) circle-squaring right triangle
(with inherent "Pi Fork" identified and highlighted).
"The '~Pi' symbol has good potential"
Rod
Nested circles all squared, simplified with salience.To square the circle one must circle the square
Looks esoteric but it's just extrapolation of lines and objects
from the foundational 2/sqrt(Pi) circle-squaring right triangle
(with inherent "Pi Fork" identified and highlighted).
"The '~Pi' symbol has good potential"
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design
and is pronounced "wiggle E" (sounds like "wiggly").
Diner: "I'll have Wiggly Pi for dessert."
Waiter: "That's served only at the House of Pi"
Rod ... ...
The new Pi symbol (for constant 2/sqrt(Pi)) is displayed in PDFTo square the circle one must circle the square
and is pronounced "wiggle E" (sounds like "wiggly").
Diner: "I'll have Wiggly Pi for dessert."
Waiter: "That's served only at the House of Pi"
Rod ... ...
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Re: Paradise Trinity Day
Re: Entangled Pi design
(see overlapping right and isosceles right triangles
on right side of this Entangled Pi composition,
colorfully simplified by salience)
2/sqrt(Pi) = 1.1283791670955125738961589031215..
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.4142135623730950488016887242097.. sqrt(2)
= 2.506628274631000502415765284811..
/ 2 = 1.2533141373155002512078826424055..
x 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.4142135623730950488016887242097.. sqrt(2)
"Say what?!" Go figure!
Rod
The math of "Wiggle E" (sounds like "wiggly").To square the circle one must circle the square
(see overlapping right and isosceles right triangles
on right side of this Entangled Pi composition,
colorfully simplified by salience)
2/sqrt(Pi) = 1.1283791670955125738961589031215..
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.4142135623730950488016887242097.. sqrt(2)
= 2.506628274631000502415765284811..
/ 2 = 1.2533141373155002512078826424055..
x 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.4142135623730950488016887242097.. sqrt(2)
"Say what?!" Go figure!
Rod
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Re: Paradise Trinity Day
Re: Entangled Pi design
2/sqrt(Pi) is the length of the shared hypotenuse of the 2 overlapping triangles
(right and isosceles right) with 2/sqrt(Pi) also that circle's diameter.
"And the other lines?" Go figure!
Rod
Geometers' secret ...To square the circle one must circle the square
2/sqrt(Pi) is the length of the shared hypotenuse of the 2 overlapping triangles
(right and isosceles right) with 2/sqrt(Pi) also that circle's diameter.
"And the other lines?" Go figure!
Rod
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Re: Paradise Trinity Day
Re: Mitosis design
Yet another 2/sqrt(Pi) construction ...
with visually confirming essence of Quadrature.
"Where's the mitosis?" Geometrically telophasic.
("ïmpossible" cell division of Quadrature ... apparently)
Rod
Yet another 2/sqrt(Pi) construction ...
with visually confirming essence of Quadrature.
"Where's the mitosis?" Geometrically telophasic.
("ïmpossible" cell division of Quadrature ... apparently)
Rod
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Re: Paradise Trinity Day
Re: Mitosis design*
"geometrically telophasic"
Geometers' secret ...
Red '7' refers to the 2 adjoined trapezoids,
inscribed in those 2 circles and share one side.
Who knew Mitosis math: 7 - 4 = 3
4 phases: Prophase, Metaphase, Anaphase, Telophase
7 trapezoid sides - 4 trapezoid sides = 3 circles
and the seven "deity" levels of "impossible" Quadrature?
(Static, Potential, Associative, Creative, Evolutional, Supreme, Ultimate)
* updated in: http://aitnaru.org/images/Quinale!.pdf
Rod ... ...
"geometrically telophasic"
Geometers' secret ...
Red '7' refers to the 2 adjoined trapezoids,
inscribed in those 2 circles and share one side.
Who knew Mitosis math: 7 - 4 = 3
4 phases: Prophase, Metaphase, Anaphase, Telophase
7 trapezoid sides - 4 trapezoid sides = 3 circles
and the seven "deity" levels of "impossible" Quadrature?
(Static, Potential, Associative, Creative, Evolutional, Supreme, Ultimate)
* updated in: http://aitnaru.org/images/Quinale!.pdf
Rod ... ...
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Re: Paradise Trinity Day
Re: Mitosis design
"geometrically telophasic"
Geometers' secret ...
Red '77' must refer to Mitosial Mitosis.
(smaller '7' appears where design might evolve)
Quadrature ever more esoteric, however ...
"Seventy times and seven" comes to mind. (139:2.5)
Rod
"geometrically telophasic"
Geometers' secret ...
Red '77' must refer to Mitosial Mitosis.
(smaller '7' appears where design might evolve)
Quadrature ever more esoteric, however ...
"Seventy times and seven" comes to mind. (139:2.5)
Rod
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Re: Paradise Trinity Day
Re: Mitosis design
"geometrically telophasic"
Geometers' secret ...
Red '77' must refer to Mitosial Mitosis.
... or maybe it's numerology:
https://trustedpsychicmediums.com/angel ... 7-meaning/
Who can tell "Be bold and brave"
Rod ... ...
"geometrically telophasic"
Geometers' secret ...
Red '77' must refer to Mitosial Mitosis.
... or maybe it's numerology:
https://trustedpsychicmediums.com/angel ... 7-meaning/
Who can tell "Be bold and brave"
Rod ... ...
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Re: Paradise Trinity Day
Re: Mitosis CQ design
"geometrically telophasic w/CQ"
Geometers' secret ...
Geometrically telophasic with a Center of Quadrature.
Rod ... ... ("CQ CQ ...")
"geometrically telophasic w/CQ"
Geometers' secret ...
Geometrically telophasic with a Center of Quadrature.
Rod ... ... ("CQ CQ ...")