Paradise Trinity Day

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Re: Paradise Trinity Day
Re: PdR Lite design*
"sans pattern begot"
Quadrature, proof that Pi is divisible by sqrt(2)
when Pi is represented as Side of Circle's Square (SoCS)
... displayed in three nested circles, all squared.
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod
"sans pattern begot"
Quadrature, proof that Pi is divisible by sqrt(2)
when Pi is represented as Side of Circle's Square (SoCS)
... displayed in three nested circles, all squared.
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod

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Re: Paradise Trinity Day
Re: Flutterbye design
Our transitions to Paradise require many planetary flutterbyes,
not necessarily beginning on Diversey in Chicago.
Rod ... ...
Our transitions to Paradise require many planetary flutterbyes,
not necessarily beginning on Diversey in Chicago.
Rod ... ...

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Re: Paradise Trinity Day
Re: Pi Are Squares design
Who knew?! Pi are squares!
... when they're not circles.
2 / sqrt(Pi) = sqrt(Pi) / (Pi/2)
= 1.1283791670955125738961589031215..
Rod
Who knew?! Pi are squares!
... when they're not circles.
2 / sqrt(Pi) = sqrt(Pi) / (Pi/2)
= 1.1283791670955125738961589031215..
Rod

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Re: Paradise Trinity Day
Re: Pi Are ABC design
Quadrature with Pythagorean je ne sais quoi
Rod ... ....
Quadrature with Pythagorean je ne sais quoi
Rod ... ....

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Re: Paradise Trinity Day
Re: Pi Are ABC design
"Quadrature with Pythagorean je ne sais quoi"
Long story short:
Several nights ago, just after 2:00 AM, I was awakened by a knock on my front door. It sounded like a knock on my wood door  not on the exterior storm door (but I didn't hear the storm door open). I decided not to answer the door unless the knock occurred again  it didn't.
Best guess: a dream.
This morning at 2:22 AM (I looked at the clock), I was again awakened by a similar knock. Best guess: A Midwayer prompt ...
so I got out of bed and created the Pi Are ABC design.
Conjecture: This was a good time for brain energies to focus on the task ... and to receive any proffered assistance.
Pythagoras said "Hello!" ... apparently.
Rod
"Quadrature with Pythagorean je ne sais quoi"
Long story short:
Several nights ago, just after 2:00 AM, I was awakened by a knock on my front door. It sounded like a knock on my wood door  not on the exterior storm door (but I didn't hear the storm door open). I decided not to answer the door unless the knock occurred again  it didn't.
Best guess: a dream.
This morning at 2:22 AM (I looked at the clock), I was again awakened by a similar knock. Best guess: A Midwayer prompt ...
so I got out of bed and created the Pi Are ABC design.
Conjecture: This was a good time for brain energies to focus on the task ... and to receive any proffered assistance.
Pythagoras said "Hello!" ... apparently.
Rod

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Re: Paradise Trinity Day
Re: Pi Are ABC design
"Quadrature with Pythagorean je ne sais quoi"
About this algebra (also in design subtitle) ...
sqrt(4Pi) x sqrt(Pi)/sqrt(4Pi) x 2/sqrt(Pi) = 2.0
0.92650275035220848584275966758914.. sqrt(4Pi)
x 1.9130583802711007947403078280205.. sqrt(Pi)/sqrt(4Pi)
x 1.1283791670955125738961589031216.. 2/sqrt(Pi)
= 2.0
Tip: hypotenuse of circlesquaring right triangle
= circle's diameter = 2.0; short side = sqrt(4Pi)
Rod ... ...
"Quadrature with Pythagorean je ne sais quoi"
About this algebra (also in design subtitle) ...
sqrt(4Pi) x sqrt(Pi)/sqrt(4Pi) x 2/sqrt(Pi) = 2.0
0.92650275035220848584275966758914.. sqrt(4Pi)
x 1.9130583802711007947403078280205.. sqrt(Pi)/sqrt(4Pi)
x 1.1283791670955125738961589031216.. 2/sqrt(Pi)
= 2.0
Tip: hypotenuse of circlesquaring right triangle
= circle's diameter = 2.0; short side = sqrt(4Pi)
Rod ... ...

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Re: Paradise Trinity Day
Re: Pi Are ABC design*
"Quadrature with Pythagorean je ne sais quoi"
Geometry nerds of Quadraturial persuasion know that
Quadrature is beyond comprehension unless it's esoteric.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod
"Quadrature with Pythagorean je ne sais quoi"
Geometry nerds of Quadraturial persuasion know that
Quadrature is beyond comprehension unless it's esoteric.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"
Similar quadrilaterals with Pythagorean perspective.
Rod ... ...
"Quadrature with Pythagorean je ne sais quoi"
Similar quadrilaterals with Pythagorean perspective.
Rod ... ...

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"
Now, with better geometric balance (esoterically speaking).
"What?! sqrt(2) is esoteric? Who knew ?!"
Rod
"Quadrature with Pythagorean je ne sais quoi"
Now, with better geometric balance (esoterically speaking).
"What?! sqrt(2) is esoteric? Who knew ?!"
Rod

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"
Voted best subtitle for this revised design:
Come explore the worlds of geometry!
"Lines and triangles and squares! Oh my!"
Rod
"Quadrature with Pythagorean je ne sais quoi"
Voted best subtitle for this revised design:
Come explore the worlds of geometry!
"Lines and triangles and squares! Oh my!"
Rod

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design*
"Quadrature with Pythagorean je ne sais quoi"
Geometer's secret ...
Lines of the outer '7' have twice the length
of the lines of the inner '7', via sqrt(2).
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...
"Quadrature with Pythagorean je ne sais quoi"
Geometer's secret ...
Lines of the outer '7' have twice the length
of the lines of the inner '7', via sqrt(2).
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"
Line length relationships in circlesquaring right triangle
sides = sqrt(Pi), sqrt(4Pi), 2.0 (hypotenuse and diameter)
sqrt(Pi) = 1.7724538509055160272981674833411..
sqrt(4Pi) = 0.92650275035220848584275966758914..
sqrt(Pi) / sqrt(4Pi) = 1.9130583802711007947403078280204..
2/sqrt(4Pi) = 2.1586552217353950788554161024245..
2/sqrt(Pi) = 1.1283791670955125738961589031215..
(2/sqrt(4Pi)) / (2/sqrt(Pi))
= 1.9130583802711007947403078280204..
Rod
"Quadrature with Pythagorean je ne sais quoi"
Line length relationships in circlesquaring right triangle
sides = sqrt(Pi), sqrt(4Pi), 2.0 (hypotenuse and diameter)
sqrt(Pi) = 1.7724538509055160272981674833411..
sqrt(4Pi) = 0.92650275035220848584275966758914..
sqrt(Pi) / sqrt(4Pi) = 1.9130583802711007947403078280204..
2/sqrt(4Pi) = 2.1586552217353950788554161024245..
2/sqrt(Pi) = 1.1283791670955125738961589031215..
(2/sqrt(4Pi)) / (2/sqrt(Pi))
= 1.9130583802711007947403078280204..
Rod

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"
"Line length relationships in circlesquaring right triangle"
are not revelatory, but this is all about "transcendental" Pi.
Rod ... ...
"Quadrature with Pythagorean je ne sais quoi"
"Line length relationships in circlesquaring right triangle"
are not revelatory, but this is all about "transcendental" Pi.
Rod ... ...

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Re: Paradise Trinity Day
Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"
"Line length relationships in circlesquaring right triangle"
are not revelatory, but this is all about "transcendental" Pi
... as nuanced by sqrt(2).
Rod
"Quadrature with Pythagorean je ne sais quoi"
"Line length relationships in circlesquaring right triangle"
are not revelatory, but this is all about "transcendental" Pi
... as nuanced by sqrt(2).
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
A circle having a diameter = 2.0, has a circumference = 2(Pi).
The two shorter sides (chords) of the circlesquaring right triangle
associate with similar circumference/2 as the two sides (chords)
of the isosceles right triangle, adjacent to the right triangle.
Either sqrt(2) is transcendental or Pi is not ... apparently.
Lines and triangles and squares! Oh my!
Rod
A circle having a diameter = 2.0, has a circumference = 2(Pi).
The two shorter sides (chords) of the circlesquaring right triangle
associate with similar circumference/2 as the two sides (chords)
of the isosceles right triangle, adjacent to the right triangle.
Either sqrt(2) is transcendental or Pi is not ... apparently.
Lines and triangles and squares! Oh my!
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
"Triangular sharing of a 2(Pi) circumference,
a confluence of transcendental Pi hosted by sqrt(2)
with the subtle ambience of a Pythagorean ABC"
Also known as a Cartesian "Whatsit?"
(of Quadraturial persuasion)
"So, whatsit ?!" Go figure!
Rod ... ...
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
"Triangular sharing of a 2(Pi) circumference,
a confluence of transcendental Pi hosted by sqrt(2)
with the subtle ambience of a Pythagorean ABC"
Also known as a Cartesian "Whatsit?"
(of Quadraturial persuasion)
"So, whatsit ?!" Go figure!
Rod ... ...

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
"Triangular sharing of a 2(Pi) circumference,
confluence of transcendental Pi hosted by sqrt(2)
with subtle ambience of Pythagorean ABC"
Geometer's secret about lower right quadrant ...
Pythagorean perspective is best represented
by a square adjoined to that circle's hypotenuse,
but the isosceles right triangle is sufficient clue
and permits enlargement of the geometry.
Rod
"Triangular sharing of a 2(Pi) circumference,
confluence of transcendental Pi hosted by sqrt(2)
with subtle ambience of Pythagorean ABC"
Geometer's secret about lower right quadrant ...
Pythagorean perspective is best represented
by a square adjoined to that circle's hypotenuse,
but the isosceles right triangle is sufficient clue
and permits enlargement of the geometry.
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
(four sqrt(2)nested circles all squared)
The equations represented by this design
(for incorrigible geometry nerds of Quadraturial persuasion):
1/(sqrt(Pi)/2)
= sqrt(2)/((sqrt(2)sqrt(Pi))/2)
= 2/sqrt(Pi)
= 2(sqrt(2))/(sqrt(2)/sqrt(Pi))
= 1.1283791670955125738961589031215..
Rod
(four sqrt(2)nested circles all squared)
The equations represented by this design
(for incorrigible geometry nerds of Quadraturial persuasion):
1/(sqrt(Pi)/2)
= sqrt(2)/((sqrt(2)sqrt(Pi))/2)
= 2/sqrt(Pi)
= 2(sqrt(2))/(sqrt(2)/sqrt(Pi))
= 1.1283791670955125738961589031215..
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
(four sqrt(2)nested circles all squared)
Lines and triangles and squares!
... and trapezoids and parallelograms! Oh my!
Rod
(four sqrt(2)nested circles all squared)
Lines and triangles and squares!
... and trapezoids and parallelograms! Oh my!
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
(five sqrt(2)nested circles, all effectively squared)
Better identification of the concentric parallelograms
(concentric relative to sqrt(2) integration).
Rod ... ...
(five sqrt(2)nested circles, all effectively squared)
Better identification of the concentric parallelograms
(concentric relative to sqrt(2) integration).
Rod ... ...

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design*
(seven sqrt(2)nested circles, all effectively squared)
Who knew
Quadrature is all about pattern
"What pattern? circle? square? triangle?
trapezoid? parallelogram? ..." Yes! Go figure!
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod
(seven sqrt(2)nested circles, all effectively squared)
Who knew
Quadrature is all about pattern
"What pattern? circle? square? triangle?
trapezoid? parallelogram? ..." Yes! Go figure!
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
(eight sqrt(2)nested circles, all effectively squared)
Now, less color noise on the perimeter of this composition.
Who knew
This composition includes a Center of Quadrature
that's obvious when you connect the points.
Rod ... ...
(eight sqrt(2)nested circles, all effectively squared)
Now, less color noise on the perimeter of this composition.
Who knew
This composition includes a Center of Quadrature
that's obvious when you connect the points.
Rod ... ...

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
(eight sqrt(2)nested circles, all effectively squared)
Tweaked daily until it squeaks (computer mouse, that is).
Who knew
This composition includes a Center of Quadrature (CoQ)
that's increasingly obvious as those points connect.
Rod
(eight sqrt(2)nested circles, all effectively squared)
Tweaked daily until it squeaks (computer mouse, that is).
Who knew
This composition includes a Center of Quadrature (CoQ)
that's increasingly obvious as those points connect.
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design
(eight sqrt(2)nested circles, all effectively squared)
Geometer's secret about red and yellow lines extending inward from perimeter of their respective circles ...
While this seems esoteric, these lines (when lengthened) help define the composition's Center of Quadrature
as well as being perpendicular to a line of similar color that represents one side of that circle's inscribed square
... and having the same length as the side of that square. HCIT ?!
Rod
(eight sqrt(2)nested circles, all effectively squared)
Geometer's secret about red and yellow lines extending inward from perimeter of their respective circles ...
While this seems esoteric, these lines (when lengthened) help define the composition's Center of Quadrature
as well as being perpendicular to a line of similar color that represents one side of that circle's inscribed square
... and having the same length as the side of that square. HCIT ?!
Rod

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Re: Paradise Trinity Day
Re: Trapezoidal Bifurcation design*
Quadrature is all about pattern
Who knew
The more spiderywebby this Cartesian composition gets,
the more it's convincing that sqrt(2) hosts Pi ...
and not viceyversey.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...
Quadrature is all about pattern
Who knew
The more spiderywebby this Cartesian composition gets,
the more it's convincing that sqrt(2) hosts Pi ...
and not viceyversey.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...