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Re: Paradise Trinity Day

Posted: Thu Jun 27, 2019 11:33 am
by Amigoo
:sunflower: Re: Pythagorean Quadrature design
You can't get there from here! (from inside the box)
:geek: 1.6 from a rounded line length is hardly transcendental, :oops:
but is still ethereal from a dreamy perspective. Nevertheless,
this new trapezoid incorporates Qphi = sqrt(Pi) / sqrt(4-Pi)
= 1.9130583802711007947403078280203..

"Quadrangularly Pythagorean" comes to mind. :roll:

Rod :)

Re: Paradise Trinity Day

Posted: Fri Jun 28, 2019 12:58 am
by Amigoo
:sunflower: Re: Pythagorean Quadrature design
You can't get there from here! (from inside the box)
:geek: Easy online trigonometry calculator:
https://www.omnicalculator.com/math/trigonometry

Verify Pythagorean circle-squaring right triangle*
by entering two of the known side lengths:
a = sqrt(4-Pi), b = sqrt(Pi), c = 2 ...
try other lengths, angles, whatever. :roll

* See Pythagorean Quadrature design in:
http://aitnaru.org/images/Pi_Fork_n_Lute.pdf

Rod :D

Re: Paradise Trinity Day

Posted: Sun Jun 30, 2019 1:33 am
by Amigoo
:sunflower: Re: Ethereal Triangulation design
(overlapping, circle-squaring scalene triangles
facilitated by Quadraturial concentricity)

:scratch: Who knew :?: :!: Another E.T. perspective: ;)
Every inhabited universe has an administrative center.

:geek: Geometers' secret ...
Red lines have equal length and sqrt(2) integration. 8)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sun Jun 30, 2019 2:38 pm
by Amigoo
Re: https://crop-circles-2019.blogspot.com/ ... ourmi.html

Fascinating :!:
The first crop circle I've seen that shows the characteristic circle-squaring right triangle
adjoined (on hypotenuse) to an isosceles right triangle, with both inscribed in the circle. 8)

:cheers: Authentic Quadrature* if short side of right triangle has length equal to sqrt(4-Pi),
long side equal to sqrt(Pi) when the shared hypotenuse has length equal to 2 ...
and hypotenuse is circle's diameter. ;) * complementing square not shown. :(

Rod :D

Re: Paradise Trinity Day

Posted: Sun Jun 30, 2019 7:34 pm
by Amigoo
Re: https://www.youtube.com/watch?v=XisPp79JVA8 (see at 46:22 in video) ,
https://temporarytemples.co.uk/project/ ... -july-2018

This aerial video and photos do not appear to show identifying two-adjoined
right + isosceles right triangles. However, the interpretation of "door ajar"
is intriguing ... and the shape seems trapezoidal, common objects
in the Ethereal Triangulation composition. ;)

Rod :)

Re: Paradise Trinity Day

Posted: Mon Jul 01, 2019 4:33 am
by Amigoo
:sunflower: Re: Ethereal Triangulation design

:geek: Geometers' secret ...

The adjoined right and isosceles right triangles represent (via shared hypotenuse)
the uniting of sqrt(Pi) and sqrt(2), long side of right and a side of the isosceles;
a uniting theorized a decade ago as the geometric secret of squared circles. ;)

Rod :)

Re: Paradise Trinity Day

Posted: Mon Jul 01, 2019 9:47 pm
by Amigoo
:shock: Is it true :?: :!: There's Pi in the Tub :?:

:idea: If the circle cannot be squared but squared circles exist,
then a cube and a sphere of the same volume (cubed sphere)
would displace the same amount of water. ;)

:bana: Said the Pi man to Simple Simon ...
"If the circle cannot be squared, you cannot cube the sphere
... but you can always have Pi in the tub." :roll

Rod :D

Re: Paradise Trinity Day

Posted: Mon Jul 01, 2019 10:39 pm
by Amigoo
:shock: Is it true :?: :!: There's Pi in the Tub :?:
Said Simple Simon to the Pi man ...

:scratch: "Then, how did Pythagoras trapezoid?"
Said the Pi man: "With his parallel bars." :roll:

Rod :)

Re: Paradise Trinity Day

Posted: Tue Jul 02, 2019 12:16 pm
by Amigoo
Re: Pythagorean Quadrature design in:
http://aitnaru.org/images/Pi_Fork_n_Lute.pdf

and: https://www.youtube.com/watch?v=XisPp79JVA8 (see at 46:22 in video) ,
https://temporarytemples.co.uk/project/ ... -july-2018

:idea: That crop circle interpretation of "door ajar" (opening to planetary enlightenment)
seems on target since PQ design's large golden trapezoid Identifies 3 overlapping,
circle-squaring scalene triangles integrated with circle-squaring right triangles. 8)

:!: There is no other geometry (in crop circles or not)
that convincingly claims "I AM Quadrature". ;)

Rod :D

Re: Paradise Trinity Day

Posted: Wed Jul 03, 2019 11:59 am
by Amigoo
:sunflower: Re: Ethereal Triangulation design
There is no other geometry (in crop circles or not)
that convincingly claims "I AM Quadrature".
:geek: Obviously, geometry that can make such a claim
is geometry that speaks for itself (but a few colors changed
to highlight the overlapping scalene and right triangles). :roll

Rod ... :bike: ... (I AM quiescing)

Re: Paradise Trinity Day

Posted: Wed Jul 03, 2019 7:22 pm
by Amigoo
:sunflower: Re: Ethereal Triangulation design
There is no other geometry (in crop circles or not)
that convincingly claims "I AM Quadrature".
:geek: Crimson Conundrum of Quadraturial Quintessence (QQ):
The two-dimensional "pyramid" in this design's center is similar
to a "pyramid" inscribed in larger circle of sqrt(2) concentricity. 8)

:idea: That such objects replicate in sqrt(2) concentricity
hint that sqrt(Pi) and sqrt(2) must be finite values ...
in the real world of Cartesian Neighborhoods. ;)

:scratch: "Those objects are red - not crimson!"
They're inscribed in crimson squares of QQ. :D

Rod ... :bike: ... (I AM quiescing ... again) :roll:

Re: Paradise Trinity Day

Posted: Fri Jul 05, 2019 12:22 am
by Amigoo
:sunflower: Re: Ethereal Triangulation design
and Crimson Conundrum of Quadraturial Quintessence

:geek: The obvious sqrt(2)^2 relationship of the inner and outer "pyramids", indicating precise
dimensional difference of 2 (or one-half), is enhanced by the 2:1 (or 1:2) grouping of angles
within that scalene triangle: inner angles formed by the pyramid's "vertical" sides are twice
the size of the corners of that scalene triangle, thereby correlating a sqrt(2)^2 grouping
of 360 degrees and 180 degrees. 8)

Rod ... :bike: ... (I AM quiescing ... supposedly) :roll:

Re: Paradise Trinity Day

Posted: Fri Jul 05, 2019 9:11 pm
by Amigoo
:sunflower: Re: Ethereal Triangulation design
and Crimson Conundrum of Quadraturial Quintessence

:geek: "A Scalience of Quadrature" (because others exist, apparently). 8)
Note that the golden scalene triangle(s) can fold in upon itself ...
or unfold into infinity? Who knew :?: :!:

"Scalience" = Scalene + salient :roll:

Rod ... :bike: ... (I AM quiescing ... yada, yada) :roll:

Re: Paradise Trinity Day

Posted: Fri Jul 05, 2019 10:04 pm
by Amigoo
:roll: New insight ...

"Multiverses" are simply the same universe as perceived by its different states of existence.
If we could, we would perceive just one universe (except that other astronomical universes
also exist, each having "multiverse" states of existence). 8)

:scratch: "Say what :?: :!:"
You can''t get there from here (yet), but you can get there from here.
:scratch: "What state are you in :?: :!:"

Rod :stars: (in that state of mind) ;)

Re: Paradise Trinity Day

Posted: Sun Jul 07, 2019 12:12 pm
by Amigoo
:sunflower: Re: Ethereal Triangulation design
golden scalene triangle(s) can fold in upon itself ...
or unfold into infinity?
:geek: This circle-squaring scalene triangle hints that sqrt(2) is a mathematical limit on Pi.

With the long sides extended and short side representing a side of an inscribed square,
a continuum of scalene triangles is created, reversing vertical position with a sqrt(2) increase.
This reversing seems to define a precise geometric wedge that creates a "Pi Corral". :o

:scratch: "Say what :?: :!: "
A wedgie of Pi, out of the box, is very tasty! :roll

Rod :)

Re: Paradise Trinity Day

Posted: Tue Jul 09, 2019 8:11 pm
by Amigoo
:sunflower: Re: Ethereal Triangulation design
A wedgie of Pi, out of the box, is very tasty! :roll
:geek: Geometers' secret ...

4 : Pi : 2 ~ sqrt(4) : sqrt(Pi) : sqrt(2)
defines the geometry where a circle of Diameter = 2
contains 3 concentric squares with largest enclosing circle,
smallest an inscribed square, and circle's area square
within the "Pi corral" of outer/inner squares. 8)

Qphi = sqrt(Pi) / sqrt(4-Pi)
= 1.9130583802711007947403078280203..

Qcue = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 2(sqrt(1/Pi))
= 1.1283791670955125738961589031215..

Rod :)

Re: Paradise Trinity Day

Posted: Wed Jul 10, 2019 3:37 am
by happyrain
Rod,

Profound insights on the multiverse...

Your works are captivating, even if I don't always understand what I'm looking at- is aitnaru your website? The welcoming on the homepage is beautiful.

Is this all sacred geometry? Sorry for my ignorance, I noticed the number 14 on the site. Do you find special meaning with this number?

Thank you.

Re: Paradise Trinity Day

Posted: Wed Jul 10, 2019 5:33 am
by Amigoo
:sunflower: Re: Parallelogram Of Ethereal Triangulation (POET) design

:geek: Geometric proof that sqrt(2) is host of the Pi Corral
with the correlating parallelogram that contrasts
the sqrt(2)-related quadrilaterals of Quadrature
(two adjoined right and isosceles right triangles).

Rod :D

Re: Paradise Trinity Day

Posted: Wed Jul 10, 2019 5:50 am
by Amigoo
Happyrain,

The Aitnaru site is the "online museum" for my decade-long study of the Greek geometry challenges from antiquity:
trisecting an angle, doubling a cube, squaring a circle (Quadrature). For closure of a study, the geometry is converted
to colorful art and posted as "design" ... since geometric proofs do not exist. ;)

My studies have always been more pattern analysis than mathematical number crunching.
I'm not aware that any of the geometry qualifies as "sacred" and to claim that Pi is finite
(at least, constrained by sqrt(2)) is a bit controversial. But the geometry suggests this! :roll:

:lol: Today, if '14' is significant it relates to:
sqrt(2) = 1.4142135623730950488016887242097..
and to Pi = 3.1415926535897932384626433832795..

Rod :)

Re: Paradise Trinity Day

Posted: Wed Jul 10, 2019 1:55 pm
by Amigoo
:sunflower: Re: Parallelogram Of Ethereal Triangulation (POET) design
(page 11 of: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf )

"A POET in the cacophony that 'the circle cannot be squared'"

:geek: Geometers' secret ...
Of the two diagonals in the related circle-squaring quadrilaterals,
one is that circle's diameter, the other proof of sqrt(2) hosting. 8)

:scratch: "How do you know that?!"
Attune to the lines of the POET. ;)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Wed Jul 10, 2019 3:11 pm
by happyrain
Thank you. It is a nice collection you have.

Re: Paradise Trinity Day

Posted: Fri Jul 12, 2019 3:22 pm
by Amigoo
:sunflower: Re: Parallelogram Of Ethereal Triangulation (POET) design

It takes a POET to discover Vesica Piscis (and Quadrature)
in the cacophony that “the circle cannot be squared”. 8)

OMG :!: This is beginning to hint that, within Quadrature,
exists all the Cartesian geometry of the universe. :o

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sat Jul 13, 2019 11:11 pm
by Amigoo
:sunflower: Re: POET VP design
This is beginning to hint that, within Quadrature,
exists all the Cartesian geometry of the universe.
:geek: Who knew :?: :!: Vesica Piscis does not exist alone
... and is obviously hosted by sqrt(2): 3 similar circles
are centered on the three vertical lines of a square
(two sides, center), further revealing that 5 circles
are in a composition hosting 4 Vesicae Piscis.

:farao: And this reveals existence of a hexagon
(or encircled cube, depending on perspective). 8)

:scratch: "How many encircled cubes exist in a squared circle?"
None (if squared circles do not exist). No "Cubes" :(
None + infinity (if they do exist). Lotsa "Qubes" ;)

Rod :stars:

Re: Paradise Trinity Day

Posted: Mon Jul 15, 2019 2:37 am
by Amigoo
:sunflower: Re: POET VP design
"Hexagonal Qube of Visicae Piscis"

:geek: Integration of 4 Visicae Piscis within Quadrature
creates the golden hexagon which reveals the Qube. :roll

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Tue Jul 16, 2019 12:22 pm
by Amigoo
:sunflower: Re: POET VP design
"Hexagonal Qube of Visicae Piscis"

:geek: Geometers' secret ...
Objects within concentric circles Quadrature
create the golden Vesicae Piscis Qube. ;)

Rod :D