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Re: Paradise Trinity Day

Posted: Wed Sep 20, 2017 3:14 pm
by Amigoo
:flower: Re: Tri Phi Pi design
Behold the geometric lock :?: :scratch:

Small, golden plus sign (+) identifies center of circle where D = 2.0;
D = 4(sqrt(1/Pi)) identifies hypotenuse of large right triangle. 8)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Thu Sep 21, 2017 2:17 am
by Amigoo
:flower: Re: Tri Phi Pi design (for the record)
Wiggly Numbers of The Right Triangle(s) :roll:

:geek: Given: SoCS = Side of Circle's Square,
3.1415926535897932384626433832795.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
1.9130583802711007947403078280203.. iPhi, long-side-to-short-side
1.1283791670955125738961589031215.. 2(sqrt(1/Pi)), hypotenuse-to-long-side

:arrow: Calculate short side a1 of largest right triangle:

Diameter = 2.256758334191025147792317806243.. 4(sqrt(1/Pi)),
side c1, hypotenuse of right triangle
SoCS = 2.0, side b1, long side of right triangle

From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a1 = 1.0454464017541266302735942239052..
iPhi = b1/a1 = 1.9130583802711007947403078280203..

:arrow: Calculate short side a2 of next largest right triangle:

Diameter = 1.27323954473516268615107010698.. 2(sqrt(1/Pi))^2,
side c2, hypotenuse of right triangle
SoCS = 1.1283791670955125738961589031215.. 2(sqrt(1/Pi)),
side b2, long side of right triangle

From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a2 = 0.58982997002716101129132484047986..
iPhi = b2/a2 = 1.9130583802711007947403078280203..

:arrow: Calculate line length ratio of short side a1 to a2:

short side a1 / short side a2
= 1.0454464017541266302735942239052..
/ 0.58982997002716101129132484047986..
= 1.7724538509055160272981674833411.. sqrt(Pi)

Rod :stars: :stars: :stars: (Tri Phi Pi overload)

Re: Paradise Trinity Day

Posted: Thu Sep 21, 2017 10:10 am
by Amigoo
:flower: Re: Tri Phi Pi design (for the record, expanded)
Wiggly Numbers of The Right Triangle(s) :roll:

:geek: Given: SoCS = Side of Circle's Square,
3.1415926535897932384626433832795.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
1.9130583802711007947403078280203.. iPhi, long-side-to-short-side
1.1283791670955125738961589031215.. 2(sqrt(1/Pi)), hypotenuse-to-long-side

:arrow: Calculate short side a1 of largest right triangle (1 of 3):

Diameter = 2.256758334191025147792317806243.. 4(sqrt(1/Pi)),
side c1, hypotenuse of right triangle
SoCS = 2.0, side b1, long side of right triangle

From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a1 = 1.0454464017541266302735942239052..
iPhi1 = b1/a1 = 1.9130583802711007947403078280203..

:arrow: Calculate short side a2 of next largest right triangle (2 of 3):

Diameter = 1.27323954473516268615107010698.. 2(sqrt(1/Pi))^2,
side c2, hypotenuse of right triangle
SoCS = 1.1283791670955125738961589031215.. 2(sqrt(1/Pi)),
side b2, long side of right triangle

From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a2 = 0.58982997002716101129132484047986..
iPhi2 = b2/a2 = 1.9130583802711007947403078280203..

:arrow: Calculate short side a3 of smallest right triangle (3 of 3):

Diameter = 0.71834848850066624675632793451077.. 4(sqrt(1/Pi)) / Pi
side c3, hypotenuse of right triangle
SoCS = 0.63661977236758134307553505349006.. 2/Pi
side b3, long side of right triangle

From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a3 = 0.33277592515360954394708556243436..
iPhi3 = b3/a3 = 1.9130583802711007947403078280203..


:arrow: Calculate line length ratio of short side a1 to a2:

short side a1 / short side a2
= 1.0454464017541266302735942239052..
/ 0.58982997002716101129132484047986..
= 1.7724538509055160272981674833411.. sqrt(Pi)

:arrow: Calculate line length ratio of short side a2 to a3:

short side a2 / short side a3
= 0.58982997002716101129132484047986..
/ 0.33277592515360954394708556243436..
= 1.7724538509055160272981674833411.. sqrt(Pi)


:farao: About the resonance of the two portals (re: SoCS) ...
2.0 / 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1.7724538509055160272981674833411.. sqrt(Pi) ;) ;)


Rod :stars: :stars: :stars: (Tri Phi Pi overload ^2)

Re: Paradise Trinity Day

Posted: Fri Sep 22, 2017 4:03 pm
by Amigoo
:flower: Re: Power of 2 design
Visual antidote to "Tri Phi Pi overload ^2" :stars: :stars: :stars:
with impressive display of similar isoceles trapezoids.

:geek: Two pairs of SoCS red lines have 2.0 or 1.0 length;
big D = 4(sqrt(Pi)), little D = 2(sqrt(Pi))

Rod (with little 'd') :roll:

Re: Paradise Trinity Day

Posted: Sat Sep 23, 2017 2:17 pm
by Amigoo
:flower: Re: Power of 2 II design
(further exploration of the Po2 geometry)

Better integration of circle-squaring scalene and right triangles :roll
... and brings the newly revealed iPhi back into focus :!:

:geek: A very interesting pattern:
These are the angles of the two geometric objects:
62.403.., 27.597.., 90.0 right triangle
62.403.., 72.597.., 45.0 scalene triangle

The only difference in these sets of angles is the 45 degree
"borrowing" by 72.597.. from 90. Such numeric assocation
may be a clue to "squaring the circle" :roll :roll

Rod :D

Re: Paradise Trinity Day

Posted: Sun Sep 24, 2017 4:44 am
by Amigoo
:flower: Re: Power of 2 II design
( http://aitnaru.org/images/The_Right_Triangle.pdf )

Now more esoteric to show that the circle-squaring scalene and right triangles
in the large circle are sqrt(2) larger than these triangles in the small circle
(where its diameter has sqrt(2) relationship to the large circle). 8)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Tue Sep 26, 2017 10:44 am
by Amigoo
:flower: Re: iCorrelation design
(impossible Correlation)

This sqrt(2)-dominated Cartesian neighborhood, a convincing "Pi Corral",
highlights influence of two conjoined trapezoids, one set per CSC circle. 8)

:geek: Note tight integration within the trapezoidal sets of the overlapping,
circle-squaring right and scalene triangles (in each trapezoidal set).

Cartesian quadrature extraordinaire :!:

Rod :D

Re: Paradise Trinity Day

Posted: Wed Sep 27, 2017 5:53 am
by Amigoo
:flower: Re: Concentric Too design
(aka "Concentric 2")

With the inclusion of one more circle (creates CSCSC), this "Pi Corral"
shouts that the dimensions of the largest circle's objects are double
the dimensions of the smallest circle's (CSCSC Geometry 101).
example: (8/sqrt(2)) / sqrt(2) = 4. :duh

From this perspective, transcendental Pi is twice as nice
... and evenly divisible by two. :shock:

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Fri Sep 29, 2017 5:55 am
by Amigoo
:flower: Re: CSCSCalenity design
(pronounced "CSC Scalenity")

Sublime geometric balance in the Pi Corral. 8)
:cheers: "Lines and triangles and squares! Oh, Pi !"

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sat Sep 30, 2017 5:22 am
by Amigoo
:flower: Re: CSCSCalenity design (new insight)

"Geometric quiescence in the Pi Corral (the circle is squared)" 8)

Rod :D

Re: Paradise Trinity Day

Posted: Sun Oct 01, 2017 2:22 am
by Amigoo
:flower: Re: Pi Double Quadrature design
(aka "PDQ Yes!")

Consistent line length ratios in two concentric circles, both squared 8)
... and with convincing presence of their geometric progeny!

Rod ... :bike: ... (cruisin' Cartesian corridors PDQ) :roll:

Re: Paradise Trinity Day

Posted: Sun Oct 01, 2017 2:33 pm
by Amigoo
:flower: Re: Pi Double Quadrature design
(aka "PDQ, Yes!") ... and created PDQ :roll:

:idea: This geometry is all about ratio 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) ) 8)

Rod :D

Re: Paradise Trinity Day

Posted: Tue Oct 03, 2017 1:07 pm
by Amigoo
:flower: Re: Foursquare Plus Four design

Four concentric* circles all squared, each with a sibling (circled radius)
... and tight integration of this entire Cartesian family! 8)

* from the center, diameters increase by 2(sqrt(1/Pi))
= 2/sqrt(Pi) = 1.1283791670955125738961589031215..

"foursquare - marked by boldness and conviction" ;)

Rod :D

Re: Paradise Trinity Day

Posted: Thu Oct 05, 2017 3:14 pm
by Amigoo
:flower: Re: Foursquare Plus Four design

"Final, final, final" ... esoterically mesmerizing ...
as would any geometry proven impossible. :roll:

Being esoterically mesmerized, now elevate your mesmery ...
and consider ratio of diameters of the two golden circles. ;)

Rod :D

Re: Paradise Trinity Day

Posted: Fri Oct 06, 2017 1:11 am
by Amigoo
:flower: Re: Foursquare Plus Four design

Blah! blah! blah! This geometry only chatters about 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 1.1283791670955125738961589031215.. )
IMO, if you've seen one squared circle, you've seen foursquare! :roll:

Rod ... :bike: ... (runnin' in foursquare circles)

Re: Paradise Trinity Day

Posted: Fri Oct 06, 2017 11:22 am
by Amigoo
:flower: Re: Foursquare Plus Four design

"Four concentric circles all squared,
each with a sibling (circled radius);
quadrarture proven by 2(sqrt(1/Pi))"
This geometry only chatters about 2(sqrt(1/Pi))
Proving quadrature geometrically is noteworthy chatter! ;)

Rod :D

Re: Paradise Trinity Day

Posted: Sat Oct 07, 2017 1:55 am
by Amigoo
:flower: Re: Foursquare Plus Four design

Geometry now so esoteric, that a disclaimer is required:
This was not found on an alien spacecraft! :roll:

Rod ... :bike: ... (once cruisin' in foursquare circles
... at warp speed ) ;)

Re: Paradise Trinity Day

Posted: Mon Oct 09, 2017 2:06 am
by Amigoo
:flower: Re: Foursquare Plus Four design
"quadrarture proven by 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) )"

:colors: Recolored to emphasize simplicity of a circle squared:

Geometers need only know diameters of two golden circles:
small golden circle - 2(sqrt(1/Pi)); large circle - 2.0
The rest of the composition is Geometry 101 8)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Tue Oct 10, 2017 10:37 am
by Amigoo
If not already declared On High, 10-10-10 presents favorable symbolism for a Paradise Trinity Day.
A quiet seven-year anniversary in the local Cartesian neighborhood
(which explains the 3:33 AM wake-up call to review the past) ;)

Rod :D

Re: Paradise Trinity Day

Posted: Tue Oct 10, 2017 10:44 pm
by Amigoo
:flower: Re: QT-1010 design (Quadrature Trinity)

Simplification of Foursquare Plus Four design, highlighting
squared-circles trinity of 2(sqrt(1/Pi)), 2.0, and sqrt(Pi)
:geek: (1.1283791670955125738961589031215.., 2.0,
and 1.7724538509055160272981674833411..).

... on the PTD 10/10/17 seven-year anniversary. 8)

Note: 2(sqrt(1/Pi)) = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2)

Rod :D

Re: Paradise Trinity Day

Posted: Wed Oct 11, 2017 5:33 am
by Amigoo
:flower: Re: QT-1010 design (Quadrature Trinity)

Long story short: 2(sqrt(1/Pi)) is the increment/decrement
from the center of the squared circles' concentricity (2.0):

x 2(sqrt(1/Pi)) =
x 2(sqrt(1/Pi)) = 2(2(sqrt(1/Pi))^2)
x 2(sqrt(1/Pi)) = 4(sqrt(1/Pi))
2.0
/ 2(sqrt(1/Pi)) = sqrt(Pi)
/ 2(sqrt(1/Pi)) = Pi/2
/ 2(sqrt(1/Pi)) =

Rod :stars:

Re: Paradise Trinity Day

Posted: Wed Oct 11, 2017 6:33 pm
by Amigoo
:flower: Re: QT-1010 design (Quadrature Trinity)
Long story short: 2(sqrt(1/Pi)) is the increment/decrement
from the center of the squared circles' concentricity
:geek: Added another set of 3 concentric circles,
(numeric continuum of both sets here below):

Set 1
2.8733939540026649870253117380431..
- - - - - - - - - - - - - - - - - - - - - - - - -
2.5464790894703253723021402139602.. 2(2(sqrt(1/Pi)))^2)
2.2567583341910251477923178062431.. 4(sqrt(1/Pi))
2.0
1.7724538509055160272981674833411.. sqrt(Pi)
1.5707963267948966192313216916398.. Pi/2
- - - - - - - - - - - - - - - - - - - - - - - - -
1.3920819992079269613212044955297..

Set 2
1.4366969770013324935126558690215..
- - - - - - - - - - - - - - - - - - - - - - - - -
1.2732395447351626861510701069801.. (2(sqrt(1/Pi)))^2
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.0
0.88622692545275801364908374167057.. sqrt(Pi)/2
0.78539816339744830961566084581988.. Pi/4
- - - - - - - - - - - - - - - - - - - - - - - - -
0.69604099960396348066060224776486..

Rod ... :bike: ... (all set for more casual cruisin') :roll

Re: Paradise Trinity Day

Posted: Thu Oct 12, 2017 2:39 pm
by Amigoo
:flower: Re: QT-1010 design (Quadrature Trinity)

In the local universe ...
squared circles do not exist without pattern :!:

:scratch: How to know the circle is squared?
Look for the Smile of Pythagoras! :roll:
(turn design clockwise 90 degrees)

Rod :D

Re: Paradise Trinity Day

Posted: Sat Oct 14, 2017 12:34 am
by Amigoo
:flower: Re: Tenetcy of Quadrature design
... highlights the other squares (inscribed), 8)
objects in the defining set of principles.

"Lines and triangles and squares! Oh, Pi!"

Rod :stars:

Re: Paradise Trinity Day

Posted: Sun Oct 15, 2017 2:31 am
by Amigoo
:flower: Re: Tenetcy of Quadrature design (aka "TQ Waldo?")
Lines and triangles and squares! Oh, Pi!

The circle-squaring scalene triangle (red) can be found in every circle,
albeit geometry expertise is required for the second largest circle. ;)
:scratch: So, where's TQ Waldo?

Speaking of "TQ Waldo?" ... Who knew :?: :!:
There are 8 similar scalene triangles in this composition! 8)

Note: "Tenetcy" is a real word
... in socially squared circles. :roll:

Rod ... :bike: ...