Page 85 of 185
Re: Paradise Trinity Day
Posted: Fri Apr 15, 2016 12:34 pm
by Amigoo

Re: Wings of Life design
"Archetype of life everlasting, beyond Paradise ascent."

First appearance of the ankh, "outside the box" ...
and intriguing symbolism of ankh plus angel wings.
Rod

Re: Paradise Trinity Day
Posted: Sat Apr 16, 2016 1:07 am
by Amigoo

Re: Wings of Life design
"Archetype of life everlasting, beyond Paradise ascent."
Online:
http://aitnaru.org/homepage/commitment.html

Interestingly, UB quotes 100:2.7-8 attached to this page
seem to apply to a period just ahead (the image and subtitle
of the original page were replaced by Wings of Life).

salient numbers ...
Largest circle's diameter = 2(sqrt(Pi))
= 3.5449077018110320545963349666823..
Side of large green square = Pi/2
= 1.5707963267948966192313216916398..
Rod ...

... (off to salient sabbatical)

Re: Paradise Trinity Day
Posted: Sat Apr 16, 2016 9:11 pm
by Amigoo

Re: Salient Nots design
"CSCSC 2:1 with 3 concentric circles"

Geometric proof that the squared circle objects* of the outer circle
have twice the dimensions of the squared circle objects of the inner circle.
Do = 2(sqrt(Pi)), Di = sqrt(Pi)
* diameter, scalene triangles, trapezoid, and their components.

Irrational analysis while packing for oft-postponed salient sabbatical:
"irrational" and "transcendental" are eclipsed by squared circle geometry.
In other words, they become salient nots.
Rod

Re: Paradise Trinity Day
Posted: Sun Apr 17, 2016 11:44 am
by Amigoo

Re: Scalene Salience CSC design

Apparently, the only way to stop this exploration will be to lock
the creativity toy box and send the key to a sabbatical destination
... where the key will be "lost in the mailroom" for weeks.
Whereas Salient Nots highlights 2:1 dimensions of CSCSC objects,
Scalene Salience highlights sqrt(2):1 dimensions of CSC objects.

Say what? Squared circle geometry (SCG) that speaks for itself
... when explored by serious SCG aficionados.
Rod ...

... (off to polish the SS CSC twin rudders*)
* permits just "a thumb and forefinger on the wheel" ...
unlike an auto steering wheel that permits a limp wrist
hung over the top (no, not that limp wrist).

Re: Paradise Trinity Day
Posted: Mon Apr 18, 2016 1:23 pm
by Amigoo

Re: Triangelation design

A technical doodle more than design, discovered in the archive,
and displays supporting geometry from the neighborhood; geometry
that includes a few parallelograms of special interest.
Rod

Re: Paradise Trinity Day
Posted: Tue Apr 19, 2016 5:07 am
by Amigoo

Re: Triangelation design
“Scalene perspective on 'trilateral impossibility'.”

Speaking of “impossibility” ...
the “parallelograms of special interest”, once identified,
easily fade back into this Cartesian camouflage,
making their discovery a challenge once again.

Geometer's riddle clue:
"Chang and Eng are sobriquet nots of two parallelograms
thrice crimson and similar in this Cartesian composition.
Rod

Re: Paradise Trinity Day
Posted: Tue Apr 19, 2016 9:11 am
by Amigoo

Re: Triangelation design
“Scalene perspective on 'trilateral impossibility'.”

These numbers have remained consistent for years:
1.7724538509055160272981674833411.. sqrt(Pi)
/ 1.5707963267948966192313216916398.. Pi/2
= 1.1283791670955125738961589031215.. 2(sqrt(1/P1))

Given: Diameter = 2.0
/ 1.1283791670955125738961589031215.. 2(sqrt(1/P1))
= 1.7724538509055160272981674833411.. sqrt(Pi), side of circle's square
Rod ...

...
Re: Paradise Trinity Day
Posted: Tue Apr 19, 2016 4:33 pm
by Amigoo

Re: Triangelation design
“Scalene perspective on 'trilateral impossibility'.”

Regarding 2(sqrt(1/Pi)) ...
1.7724538509055160272981674833411.. sqrt(Pi)
/ 1.5707963267948966192313216916398.. Pi/2
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))

Pi/2 and sqrt(Pi) define a circle-squaring right triangle,
with Pi/2 = long side and sqrt(Pi) = hypotenuse (radius);
a triangle that effectively squares all circles!

If radius = sqrt(Pi) ...
Side of square = Diameter / 2(sqrt(1/Pi))
= 3.5449077018110320545963349666823.. 2(sqrt(Pi))
/ 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 3.1415926535897932384626433832795.. Pi
Area = Pi^2 = 9.8696044010893586188344909998762..
Rod

Re: Paradise Trinity Day
Posted: Wed Apr 20, 2016 11:11 am
by Amigoo

Re: ABC 101 design
"A^2 + B^2 = C^2"

I must be traveling in a Driverless Creativity Coach! Despite my intent to embark upon sabbatical, the DC coach keeps mentoring "No pain - no gain!". And so, once a-gain, we were touring the Cartesian byways late at night (well, before morning's first light) when the Pythagorean Theorem came to mind: "Hmmm ... What geometry could be derived from a right triangle (RT) having hypotenuse = sqrt(Pi) and long side = Pi/2"

Soon, a few point-to-point lines identified a familiar trapezoid, and next a familiar scalene triangle (of circle-squaring persuasion); the rest was ABC 101 ("Even a child can do it!"). Interestingly, this right triangle also identifies the center of the circle effectively squared by this RT. But the long way is more scenic and obviously a Pythagorean perspective on how "impossible" squared circle bovines graze on premature proof.
Rod

Re: Paradise Trinity Day
Posted: Thu Apr 21, 2016 3:46 am
by Amigoo

Re: ABC 101 design
"A^2 + B^2 = C^2"

Who knew?! Squared circle geometry speaks CAPTCHA!
(two sets of lines create a U.S. state code, the state
wherein squared circle geometry speaks Captcha)
Geometer's toe tapper clue: Sing the song, beginning ...
"The stars at night are big and bright ..."

Regarding ratios of lines creating state abbreviation:
Values relate to dimensions of the geometric objects;
second letter of state abbreviation created by set
of lines within the highlighted right triangle.

First letter of abbreviation:
1.7724538509055160272981674833411.. sqrt(Pi)
/ 0.88622692545275801364908374167057.. sqrt(Pi)/2
= 2.0

Second letter (within and including Pythagorean triangle):
1.772453850905516027298167483342.. sqrt(Pi)
/ 1.5707963267948966192313216916398.. Pi/2
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1 / 0.88622692545275801364908374167057..
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))

Association of ratios in the right triangle:
2.0 / 1.1283791670955125738961589031215..
= 1.772453850905516027298167483342.. sqrt(Pi)
1.0 / 1.1283791670955125738961589031215..
= 0.88622692545275801364908374167057.. sqrt(Pi)/2
Rod ...

... (off to see the

) ...
in books at the Half-Price store, DithoT (from song)
Re: Paradise Trinity Day
Posted: Thu Apr 21, 2016 7:07 pm
by Amigoo

Re: ABC 101 design
"A^2 + B^2 = C^2"
Re: sqrt(1/Pi) = 0.56418958354775628694807945156077..
2(sqrt(1/Pi)) = 1.1283791670955125738961589031215..
While contemplating if 2(sqrt(1/Pi)) is the "new & improved" Pi constant*,
I realized that this value helps calculate the square of a circle as well as
define ratios of line lengths (in triangles) associated with squared circles.
So, yes, 2(sqrt(1/Pi)) seems a runner-up for New Pi.
The two smaller light blue circles in ABC 101 appear to contrast related
values in these squared circles with one circle-squaring right triangle
being a subset of the other.
Pi/2 and sqrt(Pi) define a circle-squaring right triangle; D = sqrt(Pi):
1.7724538509055160272981674833411.. sqrt(pi)
/ 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1.5707963267948966192313216916405.. Pi/2
* According to the internet, sqrt(1/Pi) has long been touted as the
"squared circle constant", but 2(sqrt(1/Pi)) seems more versatile,
especially when considering the line ratios.
Rod

Re: Paradise Trinity Day
Posted: Fri Apr 22, 2016 2:22 am
by Amigoo

Re: Squared Circle Geometry
How to evaluate any New Pi ...
Q: What's the last decimal digit of New Pi? “The one that is right.”
Q: What's the first decimal digit of New Pi? “The one that is right.”
Q: What digit of New Pi is left? “The whole one.”
Q: So, what's the point
“The one that is right of the whole and left of the one that is right.”
Rod ...

... (off to find the point)
Re: Paradise Trinity Day
Posted: Fri Apr 22, 2016 9:39 am
by Amigoo

Re: Squared Circle Geometry
How to evaluate any New Pi ...

While driving around, trying to "find the point", I wondered:
Q: "Just how far is it around this circle's perimeter?"
A: "The same as it has been for centuries!"
Given: Diameter (D) = 2.0 and New Pi (NP)
= 1.1283791670955125738961589031215..

For reference ...
Circumference = Pi x Diameter
= 3.1415926535897932384626433832795.. x 2.0
= 6.283185307179586476925286766559.. 2(Pi)

For New Pi contrast ...
Circumference = 2((D/NP)^2)
= 2((2.0 / 1.1283791670955125738961589031215..)^2)
= 2((1.7724538509055160272981674833411..)^2)
= 2(3.1415926535897932384626433832795..)
= 6.283185307179586476925286766559.. 2(Pi)

Hmmm ... C = 2((D/NP)^2)?! I'll just have the old Pi.
(my abacus can only do easy math: C = Pi x D)
Rod

Re: Paradise Trinity Day
Posted: Sat Apr 23, 2016 9:47 am
by Amigoo

Re: ABC 101 (DithoT) design
"2(sqrt(1/Pi)) @ A^2 + B^2 = C^2"

“To square the circle, one must circle the square.”
Who knew?! Two juxtaposed

right triangles were required to complete the story.
The center diagonal of the red triangles is both hypotenuse (of smaller triangle)
and long side (of larger triangle). "Impossible" squared circle balance.
Rod

Re: Paradise Trinity Day
Posted: Sun Apr 24, 2016 7:07 pm
by Amigoo

Re: ABC 101 (DithoT) design*
"2(sqrt(1/Pi)) @ A^2 + B^2 = C^2"
This geometry prefers a 2(sqrt(1/Pi)) focus (juxtaposed

Pythagorean red triangles).
Many of the two-line sets in these adjacent triangles have the 2(sqrt(1/Pi)) ratio.

2(sqrt(1/Pi)) = 1.1283791670955125738961589031215..
* DithoT = "Deep in the heart of Texas" (from the song)
Rod

Re: Paradise Trinity Day
Posted: Mon Apr 25, 2016 2:22 pm
by Amigoo

Re: ABC 101 (DithoT) design*
"2(sqrt(1/Pi)) @ A^2 + B^2 = C^2"
Soon, more exploration (and desire for geometric balance)
resulted in the appearance of a "Smile of Pythagoras".
(rotate design 90 degrees clockwise to see the smile;
actually a big grin!)
A dozen of these designs in recent years qualify for the
entertaining "Smile of Pythagoras" collection.
Rod ...

...
Re: Paradise Trinity Day
Posted: Mon Apr 25, 2016 11:53 pm
by Amigoo

Re: ABC 101 (DithoT) design
"2(sqrt(1/Pi)) @ A^2 + B^2 = C^2"
Re: "Smile of Pythagoras"
Well, no wonder the Smile! The diameter of the larger of the two "eyes"
has length = 2(sqrt(1/Pi)) = 1.1283791670955125738961589031215..
Rod
Re: Paradise Trinity Day
Posted: Tue Apr 26, 2016 11:03 am
by Amigoo

Re: Juxtaposition Rights (JR) design
Amusing continuance of the Texas theme (re: DithoT) ...
with the fortuitous "JR" abbreviation! - HCIT!
"Rights" alludes to the juxtaposed

red right triangles
in this study of these angles and lines in the composition.
"Sabbatical" is still planned - the recent geometry is hinting
that running in circles not so square is the next program.
Rod, self-labeled "Squared Circle Detectorist"

Re: Paradise Trinity Day
Posted: Thu Apr 28, 2016 2:17 am
by Amigoo

Re: Juxtaposition Rights (JR) design
I've been trying to persuade this geometry to display more symmetrically,
but its apparent purpose is to contrast two right triangles (smaller one
in top circle with 2 yellow lines, one red; larger one in bottom circle
with 2 red lines, one yellow) ... and with focus on the hypotenuse (h)
and long side (ls) of these triangles as proportions: ls : h ~ ls : h

Thus, Pi/4 : sqrt(Pi)/2 ~ sqrt(Pi)/2 : 1
(or Pi/2 : sqrt(Pi) ~ sqrt(Pi) : 2)
0.78539816339744830961566084581988.. Pi/4
: 0.88622692545275801364908374167057.. sqrt(Pi)/2
~ 0.88622692545275801364908374167057.. sqrt(Pi)/2
: 1.0
0.88622692545275801364908374167057.. sqrt(Pi)/2
/ 0.78539816339744830961566084581988.. Pi/4
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.0 / 0.88622692545275801364908374167057..
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
Say what? "The stars at night are big and bright ... DithoT"
Rod

Re: Paradise Trinity Day
Posted: Fri Apr 29, 2016 1:17 am
by Amigoo

Re: iTriangle design
(image replaced by My 'T' Prompt)
Another design focusing on the Texas 'T' midwayer prompt
(first mentioned on page 3 of this topic, Feb 25, 2011 1:55 pm)
was not going to tolerate simplicity - WYSIWYG.
But "right triangle symmetry in five circles, all squared"
is visually pleasing - even geometrically loquacious.
Rod

Re: Paradise Trinity Day
Posted: Fri Apr 29, 2016 6:57 am
by Amigoo

Re: iTriangle design (updated)
even geometrically loquacious.

One cannot embrace the "impossible" squared circles ...
without enlightenment from the chalice of belief and faith.
Rod ...

...
Re: Paradise Trinity Day
Posted: Mon May 02, 2016 7:46 am
by Amigoo

Re: Summa Cum Laude design

Given: hypotenuse and long side of similar right triangles:
1.2732395447351626861510701069801.. (2(sqrt(1/Pi)))^2
/ 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
/ 1.0
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.0
/ 0.88622692545275801364908374167057.. sqrt(Pi)/2
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
0.88622692545275801364908374167057.. sqrt(Pi)/2
/ 0.78539816339744830961566084581988.. Pi/4
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))

Analysis: 2(sqrt(1/Pi)) and its square define both
a circle and its circle-squaring right triangle.

Say what? 2(sqrt(1/Pi)) is Summa Cum Laude
in impossible graduation circles all squared.
Rod

Re: Paradise Trinity Day
Posted: Mon May 02, 2016 2:57 pm
by Amigoo

Re: Summa Cum Laude design (completed)
(aka "squared circle geometry that speaks for itself")

Analysis: 2(sqrt(1/Pi)) and its square define both
a circle and its circle-squaring right triangle.

Say what? 2(sqrt(1/Pi)) is Summa Cum Laude
in impossible graduation circles, all squared.
Rod

Re: Paradise Trinity Day
Posted: Tue May 03, 2016 5:55 am
by Amigoo

Re: Summa Cum Laude design
(aka "squared circle geometry that speaks for itself")
This design online:
http://aitnaru.org/homepage/harmonyofchords.html

Rod ...

... (off to find straight lines ...
after cruising in 15 circles, all squared (13 effectively)
Re: Paradise Trinity Day
Posted: Wed May 04, 2016 12:03 am
by Amigoo

Re: Summa Cum Laude design
(aka "squared circle geometry that speaks for itself")
off to find straight lines

Impressive lines were soon found!
The two sides of the light blue "butterfly"
have the 2(sqrt(1/Pi)) ratio; the other two
lines (one is red) have equal length.
Rod
