Page 110 of 182
Re: Paradise Trinity Day
Posted: Thu Aug 03, 2017 12:48 am
by Amigoo
Re: Golden rPi design
Re: The Golden Ratio by Mario Livio, 2002, p. 85
"The Golden Rectangle is the only rectangle with the property
that cutting a square from it produces a similar rectangle."
Impressive, however ...
The rPi Golden Rectangle is the only rectangle with the property
that its diagonal has length equal to the diameter of a circle and
sides a+b have length equal to a side of that circle's square.
Given that the diagonal is side d, d / (a + b) = 2(sqrt(1/Pi)),
consistent ratio of these sides in all related right triangles.
Rod
Re: Paradise Trinity Day
Posted: Thu Aug 03, 2017 1:23 pm
by Amigoo
Re: Golden rPi design
I finally comprehend
that this geometry is revealing a new "golden ratio(s)"
with another Cartesian dimension beyond the line-segmented Golden Ratio:
relationships of the lines* in a circle-squaring right triangle ... lines that
define a new spiral for consumption in believing math circles.
* one line is segmented (to allude to the old Golden Ratio?)
Rod ...
...
Re: Paradise Trinity Day
Posted: Sat Aug 05, 2017 1:55 am
by Amigoo
Re: Golden rPi design
Who knew
That a new Phi ratio would exist for the rPi rectangle!
(repeating ratio observable in similar objects/lines of Golden rPi)
Current Phi = 1.618033988749894848204586834 ..
New iPhi = 1.913067540823067208163119533.. (approx)
'i' alludes to "impossible" squaring of the circle.
Rod
Re: Paradise Trinity Day
Posted: Sat Aug 05, 2017 3:55 am
by Amigoo
Re: Golden rPi design
New iPhi = 1.913058380271100794740307828.. (approx)
Revised calculation starting with hypotenuse of right triangle
= 2,000,000,000 units and downsizing via 2(sqrt(1/Pi))
and the Pythagorean Theorem to obtain line lengths
of the golden cross in the largest rectangle.
Say what
Just ask a squared circles geometer.
Rod
Re: Paradise Trinity Day
Posted: Sat Aug 05, 2017 4:44 am
by Amigoo
Re: Golden rPi design (the wiggly numbers)
New iPhi = 1.913058380271100794740307828.. (approx)
downsize right triangles via 2(sqrt(1/Pi)):
Find length of shortest side of largest right triangle
2,000,000,000.0 = largest hypotenuse (length of longer golden line)
/ 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1772453850.9055160272981674833411.. sqrt(Pi) x 1,000,000,000
Apply Pythagorean Theorem
2,000,000,000.0^2 - 1772453850.9055160272981674833411..^2
= 858407346410206761.53735661672066..
sqrt( ) = 926502750.35220848584275966758923.. = shortest side of large right triangle
Find length of shorter golden line (another hypotenuse)
926502750.35220848584275966758923.. = shortest side of large right triangle
x 2(sqrt(1/Pi)) = 1045446401.7541266302735942239055.. length of smaller golden line
Calculate new iPhi ratio (approx)
2,000,000,000.0 / 1045446401.7541266302735942239055..
= 1.9130583802711007947403078280203.. iPhi (approx)
Rod
Re: Paradise Trinity Day
Posted: Sat Aug 05, 2017 9:11 pm
by Amigoo
Re: Golden rPi design
HCIT! After calculating the iPhi value, use it to confirm
length of shortest side from the Pythagorean Theorem:
Apply Pythagorean Theorem
2,000,000,000.0^2 - 1772453850.9055160272981674833411..^2
= 858407346410206761.53735661672066..
sqrt( ) = 926502750.35220848584275966758923.. = shortest side of large right triangle
Confirm PT calculation using iPhi
1772453850.9055160272981674833411.. length of long side
/ 1.9130583802711007947403078280203.. iPhi constant
= 926502750.35220848584275966758919.. = shortest side of large right triangle
The difference between ..8923.. and ..8919.. ?
4 (early rounding of squared circle corners)
Rod ...
... (bumping curb on quick right turns)
Re: Paradise Trinity Day
Posted: Tue Aug 08, 2017 8:37 am
by Amigoo
Re: Golden rPi design
Signature iPhi added to design (lower left, green lines) showing
two circle-squaring right triangles adjoined at their 90 degree angle
with all line pair lengths having iPhi ratio, including hypotenuse
(blue lines) of the two triangles.
Rod
Re: Paradise Trinity Day
Posted: Tue Aug 08, 2017 5:55 pm
by Amigoo
Re: Golden rPi design ("final, final")
Golden rPi “Alpha to Omega” (Aom)
"Get a clue and devour the concept"
New millennium “golden rectangle” featuring
the circle-squaring ratio 2(sqrt(1/Pi)) = d/(a+b)
and supporting ratio iPhi = c/a = (a+b)/c
= 1.91305838027110079474030782802..
About the "critter" in lower left of design ...
"The bulldog who ate Pi" (almost daily for years)
Current collection:
http://aitnaru.org/images/Squarely_Entwined.pdf
Rod ...
... (off to get a looking glass)
Re: Paradise Trinity Day
Posted: Tue Aug 08, 2017 10:37 pm
by Amigoo
Re: Golden rPi design ("final, final")
"The bulldog who ate Pi"
Trimmed its tail a bit to not look like a cat's.
About the blue trapezoid in the lower right of design ...
Probably, a doghouse from which to escape occasional
brimstone attacks on the concept of squared circles.
Rod
Re: Paradise Trinity Day
Posted: Wed Aug 09, 2017 3:22 pm
by Amigoo
Re: Golden rPi design
A quick study of the iPhi concept:
http://aitnaru.org/images/Golden_rPi.pdf
Rod
Re: Paradise Trinity Day
Posted: Thu Aug 10, 2017 3:53 am
by Amigoo
Re: Golden rPi design
(quick study of the iPhi concept)
Another page added to the PDF showing the mathematical correlation
of the circle-squaring triangle's line lengths (hypotenuse-long_side vs
hypotenuse-short_side). Who knew
... that 2(sqrt(1/Pi))
would give the most convincing evidence of this correlation!
Rod
Re: Paradise Trinity Day
Posted: Thu Aug 10, 2017 1:33 pm
by Amigoo
Re: Golden rPi design
(quick study of the iPhi concept)
Golden line 'e' is now identified. Despite all this Cartesian busyness,
the circle is not squared until a geometric proof is created.
The chess pieces are positioned - let the game begin!
Rod
Re: Paradise Trinity Day
Posted: Fri Aug 11, 2017 10:22 am
by Amigoo
Re: Golden rPi design
(quick study of the iPhi concept)
Line 'f' is now identified in Golden rPi Simplified design
(to illuminate the golden diagonals object d/e):
a = f = (a+b)/(iPhi^2). Say what
iPhi^2
iPhi = c/a = (a+b)/c = d/e
= 1.9130583802711007947403078280203..
^2 = 3.6597923663254876944787072692565..
Rod
Re: Paradise Trinity Day
Posted: Fri Aug 11, 2017 9:27 pm
by Amigoo
Re: GPSR Calculator design
(Golden Pi Square Root Calculator from Golden Pi rectangle geometry)
"Get a clue and devour the concept"
The new clue? The d/e object giving (a+b)/(iPhi^2)
iPhi supports a geometric square root calculator!
(that "calculates" both square and square root)
While GPSR geometry shows that such a calculator could exist,
its operation is a mystery (easier than creating a geometric proof
of a squared circle?) Let this next game begin!
Rod ...
...
Re: Paradise Trinity Day
Posted: Sat Aug 12, 2017 1:17 am
by Amigoo
Re: GPSR Calculator design
its operation is a mystery
Research indicates that its existence is also questionable;
the pattern locks on iPhi^2 and not just any square/root.
"Say what
" A flash in the pan.
Rod
Re: Paradise Trinity Day
Posted: Sat Aug 12, 2017 6:46 am
by Amigoo
Re: GPSR Calculator design
"Purpose and operation unknown!
"Begs the question: 'What's the point?'"
Indeed
Rod ...
...
Re: Paradise Trinity Day
Posted: Sat Aug 12, 2017 2:46 pm
by Amigoo
Re: GPSR Calculator design II
Purpose and operation unknown!
Begs the question: "What's the point?"
... but answers: "What's the point?"
Rod
OIC
Re: Paradise Trinity Day
Posted: Sun Aug 13, 2017 6:06 am
by Amigoo
Re: Golden rPi design
the circle is not squared until a geometric proof is created
Actually, this is a two-part challenge:
1. Prove that squared circles exist (geometrically)*
2. Square the circle by the Greek rules.
Methinks that Part 1 is easier than Part 2 ...
and should have the greater priority.
* Not to worry that, mathematically, this is "impossible"
(making Part 2 a futile journey ... also).
Rod
Re: Paradise Trinity Day
Posted: Sun Aug 13, 2017 6:44 am
by Amigoo
Re: Golden rPi design
Actually, this is a two-part challenge:
1. Prove that squared circles exist (geometrically)
2. Square the circle by the Greek rules.
Constant 2(sqrt(1/Pi))* suggests that sqrt(Pi) can be
represented as the long side of a circle-squaring
right triangle whose hypotenuse = 2.0 (diameter).
* 2(sqrt(1/Pi)) = 2(sqrt(Pi))/Pi = sqrt(Pi)/(Pi/2)
= 1.1283791670955125738961589031215..
Rod ...
...
Re: Paradise Trinity Day
Posted: Sun Aug 13, 2017 3:37 pm
by Amigoo
Re: The Right Triangle design
Long Story Short
(aka, "Period. End of Story"):
If this right triangle does not exist in a squared circle,
the circle is not squared
according to this geometry.
Rod
Re: Paradise Trinity Day
Posted: Mon Aug 14, 2017 2:22 pm
by Amigoo
Re: The Right Triangle design
Embellished design with geometric LOIN (Lines Of Interest Noteworthy)
since the integrated scalene triangle also squares the circle.
Rod
Re: Paradise Trinity Day
Posted: Tue Aug 15, 2017 5:43 am
by Amigoo
Re: Three Pi Vise design
"See how they run! ... in the 'impossible' House of Pi.
Diameters = Pi/2, sqrt(Pi), 2.0; Paramount ratio: 2(sqrt(1/Pi))"
Also online:
http://aitnaru.org/threepoints.html
"Answers 'What’s the point?' (about squaring circles). Right triangle (3 points)
contains defining hypotenuse-to-long-side ratio 1.128379167095512573896..
= 2(sqrt(1/Pi)) = sqrt(Pi)/(Pi/2)"
Golden diagonals of flutterby also have 2(sqrt(1/Pi)) relationship.
Rod ...
...
Re: Paradise Trinity Day
Posted: Tue Aug 15, 2017 2:17 pm
by Amigoo
Re: Three Pi Vise design
"The crown awaits the king (of a geometric proof?)"
Geometers' tip:
These line length ratios exist in the "crown":
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
^2 = 1.2732395447351626861510701069801..
^3 = 1.4366969769991930158060821722332..
Rod
Re: Paradise Trinity Day
Posted: Tue Aug 15, 2017 5:44 pm
by Amigoo
Re: Three Pi Vise design
"The crown awaits the king (of a geometric proof?)"
Geometers' tip #2:
With Diameters = Pi/2, sqrt(Pi), 2.0
and the recently revealed constants:
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.9130583802711007947403078280203.. "iPhi"
and the Pythagorean Theorem: a^2 + b^2 = c^2
Length of flutterby diagonals can be calculated
with their ratio then proven mathematically
= 1.1283791670955125738961589031215..
Easy enough once the right triangles are selected
for this short mathematical journey!
Rod
Re: Paradise Trinity Day
Posted: Wed Aug 16, 2017 12:55 am
by Amigoo
Re: Three Pi Vise design
"The crown awaits the king (of a geometric proof?)"
Who knew
The longer diagonal is the diameter of a circle
and the shorter diagonal has length equal
to a side of that circle's square!
That's why their length ratio is 2(sqrt(1/Pi))
"Three Pi Vise ... See how they run!"
Did you ever see such a sight in your life!
Rod