Paradise Trinity Day

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Re: Paradise Trinity Day
Re: Geometry Square 'n Root design
Geometers' secret ....
The inscribed square of a circle where D = 2
is the "calculator" squaring line identifier.
A Rosetta Stone of numbers squared!
... by Cartesian geometry.
Rod
Geometers' secret ....
The inscribed square of a circle where D = 2
is the "calculator" squaring line identifier.
A Rosetta Stone of numbers squared!
... by Cartesian geometry.
Rod

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Re: Paradise Trinity Day
Re: Geometry Square 'n Root design,
updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Geometers' secret about the previous geometers' secret ...
sqrt(2) and sqrt(Pi) create the green "finger" on the squaring line.
Rod ... ...
updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Geometers' secret about the previous geometers' secret ...
sqrt(2) and sqrt(Pi) create the green "finger" on the squaring line.
Rod ... ...

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Re: Paradise Trinity Day
Re: Geometry Square 'n Root design
Geometers' secret
about previous geometers' secret
about previous geometers' secret ...
Pythagorean Theorem proves line lengths of right triangles
related to their square and square root number sets.
Who knew?!
Rod
Geometers' secret
about previous geometers' secret
about previous geometers' secret ...
Pythagorean Theorem proves line lengths of right triangles
related to their square and square root number sets.
Who knew?!
Rod

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Re: Paradise Trinity Day
Re: Geometry Square 'n Root design
Pi and its square root MUST exist in this "calculator"
... but finding it is not so easy ... apparently.
Rod ... ...
Pi and its square root MUST exist in this "calculator"
... but finding it is not so easy ... apparently.
Rod ... ...

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Re: Paradise Trinity Day
Re: Geometry Square 'n Root design
Still fine tunin' the "banjo" ...
What's intriguing about this "calculator" that actually works
is that it could have been used as soon as sqrt(2) was invented
... and measurements made with knotted cords and ropes.
Rod
Still fine tunin' the "banjo" ...
What's intriguing about this "calculator" that actually works
is that it could have been used as soon as sqrt(2) was invented
... and measurements made with knotted cords and ropes.
Rod

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Re: Paradise Trinity Day
Re: Geometry Square 'n Root design
Better detail of the Pi Pickin' Quadra Lute
Geometers' secret ....
Pi Pickin' Finger (green) combines geometric objects relating to sqrt(2) and sqrt(Pi).
Its inner objects are the isosceles right triangle (defines circle squaring line) and
the adjoined right triangle, geometric signature of Quadrature.
Rod ... ...
Better detail of the Pi Pickin' Quadra Lute
Geometers' secret ....
Pi Pickin' Finger (green) combines geometric objects relating to sqrt(2) and sqrt(Pi).
Its inner objects are the isosceles right triangle (defines circle squaring line) and
the adjoined right triangle, geometric signature of Quadrature.
Rod ... ...

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Re: Paradise Trinity Day
Re: Quadra Lute design
(name changed from Geometry Square 'n Root)
Geometers' secret ...
This geometry captures a previously unknown principle of right triangles drawn in a circle,
where hypotenuse = diameter and all ends of short sides align on the circle squaring line
(as first identified by an isosceles right triangle) when lines represent a number squared.
Who knew
See also: https://www.lutesociety.org/pages/aboutthelute
"is it not strange that sheepes guts should hale soules out of mens bodies?"  Shakespeare
... and complementary that a Pythagorean Lute should hale the Square Of Numbers (SON).
Rod
(name changed from Geometry Square 'n Root)
Geometers' secret ...
This geometry captures a previously unknown principle of right triangles drawn in a circle,
where hypotenuse = diameter and all ends of short sides align on the circle squaring line
(as first identified by an isosceles right triangle) when lines represent a number squared.
Who knew
See also: https://www.lutesociety.org/pages/aboutthelute
"is it not strange that sheepes guts should hale soules out of mens bodies?"  Shakespeare
... and complementary that a Pythagorean Lute should hale the Square Of Numbers (SON).
Rod

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Re: Paradise Trinity Day
Re: Quadra Lute design
Pythagoras would be pleased with the Quadra Lute (according to his Music of the Spheres):
Re: https://en.wikipedia.org/wiki/Musica_universalis
"The discovery of the precise relation between the pitch of the musical note and the length of the string that produces it is attributed to Pythagoras. The Music of the Spheres incorporates the metaphysical principle that mathematical relationships express qualities or 'tones' of energy which manifest in numbers, visual angles, shapes and sounds – all connected within a pattern of proportion.
Pythagoras first identified that the pitch of a musical note is in inverse proportion to the length of the string that produces it, and that intervals between harmonious sound frequencies form simple numerical ratios."
"He believed that this harmony, while inaudible, could be heard by the soul, and that it gave a 'very agreeable feeling of bliss, afforded him by this music in the imitation of God.'"
Speaking of ratios ...
2:sqrt(2), 4:2, 9:3, 16:4, 25:5 ratios create the lines in Quadra Lute
and are easy to draw on the X:Y graph using intersecting circles.
Rod ... ...
Pythagoras would be pleased with the Quadra Lute (according to his Music of the Spheres):
Re: https://en.wikipedia.org/wiki/Musica_universalis
"The discovery of the precise relation between the pitch of the musical note and the length of the string that produces it is attributed to Pythagoras. The Music of the Spheres incorporates the metaphysical principle that mathematical relationships express qualities or 'tones' of energy which manifest in numbers, visual angles, shapes and sounds – all connected within a pattern of proportion.
Pythagoras first identified that the pitch of a musical note is in inverse proportion to the length of the string that produces it, and that intervals between harmonious sound frequencies form simple numerical ratios."
"He believed that this harmony, while inaudible, could be heard by the soul, and that it gave a 'very agreeable feeling of bliss, afforded him by this music in the imitation of God.'"
Speaking of ratios ...
2:sqrt(2), 4:2, 9:3, 16:4, 25:5 ratios create the lines in Quadra Lute
and are easy to draw on the X:Y graph using intersecting circles.
Rod ... ...

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Re: Paradise Trinity Day
Re: Quadra Lute design
This Quadra Lute graph proves that every number has a perfect square,
including Pi, even if advanced math goes into transcendental orbits.
Rod
Geometers' secret ...Speaking of ratios ...
2:sqrt(2), 4:2, 9:3, 16:4, 25:5 ratios create the lines in Quadra Lute
This Quadra Lute graph proves that every number has a perfect square,
including Pi, even if advanced math goes into transcendental orbits.
Rod

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Re: Paradise Trinity Day
Re: Quadra Lute design
A teaser (partial hexagon) was added to an inner circle to note that
certain diameters host a perfectly inscribed polygon, suggesting that
other Pi models exist, integrated into the Quadra Lute geometry.
Rod ... ...
Geometers' secret ...Speaking of ratios ...
2:sqrt(2), 4:2, 9:3, 16:4, 25:5 ratios create the lines in Quadra Lute
A teaser (partial hexagon) was added to an inner circle to note that
certain diameters host a perfectly inscribed polygon, suggesting that
other Pi models exist, integrated into the Quadra Lute geometry.
Rod ... ...

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Re: Paradise Trinity Day
Re: Quadra Lute design
Who knew Pi is directly related to 2(sqrt(2))
"Say what?!" Two adjoined sides of an inscribed square,
hinting that this relationship makes a good Pi model ...
especially since Pi relates to .5 Circumference.
"How does this relate to 2/sqrt(Pi)?!"
(= 1.1283791670955125738961589031215..)
Time will tell.
Rod
... and then removed for design simplicity.partial hexagon was added to an inner circle
Who knew Pi is directly related to 2(sqrt(2))
"Say what?!" Two adjoined sides of an inscribed square,
hinting that this relationship makes a good Pi model ...
especially since Pi relates to .5 Circumference.
"How does this relate to 2/sqrt(Pi)?!"
(= 1.1283791670955125738961589031215..)
Time will tell.
Rod

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Re: Paradise Trinity Day
Re: Quadra Lute design*
(aka "Pi Fork 'n Lute")
* updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Geometers' secret ...
The "Pi Fork" was added to an inner circle because this circle where D = Pi
highlights the Fork's line length ratios of sqrt(Pi), Pi/2, and 2/sqrt(Pi).
Rod
(aka "Pi Fork 'n Lute")
* updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Geometers' secret ...
The "Pi Fork" was added to an inner circle because this circle where D = Pi
highlights the Fork's line length ratios of sqrt(Pi), Pi/2, and 2/sqrt(Pi).
Rod

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Re: Paradise Trinity Day
Re: Quadra Lute design
(aka "Pi Fork 'n Lute")
Sqrt(Pi) Ratios design moved to page after Quadra Lute
for immediate reference to "Pi Fork", THE Quadrature identifier.
Rod ... ...
(aka "Pi Fork 'n Lute")
Sqrt(Pi) Ratios design moved to page after Quadra Lute
for immediate reference to "Pi Fork", THE Quadrature identifier.
Rod ... ...

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Re: Paradise Trinity Day
Re: Sqrt(Pi) Ratios design
and "Phi of Pi" ratio, a Quadraturial spiral
(SoCS = Side of Circle's Square)
Geometers' secret ...
Phi of Pi ratio developed with Pythagorean Theorem
and given: D = 2, D^2 = 4, SoCS = sqrt(Pi)
Phi of Pi = sqrt(Pi) / sqrt(4  Pi)
= 1.7724538509055160272981674833411..
/ 0.92650275035220848584275966758914..
= 1.9130583802711007947403078280203..
Rod
and "Phi of Pi" ratio, a Quadraturial spiral
(SoCS = Side of Circle's Square)
Geometers' secret ...
Phi of Pi ratio developed with Pythagorean Theorem
and given: D = 2, D^2 = 4, SoCS = sqrt(Pi)
Phi of Pi = sqrt(Pi) / sqrt(4  Pi)
= 1.7724538509055160272981674833411..
/ 0.92650275035220848584275966758914..
= 1.9130583802711007947403078280203..
Rod

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Re: Paradise Trinity Day
Oops! Forgot the algebra ...
Golden Spiral of Quadrature
Given: Circlesquaring right triangle
where c = hypotenuse = 2 (diameter),
a = long side, b = short side.
1.7724538509055160272981674833411.. a, sqrt(Pi)
+ 0.92650275035220848584275966758914.. b
= 2.6989566012577245131409271509303.. a+b
1.7724538509055160272981674833411.. a, sqrt(Pi)
/ 0.92650275035220848584275966758914.. b
= 1.9130583802711007947403078280203.. a/b, Phi of Pi
2.6989566012577245131409271509303.. a+b
/ 0.92650275035220848584275966758914.. b
= 2.9130583802711007947403078280203.. (a+b)/b
The ABCs: ((a+b)/b)  1 = a/b
... and c = See!
Rod ... ...
Golden Spiral of Quadrature
Given: Circlesquaring right triangle
where c = hypotenuse = 2 (diameter),
a = long side, b = short side.
1.7724538509055160272981674833411.. a, sqrt(Pi)
+ 0.92650275035220848584275966758914.. b
= 2.6989566012577245131409271509303.. a+b
1.7724538509055160272981674833411.. a, sqrt(Pi)
/ 0.92650275035220848584275966758914.. b
= 1.9130583802711007947403078280203.. a/b, Phi of Pi
2.6989566012577245131409271509303.. a+b
/ 0.92650275035220848584275966758914.. b
= 2.9130583802711007947403078280203.. (a+b)/b
The ABCs: ((a+b)/b)  1 = a/b
... and c = See!
Rod ... ...

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Re: Paradise Trinity Day
Re: Sqrt(Pi) Ratios design
((a+b)/b)  1 = (a/b + b/b)  1
"Nice try!"
Rod
"Yes, we see that "golden" algebra!"The ABCs: ((a+b)/b)  1 = a/b
... and c = See!
((a+b)/b)  1 = (a/b + b/b)  1
"Nice try!"
Rod

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Re: Paradise Trinity Day
Re: Quadra Lute design
(aka "Pi Fork 'n Lute")
Long Story Short: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf
Rod ... ...
(aka "Pi Fork 'n Lute")
Long Story Short: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf
Rod ... ...

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Re: Paradise Trinity Day
Re: https://www.learnreligions.com/squaring ... rcle96039
"First of all we are not saying that a square of equal area does not exist.
If the circle has area A, then a square with side A clearly has the same area.
Secondly, we are not saying that [it] is impossible, since it is possible,
but not under the restriction of using only a straightedge and compass."
Precisely what this geometry has been showing for a decade:
The circle cannot be squared but squared circles exist
And note that using a straightedge does not rule out
the lines of easily drawn degrees (30, 45, 90, etc.).
Rod
"First of all we are not saying that a square of equal area does not exist.
If the circle has area A, then a square with side A clearly has the same area.
Secondly, we are not saying that [it] is impossible, since it is possible,
but not under the restriction of using only a straightedge and compass."
Precisely what this geometry has been showing for a decade:
The circle cannot be squared but squared circles exist
And note that using a straightedge does not rule out
the lines of easily drawn degrees (30, 45, 90, etc.).
Rod

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Re: Paradise Trinity Day
Re: Pi Fork geometry (for D = 2)
Phi of Pi = sqrt(Pi) / sqrt(4Pi)
1.7724538509055160272981674833411.. sqrt(Pi)
/ 0.92650275035220848584275966758914.. sqrt(4Pi)
= 1.9130583802711007947403078280203.. sqrt(Pi)/sqrt(4Pi)
Given: circlesquaring right triangle
where a = long side = sqrt(Pi)
b = short side = sqrt(4Pi)
c = hypotenuse = 2
a^2 + b^2 = c^2 (Pythagorean Theorem)
3.1415926535897932384626433832795.. sqrt(Pi)^2
+ 0.8584073464102067615373566167205.. sqrt(4Pi)^2
= 4 = 2^2
Rod
Phi of Pi = sqrt(Pi) / sqrt(4Pi)
1.7724538509055160272981674833411.. sqrt(Pi)
/ 0.92650275035220848584275966758914.. sqrt(4Pi)
= 1.9130583802711007947403078280203.. sqrt(Pi)/sqrt(4Pi)
Given: circlesquaring right triangle
where a = long side = sqrt(Pi)
b = short side = sqrt(4Pi)
c = hypotenuse = 2
a^2 + b^2 = c^2 (Pythagorean Theorem)
3.1415926535897932384626433832795.. sqrt(Pi)^2
+ 0.8584073464102067615373566167205.. sqrt(4Pi)^2
= 4 = 2^2
Rod

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Re: Paradise Trinity Day
Re: Arcs de Triomphe design*
A peek at the Golden Rectangle of Quadrature.
* added to: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf
Rod
A peek at the Golden Rectangle of Quadrature.
* added to: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf
Rod

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Re: Paradise Trinity Day
Re: Arcs de Triomphe design
A peek at the Golden Rectangle of Quadrature.
Arc! Arc! Arc!
(it's a composition of related objects)
Rod ... ...
A peek at the Golden Rectangle of Quadrature.
Arc! Arc! Arc!
(it's a composition of related objects)
Rod ... ...

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Re: Paradise Trinity Day
Re: Arcs de Triomphe design
A peek at the Golden Rectangle of Quadrature.
Geometers' secret ...
The golden moon (lower left) is all about
the geometric signature of Quadrature:
2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 2(sqrt(1/Pi))
= 1.1283791670955125738961589031215.. Qcue
(pronounced "Q Q")
Geometers' super secret ...
Vertical green line along the base is not an arc identifier
of the bottom of the Golden Rectangle, but is a placeholder
for the arc identifying the isosceles right triangle (not shown)
that creates the number squaring line (green).
Rod
A peek at the Golden Rectangle of Quadrature.
Geometers' secret ...
The golden moon (lower left) is all about
the geometric signature of Quadrature:
2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 2(sqrt(1/Pi))
= 1.1283791670955125738961589031215.. Qcue
(pronounced "Q Q")
Geometers' super secret ...
Vertical green line along the base is not an arc identifier
of the bottom of the Golden Rectangle, but is a placeholder
for the arc identifying the isosceles right triangle (not shown)
that creates the number squaring line (green).
Rod

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Re: Paradise Trinity Day
Re: Qcue Enqueue design
"Quadrature awakens in the local universe"
Contrasting the geometric relationships
of 2/sqrt(Pi), sqrt(Pi), and 2.0 within Quadra Lute
... as hosted by sqrt(2).
"Say what?!" That's sum Pi, in deed
Rod
"Quadrature awakens in the local universe"
Contrasting the geometric relationships
of 2/sqrt(Pi), sqrt(Pi), and 2.0 within Quadra Lute
... as hosted by sqrt(2).
"Say what?!" That's sum Pi, in deed
Rod

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Re: Paradise Trinity Day
Re: Qcue Enqueue design
"Quadrature awakens in the local universe"
The salient numbers ...
1.1283791670955125738961589031215.. 2/sqrt(Pi), Qcue
^2 = 1.2732395447351626861510701069801.. 2/sqrt(Pi)^2
1.2732395447351626861510701069801.. 2/sqrt(Pi)^2
x 1.5707963267948966192313216916398.. Pi/2
= 2.0
x 1.5707963267948966192313216916398.. Pi/2
= 3.1415926535897932384626433832795.. Pi
x 1.2732395447351626861510701069801.. 2/sqrt(Pi)^2
= 4.0
Rod ... ...
"Quadrature awakens in the local universe"
The salient numbers ...
1.1283791670955125738961589031215.. 2/sqrt(Pi), Qcue
^2 = 1.2732395447351626861510701069801.. 2/sqrt(Pi)^2
1.2732395447351626861510701069801.. 2/sqrt(Pi)^2
x 1.5707963267948966192313216916398.. Pi/2
= 2.0
x 1.5707963267948966192313216916398.. Pi/2
= 3.1415926535897932384626433832795.. Pi
x 1.2732395447351626861510701069801.. 2/sqrt(Pi)^2
= 4.0
Rod ... ...

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Re: Paradise Trinity Day
Re: Qcue Enqueue design
"Quadrature awakens in the local universe"
Qcue Enqueue by the Numbers
~ Squares & Square Roots
1.2732395447351626861510701069802.. (2/sqrt(Pi))^2
1.1283791670955125738961589031215.. 2/sqrt(Pi)
2.0
1.4142135623730950488016887242097.. sqrt(2)
3.1415926535897932384626433832793.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
4.0
2.0
~ Relationships by the Numbers
1.1283791670955125738961589031215.. 2/sqrt(Pi)
^2 = 1.2732395447351626861510701069802.. (2/sqrt(Pi))^2
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.4142135623730950488016887242097.. sqrt(2)
= 2.506628274631000502415765284811.. sqrt(Pi)sqrt(2)
/ 2 = 1.2533141373155002512078826424055.. (sqrt(Pi)sqrt(2))/2
^2 = 1.5707963267948966192313216916397.. Pi/2
1.7724538509055160272981674833411.. sqrt(Pi)
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.5707963267948966192313216916398.. Pi/2
~ Continuum by the roots
1.1283791670955125738961589031215.. 2/sqrt(Pi)
x 1.2533141373155002512078826424055.. (sqrt(Pi)sqrt(2))/2
= 1.4142135623730950488016887242097.. sqrt(2)
x 1.2533141373155002512078826424055.. (sqrt(Pi)sqrt(2))/2
= 1.7724538509055160272981674833411.. sqrt(Pi)
x 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 2.0
~ Relationship of diameters
4.0
/ 1.2732395447351626861510701069802.. (2/sqrt(Pi))^2
= 3.1415926535897932384626433832795.. Pi
Rod ("That's sum Pi!")
"Quadrature awakens in the local universe"
Qcue Enqueue by the Numbers
~ Squares & Square Roots
1.2732395447351626861510701069802.. (2/sqrt(Pi))^2
1.1283791670955125738961589031215.. 2/sqrt(Pi)
2.0
1.4142135623730950488016887242097.. sqrt(2)
3.1415926535897932384626433832793.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
4.0
2.0
~ Relationships by the Numbers
1.1283791670955125738961589031215.. 2/sqrt(Pi)
^2 = 1.2732395447351626861510701069802.. (2/sqrt(Pi))^2
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.4142135623730950488016887242097.. sqrt(2)
= 2.506628274631000502415765284811.. sqrt(Pi)sqrt(2)
/ 2 = 1.2533141373155002512078826424055.. (sqrt(Pi)sqrt(2))/2
^2 = 1.5707963267948966192313216916397.. Pi/2
1.7724538509055160272981674833411.. sqrt(Pi)
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.5707963267948966192313216916398.. Pi/2
~ Continuum by the roots
1.1283791670955125738961589031215.. 2/sqrt(Pi)
x 1.2533141373155002512078826424055.. (sqrt(Pi)sqrt(2))/2
= 1.4142135623730950488016887242097.. sqrt(2)
x 1.2533141373155002512078826424055.. (sqrt(Pi)sqrt(2))/2
= 1.7724538509055160272981674833411.. sqrt(Pi)
x 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 2.0
~ Relationship of diameters
4.0
/ 1.2732395447351626861510701069802.. (2/sqrt(Pi))^2
= 3.1415926535897932384626433832795.. Pi
Rod ("That's sum Pi!")