Archimedes Plutonium
2024-01-09 09:27:24 UTC
275th Book of Science for AP:: Math Geometry Experiment: Points of Intersection of circle and Grid Coordinate Points-- Are there only 4 points?? by Archimedes Plutonium
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Archimedes Plutonium<***@gmail.com>
Jan 3, 2024, 11:32:24 PM (5 days ago)
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A problem that has been on my mind for several weeks now.
So I get out the graph paper and compass, two compasses, one with a pencil on the end.
I take a grid of 10 x-axis points and 10 y-axis points, the square has 100 small squares. I consider only the intersection points as points to think about.
Now I draw a circle from the center (5,5) of radius 5, diameter 10. Will that circle intersect only these four points (5,0) (5,10) (0,5) (10,5) or will some other points be intersected??
Proof??
Now, it gets harder. I make each of those 100 tiny squares be a grid system of 10 by 10 points.
Will the already drawn circle intersect any of these newer points????
Or, is every circle allowed to have only 4 points of intersection no matter how dense the Grid System becomes. Proof?
AP
275th Book of Science for AP:: Math Geometry Experiment: Points of Intersection of circle and Grid Coordinate Points-- Are there only 4 points?? by Archimedes Plutonium
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Archimedes Plutonium<***@gmail.com>
Jan 4, 2024, 2:49:57 AM (5 days ago)
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So here is the big question weighing on my mind. When I discovered the infinity borderline was 1*10^604 with its inverse 1*10^-604, was where pi had 3 zero digits in a row. And at infinity borderline pi had to be evenly divisible by 5 factorial or 120 as 2*3*4*5. It had to be that way to give us the regular polyhedra.
And so we take the infinity borderline as exponent 604 where pi ends in three zero digits in a row and pi ends there, for mathematics ends there at that borderline.
Now, applying that idea to this question.
If I keep dividing one tier by 10 by 10 smaller squares over 604 iterations of Grid System. The question is, instead of 4 intersections of the circle. Will I have 120 intersection points at infinity borderline of that same starting circle?
If true, it would be a reversal of the even divisibility for regular polyhedra, in that the circle would have 120 intersections with points at 1*10^-604.
I am going to spend several days, weeks thinking about the logic of that deduction. Does it work?
AP
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Archimedes Plutonium<***@gmail.com>
Jan 6, 2024, 12:05:53 PM (3 days ago)
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So we look upon the Infinity Borderline as the end of mathematics. We could say it is 10^604 and inverse or we could reach out to 10^1208 where we have even divisibility by 6 factorial where 6! = 720 = 360 x 2.
Let us take 10^1208 as the last and final border of mathematics.
Does that mean that as we make the Grid System 10 by 10 smaller successively, that we have any given circle will have 720 intersection points? That would be 720/4 = 180 points of intersection per every quarter circle.
What it then does is speak to us about what we thought was "irrational number" and that of "transcendental number". Because every circle has 4 rational points of intersection, automatically, but to retrieve the other 720 subtract 4 points we have to go to the infinity borderline 10^1208 to retrieve those 180 subtract 2 in each quarter circle.
We begin to look upon irrational number as that which needs to go to the infinity borderline to be a rational number once there.
AP
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Archimedes Plutonium<***@gmail.com>
Jan 8, 2024, 9:00:47 PM (6 hours ago)
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For the first time in my career in science, I seem to run into a problem where logic is not helpful to solving the problem, but rather the disappointment.
Suppose pi were 3.0000..... Then would we have more than 4 intersection points in the first layer of grid points?
At infinity borderline 1*10^604 where pi has 3 zero digits in a row and is evenly divisible by 5 factorial = 120. Does the circle in the 604 exponent grid have how many more intersection points? Does it have 120 intersection points instead of the 4? Does it have an infinity of intersection points as we look upon pi as ending in just all zero digits as far as you want to go?
So suppose pi were 3.0000..... We have the intersection in the 10 by 10 Grid of 0 degree, 90 degree, 180 degree, 270 degree.
With pi as 3.000.... does that not call for a equilateral triangle inside the 10 by 10 Grid that would call for 4 more points of intersection and thus 8 points in all, rather than the 4? A point of intersection at 120 degrees, 240 degrees and 60 degrees and 300 degrees?
So far, I do not seem to be able to find logic steering my way towards an answer. Very unusual, very disconcerting.
AP, King of Science
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Archimedes Plutonium<***@gmail.com>
Jan 8, 2024, 11:49:22 PM (4 hours ago)
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Upon further reflection, I overcame those difficulties.
I drew a quartercircle with a radius of 3 and then another with radius 3.14. Then I looked to see if every point of the Grid was within that band.
Then I drew a smaller quartercircle inside a 10 by 10 grid. If every point of intersection were to occur in the quarter circle, I would have 15 points of intersection for the 10 by 10. Now I wonder if each square was 10 by 10 in the 100 grid whether the quarter circle would have 15 x 10 = 150 total possible intersection points. But save that for another day.
In the smaller circle of 10 by 10, actually there are 29 points to contend with in the full circle.
I am excited now in finding the answer at infinity, how many points of intersection. All of them seems to be the answer.
When we pretend pi is 3.0000.... rather than 3.14159...... what happens is that every passing arc of the circle as it winds its way around, ensares a nearby point of the Grid and intersects with that point as trapped in between 3.0000... and 3.14159..... has a point of intersection.
This of course gives new meaning to the fractional difference of pi and 3 where pi subtract 3 = 0.14159..... which is the smallest amount of arclength needed so that a circle at infinity borderline intersects with the Grid Graph of that circle.
This can be also pictured more easily as saying a square is a circle where pi is 3.0000..... Take the midpoint of a 10 by 10 Grid as (5,5) construct a square inside instead of a circle. The square sides intersect at a diagonal, all the points of the Grid, in fact 20 intersection points.
So, now I have some Logic to apply.
AP
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Archimedes Plutonium<***@gmail.com>
12:55 AM (2 hours ago)
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And this is really nice surprising and somewhat beautiful, for we begin to picture a circle not as a curved line, but as a small band, a ring.
In Old Math, a circle was a curved line which generates irrational and transcendental numbers.
In New Math, a circle is not a curved line but a curved band or ring an thus captures all the points in a Grid system as it rotates around from a center.
In Old Math transcendental was simply the notion that the mathematician was too stupid to realize a circle cannot be a straight line that is curved. That that stupidity caused them to come up with the concept of transcendental. that you could not put pi in a linear equation nor polynomial.
It is not that pi and circles are special, and need a transcendental concept. No, it is the mathematician is too stupid to realize you cannot bend a line to be a circle. That a circle needs a tiny width. So if you want a circle whose radius is 3, then you have another circle of radius 3.14159.... to join with the radius 3 circle to make a "real circle" for a single line circle just does not exist, it is imagination gone wrong.
All circles are thin bands, thin rings, and no circle is a line that arcs around into a circle.
So in Euclid geometry, we have the axioms a straightline has no width or depth, only length is true.
But in Euclid Geometry, we must now have a circle or ellipse or oval has arc length, has arc width in 2D and has arc depth in 3D.
In New Geometry, we cannot classify circle with straightlines as being the same type of figures, for no circle exists that has only arclength and no width and no depth.
Reason:: This comes back to physics in quantum mechanics, that you need space as discrete, as Grids, and to go around in a Grid from point to point, and arc or bend, you need to be able to pick up the next point, you cannot have just 4 points of a grid produce a circle, you need 20 points out of 100 in a 10 by 10 Grid to produce that circle. And the only way you can produce that circle with 20 points of intersection is to make the circle have a width-- a band or rind.
AP
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Archimedes Plutonium<***@gmail.com>
3:23 AM (now)
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Now the spread between band width need not be as large as 3.00000... versus 3.14159..... which is a large spread. The smallest spread is that of half of the diagonal length in a square of the Grid. So the grid square is 1 by 1 and the diagonal would thus be 1^2 +1^2 = hypotenuse^2 = 2, and the sqrt of 2 is 1.4142135..... So the one circle of a two-circle band can fall halfway at most length of 1.4142135/2 = 0.7071... which happens to be also the quantity of sine 45 degrees = cosine 45 degrees.
And so what type of sigma error do we have with pi = 3.14159 and band width allowance of 0.7071? Which is 0.2% sigma error or 99.8% accurate.
Notice also, that 0.14159..... of pi when divided by 2 is 0.07079...... which tells us that isosceles right-triangles are directly related to circle curvature.
AP
AP kindly asks Google to let AP run all three, sci.math, sci.physics, PAU as he runs PAU, now--- all pure science, no spam and no govt b.s.
https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
Archimedes Plutonium
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Archimedes Plutonium's profile photo
Archimedes Plutonium<***@gmail.com>
Jan 3, 2024, 11:32:24 PM (5 days ago)
to Plutonium Atom Universe
A problem that has been on my mind for several weeks now.
So I get out the graph paper and compass, two compasses, one with a pencil on the end.
I take a grid of 10 x-axis points and 10 y-axis points, the square has 100 small squares. I consider only the intersection points as points to think about.
Now I draw a circle from the center (5,5) of radius 5, diameter 10. Will that circle intersect only these four points (5,0) (5,10) (0,5) (10,5) or will some other points be intersected??
Proof??
Now, it gets harder. I make each of those 100 tiny squares be a grid system of 10 by 10 points.
Will the already drawn circle intersect any of these newer points????
Or, is every circle allowed to have only 4 points of intersection no matter how dense the Grid System becomes. Proof?
AP
275th Book of Science for AP:: Math Geometry Experiment: Points of Intersection of circle and Grid Coordinate Points-- Are there only 4 points?? by Archimedes Plutonium
Archimedes Plutonium's profile photo
Archimedes Plutonium<***@gmail.com>
Jan 4, 2024, 2:49:57 AM (5 days ago)
to Plutonium Atom Universe
So here is the big question weighing on my mind. When I discovered the infinity borderline was 1*10^604 with its inverse 1*10^-604, was where pi had 3 zero digits in a row. And at infinity borderline pi had to be evenly divisible by 5 factorial or 120 as 2*3*4*5. It had to be that way to give us the regular polyhedra.
And so we take the infinity borderline as exponent 604 where pi ends in three zero digits in a row and pi ends there, for mathematics ends there at that borderline.
Now, applying that idea to this question.
If I keep dividing one tier by 10 by 10 smaller squares over 604 iterations of Grid System. The question is, instead of 4 intersections of the circle. Will I have 120 intersection points at infinity borderline of that same starting circle?
If true, it would be a reversal of the even divisibility for regular polyhedra, in that the circle would have 120 intersections with points at 1*10^-604.
I am going to spend several days, weeks thinking about the logic of that deduction. Does it work?
AP
Archimedes Plutonium's profile photo
Archimedes Plutonium<***@gmail.com>
Jan 6, 2024, 12:05:53 PM (3 days ago)
to Plutonium Atom Universe
So we look upon the Infinity Borderline as the end of mathematics. We could say it is 10^604 and inverse or we could reach out to 10^1208 where we have even divisibility by 6 factorial where 6! = 720 = 360 x 2.
Let us take 10^1208 as the last and final border of mathematics.
Does that mean that as we make the Grid System 10 by 10 smaller successively, that we have any given circle will have 720 intersection points? That would be 720/4 = 180 points of intersection per every quarter circle.
What it then does is speak to us about what we thought was "irrational number" and that of "transcendental number". Because every circle has 4 rational points of intersection, automatically, but to retrieve the other 720 subtract 4 points we have to go to the infinity borderline 10^1208 to retrieve those 180 subtract 2 in each quarter circle.
We begin to look upon irrational number as that which needs to go to the infinity borderline to be a rational number once there.
AP
Archimedes Plutonium's profile photo
Archimedes Plutonium<***@gmail.com>
Jan 8, 2024, 9:00:47 PM (6 hours ago)
to Plutonium Atom Universe
For the first time in my career in science, I seem to run into a problem where logic is not helpful to solving the problem, but rather the disappointment.
Suppose pi were 3.0000..... Then would we have more than 4 intersection points in the first layer of grid points?
At infinity borderline 1*10^604 where pi has 3 zero digits in a row and is evenly divisible by 5 factorial = 120. Does the circle in the 604 exponent grid have how many more intersection points? Does it have 120 intersection points instead of the 4? Does it have an infinity of intersection points as we look upon pi as ending in just all zero digits as far as you want to go?
So suppose pi were 3.0000..... We have the intersection in the 10 by 10 Grid of 0 degree, 90 degree, 180 degree, 270 degree.
With pi as 3.000.... does that not call for a equilateral triangle inside the 10 by 10 Grid that would call for 4 more points of intersection and thus 8 points in all, rather than the 4? A point of intersection at 120 degrees, 240 degrees and 60 degrees and 300 degrees?
So far, I do not seem to be able to find logic steering my way towards an answer. Very unusual, very disconcerting.
AP, King of Science
Archimedes Plutonium's profile photo
Archimedes Plutonium<***@gmail.com>
Jan 8, 2024, 11:49:22 PM (4 hours ago)
to Plutonium Atom Universe
Upon further reflection, I overcame those difficulties.
I drew a quartercircle with a radius of 3 and then another with radius 3.14. Then I looked to see if every point of the Grid was within that band.
Then I drew a smaller quartercircle inside a 10 by 10 grid. If every point of intersection were to occur in the quarter circle, I would have 15 points of intersection for the 10 by 10. Now I wonder if each square was 10 by 10 in the 100 grid whether the quarter circle would have 15 x 10 = 150 total possible intersection points. But save that for another day.
In the smaller circle of 10 by 10, actually there are 29 points to contend with in the full circle.
I am excited now in finding the answer at infinity, how many points of intersection. All of them seems to be the answer.
When we pretend pi is 3.0000.... rather than 3.14159...... what happens is that every passing arc of the circle as it winds its way around, ensares a nearby point of the Grid and intersects with that point as trapped in between 3.0000... and 3.14159..... has a point of intersection.
This of course gives new meaning to the fractional difference of pi and 3 where pi subtract 3 = 0.14159..... which is the smallest amount of arclength needed so that a circle at infinity borderline intersects with the Grid Graph of that circle.
This can be also pictured more easily as saying a square is a circle where pi is 3.0000..... Take the midpoint of a 10 by 10 Grid as (5,5) construct a square inside instead of a circle. The square sides intersect at a diagonal, all the points of the Grid, in fact 20 intersection points.
So, now I have some Logic to apply.
AP
Archimedes Plutonium's profile photo
Archimedes Plutonium<***@gmail.com>
12:55 AM (2 hours ago)
to Plutonium Atom Universe
And this is really nice surprising and somewhat beautiful, for we begin to picture a circle not as a curved line, but as a small band, a ring.
In Old Math, a circle was a curved line which generates irrational and transcendental numbers.
In New Math, a circle is not a curved line but a curved band or ring an thus captures all the points in a Grid system as it rotates around from a center.
In Old Math transcendental was simply the notion that the mathematician was too stupid to realize a circle cannot be a straight line that is curved. That that stupidity caused them to come up with the concept of transcendental. that you could not put pi in a linear equation nor polynomial.
It is not that pi and circles are special, and need a transcendental concept. No, it is the mathematician is too stupid to realize you cannot bend a line to be a circle. That a circle needs a tiny width. So if you want a circle whose radius is 3, then you have another circle of radius 3.14159.... to join with the radius 3 circle to make a "real circle" for a single line circle just does not exist, it is imagination gone wrong.
All circles are thin bands, thin rings, and no circle is a line that arcs around into a circle.
So in Euclid geometry, we have the axioms a straightline has no width or depth, only length is true.
But in Euclid Geometry, we must now have a circle or ellipse or oval has arc length, has arc width in 2D and has arc depth in 3D.
In New Geometry, we cannot classify circle with straightlines as being the same type of figures, for no circle exists that has only arclength and no width and no depth.
Reason:: This comes back to physics in quantum mechanics, that you need space as discrete, as Grids, and to go around in a Grid from point to point, and arc or bend, you need to be able to pick up the next point, you cannot have just 4 points of a grid produce a circle, you need 20 points out of 100 in a 10 by 10 Grid to produce that circle. And the only way you can produce that circle with 20 points of intersection is to make the circle have a width-- a band or rind.
AP
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Archimedes Plutonium<***@gmail.com>
3:23 AM (now)
to Plutonium Atom Universe
Now the spread between band width need not be as large as 3.00000... versus 3.14159..... which is a large spread. The smallest spread is that of half of the diagonal length in a square of the Grid. So the grid square is 1 by 1 and the diagonal would thus be 1^2 +1^2 = hypotenuse^2 = 2, and the sqrt of 2 is 1.4142135..... So the one circle of a two-circle band can fall halfway at most length of 1.4142135/2 = 0.7071... which happens to be also the quantity of sine 45 degrees = cosine 45 degrees.
And so what type of sigma error do we have with pi = 3.14159 and band width allowance of 0.7071? Which is 0.2% sigma error or 99.8% accurate.
Notice also, that 0.14159..... of pi when divided by 2 is 0.07079...... which tells us that isosceles right-triangles are directly related to circle curvature.
AP
AP kindly asks Google to let AP run all three, sci.math, sci.physics, PAU as he runs PAU, now--- all pure science, no spam and no govt b.s.
https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
Archimedes Plutonium