Does someone know this?

While I was making a small study on circle packing recently, I came across this problem that left me scratching my brains. I found this problem interesting and wanted to share. It goes something like this:

Q) We want to reduce the size of circles and as we do so, fill them up into a different and bigger circle as much as possible (maximum). As the radius of the small circles approaches to zero and fills up the mother circle,

1.Does the packing density approaches to 1 (filling up the entire mother circle)?

OR

2.Does the packing density approaches to that of the hexagonal lattice arrangement (i.e. 0.906...) and therefore not filling up the entire mother circle?

(For those who don't know, packing density is the ratio of the area covered by the circles to the total area of the space they are packed into. It's a measure of how much space is covered by the circles compared to the total space available.)

If you know calculus, in this case, the small circles will behave as infinitesimals and at first, it might seem obvious that the small circles must fill up the entire mother circle. But, if you observe carefully, circle as infinitesimals do not behave like traditional infinitesimals (rectangles) where as the breadth of the rectangle shortens, the gap between the infinitesimals and the curve decreases therefore covering up the entire area under the curve. Whereas in the case of circles, the size of the gaps (interstitial space) remains as significant as the circles themselves making up 5% of the area of the small circle. Therefore, as the size of the circles decreases, both the gaps and the circles equally compete for space. But I'm not really sure if this is still the case when the circles theoretically become point-like. It could be possible that the limiting packing density could be something else entirely.

Edited July 22Jul 22 by DavidWahl

1 hour ago, DavidWahl said:

Does someone know this?

While I was making a small study on circle packing recently, I came across this problem that left me scratching my brains. I found this problem interesting and wanted to share. It goes something like this:

Q) We want to reduce the size of circles and as we do so, fill them up into a different and bigger circle as much as possible (maximum). As the radius of the small circles approaches to zero and fills up the mother circle,

1.Does the packing density approaches to 1 (filling up the entire mother circle)?

OR

2.Does the packing density approaches to that of the hexagonal lattice arrangement (i.e. 0.906...) and therefore not filling up the entire mother circle?

(For those who don't know, packing density is the ratio of the area covered by the circles to the total area of the space they are packed into. It's a measure of how much space is covered by the circles compared to the total space available.)

If you know calculus, in this case, the small circles will behave as infinitesimals and at first, it might seem obvious that the small circles must fill up the entire mother circle. But, if you observe carefully, circle as infinitesimals do not behave like traditional infinitesimals (rectangles) where as the breadth of the rectangle shortens, the gap between the infinitesimals and the curve decreases therefore covering up the entire area under the curve. Whereas in the case of circles, the size of the gaps (interstitial space) remains as significant as the circles themselves making up 5% of the area of the small circle. Therefore, as the size of the circles decreases, both the gaps and the circles equally compete for space. But I'm not really sure if this is still the case when the circles theoretically become point-like. It could be possible that the limiting packing density could be something else entirely.

I recall from chemistry that the packing density of close-packed spheres is a fixed proportion, irrespective of the radius of the spheres. On that basis I think I would expect the same to apply to your 2D problem. Though I agree if one reduces the radius so that it ->0, one would seem to have a conundrum, since spheres of infinitesimal radius would, at least intuitively, seem to fill the space completely. Perhaps a mathematician can help.

It can be done if you rearrange your dots.

The trick is to use polar coordinates not cartesian ones and arrange you dots (which in the limit have zero diameter) in radial lines and then consider the angle between each line, taking the limit as that angle tends to zero.

You might also find the six circles and seven circles theorems interesting, along with Steiner chains.

(they do not fill all the circle.)

Just now, exchemist said:

I recall from chemistry that the packing density of close-packed spheres is a fixed proportion, irrespective of the radius of the spheres. On that basis I think I would expect the same to apply to your 2D problem. Though I agree if one reduces the radius so that it ->0, one would seem to have a conundrum, since spheres of infinitesimal radius would, at least intuitively, seem to fill the space completely. Perhaps a mathematician can help.

You may like to know that the radial method is part of the mathematics of s orbitals in quantum mechanics.

Edited July 22Jul 22 by studiot

41 minutes ago, studiot said:

It can be done if you rearrange your dots.

The trick is to use polar coordinates not cartesian ones and arrange you dots (which in the limit have zero diameter) in radial lines and then consider the angle between each line, taking the limit as that angle tends to zero.

You might also find the six circles and seven circles theorems interesting, along with Steiner chains.

(they do not fill all the circle.)

You may like to know that the radial method is part of the mathematics of s orbitals in quantum mechanics.

I remember decomposing the wave function into radial and angular parts, certainly.

But tell me, if the packing density of spheres is independent of radius, how can the packing of circles not be? Or have I perhaps misunderstood your earlier response?

Just now, exchemist said:

I remember decomposing the wave function into radial and angular parts, certainly.

But tell me, if the packing density of spheres is independent of radius, how can the packing of circles not be? Or have I perhaps misunderstood your earlier response?

The point is that polar coordinates will completely fill the plane or 3D space with the radius at infinity.

David's problem is that he has used a boundary - the circle he wishes to fill.

So you need to be able to describe that boundary in polar coordinates to use the method and further since he is filling a disk, the outer circle radius is constant.

That means we can start with a single line of that radius and sweep it around one end in infinitesimal steps, remembering that a line has no width.

As the radius sweeps around and angle of delta theta it covers all points in a sector bounded by the infinitesimal arc of the out circle.

Summing all these goes to an integral as delta theta goes to zero.

1 hour ago, studiot said:

It can be done if you rearrange your dots.

The trick is to use polar coordinates not cartesian ones and arrange you dots (which in the limit have zero diameter) in radial lines and then consider the angle between each line, taking the limit as that angle tends to zero.

You might also find the six circles and seven circles theorems interesting, along with Steiner chains.

(they do not fill all the circle.)

This is a good way to show that the dots do cover up the entire circle but what I dislike about this method is that it doesn't say much about the arrangement, it simply ignores it. Depending on the 2D-arrangement we choose, the packing density can change. Besides, I'm not sure it's safe to assume that an array of infinitesimally small circles to be approximately equivalent to a line. What about those gaps?

Edited July 22Jul 22 by DavidWahl

Just now, DavidWahl said:

This is a good way to show that the dots do cover up the entire circle but what I dislike about this method is that it doesn't say much about the arrangement, it simply ignores it. Depending on the 2D-arrangement we choose, the packing density can change. Besides, I'm not sure it's safe to assume that an array of infinitesimally small circles to be approximately equivalent to a line. What about those gaps?

That's the whole point, there are no gaps in the real number line.

Also it only works with one particular arrangement.

Other arrangements do not work , you need peano curves or go into Banach-Tarski for that.

Edited July 22Jul 22 by studiot

24 minutes ago, DavidWahl said:

This is a good way to show that the dots do cover up the entire circle but what I dislike about this method is that it doesn't say much about the arrangement, it simply ignores it. Depending on the 2D-arrangement we choose, the packing density can change. Besides, I'm not sure it's safe to assume that an array of infinitesimally small circles to be approximately equivalent to a line. What about those gaps?

OK but as I understand it that is not the problem. The problem is with circles of finite radius you have a maximum packing density <100% due to the area of the interstitial gaps. The problem is what happens to the proportion of the area filled by the circles, as the radius reduces, and in the limit r->0. I had expected that proportion to be independent of the radius.

The packing of spheres is different to the packing of circles in that one can't arrange spheres around a central sphere in a gap-free way, whereas one can arrange six circles around a central circle and this arrangement is gap-free. And because a flat plane can be tessellated with hexagons, it is clear that the maximum packing density for circles in a plane is the ratio of the area of an inscribed circle to the area of the inscribing hexagon:

Area of inscribed circle = πr² ≈ 3.1416 r²

Area of inscribing hexagon = 6r² tan π/6 ≈ 3.4641 r²

Ratio of inscribed circle to inscribing hexagon = [math]\dfrac{\pi}{6}\cot\dfrac{\pi}{6} = \dfrac{\sqrt{3}\pi}{6} \approx 0.9069[/math]

The maximum packing density for spheres is [math]\dfrac{\pi}{3\sqrt{2}} \approx 0.74048[/math] for a number of different arrangements, all of which has each sphere touching 12 neighbouring spheres. According to Wikipedia:

"In 1611, Johannes Kepler conjectured that this is the maximum possible density amongst both regular and irregular arrangements—this became known as the Kepler conjecture. Carl Friedrich Gauss proved in 1831 that these packings have the highest density amongst all possible lattice packings. In 1998, Thomas Callister Hales, following the approach suggested by László Fejes Tóth in 1953, announced a proof of the Kepler conjecture. Hales' proof is a proof by exhaustion involving checking of many individual cases using complex computer calculations. Referees said that they were "99% certain" of the correctness of Hales' proof. On 10 August 2014, Hales announced the completion of a formal proof using automated proof checking, removing any doubt."

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Just now, KJW said:

it is clear that the maximum packing density for circles in a plane is the ratio of the area of an inscribed circle to the area of the inscribing hexagon:

It is not clear to me.

A plane is of genus 0.

Which means we can shrink any circle on the surface to zero ie a point.

Which is what the OP asked us to do.

But that does not imply that any 'packing arrangment' of such points will fill the plane, you have to use the right one.

But no you cannot Tesselate a plane with finite circles.

11 minutes ago, studiot said:

Which means we can shrink any circle on the surface to zero ie a point.

Which is what the OP asked us to do.

I assumed that this was simply to eliminate boundary effects that result from trying to pack circles into a finite area. This seems unnecessary to me as one can tesselate an infinite plane with hexagons and inscribe a circle into each hexagon. Note that the hexagon is the polygon with the largest number of sides that can tesselate a plane. And given that the ratio of the area of an inscribed circle to the area of the inscribing polygon increases with the number of sides of the polygon, the hexagon is the polygon producing the maximum packing density of circles.

Edited July 22Jul 22 by KJW

It seems to me that if a circle could actually reach r=0 it would cease being a circle, meaning that there will always be space when trying to fill any space with smaller circles regardless of their size. Once r=0 it becomes a point, as stated above, (unless one wants to redefine what a circle is) and any space will be filled by an infinite number of points...