Circles in Principal Circle

There apparently is no definitive formula....it varies according to ratio...best I could do was 31...

http://answers.google.com/answers/threadview/id/597709.html

From Yahoo answers....

"This is a hard problem. There is no simple formula. The best known results are at this web site:

http://hydra.nat.uni-magdeburg.de/packin…The answer for the 1" circles in a 14" circle is 161 circles. A picture is here:http://hydra.nat.uni-magdeburg.de/packin…The answer for the 2" circles in a 14" circle is 38 circles. A picture is here:http://hydra.nat.uni-magdeburg.de/packin…"
http://answers.yahoo.com/question/index?qid=20090810112124AAEygZF

Interesting.....if you're trying to fill an infinite area a triangular pattern is best, but that may not (probably doesn't) apply when filling a circle.

Based upon the ratio (25/4 = 6.250), http://hydra.nat.uni-magdeburg.de/packing/cci/cci30.html (taken from one of your links) shows that the maximum would be 30 circles. This shows how they would be arranged within the principal circle. If kerf width were provided, this ratio could be recalculated using the method I proposed in post #10, and the table at http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html used to find the actual possible number for that method of cutting. The limit is the highest ratio LESS THAN the calculated value (column four). Clicking on the number in the left-most column takes you to a set of diagrams showing the arrangement, the count, and even the best known sequence of cutting for shortest path of CNC tools.

Suppose that the kerf was 0.5mm. Then, r₂ = 4.5 mm, and R₂ = 25.5mm. R₂/r₁ = 25.5/4.5 = 5.6666 . . . , and the next lower value is 5.6516610917654367675791591314, for 24 small circles (see http://hydra.nat.uni-magdeburg.de/packing/cci/cci24.html). This shows the value of narrow kerfs! If increased to 1mm, you'd only get 20. Getting down to 0.04mm would - theoretically! - get you the full 30, though I strongly suspect that this indicates an erroneous assumption earlier in the process, because it would require slight adjustments to most interior circle positions, I believe.

It's called a "Packing Factor" or "Packing Density" problem, here's an interesting paper from MathWorld addressing your problem. If you want a generalized solution with lots of math and examples, try this one from Cornell, Google on that term for more.

From ref 1.

The following table gives the packing densities for the circle packings corresponding to the regular and semiregular plane tessellations (Williams 1979, p. 49).

tessellation	exact	approx.
		0.9069
		0.7854
		0.6046
		0.8418
		0.8418
		0.6802
		0.7773
		0.3907
		0.5390
		0.7290
		0.4860

Not sure what diagrams at bottom are supposed to show: if I interpret them as showing "the largest single circle that can nest inside a given circle", and then, "the largest equal pair which can nest...", "the largest equal trio...", etc., then I have a problem for the sixth. The central void for "the six largest equal-sized circles that can occupy a given circular area" would accommodate a seventh circle of that identical size. Thus, if you imagine the sequence of possible packings, the numbers will go 1, 2, 3, 4, 5, 7, 8, . . . [I haven't bothered to draw it out, but it seems obvious that if seven circles are tangent to the outer circumference, with each tangent to its two adjacent neighbors, there must exist a central void capable of holding an eighth circle. As the count increases, at some point, clearly, the interior void will become large enough to hold two, or possibly three, additional circles, and there will be another "skipped" cardinal number. This will repeat periodically as count increases.]

Comparing with the Cornell link, it appears that the sixth diagram here may represent a jammed packing of the outer circle; if I understand correctly, this construction isn't trying to determine the maximum number of occupying circles, but only under what conditions "jamming" will occur.

I was browsing the older posts and came across this…

Packing factor, I never heard of that term, but have used it, not knowing the term, problem is, having it symmetrical or a pattern isn’t always the best.

i recall one of the first jobs I had was making a counter flow tube sheet, to put as many tubes in it as possible, with not consideration to minimum spacing for stresses. The layout was done by hand, and there was little symmetry.

thats for the info.

Don't forget to include the width of cut (kerf) in the layout and calculations.

Unfortunately, there is no one simple formula for how many small circles.

The area of leftover metal is simply the area of the original circle minus the total area of all the smaller circles (again taking account of cut width).

In your example, I think you can get (27) 8mm circles out of a 50mm circle.

I was to say the same.

Additionally there might be a limit of material that is necessary between two cuts to allow for precision. Not sure this goes for laser cut but a mechanical blade would probably struggle.

"Don't forget to include the width of cut (kerf) in the layout and calculations."

Good point. I would suggest adding 1/2 kerf width to the radius of the inside circles, and to the principal circle, then doing the layout based on these two new sizes [(R₂ = R₁ + 0.5k), and (r₂ = r₁ + 0.5k). For the original question, R₁ = 25 mm, and r₁ = 4 mm, and "k" is kerf width]. My reasoning is this: wherever two circles are tangent, the kerfs will overlap, regardless of whether this is a meeting between two inner circles, or between an inner one and the principal circle.

For example, suppose I use a hole saw as my tool. For a circle tangent to the circumference of the original workpiece, the teeth can hang over the edge by exactly their own width, so that the inner FINISHED piece is tangent to the actual original circumference. For the second cut, the teeth can hang over the original edge, and over the edge of the first cut, making the second finished part tangent to both the principal and first inner circles. This should generally be true for any other tooling, unless it won't handle interrupted cuts.

The center of a first circle tangent to the outer edge will be located exactly where it would have been for a no-kerf process - but each additional circle will be spaced further from its inner-circle neighbors by the width of a single kerf.

This still does not provide a working formula, but makes layout trials simpler.

If some common circular object such as a coin or metal washer is available in sufficient quantity. let its diameter be scaled to 2(r₂). Now draw a circle at the same scale using R₂. If possible, cut it out of thin cardboard, foamboard, or other suitable material, and place the hollowed-out portion of the sheet on a flat surface. The coins can be placed inside in various arrangements until no more will fit, and their positions adjusted (if they are not all tangent to each other and/or the scaled principal circle) to see whether such rearrangement permits one or more additional coins to be added. A near-optimum layout should result fairly quickly, I would think. Tilting the underlying surface (think old-fashioned drafting board) wil let gravity help compact the set of coins within the cutout.

You can tackle the problem like this:

1) draw a centre cercle of 4mm radius.

2) You can now have (only) 6 similar cercles around it, all touching. This is true for any cercle diameter!

3) the next round will take 12 circles only.

4) then 18, then 24, the 30 etc, adding 6 each round.

Now, if you join all the centres of each layer, you will notice that they form hexagones, at each layer. The side of the hexagone, at each layer, is (Layer number x diameter of cercle), starting from 0, 1, 2 layer.. == 8mm; 16mm, 24mm ...

In your case, the maximum layer that will fit entirely will be the 2nd, which gives 19 small cercles.

But, by shifting them to the side, until the centre line touches the edge of the 25mm diam. cercle, you will have part of the 3rd layer, making 26 cercles in all.

As previously stated, the formula can be complicated (if any)

Let O.D.=D and Small circle diam =d

max number of layers = (D-d) / (2xd) = Nmax

To have the best OD that will contain the maximum of d circles, Nmax must equal an integer. Example: for D = 50 and d = 8, we have Nmax = 2.625, which means that we take the integer, 2 and the number of circles is: (∫layers)x6 +1= (1+2)x6+1 = 19 circles

But since Nmax is not an integer, then we can shift the D circle such that we can add more circles (need to work out a formula to relate the remnant to the extra circles).

But if you want the whole of the 3rd layer circles, then take Nmax = 3 and Work out the D, which will be just 56mm instead of the 50mm, to have (3+2+1)x6+1 = 37 circles maximum compact.

..OOOO
.OOOOO
OOOOOO
.OOOOO
..OOOO
...OOO

(27 for sure)

..OOOO
.OOOOO
OOOOOO
OOOOOO
.OOOOO
..OOOO

(30 maybe, but I don't think they quite fit, especially if kerfs are accounted for.)

This is the construction.

The blue circle is shifted to allow more circles to be added. The red circle is the 56mm circle that will take the 37 circles.