Give ceptimus' arrangement of the Booromean rings with 60 balls around the circumference of each ring and assuming a ring radius of 1, what is the distance between the centers of the rings?
Exact number not required, some expression involving PI will do.
Edit: ring radius, 1, taken to be the radius of that circle that runs through the centers of all the balls.
Give ceptimus' arrangement of the Booromean rings with 60 balls around the circumference of each ring and assuming a ring radius of 1, what is the distance between the centers of the rings?
Exact number not required, some expression involving PI will do.
Edit: ring radius, 1, taken to be the radius of that circle that runs through the centers of all the balls.
With ballsPerRing = 60, all the overlapBalls in the range 11 to 20 give valid arrangements with perfect geometry: see summary image below (click image for large view).
An overlap of 10 balls results in all three rings crossing at one point, but the 'approach' angle of the balls is too acute and the adjacent balls to the crossing ones interfere.
An overlap of 21 balls goes too far the opposite way, and again the adjacent balls to the crossing ones interfere.
An overlap of 14 or 15 balls gives the nearest to 90 degree crossings - but as previously mentioned the arrangements with an odd overlap like 15 mean that the balls fall between the minute markers rather than on them as with the even overlaps.
For smaller number of balls per ring, 12 balls per ring with an overlap of 3 balls is (I think) the smallest one that works reasonably well.
A sweet spot that gives close-to-right-angle crossings is 18 balls per ring with an overlap of 5 (picture below).
Here's the parametric code if you wish to experiment. Just install OpenSCAD on your PC (it's free), paste the code below into the editor then tweak the three numbers in the borromean(); call and press the F5 key to render. In the code below the (6, 18, 5) means 6mm diameter balls with 18 balls per ring and an overlap of 5 balls.
I've not proved it mathematically, but with OpenScad you can zoom right in till you are looking at thousandths of a balls radius and you would see any discrepancy. I can't see the geometry working out to be that close by chance but off in the tenth decimal place. It's interesting that balls per ring that are multiples of six seem to work best which I guess is something to do with how equilateral triangles come together to form a hexagon. I've not investigated yet whether other numbers of balls per ring can be made to work.
Still, I'll be looking for some kind of actual proof. Believe it or not I don't have any bits of paper bigger enough to scribble any working out on, here in our holiday retreat.
I think we all guessed the multiple of 3 was significant here. Be great if you could find some solutions that are not multiples of 3.
I've now tried various other numbers without success. It looks like ballsPerRing has to be a multiple of 6, not just 3. So 6 bpr works in a degenerate case where all the balls in each ring all touch; 12 bpr, 18 bpr and so on all work as we've seen - but I can't make 9 bpr or 15 bpr or 21 bpr, etc. work - you can get the balls in any two of the three rings to mesh, but then the third ring is always offset by half a ball with respect to one ring when the other one is meshed. I've tried other numbers that don't have six as a factor: ten, powers of 2, prime numbers, 3 * 5 * 7 and so on - none of them seem to come close to working.
It might be possible if you're prepared to have the third ring with a different offset from the first two rings than those first two have to each other - and it's obviously possible if you're prepared to put some wiggles into the rings rather than have them as pure circles - but that wouldn't be as interesting or pretty anyway.
Those "precision" American-made marbles arrived today. They exhibit the same diameter variations and lack of sphericity as the first set. Dang it!
-Phil
One task in that film reminded me of a summer job I once had as a QC tech at Johns Manville. They made fiberglass insulation. Molten glass continuously streamed into a "spinner." The spinner was like a dish with vertical sides. The sides were peppered with tiny holes. When the glass hit the spinning spinner, it was flung through the holes, creating the thin strands we associate with fiberglass. The strands were then blown onto a slowly-moving conveyor, forming a mat. The thickness and density of the mat were determined by the air pressure blowing the fibers down and the speed of the conveyor. Downstream, the mat was sprayed with a binder to keep the strands together. Fiberglass destined for home insulation received a light spray. Insulation destined for refrigerators got a lot more. In fact, refrigerator insulation resembled a stiff board, it had so much binder applied.
Anyway, my job was to stick a long-handled crucible under the glass stream going into the spinners to extract a sample. I had to be quick, lest the spinner cool and the holes clog with solid glass. Once the sample was extracted, I set the crucible down onto the metal deck and stuck a glass rod into it and drew a long strand of glass from it, perhaps ten feet long -- just like in the video. The strands were then cut into short lengths and tested in the lab for tensile strength, among other things.
It was a fun job -- probably the best I had during my college years. 'Even got to do some programming for their engineering department over a timeshare terminal (ASR33) in BASIC.
I've been taking a glass fusion class, mostly for art but also with a notion of embedding copper wire and optics for electro-optical purposes.
About marbles, I wonder if marbles of different colors could be arranged together and fused. The important thing about glass fusion is that all of the glass that is mixed has to have the same coefficient of expansion or else it will develop stress cracks as it cools. Matched glass of different colors is rather expensive.
I was thinking what might be done with a strictly honeybee comb arrangement...
From an aesthetic standpoint, I love the honeybee comb! From a mechanical standpoint, getting those marbles to flow around the corners might be a chore. And I do see some interferences, unless the marbles are considerably undersized.
All my life I have been told that π is an irrational number. That is to say, cannot be expressed as a simple fraction p/q where p and q are integers.
Never thought so much about it. Maths dudes tell be it is so, who am I to argue? But now I find that proving that π is irrational is not so straight forward. In fact it's damn hard:
Here is one way:
Here is another:
Meanwhile, the maths professor Norman Wildberger does not like any of this irrational numbers stuff. It's all fiction as far as he is concerned:
Grief, all this just because of some marbles running around circles....
Comments
I was thinking of doing the same kind of thing with some SVG and Javascript.
Give ceptimus' arrangement of the Booromean rings with 60 balls around the circumference of each ring and assuming a ring radius of 1, what is the distance between the centers of the rings?
Exact number not required, some expression involving PI will do.
Edit: ring radius, 1, taken to be the radius of that circle that runs through the centers of all the balls.
My guess is...
distance_between_centers = 2r - difference_from_overlap
difference_from_overlap = 2(r - cos(angle))
angle = 2pi * ratio_of_half_arc_to_circumference
ratio_of_half_arc_to_circumference = 8 / 60 (based on second arrangement by ceptimus)
Assuming r = 1, combined:
distance_between_centers = 2 - 2(1 - cos(2pi * 8/60))
simplified:
distance_between_centers = 2cos(4pi/15)
Given this, I get a distance of ~1.34.
Mind you. I'm not totally convinced yet that either arrangement is a perfect fit. The first one seems more likely.
Actually that is what I'm wanting to prove. That balls that should be at coincident points on the intersection are. Exactly.
given a ball radius and number of balls per ring then the code for the radius of a ring is: Then if we say how many balls should exist in the overlap between two rings we get the code for the ring centre distance separation If you take that offset in the x direction, with zero y offset for the second ring, then the x and y offsets for the third ring are
With ballsPerRing = 60, all the overlapBalls in the range 11 to 20 give valid arrangements with perfect geometry: see summary image below (click image for large view).
An overlap of 10 balls results in all three rings crossing at one point, but the 'approach' angle of the balls is too acute and the adjacent balls to the crossing ones interfere.
An overlap of 21 balls goes too far the opposite way, and again the adjacent balls to the crossing ones interfere.
An overlap of 14 or 15 balls gives the nearest to 90 degree crossings - but as previously mentioned the arrangements with an odd overlap like 15 mean that the balls fall between the minute markers rather than on them as with the even overlaps.
A sweet spot that gives close-to-right-angle crossings is 18 balls per ring with an overlap of 5 (picture below).
Here's the parametric code if you wish to experiment. Just install OpenSCAD on your PC (it's free), paste the code below into the editor then tweak the three numbers in the borromean(); call and press the F5 key to render. In the code below the (6, 18, 5) means 6mm diameter balls with 18 balls per ring and an overlap of 5 balls.
colour1 = [1, 0, 0]; colour2 = [0, 1, 0]; colour3 = [0, 0, 1]; borromean(6, 18, 5); module borromean(ballDia, ballsPerRing, overlapBalls) { ballRad = ballDia / 2; stepAngle = 360 / ballsPerRing; ringRadius = ballRad / sin(stepAngle / 2); ringCentreDistance = 2 * ringRadius * cos(overlapBalls / ballsPerRing * 180); startAngle = overlapBalls % 2 ? stepAngle / 2 : 0; ring(ballDia, startAngle, stepAngle, ringRadius, colour1); translate([ringCentreDistance, 0, 0]) ring(ballDia, startAngle, stepAngle, ringRadius, colour2); translate([ringCentreDistance / 2, ringCentreDistance * sqrt(3) / 2, 0]) ring(ballDia, startAngle, stepAngle, ringRadius, colour3); } module ring(ballDia, startAngle, stepAngle, ringRadius, colour) { color(colour) for (angle = [startAngle : stepAngle : 360 + startAngle]) rotate([0, 0, angle]) translate([ringRadius, 0, 0]) sphere(d = ballDia, $fn = 30); }Well I never.
At least my center distance formula is the same as your :
ringCentreDistance = 2 * ringRadius * cos(180 * overlapBalls / ballsPerRing);
That is a great set of examples. Thanks.
I'm partial to the 20/60 solution for it's symmetry and visual balance. Or the 17/60 where the rings cross at more of a right angle.
These equations don't yet prove to me that the fit is exact.
Still, I'll be looking for some kind of actual proof. Believe it or not I don't have any bits of paper bigger enough to scribble any working out on, here in our holiday retreat.
I think we all guessed the multiple of 3 was significant here. Be great if you could find some solutions that are not multiples of 3.
It might be possible if you're prepared to have the third ring with a different offset from the first two rings than those first two have to each other - and it's obviously possible if you're prepared to put some wiggles into the rings rather than have them as pure circles - but that wouldn't be as interesting or pretty anyway.
No, I agree, no wiggles. No, it all most fits.
I was curious how marbles are made.
[url="https://www.google.com/search?q=how+do+they+make+marbles&ie=UTF-8&oe=UTF-8&hl=en-us&client=safari#kpvalbx=1
"]Here's one video[/url]. For plain ones an extrusion of molten glass is chopped into 1/2 inch slugs and rounded out in an ingenious spiral conveyer. Then sorted. No wonder there are variations. The video goes on to show hand forming of art marbles.
One task in that film reminded me of a summer job I once had as a QC tech at Johns Manville. They made fiberglass insulation. Molten glass continuously streamed into a "spinner." The spinner was like a dish with vertical sides. The sides were peppered with tiny holes. When the glass hit the spinning spinner, it was flung through the holes, creating the thin strands we associate with fiberglass. The strands were then blown onto a slowly-moving conveyor, forming a mat. The thickness and density of the mat were determined by the air pressure blowing the fibers down and the speed of the conveyor. Downstream, the mat was sprayed with a binder to keep the strands together. Fiberglass destined for home insulation received a light spray. Insulation destined for refrigerators got a lot more. In fact, refrigerator insulation resembled a stiff board, it had so much binder applied.
Anyway, my job was to stick a long-handled crucible under the glass stream going into the spinners to extract a sample. I had to be quick, lest the spinner cool and the holes clog with solid glass. Once the sample was extracted, I set the crucible down onto the metal deck and stuck a glass rod into it and drew a long strand of glass from it, perhaps ten feet long -- just like in the video. The strands were then cut into short lengths and tested in the lab for tensile strength, among other things.
It was a fun job -- probably the best I had during my college years. 'Even got to do some programming for their engineering department over a timeshare terminal (ASR33) in BASIC.
-Phil
About marbles, I wonder if marbles of different colors could be arranged together and fused. The important thing about glass fusion is that all of the glass that is mixed has to have the same coefficient of expansion or else it will develop stress cracks as it cools. Matched glass of different colors is rather expensive.
I was thinking what might be done with a strictly honeybee comb arrangement...
-Phil
Never thought so much about it. Maths dudes tell be it is so, who am I to argue? But now I find that proving that π is irrational is not so straight forward. In fact it's damn hard:
Here is one way:
Here is another:
Meanwhile, the maths professor Norman Wildberger does not like any of this irrational numbers stuff. It's all fiction as far as he is concerned:
Grief, all this just because of some marbles running around circles....