bobhexa Posted November 27, 2007 Share Posted November 27, 2007 [ATTACH]1664[/ATTACH] As a practising geometist I was intrigued by the attached diagram which I drew a long while ago. I endevoured to draw eight circles around a larger circle so that all the circumferences touched. By rough measurement it appeared that the ratio of the diameters of the smaller circles to the diameter of the larger centre circle was very close to the Golden Section. A good friend and mathematician advised me that it was close but no coconuts were to be won. Well I have since wondered about the fact that it was so close. I have expanded my wondering to go 3 dimensional and to consider spheres instead of circles. So the question is gentlemen would spheres which are manufactured to have their diameters exactly a golden section ratio of the central sphere be able to pack around this central sphere and make an exact fit. Link to comment Share on other sites More sharing options...

Bignose Posted November 28, 2007 Share Posted November 28, 2007 The closest packing circles of equal radius can have is [math]\frac{\pi}{\sqrt{12}}[/math]. The closest packing spheres can have is [math]\frac{\pi}{\sqrt{18}}[/math]. You might really like the article "Cannonballs and Honeycombs" by Hales in Notices of the AMS, vol 47, 2000, because it has a neat discussion of these issues and other shapes being close packed. Link to comment Share on other sites More sharing options...

bobhexa Posted November 28, 2007 Author Share Posted November 28, 2007 I thoroughly enjoyed your reference to Thomas Hales .Tis a very good informative piece of work. Absdorbing reading. I gradually understood that he was talking mainly about similar sized circles or similar sized spheres. My , I think, bottom line question is as follows. Supposing that you have a single sphere of 100 mm diameter and you have a plurality of smaller spheres with a diameter of only 62 mm (Golden Section ratio).....are you able to to distribute the smaller spheres around the surface area of the larger sphere and maintain perfect contacts between all neighbouring spheres. Is it a perfect fit? Having pondered this, I am led to the conclusion that a c60 pattern is the answer.You guys are the mathematicians. Do the ratios of the diameters of the larger sphere and the diameter of the 20 smaller spheres come anywhere close to the golden section ratio? Link to comment Share on other sites More sharing options...

bobhexa Posted December 10, 2007 Author Share Posted December 10, 2007 My attached diagram doesnt seem to be valid so I shall upload the image again. Using the C60 Buckyball as a reference, I joined 20 polystyrene foam balls together with toothpicks. In other words each of the twenty hexagons of the C60 were replaced with spheres. Surprisingly the overall appearance is of a pentagonal dodecahedron. Could it be that the truncated icosahedron can also be described as a truncated pentagonal dodecahedron? Could somebody please work out the math as to what the diameter of the inner sphere might be if the smaller surrounding spheres had a diameter of 100mm. Thanks in anticipation. Link to comment Share on other sites More sharing options...

bobhexa Posted December 20, 2007 Author Share Posted December 20, 2007 In the interim I have built a large model of a buckyball and obtained fairly precise measurements of the internal diameter...that is from the centres of the diametrically opposed hexagons. It would seem that the ratio of the diameter of the hexagon(flat to flat)to the inner sphere is certainly in the ballpark of the golden section. It is within hundredths of an inch. I am still wrestling with a method to calculate this mathematically. Any ideas Gentlemen. Mele Kalikimaka to all. Link to comment Share on other sites More sharing options...

bobhexa Posted December 30, 2007 Author Share Posted December 30, 2007 In order to clarify I made up this model of 20 foam balls. They coincidentally fitted perfectly into this buckyball framework toy. Question....What would be the diameter of a sphere that fitted perfectly inside this model if the smaller surrounding spheres had a diameter of 100mm? Link to comment Share on other sites More sharing options...

bobhexa Posted March 30, 2008 Author Share Posted March 30, 2008 Is my description too ambiguous or is this a really difficult computation. I would be so grateful if somebody would please try to calculate this out. Aloha Bobhexa Link to comment Share on other sites More sharing options...

cjohnso0 Posted April 12, 2008 Share Posted April 12, 2008 Well, on first try I came up with an maximum inner sphere size of ~180.2517mm. Basic methodology was as follows, refer to this page (Figures 2 and 2A) http://www.kjmaclean.com/Geometry/dodecahedron.html I took the 100mm sphere diameter and set this equal to the length of one of the dodecahedron sides. Using this, you can find the length of the sides of the cube which fits perfectly into the dodecahedron. I came up with cube side length of 50/cos 72. Next, I figured that the tetrahedron inside the cube would have the closest packed 4 spheres. So I found the length of it's sides, which came out to be 50*sqrt(2)/cos72. Using this, the radius of a sphere circumscribing the tetrahedron is r=sqrt(6)/4 * SideLength. Subtract 2 times the radius of the balls, or 100mm and I got my answer. Final equation was (hopefully) D = 2*(50*2*sqrt(3))/(4*cos 72) - 100 which was ~180.2517 Let me know if my method is way to confusing to read. Not that I'm 100% sure i'm even close.... Link to comment Share on other sites More sharing options...

bobhexa Posted April 14, 2008 Author Share Posted April 14, 2008 Dear cjohnso0 Hi Aloha and thank you for your reply. The Kenneth MacLean site is a golden referral...I must buy his book. I was with you all the way through to the finding the length of the side of the cube. It was at that point I got lost but I realised :doh:that the diagonal of that cube coincides with the diametrically opposed spheres of my model dodecahedron.(see earlier Attachment) And then as you say subtract from this diagonal length twice the radius of the balls, or 100mm . Does that give us the same answer? Thanks again Link to comment Share on other sites More sharing options...

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