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For those that want to EVENLY cover a sphere where the protrusions are
EQUI-distant from their neighbours, I suspect the solutions will be
derived from the Euclidean solids with valid 3D solutions at 4,6,8,12,20
as well as the trivial 2D solutions for 1,2 and 3.
It shouldn't be too hard to represent these solutions are an array of
points in spherical coordinates as the r and longitude are dead
easy to work out. The latitude will take a bit of work if you want an
exact solution
It may be possible to combine staggered patterns of one type of these
to achieve higher densities but I can't imagine these being truly
equi-distant as are the primitives.
I will see if I can develop this into a full solution when I have some spare time.
I had posed a question a few months back on how to achieve this perfect dimple spacing on a sphere "golf ball like" and the one thing I discovered is that its no different than a soccer ball meaning it should have a hexagon surrounded by pentagons i think thats why there are those small dimples left over
just curious what are your dimensions on this dimple dia vs. over all dia
I think I will pose this question to an old professor and see if he has time to give me a formula that can be plugged into pro
maybe an excel file imported into the datum point offset coordinate system
The golf ball is quite different to a soccer ball.
The golf ball is based on the related geometry of dodecahedra (12 faces/20 vertices) and icosahedra (20 faces/12 vertices).
In a typical soccer ball EVERY hexagon is surrounded by alternating
pentagons and hexagons. Your sketch shows there are 5 hexagons and 1
pentagon surrounding one hexagon.
By the way small dimples are 2,5mm surrounded by 3,8 dimples and
linearly work their way up to a 4,1 dimple in the centre of the 3 small
dimples and I found I didn't need to use ANY datum points or datums curves etc in the solution.
Do real golf balls have different size dimples? How many dimples are there in total? Coming up with an equation for equal spacing on N dimples will be a real challange, I believe. I don't think electrons actually do this, aren't they in different shell levels with only about 12 max at any one level?
Very old golf balls have regular dimples on a relatively open
octahedral/cubic pattern. More dimples grip the air better for backspin
so newer balls use different techniques to maximise the coverage and
dodge each other's patents.
Yes the dimple size does vary and the variation and patterning does
vary between manufacturers which sort of proves there is no absolute
solution. A few years ago Wilson brought out a ball with square pyramid
shaped dimples and I have also seen oval dimples oriented especially to
fit the gaps.
The Titleist ball I based my model on is not truly icosahedral. The
"vertices" are slightly biased towards the poles to allow an extra band
of dimples which doesn't look quite as tidy and avoids having dimples
on the part line yielding 392 dimples over my 362. However an "ULTRA
competition" ball I have on my desk appears to be truly icosahedral
because the fill pattern they have used in the triangles puts a natural
parting along the equator
Pairs of electrons fill s,p,d and f subshells according to the
Shroedinger wave equation. Only the s sub-shell is "spherical". From
ancient memory, the p shells look like the infinity symbol though a
doughnut and the d shells look like 3 infinity symbols mutually
orthogonal to each other so the analogy could hardly be worse.
As mentioned previously I expect the only exact solutions are the spherical forms of the Platonic (not Euclid) solids
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