I know the kissing number of a sphere in 3d space is 12, but I've also heard mention that there's a significant ammount of space left over, using a sphere of a whole fractional radius of the unit spheres, what is the largest non-whole sphere which plugs the void where a 13th whole sphere can't fit?

Thank you.

Because the kissing number is larger than 12 but smaller than 13, if we define the kissing number as the number of unit spheres that pack around it so that there is no free space left over.

Which is a slightly different definition, but I need it for a project I'm working on.

Basically the largest sphere possible is radius 1:1 and you should use the largest sphere when possible, unless you must use a smaller sphere, in which case use the largest smaller sphere possible.

That's my question/challenge.

Thank you, I couldn't fine how much empty space is left over, and I know that it is arguably infinite (bc touching spheres fractally have infinite gap between the tangent points) but I also know that what I really want to know is how much space is practically left over, in spheres.

So following the rules I laid out does anyone have a good answer? What's the largest single sphere to plug the gap, so that there are 13 spheres, 12 identical spheres and a smaller sphere, with no space to move.

Sorry if my question is unclear, I am available to clarify it

  • 1
    $\begingroup$ The 12-sphere packing around a single sphere is tight. The "excess space" comes about because spheres can only touch each other at isolated points, so they cannot fill space. Allowing a second radius of sphere raises the kissing number not to $13$ but to $32$, as you can add the same small sphere in the middle of each tetrahedron formed by the middle sphere and any three adjacent spheres on the outside. There are $20$ such groupings (sides of an icosahedron), and they are all Identical. Add a third size of sphere and you increase the kissing number by another $60$ (I think). $\endgroup$ Dec 26, 2021 at 15:31


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.