...Except that "gap" is a contiguous portion of space when spheres are dense-packed, rather than being discrete isolated sections like the two-dimensional analogue, so I'm not sure what you are asking for. Assuming you're asking for what I think you are asking for, then when taking a unit cell of a dense-packed sphere configuration, exactly (1 - π/3sqrt(2)) of that space is empty.
None.
in 3D i think, you can't fully enclose a volume with spheres, there will be holes opening to it. not sure about this one, i just thought as one progressively ads spheres there will be a window all the time to the previous area which has been attempted to close. So it can't be answered this way, if the above logic is true.
EDIT: i didn't see your post. anyway, you could elaborate what do you mean by smallest gap.
None.
Think regular tetrahedron.
None.
Think regular tetrahedron.
Tetrahedra do not tile space. The unit cell for sphere packing is a hexagonal prism.
EDIT> The fact that you're asking for the answer in terms of
r bothers me; I can define a "unit cell" differently and get a different answer. Either give us your definition of the unit cell, or ask for the fraction of space remaining rather than an answer in terms of r.
EDIT2>
http://www.staredit.net/?p=shoutbox&view=1514EDIT3> Because poison isn't willing to accept an obviously correct answer:
http://www.staredit.net/?p=shoutbox&view=1515
Post has been edited 3 time(s), last time on Nov 7 2011, 9:35 pm by Aristocrat.
None.
First we start with the volume of a tetrahedron. Which with sides 2r, would have a volume of 8r³/(6√2). This then simplifies to 4/(3√2) r³.
We then subtract the volume of the spheres inside of the tetrahedron. This may become slightly tricky with our current approach, but there is still a solution. To help visualize I'll employ my whiteboard.
So then, we simply find the volume of the tetrahedron and curved top.
The Tetrahedron-Our tetrahedron is fortunately regular, and so the volume may be found easily. The volume is then, quite easily, found to be r³/(6√2).
The Curved Top-This presents more of a challenge. Much more, actually, and probably best found with calculus. I'll leave it up to you guys (+20 minerals if you can do it without calculus).
None.
The Curved Top
-
This presents more of a challenge. Much more, actually, and probably best found with calculus. I'll leave it up to you guys (+20 minerals if you can do it without calculus).
But I already found it sans-calculus several posts back.
None.
You're on cocaine, or simply did not check the sites I linked.
From your link: 
From all of the others: 
Do they still match? I would have given you the minerals a while back, but since the second image came from multiple sites, I'm more willing to trust it than Wolfram. They match exactly, and when substituting 2r, yields exactly what I posted.
Just stop. If you are unwilling to part with your minerals, fine, but if you're going to call a correct answer incorrect to save face, that crosses the line.
None.
The reason we use Wolfram over other sources of math information is largely because Wolfram is far more complete, home to the computational engine Wolfram|Alpha, and creator of Wolfram Mathematica. For example, when defining local extrema, Wolfram is the only source I've seen to explicitly mention the use of a neighborhood.
If it disagreed with numbers of other sources I'd look very hard at what the issue is, because I find it hard to believe that the math gurus as Wolfram really made a mistake, and far more likely that I'm just overlooking something.
None.
