Inscribed sphere - Kepler Conjecture

AI Thread Summary
The discussion focuses on understanding the relationship between the radii of equal spheres in close packing and the largest inscribed sphere that can fit within the voids created by these spheres. The first radius, r1, represents the radius of the equal spheres, while the second radius, r2, pertains to the largest inscribed sphere that can fit within the arrangement. Additionally, there is interest in determining the smallest sphere, r3, that can fit within the packing and touch three of the larger spheres. A suggested approach involves drawing lines from the centers of the spheres to calculate the necessary dimensions. This topic invites further exploration and ideas from the community.
Berea81
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Newbie to the forum here. Hoping y'all can help with something that's been bugging me for a while now.

I would like to know the relationship between two characteristic radii in a close packing of equal spheres. The first radius of interest is that of the equal sphere's themselves (r1). The second radius (r2) is that of the largest inscribed sphere which would fit inside the void space created between the equal spheres of radius r1. Or as a friend put it what's the biggest (spherical) grape you could fit inside a pyramid of oranges without disturbing the pyramid.

I'm also interested in the smallest 'grape' (r3) that would fit within the close packing but be in contact with three different 'oranges'.

Ideas?
 
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Welcome to PF!

Hi Berea81! Welcome to PF!*:smile:

(try using the X2 button just above the Reply box :wink:)
Berea81 said:
… what's the biggest (spherical) grape you could fit inside a pyramid of oranges without disturbing the pyramid.

I'm also interested in the smallest 'grape' (r3) that would fit within the close packing but be in contact with three different 'oranges'.

I think the best approach would be to draw the lines joining the centre of each small sphere to the centre of each large sphere that it touches.

So each line would have length r1 + r2, and if you know the layout of the large spheres, it should be easy to find that length. :wink:
 
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