Let the radius of the sphere be r, then a=2r, and there are two hexagonal cells, so V- sphere = 8/3 π r 3. Now the key problem is to find the h of hexagonal cells, because the sphere with vertex angle 2 in tetrahedron is located at 2/h, and now the problem is transformed into finding the height of tetrahedron. ?
According to the solid geometry knowledge of mathematics, the height of tetrahedron is equal to 2/3 of the diagonal of the object. H=2√6r/3 Then h=4√6r/3 For hexagonal cells, S = 2 √ 3r 2v for hexagonal cells = SH = 8 √ 2r 3 for V-ball /V hexagonal cells =74.05%, and the available utilization rate is equal to 74.05%.
Extended data:
Simple materials with the densest face-centered cubic accumulation is:
1, calcium, strontium
2, aluminum, lead
3, neon, argon, krypton, xenon
4. Nickel, copper, rhodium, palladium, silver, iridium, platinum and gold
5, cerium, ytterbium, actinium, thorium
Baidu Encyclopedia-the densest accumulation of face-centered cubes