Cubic Tetra
Tetra quality is defined as Volume / max edge length **3 and is normalized such that the quality of the regular tetra is 1.0 . Two views of Sommerville 1-5 Tetras Sommerville 3 is a half of a pyramid. It can be divided to 8 smaller ones of the same type by "3 fold: split the longest edge". If you divide this tetra as you divide the cubic one, you get 8 tetras of 2 kinds of an average quality of 0.364 which is better than 0.25 . 2D: Every triangle can be devided to 4 smaller ones of the same type. 3D: There are at least 2 tetra, which can be devided to 8 smaller ones of the same type. 4D: Conjecture: There is no 4D simplex, which can be devided to 16 smaller ones of the same type. Is there any 4D simplex, which is bounded by 5 cubic tetras? Let P1 = (0,0,0,0) P2 = (1,0,0,0) P3 = (0.5,0.5 sqrt(2),0,0) P4 = (0.5,0,0.5 sqrt(2),0) define a 3D cubic tetra with 4th coordinate 0, let this tetra be the first of these bounds. Where is the 5th vertex? From symmetry reasons we try P5 = (0.5,0,0,0.5 sqrt(2))
The bounding tetras are T5 = P1-P2-P3-P4 T4 = P1-P2-P3-P5 T3 = P1-P2-P4-P5 T2 = P1-P3-P4-P5 T1 = P2-P3-P4-P5 edge sizes: P1 P2 = 1.0 P1 P3 = 0.5 sqrt(3) P1 P4 = 0.5 sqrt(3) P1 P5 = 0.5 sqrt(3) P2 P3 = 0.5 sqrt(3) P2 P4 = 0.5 sqrt(3) P2 P5 = 0.5 sqrt(3) P3 P4 = 1.0 P3 P5 = 1.0 P4 P5 = 1.0 T5, T4, T3 are cubic: 2 edges are 1.0 and 4 edges are 0.5 sqrt(3) T1, T2 cannot be cubic since 3 edges are 1.0 and 3 edges are 0.5 sqrt(3) => This P5 does not work. |