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Message: <blockquote data-quote="Woocash" data-source="post: 2093179" data-attributes="member: 29771">Fieari is right. It is not easy to "convert" a polyhedron from 3d to 4d. There is no way to do so with most of platonic polyhedrons (there are no 4d objects of corresponding regularity).The only dices we can make in 4d (or even more dimensions) without problems are d4 (regular simplex) and d6 (cube).n-dimensional simplex has (n+1) sides, each of them being (n-1)-dimensional simplex (3-dim d4 has 4 sides being triangles, a triangle has 3 sides being lines; in higher dimensions it works the same). n-dimensional cube has 2n sides ("top" and "bottom" in each dimension), each being (n-1)-dimensional cube (d6 has 6 sides being squares, a square has 4 sides being lines).Edit: I just found complete formulas for a simplex and a cube in n dimensions.n-dimensional simplex has (n+1)!/(k+1)!(n-k)! k-dimensional components (simplexes).ie. 3-dim d4 has (3+1)!/(2+1)!(3-2)! = 24/6 = 4 2-dim (triangle) sides, (3+1)!/(1+1)!(3-1)! = 24/(2*2) = 6 1-dim (line) edges and (3+1)!/(0+1)!(3-0)! = 24/6 = 4 0-dim (point) vortices.n-dimensional cube has (n!/k!(n-k)!)*2^(n-k) k-dimensional components (cubes).ie. 3-dim d6 has (3!/2!(3-2)!)*2^(3-2) = (6/2)*2 = 6 2-dim (square) sides, (3!/1!(3-1)!)*2^(3-1) = (6/2)*4 = 12 1-dim (line) edges and (3!/0!(3-0)!)*2^(3-0) = (6/6)*8 = 8 0-dim (point) vortices. Both above formulas are quite easy to prove, but I don't want to use too much mathematical language in this post <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":D" title="Big grin    :D"  data-smilie="8"data-shortname=":D" />.</blockquote>

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