If you start with a unit square x4o, and consider its sections vertex first, you'll start with that vertex, i.e. a point, which itself is nothing but an edge of size zero: o. Cutting deeper, that zero sized edge becomes larger and larger until it reaches the size of the diagonal of that square, i.e. sqrt(2) times the original edge of that square. This is what Wendy usually labels q. After that stage the sectioning line decreases to zero again. Thus our square is something like the sequence o => q => o. Therefore the square also could be described as a lace tower oqo&#xt : 3 layers, described by the individual node positions, those layers furthermore additionally (&) being laced (#), and that lacing edge would have metrically the size x. t then just marks that the sequence is ment as a tower. Or as a stack of segmentotopes it also could be described as o || q || o.

Next take a cube. The vertex-first axial subsymmetry here is triangular. thus it ought be described by symmetry o3o, where those nodes might get edges of varying length, depending on the depth of sectioning. In the first step we start at the vertex itself. Thus the edge size of our sectioning triangle will be zero. Accordingly we have o3o here. (Note that this symbol now is ment metrically exact, not just describing the mere symmetry group as such.) Thereafter the triangle increases in size, again until we reach the diagonals of the vertex incident squares. Thus our final shape at this stage then would be q3o. After that stage the size of the sectioning line of the still intersected squares becomes smaller again, while now inbetween in the now opening gaps, the opposite squares become interseced as well, and that sectioning line size here increases by the same amount as the former would decrease. Thus the general section in that realm would be (q-a)3a, for o < a < q. Esp. for a=q we obviously get o3q, the dual triangle. Thereafter the former squares are no longer intersected, and the sectioning line size of the latter squares now decreases again. Thus our sectioning sequence here is o3o => q3o => o3q => o3o. I.e. we could rewrite the cube as the lace tower oqoo3ooqo&#xt. Or as a stack of segmentotopes you'd have o3o || q3o || o3q || o3o.

Note furthermore, that this sectioning sequence would mark the vertex layers perpendicular to the axial direction of movement of our sectioning plane. Here we have exactly 4 such layers. The medial stage, i.e. cutting the cube right in the middle of this axial orientation, would not hit any vertices of the cube. Rather it would just intersect the edges only. From what was mentioned before that medial stage would be described by a regular hexagon y3y, where this edge size here would be derived from y = q-a = a, i.e. y = q/2.

Next consider the tesseract, vertex first. Then our axial subsymmetry would be o3o3o (tetrahedral). Our sections would be o3o3o (metrically exact), then something like a3o3o for o < a < q, then q3o3o (exact, next vertex layer), then (q-a)3a3o, then o3q3o (exact, next vertex layer), then o3(q-a)3a, then o3o3q (exact, next vertex layer), then o3o3(q-a), and finally o3o3o (exact, opposite vertex). Thus our sectioning sequence here is o3o3o => q3o3o => o3q3o => o3o3q => o3o3o. Lace tower here is oqooo3ooqoo3oooqo&#xt, segmentotopal description is o3o3o || q3o3o || o3q3o || o3o3q || o3o3o. - Btw, we could count the vertices in any of those layers individually. Then we get 1 (vertex) + 4 (tetrahedron) + 6 (octahedron) + 4 (dual tetrahedron) + 1 (opposite vertex) = 16. Thus we get the total count of the tesseract's vertices again.

Going still one dimension up, where we finally would reach your question, i.e. the vertex first sectioning sequence of the penteract, then we'd find accordingly: Subsymmetry is o3o3o3o (pentachoral). Thus o3o3o3o (metrically exact), then something described by a3o3o3o for o < a < q, then q3o3o3o (exact), then (q-a)3a3o3o, then o3q3o3o (exact), then o3(q-a)3a3o, then o3o3q3o (exact), then o3o3(q-a)3a, then o3o3o3q (exact), then o3o3o3(q-a), and finally o3o3o3o (exact: opposite vertex). Thus our section of vertex layers is o3o3o3o => q3o3o3o => o3q3o3o => o3o3q3o => o3o3o3q => o3o3o3o. The lace tower thus is oqoooo3ooqooo3oooqoo3ooooqo&#xt or the segmentopal stack description is o3o3o3o || q3o3o3o || o3q3o3o || o3o3q3o || o3o3o3q || o3o3o3o.

So what then would be the medial section here? Again this one here is not a true vertex layer. Rather it is again midway on some of the edges of the penteract. In fact it is that stage of o3(q-a)3a3o, where again y = q-a = a, i.e. y = q/2. Thus we also could write here o3y3y3o. This (up to scaling) is nothing but the bitruncated pentachoron, which also is known as decachoron. The incidence matrix of this figure is

- Code: Select all
`o3x3x3o`

. . . . | 30 | 2 2 | 1 4 1 | 2 2

--------+----+-------+----------+----

. x . . | 2 | 30 * | 1 2 0 | 2 1

. . x . | 2 | * 30 | 0 2 1 | 1 2

--------+----+-------+----------+----

o3x . . | 3 | 3 0 | 10 * * | 2 0

. x3x . | 6 | 3 3 | * 20 * | 1 1

. . x3o | 3 | 0 3 | * * 10 | 0 2

--------+----+-------+----------+----

o3x3x . | 12 | 12 6 | 4 4 0 | 5 *

. x3x3o | 12 | 6 12 | 0 4 4 | * 5

which shows that all its facets (cells) would be 10 identical truncated tetrahedra. Metrically exact we should have replaced each occurance of x by that afore defined y. The triangles here stem from sectioning a cube midway of the 3 vertex incident edges, and the hexagons stem from cubes being cut exactly halfway in their axial direction.

--- rk