Acronym | socat |
Name | hyperbolic order 4 octagon tiling |
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Circumradius | sqrt[-1/sqrt(8)] = 0.594604 i |
Vertex figure | [84] |
Dual | x4o8o |
Confer |
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Note that this tiling allows for a consistent quartering of octagons (connecting opposite edge midpoints), such as to derive a combinatorical variant of x5o4o. While for that one the edge lengths of the regular pentagons a-a-a-a-a- can be calculated as given below for comparison, in that decomposition the edge lengths of these non-regular pentagons c-c-b-a-b- would calculate (using the formulas provided here) as:
x8o4o (quartering each octagon gives a topological variant of x5o4o): P0 o-+ P1 P1 / | P0 o | P1 +-----+ P2 edges a: hypdist(P0,P0) = 2 arcosh[cos(π/8)/sin(π/4)] = 1.528571 edges b: hypdist(P0,P1) = arcosh[cos(π/8)/sin(π/4)] = 0.764285 edges c: hypdist(P1,P2) = arcosh[cos(π/4)/sin(π/8)] = 1.224226 deformed pentagons c-c-b-a-b- (to be conpared to: x5o4o all edges alike: hypdist(P0,P0) = 2 arcosh[cos(π/5)/sin(π/4)] = 1.061275 regular pentagons a-a-a-a-a- )
There exists also a series of regular modwraps of this tiling, obtained by identifying every k-th vertex on each hole (q.e. straight edge sequence here). Then it allows a representation as (possibly) infinite regular skew polyhedron, which happens to be the facial subset of octagons of the spherical tetracontoctachoron (k=3: finite modwrap), of the hyperbolic o4x4x4o tesselation (k=4), etc. This series of modwraps generally is being denoted as x8o4o|k (with according value of k).
Incidence matrix according to Dynkin symbol
o4o8x (N → ∞) . . . | 2N | 4 | 4 ------+----+----+-- . . x | 2 | 4N | 2 ------+----+----+-- . o8x | 8 | 8 | N snubbed forms: o4o8s
o8x8o (N → ∞) . . . | 4N | 4 | 2 2 ------+----+----+---- . x . | 2 | 8N | 1 1 ------+----+----+---- o8x . | 8 | 8 | N * . x8o | 8 | 8 | * N snubbed forms: o8s8o
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