Acronym ...
Name hyperbolic order 4 octagon tiling
 
  ©
Circumradius sqrt[-sqrt(2)]/2 = 0.594604 i
Vertex figure [84]
Dual x4o8o
Confer
combinatorical relatives:
x5o4o   {6}-halved x6o4o  
general polytopal classes:
regular   noble polytopes  
External
links
wikipedia  

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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