By the very definition of Waterman polyhedra, not necessarily all vertices are on the same sphere. However in here both the 8 cubical vertices and the other 24 vertices belong to the same radius.

The irregular hexagons {(h,H,H)²} have vertex angles h = arccos(-1/3) = 109.471221° resp. H = arccos[-1/sqrt(3)] = 125.264390°.

Incidence matrix according to Dynkin symbol

Qo3oo4xd&#zh   → height = 0, 
                 where Q = sqrt(8) = 2.828427 (pseudo)
                 and d = 3 (pseudo)
(tegum sum of (Q,x)-sirco and d-cube)

o.3o.4o.     | 24 * |  2  1 | 1  2
.o3.o4.o     |  * 8 |  0  3 | 0  3
-------------+------+-------+-----
.. .. x.     |  2 0 | 24  * | 1  1  x
oo3oo4oo&#h  |  1 1 |  * 24 | 0  2  h
-------------+------+-------+-----
.. o.4x.     |  4 0 |  4  0 | 6  *
Qo .. xd&#zh |  4 2 |  2  4 | * 12  {(h,H,H)2}