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}