What do Orthoscheme Tilings Look Like?

I always meant to find out. A first step it to make sure the outer tetrahedron and smaller tetrahedra have the same orientation.
In[]:=
ortho1=IntegerDigits[#,10,3]&/@#&/@{{020,111,121,022},{022,111,112,222},{022,111,121,222},{022,113,112,222},{022,113,123,024},{022,113,123,222},{111,202,212,113},{111,222,212,113}};
In[]:=
big1={{0,2,4},{2,0,2},{2,2,2},{0,2,0}};
In[]:=
Table[EuclideanDistance@@#&/@Subsets[ortho1[[k]],{2}],{k,1,8}]
Out[]=

3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2
,
3
,
2
,2,1,
3
,
2

Everything is properly oriented.
In[]:=
(EuclideanDistance@@#&/@Subsets[big1,{2}])/2
Out[]=

3
,
2
,2,1,
3
,
2

Now we can use a barycentric coordinate mapping:
In[]:=
bary1=Partition[ResourceFunction["BarycentricCoordinates"][big1,#]&/@Flatten[ortho1,1],4];
newtet1=Flatten[Table[#.ortho1[[j]]&/@bary1[[k]],{j,1,8},{k,1,8}],1];
From that, we get the orthoscheme tiling:
In[]:=
ResourceFunction["SpinShow"][Graphics3D[{Opacity[.6],Tetrahedron/@newtet1},Boxed->False,SphericalRegion->True]]
Out[]=
In[]:=
Graphics3D[{Opacity[.6],Tetrahedron/@newtet1},Boxed->False,SphericalRegion->True]
The other orthoscheme is similar:
ortho2=IntegerDigits[#,10,3]&/@#&/@{{002,022,111,113},{022,042,131,133},{022,222,111,113},{022,222,111,131},{022,222,113,133},{022,222,131,133},{111,131,220,222},{113,133,222,224}};​​big2={{0,0,2},{0,4,2},{2,2,0},{2,2,4}};​​bary2=Partition[ResourceFunction["BarycentricCoordinates"][big2,#]&/@Flatten[ortho2,1],4];​​newtet2=Flatten[Table[#.ortho2[[j]]&/@bary2[[k]],{j,1,8},{k,1,8}],1];​​Graphics3D[{Opacity[.6],Tetrahedron/@newtet2},Boxed->False,SphericalRegion->True]
Out[]=