I think you got the right feeling, quickfur, both with the relation of the 2 specific finds (according to
my second idea) to sidpith and to stawros. But both weren't exact.
First consider the
case n=m=4:
we have the lace city of sidpith
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x4o x4o
x4o x4x x4x x4o
x4o x4x x4x x4o
x4o x4o
That of quawros (quadratic wedge rosette), i.e. the corresponding Stott contraction - in one of the 2 possible displays - is
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x4o
x4o x4x x4o
x4o
But my new figure would expand in contrast rather the inverse (contracted) case
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x4x
x4x x4o x4x
x4x
thus, in fact, it happens to be
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x4x x4x
x4x x4o x4o x4x
x4x x4o x4o x4x
x4x x4x
Note how thiis display shows up quite nicely the ring of 8 ops (running along the outline) and also the orthogonal ring of 4 cubes in the center (running orthogonal to the display plane).
Now consider the
case n=5, m=5/2:
For stawros we had the lace city
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x5o
x5o x5o
x5x
x5o x5o
Note that the relavant single component here has been 5 times (in pentagrammal sequence)
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x5o x5o
x5x
But here we likewise would use the the inverse usage for those components, i.e.
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x5x x5x
x5o
i.e. in total (still for the contracted thing)
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x5x
x5x x5x
x5o
x5x x5x
Now consider first the Stott expansion of stawros (similar to going back from quawros to sidpith). It then would have the lace city
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x5o x5o
x5x
x5x x5x
x5o x5o
x5x x5x
x5o
But as you can see, the (pentagrammal) ring of 5 pips of stawros here would become a sequence of 10, outlining the outer x5o-pentagon twice. Thus that figure happens to become degenerate.
The same problem holds true in my recent n=5, m=5/2 figure, just that x5o and x5x are inversed here again
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x5x x5x
x5o
x5o x5o
x5x x5x
x5o x5o
x5x
i.e. the intended pentagonal ring of stips can be seen quite nicely in the center (orthogonal to the plane of display), but the intended ring of 10 dips would circle around at the outline twice!
But sure, there is a way out in both cases! We ought to blend out those pips (of the extended stawros), resp. the dips (in here), by identifying all completely coincident subelements, i.e. just cross-connecting the respective incident cells.
So, to conclude, what have we got in all the outlined stuff above?
Wrt. the 4-fold thingies:
Known components, being used
- there is 8-gon||cube, having for cell total: 1 cube, 2 squacues, 4 trips, and 4 tets
- there is op||4-gon, having for cell total: 1 op, 2 squacues, 4 trips, and 4 squippies
- there is op||cube, having for cell total: 1+4 cubes, 1 op, 2 squacues, and 4 trips
- there is sodip (= (4,8)-duoprism), having for cell total: 8 cubes and 4 ops
- there is tes (= (4,4)-duoprism), having for cell total: 4+4 cubes
Deduced stuff
- there is quawros (using 4 times that 8-gon||cube, connected at the squacues), having thus for cell total: 4 cubes, 16 trips, and 16 tets
- there is the Stott re-extended quawros (using in the first ring alternatingly 4 times that 8-gon||cube and 4 times op||cube, connected at the squacues, and in the second 1 sodip, interconnecting those rings at the ops), having thus for cell total: 4+4+16+4+4 cubes, 16+16 trips, and 16 tets (this in fact is nothing but sidpith, having for cell total: 32 cubes, 32 trips, and 16 tets)
- there is my new n=m=4 figure (using in the first ring alternatingly 4 times that op||4-gon and 4 times op||cube, connected at the squacues, and in the second 1 tes, interconnecting those rings at the cubes), having thus for cell total: 4+16 cubes, 4+4 ops, 16+16 trips, and 16 squippies
Wrt. the 5-fold thingies:
Known components, being used
- there is 10-gon||pip, having for cell total: 1 pip, 2 pecues, 5 trips, and 5 tets
- there is dip||5-gon, having for cell total: 1 dip, 2 pecues, 5 trips, and 5 squippies
- there is a reduced "10/2-p||5/2-gram" (blending out that 10/2-p by cross-connecting the incident cells), having for cell total: 2 5/2-cuploids, 5 trips, and 5 squippies
- there is dip||pip, having for cell total: 1 pip, 1 dip, 2 pecues, 5 trips, and 5 cubes
- there is a reduced "10/2-p||stip" (blending out that 10/2-p by cross-connecting the incident cells), having for cell total: 2 5/2-cuploids, 5 trips, and 5 cubes
- there is stardedip (= (5/2,10)-duoprism), having for cell total: 10 stips and 5 dips
- there is starpedip (=5/2,5)-duoprism), having for cell total: 5 stips and 5 pips
Deduced stuff
- there is stawros (using 5 times that 10-gon||pip, connecting at the pecues), having thus for cell total: 5 pips, 25 trips, and 25 tets
- there is the reduced Stott extended stawros (using in the first ring alternatingly 5 times that 10-gon||pip and 5 times dip||pip, connected at the pecues - but reducing the pairs of coincident pips, and in the second 1 stardedip, interconnecting those rings at dips), having thus for cell total: 10 stips, 25+25 trips, 25 tets, and 25 cubes
- there is the Stott contraction of my (un-reduced) n=5, m=5/2 figure (using 5 times that dip||5-gon, connected at the pecues), having thus for cell total: 5 dips, 25 trips, and 25 squippies
- there is the reduced version of my intended n=5, m=5/2 figure (using in the first ring alternatingly 5 times that dip||5-gon and 5 times that dip||pip, connected at the pecues - but reducing the pairs of coincident dips, and in the second 1 starpedip, interconnecting those rings at pips), having thus for cell total: 5 stips, 25+25 trips, 25 squippies, and 25 cubes
The important point here is for all that stuff to work: that the wedge angle of 2n-gon||n-p and that of n-gon||2n-p is the same!
--- rk