Duoprism and Antiprismatic Prism Formulae

[ytrepus]

Hello all,

This is my first post so my apologies if this has already been covered elsewhere!

I was hoping that someone could help me out with formulae to calculate the circumradius and bulk (4-volume) of any 4D p,q-duoprism given the values of p and q (assuming an edge length of 1).

Also, I am looking for these same values for any 4D uniform p-gonal antiprismatic prism given the value of p (again edge length=1).

I know I can get approximate values for both of these from Stella 4D on a case-by-case basis, but I am looking for the general formula. Is there a way to derive it from the formulae for the polygons / antiprisms that these polytopes are composed of by extrapolating somehow to the higher dimension?

Thank you very much.

Kind regards,
Rob

[ytrepus]

OK, I've figured out the formulas for the bulk...embarrassingly easy.

4-volume of any 4d prism of side lengths 1 = the 3-volume of the 3d prism. In retrospect that seems pretty obvious.

For a p-q-duoprism the 4-volume = the area of regular polygon p x regular polygon q, again with all side lengths = 1. I was surprised that that worked to be honest but it does make sense.

Now I just need to work on the circumradius, which I am pretty sure will be easy to figure out as well.

Thanks,
Rob

[Klitzing]

Squared circumradius of p,q-duoprism = squared circumradius of p-gon + squared circumradius of q gon.

This is because of Pythagoras, resp. because of p,q-duoprism = cartesian product of p-gon and q-gon.

--- rk

[quickfur]

The vertices of a uniform m,n-duoprism lie on the surface of a 3-sphere, so the circumradius is just the square root of the sum of the m-polygon and n-polygon's circumradii, just as Klitzing said.

[ytrepus]

Thank you both very much -- that makes perfect sense.

I've put together a little summary spreadsheet of the uniform polytopes in 2-4 dimensions if anyone would like to take a look. I am especially proud of the duoprism calc tab -- if you type in any p, q from 3-10000 it calculates various metrics for you, including the Greek name. Would love it if anyone has any comments/ideas, or especially if anyone spots any errors. [Unfortunately the file seems too large to upload but I'll post some screenshots--let me know if there is any interest.]

Regards,
Rob

[ytrepus]

Here is a link to the spreadsheet. Would love to hear your thoughts! Feel free to play around with the duoprism and antiprismatic-prism calc tabs especially.

It's nothing earth-shattering, but I thought it might provide a good summary for someone new to polytopes. I hope to add some more info on dimensions 5+ next, and add in Coxeter diagrams and potentially some images.

Note: in order to get it to work you'll have to download it -- I don't think Google Drive does a great job with the preview, but hopefully that will be enough for you to see it is legit.

https://drive.google.com/file/d/0B4ZHyL ... sp=sharing

Regards,
Rob

[wendy]

There's a spreadsheet of my design that converts the CD-diagram directly into circum-diameter. Richard makes some use of it in the segmentotopes.

[Klitzing]

There's a spreadsheet of my design that converts the CD-diagram directly into circum-diameter. Richard makes some use of it in the segmentotopes.

Haha, not only in context of the segmentotopes. Well my version of that spreadsheet was adopted such that it calculates the circum-radius rather than the diameter. And the units too are adopted to unit edge length. Even so it allows for different edge lengths as well, quite in the sense of Wythoffian variants: any node can be asigned any length you would like. - But it not only allows for circum-radii. It allows also for calculation of the height of segmentotopes, provided the top and bottom Dynkin diagrams and the lacing edge length.

By the way, it is online available: it is the first spreadsheet here.

--- rk

[wendy]

Actually i consider 'wythoffian' polytopes to include the variants: ie position polytopes in an oblique screen. The uniform wythoffian figures is a proper intersection of 'uniform' (ie all edges=1) + 'wythoffian' (derived from a position polytope in a simplex group).

So q3f is a wythoffian polygon. It's not uniform. The canonical uniform wythoffs are those with 0's and 1's (ie o and x or hugs and kisses) at each node.

I use the same process, in a more primitive form, to deal with hyperbilic polytopes, zB x4x3o8o. I think i have decoded what a negative edge means!

[Klitzing]

... I think i have decoded what a negative edge means!

Well, what would you like to be understood by those?

To me it is just a negative number as multiplicator for some x, just similar to: tau times x = f. Then it would just code its reversed orientation. That induces thus usually crossed incidences. Such thingies also occur e.g. with retrograde polygons. - Is that different from yours?

--- rk

[wendy]

wendy wrote:... I think i have decoded what a negative edge means!

Well, what would you like to be understood by those?

To me it is just a negative number as multiplicator for some x, just similar to: tau times x = f. Then it would just code its reversed orientation. That induces thus usually crossed incidences. Such thingies also occur e.g. with retrograde polygons. - Is that different from yours?

The sort of edges Richard mentions arise from using 'i' modifiers, eg replacing p/d with p/(p-d).

The sort of negative edges i am talking about is what you get when you calculate the edge of x4o3o8o. It's somehow related to the closest approach of the negative vertices (ie planes).

[Robert May]

For the p-antiprismatic prism with unit edge length:

Circumradius

1.gif (1.93 KiB) Viewed 8994 times

Bulk

2.gif (3.31 KiB) Viewed 8994 times

There may be more compact formulas, but these are what I calculated a few years ago.