Another post about geometries, because who doesn't like those.



The Bipyramid is a solid that consists of two regular-polygon-based pyramids of the same height, "glued" on their bases. When puzzles are made using this geometry, they are usually Corner-Turning prisms (for example, the Mini Dino Pent Prism by Grégoire Pfennig).

While in theory puzzles can be designed that use a bipyramid of any height, most of them will not produce very interesting puzzles, since they do not allow for very many moves that exchange pieces. However, Matt Galla recently introduced the concept of Sweet-Spots: Specific heights that allow for extra jumbling interactions to happen in the bipyramid geometry. (Note: I highly recommend you read through that topic if you're interested. I won't go into too much detail explaining everything here).

In that same topic, some users of the forums tried to study this new concept mathematically to see what could be learned from it and how it could be applied to puzzle design. From that, an expression that allowed for the calculation of how many sweet-spots existed for a bipyramid with n sides was obtained (first by will_57, then an equivalent expression by me). This expression depends on two (positive) whole-number variables, i and j, which I call the Jumbling Constants. In the past, most puzzles that had Bipyramids as their geometry used heights that provided equal dihedral angles everywhere. These are known as the "canonical" bipyramid geometries and always correspond to using i=1 and j=0.

By working more with those expressions, I managed to derive an equation for the exact height of the Sweet-Spots, and recently implemented a generator in my OnShape Tools (V9) that allows for the generation of Bipyramids of any number of sides with those heights.

During this weekend, I have been trying to apply the same concept to trapezohedra, and just today I have developed equations that spit out the Sweet-Spots for this geometry (sidenote, doing this for the bipyramids was easy, trapezohedra, on the other hand... ). I have implemented them as well in the OnShape toolkit (V9.5).

Since not everyone uses OnShape, I thought I'd leave these expressions here, so that designers can explore this new world of possible puzzles.

Sweet-Spots for the Bipyramids:
If you have a Bipyramid Geometry with:
n - Number of sides for the polygon at the base
s - Sidelength of the edges at the equator

Then to calculate the heights h for Sweet-Spots use:

Where i and j are the jumbling constants. They must be whole numbers, greater than 0, and i must always be larger than j.

Sweet-Spots for the Trapezohedra:
If you have a Trapezohedron Geometry with:
n - Number of sides for the base polygon
s - Sidelength of the edges around the equator (not the vertical ones that meet at the tips)

Then to calculate the heights h for Sweet-Spots use:


Where i and j are the jumbling constants. They must be whole numbers, greater than 0, and i must always be larger than j.

That's essentially it. I hope this provides the tools for people that want to explore these geometries. Let's make some puzzles!

A table of small values and renderings of them might be helpful

Bram wrote:A table of small values and renderings of them might be helpful

That's very true. I'm working on the best way of displaying the information (including some Jaap's Spheres for these geometries).

Thanks for your formulas! I'll try to make a complete set of these puzzles in pCubes.

Bipyramid

I made a small discovery - I accidentally found a few new proportions for all odd Bipyramid !!!
I managed to find a formula, but I'm not a mathematician ... please check this !!!
Jumble Angle is also different here. It also works better with half of the axes - this is a very strange set.

I set up all the formulas in the script, you can edit it and set the necessary parameters I and J.
NN:= 5; nAngle := Pi/NN;
i := 1; j := 0;
H := cos(nAngle) / sqrt(cos(i * nAngle) * cos(j * nAngle));

This is a new geometry - I have not seen such puzzles yet

http://twistypuzzles.com/forum/viewtopi ... 16#p380816

That's odd. According to the maths, derived in the topic linked in the first post, the formulas I gave should list all possible heights for the bipyramids. Are you sure you're not accidentally getting heights for 2*N or 3*N sided bipyramids?

I reeeeally hope I'm wrong, that would mean there are even more unknown geometries to explore

EDIT: Are you sure those puzzles don't have the cuts of a trapezohedron?
EDIT2: Where does K=0.4 come from? Can K be equal to other values in your formula?

your guess is correct !!! it turns out I accidentally found a formula for Trapezohedra!

(K - this factor affects the scale in pCubes)

for compare: pentagonal Bipyramid and pentagonal Trapezohedra
Trapezohedron

Later I will build all Trapezohedra (3-10) in pCubes - it will be very fun

grigr wrote:your guess is correct !!! it turns out I accidentally found a formula for Trapezohedra!

Ah, that makes more sense

Your formula for the trapezohedra is actually much more compact than mine. Thinking down from a 2*n-sided bipyramid and ignoring some faces is a much better approach than the one I took.

Is there a way you can share your working out? I'd love to see what you did and implement this equation on my OnShape tools.

grigr wrote:your guess is correct !!! it turns out I accidentally found a formula for Trapezohedra!

(K - this factor affects the scale in pCubes)

for compare: pentagonal Bipyramid and pentagonal Trapezohedra

pCubes_2018-07-11_09-29-10.png

Isn't the pentagonal trapezohedron just a bandaged Pentultimate?

It depends on the jumbling constants you use for the pentagonal trapezohedron. If you use (i,j) = (1,0), the canonical trapezohedron, then yes! If you use other values for (i,j), then no, you will get a different puzzle.

rubikcollector123 wrote: Isn't the pentagonal trapezohedron just a bandaged Pentultimate?

all right
but there is one more configuration - pentagonal trapezohedron. will get flattened Pentultimate
Later I will show this form

Diogo Sousa wrote:

grigr wrote:your guess is correct !!! it turns out I accidentally found a formula for Trapezohedra!

Ah, that makes more sense

Your formula for the trapezohedra is actually much more compact than mine. Thinking down from a 2*n-sided bipyramid and ignoring some faces is a much better approach than the one I took.

Is there a way you can share your working out? I'd love to see what you did and implement this equation on my OnShape tools.

I already said - I'm not a mathematician ... I took one of Boris's formulas, and by empirical experiments found this formula ...

I found the Squashed Pentultimate in my old tests of Bipyramid geometries. Here is the Pentultimate of the form (i,j) = (2,0):

Hehe
Good Image

grigr wrote:Hehe
Good Image

It looks incredibly trippy, but I imagine designing this puzzle would not be a very easy task. It would be very hard to avoid internal bandaging.

Squashed Pentiltimate
http://twistypuzzles.com/forum/viewtopi ... 32#p380932