Analyzing the Cyltrianglintigroid

[ICN5D]

I've seen this strange name appear in several places around the wiki. Also included along with its 5D pyramid shape, it's being described as ambiguous, strange and immeasurable. Is it strange? Yes! Is it immeasurable? Not sure what that means. Is it ambiguous? Well, if referring to how to define it using a notation, no. It can be expressed in the current toratope notation, along with mine. So, if I understand what this name truly means, then I will take a stab at it.

I talked about a spherated cyltrianglinder margin one time, as a possible shape. No one seemed too interested in it. Another possibility could be a tiger made from a triangle torus. But, according to the name, it's probably more likely the first one. So, let's take a step back, and point out some things with the tiger, first. The tiger is the spherated margin of a duocylinder. A duocylinder is the cartesian product of two solid disks. The tiger has a bone structure of a cartesian product of two hollow circles. Inflating this bone structure gives us the tiger.

In contrast, the cyltrianglinder is the cartesian product with a solid disk and a triangle. Its margin would then be an orthogonal superimposing of the edges of a circle and triangle. In theory, the tigroid version will give us a bone structure of a hollow circle and a hollow triangle. Inflating this margin will puff up the 1-D line into a circle cross-cut. One major radius is a circle, the other is a triangular frame.

By looking at the notation for a torus, ((II)I), where ((major)minor), a triangle torus could look like ((II)'), a spherated cone, according to one definition. We could also use ((I')I) to mean a spherated hollow triangle. The minor radius is a circle, but the major is a triangular frame, only the outside edges of a I' are inflated.

Then, we have the comparison of a duocylinder (II)(II) and a tiger ((II)(II)). If the cyltrianglinder is I'(II), or using the commutative property (II)(I'), then a spherated margin of this could look like ((I')(II)) or ((II)(I')). The cuts through either major radius would be those of a hollow circle or a hollow triangle. The symmetry of the tiger dance depends on which major radius was cut through.

Lately, I've been conversing with Marek to further understand the cut algorithm. I'd like to apply it to this theoretical shape. So, to recapitulate, the tiger is:

Tiger: ((II)(II))
3D cuts
A - ((II)(I)) - 2 vertical stacked toruses

2D cuts
A1 - ((I)(Ii)) - 4 circles in vertices of square, moving out will merge both rows into 2 circles in a row, then deflates to two points and vanishes
A2 - ((II)(i)) - origin empty, moving out will make circle appear, divide into 2 concentric circles, merge into one and vanish

* 3D cuts of the tiger are both two parallel and separated torii. Slicing through either major radius always gives circular tiger symmetry. But, this thing ((I')(II)) can have a circular or triangular tiger symmetry. Triangular as in 2-simplex symmetry, circular as is 2-sphere. So, let's cut and dice this thing up, whatever it is...



* If we place a triangle I' centered at origin, height along Y, using "i" for the axis that was cut along, cuts of a triangle are:
(x^y)
i' - cutting along X gives 2 points at medium distance, moving +Y collapses dist to zero, moving -Y grows distance to max ending in a point-prism, i.e. line
Iⁱ - cutting along Y gives 2 points at maximum dist, moving along ±X will collapse dist to zero


I' - triangle
((I')I) - spherated hollow triangle, triangle-frame torus ( triangulus?)
((II)') - spherated cone, triangle crosscut torus


Possible Cyltrianglintigroid: ((I')(II)) oriented ((x^y)(zw))
3D cuts
* A - ((i')(II)) - cutting along X gives 2 parallel circle-torii at medium dist, moving +Y will collapse distance and merge, moving -Y will grow dist then a torinder appears

* B - ((Iⁱ)(II)) - cutting along Y gives 2 parallel circle-torii at max distance, moving ±X will collapse dist and merge

* C - ((I')(iI)) and ((I')(Ii)) - cutting along Z or W makes 2 vertical stacked triangle-frame torii ( triangulii ) that have 2-sphere tiger symm, moving ±W or ±Z respectively will collapse dist and merge

Then, of course, the cyltrianglintigroidal pyramid would just be a tapering of this whole shape down to a point into 5D. It could look like this: ((I')(II))'


If the cyltrianglintigroid was made from the tiger-rotation of a triangle torus, then the current toratope notation cannot express it. This is from the minor radius not being required to be in the notation. The triangle torus will rotate making the two circular major radii of a tiger. But, the crosscut will be a triangle instead of a circle. Maybe it could be done, based off the spherated cone analogy, I'm not sure.

Let me know what you all think

--Philip

[Keiji]

Sorry to burst your bubble but the cyltrianglintigroid does not exist. It was a mistake I made when extending the rotopes many years ago.

[ICN5D]

Thanks for clearing that up! No bubble bursting happened, it's only more forward progress in understanding. My curiosity will do its thing when I find things like this.

[ICN5D]

Well Keiji, I believe I re-discovered your cyltrianglintigroid You weren't mistaken, it's a real thing! I derive my inspiration from comparing the implicit surface of duocylinder and tiger. There is a nice, direct relation in both equations, with a straightforward method to inflating a margin of some kind.

Duocylinder

|√(x² + y²) - √(z² + w²)| + |√(x² + y²) + √(z² + w²)| - a

Rotating a duocylinder, stuck halfway into 3D

I was hoping it would look a little better

Tiger

Now, if we take the two circle parameters for duocylinder, XY and ZW, assign a size variable to each, and square them adding a third diameter, we effectively inflate the margin.

|√(x² + y²) - √(z² + w²)| + |√(x² + y²) + √(z² + w²)| - a

(√(x² + y²) -a)² + (√(z² + w²) -b)² - c²

By inflating the margin, we'll get the tiger:





The 45 degree oblique section of duocylinder


The tiger cage, 45 degree oblique section. Note the continuity of the duocylinder edge and tiger ring




The Cyltrianglinder

|||x|-2y| + |x| - √(z²+w²)| + |||x|-2y| + |x| + √(z²+w²)| - a

There are two distinct ways to rotate a cyltrianglinder. We can go from the circle parameter to either X or Y of the triangle parameter.







Applying the same method as inflating the duocylinder margin,

|||x|-2y| + |x| - √(z²+w²)| + |||x|-2y| + |x| + √(z²+w²)| - a

(||x|-2y| + |x| - a)² + (√(z²+w²) - b)² - c²

Graphing this strange new function, I found it worked best to scale the circle parameter by a multiple of 3. This will round out the circular ring in both 3D sections. Basically, we're doing the same thing with tiger. It's a product of the 1-surfaces of a circle and triangle, with a set of points equally distant from the new 2-frame.

Cyltrianglintigroid
The Inflated Cyltrianglinder Margin

(||x|-2y| + |x| - a)² + 3(√(z²+w²) - b)² - c²







Note the continuity of certain edges on the cyltrianglinder, and the toric ring of cyltrianglintigroid. Very interesting.

I used these rotate functions to make the animations:

(abs(abs(x) - 2(y*sin(a))) + abs(x) - 3)^2 + 3(sqrt(z^2 + (y*cos(a))^2) - 3)^2 - 0.75^2

(abs(abs((x*sin(a))) - 2y) + abs((x*sin(a))) - 3)^2 + 3(sqrt(z^2 + (x*cos(a))^2) - 3)^2 - 0.75^2



Of particular interest are the Villarceau sections, from one of the rotation types:

The Bridge
This unique topological artifact is a bridge that suddenly appears, right when the two rings touch on the other side. This is really a 1D line, which stands above the point, when the 3-plane is bitangent to both. Hmm, point atop line ... like a triangle!



The Basket
This is the other bitangent section, before the lower basket portion splits in two, becoming triangle frames. The shape of this division is also a 1D line segment, which also corresponds with the point above it, when the ring splits.

[Marek14]

Now this is very interesting! It's an unusual shape, and it leads me back to thinking about the mantis.

Here are the parameters the mantis should have:

A plane cut looking like six circles in vertices of hexagon (either regular or with two alternating edge lengths).
A 3D cut which creates toruses above the odd-numbered edges.
A 3D cut which creates toruses above the even-numbered edges.

[ICN5D]

Pretty cool, huh? It got me thinking about mantis, too. I'm telling you man, as soon as I find an elegant way to define a hexagon, the mantis will quickly follow. These are what PolyhedronDude and Wendy were talking about in the other threads. The tigroids of non-rotatope hyperprisms. This margin inflation technique can also be applied to a tesseract, as well as many, many others. I believe the Mantis will be the inflated margin of (hexagon,circle)-prism. I might experiment more with that equation from the stack exchange response PWrong linked me to.

A while back, you derived an interesting equation for a spider, by defining something in polar coordinates, then converting into cartesian coordinates. How exactly did you do that?

[Marek14]

Pretty cool, huh? It got me thinking about mantis, too. I'm telling you man, as soon as I find an elegant way to define a hexagon, the mantis will quickly follow. These are what PolyhedronDude and Wendy were talking about in the other threads. The tigroids of non-rotatope hyperprisms. This margin inflation technique can also be applied to a tesseract, as well as many, many others. I believe the Mantis will be the inflated margin of (hexagon,circle)-prism. I might experiment more with that equation from the stack exchange response PWrong linked me to.

A while back, you derived an interesting equation for a spider, by defining something in polar coordinates, then converting into cartesian coordinates. How exactly did you do that?

Well, I don't think the spider is a margin of (hexagon,circle)-prism; that would imply a tiger is a margin of (square,circle)-prism.

As for the spider -- it wasn't probably a true spider, but the logic was about this:

Start with tiger, oriented in such a way to have four circles in the xy plane.
Convert to polar coordinates.
Replace fi with fi/2, which will cause the xy plane to contain eight circles (or rather ellipses because they will be squashed in the angular direction).
Convert back.

Theoretically, something like mantis could result if you'd replace fi with 2*fi/3, but it still wouldn't be proper circles.

Vertices of hexagon are simple to define. Since a hexagon in complex plane is a^6 - 1 = 0, the hexagon vertices in plane are obtained by simple expansion:
(x + yi)^6 = (1 + 0i) -> Re(x + yi)^6 = 1 & Im(x + yi)^6 = 0
x^6 - 15 x^4 y^2 + 15 x^2 y^4 - y^6 = 1
6 x^5 y - 20 x^3 y^3 + 6 x y^5 = 0

[ICN5D]

6 x^5 y - 20 x^3 y^3 + 6 x y^5 = 0

That looks fairly simple How would one connect those vertices into lines? I tried graphing it, makes four intersecting 2-planes. I also tried out some absolute values, but no luck.

[Marek14]

6 x^5 y - 20 x^3 y^3 + 6 x y^5 = 0

That looks fairly simple How would one connect those vertices into lines? I tried graphing it, makes four intersecting 2-planes. I also tried out some absolute values, but no luck.

Well, you need both of those equations, x^6 - 15 x^4 y^2 + 15 x^2 y^4 - y^6 = 1 and 6 x^5 y - 20 x^3 y^3 + 6 x y^5 = 0. If you want to merge them, it would look like this:

(x^6 - 15 x^4 y^2 + 15 x^2 y^4 - y^6 - 1)^2 + (6 x^5 y - 20 x^3 y^3 + 6 x y^5)^2 = 0 -- a degree-12 polynomial just for six points.

As for connecting them, one idea I get would be to start with polar coordinates, and express one of the line segments in them. Then you could use periodicity of goniometric functions to replicate it.

[ICN5D]

I have no idea what that is, but it sounds awesome Can you give me an easy example how it works, maybe with a simple triangle? Once I get the jist of it, I can experiment with a square, pentagon then hexagon.

For triangle, would we have something like (x + yi)^3 = (1 + 0i) ?

Real(x + yi)^3 = 1
Imaginary(x + yi)^3 = 0

Well, I'm not sure how you expanded that, either. I get

x^3 + 3x^2*2yi -3xy^2 - yi^3

[PWrong]

What's going on here? This seems like a very roundabout way to draw a hexagon.

[PWrong]

The vertices of a polygon are just points on a circle:
cos(2pi k/n), sin(2pi k/n) for k = 0,1,..., n-1.

The lines connecting them are just
x cos(2pi k/n) + y sin(2pi k/n) = 1

The hard part is combining the lines into one equation so that you get the polygon. But weren't we doing that with the Max function in the other thread?

[ICN5D]

Yes, I remember the max function I was hoping maybe this would lead to an easier workaround. If I wish to keep using the same rendering program, I have to expand the max into absolute values.

An equilateral triangle used a three term max function, which expanded into an enormous equation. This makes the hexagon prohibitively complex, and hence the hard part. Aside from a full hexagon, we're also looking into another type of tigroid, one that makes a 2d cut of six circles in vertices of a hexagon, the Mantis.

But, if that makes one of the line segments, then couldn't I just multiply all six together, graphing them together as one hexagon? Will try later...

[PWrong]

You can multiply the equations for the six lines, but that gives you a union of six lines, not line segments. It looks more like a star with a bunch of lines coming off it.

Single implicit equations are really not cut out for describing unions of line segments effectively. If your rendering program can only plot a single implicit equation, then you have to do a lot of manual calculations to convert a simple set of equations into a much more complicated equation. It'd be better to write a program to do the conversion for you. But if you were going to do that, it'd be quicker to write a whole new rendering program, or just find one that can plot multiple implicit equations, or parametric equations.

Out of curiosity, how long does it take your program to render one of the more complicated surfaces?

[ICN5D]

CalcPlot can do multiple implicit surfaces, a few different ways. Either open up two or more equation input fields, or use one, and combine all equations together as a product of them. It also has parametric plotting as well (something I'd like to learn more about this year).

As for plotting speed, so long as I use 30 ~ 35 cubes, it will render remarkably fast, within seconds. Calcplot uses Marching Cubes to approximate the surfaces. But, this makes everything blocky and lo-res, for a degree-64 equation for a ((((II)I)I)((II)I)). And at the same time, using the rotate function, adjusting the slider to turn the shape will render very fast, and allow real-time, dynamic exploration of it. That's how I find cool stuff to animate.

It wouldn't be feasible if it took a while to rotate 90 degrees. When I see something cool, bump it up to 70 or 90 cubes, then hit ctrl+g for smooth gradient, which slows the render time down a lot. I use this to my advantage, where I have some time between screenshots. It's anywhere between 5 to 15 seconds per animate step, at hi-res. (This is also being achieved on a worn out, 8 yr old computer. Something new and awesome will be even faster)

Here, try it out, when you get a chance. This is ((((II)I)I)((II)I)) :

Open up CalcPlot3D : http://web.monroecc.edu/calcNSF/

Input :
(sqrt((sqrt((sqrt(x^2 + (y*cos(d) - c*sin(d))^2) - 4.25)^2 + (z*cos(a))^2) - 2)^2 + (z*cos(b))^2) - 1)^2 + (sqrt((sqrt((y*sin(d) + c*cos(d))^2 + 0^2) - 2.5)^2 + (z*((sin(a))*(sin(b))))^2) - 1.25)^2 - 0.4^2 = 0

Or, another one with different positions
(sqrt((sqrt((sqrt((x*((sin(a))*(sin(b))))^2 + (z*cos(d) - c*sin(d))^2) - 4)^2) - 2)^2 + y^2) - 1)^2 + (sqrt((sqrt((z*sin(d) + c*cos(d))^2 + (x*cos(a))^2) - 2.5)^2 +(x*cos(b))^2) - 1.25)^2 - 0.4^2 = 0

Set:
35 cubes resolution
XYZ box = -9 , +9
0 < a,b,d < 1.57
-10 < c < +10

[a,b] are multi-position rotate :
[1.57,1.57] = ((((I)))((I)I))
[1.57,0] = ((((I))I)((I)))
[0,1.57] = ((((I)I))((I)))
[0,0] = invalid cut ( Though, I'm still not sure if it is a legit slice)

A - rotate
B - rotate
---------------
C - translate
D - rotate

How to use the function:
• A and B have to be both 1.57 , or 1.57 and 0, for one or the other. It's one of my new rotate functions, which have three possible positions using two parameters. If both are 0, then the 3D section is weird.

• You can freely adjust D for whichever proper values A and B are set to.

• You can freely adjust C to slide the shape at any A, B, D angle you want. Sliding C from -10 to +10 will pass the shape through the 3-plane at whichever angle you have it set to.

• Even if C is set to 0, rotating A, B, or D will sometimes make the shape completely disappear. This is when the 3-plane occupies one of the holes, making a complex 3D solution. The intercepts are above and below the 3-plane, and we see nothing.

• While in the hole, move C to slide out of the center, and you'll see those pesky, hidden intercepts pop into 3D, from a higher dimension (the complex plane). It's quite an awesome sight!

[PWrong]

That sounds a lot faster than Mathematica. If you can do parametric equations, then there's no problem. The parametric equations for a polygon are relatively simple.