non-cubic toratopes?

[student91]

Hello there !

You might know me from the CRF-polytopes section. I'm quite new to this section (and not plannig to contribute to this as much as the CRFs), and thus don't know much about toratopes. As far as I get it, toratopes are uniquely defined by the toratope notation, which works as follows:
-every I means another variable
-a set of parentheses mean you assign a radius to the distance of whatever is in these parentheses to the origin.

example: ((II)II)
start at the inside: (II)- two variables (x and y) are assigned a radius (thus x²+y²=R₁² <=> sqrt(x²+y²)-R₁=0)
next set of parentheses: the old equation (sqrt(x²+y²)-R₁) and two new variables (z and w) are assigned a radius: (sqrt(x²+y²)-R₁)²+z²+w²=R₂²

If we rewrite the "a²+b²+c²+d...=R_n²"-equation to |a,b,c,d,...|=R_n, we get the following:
||x,y|-R₁,z,w|=R₂
This is almost the same as ((II)II), so I guess this is how you derived the notation (no I am not going to read all 23 pages 'bout the tiger )

Furthermore you should know that we have, not too long ago, found an interestion function. This function works on dynkin-style notation of polytopes, and makes this notation isiomorph to the coxeter group that is used. (a Coxeter-group is a kind of symmetry-group in this context). This has shown to have interesting properties.

I think I have discovered a way to connect these two branches in some manner, yielding toratopes with other symmetries.
The symmetry of toratopes basically is .2.2.2.-symmetry. This symmetry uses the normal coordinate system with (x,y,z,w)-coordinates. This symmetry also means that if (x,y,z,w) is on the toratope, then (±x,±y,±z,±w) is also on the toratope. There are thus exactly 2⁴=16 points for every point that is on this (4-dimensional) toratope. These 16 points correspond to the elements of the coxeter-group .2.2.2. In general a coordinate in this group can be written as (x)2(y)2(z)2(w). The function that changes this coordinate into one of 16 others takes one number (say x), and change it into (-x). Furthermore the other values should be incremented with 2sin(90-180/n), where n is the number between these two values. (in this case n=2, so 2sin(90-180/n)=2sin(0)=0. and the values y,z and w shouldn't be changed at all)

Now instead of the .2.2.2.-group, one could try to use another group (say .4.3.3., the tesseractic group). Now coordinates are generally given by (a)4(b)3(c)3(d). Any coordinate (a)4(b)3(c)3(d) also implies the coordinates (-a)4(b+a*sqrt(2))3(c)3(d), (a+b*sqrt(2))4(-b)3(c+b)3(d), (a)4(b+c)3(-c)3(d+c) and (a)4(b)3(c+d)3(-d). This can then be re-used recursively to find a total of 48 implied coordinates. To define toratopes here we can take the derived definition: ||x,y|-R₁,z,w|=R₂. The ||-function should now be defined differently (or rather more generally). The distance between the origin according to a number of variables can be calculated in a way, but I don't know how. (Klitzing has a spreadsheed that does this, but that's pure magic to me (it uses matrices etc, way above my vector calculus )). The matrix that is used for the magic has the same numbers 2sin(90-180/n) as in the group-function, so if we are lucky, the new shape has the given symmetry instantly. If not, we can define the shape locally (a>0 and b>0 and c>0 and d>0), and then make the whole shape by expanding this part using the group-function

This is the general idea I had to make toratopes with other symmetries. I'm not sure if it will work directly this way, but we will see.

[Marek14]

This sounds interesting because of a mythical toratope I call "mantis". My gut tells me it exists, but I wasn't able to construct it so far.

Take a tiger ((II)(II)) and assign the variables like this: ((xy)(zw)). Now, if you take a cut by the xz plane, you will get four circles in the vertices of a rectangle (or square, if both major diameters of the tiger are the same). If you put this into 3D and add the y dimension, two pairs of the circles (say the horizontal pairs) will be connected by toruses, leading to a stack of two toruses. If you add the w dimension, the result will be similar, but it will be the other pair of circles that gets connected.

Now, the mantis starts from a plane that has six circles in the vertices of a hexagon instead. Can be either a regular hexagon or a semiregular one with all inner angles 120 degrees and two alternating edge lengths. One 3D cut would then join circles 1-2, 3-4 and 5-6 with toruses, while the other 3D cut would join circles 2-3, 4-5 and 6-1.

Mantis would therefore probably have a symmetry .3.2.2.

Similar shapes should be possible by starting with other even-sided polygons, with symmetries .n.2.2. . The octagonal one is called "spider", I don't have a good name for the decagonal one, should be a predatory crustacean of some kind.

[ICN5D]

Yeah, this sounds really interesting. Finding the Mantis equation would be a step towards the combining of toratopes and polytopes. This should yield some very interesting polypedal beasts. We might even want to develop a new way to express them.

[wendy]

Any hollow sphere, with holes let into it, will, by hopf fibulation, lead to a torotope.

The octahedral arrangement of holes derive from removing selected octagons from the octagonny.

The cubic arrangement consist of dividing the faces of the octagonny so the connections run through the triangles.

The tetrahedral arrangement passes its holes around the four sets of six octahedra of {3,4,3}.

The icosahedral arrangement passes its 12 rings of 10 through the pentagons of the twelftychoron.

The dodecahedral arrangement passes its 20 rings along the great diameter and edge of faces of the twelftychoron.

The polygons exist for all p (not just the even ones), run diagonally across the squares that make the duoprism boundary.

The tiger is a torus, no doubt.

[wendy]

Student91 mentions my spreadsheet that Klitzing rewrote.

The coordiates you enter for the polytopes are vectors in an oblique coordinate system. The matrix you enter from the dynkin symbol is the 'dynkin' matrix. It is used to calculate the stott matrix, (by simple reciprocal), the stott matrix is used to do a simple dot product between the two vectors.

[wendy]

The trouble with the 'mantis' that Marek describes, is that the image is not of four circles in the xy plane, but just 1 (in the +x +y quadrant). The rotations are into w=0,x=0 and y=0,z=0, so this would produce the necessary circles at the quadrants.

You could do something in five dimensions, i suppose, by noting that in the xy plane, put three folds, which produce the v, w, z axies coupled with one of these lines. This would produce something interesting like Marek suggests.

Given the above description, i imagine that the mantis would have a three-fold rotation or symmetry, but this would be of the nature of h{3,2,3}, the transfer of 00 to 11 to 22 in the bi-triangular prism, rather than perpendicular to the xy space. What happens is that as you rotate the xy space, you would have to rotate the vwz space too.

[Marek14]

The trouble with the 'mantis' that Marek describes, is that the image is not of four circles in the xy plane, but just 1 (in the +x +y quadrant). The rotations are into w=0,x=0 and y=0,z=0, so this would produce the necessary circles at the quadrants.

What do you mean by this? A tiger has stack of two toruses as a 3D cut, and cutting that in half gives four circles.

[wendy]

A torus in 3d yields two circles on the cross-section, but this is just one rotation. The four circles in the tiger come from two orthogonal rotations of the one circle. You can't really set them to different sizes, even though you can move the centre relative to the x-y axies.

[Marek14]

A torus in 3d yields two circles on the cross-section, but this is just one rotation. The four circles in the tiger come from two orthogonal rotations of the one circle. You can't really set them to different sizes, even though you can move the centre relative to the x-y axies.

Why would I set them to different sizes? I did nothing of the sort.

[wendy]

I did not suggest that you did. What i was hinting at was that the four circles are reflections of any one of them. Actually, the only free parameters are as the brackets in the torotope notation: viz ([xx]{yy}) where [] is the x coordinate of the centre, {} is the y coordinate, and () is the radius of it.

What i was suggesting is that with the mantis, you would have six axies, but these share one dimension with the xy plane (because it's really a comb product, and comb products share a common wrapping space). In place of say (oo)(xx)(yy) tiger in my notation: o is pondered, the normal product of 4 circles would have to replace one of xx or yy with an axis, but (o3)(xx)(yy)(zz) would take up all three pondered axies into the first circle.

[Marek14]

Why is there a (zz)? I wonder why would it be a problem if mantis can't be properly expressed in a notation system it wasn't designed for. After all, six circles can be gotten by repeated reflections with two mirrors in the plane with 60-degree angle.

Actually, it got me thinking. Some time ago, I managed to create a full implicit equation for tiger without square roots, although it was very complicated. I did that by expanding the simpler equations of its 3D cuts and combining the terms together, so maybe something like that would be possible for the mantis as well...

Let's start with the basics. Tiger is derived from a duocylinder margin. What, then, would be a mantis derived from?

Duocylinder margin can intersect a plane in four points. We're searching for a similar shape that would intersect it in six.

Duocylinder margin equation is simple:

(x^2 + y^2 = a^2) AND (z^2 + w^2 = b^2) -> (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0

For y=w=0, we get

(x^2 - a^2)^2 + (z^2 - b^2)^2 = 0, which has four solutions with x = +- a and z = +- b.

Now, let's transform this equation with x and z taken as polar coordinates:
x = r * cos(fi)
z = r * sin(fi)
(r^2 * cos(fi)^2 + y^2 - a^2)^2 + (r^2 * sin(fi)^2 + w^2 - b^2)^2 = 0

cos(fi) and sin(fi), of course, have two full periods in 360 degrees. But now let's replace them with functions that have three periods:

(r^2 * cos(2*fi/3)^2 + y^2 - a^2)^2 + (r^2 * sin(2*fi/3)^2 + w^2 - b^2)^2 = 0

We can transform this back to Cartesian coordinates:

r^2 = x^2 + z^2
fi = arctan(x/z)

((x^2 + z^2) * cos(2*arctan(x/z)/3)^2 + y^2 - a^2)^2 + ((x^2 + z^2) * sin(2*arctan(x/z)/3)^2 + w^2 - b^2)^2 = 0

So this should be a "skelet" of the mantis. What more, if this works, it's clear that it could be extended to arbitrary even number of points and yw plane could be changed in a similar, independent way. Does this relate to tegums? I seem to remember those are the shapes with polygons in two orthogonal planes.

[PWrong]

Duocylinder margin equation is simple:

(x^2 + y^2 = a^2) AND (z^2 + w^2 = b^2) -> (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0

What exactly are you doing here? Is "->" just an implication?

(x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0
is a new surface that doesn't correspond to any toratope. It does look deceptively similar to a tiger, but it's a bit droopier.

[Marek14]

Duocylinder margin equation is simple:

(x^2 + y^2 = a^2) AND (z^2 + w^2 = b^2) -> (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0

What exactly are you doing here? Is "->" just an implication?

(x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0
is a new surface that doesn't correspond to any toratope. It does look deceptively similar to a tiger, but it's a bit droopier.

Yes, it's an implication. The surface I describe here is basically a tiger with minor diameter set to 0. The standard trick is: if you want to graph something that satisfies both f(x) = 0 and g(x) = 0, you combine it as f(x)^2 + g(x)^2 = 0, because in reals the sum of squares can be 0 only if each of them is.

[ICN5D]

I graphed your equations, Marek.

skelet of Mantis
((x^2 + z^2) * cos(2*arctan(x/z)/3)^2 + y^2 - a^2)^2 + ((x^2 + z^2) * sin(2*arctan(x/z)/3)^2 + w^2 - b^2)^2 = 0

inflated skelet, I added the minor diameter C^2, to expand the skelet
((x^2 + z^2) * cos(2*arctan(x/z)/3)^2 + y^2 - a^2)^2 + ((x^2 + z^2) * sin(2*arctan(x/z)/3)^2 + w^2 - b^2)^2 -c^2 = 0


a,b,c = 3 , set diameter values, minor has to be larger than you'd think
((x^2 + z^2) * cos(2*arctan(x/z)/3)^2 + y^2 - 3^2)^2 + ((x^2 + z^2) * sin(2*arctan(x/z)/3)^2 + w^2 - 3^2)^2 - 3^2 = 0

w-cut
((x^2 + z^2) * cos(2*arctan(x/z)/3)^2 + y^2 - 3^2)^2 + ((x^2 + z^2) * sin(2*arctan(x/z)/3)^2 + w^2 - 3^2)^2 - 3^2 = 0

z-cut
((x^2 + a^2)*cos(2*arctan(x/a)/3)^2 + y^2 - 3^2)^2 + ((x^2 + a^2)*sin(2*arctan(x/a)/3)^2 + z^2 - 3^2)^2 - 3^2 = 0

y-cut
((x^2 + y^2) * cos(2*arctan(x/y)/3)^2 + a^2 - 3^2)^2 + ((x^2 + y^2) * sin(2*arctan(x/y)/3)^2 + z^2 - 3^2)^2 - 3^2 = 0

All produced only two torus-like intercepts, that deform towards each other, like a mid-tiger rotation. No three-torus intercept, or six circles in hexagon arrgmt. I was also thinking of compiling the known (theoretical) cuts to make mantis eq. Should the minor diameter term C be anywhere else? It seems to be applied the right way, just like with tiger.

[Marek14]

Not sure where the minor diameter should be, that's why I only tried the skelet at this point

[ICN5D]

Well, I'll play with that equation later. maybe it's some parentheses or something.

Also, what if you did this mirroring with a whole torus equation, compiled all cuts, then consolidated? We know it'll be the two distinct y-shaped and inverted y-shaped arrangements of 3 tori. This should make each cut a product of three tori, that are spaced and angled the proper way. One other thing is how mantis will intercept as three degree-4 surfaces, which hints at a full degree-12 equation. As you have shown, the vertices of a hexagon is also degree-12,

(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

so maybe that equation has something to do with manits....

[Marek14]

Well, I'll play with that equation later. maybe it's some parentheses or something.

Also, what if you did this mirroring with a whole torus equation, compiled all cuts, then consolidated? We know it'll be the two distinct y-shaped and inverted y-shaped arrangements of 3 tori. This should make each cut a product of three tori, that are spaced and angled the proper way. One other thing is how mantis will intercept as three degree-4 surfaces, which hints at a full degree-12 equation. As you have shown, the vertices of a hexagon is also degree-12,

(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

so maybe that equation has something to do with manits....

Well, the problem is that I only know two 3D cuts, which is probably not enough to consolidate them. The equation should be degree 12, yes.

[ICN5D]

Yeah, me neither. Well, let's explore that spider equation you wrote a while back. I parsed all cuts, and rescaled some of the parameters to round out the obliqueness. What I ended up with was this:

(sqrt((x^2 - y^2)^2/(x^2 + y^2) + 3z^2) - a)^2 + 6(sqrt((x*y)^2/(x^2 + y^2) + w^2) - b)^2 = c

a = 3
b = 1.5
c = 0.5



(sqrt((x^2 - y^2)^2/(x^2 + y^2) + 3a^2) - 3)^2 + 6(sqrt((x*y)^2/(x^2 + y^2) + z^2) - 1.5)^2 = 0.5

The Z-cut of this equation. We get 4 tori that are inward-facing towards the center. A little deformed towards the +z / -z regions. They are also right on the X, Y axes.







(sqrt((x^2 - y^2)^2/(x^2 + y^2) + 3(z*sin(a))^2) - 3)^2 + 6(sqrt((x*y)^2/(x^2 + y^2) + (z*cos(a))^2) - 1.5)^2 = 0.5

Oblique structure makes 8 Bar Spider Cage, at mid-rotation. We can clearly see the trace of eight circles in octagon array. This rotation is flipping between the Z and W dimension positions.







(sqrt((x^2 - y^2)^2/(x^2 + y^2) + 3z^2) - 3)^2 + 6(sqrt((x*y)^2/(x^2 + y^2) + a^2) - 1.5)^2 = 0.5

The W-cut, which gives us the other four torus arrangement. This time, each is in their own quadrant.





(sqrt((x^2 - a^2)^2/(x^2 + a^2) + y^2) - 3)^2 + (sqrt((x*a)^2/(x^2 + a^2) + z^2) - 1.5)^2 = 0.5

The X or Y cut, which gives a tiger cut of a vertical column of tori. Note how I removed the scaling. If plotting with the above used values, the tori will be ellispoid in the major diameter.





However, if moving out from the normal-looking tiger cut, we don't get a tiger evolution of merging tori. We get something much more wild. This is a = 2.4. I will animate this shape.





So, this equation produces something close to what we're looking for, if this isn't 'the one' . It does the correct rotate morph of the spider cage, and flips between the two arrangements of 4 tori. Mantis will have a 6 bar cage, between the two distinct, inward-facing y-shaped arrays.

[Marek14]

Yeah, it is almost the spider, except one thing -- you say "we can clearly see the trace of eight circles in octagon array", but they are actually ellipses. Because of the way I derived the equation, they are squashed in the angular direction.

Here is something you can try: The "protospider" equation is derived from tiger equation by modifying x and y coordinates. If you, in addition to it, modify the z and w coordinates in the same way, you should get something with octagon of ellipses as both xy AND zw cut. No idea how the "cage view" would look then.

[PWrong]

The pictures are amazing. Looking forward to the real thing.

Yes, it's an implication. The surface I describe here is basically a tiger with minor diameter set to 0. The standard trick is: if you want to graph something that satisfies both f(x) = 0 and g(x) = 0, you combine it as f(x)^2 + g(x)^2 = 0, because in reals the sum of squares can be 0 only if each of them is.

I see, ok. I was looking at the surface described by (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = r^2, which is definitely not a tiger.

[Marek14]

The pictures are amazing. Looking forward to the real thing.

Yes, it's an implication. The surface I describe here is basically a tiger with minor diameter set to 0. The standard trick is: if you want to graph something that satisfies both f(x) = 0 and g(x) = 0, you combine it as f(x)^2 + g(x)^2 = 0, because in reals the sum of squares can be 0 only if each of them is.

I see, ok. I was looking at the surface described by (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = r^2, which is definitely not a tiger.

And I didn't say it was -- it's a "skelet" of a tiger, or tiger with minor diameter set to 0. What it actually is is a cartesian product of two circles.

[PWrong]

Wait, isn't (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0 just exactly equal to the duocylinder margin?

(x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = r^2 is a droopy tiger, which you weren't talking about at all. I only brought it up because I misread the equation before.

[Marek14]

Wait, isn't (x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = 0 just exactly equal to the duocylinder margin?

(x^2 + y^2 - a^2)^2 + (z^2 + w^2 - b^2)^2 = r^2 is a droopy tiger, which you weren't talking about at all. I only brought it up because I misread the equation before.

Yes, sorry, missed the r^2 part

[ICN5D]

Ahh, yes, they are ellipses. That could be what's making the tori deformed, as well, maybe some missing terms. I tried graphing the double ellipse equation you described. It makes something very strange, and not recognizable. Were you referring to:

(sqrt((x^2 - y^2)^2/(x^2 + y^2) + (z^2 - w^2)^2/(z^2 + w^2)) - 3)^2 + (sqrt((x*y)^2/(x^2 + y^2) + sqrt((z*w)^2/(z^2 + w^2)) - 1.5)^2 = 0.5 ?

Other than that, it seems that your mantis skelet should be working. I also tried some combinations of square roots,

(sqrt((x^2 + z^2) * cos(2*arctan(x/z)/3)^2 + y^2) - 3^2)^2 + (sqrt((x^2 + z^2) * sin(2*arctan(x/z)/3)^2 + a^2) - 3^2)^2 -2^2 = 0

but nothing new showed up. Maybe I should look for a way to make the six circles in 2D first. The hexagon vertex equation made some interesting things when adding diameters and square roots, though:

(sqrt((x^6 - 15x^4*y^2 + 15x^2*y^4 - y^6 - 1)^2 + z^2) - a)^2 + (sqrt((6 x^5 y - 20x^3*y^3 + 6x*y^5)^2 + w^2) - b)^2 = c^2

Some combo of values for a,b, and c make six circles in hexagon. But, slightly change them, and you get a cylinder or intersecting 2-planes. This equation made six circles in hexagon as well,

(sqrt((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) - a) = b

when b=0.75 and a=0 . Some other forms of this make six discrete spikes in a hexagon.

[Marek14]

I did mean (sqrt((x^2 - y^2)^2/(x^2 + y^2) + (z^2 - w^2)^2/(z^2 + w^2)) - 3)^2 + (sqrt((x*y)^2/(x^2 + y^2) + sqrt((z*w)^2/(z^2 + w^2)) - 1.5)^2 = 0.5, maybe with some constant tweaking.

[ICN5D]

Got an interesting result working with six circles in hexagon array. I thought about how an inflated line segment will make two circles in a row. A product of two inflated line segments will make tiger cut of the 2x2 array of 4 circles. So, then, how would one make three inflated line segments, in plane XY?

two points : (sqrt(x)-a)^2
product of two points, square array of four : (sqrt(x)-a)^2 + (sqrt(y)-a)^2

• trying for three pairs of points within 2D XY plane:

(sqrt(x)-a)^2 for x-axis pair

two more point pairs will intersect betwen 180 degrees, dividing into three equal parts of 60 degrees

rotate parameter between axes:
(x*sin(b)+y*cos(b))
(x*cos(b)-y*sin(b))

3.14 = 180 degrees rotation

pi/3 = 1.0467 = 60 deg
2*pi/3 = 2.0933 = 120 deg

• Theoretical six-circle function in xy-plane:

(sqrt((x*sin(0)+y*cos(0))^2)-a)^2 + (sqrt((x*sin(1.0467)+y*cos(1.0467))^2)-a)^2 + (sqrt((x*sin(2.0933)+y*cos(2.0933))^2)-a)^2 = b^2
——— it works 100% perfectly as a hexagon of circles
——— a=2 , b=1.5

general equation
(sqrt((x*sin(pi/3)+y*cos(pi/3))^2)-a)^2 + (sqrt((x*sin(2*pi/3)+y*cos(2*pi/3))^2)-a)^2 + (sqrt((x*sin(3*pi/3)+y*cos(3*pi/3))^2)-a)^2 = b^2



I experimented a lot with trying to unite the circles into three tori, but no luck. I tried defining three circles that intersect in 4-space, using XY circle and two more that intersect at 60 deg and 120 deg towards ZW:

general equation
(sqrt((x*sin(1/3*pi)+z*cos(1/3*pi))^2 + (y*sin(1/3*pi)+w*cos(1/3*pi))^2) - a)^2 + (sqrt((x*sin(2/3*pi)+z*cos(2/3*pi))^2 + (y*sin(2/3*pi)+w*cos(2/3*pi))^2) - a)^2 + (sqrt((x*sin(3/3*pi)+z*cos(3/3*pi))^2 + (y*sin(3/3*pi)+w*cos(3/3*pi))^2) - a)^2 = b^2


z-cut
(sqrt((x*sin(1/3*pi)+c*cos(1/3*pi))^2 + (y*sin(1/3*pi)+z*cos(1/3*pi))^2) - a)^2 + (sqrt((x*sin(2/3*pi)+c*cos(2/3*pi))^2 + (y*sin(2/3*pi)+z*cos(2/3*pi))^2) - a)^2 + (sqrt((x*sin(3/3*pi)+c*cos(3/3*pi))^2 + (y*sin(3/3*pi)+z*cos(3/3*pi))^2) - a)^2 = b^2

Makes pair of genus-2 toruses with spheric bulges. Translating with ‘c’ makes strange helical twisting while merging like tiger column.


y-cut
(sqrt((x*sin(1/3*pi)+z*cos(1/3*pi))^2 + (c*sin(1/3*pi)+y*cos(1/3*pi))^2) - a)^2 + (sqrt((x*sin(2/3*pi)+z*cos(2/3*pi))^2 + (c*sin(2/3*pi)+y*cos(2/3*pi))^2) - a)^2 + (sqrt((x*sin(3/3*pi)+z*cos(3/3*pi))^2 + (c*sin(3/3*pi)+y*cos(3/3*pi))^2) - a)^2 = b^2

bizarre mantis-cage-like single structure. One well-defined torus with two intersecting deformed tori.

[ICN5D]

Here's a question:

• Do we think the mantis will have more than the two cuts of three tori, in the Y-shaped arrangement in 3D? A tiger has only two distinct arrangements in 3D, so maybe the mantis will as well.

• Would it be easier to express the mantis in polar coordinates, versus Cartesian?

• How do the toratopes look in polar coordinates, anyway? Do they take on an entirely different pattern?


I'm still trying to wrangle my head around how to mathematically create the mantis. This is one tough, bizarre shape. I got the six circles in a hexagon by spherating the ends of three intersecting line segments. But I can't seem to connect those lines into three circles.

Another thing I'd like to investigate, is if we can create the tiger equation from the expansion of the square vertices (x + yi)^4 (if that's the correct equation). If there's a systematic way to build a tiger from vertices of a square, then maybe it can be applied to the hexagon. We know what the equation of four circles in a square look like, and how they connect in 3 and 4D:

4 circles in square
(√(x²) - a)² + (√(z²) - a)² = b²

2 tori in column
(√(x² + y²) - a)² + (√(z²) - a)² = b²

1 tiger
(√(x² + y²) - a)² + (√(z² + w²) - a)² = b²


Or, maybe building torus equation from a digon? Looking for a connection in already known things could provide some new insight

First thing is, I'm not sure how that expansion works with the vertices equation. It looks very simple, but I'm missing something. From what I can see,

(x + yi)^2 = line segment
(x + yi)^3 = triangle
(x + yi)^4 = square
(x + yi)^5 = pentagon
(x + yi)^6 = hexagon

[Marek14]

Not sure how many distinct cuts the mantis would have. As for polar versus Cartesian, my suspicion is "neither"; the way I imagine mantis, it could be expressed in cubindrical coordinates (i.e. two polar and two Cartesian coordinates), with the polar coordinates used for the hexagon plane and Cartesian for the remaining two.

Now, you say three intersecting line segments, but I suspect that it works a bit differently. For the tiger, you wouldn't spherate ends of two diagonals of the square, but rather the ends of two opposite sides (and the two torus cuts depend on which pair you choose).

Now, toratopes in general are probably best expressed in some sort of mix between polar and Cartesian coordinates. Take a torus, for example. Its simplest expression would be in cylindrical coordinates, because it has a cylindrical symmetry. In (r, fi, z), you would get a generating circle:

r = a, z = 0

and then you would express the torus as (r - a)^2 + z^2 = b^2. The angular coordinate phi will disappear because the shape is symmetrical (and I think this holds in general: the simplest expression of toratope will eliminate all angular coordinates).

So tiger would be expressed in duocylindrical coordinates (r1, fi1, r2, fi2) as

(r1 - a)^2 + (r2 - b)^2 = c^2

If you look just at the r1-r2 plane, you can see how this describes four circles (you can use xy plane with r1 = abs(x) and r2 = abs(y)).

But, mantis doesn't look like it has a rotational symmetry of this kind. So, I would guess that we should describe the 3D tiger cage (the oblique cut) in cylindrical coordinates, comparing it with the torus stacks in the same coordinate system...

[ICN5D]

Something like this?

Start with tiger expressed in cubindrical coordinates (r,phi,z,w)

(r - a)^2 + (sqrt(z^2 + w^2) - b)^2 = c^2

2 Toruses in column in cylindrical coord:

(r - a)^2 + (sqrt(z^2) - b)^2 = c^2

But, I'm not sure how to make a rotate function when r = sqrt(x^2 + y^2) . One would need to split up the x and y for a 45 degree oblique equation y -> w.

[Marek14]

Something like this?

Start with tiger expressed in cubindrical coordinates (r,phi,z,w)

(r - a)^2 + (sqrt(z^2 + w^2) - b)^2 = c^2

2 Toruses in column in cylindrical coord:

(r - a)^2 + (sqrt(z^2) - b)^2 = c^2

But, I'm not sure how to make a rotate function when r = sqrt(x^2 + y^2) . One would need to split up the x and y for a 45 degree oblique equation y -> w.

Well, you use transformation:

x = r*cos(fi)
y = r*sin(fi)

But you wrote the equation wrong -- for this, the cylindrical part will not be made from x and y, but from x and z:

(sqrt(x^2 + y^2) - a)^2 + (sqrt(z^2 + w^2) - a)^2 = b^2 (if the cage is 45 degrees, both major diameters must be the same; if they are not, the cage will be at different place)
x = r*cos(fi)
z = r*sin(fi)

(sqrt((r*cos(fi))^2 + y^2) - a)^2 + (sqrt((r*sin(fi))^2 + w^2) - a)^2 = b^2

Which leads to two torus stacks:

(sqrt((r*cos(fi))^2) - a)^2 + (sqrt((r*sin(fi))^2 + w^2) - a)^2 = b^2
(sqrt((r*cos(fi))^2 + y^2) - a)^2 + (sqrt((r*sin(fi))^2) - a)^2 = b^2

and to the cage

(sqrt((r*cos(fi))^2 + (y^2 + w^2)*sqrt(2)/2) - a)^2 + (sqrt((r*sin(fi))^2 + (y^2 - w^2)*sqrt(2)/2) - a)^2 = b^2

Now, if we eliminate both y and w, we get the equation for four circles:
(sqrt((r*cos(fi))^2) - a)^2 + (sqrt((r*sin(fi))^2) - a)^2 = b^2

And what now, though? If we put b = 0 here, we get once again four points based on 2 solutions for first part x 2 solutions for second part. To get mantis, we need an expression with 6 solutions total...