non-cubic toratopes?

Discussion of shapes with curves and holes in various dimensions.

Re: non-cubic toratopes?

Postby ICN5D » Mon Jan 26, 2015 6:43 pm

Okay,

I did some checking around, and experimented with the polar -> cartesian conversion.

4 circles in square, trace of tiger
(sqrt((r*cos(fi))^2) - a)^2 + (sqrt((r*sin(fi))^2) - a)^2 = b^2


Wikipedia Conversion

r = (sqrt(x^2+y^2))
phi = (arctan(y/x))



Four circles in square, this works perfectly as trace of tiger

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


Applying your method in above post:
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



Attempting six circles with your 3-period rotation, but only makes four ellipses.

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


So, if tiger cut works with polar -> cartesian conversion, then we should look for a working 3-period equation. I previously found this function to make a hexagon array, but perhaps for a different reason than what we're looking for:

(sqrt((x*sin(1.0467)+y*cos(1.0467))^2)-a)^2 + (sqrt((x*sin(2.0933)+y*cos(2.0933))^2)-a)^2 + (sqrt((x*sin(3.14)+y*cos(3.14))^2)-a)^2 = b^2

(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(pi)+y*cos(pi))^2)-a)^2 = b^2

Is there anything similar to this in polar coordinates? Seems like it's what we're after.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Mon Jan 26, 2015 7:12 pm

Hm, intriguing. And this gives six circles?
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Mon Jan 26, 2015 7:21 pm

Sure does. Scaling parameter 'a' moves all toward/away, scaling 'b' changes circle size. But, scaling 'a' large enough, and the circle diameters collapse, not indicative of a true trace array.

Image
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Mon Jan 26, 2015 7:25 pm

And what happens if you try different values for the three instances of a? Mantis should be able to have two different values here (then the circles would be in shape of alternating hexagon with sides a1,a2,a1,a2.a1,a2), but not three...
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Mon Jan 26, 2015 7:42 pm

Using the function

(sqrt((x*sin(1.0467)+y*cos(1.0467))^2)-a)^2 + (sqrt((x*sin(2.0933)+y*cos(2.0933))^2)-b)^2 + (sqrt((x*sin(3.14)+y*cos(3.14))^2)-c)^2 = d^2

setting a,b,c = 2 / d = 1.4 , we get hexagon array of the six circles.

Adjusting a=1.6 / b,c = 2

Image

Adjusting b=1.6 / a,c=2

Image

Adjusting c=1.6 / a,b=2

Image


This is because the function is the spherated endpoints of three intersecting dyads, angled at 0, 60, and 120 degrees. So, when offsetting a,b, or c, it influences only two of the six circles.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Mon Jan 26, 2015 7:54 pm

Hm, the mantis equation should use different dyads - edges of hexagon, not diagonals.
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Mon Jan 26, 2015 10:36 pm

Yes, this is true. It brings me back to the mathexchange response, in regards to a generic polygon equation.

It's a product of line segments, as rotated around the origin. Gets me thinking: how do we express four line segments in polar? Is it the one that worked before?

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

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


It seems like a circle in polar would be equal to a straight line with slope m=0 in cartesian. So, what is an actual line segment in polar, positioned away from origin?
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Klitzing » Mon Jan 26, 2015 10:57 pm

So, what is an actual line segment in polar, positioned away from origin?

Image
Ha, you know, in the above picture OS = CP = sin(b), OC = SP = cos(b), DT = tan(b), EK = cot(b), OT = sec(b), OK = csc(b).
Thus, when you are trying to describe a line (say through T and D) within polar coordinates,
then your angle would be b, and your radius would be csc(b) = 1/sin(b), ain't it? :P

--- rk
Klitzing
Pentonian
 
Posts: 1637
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: non-cubic toratopes?

Postby ICN5D » Mon Jan 26, 2015 11:11 pm

Ah, interesting! By the looks of this, I think what we're looking for are chord segments of that circle, where the endpoints are the circumradius. It's not present, but what would be the equation of line segment DE ? An equation for six line segments, as rotated about the origin, would be a product of these lines.

EDIT : For right triangles, seems to be as simple as Pythagoras's theorem of OD^2 + OE^2 = DE^2 , but what would the eq for six lines around hexagon be? The angles between OD and OE would be 60 degrees.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby ICN5D » Tue Jan 27, 2015 12:24 am

All right! I tweaked your mantis skeleton, Marek. Originally, you had

(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



What I found had worked was,

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

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

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

Makes six ellipses in hexagon

Image

Now, to unite as a midsection in 3D and 4D

Strange structure seen before in 3-intersecting dyad function : Stack of Genus-2 Tori
(sqrt(((sqrt(x^2+z^2))*cos(3*(arctan(z/x))/2))^2 + y^2) - b)^2 + (sqrt(((sqrt(x^2+z^2))*sin(3*(arctan(z/x))/2))^2 + a^2) - b)^2 = d^2

Image


Second midsection: regular tiger cut of column of two tori
(sqrt(((sqrt(x^2+a^2))*cos(3*(arctan(a/x))/2))^2 + y^2) - b)^2 + (sqrt(((sqrt(x^2+a^2))*sin(3*(arctan(a/x))/2))^2 + z^2) - b)^2 = d^2

Third Midsection : tiger cage at 45 degree angle
(sqrt(((sqrt(a^2+x^2))*cos(3*(arctan(x/a))/2))^2 + y^2) - b)^2 + (sqrt(((sqrt(a^2+x^2))*sin(3*(arctan(x/a))/2))^2 + z^2) - b)^2 = d^2



So, it seems to be on the right track, but uniting with presumed 3D and 4D equation, makes two genus-2 tori instead of three genus-1's. Maybe we also need to set up a 3-period rotation for plane YW as well. In 3D, we expect three separate tori, in 2 distinct arrangements. So, we're looking for a degree-12 equation with three periods of a rotated copy of a torus, that share a hexagon array of circles. For tiger, the tori lie flat in XY plane, separated by Z.

This function made two tori in a column, as a cut of tiger, derived by four circles in square array

(sqrt(((sqrt(x^2+z^2))*cos(arctan(z/x)))^2 + y^2) - b)^2 + (sqrt(((sqrt(x^2+z^2))*sin(arctan(z/x)))^2 + a^2) - c)^2 = d^2



Even if we express tiger in duocylindrical coordinates, with a dual 3-period rotation in XZ and YW ( just to experiement):

Tiger in Duocylindrical coordinates:

(sqrt((r1*cos(ϕ1))^2 + (r2*cos(ϕ2))^2) - a)^2 + (sqrt((r1*sin(ϕ1))^2 + (r2*sin(ϕ2))^2) - a)^2 = b^2


Converting to Cartesian
r1 = (sqrt(x^2+z^2))
ϕ1 = (arctan(z/x))
r2 = (sqrt(y^2+w^2))
ϕ2 = (arctan(w/y))


Tigroid with dual 3-period rotation
(sqrt((r1*cos(3*ϕ1/2))^2 + (r2*cos(3*ϕ2/2))^2) - a)^2 + (sqrt((r1*sin(3*ϕ1/2))^2 + (r2*sin(3*ϕ2/2))^2) - a)^2 = b^2

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

3D midsections
W-cut : same stack of genus-2 tori, as seen in above image
(sqrt(((sqrt(x^2+z^2))*cos(3*(arctan(z/x))/2))^2 + ((sqrt(y^2+a^2))*cos(3*(arctan(a/y))/2))^2) - b)^2 + (sqrt(((sqrt(x^2+z^2))*sin(3*(arctan(z/x))/2))^2 + ((sqrt(y^2+a^2))*sin(3*(arctan(a/y))/2))^2) - b)^2 = d^2

Hmm. I feel like we're getting close here. Need to find how to make a 3-period rotation of a whole torus.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby ICN5D » Fri Jan 30, 2015 2:02 am

I've been thinking about other coordinate systems, and how we might write mantis equation. If polar -> cartesian worked for the trace arrays, then maybe we should go looking into a more complex system, like toroidal coordinates. I did some searching around, and thought about spherical and/or hyperspherical coordinates, as well. But then I came across the toroidal system. There are a few things I noticed:

• How would we express a torus and/or tiger cut in toroidal coordinates?
• Maybe look into a generalization for tiger coordinates, and the inverse transform?

http://en.wikipedia.org/wiki/Toroidal_coordinates#Definition

What I have so far:

POLAR

x = (r*cos(ϕ))
y = (r*sin(ϕ))

r = (sqrt(x^2+y^2))
ϕ = (arctan(y/x))


SPHERICAL

x = (r*cos(ϕ1)*sin(ϕ2))
y = (r*sin(ϕ1)*sin(ϕ2))
z = (r*cos(ϕ1))

r = (sqrt(x^2+y^2+z^2))
ϕ1 = (arctan(y/x))
ϕ2 = (arccos(z/(sqrt(x^2+y^2+z^2)))


N-SPHERICAL

x_1 = (r*cos(ϕ1)
x_2 = (r*sin(ϕ1)*cos(ϕ2))
x_3 = (r*sin(ϕ1)*sin(ϕ2)*cos(ϕ3))
.
.
x_n-1 = (r*sin(ϕ1) . . . *sin(ϕ_n-2)*cos(ϕ_n-1))
x_n = (r*sin(ϕ1) . . . *sin(ϕ_n-2)*sin(ϕ_n-1))


Inverse
r = (sqrt(x_n^2 + x_(n-1)^2 + . . . + x_2^2 + x_1^2))
ϕ1 = (arccos(x_1/(sqrt(x_n^2 + x_(n-1)^2 + . . . + x_1^2))))
ϕ2 = (arccos(x_(n-2)/(sqrt(x_n^2 + x_(n-1)^2 + . . . + x_2^2))))
.
.
ϕ_n-2 = (arccos(x_1/(sqrt(x_n^2 + x_(n-1)^2 + x_1^2))))
ϕ_n-1 = (2arccot((x_(n-1) + sqrt(x_n^2 + x_(n-1)^2))/x_n))



TOROIDAL (o,t,fi)

x = (a(sinh(t)/(cosh(t)-cos(o)))*cos(fi))
y = (a(sinh(t)/(cosh(t)-cos(o)))*sin(fi))
z = (a(sin(o)/(cosh(t)-cos(o))))

Inverse to (x,y,z)

fi = (arctan(y/x))
t = (ln((sqrt((x^2+y^2+a)^2 + z^2))/(sqrt((x^2+y^2-a)^2 + z^2))))
o = (arccos(-((4a^2-(sqrt((x^2+y^2+a)^2 + z^2))^2-(sqrt((x^2+y^2-a)^2 + z^2))^2)/(2(sqrt((x^2+y^2+a)^2 + z^2))*(sqrt((x^2+y^2-a)^2 + z^2))))))
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby ICN5D » Sun Feb 01, 2015 7:54 pm

Thinking a little more about this application, I feel that a great approach may be to express a tiger in tiger coordinates, then change the period of rotational symmetry. This will tweak the tiger symmetry in its most pure form of expression. Then, convert back to cartesian. It just might work, if not bring us closer!
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby ICN5D » Tue May 12, 2015 6:53 pm

Here's something cool I made recently. It's the Tiger cut expressed in spherindrical coordinates, with general plane of rotation, and translate + rotate in 4D. The idea was to experiment around with a 3D version of polar coordinates, and testing it to see if mantis can be defined. So far, polar and spherical coords can define even numbers of inward-facing elliptic tori. The spherical coordinate system allows for not just horizontal arrangement of multiple tori, but also vertical. Setting a=2 and b=2 will make 8 inward-facing elliptic tori that sit on the surface of a sphere. Adding general plane of rotation and translate+rotate in 4D makes for some wild topology changes.


1. Start with torus with major diameter parallel to xz,

(sqrt(x^2+z^2)-3)^2 + y^2 = 1


2. Set up rotation from x->y, adjusting 'c' will spin the torus around z,

x = (x*sin(c)+y*cos(c))
y = (x*cos(c)-y*sin(c))

(sqrt((x*sin(c)+y*cos(c))^2+z^2)-3)^2+(x*cos(c)-y*sin(c))^2 = 1


3. Non-intersecting rotate torus around plane yw into 4D. Adjusting 'c' will change general plane of rotation of the torus, between ditorus and tiger intercepts,

y = (sqrt(y^2+w^2)-5)

(sqrt((x*sin(c) + (sqrt(y^2+w^2)-5)*cos(c))^2 + z^2)-3)^2 + (x*cos(c) - (sqrt(y^2+w^2)-5)*sin(c))^2 = 1


4. Set up rotation from z->w, adjusting 't' will spin the tiger/ditorus around plane xy,

z = (z*sin(t)+w*cos(t))
w = (z*cos(t)-w*sin(t))

(sqrt((x*sin(c) + (sqrt(y^2+(z*cos(t)-w*sin(t))^2)-5)*cos(c))^2 + (z*sin(t)+w*cos(t))^2)-3)^2 + (x*cos(c) - (sqrt(y^2+(z*cos(t)-w*sin(t))^2)-5)*sin(c))^2 = 1


5. Convert equation to spherical coordinates, with adjustable rotational symmetry. Adjusting 'a' and/or 'b' will duplicate the number of inward-facing tori horizontally or vertically, adjusting 'd' will slide up/down in 4D,

x = (ρ*cos(a*θ)*sin(b*ϕ))
y = (ρ*sin(a*θ)*sin(b*ϕ))
z = (ρ*cos(b*ϕ))
w = d


(sqrt(((ρ*cos(a*θ)*sin(b*ϕ))*sin(c) + (sqrt((ρ*sin(a*θ)*sin(b*ϕ))^2+((ρ*cos(b*ϕ))*cos(t)-d*sin(t))^2)-5)*cos(c))^2 + ((ρ*cos(b*ϕ))*sin(t)+d*cos(t))^2)-3)^2 + ((ρ*cos(a*θ)*sin(b*ϕ))*cos(c) - (sqrt((ρ*sin(a*θ)*sin(b*ϕ))^2+((ρ*cos(b*ϕ))*cos(t)-d*sin(t))^2)-5)*sin(c))^2 = 1




Ranges:

-10 < ρ < 10
0 < θ < pi
0 < ϕ < pi
XYZbox = -10 , 10


Adjustable Parameters:

a - changes rotation period of xy (1 < a < 4)
b - changes rotation period of xz (1 < b < 4)

c - changes general plane of rotation (0 < c < 1.57)

d - slides along 4D (-10 < d < 10)
t - rotates in 4D (0 < t < 1.57)
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Tue May 12, 2015 7:10 pm

Do you have some pictures? :)
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Tue May 12, 2015 11:32 pm

Yes. Here's an animation of a=2 , b=2 , d=0 , t=0 , and animating c from 0 to 6.28. It's among the wilder things this equation makes. This is also the first gif I made using a screencapture program. The timing is a little tricky, but it's still much easier than manually one by one.

Image
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Wed May 13, 2015 5:57 am

Now that's an interesting picture. It actually looks like two interlocked shapes, as there are two groups of four tetrahedrally symmetrical toruses. Toruses from the same group touch, but not those from different groups.
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Wed May 13, 2015 5:36 pm

Yeah, it's totally bizarre. I'm not sure what to call this shape. It starts with a spider, but has two of them stacked vertically, yet still 4D, and collapses toward a common center. Rotations will alternate the 8 elliptic tori to their mirror-image counterparts. So, it's like an object with 16 faces, where only 8 of those are wrapped as a torus at any given angle. This animation is actually changing the general plane of rotation, from a tiger to a ditorus-type shape. I'll make some more pics/animations of this function to show all it can do, especially when a=1 , b=1 for analogy.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby ICN5D » Thu May 14, 2015 12:03 am

Here's some pics of what different values of a, b, and t look like.


Image

Image

Image
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Thu May 14, 2015 5:51 am

What happens when you set a to 1.5?
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Thu May 14, 2015 6:33 pm

Interestingly enough, it does make three. But, the third one is sliced in half, and reflected on opposite sides. Spherindrical coordinates does a better job at making three than cubindrical (polar x square). You can see in the above posts how cubindrical makes two genus-2 tori. Spherindrical makes three genus-1's, with only two of them intact. This is good news, nonetheless. I haven't tried out hyperspherical coords yet.

Image
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby PWrong » Fri May 22, 2015 3:42 am

I'm loving these mantis pictures. Am I right in saying they look like they have a lot of right angles?

How would we express a torus and/or tiger cut in toroidal coordinates?

Coordinate systems are not very friendly. One of the an isosurfaces of toroidal coordinates is a set of torii. So any of these torii can be described by the equation tau = constant. But it's not so easy to describe an arbitrary torus. Similarly, it will be trivial to express some tigers in tiger coordinates, but very hard to express other tigers. The same holds for any shape. I'll give you a simple example.

The unit circle is much easier to express in polar coordinates than in Cartesian. Instead of x^2 + y^2 = 1, it's simply the equation r = 1. But if you try to express a circle centred on (a,b), the equation is
(r cos th - a)^2 + (r sin th - b)^2 = 1.
You can expand and simplify that a bit, but not much.
User avatar
PWrong
Pentonian
 
Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: non-cubic toratopes?

Postby ICN5D » Fri May 22, 2015 6:46 pm

We haven't created a true mantis, quite yet! It's a much tougher challenge than it sounds. I don't know what math patterns can create three perfect tori, that are inward-facing, and morph into an identical mirror-image arrangement, while still being 4D. Using polar/spherical in the tiger cut of ((II)(I)) of a column of tori can approximate with even numbers of elliptic tori. I wonder if parametric may be better/more efficient than implicit? I'm sure it will take some tweaking of the duocylinder margin in some as-of-yet unknown way.

And, yes, the ditorus cut of (((II))I), major pair of concentric tori, does lead to intersecting right-angle tori, when multiplying the period by integers. Do that enough, and you end up with something that looks like a hollow globe, with latitude/longitude lines.

The unit circle is much easier to express in polar coordinates than in Cartesian. Instead of x^2 + y^2 = 1, it's simply the equation r = 1. But if you try to express a circle centred on (a,b), the equation is
(r cos th - a)^2 + (r sin th - b)^2 = 1.


Well, that general circle equation is more straightforward than the one I found on wikipedia. I guess it's just an expanded form, solved for r, of what you posted. I might experiment with that later, in trying to make mantis or equivalent.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Sat May 23, 2015 6:04 am

Hm, your last post got me thinking. One thing that could be explored are variants of Cartesian product.

A regular Cartesian products combines two shapes in orthogonal dimensions. So an obvious variant is to use non-orthogonal dimensions.

Here's another thought: what is the operation that creates mantis-like shapes and how many of them there are?

I'm thinking something like this: If a toratope has a cut with two shapes, create a circle passing through their centers, then redefine the toratope with respect to that circle and move the shapes on it until three fit.

We can't do this in 2D, as there is no circle to construct in the 1D cut of circle. But we can do this in 3D.

The "2 circles" cut of torus can be modified to contain 3 or more circles, leading to 3 or more half-toruses merging on "poles" in some way. So, maybe this is just an envelope of some number of half-circles or "longitude lines", like you're finding?

In that case, it's clear why the odd numbers present a problem -- to make them work, we need to consider not the equations of circle, but of a SEMICIRCLE, which might be best to do as parametric equations. I am not completely clear on how the poles of this "protomantis" would look -- would they have edges or would they be smooth?

In 4D, spheritorus has a cut of 2 spheres which can be expanded to 3 or more, and would work on the same skelet as protomantis, just in 4D. Ditorus and tiger have cuts in shape of two toruses, and here is where it gets interesting -- there are several different ways to put a circle through these toruses. All in all, a number of toruses along the edge of circle can be oriented in three different ways:

Flat (lying in the circle plane), for ditorus
Beads (threaded along the circle), for ditorus
Fence (stading along the border), for tiger

The first case is a spherated protomantis and since mantis is basically half-toruses joined together, this is joining of half-ditoruses. The second case is joining of another kind of half-ditoruses: maybe if you cut a torus into top and bottom half, attach multiples of this to the common plane in 4D and spherate that?

The third case, then, should lead to the original mantis. And seeing it from this perspective, it shows what its might truly be: multiple halves of duocylinders stuck together around a common plane.
This means that spider (8-legged) should simply have two intersecting duocylinders as its skelet and that gets spherated.

But this is still not the end of this approach, as it shows new possibilities for both spheritorus and tiger: the spheres or toruses might be considered to not lie on a circle, but on a sphere. This might potentially lead to things like a toratope based on spherating four semicircles arranged in tetrahedral configuration around their common edge in 4D, or a similar tiger-like.
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby Klitzing » Sat May 23, 2015 8:33 am

Speaking of non-cartesian coordinates, esp. wrt. mantis,
then you might want to consider Wendy's interpretation of Dynkin diagrams
as a set of oblique coordinates ...

Or applying kind a Wythoff kaleidoscopic construction to curved things -
you just would have to ensure for overall smoothness that your used partial manifold
should be everywhere orthogonal to the mirrors.

--- rk
Klitzing
Pentonian
 
Posts: 1637
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: non-cubic toratopes?

Postby ICN5D » Fri May 29, 2015 12:33 am

Still thinking about this. I experimented with half-circle equations, and making a tiger from them. Defined implicitly, it doesn't work with cartesian product of two disk edges. I also tried making a margin with 3 disk edges, embedded in 4D. It makes a hexagon of circles/ellipses, but trying to connect them as 3 tori always ends up as the two genus-2 tori.

Looking at the last pic above, if there was some way to divide the graph in half, then give one half a 180 degree flip, it would make three tori of the mantis. This would work for any odd numbers of tori. But, that would seem like we're trying to modify an approximation. A real mantis should be definable with a single elegant equation. I don't know how to do it, but would piece-wise functions be useful here?

Using different coordinate systems gets really close to what we're looking for. The rotations and translations work as expected, since we're using a tiger.

On the topic of redefining tori on a circle, the moving them so three fit: what types of equations are you thinking? How does one do that in general?
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Fri May 29, 2015 6:56 am

Well, what I meant was that perhaps we should look at the torus and ditorus relatives of mantis instead of mantis itself, as they are easier to grasp.

Here's a new thought:

http://mathworld.wolfram.com/CassiniOvals.html

The torus slices have a property which we haven't used so far: a point on a torus slice has a constant product of distance from two points (I guess those are the points where generating circle of the torus intersects the slice plane).

So, what if we tried to define a three-pronged torus by having its slices have a constant product of distance from three vertices of equilateral triangle? Then we could just move those points along the longitude lines of a sphere.

Alternately, consider the tiger cage. By comparing the tiger cage with a simple torus, something might be revealed as well.

Another thought: What use are these multipronged toruses? Well, a three-pronged torus is a surface of genus 2, like torus with two holes, but it seems a very symmetrical one. By increasing the number of prongs, we can increase the genus indefinitely. By extending this to the higher toratopes, we could construct models of more topological hypersurfaces.

So the extended family would look like this:

2D:
Circle (II)

3D:
Sphere (III)
Torus ((II)I)
Multitoruses ([II]n I) based on multiple semicircles attached to a common line.

4D:
Glome (IIII)
Torisphere ((III)I)
Multitorispheres ([III]n I) based on multiple half-spheres attached to a common circle.
Spheritorus ((II)II)
Multispheritoruses ([II]n II) based on multiple semicircles attached to a common line. This may include both 2D and 3D arrangements of semicircles, creating a mid-cut with circular or polyhedral arrangement of spheres.
Ditorus (((II)I)I)
Dimultitoruses (([II]n I)I) based on multitoruses.
Multiditoruses ([(II)I]n I) based on multiple half-toruses attached to common pair of concentric circles.
Multidimultitoruses ([[II]m I]n I) based on multiple halves of multitoruses attached to common group of circularly arranged circles.
Tiger ((II)(II))
Multitigers ([II]n (II)) based on multiple halves of duocylinder margins (circle x semicircle) attached to a common pair of parallel circles. Mantis would belong here.
Hypertigers ([II]m [II]n) based on multiple quarters of duocylinder margins (semicircle x semicircle). Their attachment seems more vague to understand.
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Fri May 29, 2015 7:02 am

I came across some interesting things on the roots of unity: http://en.wikipedia.org/wiki/Root_of_unity#Examples I think that's what we're trying to do. Considering the tiger equation,

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

making any 3D cut will produce an equivalence to

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

which factors out into two toruses as the + , - roots of the polynomial. Each torus can be considered a root of x^2 - 1 == (x-1)(x+1)

(sqrt(x^2 + y^2)-a)^2 + (z-b)^2 = c^2 , (sqrt(x^2 + y^2)-a)^2 + (z+b)^2 = c^2

So, in the case of mantis, we're aiming for an equation that has three roots of unity, of a whole torus. We can factor out a non-principle root torus, and get the primitive cube root toruses, similar to z^3 - 1 == (z-1)(z^2 + z +1) , where the value of 1 is the major diameter. I will need to work on this for a bit. It's an infectious idea, and something new to play around with. The big question (plus many more) is:

What is the equation for the primitive cube root toruses, analogous to (z^2 + z +1)?

How do we express a tiger in cubed form?
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby Marek14 » Fri May 29, 2015 7:12 am

Hm, ICN5D, once again, I think that we should first understand this for simpler shapes before trying on tiger. Multitoruses should be understood first, especially the question if order-4 multitorus is the same thing as tiger cage...
Marek14
Pentonian
 
Posts: 1191
Joined: Sat Jul 16, 2005 6:40 pm

Re: non-cubic toratopes?

Postby ICN5D » Fri May 29, 2015 7:15 am

So, what if we tried to define a three-pronged torus by having its slices have a constant product of distance from three vertices of equilateral triangle? Then we could just move those points along the longitude lines of a sphere.

Alternately, consider the tiger cage. By comparing the tiger cage with a simple torus, something might be revealed as well.



That's another good idea. It might tie in with what i'm thinking about the three roots. The centers of the 3 circles (of a 3-prong torus) are located at the roots of unity.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: non-cubic toratopes?

Postby wendy » Fri May 29, 2015 8:13 am

I sometimes wonder if ye have ways for looking at whether things are one piece or two or three etc.

The may-13 piccie looks like some sort of partial hedral spheration of oo3ox3xo3oo, that is a pair of inverted rectified pentachora (tetrahedron || octahedron by Klitzing). I think the same fate is going to befall the roots of unity, even if one manages to pull off an enigma-style coiling.

Talking of which, you can have coils in a clifford ring. This gives rise in 2D to Lassijous/Bowditch figures, but you get the result by taking w = A cos(a.t), x = A sin(a.t), y = B cos(b.t) and z = B sin(b.t). If you set eg a=1 and b=360, you get a fairly tight-looking coil. I think this one's fairly simple topologically, since all string structures undo in 4D, but it's a handy starting point for something.

Have you looked at things like say, the spherated hexagons of o3x3x3o etc, which give a contigious structure of 30 hexagons, with two 'holes' made out of the five sets of truncated tetrahedra. I often play around with o3x4x3o similarly done. It's a kind of ratrace thing, where you have "rooms" bounded by octahedral plates of the 48 tC.

I have not looked at the mantis-like structures, though.

All in all, a jolly good show, lads.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

PreviousNext

Return to Toratopes

Who is online

Users browsing this forum: No registered users and 14 guests

cron