non-cubic toratopes?

[ICN5D]

It's a tough one, I know. Slicing the 3-prong, we get a weird concentric hemi-circle surface, or 3 circles in triangle. The torus makes two concentric circles, and two displaced. But, a tiger as derived from a torus, has no concentric slices in 3D. So, by this principle, maybe a mantis has only two displaced slices, of the 3 tori.

And at the same time, a mantis would have the same structure of the the 3 hemi-circle frame (embedded with a torus in every point), which seems like it would have a weird hemi-torus concentric slicing.

I recently thought about trying to fabricate the mantis using the same method of how I approximated the 3-prong. The 3-prong can be defined as a tri-toroidal cassini system, then adjusted to self-intersection. At a certain value, the three tori converge to form a 3-prong.

I wonder if I took three tigers, multiplied them together, with a common minor diameter, into a tri-tigroidal cassini system. A self-intersecting version may approximate a mantis closer than anything I've made so far. Still not a true mantis, it would feature three perfectly formed tori in the fence arrangement, as a function of 4 variables, and may have the common hexagon of 6 circles trace array, under rotation.

This is the function that made the 3-prong:

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

Symmetrical Three-Prong Multitorus at a=2 , b=5 , c=2.5

[Keiji]

Wow, that's an amazing looking figure! Thank you for posting it

I haven't really been following this thread, just happened to notice the image.

And that got me really interested in the possibilities...

Looking back at Marek's list:

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.

I suppose we could say that sequence ([II]n I) = {(III), ((II)I), ([II]3 I), ([II]4 I), ([II]5 I), ...} follows the sequence of regular polygons including the initial degenerate cases Gn = {point, digon, trigon, square, pentagon, ...}

As I see it, the most direct extension of the 3D torus into higher dimensions is the sequence ((II)I), ((III)I), ((IIII)I), ... - so the most direct extension of this new sequence of multitoruses into higher dimensions is as ([II]n I), ([III]n I), ([IIII]n I), ... - this case seems simple enough to understand.

If we now look at the opposite case - the spheritorus instead of the torisphere, and the related sequence ((II)I), ((II)II), ((II)III), ... - I believe these would be arrangements of semicircles around a common something - not necessarily a line (digon) as Marek states. For the ((II)II) case I imagine any of the Gn sequence could replace the line. Therefore ([II]n II) isn't specific enough to represent the full array of possibilities. I believe ([II]n II) would represent the n-prisms, while other possibilities would be non-prismatic 3D figures, such as a dodecahedron, an Archimedean solid, or maybe even some (all?) of the Johnson solids.

If the above is true then the multitoruses and multitorispheres (and their analogues in higher dimensions) are perhaps better represented by replacing "n I" with "Gn" - and then the multispheritoruses can be represented by replacing "n II" with "X" - any qualifying 3D polytope. So ([II]5 I) is now ([II] G5), and potential multispheritoruses might be ([II] +G5) (pentagonal prism arrangement), ([II] Ko6) (truncated octahedral arrangement), etc.

Now, I'm a fan of powertopes, so you know where I'm going with this...

For the ((II)III) case, the family elements are identified by a 4D polytope. I imagine a duoprism would qualify. Would a duotegum also qualify? How about a square octagoltriate?

In general, what defines whether a multitorus exists for a given polytope? Or do they all exist?

Are there any restrictions on combining different qualifying polytopes at different nesting levels? For example, does ([[II] X] Y) exist for all valid ([II] X) and ([II] Y)?

As if that didn't raise enough questions already, I notice that my replacement of "n I..." with "X" doesn't work for tigers, because the "I" is not there. So, is ([II] X [II] Y) a construction that even makes sense? Or is some even more elaborate notation required to fully describe the higher-dimensional hypertigers?

[ICN5D]

It's a neat object, isn't it? I've been checking out how it intercepts a 2-plane, to get a better understanding of what to expect from mantis, or the multiditorus. This 3-prong multitorus in 3D makes a different intersection in each coordinate 2-plane:

xy : the triangle array of 3 circles, with a reverse orientation after 180 degree flip
xz : a balloned up letter C, with a reverse orientation after 180 flip : the concentric circle pair cut in half, with smaller hemi-circles closing off the open ends
yz : two disjoint ellipses , when none of the arms are in 2D

These sections have smooth and symmetrical transformations under rotation, something I've been meaning to animate. A few things I noticed about them:

The 3 circles have two distinct, reflected arrangements, which do not transform by a 90 degree rotation, but a 180 degree flip.

When a single C-shaped arm is in the real plane, there are two more that stick off at plus/minus 120 degrees into the complex plane. This is when just one of the hemi-circle lobes sits in the 2-plane.

After studying the 3-prong like this, I think I've figured out what the other sections of mantis will look like. And, I feel that the notation needs to be more elaborate to define them, if it's supposed to include hyperplane intersections as well as the whole shape.

The mantis should have something analogous to the ballooned up C , which would look like what you'd get when taking the vertical column of tori from tiger cut, chopping them in half into a column of hemi-toruses, then connecting the two together with smaller hemi-toruses at the open ends, into one continuous loop. This will be one of the hemi-circle edges sticking into the 3-plane, with a torus embedded into it in a tiger-like fashion.

I do not believe the 3 torus-in-a-triangle arrangements of a mantis are transformable into each other by a 90 degree rotation, without passing by one of the single arm sections, as described above. I base this solely on how the common frame of the 3 hemi-circle edges intercept 2D. If and when a mantis gets defined and rendered into being, it will be utterly bizarre, compared to what we see with regular toratopes!

Anyways, that's what I think I've figured out, so far.

[Teragon]

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.

Wow! Those rotatopes are crazy. What exactly is shown in this animation? A single rotation with a changing plane of rotation?

[ICN5D]

Well, firstly, this is just a wild concoction I made one day, as part of an experiment. I was trying to see if I could make a particular theoretical 4D torus called the Mantis. As you may have seen around this forum, there's a 4D torus called the tiger. This one will intercept a 3-plane as a vertical column of 2 toruses, which divide/merge when sliding along 4D. The Mantis will intercept as 3 toruses, inward-facing, in a triangular arrangement. This happens because when you set one of the 4 variables to zero, the cross-section equation becomes reducible, into perfect cube roots of a torus, in that exact arrangement.

My idea was first to approximate the Mantis, by defining a 4D tiger in a hyperspherical coordinate system (sphere x line), with the ability to adjust periodicity, and the general plane of rotation. The plane of rotation can be changed to morph from a 3-torus to a tiger (from a side-by-side pair of 2 toruses to the vertical column). If we multiply the periodicity, we'll get even numbers of inward-facing tori, like the Mantis (but not odd, which was the goal). There are two directions to change the periodicity, which will multiply the amount of inward-facing tori spaced over the surface of a sphere. That's what is shown in these pictures on this thread.

So, this animation is when both directions of periodicity are mult by 2, and rotating the general rotation plane by 360 degrees, continuously. It basically morphs between a 3-torus equivalent to a tiger equivalent. The output of the graph is pretty wild, and there's much more to it than what I made.

[Teragon]

I've read a bit about what a mantis is expected to be. It's a far more complicated to imagine than a tiger. If you know about one cross-section, you don't know about all the others automatically, let alone how the 3D shape that a 4D beeing would perceive looks like and how it transforms under rotations. Anyway cross-sections give a very limited view on 4D objects. My intuition tells me the odd toratopes are no more complicated than the even toratopes, it seems to be a problem arising from the derivation starting from the tiger. Theoretically it would be enough to stretch the whole object's angular dependence in the azimuth direction by a factor of 1.5, which is obviously not what the parameter "a" does. It's hard to figure out where the azimuth-dependence is, with all those rotations in the equation.

So, this animation is when both directions of periodicity are mult by 2, and rotating the general rotation plane by 360 degrees, continuously. It basically morphs between a 3-torus equivalent to a tiger equivalent. The output of the graph is pretty wild, and there's much more to it than what I made.

Ok, so you're not changing the point of view, but morphing the object between tiger- and the 3-torus-configuration?

[ICN5D]

Actually, I think I've figured out the mantis slices. The common geometry between a mantis and a 3-prong multitorus is a product of three half-disk edges, the tri-edge (the edges of a trigonal hosohedron), joined at two poles over the surface of a sphere. So, there should be some parallels in the slice morphings.

A 3-prong can be defined as a circle embedded into the tri-edge frame. A multi-ditorus is when you embed a torus into the tri-edge, in a similar orientation as the circle. A mantis is the other way to position a torus, when it gets embedded. The two different positions of the torus are the differences between the 3-torus and the tiger. The multi-ditorus and the mantis are analogous in this way.

After studying the rotating 2D slices of a 3-prong, you will see a finite number of distinct transformations that are unique to the symmetry of a tri-edge frame. That's the key idea behind interpreting the slices. Static images of slices on coordinate planes aren't that good of a representation. Too much handwaving is involved to make the picture clearer.

But, the dynamic rotation morph of a slice will, in fact, reveal far more information than you might expect. So long as we can indirectly interpret them correctly, to fill in the missing details. That's what I've been doing for a while. I'll probably animate the 3-prong as graphed in 2D someday, since it's interesting to see. I'm not yet sure how to manually 'build' the other proposed mantis slices, without drawing them. Maybe I can use CSG in POVray to do the trick. Describing all of these relations are best done visually.

Ok, so you're not changing the point of view, but morphing the object between tiger- and the 3-torus-configuration?

Yes, by rotating the stationary plane of rotation. This plane of rotation is the one defined in the construction process, when the starting torus was rotated into 4D around in a circle.

Theoretically it would be enough to stretch the whole object's angular dependence in the azimuth direction by a factor of 1.5, which is obviously not what the parameter "a" does. It's hard to figure out where the azimuth-dependence is, with all those rotations in the equation.

The problem with trying to use 1.5 this way, is from changing the coordinate system to approximate a shape. The periodicity in polar, cylindrical, and spherical only work with whole integers to make unbroken graphs of toruses.

There is only one rotation parameter in the equation, that rotates the 3D slice. All of the rest of the trig functions are for the stationary plane adjustment for tiger/3-torus interchange, and the spherical coordinate change.


EDIT : The periodicity in polar, cylindrical, and spherical only work with whole integers to make unbroken graphs of toruses

[mr_e_man]

Here's my approach to the mantis, spider, etc.

We know that its cross-section in the x,y-plane should be an array of 2n circles, with the symmetry of a regular n-gon. One symmetry domain contains one circle. Let's say a and b are the distances from the circle's centre to the lines of symmetry, and c is the radius of the circle. The two lines of symmetry, and the two lines through the circle's centre, form a tetragon.

mantis1.png (35.34 KiB) Viewed 36451 times

Its edge lengths and angles are A, 360°/(2n), B, 90°, a, (180° - 360°/(2n)), b, 90°; and its diagonal (the distance from the origin to the circle's centre) is d. These lengths are related by

A = (a + b cos(360°/(2n))) / sin(360°/(2n)), B = (b + a cos(360°/(2n))) / sin(360°/(2n)),
a = (A - B cos(360°/(2n))) / sin(360°/(2n)), b = (B - A cos(360°/(2n))) / sin(360°/(2n)),
d² = A² + b² = B² + a² = (a² + b² + 2ab cos(360°/(2n))) / sin²(360°/(2n)).

For the case n=3, we have

A = (2a + b)/√3, B = (a + 2b)/√3,
d² = 4/3 (a² + ab + b²).

Now, I'll assume that the equation for the 4D object has the form F(x,y,z,w)=0, where F is a polynomial. If we're considering not just one object, but a whole family of objects, with a,b,c as variables, then we may write this as F(x,y,z,w; a,b,c) = 0.

The object, and the polynomial, should have a symmetry of reflection along the z-axis, and along the w-axis: F(x,y,z,w) = F(x,y,-z,w) = F(x,y,z,-w). It should also have a symmetry of rotation in the x,y-plane: F(x,y,z,w) = F(x cos(360°/n) + y sin(360°/n), y cos(360°/n) - x sin(360°/n), z,w). And the x-axis is a polygonal symmetry line; reflecting across (perpendicular to) the x-axis is the same as reflecting along (parallel to) the y-axis, so we should also have F(x,y,z,w)=F(x,-y,z,w).

These symmetry properties imply that the polynomial can only have certain combinations of terms, so it can be rewritten in the form

F(x,y,z,w) = G((x²+y²), Re[(x + iy)ⁿ], z², w²),

where G is some lower-degree polynomial.

There is one more important symmetry. Rotating by 360°/(2n) in the x,y-plane, and swapping z and w, should have the same effect as swapping a and b :

F(x,y,z,w; a,b,c) = F(x cos(360°/(2n)) + y sin(360°/(2n)), y cos(360°/(2n)) - x sin(360°/(2n)), w,z; b,a,c).

Next, we'll get more specific about the form of F, using the equation for an array of toruses (tori?).

[mr_e_man]

Let A₀,B₀,a₀,b₀,d₀ be the vectors in the tetragon shown above, with A₀,B₀,d₀ going out from the origin, and a₀,b₀ going in to the circle. Let A_k,B_k,a_k,b_k,d_k be the corresponding vectors in the k'th symmetry domain. Some obvious relations between these:

d_k = A_k + b_k = B_k + a_k
A_k•b_k = B_k•a_k = 0
A_2k = A_2k-1, B_2k = B_2k+1
a_2k = - a_2k+1, b_2k = - b_2k-1.

The coordinates of these vectors are given by

A_2k = A (e_xcos(360°k/n) + e_ysin(360°k/n)),
B_2k = B (e_xcos(360°(2k+1)/(2n)) + e_ysin(360°(2k+1)/(2n))),
a_2k = a (e_xsin(360°(2k+1)/(2n)) - e_ycos(360°(2k+1)/(2n))),
b_2k = b (- e_xsin(360°k/n) + e_ycos(360°k/n)).

Now, the 4D object's cross-section in the x,y,z-space should be an n-gon array of tori, with major radius b and minor radius c. The centre of the k'th torus is A_2k, and this vector also represents the axis of the torus. So, with the position vector r = (x,y,z,w) = x e_x + y e_y + z e_z + w e_w (of course w=0 for the moment), the equation of the k'th torus is

0 = (r² + A² + b² - c² - 2A_2k•r)² - 4b²( ((b_2k/b)•r)² + (e_z•r)² )
= (r² + d² - c² - 2A_2k•r)² - 4(b_2k•r)² - 4b²z²
= (r²+d²-c²)² - 4b²z² - 4(r²+d²-c²)(A_2k•r) + 4(A_2k•r)² - 4(b_2k•r)²
= (x²+y²+z²+d²-c²)² - 4b²z² - 4A(x²+y²+z²+d²-c²)(x cos(360°k/n) + y sin(360°k/n)) + 4A²(x cos(360°k/n) + y sin(360°k/n))² - 4b²(y cos(360°k/n) - x sin(360°k/n))².

(Notice this difference of squares at the end, which can be factored:

(A_2k•r)² - (b_2k•r)²
= ((A_2k•r) - (b_2k•r)) ((A_2k•r) + (b_2k•r))
= ((A_2k - b_2k)•r) ((A_2k + b_2k)•r)
= (d_2k-1•r) (d_2k•r). )

The equation of the torus array is thus

0 = prod_k=0^n-1( (r²+d²-c²)² - 4b²z² - 4(r²+d²-c²)(A_2k•r) + 4(A_2k•r)² - 4(b_2k•r)² ),

which should be exactly F(x,y,z,0). It follows, from the general properties of polynomials, that F(x,y,z,w) = F(x,y,z,0) + w P(x,y,z,w) for some other polynomial P. In other words, if we want to construct F, once we have the above n-fold product (which is a degree 4n polynomial), any further added terms must have w as a factor.

The product for the trigonal array of tori, n=3, is

F(x,y,z,0) = prod_k=0²( (x²+y²+z²+d²-c²)² - 4b²z² - 4A(x²+y²+z²+d²-c²)(x cos(120°k) + y sin(120°k)) + 4A²(x cos(120°k) + y sin(120°k))² - 4b²(y cos(120°k) - x sin(120°k))² )

= ( (x²+y²+z²+d²-c²)² - 4b²z² - 4A(x²+y²+z²+d²-c²)(x) + 4A²(x)² - 4b²(y)² )
•( (x²+y²+z²+d²-c²)² - 4b²z² - 4A(x²+y²+z²+d²-c²)(- x/2 + y√3/2) + 4A²(- x/2 + y√3/2)² - 4b²(- x√3/2 - y/2)² )
•( (x²+y²+z²+d²-c²)² - 4b²z² - 4A(x²+y²+z²+d²-c²)(- x/2 - y√3/2) + 4A²(- x/2 - y√3/2)² - 4b²(x√3/2 - y/2)² );

but I think it's better to leave it in vector notation to maintain symmetry as long as possible:

F(x,y,z,0) = prod_k=0²( (r²+d²-c²)² - 4b²z² - 4(r²+d²-c²)(A_2k•r) + 4(A_2k•r)² - 4(b_2k•r)² ).

It took me a week or two, but I was able to expand this (partially, keeping some terms like (r²+d²-c²) intact) and simplify and re-factor it. Along the way, I used some identities like

sum_k=0^n-1A_2k•r = 0,
sum_k=0^n-1(A_2k•r)² = n/2 A² (x² + y²),
prod_k=0²(A_2k•r) = 1/4 A³ (x³ - 3xy²),

the first one valid for n>1, the second one valid for n>2, and the third for n=3. Here's the result:

F(x,y,z,0) =
((r²+d²-c²)² - 4b²z²)³ + 6d² ((r²+d²-c²)² - 4b²z²)² (x²+y²)
+ (r²+d²-c²)² (x²+y²) [9d⁴(x²+y²) + 48(d²+b²)b²z² - 12d²(r²+d²-c²)²]
- 64 (x²+y²) [3b⁶z⁴ + (x²+y²)(a²-ab)(a²+3ab+2b²)b²z²]
- 16/(3√3) (r²+d²-c²) (x³-3xy²) [9d²(2a+b)b²z² + (2a³+3a²b-3ab²-2b³) ((r²+d²-c²)² - 3d²(x²+y²)) ]
+ 64/27 (2a³+3a²b-3ab²-2b³)² (x³-3xy²)² - 64 a²b²(a+b)² (3x²y-y³)².

Note that the last term is (3x²y-y³)² = (x²+y²)³ - (x³-3xy²)², so this actually is a polynomial in (x²+y²), (x³-3xy²), and z², as expected. Now we just need to add some w² terms to get a mantis.

But what about symmetry? Given the other symmetries, the one that swaps a and b could have, instead of a 6-fold rotation, a reflection along the x-axis (which is the composition of the 6-fold rotation, a 3-fold rotation, and a reflection along the y-axis):

F(x,y,z,w; a,b,c) = F(-x,y,w,z; b,a,c).

So we need to add at least these terms:

F(x,y,z,w) =
((r²+d²-c²)² - 4b²z² - 4a²w²)³ + 6d² ((r²+d²-c²)² - 4b²z² - 4a²w²)² (x²+y²)
+ (r²+d²-c²)² (x²+y²) [9d⁴(x²+y²) + 48 ((d²+b²)b²z² + (d²+a²)a²w²) - 12d²(r²+d²-c²)²]
- 64 (x²+y²) [3(b⁶z⁴+a⁶w⁴) + (x²+y²) ((a²-ab)(a²+3ab+2b²)b²z² + (b²-ab)(2a²+3ab+b²)a²w²) ]
- 16/(3√3) (r²+d²-c²) (x³-3xy²) [9d² ((2a+b)b²z² - (a+2b)a²w²) + (2a³+3a²b-3ab²-2b³) ((r²+d²-c²)² - 3d²(x²+y²)) ]
+ 64/27 (2a³+3a²b-3ab²-2b³)² (x³-3xy²)² - 64 a²b²(a+b)² (3x²y-y³)²
+ w² P(x,y,z,w).

(Notice, we couldn't have done this symmetrization so easily before I had expanded it. Or could we?)

Of course r²=x²+y²+z² becomes r²=x²+y²+z²+w². And we could add anything like (2a³b³z²w²) or (a⁴b²+a²b⁴)z²w² to that term in the middle (b⁶z⁴+a⁶w⁴). In fact, now that we have this symmetry, the cross-section in the x,y,w-space is another perfect array of tori, so any added terms must have not only w² but also z² as a factor. Try first adding (or subtracting) just a constant multiple of a⁴b⁴z²w² at the end (in place of w²P), and see if the result looks like a mantis.

[ICN5D]

All right, now you're talking! I've been staring at that wonderful expression you produced there. In those 2 posts, I can follow your reasoning to some degree. Since I have little formal training, that is some deep, dark voodoo magic you just busted out there, bud I get the idea of certain symmetries allowing only certain terms in the equation. I suspected that some more powerful tools could get close to a 4-variable degree-12 equation for this thing. But I haven't seen anyone spell it all out like this, yet. So, this is fascinating. I need to check out this equation, and get it into something calcplot will recognize.

I once derived a 3d solution equation for mantis, in cylindrical coords, where, r = sqrt(x^2 + y^2) and θ = arctan(y/x) :

(r^4+2r^2z^2+62r^2+z^4+22z^2+361)^3 -1200r^4(r^4+2r^2z^2+62r^2+z^4+22z^2+361) -16000r^6*cos(6θ) -192r^2(r^2+z^2+19)^2(r^4+2r^2z^2+22r^2+z^4+22z^2+361) +64r^3(r^2+z^2+19)(r^4+2r^2z^2-22r^2+z^4+278z^2+361)*sin(3θ) = 0

This plots a triangular array of 3 toruses standing on their rim. My next, more recent idea (experiment) was to define a mirror-image (upside-down) triangle array and call the 3rd variable "w" (when it's really z for the plotter) , and then compare it to this one above. Then, try to take all the terms containing w and insert them in to the equation above. It will produce a 4-variable equation that should morph between the triangular arrays by rotation on plane zw. Or, so it seems. It's an experiment I've never tried before, and it's a total shot in the dark, but could totally work, too.

But, I'll check out your equation first. I'll be back.

[mr_e_man]

(Notice, we couldn't have done this symmetrization so easily before I had expanded it. Or could we?)

Yes, we could! No need to take a month to expand and simplify the 4-fold product for a spider; just leave the product as is. I'll explain.

This plots a triangular array of 3 toruses standing on their rim. My next, more recent idea (experiment) was to define a mirror-image (upside-down) triangle array and call the 3rd variable "w" (when it's really z for the plotter) , and then compare it to this one above. Then, try to take all the terms containing w and insert them in to the equation above. It will produce a 4-variable equation that should morph between the triangular arrays by rotation on plane zw. Or, so it seems. It's an experiment I've never tried before, and it's a total shot in the dark, but could totally work, too.

You have the right idea here. It's basically what I did with the symmetry that swaps a and b (the two major radii), and swaps z and w, and rotates in the x,y-plane by 360°/(2n) (in this case 60°, thus replacing the triangle by its mirror image). But I didn't know how to enforce this symmetry and add the appropriate w terms without expanding the product.

My new approach doesn't rely on symmetry; it can take any two 3D objects as cross-sections, defined by polynomial equations F(x,y,z,0)=0 and F(x,y,0,w)=0, provided that they have the same 2D cross-section F(x,y,0,0)=0, and construct a 4D object F(x,y,z,w)=0. Of course it's not unique; any terms can be added with zw as a factor, to get a different shape.

Clearly, from the properties of polynomials, F(x,y,z,0) is exactly all of those monomials in F(x,y,z,w) that don't have w as a factor. So F(x,y,z,w) - F(x,y,z,0) is exactly all of the terms that do have w as a factor. Similarly, F(x,y,0,0) is all of the terms that have neither z nor w as a factor, and F(x,y,z,0) - F(x,y,0,0) is all of the terms that have z but not w as a factor. So we can decompose F into four components as follows:

F(x,y,z,w) = [F(x,y,0,0)] + [F(x,y,z,0) - F(x,y,0,0)] + [F(x,y,0,w) - F(x,y,0,0)] + [zw P(x,y,z,w)],

where the first component has neither z nor w, the second component has z but not w, the third component has w but not z, and the fourth component has both z and w. This can be rearranged to

F(x,y,z,w) = F(x,y,z,0) + F(x,y,0,w) - F(x,y,0,0) + zw P(x,y,z,w).

For our 2n-legged tiger, we have an array of n tori in the x,y,z-space, a different array of n tori in the x,y,w-space, and a common array of 2n circles in the x,y-plane:

F(x,y,z,0) = prod_k=0^n-1( (x²+y²+z²+d²-c² - 2A_2k•r)² - 4b²z² - 4(b_2k•r)² )
= prod_k=0^n-1( (x²+y²+z²+d²-c² - 2A(x cos(2πk/n) + y sin(2πk/n)))² - 4b²z² - 4b²(y cos(2πk/n) - x sin(2πk/n))² ),

F(x,y,0,w) = prod_k=0^n-1( (x²+y²+w²+d²-c² - 2B_2k•r)² - 4a²w² - 4(a_2k•r)² )
= prod_k=0^n-1( (x²+y²+w²+d²-c² - 2B(x cos(π(2k+1)/n) + y sin(π(2k+1)/n)))² - 4a²w² - 4a²(x sin(π(2k+1)/n) - y cos(π(2k+1)/n))² ),

F(x,y,0,0) = prod_k=0^2n-1( x²+y²+d²-c² - 2d_k•r )
= prod_k=0^n-1 (x²+y²+d²-c² - 2d_2k•r) (x²+y²+d²-c² - 2d_2k-1•r).

(Now how can I rewrite that dot product...

d_2k•r = A_2k•r + b_2k•r
= (a + b cos(π/n))/sin(π/n) (x cos(2πk/n) + y sin(2πk/n)) + b (y cos(2πk/n) - x sin(2πk/n))
= (ax cos(2πk/n) + bx cos(2πk/n + π/n) + ay sin(2πk/n) + by sin(2πk/n + π/n)) / sin(π/n)
= (a (x cos(2kπ/n) + y sin(2kπ/n)) + b (x cos((2k+1)π/n) + y sin((2k+1)π/n))) / sin(π/n),

d_2k-1•r = A_2k•r - b_2k•r
= (a + b cos(π/n))/sin(π/n) (x cos(2πk/n) + y sin(2πk/n)) - b (y cos(2πk/n) - x sin(2πk/n))
= (ax cos(2πk/n) + bx cos(2πk/n - π/n) + ay sin(2πk/n) + by sin(2πk/n - π/n)) / sin(π/n)
= (a (x cos(2kπ/n) + y sin(2kπ/n)) + b (x cos((2k-1)π/n) + y sin((2k-1)π/n))) / sin(π/n). )

So we add the two n-fold products, and subtract the 2n-fold product, and finally add (or subtract) some z²w² terms, to get an equation for a 2n-legged tiger. But this doesn't work well for the ordinary 4-legged tiger; the "remainder" is fully 8th degree, with lots of terms that would be impossible to guess, if we want the correct exact shape:

F(x,y,z,w) = (...) + z²w² (12(x²+y²-c²)² - 8(a²-b²)(x²-y²) - 8(a²+b²)c² + 4(a²+b²)² - 8(a²-b²)² - 64a²b² + 12(x²+y²-c²)(z²+w²) + 4(a²-b²)(z²-w²) + 4z⁴ + 6z²w² + 4w⁴).

In contrast, my previous approach, of partially expanding the product and adding w² terms where they look like they should fit to match z², only requires further subtracting 64a²b²z²w² at the end to get an exact tiger. But my new approach is much easier for larger n. And we have no way to tell what a "correct" 2n-legged tiger should look like (beyond its general topology); no way to tell what terms should be added at the end, with either approach.

[Challenger007]

You have the right idea here. It's basically what I did with the symmetry that swaps a and b (the two major radii), and swaps z and w, and rotates in the x,y-plane by 360°/(2n) (in this case 60°, thus replacing the triangle by its mirror image). But I didn't know how to enforce this symmetry and add the appropriate w terms without expanding the product.

I'm not an expert, but maybe it's worth adding some variables to correctly display the volumetric body and change the directions of the vectors in dynamics, so as not to increase the scale?

[mr_e_man]

I don't know what you mean by "dynamics" or "increase the scale".

Perhaps you mean you want to see a morphing 3D object, because you can't see a 4D object all at once. Then you can just replace w with t (time) to translate the 4D object relative to the 3D cross-sectional space; or replace z with z cos(t), and replace w with z sin(t), to rotate.

That is a problem at this site, where we discuss complex concepts: Each of us has a different vocabulary and style, so it's often difficult to understand each other.

[ICN5D]

I'm not an expert, but maybe it's worth adding some variables to correctly display the volumetric body and change the directions of the vectors in dynamics, so as not to increase the scale?

We don't need any more variables for this thing The shape curves around (is "embedded") in a 4d space, so we only need 4 variables: x, y, z, w . And since it (a multi-tiger) only has 3 radius sizes, we only need 3 parameters for them: a, b, c . And we don't have an issue with the size (scale) , really. We're trying to get a hexagon array of 6 circles (on plane xy) to join as two mirror-image triangular arrays of toruses by rotation on plane zw (when we 'flip' the variables z and w back and forth by rotation). It's a hypothetical 4D donut shape that's cooler than anything I've rendered before. And we've been trying to define them with an equation for years.

[mr_e_man]

For reference, here's the polynomial for an ordinary tiger:

F(x,y,z,w) = ((x²+y²+z²+w²+a²+b²-c²)² - 4b²(y²+z²) - 4a²(x²+w²))² - 64a²b²(y²+z²)(x²+w²).

And here's the polynomial for a mantis, using the "expand product" approach (I've eliminated r² and d², and factored some more):

F(x,y,z,w) =
((x²+y²+z²+w²+4/3(a²+ab+b²)-c²)² - 4b²z² - 4a²w²)² [ (x²+y²+z²+w²+4/3(a²+ab+b²)-c²)² - 4b²z² - 4a²w² + 8(a²+ab+b²)(x²+y²) ]
+ 16 (x²+y²+z²+w²+4/3(a²+ab+b²)-c²)² (x²+y²) [ (a²+ab+b²)²(x²+y²) + (4a²+4ab+7b²)b²z² + (7a²+4ab+4b²)a²w² - (a²+ab+b²)(x²+y²+z²+w²+4/3(a²+ab+b²)-c²)² ]
- 64 (x²+y²) [ 3(b⁶z⁴+a⁶w⁴) + (x²+y²) ab(a-b) ((a²+3ab+2b²)bz² - (2a²+3ab+b²)aw²) ]
- 16/(3√3) (x²+y²+z²+w²+4/3(a²+ab+b²)-c²) x(x²-3y²) [ 12(a²+ab+b²)((2a+b)b²z²-(a+2b)a²w²) + (2a³+3a²b-3ab²-2b³)((x²+y²+z²+w²+4/3(a²+ab+b²)-c²)²-4(a²+ab+b²)(x²+y²)) ]
+ 64 [ 1/27(2a³+3a²b-3ab²-2b³)²x²(x²-3y²)² - a²b²(a+b)²y²(3x²-y²)² ]
+ z²w² [ ... ] .

Code: Select all

((x^2+y^2+z^2+w^2+4/3(a^2+ab+b^2)-c^2)^2 - (2bz)^2 - (2aw)^2)^2 ( (x^2+y^2+z^2+w^2+4/3(a^2+ab+b^2)-c^2)^2 - (2bz)^2 - (2aw)^2 + 8(a^2+ab+b^2)(x^2+y^2) )
+ 16 (x^2+y^2+z^2+w^2+4/3(a^2+ab+b^2)-c^2)^2 (x^2+y^2) ( (a^2+ab+b^2)^2(x^2+y^2) + (4a^2+4ab+7b^2)(bz)^2 + (7a^2+4ab+4b^2)(aw)^2 - (a^2+ab+b^2)(x^2+y^2+z^2+w^2+4/3(a^2+ab+b^2)-c^2)^2 )
- 64 (x^2+y^2) ( 3(b^6z^4+a^6w^4) + (x^2+y^2) ab(a-b) ((a^2+3ab+2b^2)bz^2 - (2a^2+3ab+b^2)aw^2) )
- 16/(3 sqrt(3)) (x^2+y^2+z^2+w^2+4/3(a^2+ab+b^2)-c^2) x(x^2-3y^2) ( 12(a^2+ab+b^2)((2a+b)(bz)^2-(a+2b)(aw)^2) + (2a^3+3a^2b-3ab^2-2b^3)((x^2+y^2+z^2+w^2+4/3(a^2+ab+b^2)-c^2)^2-4(a^2+ab+b^2)(x^2+y^2)) )
+ 64 ( 1/27((2a^3+3a^2b-3ab^2-2b^3)x(x^2-3y^2))^2 - (ab(a+b)y(3x^2-y^2))^2 )


And here's the polynomial for a mantis, using the "add products" approach:

F(x,y,z,w) =
[ (x²+y²+z²+4/3(a²+ab+b²)-c² - (2a+b)/√3 (2x))² - b² ((2z)² + (2y)²) ]
* [ (x²+y²+z²+4/3(a²+ab+b²)-c² - (2a+b)/√3 (-x+y√3))² - b² ((2z)² + (-y-x√3)²) ]
* [ (x²+y²+z²+4/3(a²+ab+b²)-c² - (2a+b)/√3 (-x-y√3))² - b² ((2z)² + (-y+x√3)²) ]
+
[ (x²+y²+w²+4/3(a²+ab+b²)-c² - (a+2b)/√3 (-2x))² - a² ((2w)² + (2y)²) ]
* [ (x²+y²+w²+4/3(a²+ab+b²)-c² - (a+2b)/√3 (x+y√3))² - a² ((2w)² + (-y+x√3)²) ]
* [ (x²+y²+w²+4/3(a²+ab+b²)-c² - (a+2b)/√3 (x-y√3))² - a² ((2w)² + (-y-x√3)²) ]
-
[ (x²+y²+4/3(a²+ab+b²)-c² - (2a+b)/√3 (2x))² - b²(2y)² ]
* [ (x²+y²+4/3(a²+ab+b²)-c² - (2a+b)/√3 (-x+y√3))² - b²(-y-x√3)² ]
* [ (x²+y²+4/3(a²+ab+b²)-c² - (2a+b)/√3 (-x-y√3))² - b²(-y+x√3)² ]
+
z²w² [ ... ] .

Code: Select all

( (x^2+y^2+z^2+4/3(a^2+ab+b^2)-c^2 - (2a+b)/sqrt(3) (2x))^2 - b^2 ((2z)^2 + (2y)^2) )
* ( (x^2+y^2+z^2+4/3(a^2+ab+b^2)-c^2 - (2a+b)/sqrt(3) (-x+y sqrt(3)))^2 - b^2 ((2z)^2 + (-y-x sqrt(3))^2) )
* ( (x^2+y^2+z^2+4/3(a^2+ab+b^2)-c^2 - (2a+b)/sqrt(3) (-x-y sqrt(3)))^2 - b^2 ((2z)^2 + (-y+x sqrt(3))^2) )
+
( (x^2+y^2+w^2+4/3(a^2+ab+b^2)-c^2 - (a+2b)/sqrt(3) (-2x))^2 - a^2 ((2w)^2 + (2y)^2) )
* ( (x^2+y^2+w^2+4/3(a^2+ab+b^2)-c^2 - (a+2b)/sqrt(3) (x+y sqrt(3)))^2 - a^2 ((2w)^2 + (-y+x sqrt(3))^2) )
* ( (x^2+y^2+w^2+4/3(a^2+ab+b^2)-c^2 - (a+2b)/sqrt(3) (x-y sqrt(3)))^2 - a^2 ((2w)^2 + (-y-x sqrt(3))^2) )
-
( (x^2+y^2+4/3(a^2+ab+b^2)-c^2 - (2a+b)/sqrt(3) (2x))^2 - (2by)^2 )
* ( (x^2+y^2+4/3(a^2+ab+b^2)-c^2 - (2a+b)/sqrt(3) (-x+y sqrt(3)))^2 - (b(-y-x sqrt(3)))^2 )
* ( (x^2+y^2+4/3(a^2+ab+b^2)-c^2 - (2a+b)/sqrt(3) (-x-y sqrt(3)))^2 - (b(-y+x sqrt(3)))^2 )


[mr_e_man]

No pictures yet?

I'll make my own, then. But my computer is slow; it can't handle a high resolution.

The first five images use the "expand product" approach, with the last term being -13⁴a⁴b⁴z²w², and a:b:c in the ratio 4:3:1. Here's the x,y,z cross-section:

mantis2.png (67.72 KiB) Viewed 36338 times

z,w rotated 33.75°:

mantis3.png (93.97 KiB) Viewed 36338 times

z,w rotated 45°:

mantis4.png (126.52 KiB) Viewed 36338 times

z,w rotated 60°:

mantis5.png (125.16 KiB) Viewed 36338 times

z,w rotated 90° (for x,y,w cross-section):

mantis6.png (96.27 KiB) Viewed 36338 times

The next four images use the "add products" approach, with the last term being -12⁴a⁴b⁴z²w², and a:b:c in the ratio 4:2:1.

z,w rotated 39.375°:

mantis7.png (68.22 KiB) Viewed 36338 times

z,w rotated 45°:

mantis8.png (116.26 KiB) Viewed 36338 times

z,w rotated 56.25°:

mantis9.png (149.71 KiB) Viewed 36338 times

z,w rotated 67.5°:

mantis10.png (128.27 KiB) Viewed 36338 times

[ICN5D]

Sorry dude! Went to the beach and was away from computer. But let me just say, wow! Your equation freaking works, man! That really blows my mind. I'd go with the first one you posted, using the expand product with a:b:c as 4:3:1 . All of the torus solutions are perfectly shaped, with the iconic "Mantis Cage" at the oblique angles.

I really must know everything that you know about this geometric algebra, my dude. Serious voodoo magic going on there. And thanks for putting this info here. One day, when I'm better trained at this stuff, I'll be able to come back here and understand your process!



I think a next good test is to take the other 3d slices, and try to see if it contains the 'closed-loop hemi-tiger' surface, seen here:



I made this thing by combining a family of 3 hemi-tigers into a cage-like structure, which is what we've been thinking mantis could be. It's actually not such a bad little trick, since it can help us plan the solutions of the other multi-tiger shapes. Most especially the polyhedral varieties: the simplest one being the tetrahedral-symmetric 4-prong. But the equations that define those other alien-looking slices are a mystery. Thankfully they can always be broken down into a product of four to eight separate surfaces arranged a certain way. That gives us something, but it's not elegant.

[Challenger007]

I'm not an expert, but maybe it's worth adding some variables to correctly display the volumetric body and change the directions of the vectors in dynamics, so as not to increase the scale?

We don't need any more variables for this thing The shape curves around (is "embedded") in a 4d space, so we only need 4 variables: x, y, z, w . And since it (a multi-tiger) only has 3 radius sizes, we only need 3 parameters for them: a, b, c . And we don't have an issue with the size (scale) , really. We're trying to get a hexagon array of 6 circles (on plane xy) to join as two mirror-image triangular arrays of toruses by rotation on plane zw (when we 'flip' the variables z and w back and forth by rotation). It's a hypothetical 4D donut shape that's cooler than anything I've rendered before. And we've been trying to define them with an equation for years.

Now it's clearer. It's great that we came up with the equation that creates this model.

[mr_e_man]

Sorry dude! Went to the beach and was away from computer.

Well, it looked like you were here for some time after you saw the equation, and you might have shown us the first fruits from CalcPlot3D or POV-Ray. But no matter.

I guess you were hindered by the equation being in the wrong format: d² needs to be changed to 4/3(a²+ab+b²), and r² to (x²+y²+z²+w²), and (...)² to (...)^2, and [...] to (...), and √3 to sqrt(3). (I've made those changes now; see 'code' above. ) That's not to mention converting 4D to 3D.

I really must know everything that you know about this geometric algebra, my dude. Serious voodoo magic going on there. And thanks for putting this info here. One day, when I'm better trained at this stuff, I'll be able to come back here and understand your process!

This isn't really geometric algebra; there are no multivectors or non-scalar products involved. It's just polynomials, and some vector algebra to reduce our dependence on coordinates.

Honestly, for the "expand product" approach, I'm not sure I understand the process myself! I mean, I don't know how far I have to go before I can easily convert the 3D polynomial F(x,y,z,0) to the 4D polynomial F(x,y,z,w). And the conversion is not unique; different-looking but equal forms for F(x,y,z,0) may suggest different-looking and unequal forms for F(x,y,z,w). For example (I mentioned this before), I would take b⁶z⁴ and change it to b⁶z⁴+a⁶w⁴, but (b³z²)² I would change to (b³z²+a³w²)² = b⁶z⁴+a⁶w⁴+2a³b³z²w². In any case, we must only add terms involving w (whether inside or outside parentheses), and in such a way that the result has the required symmetries.

[ICN5D]

And here's the polynomial for a mantis, using the "expand product" approach (I've eliminated r² and d², and factored some more):

What program do you use to expand and simplify these equations? I'd like to try this out. Also, now that you've established that the 4:3:1 ratio works best, the equation can be reduced even more, which is a neat little trick to overcome slow/lo-res rendering speeds. It might work better to fully expand it again then hard-set the parameters, then try to simplify it. The program usually finds other symmetries to factor out.

[ICN5D]

Sorry dude! Went to the beach and was away from computer.

Well, it looked like you were here for some time after you saw the equation, and you might have shown us the first fruits from CalcPlot3D or POV-Ray. But no matter.

I guess you were hindered by the equation being in the wrong format: d² needs to be changed to 4/3(a²+ab+b²), and r² to (x²+y²+z²+w²), and (...)² to (...)^2, and [...] to (...), and √3 to sqrt(3). (I've made those changes now; see 'code' above. ) That's not to mention converting 4D to 3D.

Well, I forgot to mention that I'm also kinda lazy I have to admit it. There's no telling how much more research and discoveries I could have made if I were more motivated. My spontaneous interest in things that can end abruptly is my main handicap in life. I'm always chasing the next shiniest object down the next rabbit hole. I don't know of any other existence without it, so I just kinda roll with it now. You know how it goes! Plus, these music video ideas have got me majorly distracted in all areas of my life right now.....

[mr_e_man]

And here's the polynomial for a mantis, using the "expand product" approach (I've eliminated r² and d², and factored some more):

What program do you use to expand and simplify these equations? I'd like to try this out.

Nothing. Just lots of paper.

The product for the trigonal array of tori, n=3, is

[...]

It took me a week or two, but I was able to expand this (partially, keeping some terms like (r²+d²-c²) intact) and simplify and re-factor it.

(Notice, we couldn't have done this symmetrization so easily before I had expanded it. Or could we?)

Yes, we could! No need to take a month to expand and simplify the 4-fold product for a spider; just leave the product as is.

Actually I've done that now, and it only took me a day. (Coincidentally, on the same day, I encountered several large 3D spiders.) I guess it was easier because the coordinate axes are aligned with the spider's symmetry axes. It may be easier in general when n is even, because each torus is exactly opposite another torus, and they can be paired together in the product, which then simplifies according to (u+v)(u-v) = u²-v².

Refer to the image of 6 circles in a mantis, and the tetragon's edge lengths. For a spider, the corresponding lengths are related by

A = a√2 + b, B = a + b√2,
d² = A² + b² = 2(a² + ab√2 + b²).

Also important to note is that A⁴-6A²b²+b⁴ gets negated when a and b are swapped, as shown by

A⁴ - 6A²b² + b⁴ = 4 (a² - b²) (a² + 2ab√2 + b²).

Relatedly, x⁴-6x²y²+y⁴ gets negated by a 45° rotation, as shown most easily by complex numbers. Let ζ₈=(1+i)/√2 be the eighth root of unity; then ζ₈(x+iy) is the rotation of x+iy. The polynomial is (x⁴-6x²y²+y⁴)=Re[(x+iy)⁴], which upon rotation becomes Re[(ζ₈(x+iy))⁴] = Re[(ζ₈)⁴(x+iy)⁴] = Re[(-1)(x+iy)⁴] = -Re[(x+iy)⁴]; thus it is negated. I point this out because of that required symmetry, which swaps a and b, and swaps z and w, and rotates x,y by 45°.

The square array of 4 tori is given by

F(x,y,z,0) =
[ (x²+y²+z²+d²-c²)² - 4b²z² - 4(x²+y²+z²+d²-c²)Ax + 4A²x² - 4b²y² ]
*[ (x²+y²+z²+d²-c²)² - 4b²z² - 4(x²+y²+z²+d²-c²)Ay + 4A²y² - 4b²x² ]
*[ (x²+y²+z²+d²-c²)² - 4b²z² + 4(x²+y²+z²+d²-c²)Ax + 4A²x² - 4b²y² ]
*[ (x²+y²+z²+d²-c²)² - 4b²z² + 4(x²+y²+z²+d²-c²)Ay + 4A²y² - 4b²x² ].

(See the first and third factors together have the form (u+v)(u-v), as do the second and fourth.) Expanding this,

F(x,y,z,0) =
((x²+y²+z²+d²-c²)² - 4b²z²)⁴ - 8d² ((x²+y²+z²+d²-c²)² - 4b²z²)³ (x²+y²)
+ 4 ((x²+y²+z²+d²-c²)² - 4b²z²)² [ 5(A²+b²)²(x²+y²)² - (A⁴-6A²b²+b⁴)(x⁴-6x²y²+y⁴) - 16A²b²z²(x²+y²) ]
+ 16 ((x²+y²+z²+d²-c²)² - 4b²z²) [ (A²+b²)(A⁴-6A²b²+b⁴)(x²+y²)(x⁴-6x²y²+y⁴) - (A²+b²)³(x²+y²)³ + 8(x²+y²)²(A²+3b²)A²b²z² - 8(x⁴-6x²y²+y⁴)(A²-b²)A²b²z² ]
+ 512 (x²+y²)² A⁴b⁴z⁴ - 512 (x⁴-6x²y²+y⁴) A⁴b⁴z⁴
- 128 (x²+y²)³ (A⁴-2A²b²+5b⁴)A²b²z² + 128 (x²+y²)(x⁴-6x²y²+y⁴) (A⁴-2A²b²-3b⁴)A²b²z²
+ 4 ( (A⁴-6A²b²+b⁴)(x²+y²)² - (A²+b²)²(x⁴-6x²y²+y⁴) )².

Hmm... Would you like to try filling this in with some w terms, to make a spider?

If you see some polynomial in A,b,z (or a,b,z), then add or subtract the corresponding polynomial in B,a,w (or b,a,w). It should be added if it's not being multiplied by (x⁴-6x²y²+y⁴), or if it has an even number of such factors. It should be subtracted if it has an odd number of factors of (x⁴-6x²y²+y⁴).

[ICN5D]

Okay, so I've been looking over your process a bit more, and I see something worth noting. Comparing our own 2 processes side by side, each seems to posses a missing link to the other process. I think if we try to find a way to make them compatible (by combining our neat little tricks), we may end up with a decent algorithm for deriving the equations. Your method of getting the solution polynomial (before adding w terms) sounds very tedious and time-consuming, where it doesn't have to be. That's where I come in. It's detailed at the very end of this lengthy post.

These polynomials (in cylindrical coords):

tiger:
X^2 +2Y^2 +2Y^2*cos(4θ) -8Z^2 +4(2Z^2-XY)*cos(2θ)

mantis:
X^3 -3XY^2 -2Y^3*cos(6θ) -12Z^2(X-2Y) +4Z(4Z^2-3XY+3Y^2)*sin(3θ)

spider:
X^4 -4X^2Y^2 +6Y^4 +2Y^4*cos(8θ) +16Z^2(2Z^2-Y^2-(X-Y)^2) -4((XY-4Z^2)^2-2(Y^2-2Z^2)^2)*cos(4θ)


are made by multiplying the solutions of a torus in multi-tiger positions, which have the form:

X(r,z) -2Y(r)*cos(2(t-θ)) +4Z(r,z)*sin(t-θ)


where,

X(r,z) => (r^2+z^2-c^2)^2 +2a^2(2r^2+z^2+b^2-c^2) -2b^2(z^2+c^2) +a^4+b^4
Y(r) => (a^2+b^2)r^2
Z(r,z) => ar(r^2+z^2+a^2+b^2-c^2)
t = angle in radians

a = translate distance from origin , R1
b= major diameter of the torus solution, R2
c= minor diameter of the torus solution, R3


Simplified hard-set values of a = 4 , b = 2 , c = 1 :

Cylindrical Conversion
X(r,z) => r^4 +z^4 +2r^2z^2 +62r^2 +22z^2 +361
Y(r) => 20r^2
Z(r,z) => 4r^3 +4rz^2 +76r

Cartesian Conversion
X(x,y,z) = (x^2+y^2)^2 +z^2(2x^2+2y^2+z^2+22) +62(x^2+y^2) +361
Y(x,y) = 20(x^2+y^2)
Z(x,y,z) = 4*sqrt(x^2+y^2)(x^2+y^2+z^2+19)
θ = arctan(y/x)


pretty much does all of the hard math for you. It's more efficient I believe. The 3 polynomials at the top are showing how those torus equations get combined when multiplied, in a highly consolidated form, by use of substitution. A highly reduced polynomial that I found, for the 3D mantis solution (in cyl coords) is:

(r^4+2r^2z^2+62r^2+z^4+22z^2+361)^3 -1200r^4(r^4+2r^2z^2+62r^2+z^4+22z^2+361) -16000r^6*cos(6θ) -192r^2(r^2+z^2+19)^2(r^4+2r^2z^2+22r^2+z^4+22z^2+361) +64r^3(r^2+z^2+19)(r^4+2r^2z^2-22r^2+z^4+278z^2+361)*sin(3θ) = 0


However, at the same time, I'm struggling to come up with the terms that contain w. This is where your process might come in Maybe a compatible part of your reasoning can add in the w terms to my equation above.

However, I still like your approach to singling out the terms we need by determining various (reflective?) symmetries. And your mantis equation does things I haven't been able to do yet, which is impressive enough. So, I'm not knocking your approach at all. I'm offering what I believe is an improvement. A matter of efficiency, sought after by a lazy guy.

What program do you use to expand and simplify these equations? I'd like to try this out.

Nothing. Just lots of paper.

I don't believe you Take a picture of those giant equations you've been scribbling down. I've got to see this.

If it's true, I recommend using wolfram alpha, and running chunks of terms through it and see what alternate expressions it comes up with. It can handle some fairly large equations.

I've made legit discoveries this way: the whole 'roots of unity/multi-torus' thread and my factoring algorithm for the algebraic solutions of toratopes. That knowledge would not even exist, were it not for our electronic friends. I highly recommend it!

F(x,y,z,0) =
((x²+y²+z²+d²-c²)² - 4b²z²)⁴ - 8d² ((x²+y²+z²+d²-c²)² - 4b²z²)³ (x²+y²)
+ 4 ((x²+y²+z²+d²-c²)² - 4b²z²)² [ 5(A²+b²)²(x²+y²)² - (A⁴-6A²b²+b⁴)(x⁴-6x²y²+y⁴) - 16A²b²z²(x²+y²) ]
+ 16 ((x²+y²+z²+d²-c²)² - 4b²z²) [ (A²+b²)(A⁴-6A²b²+b⁴)(x²+y²)(x⁴-6x²y²+y⁴) - (A²+b²)³(x²+y²)³ + 8(x²+y²)²(A²+3b²)A²b²z² - 8(x⁴-6x²y²+y⁴)(A²-b²)A²b²z² ]
+ 512 (x²+y²)² A⁴b⁴z⁴ - 512 (x⁴-6x²y²+y⁴) A⁴b⁴z⁴
- 128 (x²+y²)³ (A⁴-2A²b²+5b⁴)A²b²z² + 128 (x²+y²)(x⁴-6x²y²+y⁴) (A⁴-2A²b²-3b⁴)A²b²z²
+ 4 ( (A⁴-6A²b²+b⁴)(x²+y²)² - (A²+b²)²(x⁴-6x²y²+y⁴) )².

Hmm... Would you like to try filling this in with some w terms, to make a spider?

I'm still not sure how you're doing that. Not even with the mantis, which is a good example you've already provided. I need time to stare at it, and pick out some crucial details. You'd be surprised how well this works for me.

[mr_e_man]

Sorry, no, your cylindrical equations are not in a form to which I can easily add w terms to get that symmetry.

mantis:
X^3 -3XY^2 -2Y^3*cos(6θ) -12Z^2(X-2Y) +4Z(4Z^2-3XY+3Y^2)*sin(3θ)

spider:
X^4 -4X^2Y^2 +6Y^4 +2Y^4*cos(8θ) +16Z^2(2Z^2-Y^2-(X-Y)^2) -4((XY-4Z^2)^2-2(Y^2-2Z^2)^2)*cos(4θ)

[...]

where,

X(r,z) => (r^2+z^2-c^2)^2 +2a^2(2r^2+z^2+b^2-c^2) -2b^2(z^2+c^2) +a^4+b^4
Y(r) => (a^2+b^2)r^2
Z(r,z) => ar(r^2+z^2+a^2+b^2-c^2)

Your 'a' is my 'A', so I would write these as

X = (x²+y²+z²-c²)² + 2A²(2x²+2y²+z²+b²-c²) - 2b²(z²+c²) + A⁴+b⁴,
Y = (A²+b²)(x²+y²) = d²(x²+y²),
Z = A √(x²+y²) (x²+y²+z²+d²-c²).

Y is fine; it's already symmetric. But I can't do much with X and Z. I would start with

X = (x²+y²+z²+w²-c²)² + 2A²(2x²+2y²+b²-c²)+2A²z²+2B²w² - 2b²c²-2b²z²-2a²w² + A⁴+b⁴,

but this still doesn't have the required symmetry:

X ≟ (x²+y²+w²+z²-c²)² + 2B²(2x²+2y²+a²-c²)+2B²w²+2A²z² - 2a²c²-2a²w²-2b²z² + B⁴+a⁴.

Of course I'm just looking at X alone, not the full polynomial for the torus array. We might try to rewrite the latter, or some sub-expression such as X³ - 3XY², in a form that I can work with. But how would we do that? It seems no easier than the original problem.

The polynomial for the torus array needs to be rewritten at least to get rid of individual x's and y's, thus

F(x,y,z,0; a,b,c) = G((x²+y²), Re[(x+iy)ⁿ], z,0; a,b,c).

(Your cylindrical equations do have this form, except possibly the trigonometric terms.) That special symmetry can then be described by

G((x²+y²), Re[(x+iy)ⁿ], z,w; a,b,c) = G((x²+y²), -Re[(x+iy)ⁿ], w,z; b,a,c).

Let's abbreviate these as new variables u=x²+y², v=Re[(x+iy)ⁿ], so this becomes G(u,v,z,w; a,b,c) = G(u,-v,w,z; b,a,c).

I'm still trying to figure out how G(u,v,z,0; a,b,c) needs to be rewritten further, so that w can be added easily. Any sub-expression which doesn't involve z and is already symmetric, P(u,v; a,b,c) = P(u,-v; b,a,c), can be left as is. Any sub-expression which doesn't involve z and is not symmetric should be eliminated somehow; the array of circles definitely does have this symmetry. Any sub-expression which has z as a factor, P(u,v,z; a,b,c) = z Q(u,v,z; a,b,c), can be replaced with z Q(u,v,z; a,b,c) + w Q(u,-v,w; b,a,c).

A highly reduced polynomial that I found, for the 3D mantis solution (in cyl coords) is:

(r^4+2r^2z^2+62r^2+z^4+22z^2+361)^3 -1200r^4(r^4+2r^2z^2+62r^2+z^4+22z^2+361) -16000r^6*cos(6θ) -192r^2(r^2+z^2+19)^2(r^4+2r^2z^2+22r^2+z^4+22z^2+361) +64r^3(r^2+z^2+19)(r^4+2r^2z^2-22r^2+z^4+278z^2+361)*sin(3θ) = 0


However, at the same time, I'm struggling to come up with the terms that contain w. This is where your process might come in Maybe a compatible part of your reasoning can add in the w terms to my equation above.

I need a and b as variables, not hard-set. The only exception is when a=b, or equivalently b/A=tan(360°/(4n)); then swapping a and b has no effect. This swap is a reflection of (A,b) (Cartesian coordinates of the circle's centre) across the line at angle 360°/(4n).

If it's true, I recommend using wolfram alpha, and running chunks of terms through it and see what alternate expressions it comes up with. It can handle some fairly large equations.

I've made legit discoveries this way: the whole 'roots of unity/multi-torus' thread and my factoring algorithm for the algebraic solutions of toratopes. That knowledge would not even exist, were it not for our electronic friends. I highly recommend it!

I do use WolframAlpha occasionally, such as for factoring. But often it doesn't understand the larger expressions I give it, even when it understands smaller expressions of the same form. It may be worth a try, though.

[ICN5D]

Okay, fair enough. Thanks for looking into it. Maybe I should try to adapt my process to your beginning equations.

Also, I had a good look at your mantis equation, and here it is with hard-set parameters a:b:c = 4:3:1

f(x,y,z,w) = [(x^2+y^2+z^2+w^2+145/3)^2-4(9z^2+16w^2)]^2*[(x^2+y^2+z^2+w^2+145/3)^2+296(x^2+y^2)-4(9z^2+16w^2)] -16(x^2+y^2)(x^2+y^2+z^2+w^2+145/3)^2*[37(x^2+y^2+z^2+w^2+145/3)^2-1369(x^2+y^2)-7(225z^2+448w^2)] -16*sqrt(3)/9*(x^3-3xy^2)*[110(x^2+y^2+z^2+w^2+145/3)^3 -148(x^2+y^2+z^2+w^2+145/3)(110x^2+110y^2-297z^2+480w^2)] -64(x^2+y^2)[3(729z^4+4096w^4)+168(x^2+y^2)(15z^2-22w^2)] +64[12100/27(x^3-3xy^2)^2 -7056(3x^2y-y^3)^2] -592240896z^2w^2

A rotation on plane zw:

Calcplot3d URL:

Code: Select all

https://c3d.libretexts.org/CalcPlot3D/index.html?type=implicit;equation=((x^2+y^2+(z*cos(t)-a*sin(t))^2+(z*sin(t)+a*cos(t))^2+145/3)^2-4(9(z*cos(t)-a*sin(t))^2+16(z*sin(t)+a*cos(t))^2))^2*((x^2+y^2+(z*cos(t)-a*sin(t))^2+(z*sin(t)+a*cos(t))^2+145/3)^2+296(x^2+y^2)-4(9(z*cos(t)-a*sin(t))^2+16(z*sin(t)+a*cos(t))^2))-16(x^2+y^2)(x^2+y^2+(z*cos(t)-a*sin(t))^2+(z*sin(t)+a*cos(t))^2+145/3)^2*(37(x^2+y^2+(z*cos(t)-a*sin(t))^2+(z*sin(t)+a*cos(t))^2+145/3)^2-1369(x^2+y^2)-7(225(z*cos(t)-a*sin(t))^2+448(z*sin(t)+a*cos(t))^2))-(16*sqrt(3))/9*(x^3-3xy^2)*(110(x^2+y^2+(z*cos(t)-a*sin(t))^2+(z*sin(t)+a*cos(t))^2+145/3)^3-148(x^2+y^2+(z*cos(t)-a*sin(t))^2+(z*sin(t)+a*cos(t))^2+145/3)(110x^2+110y^2-297(z*cos(t)-a*sin(t))^2+480(z*sin(t)+a*cos(t))^2))-64(x^2+y^2)(3(4096(z*sin(t)+a*cos(t))^4+729(z*cos(t)-a*sin(t))^4)+168(15(z*cos(t)-a*sin(t))^2-22(z*sin(t)+a*cos(t))^2)(x^2+y^2))+64(12100/27(x^3-3xy^2)^2-7056(3x^2y-y^3)^2)-592240896(z*cos(t)-a*sin(t))^2(z*sin(t)+a*cos(t))^2;cubes=25;visible=true;fixdomain=false;xmin=-8;xmax=8;ymin=-8;ymax=8;zmin=-8;zmax=8;alpha=255;view=0;format=normal;constcol=rgb(153,153,153)&type=slider;slider=a;value=0;steps=100;pmin=-8;pmax=8;repeat=true;bounce=true;waittime=1;careful=false;noanimate=false;name=-1&type=slider;slider=t;value=1.54461639;steps=60;pmin=0;pmax=pi/2;repeat=true;bounce=true;waittime=1;careful=false;noanimate=false;name=-1&type=window;hsrmode=3;nomidpts=true;anaglyph=-1;center=7.058302103282751,5.808293497004737,7.102400866711125,1;focus=0,0,0,1;up=-0.45747881941680846,-0.37013375975943846,0.8085258991963988,1;transparent=false;alpha=140;twoviews=false;unlinkviews=false;axisextension=0.7;xaxislabel=x;yaxislabel=y;zaxislabel=z;edgeson=false;faceson=true;showbox=false;showaxes=false;showticks=true;perspective=true;centerxpercent=0.5;centerypercent=0.5;rotationsteps=100;autospin=false;xygrid=false;yzgrid=false;xzgrid=false;gridsonbox=true;gridplanes=false;gridcolor=rgb(128,128,128);xmin=-8;xmax=8;ymin=-8;ymax=8;zmin=-8;zmax=8;xscale=4;yscale=4;zscale=4;zcmin=-16;zcmax=16;zoom=0.293333;xscalefactor=1;yscalefactor=1;zscalefactor=1

It's pretty interesting to see how the terms get combined. It's stuff I would have never guessed at if I were tinkering around, which is how it always is with me. Really cool man. I still have to take the other 3d slices though.

So. I'm wondering what additional terms you would suggest at the end of this equation? Are there any we can use to refine the 45 degree oblique surfaces? Maybe something to smoothen out the 3 toruses as they begin touching at the poles of the cage structure - that is get rid of that little blob that appears right before they touch. I'm guessing that's what we can use them for. It all seems right so far. I wonder where Marek is, he'd get a kick out of this ......

[ICN5D]

Alright, I checked out your equation some more, made a few more rotation animations. It's not so voodoo magic after all, just some very clever use of terms I never thought of! Now I can see how you use the alternating a,b with z,w terms to add in w. And I think this is the right equation.





The other solutions are pretty much identical to the ones I got from the hemi-tiger functions, where I naively throw 3 of them together into a cage-like structure. So, let's compare them:


[old animation from above link] This is a 180 degree rotation that flips the triangle array a different way. The same 4-bar cage surface forms at the 90 degree angle between them, which is another coordinate solution.






Rotation from your equation, has exact same solution as above (the 4-bar cage).


a=0 , vary parameter t by 0 < t < 1.57

(((x*cos(t)-a*sin(t))^2+y^2+z^2+(x*sin(t)+a*cos(t))^2+145/3)^2-4(9z^2+16(x*sin(t)+a*cos(t))^2))^2*(((x*cos(t)-a*sin(t))^2+y^2+z^2+(x*sin(t)+a*cos(t))^2+145/3)^2+296((x*cos(t)-a*sin(t))^2+y^2)-4(9z^2+16(x*sin(t)+a*cos(t))^2)) -16((x*cos(t)-a*sin(t))^2+y^2)((x*cos(t)-a*sin(t))^2+y^2+z^2+(x*sin(t)+a*cos(t))^2+145/3)^2*(37((x*cos(t)-a*sin(t))^2+y^2+z^2+(x*sin(t)+a*cos(t))^2+145/3)^2-1369((x*cos(t)-a*sin(t))^2+y^2)-7(225z^2+448(x*sin(t)+a*cos(t))^2)) -16*sqrt(3)/9*((x*cos(t)-a*sin(t))^3-3(x*cos(t)-a*sin(t))y^2)*(110((x*cos(t)-a*sin(t))^2+y^2+z^2+(x*sin(t)+a*cos(t))^2+145/3)^3 -148((x*cos(t)-a*sin(t))^2+y^2+z^2+(x*sin(t)+a*cos(t))^2+145/3)(110(x*cos(t)-a*sin(t))^2+110y^2-297z^2+480(x*sin(t)+a*cos(t))^2)) -64((x*cos(t)-a*sin(t))^2+y^2)(3(729z^4+4096(x*sin(t)+a*cos(t))^4)+168((x*cos(t)-a*sin(t))^2+y^2)(15z^2-22(x*sin(t)+a*cos(t))^2)) +64(12100/27((x*cos(t)-a*sin(t))^3-3(x*cos(t)-a*sin(t))y^2)^2 -7056(3(x*cos(t)-a*sin(t))^2y-y^3)^2) -592240896z^2(x*sin(t)+a*cos(t))^2 = 0


###################################################################

[old animation from other link] This is a rotation from the 3 torus solutions to my so-called "closed-loop hemi-tiger" surface. It forms when just one of the cage bars is being 3d sliced, leaving the other two branching off at an angle, above and below the 3-plane:






Same rotation of your equation, gives pretty much exact same surface!



a=0 , vary parameter t by 0 < t < 1.57

((x^2+(y*cos(t)-a*sin(t))^2+z^2+(y*sin(t)+a*cos(t))^2+145/3)^2-4(9z^2+16(y*sin(t)+a*cos(t))^2))^2*((x^2+(y*cos(t)-a*sin(t))^2+z^2+(y*sin(t)+a*cos(t))^2+145/3)^2+296(x^2+(y*cos(t)-a*sin(t))^2)-4(9z^2+16(y*sin(t)+a*cos(t))^2)) -16(x^2+(y*cos(t)-a*sin(t))^2)(x^2+(y*cos(t)-a*sin(t))^2+z^2+(y*sin(t)+a*cos(t))^2+145/3)^2*(37(x^2+(y*cos(t)-a*sin(t))^2+z^2+(y*sin(t)+a*cos(t))^2+145/3)^2-1369(x^2+(y*cos(t)-a*sin(t))^2)-7(225z^2+448(y*sin(t)+a*cos(t))^2)) -16*sqrt(3)/9*(x^3-3x(y*cos(t)-a*sin(t))^2)*(110(x^2+(y*cos(t)-a*sin(t))^2+z^2+(y*sin(t)+a*cos(t))^2+145/3)^3 -148(x^2+(y*cos(t)-a*sin(t))^2+z^2+(y*sin(t)+a*cos(t))^2+145/3)(110x^2+110(y*cos(t)-a*sin(t))^2-297z^2+480(y*sin(t)+a*cos(t))^2)) -64(x^2+(y*cos(t)-a*sin(t))^2)(3(729z^4+4096(y*sin(t)+a*cos(t))^4)+168(x^2+(y*cos(t)-a*sin(t))^2)(15z^2-22(y*sin(t)+a*cos(t))^2)) +64(12100/27(x^3-3x(y*cos(t)-a*sin(t))^2)^2 -7056(3x^2(y*cos(t)-a*sin(t))-(y*cos(t)-a*sin(t))^3)^2) -592240896z^2(y*sin(t)+a*cos(t))^2 = 0


######################################################


Which brings me to my next point: The next two pics below show these unique solutions, where the plane that we see slicing them is the zw plane. Both have a common solution, defined by only the zw terms: a square array of 4 circles.


Solution XZW with ZW plane






Solution YZW with ZW plane



Above, we can see right away how the sharp, 180 degree bend sections are thinner than the 4 fatter tubes (also identical to the first img). So, I believe that by adding in more zw terms the right way, we can reduce the diameters to where the snaking tube is uniform. Knowing the zw solution is a product of 4 circles in a square array, that should make the job much simpler, I would hope! Maybe by comparing the terms in a standalone degree-8 polynomial of 4 circles in sq array, we can find the missing ones, while hoping they don't mess with the other solutions.

[Marek14]

Looks great!

[mr_e_man]

So. I'm wondering what additional terms you would suggest at the end of this equation? Are there any we can use to refine the 45 degree oblique surfaces? Maybe something to smoothen out the 3 toruses as they begin touching at the poles of the cage structure - that is get rid of that little blob that appears right before they touch. I'm guessing that's what we can use them for. It all seems right so far. I wonder where Marek is, he'd get a kick out of this ......

Yes, I noticed that blob. I don't know what to do about it. Just try adding any polynomial with the required symmetry (and with factors of z²w²), and see what happens when you vary the coefficients.

Also, my spider equation gives the wrong topology (too many holes in some oblique cross-sections) with the low-degree end terms that I tried.

Alright, I checked out your equation some more, made a few more rotation animations. It's not so voodoo magic after all, just some very clever use of terms I never thought of! Now I can see how you use the alternating a,b with z,w terms to add in w. And I think this is the right equation.

I'm glad it makes sense.

Which brings me to my next point: The next two pics below show these unique solutions, where the plane that we see slicing them is the zw plane. Both have a common solution, defined by only the zw terms: a square array of 4 circles.

I never thought about the z,w cross-sections. Should they be exactly circles? My equation gives a 12th degree polynomial in z,w when x=y=0. So some extra terms at the end would have to also be 12th degree, to cancel and reduce it to 8th degree.