The Hemi-Torus Function

Discussion of shapes with curves and holes in various dimensions.

The Hemi-Torus Function

Postby ICN5D » Sun Aug 04, 2019 8:11 pm

Here's another little discovery I made, before the multi-torus polynomial breakthrough. It's also another function that's been sitting right in front of me for many years, not knowing if I should go tinkering to see how it works. It's identical to the taper function from STEMP notation, but doesn't add dimensions.

The one and only drawback is the self-intersection artifacts where the hemi-torus ends meet (for some slices). It's a minor cosmetic shortcoming to the otherwise many advantages:

- has an efficiently low number of symbols, easy to combine family of eq
- builds highly accurate models, with very precise x-sections
- extends very easily to polyhedral multi-tigers
- extends easily beyond 4D
- combined with the multi-torus polynomials, we can easily create HYBRIDS : 3-prong of the 3-prong , tetrahedral array of 7-prong multitoruses, etc.

An implicit equation for a finite hemi-torus :

Image

abs((sqrt(x^2 + y^2) -3)^2 + z^2 +x/6 -1) + (sqrt(x^2 + y^2) -3)^2 + z^2 = 1




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Implicit equation for a finite hemi-tiger:

Image


abs((sqrt(x^2 + y^2) -2)^2 + (sqrt(z^2 + w^2) -4)^2 +z/6 -1) + (sqrt(x^2 + y^2) -2)^2 + (sqrt(z^2 + w^2) -4)^2 = 1




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Application of the hemi-tiger equation to form a 3-prong multi-tiger, the Mantis. This is the most accurate model of the mantis to date. It has shed some light on what direction to take with adding the 4th variable to the reduced cylindrical polynomial P(X,Y,Z) on the multi-torus development thread.


Hemi-tiger equation:

abs((sqrt(x^2 + y^2) -2)^2 + (sqrt(z^2 + w^2) -4)^2 +z/6 -1) + (sqrt(x^2 + y^2) -2)^2 + (sqrt(z^2 + w^2) -4)^2 -1 = 0



Rotate/translate equation that places the single torus into triangular array of mantis solutions

abs((sqrt((x*cos(b)-a*sin(b))^2 + y^2) -2)^2 + (sqrt(z^2 + (x*sin(b)+a*cos(b))^2) -4)^2 +z/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + y^2) -2)^2 + (sqrt(z^2 + (x*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

x = (x*cos(b)-a*sin(b))
w = (x*sin(b)+a*cos(b))



Next step is to add 2 more, rotated about the yz plane. End result is the Mantis, as a family of 3 equations. Here, I just used the cartesian forms of 2n*pi/3 radian intervals.

- Eq 1
abs((sqrt(x^2 + y^2) -2)^2 + (sqrt(z^2 + w^2) -4)^2 +z/6 -1) + (sqrt(x^2 + y^2) -2)^2 + (sqrt(z^2 + w^2) -4)^2 -1 = 0

- Eq 2
abs((sqrt(x^2 + (-(y+sqrt(3)z)/2)^2) -2)^2 + (sqrt(((sqrt(3)y-z)/2)^2 + w^2) -4)^2 +((sqrt(3)y-z)/2)/6 -1) + (sqrt(x^2 + (-(y+sqrt(3)z)/2)^2) -2)^2 + (sqrt(((sqrt(3)y-z)/2)^2 + w^2) -4)^2 -1 = 0

- Eq 3
abs((sqrt(x^2 + ((sqrt(3)z-y)/2)^2) -2)^2 + (sqrt((-(sqrt(3)y+z)/2)^2 + w^2) -4)^2 +(-(sqrt(3)y+z)/2)/6 -1) + (sqrt(x^2 + ((sqrt(3)z-y)/2)^2) -2)^2 + (sqrt((-(sqrt(3)y+z)/2)^2 + w^2) -4)^2 -1 = 0

*** Note : the R1b radius on plane zw should be 2x or larger than R1a for a 3-prong, and perhaps 3x for 4-prong shapes.


Slicing on plane XYW , translating along z-axis from -5 < z < 5

Image










Single rotate + translate on XW:

x = (x*cos(b)-a*sin(b))
w = (x*sin(b)+a*cos(b))

- Eq 1
abs((sqrt((x*cos(b)-a*sin(b))^2 + y^2) -2)^2 + (sqrt(z^2 + (x*sin(b)+a*cos(b))^2) -4)^2 +z/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + y^2) -2)^2 + (sqrt(z^2 + (x*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

- Eq 2
abs((sqrt((x*cos(b)-a*sin(b))^2 + (-(y+sqrt(3)z)/2)^2) -2)^2 + (sqrt(((sqrt(3)y-z)/2)^2 + (x*sin(b)+a*cos(b))^2) -4)^2 +((sqrt(3)y-z)/2)/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + (-(y+sqrt(3)z)/2)^2) -2)^2 + (sqrt(((sqrt(3)y-z)/2)^2 + (x*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

- Eq 3
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((sqrt(3)z-y)/2)^2) -2)^2 + (sqrt((-(sqrt(3)y+z)/2)^2 + (x*sin(b)+a*cos(b))^2) -4)^2 +(-(sqrt(3)y+z)/2)/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((sqrt(3)z-y)/2)^2) -2)^2 + (sqrt((-(sqrt(3)y+z)/2)^2 + (x*sin(b)+a*cos(b))^2) -4)^2 -1 = 0


Morphs from the triangle array of 3 toruses (b=0), to the 6-bar cage (b=pi/4), to the alt array of 3 tori (b=pi/2).

Image

A 90 degree turn back and forth. Morphs from the triangular array of 3 toruses to an alternate array of 3. The alt array shows some of the self-intersection artifacts, which is a shortcoming of the hemi-torus function. The true mantis equation will likely make a smooth toroidal connection in those areas.










Rotate + Translate on YW

y = (y*cos(b)-a*sin(b))
w = (y*sin(b)+a*cos(b))

- Eq 1
abs((sqrt(x^2 + (y*cos(b)-a*sin(b))^2) -2)^2 + (sqrt(z^2 + (y*sin(b)+a*cos(b))^2) -4)^2 +z/6 -1) + (sqrt(x^2 + (y*cos(b)-a*sin(b))^2) -2)^2 + (sqrt(z^2 + (y*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

- Eq 2
abs((sqrt(x^2 + (-((y*cos(b)-a*sin(b))+sqrt(3)z)/2)^2) -2)^2 + (sqrt(((sqrt(3)(y*cos(b)-a*sin(b))-z)/2)^2 + (y*sin(b)+a*cos(b))^2) -4)^2 +((sqrt(3)(y*cos(b)-a*sin(b))-z)/2)/6 -1) + (sqrt(x^2 + (-((y*cos(b)-a*sin(b))+sqrt(3)z)/2)^2) -2)^2 + (sqrt(((sqrt(3)(y*cos(b)-a*sin(b))-z)/2)^2 + (y*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

- Eq 3
abs((sqrt(x^2 + ((sqrt(3)z-(y*cos(b)-a*sin(b)))/2)^2) -2)^2 + (sqrt((-(sqrt(3)(y*cos(b)-a*sin(b))+z)/2)^2 + (y*sin(b)+a*cos(b))^2) -4)^2 +(-(sqrt(3)(y*cos(b)-a*sin(b))+z)/2)/6 -1) + (sqrt(x^2 + ((sqrt(3)z-(y*cos(b)-a*sin(b)))/2)^2) -2)^2 + (sqrt((-(sqrt(3)(y*cos(b)-a*sin(b))+z)/2)^2 + (y*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

Morphs from triangle array of 3 toruses (b=0) to the hemi-tiger slice (b=pi/2). This is the most perfectly formed, closed-loop hemi-tiger I've ever seen, and came as a surprise. It's exactly as I imagined it would be theoretically, assuming I finally constructed one. This is far more accurate than the tri-tigroidal cassini surface hemi-tiger, and makes me think I'm finally on track to unlocking the rest of them.

Image










Rotate + Translate on ZW

z = (z*cos(b)-a*sin(b))
w = (z*sin(b)+a*cos(b))

- Eq 1
abs((sqrt(x^2 + y^2) -2)^2 + (sqrt((z*cos(b)-a*sin(b))^2 + (z*sin(b)+a*cos(b))^2) -4)^2 +(z*cos(b)-a*sin(b))/6 -1) + (sqrt(x^2 + y^2) -2)^2 + (sqrt((z*cos(b)-a*sin(b))^2 + (z*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

- Eq 2
abs((sqrt(x^2 + (-(y+sqrt(3)(z*cos(b)-a*sin(b)))/2)^2) -2)^2 + (sqrt(((sqrt(3)y-(z*cos(b)-a*sin(b)))/2)^2 + (z*sin(b)+a*cos(b))^2) -4)^2 +((sqrt(3)y-(z*cos(b)-a*sin(b)))/2)/6 -1) + (sqrt(x^2 + (-(y+sqrt(3)(z*cos(b)-a*sin(b)))/2)^2) -2)^2 + (sqrt(((sqrt(3)y-(z*cos(b)-a*sin(b)))/2)^2 + (z*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

- Eq 3
abs((sqrt(x^2 + ((sqrt(3)(z*cos(b)-a*sin(b))-y)/2)^2) -2)^2 + (sqrt((-(sqrt(3)y+(z*cos(b)-a*sin(b)))/2)^2 + (z*sin(b)+a*cos(b))^2) -4)^2 +(-(sqrt(3)y+(z*cos(b)-a*sin(b)))/2)/6 -1) + (sqrt(x^2 + ((sqrt(3)(z*cos(b)-a*sin(b))-y)/2)^2) -2)^2 + (sqrt((-(sqrt(3)y+(z*cos(b)-a*sin(b)))/2)^2 + (z*sin(b)+a*cos(b))^2) -4)^2 -1 = 0

Morphs between the two flip-flopped triangular arrays (b=0,pi) of 3 toruses. Still requires a full 180 deg turn (b=pi) to flip the triangle, just like the tri-tigroidal cassini surface, and has a very similar (b=pi/2) surface. The surface seen in the middle of the pauses is the translate gif at the top.

Image
in search of combinatorial objects of finite extent
ICN5D
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Re: The Hemi-Torus Function

Postby ICN5D » Mon Aug 05, 2019 9:42 pm

Now for the long awaited ....

4D Tetrahedron-Symmetric 4-Prong Multi-tiger

For this shape, I decided to start with a square-symmetric 4-prong Spider with toruses in the xz quadrants, rather than a rhombus-symmetric that has toruses on the x and z axes. That way, we define each pair on the xz and yz planes with similar, easy to derive equations. Once constructed, it does not appear to be a unit tetrahedron array, but good enough for the proof of concept.


Base Hemi-tiger, adjustable radius parameters, a > b > c for ring torus

abs((sqrt(x^2 + y^2) -b)^2 + (sqrt(z^2 + w^2) -a)^2 +z/6 -c^2) + (sqrt(x^2 + y^2) -b)^2 + (sqrt(z^2 + w^2) -a)^2 -c^2 = 0


Prong 1:
--------
getting the 45 deg angle on plane xz,

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

t = pi/4

x = ((x-z)/(sqrt(2)))
z = ((x+z)/(sqrt(2)))

abs((sqrt(((x-z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((x+z)/(sqrt(2)))^2 + w^2) -a)^2 +((x+z)/(sqrt(2)))/6 -c^2) + (sqrt(((x-z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((x+z)/(sqrt(2)))^2 + w^2) -a)^2 -c^2 = 0

For each eq, set w=d to 3D slice with translation,

abs((sqrt(((x-z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((x+z)/(sqrt(2)))^2 + d^2) -a)^2 +((x+z)/(sqrt(2)))/6 -c^2) + (sqrt(((x-z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((x+z)/(sqrt(2)))^2 + d^2) -a)^2 -c^2 = 0



Prong 2:
--------
x = (x*cos(t)-z*sin(t))
z = (x*sin(t)+z*cos(t))

t = -pi/4

x = ((x+z)/(sqrt(2)))
z = ((-x+z)/(sqrt(2)))

abs((sqrt(((x+z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((-x+z)/(sqrt(2)))^2 + w^2) -a)^2 +((-x+z)/(sqrt(2)))/6 -c^2) + (sqrt(((x+z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((-x+z)/(sqrt(2)))^2 + w^2) -a)^2 -c^2 = 0

abs((sqrt(((x+z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((-x+z)/(sqrt(2)))^2 + d^2) -a)^2 +((-x+z)/(sqrt(2)))/6 -c^2) + (sqrt(((x+z)/(sqrt(2)))^2 + y^2) -b)^2 + (sqrt(((-x+z)/(sqrt(2)))^2 + d^2) -a)^2 -c^2 = 0



Prong 3:
--------
y = (y*cos(t)-z*sin(t))
z = (y*sin(t)+z*cos(t))

t= pi/2+pi/4

y = ((-y-z)/(sqrt(2)))
z = ((y-z)/(sqrt(2)))

abs((sqrt(x^2 + ((-y-z)/(sqrt(2)))^2) -b)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + w^2) -a)^2 +((y-z)/(sqrt(2)))/6 -c^2) + (sqrt(x^2 + ((-y-z)/(sqrt(2)))^2) -b)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + w^2) -a)^2 -c^2 = 0

abs((sqrt(x^2 + ((-y-z)/(sqrt(2)))^2) -b)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + d^2) -a)^2 +((y-z)/(sqrt(2)))/6 -c^2) + (sqrt(x^2 + ((-y-z)/(sqrt(2)))^2) -b)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + d^2) -a)^2 -c^2 = 0


Prong 4:
--------
y = (y*cos(t)-z*sin(t))
z = (y*sin(t)+z*cos(t))

t= -pi/2-pi/4

y = ((-y+z)/(sqrt(2)))
z = ((-y-z)/(sqrt(2)))

abs((sqrt(x^2 + ((-y+z)/(sqrt(2)))^2) -b)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + w^2) -a)^2 +((-y-z)/(sqrt(2)))/6 -c^2) + (sqrt(x^2 + ((-y+z)/(sqrt(2)))^2) -b)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + w^2) -a)^2 -c^2 = 0

abs((sqrt(x^2 + ((-y+z)/(sqrt(2)))^2) -b)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + d^2) -a)^2 +((-y-z)/(sqrt(2)))/6 -c^2) + (sqrt(x^2 + ((-y+z)/(sqrt(2)))^2) -b)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + d^2) -a)^2 -c^2 = 0


---------------------------------

Some good values for the radius parameters are a = 5 , b = 2 , c = 1

Eq1
abs((sqrt(((x-z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((x+z)/(sqrt(2)))^2 + w^2) -5)^2 +((x+z)/(sqrt(2)))/6 -1) + (sqrt(((x-z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((x+z)/(sqrt(2)))^2 + w^2) -5)^2 -1 = 0

Eq2
abs((sqrt(((x+z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-x+z)/(sqrt(2)))^2 + w^2) -5)^2 +((-x+z)/(sqrt(2)))/6 -1) + (sqrt(((x+z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-x+z)/(sqrt(2)))^2 + w^2) -5)^2 -1 = 0

Eq3
abs((sqrt(x^2 + ((-y-z)/(sqrt(2)))^2) -2)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + w^2) -5)^2 +((y-z)/(sqrt(2)))/6 -1) + (sqrt(x^2 + ((-y-z)/(sqrt(2)))^2) -2)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + w^2) -5)^2 -1 = 0

Eq4
abs((sqrt(x^2 + ((-y+z)/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + w^2) -5)^2 +((-y-z)/(sqrt(2)))/6 -1) + (sqrt(x^2 + ((-y+z)/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + w^2) -5)^2 -1 = 0

This family of 4 hemi-tigers will plot a nice tetrahedral array of 4 toruses when w=0

----------------------------------

Single Rotate + Translate Slices

XW Rotation , W=0
------------------
x = (x*cos(b)-a*sin(b))
w = (x*sin(b)+a*cos(b))

Also identical to the YW rotation

Eq1
abs((sqrt((((x*cos(b)-a*sin(b))-z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 +(((x*cos(b)-a*sin(b))+z)/(sqrt(2)))/6 -1) + (sqrt((((x*cos(b)-a*sin(b))-z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Eq2
abs((sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-(x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 +((-(x*cos(b)-a*sin(b))+z)/(sqrt(2)))/6 -1) + (sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-(x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Eq3
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((-y-z)/(sqrt(2)))^2) -2)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 +((y-z)/(sqrt(2)))/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((-y-z)/(sqrt(2)))^2) -2)^2 + (sqrt(((y-z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Eq4
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((-y+z)/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 +((-y-z)/(sqrt(2)))/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((-y+z)/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Morphs from tet array of 4 toruses (b=0) to some kind of concentric pair of closed-loop hemi-tigers (b=pi/2), approximated by intersecting hemi-toruses. I have no idea what to expect from this thing. This is completely unexplored territory.

Image




Slicing on plane XYZ , translating on w-axis from -6 < w < 0 , then rotating to plane YZW , then translate along x-axis from 0 < x < 6

Image







ZW Rotation , W=0
------------------
z = (z*cos(b)-a*sin(b))
w = (z*sin(b)+a*cos(b))

Eq1
abs((sqrt(((x-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((x+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 +((x+(z*cos(b)-a*sin(b)))/(sqrt(2)))/6 -1) + (sqrt(((x-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((x+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Eq2
abs((sqrt(((x+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-x+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 +((-x+(z*cos(b)-a*sin(b)))/(sqrt(2)))/6 -1) + (sqrt(((x+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-x+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Eq3
abs((sqrt(x^2 + ((-y-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((y-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 +((y-(z*cos(b)-a*sin(b)))/(sqrt(2)))/6 -1) + (sqrt(x^2 + ((-y-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((y-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 -1 = 0

Eq4
abs((sqrt(x^2 + ((-y+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 +((-y-(z*cos(b)-a*sin(b)))/(sqrt(2)))/6 -1) + (sqrt(x^2 + ((-y+(z*cos(b)-a*sin(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-(z*cos(b)-a*sin(b)))/(sqrt(2)))^2 + (z*sin(b)+a*cos(b))^2) -5)^2 -1 = 0


Morphs between 2 distinct tet arrays from 360 deg spin. Mirror image tet array at b=pi. This must be the analogue of the mirror image triangular array of Mantis flipping back and forth from a 360 spin. The flip-flopping effect must be a common feature of n-prong m-tigers, of a certain type. I wasn't expecting it with a polyhedral array, though.

Image




Slicing on XYW , Full translation on z-axis from -6 < z < 6

Image






XZ Rotation , Z=0
------------------
x = (x*cos(b)-a*sin(b))
z = (x*sin(b)+a*cos(b))

Also identical to YZ rotation

Eq1
abs((sqrt((((x*cos(b)-a*sin(b))-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt((((x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 +(((x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))/6 -1) + (sqrt((((x*cos(b)-a*sin(b))-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt((((x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 -1 = 0

Eq2
abs((sqrt((((x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-(x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 +((-(x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))/6 -1) + (sqrt((((x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + y^2) -2)^2 + (sqrt(((-(x*cos(b)-a*sin(b))+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 -1 = 0

Eq3
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((-y-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((y-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 +((y-(x*sin(b)+a*cos(b)))/(sqrt(2)))/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((-y-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((y-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 -1 = 0

Eq4
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((-y+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 +((-y-(x*sin(b)+a*cos(b)))/(sqrt(2)))/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((-y+(x*sin(b)+a*cos(b)))/(sqrt(2)))^2) -2)^2 + (sqrt(((-y-(x*sin(b)+a*cos(b)))/(sqrt(2)))^2 + z^2) -5)^2 -1 = 0


Morphs from:

(b=0) : 4-lobed torus surfaces in vertical stack of 2, slice equal to ZW's b=pi/2
(b=pi/2) : concentric pair of closed-loop hemi-tigers
(b=3pi/2) : mirror-image conc pair of closed-loop ht

Image











XY Rotation , Y=0
------------------
x = (x*cos(b)-a*sin(b))
y = (x*sin(b)+a*cos(b))

Eq1
abs((sqrt((((x*cos(b)-a*sin(b))-z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -2)^2 + (sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + y^2) -5)^2 +(((x*cos(b)-a*sin(b))+z)/(sqrt(2)))/6 -1) + (sqrt((((x*cos(b)-a*sin(b))-z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -2)^2 + (sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + y^2) -5)^2 -1 = 0

Eq2
abs((sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -2)^2 + (sqrt(((-(x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + y^2) -5)^2 +((-(x*cos(b)-a*sin(b))+z)/(sqrt(2)))/6 -1) + (sqrt((((x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + (x*sin(b)+a*cos(b))^2) -2)^2 + (sqrt(((-(x*cos(b)-a*sin(b))+z)/(sqrt(2)))^2 + y^2) -5)^2 -1 = 0

Eq3
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((-(x*sin(b)+a*cos(b))-z)/(sqrt(2)))^2) -2)^2 + (sqrt((((x*sin(b)+a*cos(b))-z)/(sqrt(2)))^2 + y^2) -5)^2 +(((x*sin(b)+a*cos(b))-z)/(sqrt(2)))/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((-(x*sin(b)+a*cos(b))-z)/(sqrt(2)))^2) -2)^2 + (sqrt((((x*sin(b)+a*cos(b))-z)/(sqrt(2)))^2 + y^2) -5)^2 -1 = 0

Eq4
abs((sqrt((x*cos(b)-a*sin(b))^2 + ((-(x*sin(b)+a*cos(b))+z)/(sqrt(2)))^2) -2)^2 + (sqrt(((-(x*sin(b)+a*cos(b))-z)/(sqrt(2)))^2 + y^2) -5)^2 +((-(x*sin(b)+a*cos(b))-z)/(sqrt(2)))/6 -1) + (sqrt((x*cos(b)-a*sin(b))^2 + ((-(x*sin(b)+a*cos(b))+z)/(sqrt(2)))^2) -2)^2 + (sqrt(((-(x*sin(b)+a*cos(b))-z)/(sqrt(2)))^2 + y^2) -5)^2 -1 = 0


Morphs between alternating conc pairs of closed loop hemi-tigers every b = pi/2

Image
in search of combinatorial objects of finite extent
ICN5D
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