Had some amazing insights today, after lots of hard work with these functions. So many experiments, so many equations, so much testing and extrapolating one result to another. But, I believe I found something very worthy : a factored out equation for the cut of mantis. It makes three inward facing tori, with some peculiar properties in the function. Elegant equations are one way of seeing yourself on the right track. I'll make pics when I can, got things to do tomorrow.

Third Roots of Unity

x^3 = 1

solve for x:

x = 1

x = -1/2 - i*sqrt(3)/2

x = -1/2 + i*sqrt(3)/2

Circle equation

(x^2 + y^2 -b^2) = 0

Applying coordinate translation of circles, based on third roots of 1 , from (Re,Im) to (x,y):

x = a

x = -a/2 , y = a sqrt(3)/2

x = -a/2 , y = -a sqrt(3)/2

equals,

((x-a)^2 + y^2 -b^2)((x+a/2)^2 + (y- a sqrt(3)/2)^2 -b^2) ((x+a/2)^2 + (y+ a sqrt(3)/2)^2 -b^2) = 0

* Produces three circles in vertices of triangle, whose centers are derived from x^3 = 1

Torus equation of ((I)I), two displaced circles in xy plane

((sqrt(y^2)-3)^2 + x^2 -1) = 0

Applying coordinate translations of torus cut, based on third roots of 1 , from (Re,Im) to (x,y):

x = a

x = -a/2 , y = a sqrt(3)/2

x = -a/2 , y = -a sqrt(3)/2

equals,

((sqrt(y^2)-3)^2 + (x-a)^2 -1)((sqrt((y- a sqrt(3)/2)^2)-3)^2 + (x+a/2)^2 -1)((sqrt((y+ a sqrt(3)/2)^2)-3)^2 + (x+a/2)^2 -1) = 0

* Produces three torus cuts of ((I)I) in triangle array, all facing same direction as principle root torus.

For rotating the two primitive root tori, so they face inwards, I set up an x,y rotate function, with n*pi/6 steps, for aligning the hexagonal array of circles:

Primitive in negative y region

x = (x*sin(pi/6)+y*cos(pi/6))

y = (x*cos(pi/6)-y*sin(pi/6))

Primitive in positive y region

x = (x*sin(5pi/6)+y*cos(5pi/6))

y = (x*cos(5pi/6)-y*sin(5pi/6))

Equals the product of six circles in hexagon array, from triangle array of 3 tori, of mantis cut:

((sqrt(y^2)-3)^2+(x-a)^2-1) = 0

((sqrt(((x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6))^2)-3)^2+((x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6))^2-1) = 0

((sqrt(((x+a/2)*cos(5pi/6)-(y-a*sqrt(3)/2)*sin(5pi/6))^2)-3)^2+((x+a/2)*sin(5pi/6)+(y-a*sqrt(3)/2)*cos(5pi/6))^2-1) = 0

Trying for 3D version of array. Apply +z^2 inside square root term, along with y^2:

Principle Root Torus

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

Primitive Root Torus A

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

Primitive Root Torus B

((sqrt(((x+a/2)*cos(5*pi/6)-(y-asqrt(3)/2)*sin(5*pi/6))^2+z^2)-b)^2+((x+a/2)*sin(5*pi/6)+(y-a*sqrt(3)/2)*cos(5*pi/6))^2) = c^2

Simplifying the forms of the primitive root tori, eliminating the sin/cos:

First Primitive root torus

((sqrt(((x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6))^2+z^2)-b)^2+((x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6))^2)

(x+a/2)*cos(pi/6)-(y+a*sqrt(3)/2)*sin(pi/6) == (sqrt(3)x -y)/2

(x+a/2)*sin(pi/6)+(y+a*sqrt(3)/2)*cos(pi/6) == (x +sqrt(3)y)/2 + a

equals,

((sqrt(((sqrt(3)x-y)/2)^2+z^2)-b)^2+((x+sqrt(3)y)/2 + a)^2)

Second Primitive root torus

((sqrt(((x+a/2)*cos(5*pi/6)-(y-asqrt(3)/2)*sin(5*pi/6))^2+z^2)-b)^2+((x+a/2)*sin(5*pi/6)+(y-a*sqrt(3)/2)*cos(5*pi/6))^2)

(x+a/2)*cos(5*pi/6)-(y-asqrt(3)/2)*sin(5*pi/6) == -(sqrt(3)x +y)/2

(x+a/2)*sin(5*pi/6)+(y-asqrt(3)/2)*cos(5*pi/6) == (x -sqrt(3)y)/2 + a

equals,

((sqrt((-(sqrt(3)x+y)/2)^2+z^2)-b)^2+((x-sqrt(3)y)/2 + a)^2)

Rewrite with simplified principle and primitive root tori,

Tri-toroidal Cassini Coordinate System

((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

Mantis Cut, factored out as product of three tori.

Note the constants of x and y, in the root tori : the oblique plane equation terms are the third roots of unity! This part gave me the chills....

((sqrt(y^2+z^2)-b)^2+(x-a)^2-c^2)((sqrt(((sqrt(3)x-y)/2)^2+z^2)-b)^2+((x+sqrt(3)y)/2+a)^2-c^2)((sqrt((-(sqrt(3)x+y)/2)^2+z^2)-b)^2+((x-sqrt(3)y)/2+a)^2-c^2) = 0

Oblique terms pulled out:

Principle

x-a

y-0

Primitive A

(sqrt(3)x -y)/2

(x +sqrt(3)y)/2 +a

Primitive B

-(sqrt(3)x +y)/2

(x -sqrt(3)y)/2 +a

compare to x^3 = 1 :

x = 1

x = -1/2 - i*sqrt(3)/2

x = -1/2 + i*sqrt(3)/2

Next step is to take the relationship with x and y , and apply it to z with w, in the same equation. The aim would be for a mirror-image arrangement of tori, when flipping from z-cut to w-cut, as the equation is currently written. Not sure how to add in w in the root tori, may be just as simple as tiger terms...