The key to constructing roots is recognizing the power of the pattern.
First off, notice that every point is the center of a circle.
Notice that the points in the pattern can be considered to be part of a square/cube. Note the lines C1 - O1 - W. The pattern can be considered to be laid out in a square/cube.
Notice the 6 directions of the pattern, originating from any point. I have highlighted the 6 directions in cyan from the circle at Y, and have drawn a hexagon around the points.
This pattern is primarily heaxagonal, with the square/cube fitting into it.
From 3D Basics we saw that first comes the hexagon and the equilateral triangle, then the square. So the \/¯3 is the primary pattern with the \/¯2 contained within it.
The geometry of this pattern is \/¯3 and \/¯2. It is the geometry of crystals, minerals, silicon, snowflakes, etc.
This is Buckminster Fuller's Isotropic Vector Matrix, with the addition of the intersecting circle/sphere. Bucky probably considered this pattern and rejected it, because there is no practical way to get to solids to intersect! So he left out the intersecting spheres and worked only with the tangent ones.
Apologies to Bucky, but I believe that the circle/sphere is the primary geometric pattern in this universe.
From the intersecting spheres, and lines drawn between the points, any imaginable shape can be designed.
Working With The Pattern
Single integers are easy to determine.
The distance from one end of any of the smaller vesicas to the other
end is precisely 1.
Notice OA and OF, which are radii of the circle centered at O.
The \/¯3 and \/¯3 / 2 are also trivial to determine. These
can be eyeballed.
From one end of any of the large vesicas to the other end is precisely \/¯3!
This is CB, marked in yellow.
The distance CB is easy to determine. Notice that OA, OB, and AB are all the same distance. That means triangle OAB is equilateral. The height of the triangle is BK1.
From Equilateral Triangle we know this distance to be \/¯3 / 2.
Sine OA is a perpendicular bisector of CB, BK1 is one-half of CB.
Therefore CB is 2 * \/¯3 / 2, = \/¯3.
The \/¯2 is also easy to determine, although somewhat less so.
Nevertheless, it also fits elegantly into the pattern.
Notice that the distance IN is 1, it being a radius of the circle centered at I.
Notice that the distance NQ1 is 1, it being a radius of the circle centered at N.
Notice that the distance IP1 is 1, it being a radius of the circle centered at I.
I,N,Q1,P1 is a square and Q1P1 is also equal to 1.
Notice that Q1 is on a straight line horizontally to N, and P1 is on a straight line horizontally to I. The line Q1P1 is tangent to the circles with centers at I and N.
The \/¯2 is the line IQ1 (or NP1) which is the diagonal of a square.
So the \/¯2 can always be eyeballed in the pattern by starting at any point (all points are circle/sphere centers in the pattern), going up one circle, and finding a point on the
circumference of that circle directly opposite the center of that circle.
The \/¯5 can also be easily determined. It also fits into the pattern,
although \/¯5 is not necessary to determine any of the roots!
For example, look at the triangle PRR1.
PJR is 2 radial distances. RR1 is 1 radial distance. So,
PR1² = PJR² + RR1² = 2² + 1² = 5.
PR1 = \/¯5.
This can also be eyeballed on the pattern. Go 2, then over 1 and connect.
You can see that if you started at P and observed the 2 intersecting large vesicas PE and JL,
the \/¯5 would be in the middle of the vesica JL, horizontally over from R.
R1 is on the middle of the arc EL.
Now you know everything there is to know about how to determine roots
in the pattern.
\/¯ ° ¹ ² ³ × ½ ¼ Ø \/¯(Ø² + 1)