This following may lead to most definitive Pi-Time { 66.4 } associated with a single electron
Diameter of hydrogen atom --ergo one electron-- is 1.06 * 10 to negative 10. i.e. 1.06 * 0.1 =
0.106m and that is the value I will begin with via the torus for diameter { <--D--> } of electron.
This 0.106 is the value of the diameter of the larger circle of my numerical torus. So we want to use half of that value ---ergo 0.053--- to attain the radius { R--> } and then the resultant surface area of
one torus that may be involved with the hydrogen's, single electron, fuzzy cloud.
0.053 * 12.56637061{ R--> }... = 0.66601764233
The calculation can be done different ways with different kinds of radii being used. I'm going with this following pathway since it does the math for me if I can properly assign their radii to my numerical tori. I also like this method because it approximates what I believe the tubes radius to be, however,
it does subtract the value of diameter of hydrogen atoms single proton.
I assume my numerical torus inside inner radius ---via the following link--- begins at the outside surface of diameter of the single hydrogen proton ergo I will have to double the value of the inner radius in the following link so as the inner circle of the diagram.
S = π2 * (R2 - r2)
LINK to calculator that does the math for us.
R{ R--> } = 0.053
r{ r--> } = 0.0 minus radius of proton and we will
exclude proton values for following resultants
and the resultant
surface area for my numerical torus is 0.02772372 ---and that value includes protons diameter---.
Next I need to establish a connection/association to Pi-Time 66.4 ie. 66.4 revolutions { inversions } and subtract that from volume{?}?