S&E Modules

"""
The concentric hierarchy can be described in terms of 
phi-scaled S modules: 

Tetrahedron: 21S+5s3 = 5S3+1S = 1S6+1S3 
Cube: 72S+15s3 = 15S3+3S = 3S6+3S3 
Octahedron: 84S+20s3 = 20S3+4S = 4S6+4S3 
Rh Triac: 105S+25s3 = 25s3+5S = 5S6+5S3 (volume = 5) 
Rh Dodec: 126S+30s3 = 30S3+6S = 6S6+6S3
"""
from math import sqrt as rt2

φ  = (rt2(5)+1)/2
S3 = rt2(9/8)

Smod = (φ **-5)/2
smod3 = Smod * φ ** -3
Smod3 = Smod * φ **  3
Smod6 = Smod3 * φ **  3

cubocta = 20
SuperRT = S3 * cubocta

Emod = (rt2(2)/8) * (φ ** -3)
emod3 = Emod * φ ** -3
Emod3 = SuperRT / 120

print("Emod3:", Emod3) 
print("Emod:",  Emod3 * φ ** -3)

S_factor = Smod / Emod

print("Volume of SuperRT : {:10.6f}".format(SuperRT))
print("Volume of Cubocta : {:10.6f}".format(cubocta))
print("Volume of Icosa   : {:10.6f}".format(20 * 1/S_factor))
print("Volume of Icosa   : {:10.6f}".format(420 * Emod + 100 * emod3))
print("Volume of Icosa   : {:10.6f}".format(100*Emod3 + 20*Emod))
print("Volume of Octa    : {:10.6f}".format(4*Smod6 + 4*Smod3))
print("Volume of Cube    : {:10.6f}".format(3*Smod6 + 3*Smod3))
print("6 * Emod3:", 6 * Emod3)
print("UVT: ", 5*Smod3 + Smod)


"""
Area of sphere is 4πr^2
Volume of sphere is (4/3)πr^3
Square and Cubic units
URS S.A. = 4π = 12.566
URS Vol = (4/3)π = 4.188
Rh Dodec S.A. = 12√2
Rh Dodec Vol = 4√2
4π/12√2 = .740
((4/3)π)/(4√2) = .740
"""

"""
Obviously, we can play the "Tuna Can" game 
with our Unit Radius Sphere.
URS = 120 Icosa Spherical Module (ISM) or 
48 VE Spherical Modules (VESM)
Cone = 30 ISM or 12 VESM
Tuna Can = 90 ISM or 36 VESM
Double Tuna Can = 180 ISM or 72 VESM

If you noticed on the great circle grid 
I am working with, there are 12 small 
pentagons in the middle of each overall 
pentagon.

Using PhiPi, the 1/20th of the URS, 
equates to the Icosahedron, the VE triangle 
is smaller than it, so the volume is less.

The difference in the volume between the 
two is the Super RT/1000 or .021213 or 
(15(sqrt 2))/1000.

This is easy to remember 🤓
"""