#### 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 🤓 """ ```