When we consider, a regular icosahedron in the time dimension and a sphere in the space dimension.
From the post "Freaking Out Entanglement" dated 14 Dec 2017 (corrected as),
\(\cfrac{8^3}{6}\pi^2*f^2*m_ac^2=1*\cfrac{3}{4\pi}*f^3*m_ac^2\)
instead of a sphere in time we use the regular icosahedron,
\(\cfrac{8^3}{6}\pi^2*f^2*m_ac^2=1*\cfrac{1}{Vol_{icosa}}*m_ac^2\)
The centroid to vertex distance, \(r\) of an isocahedron of edge \(a\) is,
\( r=\cfrac {a}{4}{\sqrt {10+2{\sqrt {5}}}}=T\)
if this is equal to \(T\) (a unit of time) then its volume is given by,
with \(a=\cfrac {4}{\sqrt {10+2{\sqrt {5}}}}T\)
\(Vol_{icosa}={\cfrac {5}{12}}(3+{\sqrt {5}})\left(\cfrac {4}{\sqrt {10+2{\sqrt {5}}}}T\right)^{3}\)
and so,
\(\cfrac{8^3}{6}\pi^2*f^2*m_ac^2=1*\left[{\cfrac {5}{12}}(3+{\sqrt {5}})\left(\cfrac {4}{\sqrt {10+2{\sqrt {5}}}}\right)^{3}\right]^{-1}*f^3*m_ac^2\)
where \(T=\cfrac{1}{f}\),
\(f=\cfrac{8^3}{6}{\cfrac {5}{12}}(3+{\sqrt {5}})\left(\cfrac {4}{\sqrt {10+2{\sqrt {5}}}}\right)^{3}\pi^2=216.418\pi^2=2135.962\,\,Hz\)
What is this frequency? icosahedron 2135-962 Hz
