When we consider, a regular dodecahedron 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 octahedron,
\(\cfrac{8^3}{6}\pi^2*f^2*m_ac^2=1*\cfrac{1}{Vol_{dodeca}}*m_ac^2\)
The centroid to vertex distance, \(r\) of an dodecahedron of edge \(a\) is,
\( r=a{\frac {\sqrt {3}}{4}}\left(1+{\sqrt {5}}\right)=T\)
if this is equal to \(T\) (a unit of time) then its volume is given by,
with \(a=\frac{4}{\sqrt {3}\left(1+{\sqrt {5}}\right)}T\)
\(Vol_{dodeca}=\frac {1}{4}(15+7{\sqrt {5}})a^{3}={\frac {1}{4}}(15+7{\sqrt {5}})\left(\frac{4}{\sqrt {3}\left(1+{\sqrt {5}}\right)}T\right)^{3}\)
and so,
\(\cfrac{8^3}{6}\pi^2*f^2*m_ac^2=1*\frac{1}{16} \frac{\left({\sqrt {3}\left(1+{\sqrt {5}}\right)f}\right)^{3}}{(15+7{\sqrt {5}})}*m_ac^2\)
\(\)
where \(T=\cfrac{1}{f}\),
\(f=\cfrac{8^3}{24}(15+7{\sqrt {5}})\left(\frac{4}{\sqrt {3}\left(1+{\sqrt {5}}\right)}\right)^{3}\pi^2=237.667\pi^2=2345.682\,\,Hz\)
What is this frequency? dodecahedron 2345-682 Hz
Telekinesis? Poor spoons...
