User:Hamster-baldwin/sandbox

\begin{document} \maketitle

$\frac{e}{on/off}=mc^2

$$

$e=\frac{m}{on/off}c^2

$$

To be clear; on/off is a literal referance to computational on switches divided by computational off switches.

$$

$e2=(\frac{\frac{m}{3}}{m_e}+\frac{\frac{m}{3}}{m_p}+\frac{\frac{m}{3}}{m_n})\hbar[\frac{h(\frac{c}{2\pi5.35317245x10^-11})^4}{c^2}]

$$

$m2=\frac{[(\frac{\frac{m}{3}}{m_e}+\frac{\frac{m}{3}}{m_p}+\frac{\frac{m}{3}}{m_n})2\frac{2Gm}{c^2}][\frac{h^2(\frac{c}{2\pi5.35317245x10^-11})^4}{c^2}]}{[\frac{G8\pi}{[\frac{1}{(2c^2)}]}]}

$$

$m2=\frac{[(\frac{\frac{m}{3}}{m_e}+\frac{\frac{m}{3}}{m_p}+\frac{\frac{m}{3}}{m_n})2\frac{2Gm}{c^2}][\frac{h^2(\frac{c}{2\pi9.26875894x10^-7})^4}{c^2}]}{[{G8\pi}]}

$$

$m2=(\frac{\frac{m}{3}}{m_e}+\frac{\frac{m}{3}}{m_p}+\frac{\frac{m}{3}}{m_n})\hbar[\frac{h(\frac{c}{2\pi9.26875894x10^-7})^4}{c^2}]

$$

Electrostatic acceleration of small masses;

$$

$f=ma

$$

Let l_p$ represent the Planck Mass, and cl_p$ represent c times the planck mass.

$$

$2c^2=(\frac{\frac{m}{3}}{m_e}+\frac{\frac{cl_p140.938613}{3}}{m_p}+\frac{\frac{cl_p140.938613}{3}}{m_n})\hbar[\frac{h(\frac{c}{2\pi5.35317245x10^-11})^4}{c^2}]

$$

$cl_p140.938613=0.00000306781079

$$

Let $m_p_na$ represent the acceleration of the protons and neutrons (for teleportation).

$$

$m_p_na=c\frac{\frac{2G(cl_p140.938613)}{c^2}}{[\frac{2Gm}{c^2}]}

$$

Teleportation; the electron masses contain two of the entire masses of the object.

$$

$m4=\frac{\frac{(\frac{\frac{m}{3}}{m_e}+\frac{\frac{c[\frac{\frac{2G(cl_p140.938613)}{c^2}}{[\frac{2Gm}{c^2}]}]}{3}}{m_p}+\frac{\frac{c[\frac{\frac{2G(cl_p140.938613)}{c^2}}{[\frac{2Gm}{c^2}]}]}{3}}{m_n})\hbar[\frac{h(\frac{c}{2\pi9.720723344x10^-12})^4}{c^2}]}{2c^2}}{m_e}

$$

$6=\frac{(\frac{1}{m_p}+\frac{1}{m_n})\hbar[\frac{h(\frac{c}{2\pi9.720723344x10^-12})^4}{c^2}]}{c^2}

$$

$e2=(m/3/m_e)/[1/(m_ec^2)/6]

$$

$Cpu Hz=\frac{c}{r\pi2}

$$

The Bohr Radius (r_0)$ of hydrogen 5.29177210903x10^-11$ meters

was modified (r_m)$ in this theory to 5.35317245x10^-11$ meters

because hydrogen didn't carry sufficent charge for computation.

$$

$ Duk-Ro

$$

I regret that I would be lying if I had said I had found a physical or technical solution to the Hubbert Peak, yet it was worth noting so I have done so.

$$

https://en.wikipedia.org/wiki/Hubbert\_peak\_theory

$$

Bradley William Busch

\end{document}