Where Did The Pound Come From?

This post is defunct the post "Into A Pile Of Deep Shit" dated 09 Jun 2016, from which this discussion originates has been corrected.

From the post "Equating To A Indefinite Integral" dated 10 Jun 2016,

$q_{ a_{ \psi  } } =i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }$

If we compare the force due to $\psi$ and Newton's Gravitational force,

$m=M_e$

$\cfrac{1}{4\pi a^2_{\psi}}.i4\pi{ \cfrac { \dot { x }  }{ a_{ \psi  } }M_ec^{ 2 } }=G\cfrac{M_e}{a^2_{\psi}}$ --- (*)

we have,

$G={ \cfrac { i\dot { x }  }{ a_{ \psi  } } c^{ 2 } }$

The Gravitational constant, $G$ depended on the size of the planet.

if $i\dot{x}=c$,

$G={ \cfrac { c^{ 3 }  }{ a_{ \psi  } }  }$

For earth, $a_{\psi}=6371000$,

$G=\cfrac{299792458^3}{6371000}$

$G=4.229164e18$

If we consider entanglement by dividing by the Durian Constant,

$\cfrac{G}{A_D}=\cfrac{4.229164e18}{9.029022e26 }=4.683967e(-9)$

In the calculation for $\varepsilon_o$ we did not divide the result by the Durian Constant because the particle involved is not interacting as wave and is entangled to other particles.  In the case of gravity above where like particles attract, they are interacting as waves and are entangled to each other.

When we correct for the fact that the quoted $G$ is actually $\pi G$ because of the spin of the planets,

$\pi G=\pi*4.683967e(-9)=1.4715116317e-8$

What happened?  Maybe it should have been,

$m.2c^2ln(cosh(\pi))=G\cfrac{m}{a^2_{\psi}}$ --- (*)

at $x=a_{\psi}$.

$G=2a^2_{\psi}c^2ln(cosh(\pi))$

$G$ is dependent on $a^2_{\psi}$ and I am in deeper....

$G=2*6371000^2*299792458^2*ln(cosh(\pi))$

$G=1.787754e31$

If we replace $c^2$ with $\dot{x}^2$ and interpret it to be the spin of the planet,

$\dot{x}=\cfrac{2\pi a_{\psi}}{24*60*60}$ 

$G=2a^4_{\psi}\left(\cfrac{2\pi }{24*60*60}\right)^2ln(cosh(\pi))$

$G=a^4_{\psi}*2.5916926e(-8)$

$G$ is dependent on $a^4_{\psi}$, for earth $a_{\psi}=6371000$,

$G=4.269863e19$

We are wrong; on both counts, expression (*) and $i\dot{x}=v_s$.

And lastly, if we have,

${ q_{a_{\psi}} }=2\dot { x }F_{\rho}|_{x}=4\pi GM$

as

$4\pi x^2.G\cfrac{M}{x^2}=4\pi GM$

where the attractive force on the gravity particle interacting as wave is due only to the flux emanating below it.

$G=\cfrac{\dot { x }F_{\rho}|_{x}}{2\pi M}$

where,

$F_{ \rho  }=i\sqrt { 2{Mc^{ 2 } } } \, G_o.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x-x_{ z }) \right)$

and so,

$G=\cfrac{\dot { x }}{2\pi M}i\sqrt { 2{Mc^{ 2 } } } \, G_o.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)$

with $x_z=0$

$G=\cfrac{i\dot { x }c}{\pi \sqrt{2M}}\, G_o.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)$

$\cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } a_{\psi}=\pi$

$G=\cfrac{i\dot { x }}{a_{\psi}}c^2.tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)$

and $G$ becomes a running target...

Because at $x=a_{\psi}$,  

$tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (a_{\psi}) \right)=1$

we can approximate,

$tanh\left( \cfrac { G_o }{ \sqrt { 2{ Mc^{ 2 } } }  } (x) \right)\approx\cfrac{x}{a_{\psi}}$

$G\approx\cfrac{i\dot { x }}{a_{\psi}}c^2.\cfrac{x}{a_{\psi}}$

$F=G\cfrac{M}{x^2}=\cfrac{i\dot { x }}{a^2_{\psi}}c^2.\cfrac{M}{x}$

when $i\dot{x}=c$,

$F=\cfrac{c^3}{a^2_{\psi}}.\cfrac{M}{x}$

$G_{l}=\cfrac{c^3}{a^2_{\psi}}$

$G_{l}=\cfrac{299792458^3}{6371000^2}=6.638148e11$

$\cfrac{G_{l}}{A_D}=\cfrac{6.638148e11}{9.029022e26 }=7.352e(-16)$

And deep blue is brown...

Note:  What if $i\dot{x}$ is interpreted as the spin of the planet?  Instead of  $i\dot{x}=c$, we have $i\dot{x}=v_s$ and we replace $c$ with $t\dot{x}$. So,

$F=\cfrac{c^3}{a^2_{\psi}}.\cfrac{M}{x}$

$F=\cfrac{v_s^3}{a^2_{\psi}}.\cfrac{M}{x}$

$G_{l}=\cfrac{v_s^3}{a^2_{\psi}}$

$G_{l}=\cfrac{465^3}{6371000^2}$

$G_{l}=2.4771006228e(-6)$

$\psi$ at light speed wrap around a sphere manifest as gravitational mass that spins independently of $\psi$.  $i{x}\ne v_s$ the speed of the wave is not the spin of the mass.