Anti-gravity

Consider a body in circular motion in a circular tube at high velocity, $v$.   Where

$mg=m\cfrac{v_g^2}{r}$,    $v_g=\sqrt{g.r}$

$v>>v_g$

The body pushes against the outer wall of the tube and a reaction force from the wall pushes it towards the center of the circular path.  It is this force on the body together with gravity that provides for the centripetal force.

$m\cfrac{v^2}{r}=F_b+mgcos(\theta)$

Numerically,

$F_w=F_b=m\cfrac{v^2}{r}-mgcos(\theta)$

where $F_b$ and $F_w$ are action and reaction pair.  The vertical lift created by the reaction force on the tube wall by $F_w$ is,

$F_l=F_wcos(\theta)=m\left\{\cfrac{v^2}{r}-gcos(\theta)\right\}cos(\theta)$

The horizontal component that comes with this lift can be counter-balanced by another similar tube with the body rotating in the opposite sense.  At high velocity $v>>v_g$ such that we can ignore the term $mgcos^2(\theta)$, then

$F_l=m\cfrac{v^2}{r}cos(\theta)$

If the body is a ferromagnetic fluid or mercury driven by a electromagnetic force, then it is possible to change its velocity $v$ by changing the hollow radius of the tube or $r$ the path of the tube, since velocity is inversely proportional to the radius squared,

$\cfrac{v_o}{v}=\cfrac{a^2}{{a_o}^2}$

If

$\cfrac{a^2}{{a_o}^2}=\sqrt{cos(\theta)}$,    then

$v^2={v_o}^2\left\{\cfrac{{a_o}^2}{a^2}\right\}^2={v_o}^2\cfrac{1}{cos(\theta)}$

and so,

$F_l=m\cfrac{v^2}{r}cos(\theta)=m\cfrac{{v_o}^2}{r}$

which is a constant provided $r$ is a constant, ie. a circular path.  $F_l$ can then be varied by changing $v_o$.  A pair of such a device that compensate each other laterally, will then generate a total lift of $2*F_l$.  It is not possible to make the tube with a hollow radius of zero at $\theta=\cfrac{\pi}{2}$, but it can be at a practical minimum, $a_f$ at some $\alpha_f$.

On the lower half of the tube,

$m\cfrac{v^2}{r}=F_b-mgsin(\alpha)$

$F_w=F_b=m\cfrac{v^2}{r}+mgsin(\alpha)$

When $\theta$ is measured from the vertical axis, then

$\theta=\alpha+\cfrac{\pi}{2}$

So,

$sin(\alpha)=sin(\theta-\cfrac{\pi}{2})=sin(-(\cfrac{\pi}{2}-\theta))=-sin(\cfrac{\pi}{2}-\theta)=-cos(\theta)$

Still numerically,

$F_w=m\cfrac{v^2}{r}-mgcos(\theta)$    for     $\cfrac{\pi}{2}<\theta<\pi$

This force is downwards, an expression for vertical lift is given by,

$F_l=-F_wcos(\theta)=mcos(\theta)\left\{-\cfrac{v^2}{r}+gcos(\theta)\right\}$

which is always negative given the value of $r$ and $\theta$.  It is possible to change $r$ with $\theta$ to obtain a path over which the body in motion inside is always experiencing a positive lift.  In this case however we will bent the tube such that there is no lower half and at the same time provide for a tube that runs in the opposite sense to compensate for the horizontal component of $F_w$.

Anti Gravity Device I

And we consider the whole tube filled with fluid in circular motion, then

$\Delta F_l=\Delta m\cfrac{v^2}{r}cos(\theta)$

where $\Delta m=\rho.\pi {a^2}r\Delta \theta$ and $\rho$, the density of the fluid.

$F_l=2\int^{\pi/2}_{\alpha_f}{\rho.\pi a^2r\cfrac{v^2}{r}cos(\theta)}d\theta$

$F_l=2\rho.\pi a^2{v^2}\int^{\pi/2}_{\alpha_f}{cos(\theta)}d\theta$

$F_l=2\rho.\pi a^2{v^2}\{1-sin(\alpha_f)\}$

$F_l=2\rho.\pi a^2{v^2}$    when $\alpha_f=0$

In the picture above, the device can then provide a constant lift of $6*F_l$ at all time.  Unfortunately,  nobody can patent this but me.

In the case where $v$ is varied, for a tube filled with fluid in circular motion, we let,

$\cfrac{{a_o}^2}{a^2}={cos^{-1}(\theta)}$,    then since,

$\cfrac{v_o}{v}=\cfrac{a^2}{{a_o}^2}$

$v^2={v_o}^2\left\{\cfrac{{a_o}^2}{a^2}\right\}^2={v_o}^2cos^{-2}(\theta)$

and so,

$\Delta F_l=\Delta m\cfrac{v^2}{r}cos(\theta)=\Delta m\cfrac{v^2_o}{r}cos^{-1}(\theta)$

where $\Delta m=\rho.\pi a^2 r\Delta \theta$ and $\rho$, the density of the fluid.

$F_l=2\int^{\pi/2}_{\alpha_f}{\rho.\pi a^2 r\cfrac{v^2_o}{r}cos^{-1}(\theta)}d\theta$

Since $a^2=a_o^2{cos(\theta)}$,

$F_l=2\rho.\pi a^2_o{v^2_o}\int^{\pi/2}_{\alpha_f}{}d\theta$

$F_l=2\rho.\pi a^2_o{v^2_o}\{\cfrac{\pi}{2}-\alpha_f\}$

$F_l=\rho.\pi^2 a^2_o{v^2_o}$    when $\alpha_f=0$

It is not necessary to vary the hollow radius of the tube.  However $F_l$ independent of $\theta$ would mean that lift is uniform throughout the length of the tube.

If we do not ignore the effect of gravity, there is a negative term given by,

$F_{ g }=-2\rho g\pi a^2_o\int _{ 0 }^{ \cfrac{\pi}{2}  }{ rcos^{ 2 }(\theta ) } d\theta$

When $r$ is a constant $r_o$

$F_{ g }=-2\rho g\pi a^2_o r_o\int _{ 0 }^{ \cfrac{\pi}{2}  }{ cos^{ 2 }(\theta ) } d\theta$

$F_{ g }=-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o$

And the total lifts for the two cases are given by

$F_l=2\rho\pi a^2_o{v^2}-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o=\rho\pi a^2_o(2v^2-\cfrac{1}{2}g\pi r_o)$

$F_l=\rho\pi^2 a^2_o{v^2_o}-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o=\rho\pi^2a^2_o(v^2_o-\cfrac{1}{2}g r_o)$

respectively.