\(mg=m\cfrac{v_g^2}{r}\), \(v_g=\sqrt{g.r}\)
\(v>>v_g\)
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.
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
\(\theta=\alpha+\cfrac{\pi}{2}\)
\(sin(\alpha)=sin(\theta-\cfrac{\pi}{2})=sin(-(\cfrac{\pi}{2}-\theta))=-sin(\cfrac{\pi}{2}-\theta)=-cos(\theta)\)
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 \(F_w\).
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| 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.
