More Anti Gravity, Have a Heart Too

Consider a elemental fluid mass, $\Delta m$ in circular motion inside a tube, where the centripetal force is provided by the reaction force from the surface of the tube on the rotating mass,

$\Delta m = \rho\pi a^2_o r\Delta\theta$

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

The vertical lift on the wall of the tube is given by,

$F_wcos(\theta)=F_ccos(\theta)=\Delta m\{\cfrac{v^2}{r}-gcos(\theta)\}cos(\theta)$

Over the full length of the tube, the total upward force is given by

$F_{ l }=2\int _{ 0 }^{ \pi  }{ \rho \pi a^2_o r\{ \cfrac { v^{ 2 } }{ r } -gcos(\theta )\} cos(\theta ) } d\theta -m_{ t }g$

where $m_t$ is the total mass of the contraption.

$F_{ l }=2\rho \pi a^2_o\int _{ 0 }^{ \pi  }{ r\{ \cfrac { v^{ 2 } }{ r } -gcos(\theta )\} cos(\theta ) } d\theta -m_{ t }g$

$F_{ l }=2\rho \pi a^2_o\int _{ 0 }^{ \pi  }{ v^{ 2 }cos(\theta )-grcos^{ 2 }(\theta ) } d\theta -m_{ t }g$

$F_{ l }=2\rho \pi a^2_o\{ v^{ 2 }sin(\theta )|^{ \pi  }_{ 0 }-\int _{ 0 }^{ \pi  }{ grcos^{ 2 }(\theta ) } d\theta \} -m_{ t }g$

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

For positive overall lift,

$F_l≥0$

$\int _{ 0 }^{ \pi  }{ rcos^{ 2 }(\theta ) } d\theta \le \cfrac { -m_{ t } }{ 2\rho \pi a^2_o }$

$\int _{ 0 }^{ \pi  }{ rcos^{ 2 }(\theta ) } d\theta \le -A$

$A=\cfrac { m_{ t } }{ 2\rho \pi a^2_o }$

Let,

$\int { rcos^{ 2 }(\theta ) } d\theta =B(\theta )$

where    $B(\pi )-B(0)\le -A$ and is symmetrical about $\theta=0$,

$rcos^{ 2 }(\theta )=\cfrac { dB(\theta ) }{ d\theta  }$

Therefore,

$r=\cfrac { 1 }{ cos^{ 2 }(\theta ) }  \cfrac { dB(\theta ) }{ d\theta  }$

Consider,

$B(\theta )=\cfrac { a }{ 2 }\theta +\cfrac { 3b }{ 4 }sin(\theta) +\cfrac { a }{ 4 }sin(2\theta) +\cfrac { b }{ 12 }sin(3\theta)$

$B(\pi )-B(0)=\cfrac { a\pi }{ 2 }≤-A$

$a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^2_o }$

$\cfrac{dB(\theta)}{d(\theta)}=acos^{ 2 }(\theta )+bcos^{ 3 }(\theta )$

So,

$r=\cfrac { 1 }{ cos^{ 2 }(\theta ) } \cfrac { dB(\theta ) }{ d\theta  }=a+bcos(\theta)$

where $a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }$ and $b$ is arbitrary.  $b$ is in the expression for total length of the tube and so effects the total mass $m_t$ of the device.  In this way $b$ effects lift.

A example polar plot of such a Limaçon, r=-5+7cos(t) is shown below,

This shape is symmetric about the center line and so we know that the resultant will be acting through the C.G. of the tube along the center line and that there is only one upward resultant force.  All horizontal force components cancel.  To counter the tendency to rotate, two such tubes with counter flowing fluid are used.

Have a nice day.