And Some More Auntie Gravy

From the post "More Anti Gravity,  Have a Heart Too",

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

If much of the weight of the device is in the fluid then

$m_t=\rho L.\pi a^2_o$

where $L$ is the total length of the tube.  Then,

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

and so,

$|a|≥\cfrac{L}{\pi}$

But the total length of a polar curve is,

$L=\int{\sqrt{r^2+(\cfrac{dr}{d\theta})^2}}d\theta$

In this case,

$r=a+bcos(\theta)$

$\cfrac{dr}{d\theta}=-bsin(\theta)$

So,

$L=2\int^{\pi}_{0}{\sqrt{(a+bcos(\theta))^2+b^2sin^2(\theta)}}d\theta$

$L=2\int^{\pi}_{0}{\sqrt{a^2+b^2+2abcos(\theta)}}d\theta$

For positive lift,

$\pi|a|≥2\int^{\pi}_{0}{\sqrt{a^2+b^2+2abcos(\theta)}}d\theta$

The expression on the Right Hand Side cannot be easily evaluated, but we know that $a$ and $b$ are of opposite sign,

$\pi|a|≥2\int^{\pi}_{0}{\sqrt{a^2+b^2-2|a||b|cos(\theta)}}d\theta$

When $b=0$,

$RHS=2|^{\pi}_0a=2\pi a$   which the perimeter of a circle but is greater than LHS.

So, $b≠0$; the tube in the shape of a circle does not provide a net lift.

If we consider, $\cfrac{dL}{db}=0$ for a minimum $L$ given $a$

$\cfrac{dL}{db}=\int^{\pi}_{0}{(a^2+b^2-2|a||b|cos(\theta))^{-\cfrac{1}{2}}(2b-2{|a|cos(\theta))}}d\theta$

$\cfrac{dL}{db}=2\int^{\pi}_{0}{(a^2+b^2-2|a||b|cos(\theta))^{-\cfrac{1}{2}}(b-{|a|cos(\theta))}}d\theta$

For any possibility of $\cfrac{dL}{db}=0$, $b<|a|$ such that the expression,

$(b-|a|cos(\theta))$ can change sign.

So, $|a|>b>0$

For the case of  $|a|=b$,

$\cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi  }{ (b^{ 2 }-b^{ 2 }cos(\theta ))^{ -\cfrac { 1 }{ 2 }  }(b-{ bcos(\theta )) } } d\theta$

$\cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi  }{ (1-cos(\theta ))^{ -\cfrac { 1 }{ 2 }  }(1-{ cos(\theta )) } } d\theta$

$\cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi  }{ (1-cos(\theta ))^{ \cfrac { 1 }{ 2 }  } } d\theta$

$\cfrac { dL }{ db }=\sqrt{2}|^{\pi}_{0}\left(-\sqrt{2}{\cdot}\cos\left(x\right)-\sqrt{2}\right){\cdot}\sin\left(\dfrac{\mathrm{arctan2}\left(\sin\left(x\right), \cos\left(x\right)\right)+{\pi}}{2}\right)+\sqrt{2}{\cdot}\sin\left(x\right){\cdot}\cos\left(\dfrac{\mathrm{arctan2}\left(\sin\left(x\right), \cos\left(x\right)\right)+{\pi}}{2}\right)$

$\cfrac { dL }{ db }=4≠0$ but $L=(2\sqrt{2})^2a=8a$.

For the case of  $b=0$,

$\cfrac { dL }{ db } =2\int^{\pi}_{0}{-cos(\theta)}d\theta=0$

This means for a given $a$, $b=0$ is a turning point for $L$ which is $2\pi a$, the perimeter of a circle of radius $a$.  This is a local maximum value for $L$ because,
 
$\cfrac { d^{ 2 }L }{ db^{ 2 } } =\int _{ 0 }^{ \pi  }{ \cfrac { 2a^{ 2 }sin^{ 2 }(\theta ) }{ (a^{ 2 }+b^{ 2 }-2abcos(\theta ))^{ \cfrac { 3 }{ 2 }  } }  } d\theta$

When $b=0$,

$\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{2}{a}\int _{ 0 }^{ \pi  }{sin^{ 2 }(\theta ) }d\theta$

$\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{1}{a}|^{\pi}_{0}\{\theta-\cfrac{1}{2}sin(2\theta)\}$

$\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{\pi}{a}$

Since, $a<0$, $\cfrac { d^{ 2 }L }{ db^{ 2 } }<0$ for $b=0$, $L$ is a maxima.

When $b=|a|$, $L=8a$, when $b=0$, $L=2\pi a<8a$ and at $b=|a|$,$\cfrac{dL}{db}>0$.  This means there is at least one minimum before $b=|a|$ so that the curve turns upward again after the maximum at $b=0$.  So, there is still the possibility of satisfying the positive lift condition.  Hold the gravy.