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.
