The following formula is for the force, \(F\) on a sphere moving through a fluid of density, \(\rho\) at a velocity of \(v\) and a spin about a axis, \(\omega\),
\(F=\pi^2r^3\rho\omega\times v=\pi^2r^3\rho\omega v cos(\theta)\)
from the book "The Math Behind..." by Colin Beveridge, Cassell, Octopus Publishing Group Ltd.
For \(\theta=0\), the spiral flattens out and the particle is just in circular motion. For a spin axis through the center of the particle, \(r=a_{\psi}\); \( r\) is the size of the particle.
Since, \(cos(\theta=0)=1\), \(F=F_{max}=\pi^2r^3\rho\omega v\)
If \(v=c\) and \(v=r\omega=c\)
\(F_{max}=\pi^2r^3.\rho.\cfrac{c^2}{r}=\pi^2r^2.c^2.\rho\)
And if this force is the drag force, which is proportional to velocity squared,
\(F_{drag}=F_{max}=k.c^2=\pi^2r^2.c^2.\rho\)
\(k=\pi^2r^2.\rho\)
And in general, \(\theta\ne0\), the spin axis is not through the center of the particle, and the spiral stretches out,
\(k=\pi^2r^2.\rho.cos(\theta)\)
where \(\pi r^2.\rho\) is the mass per unit length in direction perpendicular to the area \(\pi r^2\). This value multiplied by \(v\), \(\pi r^2.\rho.|v|\) is the mass of the fluid displaced by the particle of radius \(r\) with velocity \(v\) in one unit second.
\(k=\pi.cos(\theta).m_{fluid,\,v}\)
\(m_{fluid,\,v}=\pi r^2.\rho.|v|\)
What is \(\theta\)?
