 Axial coupling of an elliptic helix

For a slender centreline $\mathbf{X}(\theta )=(A\cos \theta ,B\sin \theta ,b\theta )$, $-\pi \le \theta \le \pi$, let $s^{\prime }(\theta )=(A^2{\sin }^2\theta +B^2{\cos }^2\theta +b^2{)}^{1/2}$ and use local force density $\mathbf{f}=c(2I-\mathbf{tt})\mathbf{v}$, $\mathbf{t}=d\mathbf{X}/ds$, $c=2\pi \mu /\ln (L/R)$. Rotation with angular speed $\Omega$ gives $\mathbf{t}\cdot \mathbf{v}=\Omega AB/s^{\prime }$, henceAxial translation with speed $W$ gives exactly the same coefficient for the torque: $G_z=-cWABb\int d\theta /s^{\prime }$. Indeed $(\mathbf{X}\times \mathbf{t}{)}_z=AB/s^{\prime }$. Both forces and torques here are exerted on the fluid. In the circular case $A=B=a$, this reduces to $F_z/\Omega =G_z/W=-2\pi c{a}^{2}b/\sqrt{a^2+b^2}$. The equality is also a consequence of the Lorentz reciprocal theorem, since the local resistance tensor is symmetric.