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In the oscillating rigid rod in resistive-force theory, $\sin \theta \cos \theta =\theta +O({ϵ}^3)$ and $\dot{x}=-ϵa\omega \sin \omega t$. Sincethe mean transverse force from a rocking rod, on the fluid in the tilt convention of part (a), isThe parallel and perpendicular drag coefficients of a slender filament are per unit length here. Their difference is essential: isotropic local drag gives zero transverse force. The hydrodynamic force on the rod is the negative of the displayed answer.

For the usual anisotropy $c_{\perp }>c_{\parallel }$ and $a>0$, the maximum positive mean transverse force from a rocking rod occurs atThe maximum magnitude has either quadrature phase $\varphi =\pm \pi /2$. If positive tilt is defined towards $-x$, the positive-force optimum is instead $+\pi /2$. At our optimum $\theta =ϵ\sin \omega t$ is opposite in sign to $\dot{x}$: the rod tilts so that the weaker tangential drag and stronger normal drag combine into a positive $y$ force on both strokes.

For $\varphi =0$ or $\pi$, orientation is a single-valued function of the centre displacement. The configuration retraces its path under time reversal, and kinematic reversibility of Stokes flow makes the force impulse cancel. Equivalently, $⟨\theta \dot{x}⟩=0$. The enclosed loop in the displacement-orientation plane is largest in quadrature and vanishes for a reciprocal stroke, illustrating the principle behind the scallop theorem.