Opposite-handed counterrotating helical swimmer ($U=-\frac{2nBD\omega }{AD(1+n{)}^2-B^2(1-n{)}^2}$)

Two coaxial opposite-handed helices linked by an internal motor share a translation speed but counterrotate. In an additive local-drag model, a length ratio $n$ multiplies the second helix's diagonal resistances by $n$ and reverses its coupling. With ${\Omega }_1-{\Omega }_2=\omega$, force and torque balance give $U=-2nBD\omega /[AD(1+n{)}^2-B^2(1-n{)}^2]$.

As the length ratio of an opposite-handed reaction helix tends to infinity, its translation and rotation tend to zero as $1/n$, while the finite first helix rotates at almost the full motor rate. Small motions of the long rotor balance finite force and torque. The common propulsion vanishes in this limit of the additive model.

If the second rotor has zero drag, it cannot supply reaction torque. The single remaining helix must be both force-free and torque-free; a positive-definite resistance matrix forces its translation and rotation to vanish. The relative motor motion is absorbed by the zero-resistance rotor.

Equal local resistance magnitudes and opposite coupling signs give ${\Omega }_1=\omega /2$, ${\Omega }_2=-\omega /2$ and $U=-B\omega /(2A)$. Translation-induced torques cancel, while propulsive forces add because handedness and rotation both reverse.