Optimal phase of a two-state rotary motor ($s_{\max }$)

For the driven two-state rotary motor with a triangular bound potential of height $U_0$, the torque scale is $T_0=U_0/\pi$. Differentiating the phase-averaged torque gives ${\overline{T}}^{\prime }(s)=T_0[2P_A(s)-1]/\pi$. The decreasing crossing of $P_A=1/2$ therefore gives $s_{\max }=\ln (2\alpha )/\beta$ and ${\overline{T}}_{\max }=T_0\ln \cosh (\pi \beta /2)/(\pi \beta )$. The slow-speed limit is $T_0/2$; the fast-speed limit vanishes as $T_0\pi k_0/(8\omega )$.