= Optimal phase of a two-state rotary motor
{title2=$s_{\max}$}
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 $\overline T'(s)=T_0[2P_A(s)-1]/\pi$. 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 $T_0\pi k_0/(8\omega)$.
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