 Monodisperse circular ash runout

With $A=\mathcal{V}/\pi$, $g_0^{\prime }=g_0{c}_0$ and initial radius $R_0$, eliminating time from the axisymmetric settling box model gives $\sqrt{c/c_0}=(R_{\infty }^4-R^4)/(R_{\infty }^4-R_0^4)$. Put $s_0=R_0^2/R_{\infty }^2$ and $\tau =4A/(w_s{R}_{\infty }^2)$. Then $R^2/R_{\infty }^2=\tanh [t/\tau +\mathrm{artanh}{s}_0]$. Thus the finite limiting runout is approached at infinite time; treating it as an exactly attained finite-time stopping point contradicts these balances.