= Monodisperse circular ash runout
{title2=$R_\infty^4=R_0^4+8\operatorname{Fr}\sqrt{g'_0}(\mathcal V/\pi)^{3/2}/w_s$}
With $A=\mathcal V/\pi$, $g'_0=g_0c_0$ 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 $\tau=4A/(w_sR_\infty^2)$. Then $R^2/R_\infty^2=\tanh[t/\tau+\operatorname{artanh}s_0]$. Thus the finite limiting runout is approached at infinite time; treating it as an exactly attained finite-time stopping point contradicts these balances.
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