Monodisperse circular ash runout ($R_{\infty }^4=R_0^4+8\mathrm{Fr}\sqrt{g_0^{\prime }}(\mathcal{V}/\pi {)}^{3/2}/w_s$)

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.

For the monodisperse circular ash runout, integrate the local particle deposition flux ${\rho }_p{w}_{s}c(t)$ from first arrival to infinite time. Here $s=\max (s_0,r^2/R_{\infty }^2)$ for $0\le r<R_{\infty }$ and $s_0=R_0^2/R_{\infty }^2$. The displayed $\Sigma$ is deposited particle mass per horizontal area, and vanishes outside the limiting current. It is constant under the initial disk, then decreases to zero; integrating $2\pi r\Sigma$ gives the initial ash mass ${\rho }_p{c}_0\mathcal{V}$. A deposit thickness additionally requires its packing fraction or bulk deposit density.