Hindered-settling runout in a rectangular channel ($L_{\infty }^{5/2}=L_0^{5/2}+\frac{5F{M}^{3/2}\sqrt{g_0^{\prime }}}{V_s}\mathrm{artanh}\sqrt{{\varphi }_0}$)

For a constant-volume unit-width particle-laden gravity current with $M=Lh$, define the particle-fluid reference reduced gravity $g_0^{\prime }=g({\rho }_p-{\rho }_0)/{\rho }_0$, so the current has $g^{\prime }=g_0^{\prime }\varphi$. A constant front Froude number and absorbing-bed hindered settling give $\dot{L}=F\sqrt{g_0^{\prime }\varphi M/L}$ and $\dot{\varphi }=-(V_{s}L/M)\varphi (1-\varphi )$. Eliminating time and integrating gives the displayed runout length of a gravity current. For large runout and dilute initial concentration, $L_{\infty }^5\sim 25F^2{M}^3{g}_0^{\prime }({\varphi }_0+2{\varphi }_0^2/3+\cdots )/V_s^2$. This $g_0^{\prime }$ is buoyancy per unit particle fraction, not the initial current reduced gravity $g_0^{\prime }{\varphi }_0$.