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Use an inertial gravity-current box model in a horizontal prismatic V-shaped valley. Let its section have area $A(h)=\lambda h^2$ and width $2\lambda h$, where the dimensionless constant $\lambda >0$ is half-width divided by depth. Let $X(t)$ be front position and $h(t)$ the representative uniform current depth. Assume negligible entrainment and volume loss, hydrostatic cross-sections, large Reynolds number, and negligible bed friction and axial bed slope. These assumptions describe the bulk spreading after the short dam-break adjustment; they do not give a universal law for a steep or friction-dominated valley.

For water of fixed volume $W$, conservation gives $W=\lambda h^{2}X$. Close the front speed by the gravity-current front condition $\dot{X}=\mathrm{Fr}\sqrt{gh}$, where $\mathrm{Fr}$ is a constant order-one Froude number for the chosen inertial front model. ThereforeIntegration gives $X^{5/4}=X_0^{5/4}+5K_{w}t/4$. Once the initial reservoir length $X_0$ is negligible, the inertial dam-break current in a triangular valley satisfiesThe time exponent differs from a flat-bottomed channel because a V-shaped section stores volume proportional to $h^2$. A viscous V-shaped channel gravity current obeys a different spreading law, so retaining the inertial assumption is essential.

For the short-lived eruption, interpret $W$ as the bulk mixture volume released during the event. Let $\varphi (t)$ be the well-mixed particle volume fraction, initially ${\varphi }_0$, let ${\rho }_p$ and ${\rho }_a$ be particle and ambient carrier densities, and assume a dilute Boussinesq approximation for the mixture. The particles provide reduced gravityAssume thermal buoyancy, compressibility, erosion and continued injection are negligible, and that the carrier volume remains $W$. Turbulent mixing keeps the particles suspended uniformly while they settle vertically at a constant terminal settling velocity $w_s$. For small spherical particles of radius $a_p$ in carrier viscosity ${\mu }_a$, the Stokes settling velocity is $w_s=2({\rho }_p-{\rho }_a)g{a}_p^2/(9{\mu }_a)$ when the particle Reynolds number is small. If an eruption volume is quoted as solid particle volume rather than mixture volume, replace the $W$ used here by that volume divided by ${\varphi }_0$.

The particle deposition flux per horizontal projected bed area is $w_s\varphi$. The current's projected settling area is $2\lambda hX$, soThe factor two reflects the triangular section's mean depth $h/2$. Put $s=\sqrt{\varphi /{\varphi }_0}$ and $K=\mathrm{Fr}\sqrt{g_0^{\prime }}(W/\lambda {)}^{1/4}$. The front and concentration balances becomeTaking the initial release length negligible compared with the final runout, integration yields the constant-settling runout in a triangular valley:In particular $R\propto W^{3/7}(g_0^{\prime }{)}^{2/7}{w}_s^{-4/7}$ at fixed valley shape and front coefficient. With a finite initial length, replace $R^{7/4}$ by $X_0^{7/4}+7K\sqrt{W/\lambda }/(4w_s)$ in the corresponding integrated relation.

The whole time dependence is determined analytically by the quadratureEarly spreading has $X\simeq (5Kt/4{)}^{4/5}$, the water law with $g$ replaced by $g_0^{\prime }$. At late times the concentration loss slows the flow, and $R-X$ decays exponentially with rate $7K/(4R^{5/4})=w_s/h_R$, where $h_R=\sqrt{W/(\lambda R)}$. The final length $R$ is a limiting runout reached as $t\to \infty$, not a finite-time stopping event in this ideal model.

To calculate the deposit profile of a finite triangular-valley particle current, let $m_d(x)$ be final deposited particle mass per unit downstream length, and $M_0={\rho }_p{\varphi }_{0}W$ the initial particle mass. Deposition at a fixed section starts at arrival time $t_a(x)$, when the front first reaches $x$. With the instantaneous front coordinate denoted $\xi$, use $dt={\xi }^{1/4}d\xi /[K(1-(\xi /R{)}^{7/4})]$ to obtainWriting $\eta =x/R$, this evaluates toand $m_d=0$ beyond $R$. The profile decreases downstream, vanishes quadratically at the runout, and obeys ${\int }_0^R{m}_d(x)dx=M_0$. This integral check ensures that all initially suspended particle mass is accounted for.

For an areal rather than a linear deposit density, the changing valley width must also be specified. At lateral coordinate $y_{\perp }$, the current covers the bed only while $|y_{\perp }|\le \lambda h$. Set $Z=\min (R,\lambda W/y_{\perp }^2)$, with $Z=R$ on the valley axis. The mass per horizontal projected bed area is zero if $Z<x$, and otherwise isOn the axis this reduces to ${\rho }_p{\varphi }_0{h}_R[\frac{7}{5}(1-{\eta }^{5/4})-\frac{7}{12}(1-{\eta }^3)]$. Divide projected areal density by $\sqrt{1+{\lambda }^{-2}}$ to obtain mass per actual sloping bed area. A specified packed deposit density then converts this areal mass to local deposit thickness; the spreading model itself predicts deposited mass, not a packing fraction.