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An axisymmetric gravity-current box model with no entrainment and negligible suspended-particle volume loss has constant bulk volume $\mathcal{V}$, uniform height $h$, and uniform particle volume fraction $c$. For an absorbing horizontal bed and downward settling velocity $w_s>0$, particle mass conservation gives $\dot{c}=-w_{s}c/h$. A constant front Froude number closes spreading by $\dot{R}=\mathrm{Fr}\sqrt{g_{0}ch}$, where $g_0=g({\rho }_p-{\rho }_a)/{\rho }_a$ is buoyancy per unit volume fraction. This model describes the mixed bulk and its head closure, not the initial inertial blast or a resolved head shape.

A gravity-current box model in a prismatic channel with width equal to height has volume $V_0=L{h}^2/2$. Closing the front speed by $\dot{L}=F\sqrt{g^{\prime }h}$ gives the displayed law and long-time spreading $L\propto t^{4/5}$. This is an integral approximation with constant volume and reduced gravity, not an exact uniform-depth solution of the local shallow water equations.

In a constant-volume gravity-current box model without settling, $\varphi ={\varphi }_0$ and $g^{\prime }=g{\varphi }_0(\gamma -\beta t)$. A constant Froude number then gives finite runout length of a gravity current at the neutral-buoyancy time $\gamma /\beta$. Further motion requires a different physical model once the stable density stratification is lost.

A heated particle-laden layer released along a bed forms a gravity current while $g^{\prime }=g(\gamma \varphi -\theta )>0$. In a gravity-current box model, heating can make $g^{\prime }$ vanish in finite time, whereas settling velocity alone can give finite runout length of a gravity current approached asymptotically.

In a horizontal prismatic V-shaped valley with section area $A=\lambda h^2$, a fixed-volume high-Reynolds-number gravity-current box model has $h=(W/(\lambda X){)}^{1/2}$ and front speed $\dot{X}=\mathrm{Fr}\sqrt{gh}$. Integrating gives $X=[5\mathrm{Fr}\sqrt{g}(W/\lambda {)}^{1/4}t/4{]}^{4/5}$ after the initial release scale is neglected. This is an inertial closure, distinct from the viscous V-shaped channel gravity current, whose exponent is different.

Use a gravity-current box model: approximate the current by uniform depth $h(t)$ and length $L(t)$, with constant reduced gravity, no deposition and no fluid entrainment. The triangular cross-sectional area is $h^2/2$, so the fixed released volume isThe Benjamin deep-ambient front condition closes the integral model:For a finite initial length $L_0$, direct integration givesThe ideal point-release limit takes $L_0=0$ and has $L\propto t^{4/5}$, $h\propto t^{-2/5}$. Its divergent initial depth is outside the shallow water approximation, so a real finite reservoir regularizes the early stage. This is the finite-volume triangular-channel current integral law; it does not assert a spatially uniform exact solution of the local equations.

Model the post-eruption flow as an axisymmetric, shallow, hydrostatic, turbulently mixed layer on a horizontal bed. Neglect entrainment, erosion, heat exchange, rotation and large-scale viscous drag. Particles are dilute, with density ${\rho }_p$, uniform-in-depth particle volume fraction $c(r,t)$ and constant downward settling speed $w_s>0$ relative to the carrier gas. Deposited particles are not resuspended. Their volume is small enough that settling does not significantly change the bulk carrier volume. Let ${\rho }_a$ be ambient air density and ${\rho }_g$ carrier-gas density at its local temperature, and use a leading Boussinesq approximation for the mixture motion. DefineThe hot gas reduces the density excess; the ground-hugging current model requires $g^{\prime }>0$. If density contrasts are too large for Boussinesq dynamics, density-weighted balances would be needed instead. The equations below are a stated shallow-layer model, not a description of the initial explosive vent jet.

For layer height $h$ and radial depth-averaged speed $u$, the appropriate shallow-water balances areThe hydrostatic force gives the nonconservative momentum equation $u_t+u{u}_r=-g^{\prime }{h}_r-(h/2)g_r^{\prime }$. In particular a spatially varying particle fraction must not be treated as constant reduced gravity when differentiating pressure. With no heat exchange or entrainment, a carrier-temperature anomaly $\Theta$ is advected:At this order $b_T$ is proportional to $-\Theta$. At the bed the carrier has no normal penetration and particles have absorbing deposition flux $w_{s}c$; at the top the pressure equals ambient pressure. Axis regularity requires $u(0,t)=0$ and smooth scalar fields, with no further source after the initial pulse. The initial height, particle inventory and temperature anomaly specify the released volume and ash content. Outside the front $r=R(t)$, the current layer is absent. At a resolved advancing head, use the material-front condition and the gravity-current front condition $\dot{R}=\mathsf{F}\sqrt{g_f^{\prime }{h}_f}$, where $\mathsf{F}$ is an order-one front Froude number and $h_f,g_f^{\prime }$ refer to the fluid feeding the head. The front closure replaces unresolved head dynamics; simply prescribing both a zero-depth mathematical edge and a finite-depth box-head speed would be inconsistent.

To obtain a gravity-current box model, assume horizontal as well as vertical mixing, uniform $h(t),c(t),\Theta (t)$ in a disk of radius $R(t)$, and a negligible-volume head whose feeding depth is the box depth. The conserved bulk volume $\mathcal{V}$ givesA bulk velocity $u=(\dot{R}/R)r$ satisfies interior continuity for this box. Rather than impose a uniform pressure on the full shallow-water equations and thereby lose the head force, replace their momentum dynamics by the front closure. Total suspended particle volume is $\mathcal{V}c$, so bed deposition givesThese ordinary differential equations, together with initial $R_0,c_0,\mathcal{V},{\Theta }_0$, are the axisymmetric settling box model. The maintained negative thermal term may eventually cancel particle buoyancy and cause lift-off; beyond $g^{\prime }=0$ this horizontal gravity-current model does not apply.

For the requested particle-only calculation set $b_T=0$ and define the initial reduced gravity $g_0^{\prime }=b_p{c}_0$. The equations reduce toEliminating time gives $d\sqrt{c}/dR=-w_s{R}^3/[2\mathsf{F}\sqrt{b_p}{A}^{3/2}]$. Integrating from the release state yieldsThus the monodisperse circular ash runout has limiting radiusThe maximum outward distance beyond the initial release edge is $R_{\infty }-R_0$. When the vent scale is negligible, set $R_0=0$ in this late-time source idealization. The resulting radius scales as ${\mathcal{V}}^{3/8}(g_0^{\prime }{)}^{1/8}{w}_s^{-1/4}$; it is not obtained by conserving reduced gravity after particles settle out.

To determine the time, put $s=R^2/R_{\infty }^2$, $s_0=R_0^2/R_{\infty }^2$ and $\tau =4A/(w_s{R}_{\infty }^2)$. The concentration relation becomes $\sqrt{c/c_0}=(1-s^2)/(1-s_0^2)$, and the front equation reduces toConsequentlyThe furthest reach is a limiting radius, approached asymptotically rather than attained at a finite stopping time. As the residual particle concentration vanishes, the front slows exponentially. A finite observational threshold for remaining ash would give a finite practical stopping time, but no such threshold is part of these balances.

The local ash deposit density is most naturally its mass per horizontal area, $\Sigma (r)$. Once the front has arrived at $r$, its local deposition flux is ${\rho }_p{w}_{s}c(t)$. For $r<R_{\infty }$, define $s_a=\max (s_0,r^2/R_{\infty }^2)$; the arrival time is zero inside the initial footprint and otherwise $t_a=\tau [\mathrm{artanh}{s}_a-\mathrm{artanh}{s}_0]$. Sincewe can integrate the flux exactly. Using $\int {\mathrm{sech}}^4\xi d\xi =\tanh \xi -({\tanh }^3\xi )/3$ gives the circular ash deposit profileIt is constant under the initial disk, then decreases to zero at the limiting edge. Particle solid volume per area is $\Sigma /{\rho }_p$; deposit thickness is $\Sigma /{\rho }_d$ if a bulk deposit density ${\rho }_d$ including voids is supplied. The problem does not determine that packing density.

For a negligible initial radius this simplifies toIts integral $2\pi {\int }_0^{R_{\infty }}r\Sigma (r)dr={\rho }_p{c}_0\mathcal{V}$ confirms that the complete initial ash inventory is deposited. The finite-$R_0$ formula gives the same budget including its central flat portion.

The diagram uses the negligible-vent-radius idealization and time normalized by $\tau$. It illustrates both the asymptotic approach to $R_{\infty }$ and the smoothly vanishing outer deposit. All these predictions depend on the mixing and constant-Froude head closure; they do not claim to resolve the initial blast or the ultimate inertia-dominated motion of almost particle-free gas.

Assume the carrier fluid has ambient mass density ${\rho }_0$, the particles have mass density ${\rho }_p>{\rho }_0$, and volumes add. The well-mixed current then hasHere $g_0^{\prime }$ is the particle-fluid reference reduced gravity per unit volume fraction, which is the convention needed in the requested expansion. The initial current reduced gravity is $g_0^{\prime }{\varphi }_0$; it should not be confused with $g_0^{\prime }$ itself. The Boussinesq approximation requires $\varphi ({\rho }_p-{\rho }_0)/{\rho }_0\ll 1$.

For particles to remain well mixed, the turbulent vertical mixing time must be short compared with the settling time. In a turbulent diffusion description with diffusivity $K_z$, require $h^2/K_z\ll h/[V_s(1-\varphi )]$, or $K_z\gg h{V}_s(1-\varphi )$. Equivalently, characteristic turbulent vertical speeds should substantially exceed the particle settling velocity. Turbulence can maintain an approximately uniform concentration in the body while particles still settle relative to the fluid. At the impermeable bed, fluid has no normal transport into the solid boundary, but particles can reach and be captured by it. Assume no resuspension after capture.

Use a gravity-current box model of length $L(t)$, uniform depth $h(t)$ and uniform concentration $\varphi (t)$. Neglect ambient entrainment and the small volume change caused by depositing dilute solids, so $Lh=M$ remains constant. Take a constant front Froude number $F>0$ and the gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }h}$. With ${\varphi }_{\max }=1$, the deposition flux is $V_s\varphi (1-\varphi )$ per unit bed area. The particle-volume budget isHence the integral model isThe model assumes a fixed rear boundary, one advancing front, a deep stationary ambient, negligible drag during the inertial spreading regime, and a perfectly absorbing bed. The supplied hindered-settling law is retained in the deposition closure.

For $0<\varphi <1$, eliminating time givesThe concentration integral is $2\mathrm{artanh}\sqrt{\varphi }$. ThereforeAs the particle concentration tends to zero, the current reduced gravity and front speed tend to zero. The hindered-settling runout in a rectangular channel isIt is a finite limiting distance; in this idealized model, the exponential tail of particle loss makes complete stopping asymptotic in time. For $L_{\infty }\gg L_0$, omit $L_0^{5/2}$. Using $\mathrm{artanh}x=x+x^3/3+O(x^5)$ givesThe original PDF has $V_s^2$ in this expression; the TeX aid's $V_s^3$ is a transcription error. Dimensions independently confirm the square: $M^3{g}_0^{\prime }/V_s^2$ has dimensions of length to the fifth power. If one instead names the initial current reduced gravity $G_{\mathrm{init}}=g_0^{\prime }{\varphi }_0$, the same expansion is $25F^2{M}^3{G}_{\mathrm{init}}[1+2{\varphi }_0/3+O({\varphi }_0^2)]/V_s^2$. Late loss of turbulent mixing, viscous resistance or bed resuspension can change the physical runout beyond this specified box model.

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.

At the gravity current nose, the depth and velocity change over a distance comparable with the depth, with appreciable vertical acceleration and mixing. The hydrostatic approximation and the depth-uniform interior model do not resolve that structure. A suitable gravity-current front condition iswith a positive, order-one Froude number determined by the nose closure and geometry. It is not fixed by the interior equations alone. This definition uses $\sqrt{g^{\prime }h}$; if the front condition is expressed using $c=\sqrt{g^{\prime }h/2}$, its numerical coefficient changes by $\sqrt{2}$.

For a late-time gravity-current box model, take the depth and particle volume fraction to be uniform over $0<x<X(t)$, and neglect entrainment and resuspension. The initial volume is $\mathcal{V}=\beta h_0^{2}L/2$, soThe total particle volume is $\mathcal{V}\varphi$. Deposition over the projected boundary width gives $d(\mathcal{V}\varphi )/dt=-\beta hXw\varphi$. With the specified hindered settling factor, the prismatic triangular-channel gravity-current box model isInitializing this approximate model with $X(0)=L$ and $\varphi (0)={\varphi }_0$ matches the original volume and concentration; it does not assert that the immediate dam-break flow is a uniform box.

Eliminate time to obtainSince the left primitive is $2\mathrm{artanh}\sqrt{\varphi }$, the hindered-settling runout invariant in a prismatic triangular channel isThis relation, the volume constraint and $\dot{X}$ determine the late-time evolution. If desired, the time is given by the quadrature $t={\int }_L^{X(t)}[\mathrm{Fr}\sqrt{G\varphi (s)h_0\sqrt{L/s}}{]}^{-1}ds$. The physical branch has nonnegative right-hand side in the boxed relation; it ends when that expression reaches zero.

Thus the limiting runout length of a gravity current, measured from the closed end, isThe advance beyond the dam is $X_{\infty }-L$. In the dilute limit, $\mathrm{artanh}\sqrt{{\varphi }_0}\simeq \sqrt{{\varphi }_0}$. The ideal model approaches its finite runout length of a gravity current only as $t\to \infty$: near the limit $\varphi$ decays exponentially with rate $2w_0/h_{\infty }$ and the front speed tends to zero. The exponent $7/4$ uses the prismatic triangular geometry; a channel widening with $x$ has a different volume constraint and runout law.

Let $\mathcal{V}=H{L}_0$ be the layer volume per unit transverse width. The gravity-current box model imposes $hL=\mathcal{V}$; it approximates the front and scalar balances, rather than an exactly uniform solution of the local momentum equation. For a specified constant front Froude number $\mathrm{Fr}$, the heated particle-laden gravity current satisfiesThe initial data are $L=L_0$, $\varphi ={\varphi }_0$, and $\theta =0$. The gravity-current front condition applies only while $\gamma \varphi -\theta \ge 0$.

For heating without settling, $\varphi ={\varphi }_0$ and $\theta =\beta {\varphi }_{0}t$. Integration of the front equation givesHence the heated gravity-current runout isIt is reached at the neutral-buoyancy time $\gamma /\beta$ within this front model. A finite-momentum current could subsequently coast or lift off: the formula is the maximum predicted by the prescribed gravity-current front condition, which contains no independent front inertia.

For settling without heating, $\theta =0$. Set $K=\mathrm{Fr}\sqrt{g\gamma \mathcal{V}}$, so $\dot{L}=K\sqrt{\varphi }{L}^{-1/2}$. Eliminating time yieldsThus the runout length of a gravity current isThis length is approached as $t\to \infty$, because positive concentration cannot disappear at a finite time under the settling law. If distance means displacement from the removed barrier, the answer is $L_{\max }-L_0$ in either case.

Use a well-mixed gravity-current box model of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard gravity-current front condition areThese two ordinary differential equations are the required integral model.

First suppose $R_1>0$ and putEliminating time givesWithintegration from $(L,\varphi )=(L_0,1)$ yieldsAs $t\to \infty$, the concentration tends to zero and the runout length of a gravity current isThe formula has a regular $a\to 0$ limit. If $R_1=0<R_2$, direct integration instead giveswhose value at $\varphi =0$ gives $L_{\infty }$.

Let the fixed current volume per unit channel width beA suitable high-Reynolds number deep-ambient gravity-current front condition iswhere the order-one Froude number $\mathrm{Fr}$ records the selected front closure. In the dilute limit the mixture's reduced gravity iswith ambient and carrier-fluid density ${\rho }_a$.

The well-mixed particle volume of species $i$ is $V_0{\varphi }_i$. Its deposition rate through the base of length $L$ is $L{\hat{W}}_i{\varphi }_i$, where ${\hat{W}}_i<0$. The resulting gravity-current box model is thereforeIt conserves fluid volume while suspended particle volume, and therefore the driving buoyancy, decreases by deposition.

A finite-volume gravity-current box model in a widening triangular channel has volume proportional to $h^2{L}^2$, so $hL=K$ is constant. A gravity-current front condition and absorbing-bed particle deposition flux giveEliminating time makes $\sqrt{\varphi }$ affine in $L^{5/2}$, producing a finite limiting runout length of a gravity current despite an infinite idealized stopping time.