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If a finite-volume gravity current loses or dilutes the scalar that supplies its reduced gravity, its front can approach a finite runout length of a gravity current. A box model couples the scalar balance to a gravity-current front condition; eliminating time then gives the front position directly as a function of the remaining scalar concentration.

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 $L_{\infty }$.

When ${\hat{W}}_1={\hat{W}}_2=\hat{W}$, both concentrations acquire the same decay factor. The total reduced gravity consequently obeysEliminating time with the front equation givesIntegration from the initial state yieldswhereThe particle-laden gravity current reaches its runout length of a gravity current when $g^{\prime }$ vanishes. Since $\hat{W}<0$,Thus $C=5\mathrm{Fr}$ for the stated front condition; the common normalization $\mathrm{Fr}=1$ gives $C=5$.