Triple-shock sedimentation point
= Triple-shock sedimentation point
{title2=$(z_m,t_c,h_c)=\left(\frac{H(1-a)}{1+a},\frac{H}{u_s(1+a)},\frac{2aH}{1+a}\right)$}
For initial concentrations $a$ above $z_m$ and $2a$ below it, all three straight <shocks> meet when $z_m=H(1-a)/(1+a)$. The common time is $H/[u_s(1+a)]$ and height is $2aH/(1+a)$. Afterwards only a stationary deposit/clear-fluid boundary remains.