= Triangular-channel dry-bed dam break
{title2=$h/h_0=[(4-x/(c_0t))/5]^2,\quad c_0=\sqrt{gh_0/2}$}
In a prismatic triangular channel with wetted area proportional to $h^2$, the <Riemann invariants> are $u\pm4c$, with $c^2=gh/2$. Releasing water initially at rest into a dry channel produces a <rarefaction wave> with $u+4c=4c_0$ and $x/t=u-c$. Hence $c=(4c_0-x/t)/5$, $u=4(c_0+x/t)/5$ and the displayed depth profile for $-c_0t<x<4c_0t$. The dry front advances at $4c_0$ before finite-reservoir or varying-channel effects become important.
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