= Triangular-channel hydrostatic pressure head
{title2=$I=AH/3,\quad I_x/A=H_x$}
For a fixed symmetric triangular channel with depth $H$ and <area> $A=kH^2$, the hydrostatic first moment is $I=\int_A(H-z)\,dA=kH^3/3=AH/3$. Since $dI/dH=A$, the depth-averaged <pressure gradient> is $gI_x/A=gH_x$. A mean geometric depth $H/2$ cannot replace $H$ in that gradient without an extra factor two. This distinction prevents confusion between <pressure> head, centroid depth, and <hydraulic radius>.
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