= Gravity-capillary drop with a meridional curvature approximation
{title2=$h=\alpha l[\cosh(R/l)-\cosh(r/l)]$}
If an axisymmetric <thin-film equation> is deliberately closed using only the meridional <curvature> $\kappa=-h''$, its regular equilibrium obeys $h'''=h'/l^2$, with <capillary length> $l=\sqrt{\gamma/(\rho g)}$. The <boundary conditions> $h'(0)=0$, $h(R)=0$, $h'(R)=-\tan\theta_c$ give the displayed profile and $\alpha=\tan\theta_c/\sinh(R/l)$. Its <volume> is $2\pi\alpha l^3[s^2\cosh s/2-s\sinh s+\cosh s-1]$, $s=R/l$. This model omits the azimuthal contribution of <axisymmetric free-surface curvature>. Retaining that term instead gives $h=l\tan\theta_c[I_0(R/l)-I_0(r/l)]/I_1(R/l)$ in terms of <modified Bessel functions>.
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