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In a cross-section normal to the cylinder axes, measure $y$ from the line of contact. The two circular boundaries have the parabolic approximationfor their separation. The contact point of the meniscus is at $y=w+o(w)$, so the leading cross-sectional area isThe area of the small meniscus cap is $O(w^4/a^2)$ and is lower order. At its upper end the gap has width $h(w)=w^2/a$. A tangent semicircle therefore has radius $w^2/(2a)$ and curvature

The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,At a fixed $y$, lubrication theory gives a planar Poiseuille flow through a gap of width $h(y)$, with axial flux per unit $y$Integrating across the cusp,The continuity equation $A_t+Q_z=0$ now gives

Set $q=w^3$. Since $w^4{w}_z=\frac{1}{5}(q^{5/3}{)}_z$, this is the porous medium equationConservation of the fixed volumeand dimensional analysis give the self-similar solution $q=t^{-3/8}F(z/t^{3/8})$. One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solutionEquivalently,Its tip is $z_N=(C/B{)}^{1/2}{t}^{3/8}$. Usingin the volume constraint gives $C^2=8aV\sqrt{B}/\pi$. Therefore

For vertical cylinders at equilibrium, hydrostatic pressure gives $p-p_{\mathrm{atm}}=-\rho gz$. Balancing this with the capillary pressure $-2\gamma a/w^2$ gives the large-height profile

The signed curvature isAn orientation-preserving isometry sends $\mathbf{t},\mathbf{n}$ to $R\mathbf{t},R\mathbf{n}$ and leaves $s$ unchanged, soAn orientation-reversing isometry reverses the chosen normal and hence the sign convention for signed curvature, while the geometric curvature $|k|$ remains invariant.