Drops, menisci, and lenses

Thickness of a large drop

Small drops are spherical segments. Large drops are flattened.

de Gennes, 2004, Fig. 2.4

To “spread” the drop requires an force per unit length:$\sigma _{sv}-\left( \sigma _{lv}+\sigma _{sl} \right)$

The hydrostatic pressure integrated over the depth of the drop is a force per unit length pushing to “spread” the drop: $\tilde{P}=\int\limits_{0}^{e}{\rho g\left( e-\tilde{z} \right)d\tilde{z}}=\frac{1}{2}\rho ge^{2}$

At equilibrium the sum of the two is zero: $\sigma _{sv}-\left( \sigma _{lv}+\sigma _{sl} \right)+\frac{1}{2}\rho ge^{2}=0$

Substituting the Young-Dupré equation: $\sigma _{lv}\left( 1-\cos \theta _{e} \right)=\frac{1}{2}\rho ge^{2}$

Re-arranging gives: $\text{ }e=2\kappa ^{-1}\sin \left( \frac{\theta _{e}}{2} \right)$

Profile of a large drop

The general ideas used to calculate the thickness are used again to calculate the profile. With a couple of modifications.

de Gennes, 2004, Fig. 2.4

The limits on the integration are changed: $\tilde{P}=\int\limits_{0}^{z}{\rho g\left( e-\tilde{z} \right)d\tilde{z}}=\rho g\left( ez-\frac{z^{2}}{2} \right)$

The “spreading” force per unit length is now: Where q$\theta$ is the angle marked in the diagram: $\sigma _{sv}-\left( \sigma _{lv}\cos \theta +\sigma _{sl} \right)$

“It can be shown” from the diagram that: $\cos \theta =\sqrt{1+\dot{z}^{2}}$

This results in a differential equation for the shape consistent to the algebraic equation for the drop thickness: $\sigma _{lv}\left( \sqrt{1+\dot{z}^{2}}-\cos \theta _{e} \right)=\frac{1}{2}\rho g\left( 2ez-z^{2} \right)$

Menisci shapes

The ascending meniscus against a vertical wall (de Gennes, 2004, pp. 45f). The Laplace equation, shown on the diagram, is a differential equation that describes the shape of the meniscus. The curvature increases with height.

$\frac{1}{R\left( z \right)}=-\frac{{\ddot{z}}}{\left[ 1+\dot{z}^{2} \right]^{3/2}}$

Substituting the curvature into the Laplace equation and integrating twice gives: $x-x_{0}=\kappa ^{-1}\cosh ^{-1}\left( \frac{2\kappa ^{-1}}{z} \right)-2\kappa ^{-1}\left( 1-\frac{z^{2}}{4\kappa ^{-2}} \right)^{1/2}$

(Where x0 makes z = h at x = 0)

A correct, but not illuminating, result.