When you fill a glass with water, water forms a concave meniscus with constant contact angle $\theta$ (typically $\theta=20^\circ$ for tap water):
Once you reach the top of the glass, the water-air interface becomes convex and water rises up to a height $\Delta h$ above the edge of the glass, allowing you to fill the glass beyond the naive capacity $\pi r^2 h$:
So when getting myself a glass of water, I came to wonder exactly how much this increases the capacity of a glass, and what physical constants are involved.
My intuition would be that for a very large glass, $\Delta h$ converges to a constant so that the effective water capacity of the glass grows like $\pi r^2 (h+\Delta h)$ (to make things simple I'm assuming that the glass is very thin: $\Delta r\ll r$). Perhaps such a constant depends on the precise shape of the rim of the glass. But if not, perhaps it is a constant multiple of the capillary length?
So, what can we say about $\Delta h$, the "rim contact angle" $\alpha$, or the shape of the water-air interface when the glass is filled at maximum capacity?
Best Answer
As is given in Jamie's answer I'll assume the surface is a revolution about $r=0$, that the mean curvature is proportional to the pressure difference, and that the radius of the cup is much larger than the inverse of this mean curvature. In this case the mean curvature can be specified as $$ K_m = \frac{r''}{2(1+r'^2)^{\frac32}}$$
As in Jamie's answer the Young-Laplace Equation and hydrostatic pressure give $$ 2\,\gamma\,K_m = \Delta P = -\rho\,g\,z$$ Placing the origin at the surface and positive z in the direction of gravity.
Combining yields
$$ -\rho\,g\,z = \frac{\gamma\,r''}{(1+r'^2)^{\frac32}}$$
Substituting $q=r'$ yields a first order differential equation
$$ -\rho\,g\,z = \frac{\gamma\,q'}{(1+q^2)^{\frac32}}$$
Integrating
$$ -\frac12\,\rho\,g\,z^2 = \frac{\gamma\,q}{\sqrt{1+q^2}}+C$$
We know that at the very top of the water the surface is flat which would correspond to $q=r'=\infty$ this condition is gives $C=-\gamma$
$$ z= \sqrt{\frac{2\,\gamma}{\rho\,g}(1-\frac{q}{\sqrt{1+q^2}})}$$
Now since $q=r'=tan\,\alpha$ where $\alpha$ is described in the question, $z$ simplifies to
$$ z= \sqrt{\frac{2\,\gamma}{\rho\,g}(1-sin\,\alpha)}$$
Which is indeed the formula given in John Rennie's answer.
So now the question is what alpha to use. I propose that the water will continue expanding around the curved lip of a glass maintaining its contact angle until the point at which going farther along the lip would lower the top of the surface according to the above equation as at that point the surface would be unstable. This depends on the curve of the lip of the glass $r_l$.
If the water edge on the lip of the glass is at polar coordinate $\phi$ and the liquid has a contact angle $\theta$, then $\alpha=\phi-\theta$, and my total height $h$ will be given as $$ h=\sqrt{\frac{2\,\gamma}{\rho\,g}(1-sin(\phi-\theta))}+r_l\,(sin\,\phi-1) $$
Unfortunately this does not have a closed form maximum over theta, but we can see that for small values of $r$ the maximum will be when $\phi\lt 0$. This is not physical as the liquid would start creeping down the side of the glass and become unstable first. We can solve for the value of r at which this occurs $$r_l=cos\,\theta\sqrt{\frac{\gamma}{2\,\rho\,g\,(1+sin\,\theta)}}$$ For any lip radius under this value the maximum liquid height would be $$ h=\sqrt{\frac{2\,\gamma}{\rho\,g}(1+sin\,\theta)}-r_l $$
For water this radius calculates out to about $1mm$ and for a glass with a very small radius the height calculates out to about $4mm$ a bit higher than I've managed, but not unreasonable for a theoretical upper bound.
For glasses with a larger radius the maximum height can be solved numerically. Here's a plot.
And the corresponding "lip contact angles"