# Shape of a water drop

Some years ago (1987 time frame), a mechanical engineering professor asked a question in a graduate level heat transfer class that I have never been able to solve. The questions is:

"Given the surface tension of water, provide a mathematical description for the shape of a water drop that is sticking to the bottom side of a horizontal surface".

If anyone knows how to model a water drop mathematically, I would be interested in seeing it. In addition, if the equation(s) also allow for the changing shape of the water drop as it slowly accumulates mass but before it falls from the surface (e.g., think of a water faucet with a slow drip rate), I would appreciate seeing it.

(Edited on 11/15 for clarity)

• I think the follow link is highly relevant: en.wikipedia.org/wiki/Surface_energy#Wetting – AccidentalBismuthTransform Nov 14 '15 at 19:21
• I'm not looking for the shape of a water drop on top of a horizontal surface ... I'm looking for the shape of a water drop on the bottom of a horizontal surface. – David White Nov 14 '15 at 23:54
• @DavidWhite When you have a clarification it's usually a really good idea to edit the original post. – DanielSank Nov 15 '15 at 2:45
• @DavidWhite in this case I think that cohesion and adhesion will be involved in the mathematical formula. I'm almost sure that the shape can be pretty "complicated" mathematically, I don't know whether it's been derived yet. – AccidentalBismuthTransform Nov 15 '15 at 3:23

The way to proceed is as follows. Take a coordinate system with $x$ pointing to the right and $y$ pointing down. Imagine the inverted bell-shaped drop hanging from the $y = 0$ plane. Now "cut" the bell horizontally at at $y$. The circumference of the bell at the cut is $2\pi x$ and the total force on the cut part is $2\pi xc$, where $c$ is the surface tension per unit length (a constant). The surface tension is parallel to the tangent and the total upwards force on the cut part has the component $$F_y = -2\pi xc\sin(a),$$ where $a$ is the angle (a negative value) of the tangent at $(x,y)$.
Now, this force must equal the weight of the cut part of the bell. The volume $V$ of this part is the integral $$V = \int_y^{y_{tip}}\pi x^2dy.$$ The mass of this volume is $mV$, where $m$ is the specific mass, and the force of gravity thus $gmV$. Hence the basic equation is $$-2pcx\sin(a) = gmV.$$
Now, $V$ is an integral of $x^2$ versus $y$ and $\tan(a) = dy/dx = y'$, so $$sin(a) = \frac{y'}{\sqrt{1+y'^2}}.$$ Differentiating with respect to $y$ on both sides gives the final differential equation: $$2\pi c\frac{d(xy'/\sqrt{1+y'^2})}{dy} = gm\pi x^2$$ or $$\frac{d(xy'/\sqrt{1+y'^2})}{dy} = kx^2,$$ where $k = gm/2c$. This can be solved exactly by a change of variables, but I wont go into the technicalities (I don't have a math texter) but just say that the shape of the bell (surface of the drop) involves elliptic functions.
Further, if you add the acceleration $mV\,d^2Y/dt^2$ (where $Y$ is the center of gravity of the drop, which is simple to express from the quasi-stationary shape) to the force difference at $y = 0$ and slowly increase its volume at a constant rate $dV/dt$, then you get an equation which can be solved to give the time of pinch-off, where the drop falls, as well as the maximum volume the surface tension can support. One day I will repeat the math, which I actually did in my student days.