On a day after a rainfall, some raindrops from a tree branch had fallen onto a car. It was there when I found an unusual pattern formed when the raindrops had hit the the car panel.

Around each raindrop, a smaller ring of smaller droplets had formed around the raindrop. I first thought that this was a coincidence because depending on the location of where a smaller droplet was, it would fly off at a farther/closer distance than one that is located at a different location (assuming they are projected at the same energy). I then went back home and tested it on a table surface (a picture is shown below):

enter image description here

Under what conditions/parameters could we find and measure the dimensions of such a ring?

If we were to assign variables $h$ to the height dropped, $m$ and $r$ to the radius and mass of the initial raindrop, and $\sigma$ to denote the surface tension, we already have too many dimensions to accurately measure the radius $d$ of this ring. Furthermore, even if we could find a dimensional formula, I wouldn't be satisfied as there are probably some conditions on when such a ring can form; I would like to know how/why it does.


2 Answers 2


Why it forms: When the drop lands, it can sometimes do this

Harold Eugene Edgerton (American, 1903–1990) Title: Milk Drop Coronet , 1957

As mentioned in the comments, energy conservation could be used to find the radius of the ring...

Initial energy of drop, mass $m$, radius $R$, density $\rho$, released from height $h$, surface tension $\sigma$ is $$mgh + 4\pi R^2 \sigma$$

energy after landing and splitting into $n$ smaller drops is

$$ n^{1/3}4\pi R^2 \sigma + \frac{1}{2}mv^2$$

using conservation of energy and solving for $v^2$ gives a maximum range $d$, (radius of the ring) of $\frac{v^2}{g}$ of

$$d = 2h + k(1-n^{1/3})$$

where $$k = \frac{6 \sigma}{\rho R g}$$

for a 2mm drop, using g=9.8, $\sigma$ of 0.0072N/m and $\rho$ = 1000 gives k=0.022

This modelling gives the maximum range and depends on $n$, it also assumes all the big drop becomes $n$ smaller drops.

Using $n$ = 20: For a drop height of 10cm the range is 16cm, for height 5cm it's 6.4cm and for a height of 2cm it's only 3mm.

Maybe this can serve as a starting point.

An improved model would have to find the number of smaller drops $n$ and also what proportion of the big drop becomes smaller drops.

  • $\begingroup$ Ok so, here this is a droplet that is being dropped into another fluid (milk) right? My scenario is a bit different where this is a droplet just landing ontop of a table surface. BUT I can see how this is related as there will be equal momentum imparted in all equal directions when the raindrop hits the tabletop (as the raindrop is spherical) resulting in a ring shape of the smaller drops. This answer would only answer half of my question though. $\endgroup$ Jul 14, 2021 at 0:35
  • $\begingroup$ Yes, the other half - details of the radius of the ring, number of drops etc...might need an expert in fluid dynamics to post an answer (or computer modelling). Could be that the gravitational potential energy lost + original surface energy (due to surface tension) of the original drop gets turned into energy stored in the new surface area of the smaller drops + kinetic energy. Then if the kinetic energy of the smaller drops were known, it could be modelled as a projectile problem to find the radius of the ring. but it seems complicated, let's see if someone else can add something... $\endgroup$ Jul 14, 2021 at 5:25
  • $\begingroup$ something on this now added, but an improved model would have to find the number of smaller drops $n$ and also what proportion of the big drop becomes smaller drops $\endgroup$ Jul 14, 2021 at 6:48
  • $\begingroup$ Nice, although unless I'm reading wrong, did you neglect the energy of the large droplet that remains? Also would energy really be conserved? Because this seems like the droplet performs a half inelastic collision because of the larger droplet. $\endgroup$ Jul 14, 2021 at 12:17
  • 1
    $\begingroup$ Yes, it was mentioned earlier that it depends what proportion of the large drop is turned into small drops. The answer could serve as a starting point. Then (depending how far people want to go with it) - you could include other factors to refine it, such as the inelastic collision etc...to decide how much energy is left over for the small drops motion. Also quite important would be the number of small drops formed, the model doesn't answer that, although it can be put in as a number. Best of luck with it. $\endgroup$ Jul 14, 2021 at 13:14

I believe I have now found a suitable answer to my question. My physics/math may be slightly shaky here so please point out anything if you see something wrong.

How Are The Smaller Droplets Being Released?

The smaller droplets, in a way, act to stop the moving inertia of the large water droplet when it collides with the surface. If you see John Hunter's answer, drops of milk spread out equally in all directions so it can stop the continued motion of the the drop that falls inside it. Another good example that highlights this can be found in this slow motion video of a water droplet disturbing a surface. (Although this isn't exactly the same process as what is in my question, I believe it is suitable enough for one to make at least a little sense of what is going on.)

enter image description here

But where and when are the droplets being released? The most reasonable answer is that they are released immediately after the collision (or maybe somewhere along the process of moving to a new equipotential surface. More specifically, the droplet goes from a sphere to a new hyperbolic curve when on the table surface. This process will result in some loss of energy (due to both an inelastic collision and changing into a new surface) which is given to the droplets.

Modelling Loss of Energy

For the cause of simplicity, we are going to view the droplet as a spherical drop that becomes squished (i.e. the bottom becomes flat). In actuality, this is not true because the equipotential surface will change (here nothing changes except for the part of the droplet in contact with the ground), but the amount of energy that is lost will be around the same. Consider the following image:

The droplet hitting the ground with relative parameter <span class=$x$" />

We define the relative parameter $x$ by the distance the droplet moves into the ground. Then Young-Laplace tells us that $\Delta p = \frac{2\sigma}{R}$ which means that the force acting through that little segment is $\Delta F = \Delta p S$ where $S = \pi r^2$. Pythagorean theorem easily suggests that $(R - x)^2 + r^2 = R^2 \implies r^2 = x(2R - x)$. Assuming that the relative deformation is $x \ll R$, we can find that the force acting is linear such as $F = -kx = -2\pi R\Delta p x$. So we have an analogy to a spring with spring constant $k = 2\pi R \Delta p = 4\pi \sigma$ further implying an oscillation period of $$ T = 2\pi \sqrt{\frac{k}{m}} = 2\pi \sqrt{\frac{4\pi \sigma}{\rho (\frac{4}{3}\pi R^3)}} = \sqrt{\frac{3\sigma}{\rho R^3}}. $$ Now modelling the energy loss will be relatively straightforward. When a part of the water droplet hits the table surface, it loses a kinetic energy of $\text{d}E = \frac{1}{2} v^2\text{d}m.$ The amount of differential mass can be modelled from a cylinder of changing radius $\text{d}r$ and changing length $v\text{d}t$ where $v(t) = -\omega A(t)\sin (\omega t)$. By applied differentiation, the differential volume is then $\text{d}V = \pi v \text{d}t (r^2\text{d}t + 2r\text{d}r)$. So the total loss is $\left|\frac{\pi}{2}\rho v^3 (r^2 \text{d}t + 2r\text{d}r)\right|$. Thus, the change in energy over half of its oscillation can simply be $$\Delta E = \int_{0}^{r_0} \int_{0}^{\pi/\omega} \frac{\pi}{2} \rho \omega^3 A^3 \sin^3 (\omega t) (r^2 \text{d}t + 2r\text{d}r).$$ Here $h = \sqrt{\frac{\sigma}{\rho g}}$ (assuming contact angle is 90 degrees) so you can again find $r_0$ from Pythagorean theorem.

Now, this energy will be distributed to $n$ droplets, so the energy each droplet contains is $\Delta E/n$. Similarly, if the amount of volume all the droplets take from the larger water droplet is $V'$, then each droplet has a mass $\rho V'/n$. So the velocity move outwards at is $v = \sqrt{\frac{2\Delta E}{\rho V'}}$. So the radius can be described as $\mathcal R = \frac{2\Delta E\sin (2\theta)}{\rho V' g}$ where $\theta$ is the angle the droplets fly from.


  • Radius $R$ of falling droplet can be found from this StackExchange post.

  • This seems like a good way to model the energy loss, but I still feel like there would be a better way to end this analysis with less parameters (we introduced two more relative parameters of $V'$ and $\theta$ at the end).


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