How do liquid jets break up? So when I turn on a tap and let out a thin laminar flowing jet, I notice that the jet eventually breaks up into droplets after a certain height. Is there a way to model/calculate this height using an equation? 
I've done some research on the Rayleigh-Plateau Instability but I still can't find a way to calculate this height given inputs such as jet velocity, radius, and viscosity.
Any help at all is greatly appreciated!
 A: I am no fluid expert but I will try to give a rough estimate based on dimensional analysis.
Assuming that the section of the flow tube will be circular, we can compute its radius at a given height $z$ using the continuity equation:


*

*the speed is $v=\sqrt{2gz}$ (starting from speed zero);

*$v\cdot\pi r^2$ is a constant, so $r=\frac{R_0}{z^{1/4}}$;


The forces in action will be gravity and the surface tension; for an element of fluid with volume $V$ and external surface $S$ we will have:


*

*$F_{grav}=\rho V g$ ;

*$F_{surf}=\tau S$ ;


Where $\rho$ is the density and $\tau$ is a proportionality coefficient, representing the force per unit of surface due to the surface tension (here maybe there are some more standard quantities to define).
Now let's try to see the dimensional analysis. Our dimensional parameters are $\rho$ , $g$ , and $\tau$ , which have dimensions:
$$[\tau]=\frac{[M][L]}{[T]^2[L]^{2}}\qquad[g]=\frac{[L]}{[T]^{2}}\qquad[\rho]=\frac{[M]}{[L]^{3}}$$
So we can only build one length scale:
$$R=\tau^{\alpha}g^{\beta}\rho^{\gamma}\qquad[R]\equiv[L]=[M]^{\alpha+\gamma}[L]^{-\alpha+\beta-3\gamma}[T]^{-2\alpha-2\beta}$$
So that it must be:
\begin{align*}
&\alpha=-\beta\qquad\alpha=-\gamma\qquad-\alpha+\beta-3\gamma=1\\
\Longrightarrow\quad&\alpha=1\quad\beta=-1\quad\gamma=-1\\
&\text{and in conclusion}\qquad R=\frac{\tau}{g\rho}
\end{align*}
Now, what is the meaning of this scale?
If we compute the forces acting on an element of fluid of size $R$ (here we have no control over the shape) we get:
$$F_{grav}\sim\rho g R^3\qquad\qquad F_{surf}\sim \tau R^2$$
and substituting the value of $R$ we get:
$$F_{grav}\sim \frac{\tau^3}{(g\rho)^2} \sim F_{surf}$$
So that $R$ is the size of an element of fluid for which the force of gravity and the surface tension are of the same order. For elements of smaller size the surface tension (that goes as $r^2$) will be dominant, while the contrary happens for larger fluid elements.
In conclusion, as the flow tube shrinks, it reaches a size for which (locally) the surface tension is as strong as of the gravitational force. From that point on it should become more and more likely for the flow tube to split in droplets.
It is a bit difficult to motivate this in a rigorous way, but I can think of a somewhat intuitive argument: when a droplet is formed basically a slice of the fluid is collapsing under the surface tension. Ideally, half of the slice gets pushed up against the gravitational force, as a result of the surface tension. This would not happen if the surface tension was weaker than the gravitational force.
Maybe this argument can be improved, and surely the estimate of the radius is very rough. Anyway, despite being rough the estimate should give at least a basic insight of when droplets are nucleated. Surely viscosity may play a role, but that should be related to dynamical effects and bring less important corrections. Still I might be wrong and I could have forgotten some important effect. Please comment or edit if this is the case.
A: rayleigh instability is indeed the phenomenon responsible for jet breakup into droplets. There is a nice paper by Donnelly from the 1950's that describes this; it is also a topic of great interest in the design of ink jet printheads. Try searching on "inkjet breakup" to find papers written about it in that field.
