4
$\begingroup$

I noticed several times while washing my hands that accumulated water when starts streaming vertically does not flow vertically in straight line on vertical porcelain surface under the force of gravity. It rather flows in strange zig-zag like path and the flow changes its direction rapidly and erratically from left to right and vice versa. Intuitively, it seems that the straight vertical path for the stream is the most physically economic path.

Is this well studied phenomena? What is the physical explanation? Is there a mathematical model that can predict any aspect of this type of flowing stream?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Any video or image will be appreciable. $\endgroup$
    – Sensebe
    Jan 19, 2014 at 23:16
  • $\begingroup$ Most surfaces are not as smooth as you might be able to feel. It is likely the surface bumps are causing the path you see. $\endgroup$
    – Kyle Kanos
    Jan 20, 2014 at 1:27
  • $\begingroup$ There's probably a better answer, but the water streaming down the surface has surface tension, that apparently affects it chaotically, the same way rivers meander. Flow in a straight line is unstable, because the slightest curve tends to grow on the outside. $\endgroup$
    – Mike Dunlavey
    Jan 20, 2014 at 15:16

1 Answer 1

Reset to default
5
$\begingroup$

This phenomenon is known as a meandering instability of the rivulets and is due to an interplay of surface tension, microscopic impurities/dirty spots and inertia.

It is well understood and described beautifully in the 2006 and 2011 PRL papers by Daerr et al.

To summarize the main message briefly: for low velocities the water will flow straight down. Above a certain threshold value of the velocity the stream becomes unstable and perturbations caused by the microscopic impurities form small bends that can be sustained based on a balance of surface tension and inertia. In the papers they present a mathematical model based on this balance that can be used to predict the threshold velocity and the radius of curvature of the meanders.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.