I'm looking at this picture enter image description here from this site and I'm curious about what is the curve described by the water. The involutes $$x=r\left[\cos(\theta+n)+\theta\sin(\theta+n)\right]\\ y=r\left[\sin(\theta+n)-\theta\cos(\theta+n)\right]$$ for different $n$ seem to make it right at least near the ball:

enter image description here

But I'd like to know precisely the idea of how the curve develops. What I'm thinking is that the water leaves off the ball about the normal at each point because of the adhesion to it and then it should just spread 'freely'. Can the evolution of the water be described easily or has it been done accurately before?


2 Answers 2


I think you are right. (Involute spiral: Wikipedia)

If you take a string wound around a stationary ball and unwind it, its end traces an involute spiral.

Similarly, if the ball is rotating and the string runs out in one direction, it is the same curve with respect to the ball. A drop of water leaving the surface of the ball should travel in a straight line out from the center of the ball, at a velocity equal to its tangential velocity when it was on the surface.


If you assume that the water leaves the ball tangentially and ignore the effects of gravity you get exactly those equations. To see it look at the trajectory of a drop released from a given point at time t (ball rotating clockwise):

$x=x_0+v_{0x}t=r \cos\omega t+ r\omega t \sin(\omega t) $

$y=y_0+v_{0y}t=r \sin\omega t- r\omega t \cos(\omega t)$

The shape of the curve of water that leaves from a fixed point at different times given by $t= \theta/\omega$ then you get

$x=r \cos\theta+ r \theta \sin\theta $

$y=r \sin\theta- r\theta\cos\theta$

each $n$ (not necessarily a natural number) corresponds to a different point on the sphere through which water escapes, so they do not need to be equally separated, as you can see them in the picture, they are located more randomly than uniformly.

  • $\begingroup$ I'm confused with "the trajectory of a drop". Shouldn't that be a straight line? And what if the ball rotates anticlockwise? $\endgroup$
    – Cal Gibson
    Jun 3, 2015 at 17:37
  • $\begingroup$ the trajectory of an individual drop is a straight line, but the picture and the last equations shows all the drops at a given time, so each drop in the curve left the ball at a different time and at at a different angle. Individual drops do not make up that trajectory $\endgroup$
    – user66432
    Jun 3, 2015 at 18:29

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