Our coffee machine catches the last couple of droplets, after your cup is removed on a shape to reduce plash of the coffee droplets.

These shapes are placed inside the spill reservoir.
The shape used by our coffee machine is shape nr 2 with the droplet falling at position 1, this works well but not perfect.
What shape would be even better for this purpose? And at which position should the droplet fall?

I've included some shapes I came up with but I'm open for other suggestions.

The image displays a 2D version of the shapes, the first 2 shapes are cone shaped and the last 2 would have to be open on the ends (or sides) to release the coffee into the spill reservoir.

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    $\begingroup$ $4$th one! will carry droplet in it and droplet will flow into the space! $\endgroup$ – ABC Mar 22 '13 at 12:59
  • $\begingroup$ I vote for the 4th $\endgroup$ – Anixx Mar 22 '13 at 13:09
  • 1
    $\begingroup$ I'll suggest choice 3 and 4 (or 2-2). 3rd (3-1) and 4th are amazing. As the shapes have a path nearly the same as that of the falling drop, it will be dragged sliding towards the bottom and hence --> easy landing... Tadaaa ;-) $\endgroup$ – Waffle's Crazy Peanut Mar 22 '13 at 13:19
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    $\begingroup$ For the question of how to redesign this machine I think you need to take into account that shape 3 and 4 will fill up and therefore will give different splashing over time. Moreover, there is an additional strategy: catching the splash. A fishbowl shape would work perfectly well for that $\endgroup$ – Michiel Mar 22 '13 at 13:59

Obvouusly the highest number of splashes will be when the impact happens at right angle. This is because at such impact all the kinetic energy of the drop at one moment goes into the forces directed into different directions and tears the drop apart.

Conversely the least splashes will be when the drop impacts at narrow angle. In this case the drop continues sliding over the surface and keeps its integrity while kinetic energy slowly transforms into heat due to friction. The surface tension keeps the drop intact because all parts of the drop keep the same movement direction.

As such I would recommend a form which meets the drop in the most probable impact place with a sloppy angle. Evidently, such form should not be symmetric (because all even functions have zero derivative at zero).

The only variant of yours that has such property is the fourth. Additionally it changes the drop's velocity vector to the right direction so that any splaches that can occur will go to the right where it has a border that covers the greatest angle of possible splashes path out than all other variants.

The slope's curvature should be such that the amount of energy converted into heat was uniform along the drop's path and the pressure never exceed the surface tension. So when the drop has the highest speed, the angle between the forces acting on the drop (inertia and gravity combined) and the surface should be the smallest, but as long as the drop slows, the angle slows the angle should rise so to keep the component which is perpendicular to the surface constant.

The fourth picture roughly satisfies this criterion.

Thus the fourth variant in the best.

  • $\begingroup$ Accepted until someone comes up with a better shape :D $\endgroup$ – user17615 Mar 22 '13 at 14:29
  • $\begingroup$ Hi Anixx. What do you think of 3-1? It's almost the same as 4, but somewhat less wider slope. The path is still more or less the same. I think it could also satisfy the condition. Am I wrong about the reservoir? ;-) $\endgroup$ – Waffle's Crazy Peanut Mar 22 '13 at 16:25
  • $\begingroup$ @Crazy Buddy It has several disadvantages: the place of the first hit is higher (and as such the possibility the splashes from the first hit go over the border is higher), also the slope in the place of the first hit has greater angle with the speed vector so the burst in the first hit will be greater, also the overall slope is shorter and it is steeper so there is possibility the drop will not slow enough till the lowest point where the angle of attack again rises abruptly so giving possibility of another burst. $\endgroup$ – Anixx Mar 22 '13 at 16:53
  • $\begingroup$ Notisce also that even if the drop still has high speed at the rightest part of the path on figure 4, so there will be a burst, the splashes have virtually no chance to escape: nearly all the straight lines drawn from the vertex of the curve end on the border. Given the parabolic character of the splashes trajectory, their chance to escape is even smaller. $\endgroup$ – Anixx Mar 22 '13 at 16:58
  • $\begingroup$ Hmm... That makes sense. I definitely agree that #4 is more efficient than any other. If the drop fell slightly inward (near the bottom), it can be considered efficient. Isn't it? $\endgroup$ – Waffle's Crazy Peanut Mar 22 '13 at 17:12

The other answers may be correct but claims that the solution to this is "obvious" for a real fluid are, in my opinion, questionable.

However, people have studied similar problems empirically. See for example the 2009 paper "Experimental Splash Studies of Monodisperse Sprays Impacting Variously Shaped Surfaces" by Yoon, et al. (DOI:10.1080/07373930802606188).

  • $\begingroup$ It is not just "obvious", it is substantiated by the fact that the kinetic energy of the drop completely converts goest to tear the drop apart in case of the right angle impact. If you disagree, please rise your objections. Or did you read only the first word? $\endgroup$ – Anixx Mar 22 '13 at 13:53
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    $\begingroup$ See for example "Effect of impingement angle on the outcome of single water drop impact onto a plane water surface" by Okawa et al. (DOI: 10.1007/s00348-007-0406-z). Your argument is not invalid, but it is merely suggestive for something as complex as the splash dynamics of a real fluid. $\endgroup$ – Joshua Barr Mar 22 '13 at 14:07
  • $\begingroup$ there is no need for complex calculations so to see that to keep the drop intact, in any moment there should be no force acting on the drop exceeding its surface tension. $\endgroup$ – Anixx Mar 22 '13 at 14:12
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    $\begingroup$ Both those papers are experimental (no complex calculations). $\endgroup$ – Joshua Barr Mar 22 '13 at 14:13

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