[kprice (OP)]

I noticed that when I drop a round rock in the lake it splashes higher than a flat rock (I was surprised--I thought it would be the other way around). I decided to make this into a home school science project. I made spheres and flat shapes the same mass out of modeling clay and took videos of my mom dropping them at the same time. The sphere splash was higher (I did 10 replicates). I also tried hemispheres--again the round side down made a higher splash. I tried looking up fluid dynamics on the web but the math was too difficult. It must have something to do with the way the water moves around the shape, but I'm not sure how--is it easier to move around the sphere and so it loses less energy? Thank you for your help.

[Vern]

I will hazard a guess; the sphere encounters less water resistance and so makes a bigger hole in the water. The bigger hole fills back with water and the inertia of the filling water makes the splash.

[lightarrow]

I will hazard a guess; the sphere encounters less water resistance and so makes a bigger hole in the water. The bigger hole fills back with water and the inertia of the filling water makes the splash.

I would tend to agree with you.

[LeeE]

While I think that the inertia of the back-filling water may be a factor, I don't think that's the main reason.

When a flattened object hits the surface of the water, the water directly beneath it can only move sideways as the object forces the water out of its way, and once the water has acquired this sideways momentum it will tend to force the surrounding water sideways too.  When any object drops into water, it is actually easier for the water that is displaced by the object to rise up against gravity than for the water to compress, which is why we get a splash in the first place, but because of the sideways momentum of the water displaced from beneath the flattened object, much of the splash will go sideways too.

When a rounded object hits the water though, it can pass around the object, and in fact, due to hydrodynamics, it will tend to flow around the object, changing the direction and momentum of the flow from sideways to upwards.  Combined with the water compressibility issue, the displaced water goes up instead of sideways, so it goes higher than the sideways splash from the flattened object.

If instead of asking "Why is the splash from a rounded object higher than that from a flattened object? you could have equally validly asked "Why is the splash from a flattened object wider than that from a rounded object? and got the same answer.  Well, from me, at least  []

[Vern]

I wonder if the back-filling splash goes higher than the side displaced redirected upward moving splash. I seem to remember from freeze frame studies of splashes forming that the back filling centre portion produces a nodule of water that goes higher. I'll look around to see if I can find such a study.

[yor_on]

Cool subject.

Isn't the force displaced over a greater surface when you have a flat stone. That will mean that the edges of that flat stone will have a lesser surface force to create waves with than our spheroid stone, if I'm thinking right here, who will, as Vern said, make a bigger hole, assuming we are discussing waves? If it is the sound you're referring to? I would expect it to be the opposite

But the total 'force' of the stones should be the same, and as you suspect it must have a lot to do with how fluids behave. You might illustrate it with arrows pointing in the direction the stone falls. for a 'pointy' stone there will be a small amount of arrows fitting the 'point' but each containing a lot of 'force', and for our flat one a lot off arrows but with all of them containing a lesser force. I think that the flat surface 'flattens' out the force and as the waves only can be created at the edges of the stone that might be your answer. To that you have to add the direction of 'force' on the water, first down and then extending to the sides and their (waters) possible interaction.

---
What I mean with the last is that in the case of our 'flat' stone the time for those 'water'-forces' created will come 'extended' in time, as compared to our round one where the 'forces' will be able to act closer in time..

But I would still expect a conical stone, dropped point first, to make a smaller wave than a round one?

[LeeE]

Ultimately, it's an energy equation.  If most of the energy directs the water upwards instead of sideways then it will go higher than water where most of the energy directs it sideways, and visa-versa.

Waves are a related phenomenon but don't get too bogged down in them Yor_on.  Waves are just splashes where the energy is insufficient to break the surface tension.  Also, be careful of referring to something like 'surface-force' unless you can define it clearly.

In the end it does comes down to fluid dynamics and how fluids flow around different shapes.  A fluid will flow and change direction around a streamlined object whereas it will not flow so smoothly and change direction when it is forced aside by a non-streamlined object.

Google for 'milk droplet' in image search to get a range of different shaped splashes; you'll find pics of all types of splash.  This is because the droplets tend to become flattened, depending on how far they have fallen, with the flattened droplets forming the 'ring' type sideways splashes, and the non-flattened droplets forming a central column of water with relatively little disturbance to the surrounding fluid.

[yor_on]

Nice one LeeE
"If most of the energy directs the water upwards instead of sideways then it will go higher than water where most of the energy directs it sideways"

That's what I think too, but also that the larger the surface hitting the water the weaker the impact per mm2 (f ex.) Which then should mean a lesser 'force per time segment' reaching the edge of that flat stone creating those waves. But that seems to me as the opposite side of what you are saying reaching the same conclusion? And water will act as a solid ground (concrete) when objects are falling from a higher height ( 30 m or higher? ) which also should influence the results. As for the 'milk droplets' I'll check them up, but we seem to agree as I see it.

----

Thinking somewhat..

Reading you again I'm not sure I truly understood what you meant by that quote I used from you.

Take a swimmer diving. The 'cleaner' the dive meaning the smaller the circumference of 'full penetration' of the waters surface the less waves, right. We all have seen them competing in the telly.

In this case we can see one 'idealized arrow' of force as the ultimate dive according to how I thought before with my 'arrows'.

The 'force' will in this case be directed straight down causing the fluid to release its energy in what I guess would be small 'vortexes' along and in the direction of the divers body moving.

in the case of a ball the force will probably also create those vortexes but as the displacement of water will grow as more and more of the ball penetrates the surface the vortexes will create our wave as they are being pushed in a tangent out from that sinking ball, up to its middle where they should change direction again inward toward the center, if I'm thinking right?

Otherwise it may seem as if this wouldn't be true and I think it is.

'But I would still expect a conical stone, dropped point first, to make a smaller wave than a round one?' as that is a idealized diver 'opening' the surface?

Well 'conical' within limits thinking of it
To 'conical' relative its length, and you will see a similar effect as with the ball I guess?

--

And that there in a perfect dive won't be any waves to speak of?
Suction, could that be it? The energy is placed under the water still working its way up (bubbles?) or maybe just 'dies out' while still being submerged?, and instead of waves the water around that 'hole' will fall in on it?

[LeeE]

Ultimately, I think we're dealing with two different effects, or types of 'splash', here.

There is the splash that's created by the water that is immediately displaced when the object first enters the water and then there's the central column of water that's raised by the water that fills the void left by the object once it has fallen below the surface.

When a highly streamlined shape enters the water, the water is displaced progressively as the shape widens and the volume of water displaced increases.  If the rate at which water is displaced is low enough the surface tension of the water won't be broken and all you'll get is waves, but when a flattened object enters the water an entire diameters worth of water has to be displaced immediately.  In both cases, the displaced water flows across the surface of the object but with the flattened object the water flows sideways until it reaches the sides of the object, at which point it finds a sharp edge i.e. where the radius of curvature becomes too small, and the flow detaches from the surface.  With a streamlined shape though, the water can continue to flow around the object without detaching.

With the flattened object, once the flow of displaced water has detached from the surface it will continue moving in the direction it was flowing when it detached i.e. sideways, but with the more streamlined object the water will end up flowing back in again.  Eventually, as the water flows back in around the streamlined object it will collide with the water flowing in from the opposite side, and the only way, as they say, is up, resulting in the central column type of 'high' splash.  With the flattened object though, most of the water displaced by the entry of the object has detached and gone on its way and the water flowing into the void behind the object, once it's sunk beneath the surface, won't be doing so with as much energy as with the streamlined object, so although you'll still get a central splash it won't have the same amount of energy and won't be so high; much of the energy has been spent on sending the detached water sideways.

If the shape is streamlined enough, i.e. long enough and pointy enough (which is what the divers are trying to achieve), it should be possible for it to enter the water without any splash, for as long as the rate of displacement of water stays within what the surface tension can cope with, all you should get is waves.  In effect, increasing the degree of streamlining (which for a given diameter, and therefore cross-sectional area, means increasing the length) is equivalent to slowing the speed of entry (although the total volume will increase too).  You'll still get waves, of course, because which ever way you look at it, you're displacing some of the water, and that water has got to go somewhere.

The topic is further complicated by the fact that the flattened object, especially if it has a greater cross section than the streamlined object, will slowed by the water more quickly, which will in turn affect the rate at which the water is being displaced and moved around, but I think that essentially what I've said above is basically correct.

[kprice (OP)]

Thanks leeE and everyone else,
Your answers have been very useful. We had already tried your one suggestion of dropping a cone, and, as you suspected, it made a tiny splash (I've been measuring the central splash). To test the idea of flow being limited around flat objects, we also tried dropping two objects of the same shape and mass into the water only one bigger than the other. We tried this with round and flat shapes. With the spheres, size made hardly any difference at all, whereas with the flat shapes, the bigger one had a much lower splash than the small one. This completely fits with your hypothesis that the flat ones will break the flow of the water, therefore producing a lower splash. It seems that the molecules don't have as far to travel to make a splash with the small flat shapes, but with the big flat ones, they have much farther to go, and have less energy left to make a splash. With the round ones, as you suggest, the molecules are likely flowing around the shape.
Once again, Thanks so much for the interesting discussion.