The Water Jumper

*Sniff, sniff..* You guys smell that? The smell of awesome in the air.

Enjoy the video first…

About four years ago, during an epic sunset, I was standing at a windy seashore throwing stones. The feeling when I watch the stones skipping was like HOLY! Holy! Hooolholy! holy cow! It could be longer, It’s just because my record was only five skips.  I’m sure you could do it much better than me, but beware when you start doing this thing it’s kind of hard to stop. Once you start, you won’t stop until the sun says byebye to you.

I promised myself to think about the physics behind it once I get back to my room. But somehow it ran away from my mind, it takes about four years for me to remind myself about it. And unfortunately I haven’t got a chance to skip stones again since then. However I still remember some important details. From my experiences, I intuitively think that a good stone for stone skipping is a flat one. After trying many times I concluded that the only way to make stone skips is to give the stone enough horizontal velocity with a correct orientation and also a quick flick at the end of the throwing process to give the stone a sufficiently fast spin. So I will suppose that these three factors, the horizontal velocity, the orientation and the spin are the main factors to give an awesome throw.

My first naive suspect was the surface tension but I think I can get rid of it pretty quickly. First of all, it is extremely weak. Here’s a quick estimation. The mass of a good skipping rock is about 300 g, so its weight is W=3 N. The surface tension of water at 25 ^\circ C is \gamma=7.2\times 10^-2 N/m. Thus if the planar dimension of the stone is about L=10 cm, then the order of the surface tension force is roughly F_{st}=7.2 \times 10^-3 N which is only sufficient to cause the stone feels itchy. And no matter how hard I tried, I could not find any possibility of surface tension force having dependence with the three factors above, so I guess we can safely forget about surface tension here. I also remember the motion of the water where I throw the stones at was very unfavorable for surface tension. Perhaps we can test this putting some detergent into a pool of water, to see if the stone can still skip on it.

Okay so what else could possibly cause the bouncing effect? In order to bounce the stone we need a force that can flip the vertical velocity of the stone. It means, we need a force that continues to exist even when there is no vertical velocity. For a typical object bouncing on the floor, the one that makes sure the flipping to happen is the elasticity of the object. Some of the kinetic energy is stored as potential energy which later will be used to renew the kinetic energy again. Okay, but what kind of elasticity can we expect from a collision of a solid with a fluid?

But it turns out that we don’t need one, maybe it is just a simple Newton’s third law. As the stone hits the water surface, it displaced an amount of mass downward which gives a lift force. It requires the stone to be in a correct orientation so that as the stone travels a distance on the surface of water it pushes some water downwards along the way. As we can see, the higher horizontal velocity of the stone , the more mass it can deflects in a time interval. Which means that if the stone  has a sufficiently high velocity, the vertical momentum of the water pushed maybe enough to fling the stone back to the air. Also most importantly the stone can still deflect water downward even when it has no vertical velocity. It is somewhat similar to how airplanes fly but this time water is deflected instead of air. But wait, how about the spin? We know that it won’t work without the spin, so it must contributes somewhere. I think most likely the spin plays a role as stabilizer, maintaining the correct orientation favorable for the skipping. Otherwise if there is no spin, the stone may easily change its orientation and hit the water with bad orientations, unfavorable for skipping. If the stone gets really lucky all the time, it may looks like this

Okay let’s analyze the collision process quantitatively. To get some idea about the flow of water around the rock, let’s estimate the Reynold’s number. It is not hard to find the viscosity of water at  20^\circ C, it is about \mu=1.002\times 10^-3 Pa.s. I suppose typical velocity of the stone would be about several meter per second, so let’s just take v=1 m/s. And then the planar dimensions of the stone are assumed to be about L=10 cm. Finally the density of water is of course \rho=1000 kg/m^3. Using these numerical values we can calculate the Reynolds numer

Re=\frac{\rho v L}{\mu}=9.98 \times 10^4\approx 10^5

Which says that in this regime there would be a big gap behind the stone as it moves on the water surface. Roughly it should looks like this

stone

After looking at the video in the begining, it seems reasonable to assume that the velocity of the rock to be horizontal, that is v\approx v_{x}>>v_{y}. I guess I can safely ignore the adhesion force between the rock and the surface of water which cause a portion of water to move with the velocity of the stone surface in contact, this is somewhat similar to friction. Also note that there could be some water being deflected upwards following the surface of the stone , take a look at the red arrow above. We can observe this as the splashes around the water, I am also going to ignore this effect mainly because it is too complicated. Or perhaps I could say I will only consider the case where the these effects are small. So the only forces that we will consider here is the force required to deflect an amount of water coming from the front of the stone and the gravitational force.

Since the force is independent of frame reference, it is a good idea to move to the rock’s frame to simplify things. In this frame, the water is moving with velocity v towards the rock before being deflected downwards at an angle \alpha. The amount of water being deflected depends on the cross sectional area of the stone A(y) below the surface of  water as shown above. During the collision, the equations of motion are

\frac{dp_{y}}{dt}=m \frac{d^2 y}{dt^2}=C_{1}\rho A(y) (\frac{dx}{dt})^2 sin\alpha-mg

\frac{dp_{x}}{dt}=m\frac{d^2 x}{dt^2}=-C _{2}\rho A(y) (\frac{dx}{dt})^2 (1-cos\alpha)

I put the factor C_{1}  and C{2} to account for a portion of water outside the cross-sectional area that is also get pushed downward by the neighbors and also the portion of water that is pushed upwards as a splash. For  C_{1} the two effects are competing, and we don’t know who win the fight. But we can guess that the value won’t differ too much from 1, thus for simplicity let’s just choose C_{1}=1. On the other hand the two effects work together in the case of  C_{2}, so C_{2} must be bigger than one. Let’s say  C_{2}=1.5. Now the A(y) depends on the shape of the stone, and it can be really wild if the shape is not alright. Again, to simplify things let’s assume that the stone is coin shaped with radius R so that the immersed area depends only on y, the vertical position of the lower edge of the stone. We can obtain A(y) by integration

A(y)=R^2[-(1-\frac{y}{R})\sqrt{1-(1-\frac{y}{R})^2}+\frac{\pi}{2}-arcsin(1-\frac{y}{R})]sin\alpha

Finally we have to assume that the the spin is good enough at stabilizing the angle \alpha that we can leave the angle as a constant throughout the collision. Now the two equations of motion are complete, however they are nonlinear and unsolvable.

Thus we have no choice but to deal with it numerically. Using the following numerical values:

m=0.5 kgC_{1}=1C_{2}=1.5R=10 cm\alpha_{0}=10^\circ\rho=1000 kg/m^3g=10 m/s^2

And the following initial conditions(The moment the stone leaves the thrower’s hand):

y_{0}=0.5 m

x_{0}=0

vx_{0}=10 m/s

vy_{0}=0

Python simulation shows the following beautiful trajectory of the stone’s CoM before sinking:

stonesimulation

With these initial conditions, the stone bounces 14 times before eventually sinking. I think this result seems to be quite reasonable.

It is also interesting to check out the minimum horizontal velocity we need to give in order to bounce at least once. My laptop says that we should at least throw the stone faster than v_{x}=5.43 m/s in order for it to bounce once. Seems easy right? Yeah it’s not the hard part, actually the hard part is to make the right orientation and to give enough spin to the stone to maintain the orientation.

I think this is good enough for now, I will refine my model if I get a chance to play at a seashore.

Edit:

Here are some thoughts to estimate C_{1} and C_{2}:

Okay soo we want to see, if the main portion of water get pushed aside, how much extra water will also get pushed together with it. Close to the main stream(inside the cross section of the stone), the streamlines would be deflected strongly, as we move farther from the surface the streamline becomes more and more straight(less deflected). I think this cannot occur if the velocity of water flow is always constant along the streamlines. I think the only way to make the streamline less deflected is that the water streamlines should get thinner as it flows around the stone. And in order to be thinner the water flow must accelerate in the direction of the flow(continuity). This can only works if water pressure in front of the stone is higher than behind the stone, I think intuitively it’s obvious that the front pressure is indeed higher, the surface of water in front of the stone is also higher. So from Bernoulli we can see how much the water flows can accelerate and how much it’s area would decrease, and thus how much the deflection will decrease. And from these we can obtain the correction terms.

All of this can be calculated analytically, but think this involves a lot of messy works. While the effect might not be significant and interesting, so I decided not to work on the details.

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