# Is mid-water bouyancy a classic example of a balanced but unstable system?

I came to this thought experiment as I was pondering good teaching examples of stable and unstable systems. It occurred to me that stable systems are really quite abundant. For a shoot-from-the-hip example, the speed of a car is stable about a given speed given a constant rate of fuel injection, since any perturbation will disappear over time. To be perfectly illustrative, here is a ball on a hill. Note that the forces are balanced in both these cases, but stability is different.

Now, the above illustrates the concepts well enough, but I like to present something that's slightly more non-trivial and try to get people to exercise real physical understanding.

Consider:

A submersible with a compressible cavity. A model sufficient for this will be that of a weight attached to a balloon. Saying it's an airtight bag instead of a balloon might be more straightforward, as it avoids the contribution to pressure from the balloon elasticity itself.

I want to make the qualitative argument that given the buoyancy and gravitational forces are balanced at some underwater point (obviously below the surface and above the ocean floor), this point is unstable.

The logic is that the balloon will increase in volume and decrease in density as it gets closer to the surface, where the pressure is lesser. Correspondingly, increasing depth and pressure contracts the balloon. That means that upward movement causes a net upward force and moving down causes a net downward force. Unstable.

Several questions

• Does this imply that any submersible (below the surface and above the floor) must actively maintain its depth level? Is the condition associated with this claim that the submersible be more compressible than the liquid? What about a compressible atmosphere?
• Are there passive control systems that could maintain a depth setpoint?
• Would anyone like to address the problem and questions with equations, derivatives, and all that good stuff?
• This might be of some interrest: en.wikipedia.org/wiki/Cartesian_diver – Georg Oct 13 '11 at 17:52
• @Georg Indeed, that link does mention the exact phenomena I alluded to. It's rather wordy about it too: "It might be thought that if the weight of displaced water exactly matched the weight of the diver, it would neither rise nor sink, but float in the middle of the container, however, this does not occur in practice." – Alan Rominger Oct 13 '11 at 19:24
• To complicate matters, even if the balloon remained unaltered, the seawater salinity and density could alter over time and the point of balanced forces will change. This is Metastability..much like the upright beer can in front of me :) – osknows Oct 13 '11 at 22:56

## 2 Answers

The assumption here is that the water is more incompressible than a balloon, so that the density variations in the water are less than the density variation in a balloon. This is true for a balloon, but false for a submarine made of steel. The compressibility of steel is roughly 80 times less than the compressibility of water, so if the submarine has thick steel walls, it will make a stable depth equilibrium using bouyancy alone. You can maintain a depth with a submarine by adjusting the density very slightly to the one appropriate for water at a given depth.

For the balloon, the equilibrium is unstable, but for there to be an equilibrium at all requires that the balloon skin be made of a material significantly denser than water, and the depth where there is balance, the air will be compressed enormously.

### EDIT: Submarine compressibility

(In response to the comments of Zassounotsukushi)

When a submarine is submerged in water, the compressibility is not determined by the bulk compressibility of steel, because the pressure stress has to travel through the much thinner hull to get from one side of the submarine to the other. The actual stress in the hull is increased.

To see how much it is increased, roughly, consider a model submarine which is a square-box-cylinder, with walls of thickness w and side-length L. The pressure is a flow of momentum per unit length equal to PL from one side (times the length) of the box to the other, through the sides of the wall, and if the thing were perfectly solid, the compressibility would be roughly 80 times less than water.

But the momentum must flow from right to left through a region of width 2w instead of the full width L, as it would be in a solid steel cylinder, so the actual momentum current in the side of the box is increased by a factor of L/2w. For a round cylinder, up to a factor of order unity (it will be between .5 and 2), the analogous factor is R/w.

For a typical submarine, I found a hull width w of 30 cm, while a spatious radius is 6m. The ratio is 1/20, so the submarine is about 20 times more compressible than bulk steel. But this still makes the submarine 4 times less compressible than water, and makes the equilibrium stable.

### Steel balloon

In order to make a hollow steel balloon neutrally bouyant, ignoring the density of air, the ratio of unoccupied volume to occupied volume is as the ratio of the density of steel to air volume is as the ratio of the density of steel to water, about 8 to 1. Since this ratio is much smaller than the bulk modulus ratio of 80 to 1, a neutrally bouyant steel balloon will acheive a stable equilibrium. This was the model I originally had in mind for a submarine, and for this model, the walls are extremely thick, and the bulk compressibility is only changed by a factor of order 10 at most, not 80.

But this model is nonsense: it is assuming an empty submarine! The accurate description is above.

• A major point I will take issue on is equating a submarine to steel. A submarine is much more than that, and although steel is less compressible, I think a hollow steel can would be more compressible than water. – Alan Rominger Oct 14 '11 at 4:43
• I considered this, but a sufficiently thick hull has a compressibility which is indistinguishable from that of bulk steel. I believe submarines are in this limit, but if you have better information let me know. There is no reason to downvote when in doubt. – Ron Maimon Oct 14 '11 at 5:03
• I actually think that's not the case. Consider two different thicknesses. The steel in the thinner case will be subject to more material stress, and it will contract more. I see no reason this shouldn't continue to hold when comparing a 100% steel volume to that same volume with a cavity in it, no matter how small. Practically, you could assess that change from the change in pressure in submarines as they dive, which is known to change. But I thought they also had some variable volume bags that they used to control buoyancy anyway. – Alan Rominger Oct 14 '11 at 12:16
• @Zassounotsukushi: You are right, of course, I had a bogus intuition. The extra compressiblity of a cylindrical hull with thickness w is approximately w/R, where R is the cylinder radius. For a 30cm thick pressure hull (this is what I found by google), and an 6m radius, this ratio is 1/20, which still gives the submarine a factor of 4 less compressibility than water (but not 80). I will modify the answer. – Ron Maimon Oct 14 '11 at 19:36

To obtain the highest degree of generalism, I will refer to a reference state at the buoyant point. The volume of the object at that point will be $V_0$ because it is a reference value, and mass will just be $M$, because it is constant. Since this is a discussion about stability, the non-linearity of compressibility is not of great interest, so I'll write the following form to relate volume and external pressure for the object. Note that if we compare, say, a rock in water, this is meaningless because the pressure that makes the density of the object and the fluid the same is absurd.

$$V = V_0 - \rho_w g h \frac{dV}{dP}$$

Although other contributors can relax this assumption, I'm interested in assuming a truly incompressible fluid. For experiments in the bathtub, that's sufficiently correct. The implication of this follows:

$$F_b = \rho_w g V = \rho_w g \left( V_0 - \rho_w g h \frac{dV}{dP} \right)$$

$$F_g = -Mg$$

$$a = \frac{F_b+F_g}{M} = \frac{g}{M} \left(\rho_w V_0 - \rho_w^2 g h \frac{dV}{dP} -M \right)$$

Also, for the stable point, it is buoyant.

$$0 = \frac{g}{M} \left(\rho_w V_0 -M \right)$$

Thus, we have the much more simple form:

$$a = -\frac{ (\rho_w g )^2 }{M} \frac{dV}{dP} h$$

Thus as long as the object is a constant mass, the only way for it to have a stable point is for $\frac{dV}{dP}$ to be positive, and this is very difficult. I do not know of ordinary thing that fits this criteria. In order to accomplish my desire to have a passive system that gives a stable buoyant point, I will need to relax the assumption that $M$ is constant, and use that to make something useful.

$$V = V_0 - \rho_w g h \frac{dV}{dP}$$

$$M = M_0 - \rho_w g h \frac{dM}{dP}$$

$$a = \frac{\rho_w g^2}{M} \left( - \rho_w \frac{dV}{dP} + \frac{dM}{dP} \right) h$$

I hope you can follow the algebra, it's just straight from prior equations. Now, can $\frac{dM}{dP}$ be positive? Why yes it can. It can also be big enough to counteract the $\rho_w \frac{dV}{dP}$ term.

Say what?

Experiment

Have 2 tanks, one of fresh water, one of salt water (higher density). Have a submersible consisting of a bag of salt water with a tube connecting it to the salt water tank attached to a floatie.

Here it is in the fresh water tank. It's stable!

When the device rises above the buoyant point, the fresh water pressure is lower than what the head should be for a salt water column. Now, hypothetically, the entire volume of the salt water bag rushes for any differential movement above this point. Hilariously, the buoyant force is literally a step function, as I tried to illustrate in the image.

So now school kids can make a diver that is unstable and one that's stable!

To add some equations for this, let's note that the operating principle is exchange of a different kind of fluid. Then we can relate the change in volume and mass.

$$dV \rho = dM$$

This simplifies the prior equation.

$$a = \frac{\rho_w g^2}{M} \left( - \frac{\rho_w}{ \rho} \frac{dM}{dP} + \frac{dM}{dP} \right) h = \frac{\rho_w g^2}{M} \frac{dM}{dP} \left( 1 - \frac{\rho_w}{ \rho} \right) h$$

Now this gets us very close to formulating a useful general formula. Note first that it's very easy to get a $dM/dP$ that is negative. It's easy to obtain a matter outflow with increased pressure. Because of that, it's preferable to use a working fluid of a higher density than water, making the terms with the densities negative and the system stable. In the system I described, $dM/dP$ is really a Dirac function provided a large reservoir of salt water. I guess you could call for a stationary balloon instead. I still don't know how to achieve this with a passive system on-board the device.