# Logarithmic water pressures

1. Jul 1, 2010

### DaveC426913

I'm discussing decompression in shallow water. I never studied logarithms.

At sea level, a diver is at 1 atm,
at 33ft: 2atm,
at 66ft: 3atm.

How does it work between 33 and 0? Specifically, at what depth is 1.25atm?

Is it less than 8.25ft or more than 8.25ft?

2. Jul 1, 2010

### rcgldr

It's basically linear since water is nearly incompressable, so 8.25 ft. From an old post:

If you increase the pressure of water from ambient at 14.7 psi to 1000 psi, it's density only increases by .32% (by a factor of 1.0032). A gallon of water at 1000 psi reduced to 14.7 psi would only expand by 2.5 teaspoons (768 teaspoons per gallon x .0032 = 2.5 teaspoons).

Last edited: Jul 1, 2010
3. Jul 1, 2010

### DaveC426913

Thank you. I thought so.

4. Jul 1, 2010

### rcgldr

The amount of nitrogen absorption depends on both pressure and time, but I don't know the time factors, other that long periods at somewhat shallow depths can be an issue.

5. Jul 2, 2010

### DaveC426913

What we were discussing wasn't N absorption, it was overpressure in the lungs. If my understanding is correct, the lungs can withstand a 25% overpressure before the alveoli begin erupting. My argument was that this equates to an 8.25ft depth. The counterargument was that a] pressure is logarithmic, and b] the biggest changes in pressure are nearest the surface, thus 1.25atm would be reached shallower than 8.25 ft.

My opponent has a good case though, a friend of his died from a blown lung at three feet...

6. Jul 2, 2010

### rcgldr

I recall one anecdote that if it's too deep to breathe with a snorkle, which is limited by how much inhaling force you can exert to oppose pressure differential, restricting the level to very shallow depths, then overpressure issues are possible. I don't know what this limit is but a 3 foot case sounds like an inhale, hold air, and surface problem. I also don't know how true that anecdote is.

7. Jul 2, 2010

### TubbaBlubba

I've heard that your lungs can sustain permanent damage from as much as 50 cm.

8. Jul 2, 2010

### broean01

Slightly tangential, but I have to ask what the circumstances of this injury were

9. Jul 2, 2010

### arildno

Well, Dave:

Why human lungs are vulnerable has more to do with how they are constructed than with whether or not water pressure varies in a linear manner or not..

Perhaps the vulnerability just shows that natural selection has opted for a lung design that is excellent for living in, and on, air pressure, but that precisely because it is rare to experience other than that, the lungs will fail miserably if we try to make ourselves into fish.

10. Jul 2, 2010

### my_wan

I wonder if this kind of thing may be triggered as much by hiccups, sneezes, or something similar at such minimal depths, with the aquatic conditions not only exacerbating but also preventing the capacity to properly compensate. I'm not much on medical knowledge.

Unfortunate

11. Jul 2, 2010

### DaveC426913

Definitely. I was not suggesting anything else. See below.
The diver was bent over at the waist fiddling with some equipment. When he straightened out, he blew and died. I am pretty sure he'd have been holding his breath while bent over like that, putting added pressure on his diaphragm.

The guy I am discussing this with is the safety diver for our club and it was his friend that died. It's his job to be accurate about this stuff.

Perhaps, but that's not related to my question.

I simply wanted to establish, given the parameters mentioned, at what depth (in an ideal case, no mitigating factors) one might expect to blow a lung by holding one's breath.

12. Jul 2, 2010

### DaveC426913

The reason this is coming up is because there is currently contention at our club as to whether a pool dive counts as a dive as per safety considerations.

Yes, you can die by doing a free ascent from only 8 feet, even in a pool. So yes, so a pool dive is still a dive.

And yes, an inexperienced diver blowing his BCD to rocket to the surface fast enough to rise chest-high out of the water is very much in danger of dying.

13. Jul 2, 2010

### Staff: Mentor

Pressure DIFFERENCE is not linear: 2 atm is twice 1 but 3 atm is only 1.5x 2. So the percentage change is indeed greater the closer you are to the surface.

So consider that when you swim from 66 to 33 feet the air in your lungs wants to occupy 33% more volume but when you go from 33 to 0 it wants to occupy twice the volume.

Last edited: Jul 2, 2010
14. Jul 3, 2010

### Studiot

Russ has put his finger on the issue. Volume expansion increases geometrically with ascent.

But why did you state this and then introduce pressure difference, Russ?
Whatever units you measure in, pressure difference is is linear. There is exactly the same pressure difference between 1 atm and 2 atm and 1000 atm and 1001 atm.

For diving purposes we can treat water as an incompressible fluid so pressure of water increases linearly with depth in accordance with the elementary equation

Edit add surface pressure to equation.

P=pgD + Ps

Where Ps is the pressure at the surface , ie atmospheric pressure.

Now we introduce air, which is compressible and dive.

There are two different scenarios.

1) If we take a breath at the surface and dive, we breath air at surface pressure and rely on the structure of our bodies to resist the increased pressure at depth. The air in our lungs is not in pressure equilibrium with the water pressure, even at very shallow depths. this is why if we open our mouths water will enter. There may be some slight pressurisation of the lung air, but it never deviates far from the surface pressure and returns to surface on our return.

2) At some depth if we breath air pressurised to be in equilibrium with the water pressure and open our mouths under water at that depth, water will not enter. This is in fact a divers show off trick (not recommended).
As we ascend the air in our lungs will try to expand as it is now at higher pressure than the surrounding water. this is why divers are taught to breath out on ascent.

Given the above information I conclude there is more to Dave's story yet to be told.

Last edited: Jul 3, 2010
15. Jul 4, 2010

### DaveC426913

Yes, this is precisely why I posted. I know that the biggest pressure changes occur in the top few feet, which sounds like the atm value would change nonlinearly.

But no, I don't think that's the crux. If lung damage were reported as a factor of expansion, that would be true, but it's not, it's reported as a simple atmospheric pressure, which is linear.

16. Jul 4, 2010

### DaveC426913

You are correct that there the two scenarios, but your surface air scenario is not accurate:

In fact, you body does compress markedly when you dive with surface air. Your body does not resist this compression and lungs do stay in pressure equilibrium with the water.

As with any non-rigid cavity filled with 1atm of air, your lungs will shrink as you descend and will expand back to normal size as you ascend.

17. Jul 5, 2010

### Staff: Mentor

I suppose you can express pressure difference in terms of psi or %, so maybe the wording could have been different. Perhaps I should call it "fractional change in pressure".

18. Jul 5, 2010

### Staff: Mentor

Where is it reported that way? I seriously doubt the human lung has much capability to hold pressure - more than a few psi would surprise me. And you can't fill completely and pressurize your lungs yourself, so the volume change is how the pressure is generated. Ie, when you go from 66 to 33 feet, your lungs aren't really taking-on 5psi, they are probably just expanding by [almost] 33%.

19. Jul 5, 2010

### Dickfore

Your dependence is not geometric, but a linear one. A good indicator of this is the calculation of the slope:

between 0 ft and 33 ft: 1 atm to 2 atm -> k = 1 atm/33 ft

between 33 ft and 66 ft: 2 atm to 3 atm -> k = (3 - 2) atm/(66 - 33) ft = 1 atm/33 ft

The general equation for a linear dependence if you know one point $(x_{0}, y_{0})$ and the slope k is:

$$y - y_{0} = k (x - x_{0})$$

Taking $x = h/1 \mathrm{ft}$ and $y = P/1 \mathrm{atm}$, you have:

$$x_{0} = 0, y_{0} = 1, k = 1/33$$

so:

$$\frac{P}{\mathrm{atm}} - 1 = \frac{1}{33} \, \frac{h}{\mathrm{ft}}$$

$$\frac{P}{\mathrm{atm}} = 1 + \frac{h}{33 \, \mathrm{ft}}$$

Taking P = 1.25 atm and solving for h, we get:

$$1.25 = 1 + \frac{h}{33 \, \mathrm{ft}}$$

$$h = 0.25 \times 33 \, \mathrm{ft} = 8.25 \, \mathrm{ft}$$

20. Jul 5, 2010

### DaveC426913

I'll have to dig it up again. It was a number I got a while back and have had no reason to question until now.

Meanwhile, I didn't find a lot of references but here is an article that on page 2, says

"...an over-pressure of only 73-80 mmHg is required to rupture the lung. However, if the chest wall is splinted, an over-pressure of between 133 and 190 mmHg is needed to rupture the lung..."

which equates to a mere 9-10.5% and 17.5-25% respectively.

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1296705/pdf/jrsocmed00025-0040.pdf