Why is equalizing easier after approx. 10m/33ft?

I understand that when dropping to 10 m in water we are then at 2 atm of pressure. If you would not have equalized by then (yikes) then the absolute pressure difference is 2-1=1 atm (~1e5 Pa), relative difference is 100% (doubling). Surface area of ear drum ~5e-5 m sq. Gives a force of 5 N (yikes again, but thankfully hypothetical).

Now I equalize at 2 atm and dive 10 more meters to 3 atm without equalizing. The absolute pressure difference is again 3-2=1 atm, same force of 5 N on ear drum. But the relative pressure difference is 50% (from 2 atm to 3 atm is +50%).

However, in practice I (and most?) find it mory easy to equalize, the deeper we are. Online sources say that this is because the relative pressure decreases (which we just showed it does). But the relative pressure (i.e. pressure ratio) does not determine the force on the ear drum! Force = absolute pressure difference * area.

I must be missing some basic (bio)physics here. Any suggestions?

in scubaby Retinite

82%

20d

Revolutionary_Bid595

UW Photography

18d

Im the only one feeling so dumb reading all those comments???

Retinite

19d

A lot of posts jump/refer to the ideal gas law and the typical balloon shrinking example that p1* V1=p2* V2, but in my view that is too simplistic a model that does not explain enough here. It either pretends your ear drum is like a piston sliding into your inner ear, or your inner ear with ear drum is like a shrinking balloon. I think stiffness (or flexibility/compliance) of everything involved does play the major role, as one contributor mentioned. The water is incompressible and can be assumed an infinite volume. The eardrum and tissues have stiffness and the air inside the inner ear cavity has stiffness that is a function of pressure. All these stiffnesses more or less add (i.e. parallel). If p=nRT/V and if the volume depends on "piston" (airdrum) displacement x then first order stiffness assuming ideal gas law = -dF/dx= -A* dp/dx= p² /(nRT)* (dV/dx) (or something like this, I think, hard to do math on mobile, could not easily find piston air stiffness formula). Showing that the air compartment most likely becomes stiffer when its pressure is higher. This could imply that less flexion of the ear drum happens for the same absolute pressure difference, when the gass pressure in the inner ear is higher. This possibly explains why going from 15m to 18m I feel almost nothing but from 0m to 3m I feel a lot of squeeze (when not eq.). This would explain the difference in felt pain/squeeze, but not fully why the ET's would open more easily (if that is even a contributor to easier equalization).

Bonne_Journee

19d

"Everything's relative" - Ever heard that phrase before? That's because it's true.

The deeper you go, the lower the relative pressure around you gets. To 10 meters the pressure increases by 1 bar, which is double the current pressure.
To double the pressure again, you have to go to 30m.
So the first ten meters are the hardest because that's when the biggest differential exists.

ItsYaBoiAnatoman

19d

As you said, 10 meters is 1 atm.

If we start at -0 m and go to -10 m, we go from 1 atm to 2 atm.

If we start at -10 m and go to -20 m, we go from 2 atm to 3 atm.

"So what?!" you say "The nominal difference is still 1 atm"

The ideal gas law states "pV = nR*T". p: absolute pressure; V: Volume; n: amount of gas in moles (matter); R: gas constant; T: temperature.

As you can tell, the right hand side doesn't really change much. The number of atoms in your gas stays the same, the gas constant is a constant, and the temperature change only has a negligible effect at that size. This is now pretty much Boyle's Law, and we can use P1V1 = P2V2 and reform it to (P1*V1)/P2 = V2.

In other words, we don't care about the nominal difference in pressure, we care about the relative difference to the already existing pressure because pressure and volume have an inverse relationship.

Simple example to visualize this:

Take a balloon with a volume of 1 liter at 1 atm (-0 m).

1.1. Bring the balloon down from 0- m to -10 m, so 2 atm. The pressure effectively doubled.

1.2 Notice the balloon shrunk to half it's volume. So if you could open it up and fill it with water, only half a liter would fit. (P1*V1)/P2 = V2; (1 atm * 1 l) / 2 atm = 0.5 l)

1.3 Bring the balloon down from -10 m to -20 m, so 3 atm. The pressure effectively increased by half, it didn't double (2 atm to 3 atm).

1.4 Notice the balloon shrunk to a third of it's full volume at 1 atm, so 0.33 liters (P1*V1)/P2 = V2; (1 atm * 1 l) / 3 atm = 0.33 l). The step size decreased. Compared to the volume at -10 m it's this "(2 atm * 0.5 l) / 3 atm = 0.33 l", so it holds up. To get half the volume compared to the volume at 2 atm (-10 m), we'd have to go to 4 atm (-30 m). Double the distance, because of the inverse relationship (forming a hyperbollic curve on a plot).

Now to see if we understood the concept of inverse relationship between volume and pressure:

What happens if we take an infinitely stretchy balloon with 1 liter in volume at 1 atm, put it in a vacuum chamber and reduce the pressure?

That's right, it grows in size (volume).

By how much?

New volume = (old pressure * old volume) / new pressure

Then, what happens if we go down to 0.5 atm? The balloon grows by the factor of atmosphere difference, so it doubles!

Now, what if we go all the way to incredible and impossible 0.0 atm? OH NO, we're dividing by zero! The balloon becomes infinitely big on paper, and that's actually what happens in real life too (except we don't have the materials or the space to replicate it). All that, even though the total difference is just 1 atmosphere.

So, in diving terms: When fully ascending from -10 meters, the gas in your body expands to double it's volume. From -20 meters it triples, from -30 it's factor 4, and so on.

Ok-Spell-3728

19d

If you typed this out and not copied, you deserve an award, I haven't seen it explained this well and captivating

workact

19d

So the other way to look at it is, you want to set the pressure in a known volume, known temperature, with a constant gas.

which reduces pV = nRT to basically pressure scaling linearly with moles of gas.

Its easier to equalize at higher pressure, because the amount of presser that needs to be equalized is the same (1 atm or whatever) but density of the gas "transferred" in is much higher, so less volume needs to be added to achieve the same effect.

CidewayAu

20d

The percentage change in pressure from 0-5m (50% increase 1-1.5 bar) is the same as the change from 10-20m (50% increase 2-3 bar).

EDIT: Because you have already equalised for the pressure you are at, you are only dealing with relative pressures, at 10m assuming you have equalised properly your air space is pushing back at 2 bar while the water is pushing in at 2 bar so effectively you have 0 bar change between them.

Retinite

-4

20d

You are not explaining why the relative (maybe I should call it rational) pressure difference physically matters. Relative/rational pressure difference does not describe the force on the eardrum, or any air space, the absolute pressure difference does. 2 atm outside 1 atm inside is the same compressive force as 101 atm outside 100 atm inside an air space.

Chasman1965

19d

Yes, but when you are at 10m assuming clearing on the way down, you only need to clear more for the increased pressure, as you have pressurized air in your air spaces from equalizing . It’s the relative changes that are important. What you are missing is that if you are static at 10m after equalizing you already have 2 atmosphere pressure pushing out on your eardrum, as well as the 2 atm pushing from the outside.

Retrogradefoco

19d

Think about it this way. Let’s say you’re playing a game where 5 people are on each team. The goal is to push a board over the line (think tug of war, but pushing instead of pulling).

At first it’s 5 v 5 even and difficult. But if you can get past the first 10m, the other team loses a person. Now it’s 5 v 4 and it’s easier to push. Same thing happens at every 10m, so at 20m it’s now 5 v 3.

The other team, is the water pressure. You were able to push out 1 atm of “equalization” by moving yourself 10m and being able to equalize. Now when you go the next 10m you still need to push out 1 atmospheres worth, but there overall strength is diminished.

If each player/atmosphere has a strength rating of 20, then first you are pushing against 100 strength. At 10m, you’ve eliminated one player, thus only needing to push against a strength of 80. The pressure above you and below you (in front of you/out of the game) hasn’t changed, but your position and strength against it has as you’ve already competed 1/5th of the work toward full equalization at your max depth.

(Crazy weird analogy that I don’t feel like explained it super well, but hopefully was enough to help make some sense of how it works)

rdvr193

20d

You answered your own question. It’s about differential. 0-33ft is the biggest single differential. Hence why it’s the biggest hurdle.

jw_622

20d

Your entire initial post is correct but once your ears equalize to 2ATM, as you descend to 3ATM, the force to clear isnt an absolute pressure of delta 3ATM. It’s delta 1ATM. It’s still the relative pressure change of the delta 1ATM because you already have an internal 2ATM existing from your prior equalization stage, which aids in further equalizations. This is why it is easier to clear as you get deeper. If it was always absolute pressure relative to the surface, it would become near impossible to go beyond a certain based only on your ears alone. Past 120’ I don’t notice much equalization occurring.

TomvdZ

20d

What you're missing might be flexibility. The ear(-drum) is not perfectly rigid. It will deform slightly under pressure, and the amount of deformation depends on the relative pressure difference.

If you imagine taking a baloon down from 0 to 10 meters, its volume will halve. But if you take it down from 10 to 20 meters, its volume will only decrease by 33%.

I would speculate that this is what makes it easier to equalize. As the pressure on the ear(-drum) increases, it flexes, which decreases the volume inside the ear, increasing the pressure. That means that the force you need to overcome when equalizing is lower for lower relative pressure differences.

caversluis

20d

It is an interesting question. Let me take a stab at it

First of all, Divers Alert Network (DAN) has described the effects of depth in the article [How Your Ears Respond to Pressure](https://dan.org/health-medicine/health-resource/smart-guides/beat-the-squeeze-equalize-like-a-pro/how-your-ears-respond-to-pressure/) They include a table which describes the effects of not equalizing. Already at 10 ft (3 meters!), you risk eardrum break.

So, my guess is that most of us do equalize somewhat in some way; (just) enough to avoid (significant) damages.

Having said that, you might be right that it feels easier at greater depth (but definitely not 20m). My guess is that equalization with a small difference is not easy to notice. Whereas equalization with a larger difference is much more noticeable. It might be a matter of perception, it feels easier because it is easier to notice.

I was told that good equalization starts early on in the dive. As early as when you get your head down in the water.

gnarliest_gnome

Nx Rescue

20d

But the relative pressure (i.e. pressure ratio) does not determine the force on the ear drum! Force = absolute pressure difference * area

It's the % change in pressure that you need to think of. From surface to 10m the pressure doubles (1ata to 2ata). From 20m to 30m it's a 25% increase in pressure (3ata to 4ata). The gasses in your ears & sinuses compress according to the ideal gas law: pV=nRT. n, R, & T are constants in this equation. So if the pressure increase is 25%, the volume of the air decreases by 25%.

2 missing replies