**Diving at Altitude**
__DIVING AT ALTITUDE__
Frank was posing the same question at the Austrian freediving forum last week. I feel it's more appropriate to answer here, so a wider audience would benefit. First a direct answer to Your question: As You stated correctly: Air pressure at 2600mt altitude is 560mm Hg. More practical: 0,737 atm.

Just a brief reminder:

**1 atm** = 1.01325 bar which is equivalent to 10,33 mt of fresh water or 10,13 mt of salt water with average salinity.

**1 bar** is equivalent to a pressure caused by a layer of 10 metres of saltwater which is equivalent to a pressure of 1.02 kg/cm².

**1 kg/cm²** is equivalent to a pressure caused by a layer of 10 metres of freshwater.

For the sake of convenience all units may be rounded and used interchangeably without loosing essential accuracy.

Coming back to Your 0,737 atm, where O2-partial pressure is approximately ¼ (25%) less than at sea level.

**The consequences are twofold:**
**1) Oxygen storage **
Obviously You cannot store as much oxygen in Your lungs at an altitude of 2600mt compared to sea level. Straight proportionally up there You can only store a little less than ¾ (75%) of oxygen in Your lungs. You may expect a loss of roughly ¼ (25%) of Your performance - in depth as well as in distance. This could also be applicable to static times!

By the way: observing the ratio between CW and Dynamic in a wide range of athlethes the following simple formula is reliable:

[1]

** 2×CW = D**
Example: 50 mt in depth equals 100 mt in dynamic. Under a pressure of 0.737 atm (=2600 mt altitude) the same performance may be reduced to 37 mt in CW and to 74 mt in Dynamic, respectively.

**2) Pressure increase and associated equalization problems**
Consider the formula:

[2]

**P(A) / P(DA) = P(SL) / P(DSL)**
It helps me to calculate how deep I have to dive (

**X**) at sea-level to experience the same (relative) pressure increase compared to a dive at a given altitude

**A** to a given depth

**D**.

**P(A)** = Pressure at the surface of a lake at altitude

**P(DA)** = (Total) pressure at a depth of a lake at altitude

**P(SL)** = Pressure at sea level (

**always 1**)

**P(DSL)** = (Total) pressure at an equivalent depth at (or below) sea level. (

**always X**)

Practical application: how radical does pressure change for You when You dive to -30mt at an altitude of 2600mt compared to a dive that is started at sea-level?

**P(A)** = 0,737 atmospheres (or bar, doesn't matter)

**P(DA)** = 0,737 + 3 = 3,737 (total pressure in -30mt at 2600mt altitude)

We apply [2]:

**P(A) / P(DA) = P(SL) / P(DSL)**
0,737/3,737 = 1/

**X**
3,737 = 0,737

**X**
**X** = (approx) 5,07.

What's left to do is to convert a relative pressure increase into an equivalent pressure increase at sea level:

**VD** = (

**X**-1)×10 = 4,07 × 10 = 40,7 Metres.

Diving to a depth of -30mt at an altitude of 2600 mt will be equal to a

**virtual depth** of 40,7 metres at sea level.

If You have already been packing at sea level, a potential advantage of packing at altitude would be compensated. You will end up (roughly) with the same disadvantage (less pressure and less oxygen storage).

__General formula to calculate air pressure at altitude__
Here is the barometric formula to calculate an air pressure

**P** at arbitrary heights:

**P = P° × exp (-rgh) ** with:

**P°** as ground pressure, this time I choose the average sea level pressure of 1013 mb

**r** as the specific weight of air (expanded to a pillar of air with a height of 1 km)

**g** as the gravitational acceleration (weight = mass×gravitation)

**h** as my (relative) target height above the ground in km's.

**rg** works best with a value of

**0,122.**
Example: last year Herbert Nitsch dove to a depth of -75mt at Millstatt Lake (Altitude = 588mt)

To obtain the air pressure

**P** at the surface of Millstatt Lake I set the following equation:

**P = 1013 mbar × exp (-0,122×0,588) = 931 mbar ** (rounded to mb)

**P(A)** = 931 mbar

**P(DA)** = 931 + 7350 mbar = 8281 mbar

(75 mt of fresh water cause a pressure of 7.5×980 mb = 7350 mb, this time I want to be precise!)

Now we apply [2]:

**P(A) / P(DA) = P(SL) / P(DSL)**
931/8281 = 1/

**x**
8282 = 931

**X**
**X** = approx. 8,9

**VD**= (

**X**-1)×10 = 7,9 × 10 = 79 Metres

**virtual depth** for an equivalent dive at sea level.

__General Formula to calculate relative water pressure increase__
Let's assume I manage to equalize at -30mt for the last time efficiently. (being able to pop my eardrums back into a relaxed position).

How much further can I go without damaging my ears?

I know I can go to -3mt from the surface without having too much pain.

Here comes a slight moderation of [2]:

[3]

**P(SL) / P(DSL) = P(LEQ) / P(PD)**
**P(SL)** = surface pressure at sea level (to keep it simple) =

**1** (bar, atm, ...)

**P(DSL)** = total pressure at -3mt =

**1,3 **
**P(LEQ)** = total pressure at point of last equalization, for 30mt:

**4**.

**P(PD)** = total pressure at potential depth =

**X**
Now we apply [3]:

**P(SL) / P(DSL) = P(LEQ) / P(PD)**
1/1,3 = 4/

**X**
**x** = 5,2

**PD** = (X-1) × 10 =

**42 Metres**
It's amazing how far we can go below the point of our last efficient, good equalization. Thanks to the Frenzel Fattah Mouthfill technique this should be only a theoretical problem, providing there are no other hindrances (sinuses, ...)

Hope that answers some questions ...

Gerald