# constant weight in lake

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#### fpernett

##### Well-Known Member
Hello
I know that the AIDA score for constant weight in lake is 1 meter=1.32 points
This apply in despite of the lake altitude?
I'm living far away from my beloved sea so I usually train constant weight in a lake that
is at 2600 over the sea level. The barometric pressure at this altitude is 560 mm Hg, and the PAO2 is usually between 60-65 mm Hg. I think is not the same if the lake is at sea level.
Is there another score?
We should create another score with the barometric pressure?

Sincerely

Frank Pernett

• cjborgert
Acclimation ?

This is a very good question, IMO.

We know from other sports that training at altitude will produce acclimation to many of the O2-related challenges, but I don't know that acclimation is ever complete nor how well it would transfer to apnea performance at altitude. In addition, reduced atmospheric pressure per se would alter some physiological parameters that affect freediving performance, but the question is, to what extent?.

Dive tables have been adjusted for altitude, so it should be possible to develop an algorithm for approximating the differences that apply to freediving, and thus produce a correction factor for altitude, but I don't know if one already exists.

OK, so I've mused on the details of your question without giving you any help at all . . . now it's time for someone with some real knowledge on the subject to weigh in (I'm heading for my physiology and sports medicine reference texts for some help here - maybe I'll be back later).

AIDA score in lake / altitude

Hello Frank,

The AIDA coefficient for lake performances (1.16 in 2001 and not yet 1.32) is based only on the World Records of depht-disciplines (both male & female) and not regarding the altitude.

Otherwhise, when starts the "altitude performances" ?... 1000 m, 2000 m, 3000 m, ???... We should have many different coefficient :

0-500 m
500-1000 m
1000-1500 m
etc...

It's too hard to manage...

Finally, for information, the AIDA lake coefficient will probabely be erase since January 1st, 2002.

Sincerely.

• Stephan Whelan
Thank´s

Thank´s for the correction Sébastien.
cjb, I appreciate your good intentions and I´ll make some investigation too.
Sincerely

Frank Pernett

equivalent oxygen depth?

I was thinking a lot about this problem. I know that the score system is based in the world records (herbert), but I dont know the altitude of the lake where Herbert made his fresh water record. I´m sure that the altitude afects the performance. And for the people (like me) who trains in high altitude lakes is important to know in what level we are. When we are using nitrox we converted our real depth to equivalent air depth to use the air tables. Why don´t we use the barometric pressure and the oxygen concentration to make a diferent score. for example:

The lake where I train is at 2600 meters over sea level, the barometric pressure is 560 mmHg and the O2 concentration is: 117,6 mmHg. The O2 concentration at sea level is 159,6 mmHg so the O2 concentration at altitude is 73,6% of Sea Level.
Can we use this value to calculate the score?
Any suggestion?

Frank Pernett

Equivalent Depth Formula

I think, I have been very inquisitive with this thread. But this affects my training.
I can't train in the sea so I have to develop a way to extrapolate my performance to the sea.
I invented a formula to calculate the score of constant weight in lake.
It's based on the barometric pressure.
This is the formula:

Equivalent Depth=(Dl/0.131 x Bp) x 100%

Where Dl is the Depth achieved in lake in meters
0.131 comes from 760 mmHg/100%
And Bp is the Barometric Pressure at the lake altitude in mmHg

The Equivalent Depth is in meters of sea water.

If we apply the formula the World record in CW in lake and sea, and fortunately the same freediver holds the records, we have:

ED=72 meters/(0.131 x 612.22)x 100
ED=89

As We can see It's not so far from his CW record in sea, so I think it can work.

What do you think, because the lake I train (Is more than I was thinking) is at 3020 meters osl (Pb=505.32mmHg)

sincerely

Frank Pernett

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:

 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:
 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 : 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:

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 :

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 :

 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 : 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

Last edited:
• fpernett
Extensive calculations