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The narcosis of trimix and the calculation of END   Author and copyright owner: Matti Anttila. Copying etc. prohibited without prior written permission by the author. Comments: matti@antti.la. Last update 12th of September, 2001. In case the reader of this page has comments, questions etc, I'm happy to receive them by e-mail or web form.

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Table of contents:  1.  What is this page for?  2.  Definitions  3.  Defining the gas  4.  Calculating the END  5.  Summary

!   Warning: Scuba diving is a safe hobby, but without proper training, equipment and attitude it may endanger yours and your dive buddy's life. The author takes NO responsibility about the information of this page.

Contents:

1. What is this page for?

2. Definitions

In this example a diver dives to 100 meters of fresh water (328 feet) using so called "ideal gas blend" (p.p.O₂=1.4 ATA and END = 40 m (131 feet), which means that p.p.N₂= 3.95 ATA).

Acronyms, abbreviations and constants: END = equivalent narcosis depth p.p. = partial pressure of gas F = fraction of gas P = pressure F Narc = fraction of narcotic gases (transformed to nitrogen narcosity) mfw = meters of fresh water (10 m is equal to 1 ATA change) ATA = absolute pressure in atmospheres Air consists of 21% of oxygen and 79% of nitrogen (real components: nitrogen N2: 78.08%, oxygen O2: 20.95%, argon Ar: 0.934%) Narcotic level of nitrogen and oxygen = 1.00, helium = 0.23. Density of fresh water is 1000.0 kg/m3 not depending of depth/tempereture etc. (in this example), and change of surrounding pressure is +1.00 ATA / 10 m of depth. Pressure on surface is 1.00 ATA. 1 f = 0.3048 m and 1 m = 3.281 f

3. Defining the gas

Fraction of oxygen:

FO₂ = p.p.O₂ / P = 1.4 ATA / 11 ATA = 0.12727 = 12.73 % O₂

Fraction of nitrogen:

p.p.N₂ at 40 m, using air = 0.79 x 5 ATA = 3.95 ATA p.p.N₂
100 m of depth: surrounding pressure is 11 ATA:
FN₂ = 3.95 ATA / 11 ATA = 0.3591 = 35.91 % N₂

Fraction of helium:

100 % - ( FO₂ + FN₂) = 100 % - 12.73 % - 35.91 % = 51.36 % He

Result:

Trimix 12.7 / 51.4 , and if the decimals are rounded: Trimix 12 51

4. Calculating the END

Method 1: Only nitrogen is narcotic

Formula (metric): END = [(FN₂ * (Depth + 10))/ 0.79 ] - 10

The gas determinition above was made this way, so END = 40 mfw.

Method 2: Nitrogen and oxygen are narcotic

There isn't really exact values or numbers to determine the ratio of oxygen's narcotic effects, but several sources claim it to be about as narcotic as nitrogen, ie.: The Physiology and Medicine of Diving" by Peter Bennett and David Elliott, 4th edition, 1993, W.B.Saunders Company Ltd, London. From here on, we use factor 1 for both nitrogen and oxygen:

F Narc = F N₂+ F O₂= 35.91% + 12.73% = 48.64%.

The following equation is common way to calculate the END, but it's wrong (look further to see the right one):
[With p.p.N₂ of 3.95 ATA, x (48.64% / 35.91%) = p.p.N₂ = 5.35 ATA (END = 57.7 mfw with air, which we get of: 10*((5.35/0.79)-1)]

NOTE! We have to take into account, that in equivalent narcosis (air) depth the oxygen have to be counted also from the gas that we compare to (air):

Formula: END = [((1 - F He) * surrounding_pressure * 10) - 10]

Then: 11 ATA x 48.64% (FN₂+FO₂) = p.p.(N₂+O₂) = 5.35 ATA (END = 43.5 mfw, which comes from: 10*((5.35/(0.79+0.21))-1).)

Method 3: Nitrogen and helium are narcotic

Bennett's book (mentioned above) determines the narcosis factor of helium 0.23 times of nitrogen. Using this value, we can count the sum effect of both nitrogen and helium:

F Narc = F N₂+ (0.23 x F He ) = 35.91 % + (0.23 x 51.36 %) = 35.91 % + 11.81 % = 47.72 %
Then p.p.N₂ 3.95 ATA x (47.72% / 35.91%) = p.p.N₂ 5.25 ATA (END = 56.4 mfw)
However, this is not realistic, since the oxygen component has to be counted also.

Method 4: Nitrogen, oxygen and helium are narcotic

F Narc = F N₂+ F O₂ + (0.23 x F He) = 35.91 % + 12.73% + (0.23 x 51.36 %) = 35.91 % + 12.73% + 11.81 % = 60.45%
Then: 11 ATA x 60.45% = p.p.(N₂+O₂+0.23He) 6.65 ATA (END = 56.5 mfw)
Note: Both oxygen and nitrogen were treated as narcotic gases also in the air that we compare this to (with factor of 1 for both of them), so 6.65 ATA can directly be transformed to depth of 56.5 mfw (instead of dividing it by 0.79).

Method 5: Mathematical way

We'll deteremine the END for gas mix: Trimix 12.7 / 51.4
(NOTE! We'll assume here, that our trimix is indeed ONLY oxygen, nitrogen and helium, and there are no traces of other compounds (such as argon from air) due possible top-up-fill with air etc.)

Air consists of: Nitrogen N₂: 78.08%, Oxygen O₂: 20.95%, Argon Ar: 0.934%
(NOTE! Normally this would be accurate enough: N₂: 79%, O₂: 21%)
The density of water in 4 °C temperature is 0.99997 g/cm³, but because of the compressibility of water as a liquid in high pressure, the density will be 0.99998 g/cm³ in 100 meters. However, this is really irrelevant, so we use the value 1 g/cm³.
10 m H₂O (mfw) = 98066.5 Pa, g = 9.81 m/s², 1 atm = 101325 Pa (Pa = Pascal = 1 N / m², N = Newton), air pressure on surface = 101.325 kPa

Argon (Ar): narcotic value = 2.33, helium (He) has a value 0.23. Both nitrogen (N₂) and oxygen (O₂) have the value 1.00.

Absolute pressure in 100 meters:

101.325 kPa + (100 m * 9806.65 Pa / m) = 1081.990 kPa = 10.678 ATA (Atmospheres of absolut pressure)

Partial pressures of the components of our trimix gas in 100 meters are:

Oxygen (p.p.O₂): 10.678 ATA * 12.7 % = 1.356 ATA
Nitrogen: (p.p.N₂): 10.678 ATA * 35.9 % = 3.833 ATA
Helium: (p.p.He): 10.678 ATA * 51.4 % = 5.488 ATA

Sum ( Narc.factor * p.p.Gas ) =
(1 * p.p.O₂) + (1 * p.p.N₂) + (0.23 * p.p.He) = 6.451

Sum ( Narc.factor * p.p.Gas ) =
(1 * p.p.O₂) + (1 * p.p.N₂) + (2.33 * p.p.Ar) =
0.2095 + 0.7808 + 0.0218 = 1.0121

6.451 / 1.0121 = 6.374

END = (6.374 - 1) * 10
END = 53.74 m

5. Summary

So, what is the END for this dive?

Hard to tell, really! For example Bennett describes the sum effect of nitrogen and oxygen to be stronger than these two would be by calculated separately and then added to the equation. There aren't exact values for the narcotic effects of oxygen, and ie. Richard Pyle says that the relation of the narcocity value of oxygen vs. nitrogen depends on their fraction in current gas mix.

What makes this even more uncertain, is the variation of narcosis depending on day, diver's feelings, conditions, stress, psychological factors etc.

Method #2 is widely used, but the main reason for this page really is to show the differencies of these methods. Too many people use only the method #1.

Summary of END values of methods shown here:

Method	END
Method 1: Nitrogen is narcotic	40 m
Method 2: Nitrogen and oxygen are narcotic	43.5
Method 3: Nitrogen and helium are narcotic	56.4 m
Method 4: Nitrogen, oxygen and helium are narcotic	56.5 m
Method 5: Mathematical way	53.74 m

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