Equivalent Narcotic Depth
(END) Calculator
meters feet
What is END or Equivalent Narcotic Depth?
END, or Equivalent Narcotic Depth, is used for gas blends containing Helium (and potentially Hydrogen): It describes, while breathing a certain gas blend at a certain depth, at which depth air would have the same narcotic potential.
As Helium doesn't have any narcotic potential (or at least very limited), adding some in a gas blend (trimix) reduces the overall narcotic potential of the blend, allowing the diver to go deeper while reducing the effect of narcosis. Completely replacing the nitrogen with Helium (heliox) will diminish the narcotic potential of the blend even further. There are two schools of thought: those who think that Oxygen isn't narcotic, and that only the nitrogen part should be taken into account; and those who think Oxygen is a narcotic and should be treated as such. Because opinions vary, we (and a lot of other calculators) let the user decide if they want to consider oxygen as a narcotic or not.
To give an example, imagine we consider Oxygen to not be Narcotic, and we are breathing a 19% Oxygen, 50% Helium, 31% Nitrogen mix at 50 meters. We are subjected to 0.31×6 bar = 1.86 bar ppN2. If we were breathing air, we would need to be at about 13.5m to experience a ppN2 of 1.86 bar. This means that the Equivalent Narcotic Depth of this specific mix at 50 meters is 13.5 meters. If we consider Oxygen to be narcotic, then at 50 meters, the mix would have a narcotic potential of (0.19+0.31)×6 bar = 3 bar; in this case, the END would be 20 meters as we consider the total pressure of air.
Formulas and calculating the END
Metric
To find the END in a trimix blend using meters, you can use the following formula if you consider O2 narcotic:
\(END =(1 - f_{\text{He}}) \cdot (d+10) -10 \)And if you consider O2 not narcotic:
\(END =(1 - f_{O_2} - f_{\text{He}}) \cdot \frac{(d+10)}{0.79} -10 \)To find the END in a Hydreliox blend using meters, you can use the following formula if you consider O2 narcotic:
\(END =(1 - f_{\text{He}} - 0.4f_{H_2}) \cdot (d+10) -10 \)And if you consider O2 not narcotic:
\(END =(1 - f_{O_2} - f_{\text{He}} - 0.4f_{H_2}) \cdot \frac{(d+10)}{0.79} -10 \) \( f_{\text{He}}: \text{fraction of Helium present in the mix (value between 0-1)} \) \( f_{H_2}: \text{fraction of Hydrogen present in the mix (value between 0-1)} \) \( f_{O_2}: \text{fraction of Oxygen present in the mix (value between 0-1)} \) \( d: \text{depth (in meters)} \)Imperial
To find the END in a trimix blend using feet, you can use the following formula if you consider O2 narcotic:
\(END =(1 - f_{\text{He}}) \cdot (d+33) -33 \)And if you consider O2 not narcotic:
\(END =(1 - f_{O_2} - f_{\text{He}}) \cdot \frac{(d+33)}{0.79} -33 \)To find the END in a Hydreliox blend using meters, you can use the following formula if you consider O2 narcotic:
\(END =(1 - f_{\text{He}} - 0.4f_{H_2}) \cdot (d+33) -33 \)And if you consider O2 not narcotic:
\(END =(1 - f_{O_2} - f_{\text{He}} - 0.4f_{H_2}) \cdot \frac{(d+33)}{0.79} -33 \)Common blends and their END
Here is a table of three trimix blends:
- 21% O2 and 35% He
- 18% O2 and 45% He
- 15% O2 and 55% He
and their respective END at various depths if O2 is considered narcotic (in red) and if O2 is not considered narcotic (in green):
| 21/35 | 18/45 | 15/55 | ||||
|---|---|---|---|---|---|---|
| 18m | 8m | 6m | 5m | 3m | 3m | 1m |
| 30m | 16m | 12m | 12m | 9m | 8m | 5m | 40m | 23m | 18m | 18m | 14m | 13m | 9m | 50m | 29m | 23m | 23m | 18m | 17m | 13m | 60m | 36m | 29m | 29m | 23m | 22m | 17m |
Here is a table of three hydreliox blends:
- 2% O2, 49% He, and 49% H2
- 2% O2, 24% He, and 74% H2
- 2% O2, 80% He, and 18% H2
and their respective END at various depths if O2 is considered narcotic (in red) and if O2 is not considered narcotic (in green):
| 2/49/49 | 2/24/74 | 2/80/18 | ||||
|---|---|---|---|---|---|---|
| 100m | 25m | 31m | 41m | 51m | 4m | 5m |
| 200m | 56m | 68m | 87m | 108m | 17m | 19m |
| 300m | 87m | 105m | 133m | 164m | 30m | 32m |
Using the END to plan a dive
Using the END to plan a dive is useful for deep dives, when narcosis can impair good judgment at depth, and the diver wants to stay lucid at depth. For example, say a diver is planning to go to a depth of 55m, but doesn't want to exceed a narcotic depth of 30m, he can use the formulas above to calculate how much helium he will need to add to his mix (assuming he considers O2 to be narcotic):
\(END =(1 - f_{\text{He}}) \cdot (d+10) -10 \) \(\Leftrightarrow 30 =(1 - f_{\text{He}}) \cdot (55+10) -10\) \(\Leftrightarrow 30+10 =(1 - f_{\text{He}}) \cdot 65\) \(\Leftrightarrow \frac{40}{65} =1 - f_{\text{He}}\) \(\Leftrightarrow f_{\text{He}} =1 - \frac{40}{65}\) \(\Leftrightarrow f_{\text{He}} \approx 0.38\)In this case, the diver can plan his dive by using a 21/38 trimix to respect his chosen equivalent narcotic depth.
Why include hydrogen?
Hydrogen is often not talked about when talking about reducing the narcotic effect of a gas, because it is rather used as a substitute (or as a complement) for helium on super-deep dives to reduce or slow the onset of HPNS. Nevertheless, Hydrogen is still narcotic, although less than nitrogen: according to researchers, Hydrogen has a Relative Narcotic potency of 0.6 (compared to Nitrogen, which has an RNP of 1). I believe that as diving on hydrogen may become more and more popular, it is important for anyone to be able to know the END of the gas they plan on using.
Deriving the general formula
To calculate the END of any mix at any depth, we can start by expressing the narcotic potential N of a gas mix at a depth d as a function of the fraction and narcotic potential of the individual gases that make it up
\(N_{\text{mix}}(d) =(\frac{d}{10}+1) \cdot \sum {(f_{\text{gas}} N_{\text{gas}})} \)The narcotic potential of the gases used in this context is:
| O2 | H2 | N2 | He |
|---|---|---|---|
| 1 if narcotic, 0 if not | 0.6 | 1 | 0 |
By definition, we have:
\(N_{\text{mix}}(d) =N_{\text{air}}(\text{END}) \)which gives us, for a generalized mix containing Oxygen, Nitrogen, Helium, and Hydrogen:
\[ (f_{O_2}N_{O_2} + f_{N_2}N_{N_2} + f_{H_2}N_{H_2} + f_{\text{He}}N_{\text{He}}) \cdot (\frac{d}{10}+1) = (f_{O_2}^{\text{air}}N_{O_2} + f_{N_2}^{\text{air}}N_{N_2}) \cdot (\frac{END}{10}+1)\]We can simplify the values that stay constant:
\(N_{N_2} = 1 \\ N_{\text{He}} = 0 \\ N_{H_2} = 0.6 \\ f_{O_2}^{\text{air}} = 0.21 \\ f_{N_2}^{\text{air}} = 0.79 \)As NO2 changes depending on whether Oxygen is considered narcotic or not, we cannot simplify that yet.
\[ (f_{O_2}N_{O_2} + f_{N_2} \cdot 1 + f_{H_2} \cdot 0.6 + f_{\text{He}}\cdot 0) \cdot (\frac{d}{10}+1) = (0.21 \cdot N_{O_2} + 0.79 \cdot 1) \cdot (\frac{END}{10}+1)\] \[\Leftrightarrow (f_{O_2}N_{O_2} + f_{N_2} + 0.6f_{H_2}) \cdot (\frac{d}{10}+1) = (0.21 \cdot N_{O_2} + 0.79) \cdot (\frac{END}{10}+1)\]The goal is now to isolate END:
\[\Leftrightarrow \frac{(f_{O_2}N_{O_2} + f_{N_2} + 0.6f_{H_2})}{(0.21 \cdot N_{O_2} + 0.79)} \cdot (\frac{d}{10}+1) = \frac{END}{10}+1\] \[\Leftrightarrow \frac{(f_{O_2}N_{O_2} + f_{N_2} + 0.6f_{H_2})}{(0.21 \cdot N_{O_2} + 0.79)} \cdot (\frac{d}{10}+1) -1= \frac{END}{10}\] \[\Leftrightarrow \frac{(f_{O_2}N_{O_2} + f_{N_2} + 0.6f_{H_2})}{(0.21 \cdot N_{O_2} + 0.79)} \cdot (d+10) -10= END\]As we want to find the END as a function of Oxygen, Helium, and Hydrogen fraction, we can substitute the nitrogen fraction, knowing:
\[f_{N_2} = 1 - f_{O_2} - f_{H_2} - f_{\text{He}}\]We then have:
\[\frac{(f_{O_2}N_{O_2} + 1 - f_{O_2} - f_{H_2} - f_{\text{He}} + 0.6f_{H_2})}{(0.21 \cdot N_{O_2} + 0.79)} \cdot (d+10) -10= END\] \[\Leftrightarrow \frac{(1 + (f_{O_2}N_{O_2} - f_{O_2}) - f_{\text{He}} - 0.4f_{H_2})}{(0.21 \cdot N_{O_2} + 0.79)} \cdot (d+10) -10= END\]We now have to possibilities, if we consider O2 Narcotic (NO2 = 1):
\[END = \frac{(1 + (f_{O_2} \cdot 1 - f_{O_2}) - f_{\text{He}} - 0.4f_{H_2})}{(0.21 \cdot 1 + 0.79)} \cdot (d+10) -10\] \[\Leftrightarrow END = \frac{(1 - f_{\text{He}} - 0.4f_{H_2})}{(1)} \cdot (d+10) -10\] \[\Leftrightarrow END = (1 - f_{\text{He}} - 0.4f_{H_2}) \cdot (d+10) -10\]And if we consider O2 not Narcotic (NO2 = 0):
\[END = \frac{(1 + (f_{O_2} \cdot 0 - f_{O_2}) - f_{\text{He}} - 0.4f_{H_2})}{(0.21 \cdot 0 + 0.79)} \cdot (d+10) -10\] \[END = \frac{(1 - f_{O_2} - f_{\text{He}} - 0.4f_{H_2})}{0.79} \cdot (d+10) -10\]All the calculations were done in metric. If you prefer using imperial (feet) instead, simply change the 10 by 33.