Editing Equivalent air depth (section)

==Derivation of the formulas==
For a given nitrox mixture and a given depth, the equivalent air depth expresses the theoretical depth that would produce the same [[partial pressure]] of nitrogen if regular air (79% nitrogen) was used instead:
:<math>ppN_2(nitrox, depth) = ppN_2(air, EAD)</math>

Hence, following the definition of partial pressure:
:<math>FN_2(nitrox) \cdot P_{depth} = FN_2(air) \cdot P_{EAD} </math>

with <math>FN_2</math> expressing the fraction of nitrogen and <math>P_{depth}</math> expressing the pressure at the given depth. Solving for <math>P_{EAD}</math> then yields a general formula:
:<math>P_{EAD} = {FN_2(nitrox) \over FN_2(air)} \cdot P_{depth} </math>

In this formula, <math>P_{EAD}\,</math> and <math>P_{depth}\,</math> are absolute pressures. In practice, it is much more convenient to work with the equivalent columns of [[seawater]] depth, because the depth can be read off directly from the [[depth gauge]] or [[dive computer]]. The relationship between pressure and depth is governed by [[Pascal's law]]:
:<math> P_{depth} = P_{atmosphere} + \rho_{seawater} \cdot g \cdot h_{depth}\,</math>

Using the SI system with pressures expressed in [[pascal (unit)|pascal]], we have:
:<math> P_{depth}(Pa) = P_{atmosphere}(Pa) + \rho_{seawater} \cdot g \cdot h_{depth}(m)\,</math>

Expressing the pressures in [[atmosphere (unit)|atmospheres]] yields a convenient formula (1 atm ≡ 101325 Pa):
:<math> P_{depth}(atm) = 1 + \frac{\rho_{seawater} \cdot g \cdot h_{depth}}{P_{atmosphere}(Pa)} = 1 + \frac{1027 \cdot 9.8 \cdot h_{depth}}{101325}\ \approx 1 + \frac{h_{depth}(m)}{10}</math>

To simplify the algebra we will define <math>\frac{FN_2(nitrox)}{FN_2(air)} = R</math>. Combining the general formula and Pascal's law, we have:
:<math>1 + \frac{h_{EAD}}{10} = R \cdot (1 + \frac{h_{depth}}{10})</math>
so that
:<math>h_{EAD} = 10 \cdot (R + R \cdot \frac{h_{depth}}{10} - 1) = R \cdot (h_{depth} + 10) - 10</math>

Since <math>h(ft) \approx 3.3 \cdot h (m)\,</math>, the equivalent formula for the imperial system becomes
:<math>h_{EAD}(ft) = 3.3 \cdot \Bigl(R \cdot (\frac{h_{depth}(ft)}{3.3} + 10) - 10 \Bigr) = R \cdot (h_{depth}(ft) + 33) - 33</math>

Substituting R again, and noting that <math>FN_2(air) = 0.79</math>, we have the concrete formulas:

:<math>h_{EAD}(m) = \frac{FN_2(nitrox)}{0.79} \cdot (h_{depth}(m) + 10) - 10</math>

:<math>h_{EAD}(ft) = \frac{FN_2(nitrox)}{0.79} \cdot (h_{depth}(ft) + 33) - 33</math>