Doesn't that kind of go against decompression theory? The slowest ascent you can make is a stop. And don't people make stops to off-gas?
I get the whole idea of different tissue loads and unloads at different rates. But the ones that are slow to load and unload also won't have as much nitrogen in them.
My understanding is that you can't add a lower partial pressure of nitrogen that has a higher partial pressure. I can go along with the instant you make your ascent you may take on a little more but that should quickly reverse. So my thinking is that you can't ascend too slow unless you only have a couple seconds before your no-deco time is up and you can't get to a shallower depth before you go into deco.
I really want to understand this as it is very interesting to me. So, please don't think I am trolling or starting an argument.
As mentioned in previous posts some slower tissue compartments may still on-gas after ascending from depth. Here is the mathematical example you requested.
P = Po + (Pi - Po)(1 - 2^(-t/HT))
P = final tissue compartment pressure
Po = initial tissue compartment pressure
Pi = Inspired air pressure
t = time at depth (or surface)
HT = tissue compartment half-time
From the Buhlmann ZH-L16 chart in Erik C. Baker's "Understanding M Values" the half-times from compartments 2 and 10 are:
2 = 8.0 minutes
10 = 146 minutes
Using the constant depth version of the Schreiner equation (shown above) we can calculate the P values for the following dive profile: descend to 99 ft for 15 minutes, then ascend to 33 ft for 15 minutes. I will work out the solution using minutes for t and atmospheres absolute for pressure. For this example I will neglect loadings on descent and ascent and assume an instantaneous depth change. To keep things simple I will ignore the water vapor pressure and make the inspired pressure equal to ambient pressure. P(n) = pressure for tissue compartment n.
At 99 ft Po = 1 atm (we were at the surface), Pi = the pressure at 99 ft. Solving for P:
P(2) = 1 + (4 - 1)(1 - 2^(-15/8)) = 1 + 5(1 - 0.27) = 1 + 3.65 = 4.65 atm
P(10) = 1 + (4 - 1)(1 - 2^(-15/146)) = 1 + 5(1 - 0.93) = 1 + 0.35 = 1.35 atm
Ascend to 33 ft for 15 minutes. Po becomes P from the previous segment of the dive. Pi = the pressure at 33 ft. Solving for P:
P(2) = 4.65 + (2 - 4.65)(1 - 2^(-15/8)) = 4.65 - 2.65(1 - 0.27) = 4.65 - 1.93 = 2.72 atm
P(10) = 1.35 + (2 - 1.35)(1 - 2^(-15/146)) = 1.35 + 0.65(1 - 0.93) = 1.35 + 0.11 = 1.46 atm
After ascending to 33 ft from 99 ft compartment 2 off-gassed from 4.65 to 2.72 atm. However, compartment 10, a much slower tissue compartment is still on-gassing from 1.35 to 1.46 atm.
