Nitrogen (N2), Oxygen (O2), Carbondioxid (CO2), Argon (Ar), Helium (He), ...
Index t (Index t) for tissue = actually this is not a real body tissue but rather a compartment!
That is, a mathematical entity which symbolises a group of various body tissues which share common properties,
namely the halftime (that is basically the inverse of the perfusion) and solubilities.
Index amb (index amb) for ambient
Relationship between partial pressures of various gases and the total ambient pressure Pamb = P
|P = ∑ pi = p1 + p2 + p3 + ...||(51)|
At sealevel this is approx.:
|P = ∑ pi = pO2 + pN2 + pRest = 1 Bar (ca.)|
In order to calculate the partial pressures pi we need the portion in the volume f (= fraction) of each of these gases:
|pi = Pamb * fi|
Example for regular air at sealevel (ca.):
pO2 = 1 Bar * 0,2
pN2 = 1 Bar * 0,8
1 m = 1,094 yard = 3,28 feet = 39,37 inches
1 foot = 0,3048 m = 12 inch
1 inch = 2,54 cm
1 fathom = 6 ft = 2 yard
3 feet = 1 yard = 0,91439 m
1 Atmosphere (atm) = 760 mm Hg (Torr) = 10,080 Meter Sea Water (msw) = 10,33227 m WS =
33,071 feet sea water (fsw) = 33,8995 feet fresh water = 14,6960 psi
1 fsw = 0,0306391 Bar
1 foot of pure water = 0,0298898 Bar
1 Atmosphere (atm) = 1,03323 kg/cm2 (bzw.: 1.033,23 cm H2O) = 1,01325 Bar (= 1013,25 mbar) =
760 Torr (mmHg) = 101,325 Kilopascal (kPa)
1 fsw = 22,9809 mm Hg
1 psi = 0,0689474 Bar
Relationship between a diving depth d, measured in a pressure unit for eg. [bar] or [fsw]
and the absolute ambient pressure Pamb and the pressure at sea level (p0):
Pamb = p0 + d
Pamb = p0 + d * const.
This constant is for conversion from units of length (diving depth) to units of pressure.
If you have the diving depth d in meters [m], there is approx. this relation with the pressure [Bar]:
const. = 0,0980665 [Bar/m], for fresh water and
const. = 0,100522 [Bar/m], for sea water, and thus for eg.:
Pamb = p0 + d / 10,197
Pamb = p0 + d / 9,948
References for the conversion factors, for e.g.:
, p. 575; , p. 457; and especially , p. 893! As well: , p. 354 - 358
The various citations of books here in this file, like [yxz] refer to:
the corresponding entry in our literature list here on divetable.info
|dPt(t)/dt = k[Palv(t) - Pt(t)]||(1a)|
Descriptively this formula means:
the pressure change over time dPt(t)/dt within a compartment is
dependant of the actual pressure gradient Δ P.
So the basic mathematical idea is simply like that:
∂P / ∂ t ∼ Δ P.
From this proportionality we make a real equation with the help of a simple constant; and the ∂s become real differentials:
dP(t)/dt = k * Δ P
The central idea of all so-called perfusion models is an instantaneous pressure equalization from the alveoli in the lung with the arterias!
I.e..: Palv(t) = Parterial(t) , for all t
|Pt(t)||Partial Pressure of an inert gas within a compartment at time t, in [Bar]|
|Palv(t)||Partial Pressure of an inert gas within the alveoli [Bar]|
|k||a constant, for each compartment [min-1]|
|t||time (as well dive time) [min]|
Step 1: We are separating the variables in (1a) and find the inhomogeneous differential equation (1b):
|dPt(t)/dt + kPt(t) = kPalv(t)||(1b)|
Step 2: The homogeneous equation (2):
|dPth(t)/dt + kPth(t) = 0||(2)|
For the solution we take the following approach: Pth(t) = C0e-λt
The subscript 'h' in Pth means the homogeneous equation. We put this into (2) and get:
-λC0e-λt + kC0e-λt = 0 => (k - λ)C0e-λt = 0,
only if k = λ or C0 = 0 (uninteresting: it means no change in pressure, i.e.: no dive),
and because e-λ*t ≠ 0 for all t, we take k = λ and thus getting the homogenous solution for (2):
|Pth(t) = C0e-kt.||(3)|
Step 3: Find a special solution of the inhomogeneous equation (1b) through the boundary conditions. We are taking two special situations concerning Palv(t).
1) Palv(t) = const., i.e. the depth as well is constant!
2) Palv(t) varies linearly in time, i.e. the ascent- and descent-ramps are constant, that is the ascent- and descent velocities remain constant.
constant depth = constant ambient pressure
In this case the alveolar inertgas pressure will take a constant value: Palv(t)=Palv0 . So the equation becomes (1b):
|dPt(t)/dt + kPt(t) = kPalv0||(4)|
We are taking the 'solution':
|Pt(t) = C0e-kt + C1||(5)|
we put (5) into (4) and the exponential terms are cancelling themselves:
- kC0e-kt + kC0e-kt + kC1 = kPalv0
thus we recieve: C1=Palv0.
We are searching another boundary condition in order to determine C0.
We assume a certain but constant partial pressure at start in the compartment: Pt(0) = Pt0 for time t=0. This means for eg. a certain reached saturation pressure or as well a starting pressure for mountain lake diving. We put his into equation (5):
Pt0=C0e-0 + Palv0
and thus follows: C0=[Pt0 - Palv0].
We will have then the Haldane equation for the partial pressure of an inert gas in a specific compartment k:
|Pt(t) = Palv0 + [Pt0-Palv0] e-kt||(6a)|
|Pt(t)||partial pressure of an inert gas within a compartment with the constant k [Bar]|
|Pt0||initial partial pressure of an inert gas within a compartment at time t=0 [Bar]|
|Palv0||the constant partial pressure of an inert gas within the alveoli [Bar], for t = 0 and therefore for all times t due to the boundary condition|
|k||a constant, dependant of compartment [min-1]|
(6a) is the famous Haldane equation, we re-arrange and arrive at (6b);we simply add at the left and right side Pt0 and group according Palv0 - Pt0:
|Pt(t) = Pt0 + [Palv0-Pt0] [1 - e-kt]||(6b)|
This is the equation which is regularly cited, e.g.:
Bühlmann,  p. 14, 1983;  p. 64, 1993;  p. 96, 2002,
with slightly changed indices.
For e.g. PI is the inspiratory inert gas pressure, this means the respiratory coefficient Rq (pls. cf. equation (13)) is implicitly = 1,
thus we have variances across different tables/models/researchers: have a look at the next chapter.
In  on p. 60 we could see this very clearly.
the ambient pressure changes linearly with time
During ascents and descents with constant velocity the inspiratory partial pressure of the inert gases changes linearly with time (with an open circuit SCUBA! With rebreathers (CCR = closed circuit rebreather) we have, according to make, other dependencies!). In equation (1) this implies:
Palv(t)=Palv0 + R*t.
Palv0 is the partial pressure of the inert gases from start, i.e.: at time t=0, and R is the rate of change (in Bar/minute) of the partial pressure of the inert gases in the alveoli. R is positiv for descent (increase of pressure) and negativ for ascent (decrease of pressure). We put it into equation (1b):
|dPt(t)/dt + k Pt(t) = k Palv0 + k R t||(7)|
We try the solution:
|Pt(t) = C0e-kt + C1 t + C2||(8)|
We plug solution (8) into equation (7) and recieve:-k C0e-kt + C1 + k C0e-kt + k C1 t + k C2 = k Palv0 + k R t
|[k C1 - k R] t + [ C1 + k C2 - k Palv0 ] = 0||(9)|
In order to find a solution for C1 and C2 which should be valid for all times t, we put the terms in the squared brackets in (9) equal to 0. Thus we have:
C1 = R and C2 = Palv0 - R/k. Thus we have the folowing:
|Pt(t) = C0e-kt + R t + Palv0 - R/k||(10)|
Once more we exploit the boundary condition Pt(0) = Pt0 for t=0 in order to calculate C0. We put this into (10) and get:
Pt0 = C0e-0 + Palv0 - R/k
and thus: C0 = Pt0 - Palv0 + R/k. As a definite solution we find:Pt(t) = Palv0 + R*t - R/k + [Pt0 - Palv0 + R/k] e-kt
|Pt(t) = Palv0 + R * [t - 1/k] - [Palv0 - Pt0 - R/k] e-kt||(11)|
|Pt(t)||partial pressures of the inert gases within compartment [Bar]|
|Pt0||initial partial pressures of the inert gases in compartment at time t=0 [Bar]|
|Palv0||initial (alveolar) partial pressure at time t=0 [Bar]|
|k||a constant, dependant of compartment|
|R||rate of change of the partial pressure of an inert gas in the alveoli (Bar/min) R = f * Ramb, f is the fraction, i.e.: volume part of inert gase and Ramb is the rate of change of the ambient pressure|
the so-called Schreiner equation.
By putting the rate of change R = 0 (constant diving depth) the Schreiner equation (11) becomes again the Haldane equation (6a).
The variable τ (a greek Tau) is designated as a 'Half Time' (HT) and is specific for each compartment:
-k * τ = ln(1/2) = -ln(2)
The relation between k and the HT is:
τ = ln(2) / k
resp.: k = ln(2) / τ
The stringent physiological relation between the HT and the compartments is through the solubility and the perfusion (blood supply):
|τ = 0,693 * αti / (αbl * dQ/dt)||(12a)|
with the following definitions:
αti = solubility of an inert gas in tissue,
ml(S)gas * mlti -1 * (100 kPa) -1
αbl = solubility of an inert gas in blood, ml(S)gas * mlblood -1 * (100 kPa) -1
dQ/dt = compartment perfusion, mlblood * mlti -1 * min -1
pls. cf. further down for a rationale of this relationship
we want to determine the alveolar partial pressure Palv more exactly. The gas composition and thus the partial pressures of the component gases are dependant of the ambient pressure Pamb, water vapour, carbon dioxid, and the other components of the breathing mix, i.e. the chemical composition.
The ambient pressure Pamb is the sum of the atmospherical pressure (ca. 1 Bar at sea level) + statical pressure of the water column (from the diving depth), thus it is increased ca. 1 Bar per every 10 m increase in depth (not considering the differences between fresh and sea water ...). The ambient pressure and the absolute pressure within the lung have to be (roughly) equal, only through in- and exhalation there is a small pressure difference of approx. and up to max. 30 cm of water column.
The partial pressure of an inert gas in the lung Palv can be roughly estimated through:
The relation of the oxygen consumption to the carbon dioxid production is designated with the
respiratory quotient Rq.
Rq is the proportion in volumes of carbondioxid production to oxygen consumption in ml, average values are:
200 ml carbondioxid production / oxygen consumption 250 ml per minute, i.e. ca.:
Rq = 200 / 250 = 0,8
Rq is dependant on nutrition and on the workload:
We have Rq = 0.7 for metabolism mainly on fat, ca. 0.8 for proteins, ca. 0.9 – 1 with carbohydrates;
With average workload and nutrition we have 0,8 -> 0,85, with CO2 retainers < 0,7
and > 1,0 with heavy workloads or acidosis/enrichment with lactic acid.
The equation of the alveolar ventilation describes this:
Palv=[Pamb - PH2O -
PCO2 + ΔPO2] * f
|Palv=[Pamb - PH2O + (1 - Rq)/Rq * PCO2 ] * f||(13)|
|Palv||partial pressure of the inert gas in the alveoli [Bar]|
|Pamb||ambient pressure, i.e. absolute pressure of breathing mix [Bar]|
|PH2O||water vapor partial pressure, at 37 degrees of Celsius ca. 0.0627 Bar (47 mm Hg)|
|PCO2||carbon dioxid partial pressure, ca. 0.0534 Bar (40 mm Hg)|
|ΔPO2||Delta = change of oxygen partial pressure through metabolism / ventilation in the lung|
|Rq||the respiratory quotient|
|f||fraction of inert gases in the breathing mx; e.g.: N2 in dry air = ca. 0.78. Normally for diving we put f = 0,79xx, i.e.: we take into account the various trace gases|
Across the various models/researchers we have as well various settings for the Rq,
even if not explicitly stated:
Schreiner: Rq = 0.8
Workman: Rq = 0.9
Buehlmann / Hahn: Rq = 1.0
2 * 0,78 = 1,56, i.e.: 1,56 : 1 (elucidation from Workman)
and is a constant for all compartments.
The linear Workman equation, is a simple straight line in a pressure coordinate system. It looks like that for each compartment:
|M = M0 + ΔM * d||(14)|
|M||M-Value, maximal partial pressure of an inert gas in a compartment [fsw]|
|M0||M-0 Value, for sea level or diving depth = 0 ft for each compartment [fsw]|
|ΔM||Delta M, increase of M per each foot of diving depth, defined for each compartment [fsw/ft]|
|d||diving depth [ft]|
We have compiled all these coefficients in a separate file so that you can copy-&-paste it directly to your own DIY deco procedures.
Bottom Line: Workman had the M-Values decreasing with increasing HT, thus the fast compartments (short HT) can tolerate a higher inert gas overpressure than the slower compartments (longer HT).
With equation (14) we could determine the minimal depth dmin where the diver has to stay during his/her deco stops. The calculation we have to do for each of the compartments and is dependant of the topical inert gas overpressures:
|dmin = (Pt - M0) / ΔM||(15)|
So each compartment has got its own saturation and thus its own minimal depth. We simply have to take the biggest value (greatest depth) out of the family of these values.Between the M-values and the HT we have approximately the following empirical relation according to Workman:
|M = 152.7 τ -1/4 + 3.25 d τ -1/4 = M0 + ΔM d||(16)|
According to Hempleman we have an approximate empirical relation between the no-decompression limits tNDL [min] and the diving depth d [fsw]:
|d * tNDL1/2 = 475 fsw min1/2||(16.1)|
The PADI/DSAT RDP table has been generated with such a formula, but with a couple of more conservative modifications:
|d - A = C * tNDL- x||(16.2)|
Buehlmann offers, very like Workman, a linear relationship between the tolerated tissue overpressure and ambient pressure (, p. 117):
|Pt.tol.ig = Pamb / b + a||(17)|
|Pt.tol.ig||tolerated inert gas pressure, for each compartment, (analogous M) [Bar],
sum of all partial pressures of the present inert gases
|a||boundary value at a theoretical ambient pressure of 0 Bar, i.e. the axis intercept [Bar]|
|Pamb||ambient pressure, absolute pressure of the breathing mix [Bar]|
|b||1/b pressure gradient: value of increase per pressure unit depth (dimensionless), i.e. the slope of the straight line|
Basically there is no difference between Workman and Buehlmann / Hahn exept that the M-values are referenced to the ambient pressure at sea level whereas Buehlmann / Hahn are extrapolating against 0 Bar ambient pressure and thus reaching the region between 1 and 0 Bar automatically, i.e. mountain lake diving. You could convert both systems into eachother:
|Pt.tol.ig = M|
|ΔM = const. * 1/b||(18)|
|M0 = a + p0/b|
According to Buehlmann and co-workers (Keller et al ...)
there is the following empirical relationsship
between the a- and b- coefficients to the HT.
This one is valid only for nitrogen. (, p. 129)!
For helium it looks a little bit different ... (pls. cf. , S. 131).
|a = 2 Bar * τ-1/3|
|b = 1.005 - τ-1/2||(19)|
This is the so-called set "A"; further on there is a set called
"B" (more conservative, for printed tables) and as well a "C" set for on-line calculations
with a diving computer (, p. 158).
The ratio of the HT of two immediately sequenced compartmens is roughly 1.2 to 1.4.
(Workman for eg. puts this to another value: 5 * 2n, n being the compartment no. ...)
This is not a strict physiological law but repects the desire to have a narrowly cramped network of the HT. Only the very fast and the extremly slow HT are representing compartments which have been verified through a lot of experiments on humans in the pressure chamber in Zuerich in the 60's. Nevertheless there is a coarse one-to-one mapping of HT with "real" body tissues from p. 115 in :
Pamb, tol = ( Pt, ig - a ) * b
DCAP (Decompression Computation and Analysis Program) is using the M11F6 M-Values,
developped by Bill Hamilton et al. for the swedish navy.
The 11 HT for nitrogen are from 5 - 670 min., and therefor these for helium: 5 - 240 min.
The PADI RDPTM coeffcient set uses the 14 M-Values developped by Raymond E. Rogers and Michael R. Powell, Diving Science and Technology Corp (DSAT), With this model, YOU KNOW IT, you are restricted to only NDL dives! There are no validated data outside the recreational NDL enveloppe! I.e.: these coefficients are NOT for decompression diving and as well not for mountain lakes!
|P1 = Pambient - 8 / π2 * (Pambient - P0) * (e-k*t + 1/9 * e-9*k*t + 1/25 * e-25*k*t)|
r = 27.5714 / P1 + 12.407
r = Supersaturation Ratio, 1.6 < r < 1.9
P1 in Bar
k = 0.007928
This is a special solution of the 2nd Fick equation (yes, yes: that was really the guys name ...)
For the diffusion coefficient D we have:
k = D * π2 / 4 * b2, b is the thickness of the slab.
Das DCIEM Kidd-Stubbs model:
DCIEM = Defence and Civil Institute of Environmental Medicine, Canada:
In this model we have 4 compartments in series , i = 1 - 5.
This is the model of Kidd and Stubbs from 1962 resp. the "KS-1971".
The DCIEM tables from 1983 (R.Y. Nishi and G.R. Lauckner) a relying on it.
|dPi / dt = A * ((B + Pi-1 + Pi) * (Pi-1 - Pi) - (B + Pi + Pi+1)(Pi - Pi+1))|
with: P0 = Pambient
and: Pi = P in the compartment i
and: P5 = 0
A = 0.0002596
B = 83.67
A and B are the so-called "flow constants" for air, in msw
The HT themselves are defined via the A- and B-coefficients and pressure- resp. time-dependant.
The asymmetry between inert gas absorption (faster) and -release (slower) can be seen here:
T1/2 = ( ln ( 2 - ΔP / (B + Pi + Pf)))/(A(B + 2Pf))
with: Pi initial pressure
mit: Pf finale pressure
ΔP = Pf - Pi
with: SAD = Safe Ascent Depth
SAD = Pmax, Komp.i / 1.8 - 10.06
for the DCIEM table we have the following:
SAD = Pmax, Komp.i / R - OFF - Psealevel
i = 1, R = 1.300 and OFF = 4.8
i = 2, R = 1.385 and OFF = 2.5
Psealevel = 10.06 msw
and for i = 3; 4; 1/R = 0.0 and, as well OFF = 0.0
|e- k * t = [ (Pt(t) - Palv0) / (Pt0 - Palv0) ]||(20)|
The fraction may be the expression in the squared brackets [ ... ]. We put the equation to logarithms:
ln [...] = ln ( e- k * t ) = - k t
We are solving for t:
t = - 1 / k * ln [ ... ]
and from (12) with k = ln(2) / τ we have the times t in question with:
|t = - τ / ln2 * ln[ (Pt(t) - Palv0) / (Pt0 - Palv0) ]||(21)|
With (21) and reasonable assumptions for the each of the pressure terms we are able to calculate now:
Buehlmann offers in , on p. 119:
Pt(t) = Pt, He(t) + Pt, N2(t)
with the a- and b- coefficients being normalized with the respective partial pressures in each compartment
(pls. cf.  on p. 86):
So for each compartment at any time t and for each combination of a- & b-values we have:
|a (He + N2) = [( Pt, He * aHe ) + ( Pt, N2 * aN2)] / ( Pt, He + Pt, N2 )|
|b (He + N2) = [( Pt, He * bHe ) + ( Pt, N2 * bN2)] / ( Pt, He + Pt, N2 )||(22)|
(pls. cf. the examples in , p. 27 and as well in , p. 80.)
These are just basic remarks found in the sources and no one offers the full blown theory behind it.
|a -> a * GF and as well b -> b / (GF - GF * b + b)||( 22.a )|
For all GF per compartment we have: 0 < GF ≤ 1
by calculating more conservatively.
If calculated more aggressively these GF become: GF > 1. Thus we end up at the VGM method, which allows this.
The conventional desktop deco softwares permit the input of 2 gradient factors: GF Hi and GF Lo.
GF Hi (= High) means the reduction of the leading M-value, or, as you would have it, the respective a- & b values pair, and thus resulting in a prolonged, last shallow stop.
GF Lo (= Low) results in a deeper, first stop.
With a simple linear relationsship you can use the GFs automated over all required deco stages:
with: GF Hi > GF Lo
GFm = ( GF Hi - GF Lo ) / first stop depth
GF = GF Hi - GFm * actual stop depth
|( 22.b )|
Within the VGM (Variable Gradient Method) framework you could, instead of using ONE PAIR of
GF Hi/Lo for every compartment,
use a different pair of GF Hi/Lo individually per EACH compartment.
More and other sources in the german version of:
r1min = 1 / [ (1/r0min) + (P1 - P0)/
2 * (γc - γ) ]
r(tr) = r1min + ( r0min - r1min) * [1 - exp(-tr / τr) ]
Pssmin = 2 * ( γ / γc ) * ( γc - γ ) / r(tr)
Pssnew = [ b + ( b2 - 4 * c )1/2 ] / 2
b = Pssmin + λ * γ / [ γc * (tD + H / ln(2)) ]
c = ( γ / γc )2 * λ * (P1 - P0) / (tD + H / ln(2))
|(Eq. 1 -> 4c)
on p. 150 - 151
These 5 "free" parameters have been established through numerical "best fits" of the TTS (time-to-surface) from the USN- and the RNPL-Tables:
γc = 257 dyn / cm ( = mN/m)
γ = 17,9 dyn / cm ( = mN/m)
r0min = 0,8 μm
τr = 20160 min (= 14 days)
λ = 7500 feet * min (ca. 227,3 Bar * min)
|(on p. 151 & 155)|
creative labels for it:
This model you will find in all RGBM-simulators like:
GAP, ABYSS, etc.-softwares and as well in all MARES and SUUNTO computers.
This is accomplished with the so-called "reduction factors" f. In plain language:
the 16 ZH-L compartments and the corresponding HT with the central equation (17) is still used.
With these f, ( f < 1), we will have (17):
|P min = ( p - af ) * bf|
af = a * f
bf = b / f * ( 1 - b ) + b
f is dependant on the HT and defined only for air/nitrox for a HT > 180 min.
On closer inspection this is just identical with (22.a) )
As well all these factors which are missing up to now in the standard ZH-L model
are ostensibly keyed in:
|f = (1 - f 0 ) * τ / 180 + f 0|
f < 1
τ > 180 min.
For the factors: repetitive dives (rp), deeper than previous, i.e. reversed profiles (dp) and multiday-non-limit diving (dy) there is the following for f0:
|f 0 = 0.45 * frp + 0.30 * f dp + 0.25 * f dy|
rp = repetitive
dp = deeper than previous (reversed profile), i.e. shallow first
dy = multiday, over a time span > 30 h
For these factors there are the following relationships:
|frp = 1 - 0.45 exp [ - (tsur-ηrp)2 / 4 * ηrp2||(23)|
|fdp = 1 - 0.45 * [ 1 - exp (- ΔPmax / Pmax)] * exp [ - (tsur - ηdp)2 / 4 * ηdp2|
|fdy = 0.70 + 0.30 exp ( - n / ηdy)|
|10 min. < ηrp < 90 min.|
|30 min. < ηdp < 120 min.|
|12 h. < ηdy < 18 h.|
|tsur||Surface Interval (SI) in [min.]|
|ΔPmax||the maximal pressure difference for reversed profiles|
|Pmax||maximal ambient pressure|
|n||1. multi day-diving frequency within 24 h, p.36|
|2. number of diving-days within 30 h, p. 37|
1) the factor f dy somehow fluctuates between 0.25 and 1.0 ... (p. 36)
2) n is obviously not defined soundly, as well the units (1/time or dimensionless)
do not fit to the units of η !!!)
For helium we have analogous:
|f = (1 - f 0 ) * τ / 67.8 + f 0|
f < 1
τ > 67.8 min.
For trimix we have for the corresponding partial pressures: fO2 + fN2 + fHe = 1.
For the complete inert gas pressure Π in the compartments we have:
|Π = ( Pamb, N2 + Pamb, He ) + ( Pt, N2 - Pamb, N2) * e(- kN2 * t) + ( Pt, He - Pamb, He ) * e(- kHe * t)|
Now, however, the fractions f for the breathing gas are used for normalization: af, N2 and af, He and as well bf, N2 and bf, He in order to calculate the critical a, b - coefficients for the trimix:
Af = (fN2 * af, N2 + fHe * af, He) / ( fN2 + fHe )
and, as well:
Bf = (fN2 * bf, N2 + fHe * bf, He) / ( fN2 + fHe )
and thus the critical pressure:
Pmin = ( Π - Af ) * Bf
1) the formula for Π on p. 39 is wrong, but here we have it corrected!
2) Π is being calculated obviously in a different manner as per (22). The question is: what happens during a change of the deco mix with pure oxygen?
the fractions of He and N2 become identical to 0 in the breathing gas, und thus Af = 0
and as well Bf = 0 and thus also Pmin = 0 ... ???)
|v = fO2 * P - 2.04 * ( 1 - fO2 ) - 5.47||(24)|
But the oxygen window is non-linear, it reaches its maximum opening at ca. 1600 mm Hg! And stays there ...)
How is now the physiological relationship between k and the compartments, i.e. the various rates of perfusion and the various solubilities?
A very simple mass-balance for an inert gas within a compartment of tissue, solely feed by blood perfusion:
|N2stored = N2in - N2out|
The inert gas being transported via arterial blood to the tissue and leaving this region with the venous return.
Hereby it is assumed that the arterial (pNa) and the alveolar inert gas partial pressures (pNA) are equal, and that the
diffusion between adjacent capillaries is instantaneously, i.e.: as well the venous (pNv) and the compartment partial pressure (pNt) are equal for this particular inert gas.
The solubility coefficients for this inert gas for blood (b) and tissue (t) are: αb
and as well αt.
Q is the blood flow, Vt is the tissue volume. Starting at time t = 0 it is assumed that pNa reaches immediately a constant value, pa.
The rate of change of the pressure in the compartment dictates the rate of change for storage in this compartment.
The little drawing may help. We have the following:
|αt * Vt * dpt/dt = αb * dQ/dt * pNa - αb * dQ/dt * pNv||(25)|
|with: dpt/dt + k * pt = k * pa|
|as well: k = αb * dQ/dt / ( αt * Vt )|
The solution of (25) is either (6a) or (6b). Thus we have in the end the relationsship between the HT and the tissue perfusion and the solubilities:
|τ = 0.693 / k = 0.693 / ( αb * dQ/dt / αt * Vt )||(26)|
The total partial pressure over time ( pt(t) ) is the sum of of the decrease of p0 and the increase of the alveolar pressure pNA:
The small stirrer is symbolizing that we have a "well stirred" compartment
EE means: exponential for saturation AND desaturation
LE means: exponential for saturation, linear for desaturation
LEM means: exponential for saturation, linear for desaturation, M for multigas
At first the arterial inert gas partialpressure pa is corrected:
|pa,N2 = Pamb - PI,O2 -1.5||(30)|
|pa,N2 = arterial inert gas partialpressure|
|Pamb = absolut, i.e. ambient pressure|
|PI,O2 = inspiratory oxygen partialpressure (in FSW)|
|1.5 = arterial CO2 partial pressure in FSW (35 mmHg)|
Starting from equation (6a) a linear behavior is defined for the desaturation.
This linear desaturation is being calculated until the "exponential cross over point" is reached,
after that we have the usual exponential behavior.
Is for any time in any compartment during the desaturation-phase the inert gas-
partial pressure greater than the ambient pressure the following linear equation has to be used:
|pT,N2 = PTi,N2 + (2.8 - PI,O2) * K * T||(31)|
|pT,N2 = inert gas partialpressure in compartment at time T|
|PTi,N2 = initial inert gas partialpressure in compartment|
|T = time on actual depth stage|
|2.8 = sum of venous O2 (46 mmHg) and CO2 (53 mmHg) minus arterial CO2 (35 mmHg) = 64 mmHg in FSW|
|pT,N2 ≥ Pamb - 4.3||(32)|
The criteria for safe ascent are no longer called M-Values but instead:
MPTT = Maximum Permissible Tissue Tensions.
You will find various MPTT-tables there....
The calibration for mixed gases and as well the adaption via the NMRI model is taking place according to further parameters of gases in the blood.