Arie0510
Registered
Educated as a mathematician, and being an enthousiastic scuba diver for 5 years now, I am interesting in reconstructing the numbers in the RDP. Mainly just out of curiosity, true to my education, but also to better understand the tables I am teaching my students as a DM.
Via this forum I found a reference to the paper [1], where it was explained that for deriving the RDP a Haldane model was used, which can be described by the ordinary differential equation
dP(t)/dt = -k ( P_I(t) - P(t) )
where P(t) denotes the partial pressure of the gas in the compartment, P_I(t) the ambient partial pressure of the gas, and k = ln(2) / t_{0.5} is a constant with t_{0.5} denoting the half time of the compartment.
Furthermore, the following no-stop relation
D - A = C * t^(-x)
was used with constants C = 803 [fsw], A = 20.15 [fsw], and x = 0.7476, where D denotes the depth [fsw] and t the no-stop time [minutes].
In [1] it was mentioned that "The RDP M0-values were back-calculated from the no-stop curve". Furthermore, it was mentioned that "For the record, when M-values are calculated from the no-stop times the shape of the no-stop dive profile used for the calculation has an influence on the resulting M0-value. For the RDP we calculated these assuming a descent to depth at 60 fsw/min beginning at the beginning of bottom time, with the bottom time ending at the end of the time on bottom; the timed descent was used, not an instantaneous one." A final remark of importance in [1] was "Table III shows the M-values in the modern perspective which considers the atmosphere to be made up of 79.1% nitrogen".
Unfortunately it was not explained in [1] how the 'back-calculation' was done, and that is the reason for this question, as I wonder how to do this back-calculation.
My first thought was to consider the no-stop relation as a relation that 'guarantees' that when going back to the surface (at an ascent rate not faster than 60 fsw/min as was the standard in 1994) no DCS would occur. For any depth D one can determine the supersaturation upon surfacing after the above mentioned profile when the end of bottom time is given by the no-stop relation. The M0-value for a compartment then would be the maximal achievable partial pressure of the gas upon surfacing over all possible depths.
To be more precise, I considered the following evolution of the ambient pressure:
Descent: P_I(t) = 0.791 * ( 33 + 60 * t ) for 0 <= t <= D/60
Bottom: P_I(t) = 0.791 * ( 33 + D ) for D/60 <= t <= tstop
Ascent: P_I(t) = 0.791 * ( 33 + D + 60 * (tstop-t)) for tstop <= t <= tstop + D/60
where for any depth D the value of tstop is given by the no-stop relation. Also the initial condition P(0)=33 [fsw] was taken.
[taking this profile one obtains a Schreiner equation, followed by a Haldane equation, followed by an other Schreiner equation]
For any depth D, using the above expression for the ambient pressure, one can determine P(tstop+D/60). Maximizing this over all possible values for D might give the M0-value I conjectured. However, doing so (differentiating w.r.t. D, setting this to zero and solving for D, and substituting that value of D) leads to the values
107.57, 86.16, 67.71, 60.01, 55.85, 51.48, 49.22, 47.85, 46.93, 45.78, 45.07, 44.60, 43.81, 43.40
associated with the half times of
5, 10, 20, 30, 40, 60, 80, 100, 120, 160, 200, 240, 360, 480.
This perfectly reconstructs the M0-values as mentioned in [1] for half-times of 120 minutes and longer, but yields too high values for smaller half times.
Since the values from my first attempt are too high, I reasoned that slightly shortening the dive would reduce the values for the fast compartments, and hardly effect the slow compartments. As a second attempt I therefore decided to modify the profile to ensure that the dive was ended at tstop.
To be more precise, I considered the following evolution of the ambient pressure:
Descent: P_I(t) = 0.791 * ( 33 + 60 * t ) for 0 <= t <= D/60
Bottom: P_I(t) = 0.791 * ( 33 + D ) for D/60 <= t <= tstop - D/60
Ascent: P_I(t) = 0.791 * ( 33 + D + 60 * (tstop-t)) for tstop - D/60 <= t <= tstop
where again for any depth D the value of tstop is given by the no-stop relation. Also with the initial condition P(0)=33 [fsw].
Using the modified profile, I obtained the values
98.50, 82.56, 66.88*, 59.74, 55.73, 51.44, 49.21, 47.85, 46.93, 45.77*, 45.07, 44.60, 43.81, 43.40
which perfectly reconstructs the M0-values as mentioned in [1] (with the small remark that for compartments with half times of 20 and 160 minutes I would have to round instead of floor to reproduce the numbers in [1]), except of those of the two fastest compartments with half-times of 5 and 10 minutes. In this second attempt the values for the two slowest tissues are too small (should have been 99.08 and 82.63 according to [1]).
Though my second attempt better reconstructs the M0-values in [1], it is less consistent with the dive profile as described in [1]. I conjecture that I have to reduce the values in a different way. But how? Note that assuming an instant ascent leads even to higher values.
An other thing that is still puzzling me, which might be related, is that in [1], when discussing the no-stop relation it was mentioned that "A deep asymptote of 262 fsw was also calculated. This was the theoretical depth to which a dive could be made with zero bottom time". If I solve the no-stop relation together with the equation
t = D/60
I obtain D=276 [fsw]. So also the value of 262 fsw as mentioned in [1] I also could not explain...
[1] Hamilton Jr RW, Rogers RE, Powell MR (1994). "Development and validation of no-stop decompression procedures for recreational diving: the DSAT recreational dive planner"
Via this forum I found a reference to the paper [1], where it was explained that for deriving the RDP a Haldane model was used, which can be described by the ordinary differential equation
dP(t)/dt = -k ( P_I(t) - P(t) )
where P(t) denotes the partial pressure of the gas in the compartment, P_I(t) the ambient partial pressure of the gas, and k = ln(2) / t_{0.5} is a constant with t_{0.5} denoting the half time of the compartment.
Furthermore, the following no-stop relation
D - A = C * t^(-x)
was used with constants C = 803 [fsw], A = 20.15 [fsw], and x = 0.7476, where D denotes the depth [fsw] and t the no-stop time [minutes].
In [1] it was mentioned that "The RDP M0-values were back-calculated from the no-stop curve". Furthermore, it was mentioned that "For the record, when M-values are calculated from the no-stop times the shape of the no-stop dive profile used for the calculation has an influence on the resulting M0-value. For the RDP we calculated these assuming a descent to depth at 60 fsw/min beginning at the beginning of bottom time, with the bottom time ending at the end of the time on bottom; the timed descent was used, not an instantaneous one." A final remark of importance in [1] was "Table III shows the M-values in the modern perspective which considers the atmosphere to be made up of 79.1% nitrogen".
Unfortunately it was not explained in [1] how the 'back-calculation' was done, and that is the reason for this question, as I wonder how to do this back-calculation.
My first thought was to consider the no-stop relation as a relation that 'guarantees' that when going back to the surface (at an ascent rate not faster than 60 fsw/min as was the standard in 1994) no DCS would occur. For any depth D one can determine the supersaturation upon surfacing after the above mentioned profile when the end of bottom time is given by the no-stop relation. The M0-value for a compartment then would be the maximal achievable partial pressure of the gas upon surfacing over all possible depths.
To be more precise, I considered the following evolution of the ambient pressure:
Descent: P_I(t) = 0.791 * ( 33 + 60 * t ) for 0 <= t <= D/60
Bottom: P_I(t) = 0.791 * ( 33 + D ) for D/60 <= t <= tstop
Ascent: P_I(t) = 0.791 * ( 33 + D + 60 * (tstop-t)) for tstop <= t <= tstop + D/60
where for any depth D the value of tstop is given by the no-stop relation. Also the initial condition P(0)=33 [fsw] was taken.
[taking this profile one obtains a Schreiner equation, followed by a Haldane equation, followed by an other Schreiner equation]
For any depth D, using the above expression for the ambient pressure, one can determine P(tstop+D/60). Maximizing this over all possible values for D might give the M0-value I conjectured. However, doing so (differentiating w.r.t. D, setting this to zero and solving for D, and substituting that value of D) leads to the values
107.57, 86.16, 67.71, 60.01, 55.85, 51.48, 49.22, 47.85, 46.93, 45.78, 45.07, 44.60, 43.81, 43.40
associated with the half times of
5, 10, 20, 30, 40, 60, 80, 100, 120, 160, 200, 240, 360, 480.
This perfectly reconstructs the M0-values as mentioned in [1] for half-times of 120 minutes and longer, but yields too high values for smaller half times.
Since the values from my first attempt are too high, I reasoned that slightly shortening the dive would reduce the values for the fast compartments, and hardly effect the slow compartments. As a second attempt I therefore decided to modify the profile to ensure that the dive was ended at tstop.
To be more precise, I considered the following evolution of the ambient pressure:
Descent: P_I(t) = 0.791 * ( 33 + 60 * t ) for 0 <= t <= D/60
Bottom: P_I(t) = 0.791 * ( 33 + D ) for D/60 <= t <= tstop - D/60
Ascent: P_I(t) = 0.791 * ( 33 + D + 60 * (tstop-t)) for tstop - D/60 <= t <= tstop
where again for any depth D the value of tstop is given by the no-stop relation. Also with the initial condition P(0)=33 [fsw].
Using the modified profile, I obtained the values
98.50, 82.56, 66.88*, 59.74, 55.73, 51.44, 49.21, 47.85, 46.93, 45.77*, 45.07, 44.60, 43.81, 43.40
which perfectly reconstructs the M0-values as mentioned in [1] (with the small remark that for compartments with half times of 20 and 160 minutes I would have to round instead of floor to reproduce the numbers in [1]), except of those of the two fastest compartments with half-times of 5 and 10 minutes. In this second attempt the values for the two slowest tissues are too small (should have been 99.08 and 82.63 according to [1]).
Though my second attempt better reconstructs the M0-values in [1], it is less consistent with the dive profile as described in [1]. I conjecture that I have to reduce the values in a different way. But how? Note that assuming an instant ascent leads even to higher values.
An other thing that is still puzzling me, which might be related, is that in [1], when discussing the no-stop relation it was mentioned that "A deep asymptote of 262 fsw was also calculated. This was the theoretical depth to which a dive could be made with zero bottom time". If I solve the no-stop relation together with the equation
t = D/60
I obtain D=276 [fsw]. So also the value of 262 fsw as mentioned in [1] I also could not explain...
[1] Hamilton Jr RW, Rogers RE, Powell MR (1994). "Development and validation of no-stop decompression procedures for recreational diving: the DSAT recreational dive planner"