OP
Arie0510
Registered
By now I have been able to exactly reproduce all M0-values for tissues with half times of 20 minutes and more. Only for tissues with half times of 10 and 5 minutes I obtain slightly smaller values than presented in [1].
For those of you interested in reproducing (almost all of) the M0-values presented in [1], I summarize below the approach I followed.
Starting point is the differential equation dP(t)/dt = -k ( P_I(t) - P(t) ) where P(t) denotes the pressure in the compartment, P_I(t) the ambient pressure, and k = ln(2) / t_{0.5} is a constant with t_{0.5} denoting the half time of the compartment.
Start with initial pressure P(0)=33 [fsw], descent at a rate of 60 [fsw/min] to a depth D. Stay at depth D for some time, and subsequently ascent at a rate of 60 [fsw/min], surfacing at time t=tsurf. Using a Schreiner equation, followed by a Haldane equation, followed by an other Schreiner equation one can derive that the pressure at surfacing is given by:
P(tsurf) = 33 + 60 * t_{0.5} / ln(2) * ( 2^( - tsurf / t_{0.5} ) - 2^( D / ( 60 * t_{0.5}) ) * ( 1 - 2^( D / ( 60 * t_{0.5} ) ) )
If in this expression one takes for the surfacing time the maximal no-stop time, i.e.,
tsurf = ( ( D - A ) / C )^( - 1 / x ), where A=20.15, C=803, x=0.7476
one can consider P(tsurf) as a function of the depth D (for a given tissue half time t_{0.5} ).
If you want, you can plot this function for D>20.15. This function is increasing for increasing D, until a maximum is reached, after which the function decreases again. In particular for a given half time t_{0.5} one can therefore determine the depth D for which this function is maximal. Determine this optimal depth D and round the value to two digits behind the comma. For this optimal depth, determine P(tsurf) and floor the outcome to two digits behind the comma. This could be interpreted as an M0-value for air, so to obtain an M0-value for nitrogen, multiply the outcome by 0.791 and round again to two digits behind the comma.
Following this procedure on gets
As you can see the M0-values are exactly the same as presented in [1] for tissues with half times of 20 minutes and more. However, in [1] they obtained for a half time of 5 minutes a value M0=99.08, and for 10 minutes: M0=82.63. I was unable to explain this difference...
Any other approach I tried was not able to exactly reproduce the M0-values of the other tissues.
[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"
For those of you interested in reproducing (almost all of) the M0-values presented in [1], I summarize below the approach I followed.
Starting point is the differential equation dP(t)/dt = -k ( P_I(t) - P(t) ) where P(t) denotes the pressure in the compartment, P_I(t) the ambient pressure, and k = ln(2) / t_{0.5} is a constant with t_{0.5} denoting the half time of the compartment.
Start with initial pressure P(0)=33 [fsw], descent at a rate of 60 [fsw/min] to a depth D. Stay at depth D for some time, and subsequently ascent at a rate of 60 [fsw/min], surfacing at time t=tsurf. Using a Schreiner equation, followed by a Haldane equation, followed by an other Schreiner equation one can derive that the pressure at surfacing is given by:
P(tsurf) = 33 + 60 * t_{0.5} / ln(2) * ( 2^( - tsurf / t_{0.5} ) - 2^( D / ( 60 * t_{0.5}) ) * ( 1 - 2^( D / ( 60 * t_{0.5} ) ) )
If in this expression one takes for the surfacing time the maximal no-stop time, i.e.,
tsurf = ( ( D - A ) / C )^( - 1 / x ), where A=20.15, C=803, x=0.7476
one can consider P(tsurf) as a function of the depth D (for a given tissue half time t_{0.5} ).
If you want, you can plot this function for D>20.15. This function is increasing for increasing D, until a maximum is reached, after which the function decreases again. In particular for a given half time t_{0.5} one can therefore determine the depth D for which this function is maximal. Determine this optimal depth D and round the value to two digits behind the comma. For this optimal depth, determine P(tsurf) and floor the outcome to two digits behind the comma. This could be interpreted as an M0-value for air, so to obtain an M0-value for nitrogen, multiply the outcome by 0.791 and round again to two digits behind the comma.
Following this procedure on gets
| t_{0.5} | D | P(tsurf) | M0 |
| 5 | 134.11 | 124.53 | 98.50 (*) |
| 10 | 105.02 | 104.37 | 82.56 (*) |
| 20 | 74.23 | 84.56 | 66.89 |
| 30 | 57.92 | 75.52 | 59.74 |
| 40 | 48.38 | 70.45 | 55.73 |
| 60 | 38.37 | 65.03 | 51.44 |
| 80 | 33.43 | 62.21 | 49.21 |
| 100 | 30.56 | 60.49 | 47.85 |
| 120 | 28.71 | 59.33 | 46.93 |
| 160 | 26.47 | 57.87 | 45.78 |
| 200 | 25.17 | 56.98 | 45.07 |
| 240 | 24.32 | 56.39 | 44.60 |
| 360 | 22.94 | 55.39 | 43.81 |
| 480 | 22.27 | 54.87 | 43.40 |
As you can see the M0-values are exactly the same as presented in [1] for tissues with half times of 20 minutes and more. However, in [1] they obtained for a half time of 5 minutes a value M0=99.08, and for 10 minutes: M0=82.63. I was unable to explain this difference...
Any other approach I tried was not able to exactly reproduce the M0-values of the other tissues.
[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"
