buhlman msw to bar problem?

[marc.collin]

hi

i'm trying to convert this exemple in bar

A diver begins an air dive by descending to a depth of 30 meters of seawater (msw)very rapidly.
Let's assume that it was an "instantaneous" descent. Now he is onthe bottom at 30 msw.
What is the "No Stop Time" or "No Decompression Limit" forthis depth?

Note: Remember all pressures are ABSOLUTE!

Assuming the dive is from sea level, the ambient pressure is the depth pressure of30 msw plus the surface barometric pressure of 10 msw:
Pamb = 30 msw + 10 msw = 40 msw

The inspired pressure is the ambient pressure minus the water vapor pressure(Buhlmann value) times the fraction of inert gas:
Pi = (Pamb - PH2O)*FN2 = (40 msw - 0.627 msw)*0.79 = 31.1 msw

The initial compartment pressure will be due to breathing atmospheric air at thesurface for a long time (first dive of the day):

Po = (Pamb - PH20)*FN2 = (10 msw - 0.627 msw)*0.79 = 7.4 msw

The surfacing M-values will vary by compartment.

See Table 2 in my"Understanding M-values" article.

Example: for Compartment No. 2 using theBuhlmann ZH-L16B M-values, Mo = 25.4 msw.

The time constant, k, for Compartment No. 2 is the natural logarithm of 2 dividedby the half-time:

k (Cpt 2) = ln2/half-time = 0.693/8 min = 0.086625

Now we can compute the "No Stop Time" or NDL for this compartment: t = (-1/k)*ln[(Pi - Mo)/(Pi - Po)] t (Cpt 2) = (-1.0/0.086625)*ln[(31.1 - 25.4)/(31.1 - 7.4)] t (Cpt 2) = 16.45 minutes

bulhman zhl16c


bulhman zhl16c mvalue in bar



bulhman zhl16c mvalue in msw


coefficient a same to be the mvalue

if we take compartment 2

in msw we have 25,4
in bar we have 1

1 msw = 0.100693064 bar

so instead of 1 bar we shoud had 2.5576038256 bar, no?

so i tried to convert the previous example in bar


via C++ Shell i tried

// Example program
#include <iostream>
#include <string>
#include <math.h>


int main()
{
float seaAtmPressure=1.00693064;
float sea30Pressure= 4.02079192;
float waterPressure = 0.0627;
float n2=0.79;
float pi = (sea30Pressure - waterPressure) *n2;
float mo=2.55;
float po=(seaAtmPressure - waterPressure) * n2;

//log(2)=0.693
//8 value of compartment 2
float k = log(2) / 8;

float t = (-1/k) * log((pi-mo)/(pi-po));
printf("k = %f\n ", k);
printf("pi = %f\n ", pi);
printf("mo = %f\n ", mo);
printf("po = %f\n ", po);
printf("t = %f\n ", t);

}

i get t = 16.361300 because i used 2.55 bar instead of the 1 bar from buhlman table.

my calcul is bad or is the bulhman table who have a problem?

 

[EFX]

i'm trying to convert this exemple in bar
coefficient a same to be the mvalue
if we take compartment 2:in msw we have 25,4; in bar we have 1

1 msw = 0.100693064 bar

so instead of 1 bar we shoud had 2.5576038256 bar, no?

so i tried to convert the previous example in bar
via C++ Shell i tried

From Stuart Morrison's DIY "Decompression" the A, B, and C Mo values are: A are the original set not intended for practical use, B is used for software and table generation, and C is used for incorporation into dive computers. With your calculation I would use the B set which for compartment 2 is 25.4 m.

Using 0.101325 bar/msw I get:

1 msw = 0.101325 bar (you got 0.100693064 bar).
25.4 msw = 2.57 bar (you got 2.55, probably close enough).
sea30Pressure = 4.053 (you got 4.02079192, close).

i used 2.55 bar instead of the 1 bar from buhlman table.

The 2.55 bar (25.4 msw) in your program is the Mo for compartment 2 and is correct. The slight difference in time is probably due to the slight difference in variables used.

I used your program with my variables and got t = 16.351606.

 

[Basking Ridge Diver]

Not sure if you have read this but it might help - http://www.frogkick.dk/library/deco/Erik Baker/decolessons.pdf

I pulled it from - Decompression Theory | Links and Resources | Technical Scuba Diving - Andy Davis website

 

[EFX]

Yes. I've read it before. Thanks.

Your article and a few others (ex. DIY Decompression) really helped me understand deco theory. Fortunately, I don't have any trouble with the math which is just algebra. I've tried to get through Bruce Wienke and Timothy O'Leary's paper on the RGBM with equations but the calculus is too much. I had it in college but haven't used it for 40 years. There is the same paper without formulas which is quite informative. A variation of the RGBM (Reduced Gradient Bubble Model) is used in my Mares Puck dive computer.

 

[Basking Ridge Diver]

Sorry that was aimed at Marc.Collin...