Decompression Model [FR]MN90 - SC?

[Hcklo]

Hey guys, I am quite a noobie in theory of decompression. I arrived into a point where the tables of the French National Marine provides some SC for the differents compartiments. And they use those SC to caculate the deco stops.
My question is, how did they find the SC for each compartment? What kind of calculation and empirical research they have done? Maybe some of you would have some litterature about this?

Thank you for your help and kindness !

 

[dmaziuk]

What is "SC"? The numbers look similar to M0 values in bar, close to DSAT only noticeably more conservative in the fast compartments.

DSAT:
1 : { "t" : 5.0, "M0" : 3.042 },
2 : { "t" : 10.0, "M0" : 2.537 },
3 : { "t" : 20.0, "M0" : 2.054 },
4 : { "t" : 30.0, "M0" : 1.834 },
5 : { "t" : 40.0, "M0" : 1.711 },
6 : { "t" : 60.0, "M0" : 1.579 },
7 : { "t" : 80.0, "M0" : 1.511 },
8 : { "t" : 100.0, "M0" : 1.469 },
9 : { "t" : 120.0, "M0" : 1.441 },

However to calculate deco stops you also need the "delta-M": the M-value is a line "M = M0 + Depth * delta-M"

 

[Hcklo]

So in the N4 french level certification to calculate the deco stop they teach


tn2 initial + (gradient) x périod% =
tn2 final after périod

Sc = tn2fin / PA
PA = tn2fin / sc

The gradient is defined as
G= Tension Final - Tension initial
Sc is m-value
PA is absolute pressure

In the picture that I sent the MN90 stated for example the m value of c5 as 2.72.
My question is what kind of research is done to reach this value ?

Indeed it seems to be similar as the M0 values of DSAT, maybe would you have some information about how DSAT reached those values ?

I am completly noobie in theory of decompression, just here because of curiosity.

Thank you for your time !


(Example of calculation so maybe it's easy to translate the terms and understand how they proceed....

Dive at 50m during 20min

T N2 initial 80% x 1 bar = 0.8b
T n2 final 80% x 6 bar = 4.8b
Gradient = 4.8 - 0.8 = 4b


C5 = 4 periods 93.75%
0.8 + 4 x 0.9375 = 4.55b tension n2 after 4 periods

C10 = 2 periods 75%
0.8 + 4 x 0.75 = 3.8b tension n2 after 2 periods

C20 = 1 period 50%
0.8 + 4 x 0.5 = 2.8 tension n2 after 2 periods

M.value = tn2 final ÷ absolute pressure
Absolute pressure = tn2 final ÷ m.value


For C5:
Absolute pressure = 4.55÷2.72= 1.67b
= 6.7m
= stop at 9m

For c10
Abs press = 3.8÷2.38=1.59b
=5.9m
= stop at 6m

For c20
Abs press = 2.8÷2.04 = 1.37b
= 3.7m
= stop at 6m

 

[dmaziuk]

DSAT report is available from Rubicon foundation and is, IMO, a required read. However for M-values specifically you should read Erik Baker's "understanding M-values" (google it, pdf is available from multiple sources).

As to how they actually arrived at those particular values: good question. The only answer I know is for Buhlmann's ZH-L16, no idea how others calculated theirs.

 

[Hcklo]

Thank you for your recommendations I will take a look about those publications ! Cheers mate !!