Quote: (Originally Posted by
Freef)
You're not wrong, just not quite right. As you said in the first post, for a given pO2 there is a certain volume of O2 in the loop-22.05L for a pO2 of 2.1. You should have gone on to see that this is also true for the lower pO2 I used in my calculations [1.3] which gives a loop volume of O2 of 13.65L.
So at whatever depth you are at, to get down from 2.1 [22.05L] to 1.3 [13.65L] will take the 8 min 24 sec at a vO2 of 1. The amount of N2 in the loop makes no difference in this case, and you have to use 8.4 L of O2 to drop the pO2.
When you were looking at the lower pO2/fO2 figure did you remember to multiply back up to get the depth? I forgot when I first started running the figures
Sorry Freef, but my calculations are correct, I dont know if we misunderstand each otherbut......
The whole point is that the amount of O2 you have to metabolize is not canstant, it is dependant on the
FIXED AMOUNT of N2 in the loop, since this is a
"diluter" of the O2 and therefore also the PO2.
We agree on the the PPO-calculations for the high loop, wich contains the same amount of O2 at each depth. For illustration I added the N2 (Could also just be Total inertgasses, makes no difference). This is very simple, since we have the amount of O2:
2.1 at 15m = fO2 of 84.0% = 22.05L of O2. + 4,2 l N2
2.1 at 20m = fO2 of 70.0% = 22.05L of O2 + 9,45l N2.
2.1 at 30m = fO2 of 52.5% = 22.05L of O2 + 19,95l N2.
2.1 at 60m = fO2 of 30.0% = 22.05L of O2. + 51,45l N2
Now we can calculate the fO2 at PPO=1.3 easily enough,
but the main point is that this does not scale linearly with volume of O2. Note: You cannot just cut the amount of O2 by half and then exspect the drop in PPO to be halfed also. It nearly works when the FO2 your reducing is very low (at depth), but it does not work when the mix is rich! Case in point: using 11 l. of a pure O2 loop, does not affect the loop-PPO at all.
For at given FO2 with a
FIXED AMOUNT of N2 you actaully have to do a little math. I use the formula FO2 = VO2 / Vol_Total = VO2 / (VO2+VN2).
So we start by calculating Fo2 at PPO=1.3:
1.3 at 15m = fO2 of 52,0%
1.3 at 20m = fO2 of 43,3%
1.3 at 30m = fO2 of 32,5%
1.3 at 60m = fO2 of 18,5%
For each depth the following will hold true:
FO2 = VO2 / (VO2+VN2), where FO2 and VN2 is known, you therefor solve for VO2:
VO2 = FO2*(VO2+VN2) <=> VO2 = FO2*VN2 /(1-FO2)
So you end up with:
1.3 at 15m = fO2 of 52,0% => 4,55l O2 + 4,2 LN2 (ie. FO2=52%)
Therefor you have to Metabolise 17,5 l O2
1.3 at 20m = fO2 of 43,3% => 7,25l O2 + 9,45l N2 (ie. FO2=43,3%)
Therefor you have to Metabolise 14,8 l O2
etc.
Still think you are wrong freef, but perhaps we are misunderstanding each other?
Its actually quite logical when you think about it...
Best regards
Nicolai