Can someone give a very very layman's explanation on how the halftime compartment works? I have read the encylopedia and the DM manual 3 times and still cannot full undestand it.
Halftime compartments
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Axua:
Hmm. Could you perhaps narrow down your request to specificly what you don't understand?
James
miketsp
Contributor
Try this article:
http://scuba-doc.com/halftimesreal.html
http://scuba-doc.com/halftimesreal.html
Imagine a cylindrical bucket a foot tall with a small hole in the bottom.
Imagine shoving that bucket 8 inches into the water.
Time when the water reaches 4 inches deep in the bucket - that is the bucket's "half-time" - the time it takes the bucket to get half the water in it that it can get in it. For this example, let's assume that the time it took was 10 minutes.
If you leave the bucket in the same position for another half-time - that is, another 10 minutes - it'll fill half of the distance that's left between the water level outside the bucket and the water level inside the bucket, or another two inches; another 10 minutes and it'll fill half of the two inches that's left, or another inch. So with an initial pressure difference of 8 inches of water, after one half-time you'll have 4 inches of water in the bucket, after two half-times you'll have 6 inches, and after three half-times you'll have 7 inches. After six half-times you'll only be 1/8 inch shy of the 8 inches possible, or your bucket is essentially as full as it can get at that depth - "saturated."
---
Ok, let's look at going the other direction after only a single half-time of submersion. If after 10 minutes you then take the bucket out of the water, there will be four inches of water in it, and it'll start draining out the same hole where it filled. But the rate will be lower, because you'll be starting with only 4 inches of water pressure forcing water through the hole instead of eight inches. So, after 10 minutes of draining, instead of having an empty bucket, you'll have drained only half the water, and still have 2 inches left - after two half-times (20 minutes) you'll still have one inch left, and so on.
As you can see, even though it only took 10 minutes to get 4 inches of water into the bucket, it'll take quite a bit longer than that to drain it all back out, because the relative pressure forcing it out is much lower than the pressure that forced in in.
Now, substitute "10 minute tissue compartment" for the bucket, and substitute "Nitrogen" for the water, and you have the theory of tissue half-times down pat.
Rick
Imagine shoving that bucket 8 inches into the water.
Time when the water reaches 4 inches deep in the bucket - that is the bucket's "half-time" - the time it takes the bucket to get half the water in it that it can get in it. For this example, let's assume that the time it took was 10 minutes.
If you leave the bucket in the same position for another half-time - that is, another 10 minutes - it'll fill half of the distance that's left between the water level outside the bucket and the water level inside the bucket, or another two inches; another 10 minutes and it'll fill half of the two inches that's left, or another inch. So with an initial pressure difference of 8 inches of water, after one half-time you'll have 4 inches of water in the bucket, after two half-times you'll have 6 inches, and after three half-times you'll have 7 inches. After six half-times you'll only be 1/8 inch shy of the 8 inches possible, or your bucket is essentially as full as it can get at that depth - "saturated."
---
Ok, let's look at going the other direction after only a single half-time of submersion. If after 10 minutes you then take the bucket out of the water, there will be four inches of water in it, and it'll start draining out the same hole where it filled. But the rate will be lower, because you'll be starting with only 4 inches of water pressure forcing water through the hole instead of eight inches. So, after 10 minutes of draining, instead of having an empty bucket, you'll have drained only half the water, and still have 2 inches left - after two half-times (20 minutes) you'll still have one inch left, and so on.
As you can see, even though it only took 10 minutes to get 4 inches of water into the bucket, it'll take quite a bit longer than that to drain it all back out, because the relative pressure forcing it out is much lower than the pressure that forced in in.
Now, substitute "10 minute tissue compartment" for the bucket, and substitute "Nitrogen" for the water, and you have the theory of tissue half-times down pat.
Rick
Great analogy, Rick. I visualize half times as exponential curves on cartesian planes.. it's sometimes difficult for me to realize that not everyone works that way 
R
redacted
Guest
Rick Murchison:
I would not have thought of this example as an exponential process, but it sounds right. Now now wife is probably going to be ticked when she sees her bucket.
pete340
Contributor
awap:
They're essentially the same thing: the rate of flow is proportional to the pressure difference in this example and in tissue ongassing and offgassing.
miketsp
Contributor
pete340:
The analogy will serve but if I remember rightly, flow through a single orifice is proportional to the square root of the pressure difference.
To a first order approximation:
Flow rate = (flow coeff) x (orifice area) x sqroot(2 x (pressure drop)/(fluid density))
Don't ask me for the units. The last time I played with this was about 36 years ago.
miketsp
Contributor
As for gas diffusion, there is a very nice presentation here:
http://physiology.usouthal.edu/Handouts/Gas_exchange.ppt
really worth looking at.
http://physiology.usouthal.edu/Handouts/Gas_exchange.ppt
really worth looking at.
pete340
Contributor
miketsp:
Could be. But that's for a small orifice, isn't it?
[Edit:] Never mind. I shouldn't try to do physics when I'm tired.
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