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mouthfill tips?

Thread Status: Hello , There was no answer in this thread for more than 60 days.
It can take a long time to get an up-to-date response or contact with relevant users.
Bill

Bill

Baron of Breathold
Oct 17, 2001
1,805
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Jim
I worked out two sets of numbers based on pressure differential. The first set is pushing as much as I care to and the second is a better plan. The deeper you go, the slower the pressure changes but there are many other factors in clearing.
max til clear 8', 19', 32', 48', 68', 93', 125', 165Õ, 215 & 277'
([email protected] good for 180.)
clr 5 11 17 25 34 44 56 69 84 102 123 146 174 204 241 282 330
Aloha
Bill
 
J

JimGlynn

New Member
Jan 16, 2002
278
19
0
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Bill and Eric, thanks for the info. I have always known that as one descends the need for mask eq. becomes less, but was never really sure if the same would hold true for ear eq. Is there a formula I can use based on the number of times I need to eq. the ears from the surface to say 100', that I can then extrapolate my eq. needs for deeper depths?
Jim
 
U

Uli

New Member
Sep 28, 2001
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so lets assume the airspace in your mask is "10" at the surface. At -10 metres, if you do not equalise, it will be "5". Then at -20 metres it will be "2.5", and -30 metres it will be "1.25". The volume is reduced by half at every 10 metre increment.

Oops, someone didn't read his SCUBA manual properly...
:naughty :D
Shouldn't it be 3.3 at -20m and 2.5 at -30m?

And the volume is of course not reduced by half at every 10 metre increment. This is only true for the first ten metres, all the deeper 10 metres increments will reduce the volume by less than half...

So Jim is perfectly right

Cheers
Uli
P.S.: Have sat in the back row of my elder brother's scuba course when I was ten years old... :cool:
 
Erik

Erik

Well-Known Member
Jan 21, 2001
4,731
753
218
Maybe I screwed up in my explanation? I see my numbers are wrong, sorrry.
From a depth of 10 metres to 20 metres, the decrease in volume is half of the volume at 10 metres. If you have a volume in your body of "10" at -10 metres, the volume will be "5" at -20 metres.
From a depth of 40 metres to 50 metres, the volume in the space AT 40 METRES will be reduced to .5 of the amount that you had at 40 metres, etc.

Every ATM of pressure reduces the volume by half every 10 metres because of the incrompressability of water. At 30 metres (99', 4 ATMS), you end up with either 4 times the amount of air used to equalise a space, or 1/4 the volume of the original size if you didn't equalise.
I am not saying that the decrease in volume is half of the original 1 ATM amount every 10 metres.....just half of the ceiling at 10 metres above you. It;s just that my math sucks ;)
If I believed that the amount was reduced as you suggested I said, then there would be "0" volume at -20 metres, which is obviously not the case.
Therefore, Jim (and all of us) will find that the amount needed to equalise becomes less and less as he (and we) descend.
Cheers,
Erik Y.
 
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E

efattah

Well-Known Member
Mar 2, 2001
3,294
488
173
Erik,

I think you should check your numbers...

As always we assume that the air is an ideal gas:
PV = nRT (ideal gas law)

V = nRT / P = K / P (K is just a constant)

Assume V = 100L at 1atm
V = K / P = K / 1atm = 100, thus K in this case is 100atm*L
V = 100atm*L / 1atm = 100L

Then at 2atm (-10m)
V = 100atm*L / 2atm = 50L

Then at 3atm (-20m)
V = 100atm*L/ 3 atm = 33.3L

Then at 4atm (-30m)
V = 100atm*L / 4 atm = 25L

Then at 5atm (-40m)
V = 100atm*L / 5 atm = 20L

Then at 6atm (-50m)
V = 100atm*L / 6 atm = 16.7L

Then at 7atm (-60m)
V = 100atm*L / 7atm = 14.3L


So, going from 40m to 50m, the volume change is 20L down to 16.7L, or a decrease of 3.3L:
3.3L / 20L = 0.165

Thus, going from 40m to 50m, the volume of air decreases by 16.5%. In fact, the decrease will be slightly less than that because at that depth air is no longer an ideal gas.

Notice that from 40m to 50m, the pressure doesn't double, but the pressure increases from 5atm to 6atm, at 20% increase in pressure. As the pressure increases by a factor of 6/5, the volume changes by a factor of 5/6.


Eric Fattah
BC, Canada
 
J

JimGlynn

New Member
Jan 16, 2002
278
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Does this law then follow that the E-Tubes need to be cleared less as one decends? Eric F., you have been deeper than anyone else here, if you were wearing a mask down to 300', do you have a formula you would use to compute how many times you would need to clear your ears?
Jim
 
Erik

Erik

Well-Known Member
Jan 21, 2001
4,731
753
218
Please form another line to the left to kick me in the nuts.
As I said earlier, my spatial thinking is lacking....I understand and can "feel" the theory, but apparently can't apply it to mathematics. It's like quantum physics....I can "see" it, but would never be able to map it out.
My apologies for bad info,
Erik Y.
 
A

Alun

Well-Known Member
Oct 5, 2001
763
83
118
yes that's true.
you can use the formula i gave you for the mask to work it out. you would have to do it iteratively, so it may take you a few mins to work out all the numbers.... to start you off....

d_1 = 3m
d_2 = 3m
d = 6.9

d_1 = 3
d_2 = 6.9
d = 11.97m

surface to first eq = 3m
1st to 2nd eq = 3.9m
2nd to 3rd = 5.07m

you can see how the spacings are increasing with depth

for the next one we have
d_1 = 3m
d_2 = 11.97m
d = 18.56m

the number and spacings depend only on d_1 - the depth you can reach from the surface when you need to equalise. it has to be a sensible depth, where you dont risk rupture

this is all theory remember...


....come to think of it, 3m is way too much for the first equalisation. it should be more like 1m really. 3m without equalising would bend your ears quite a lot. obviously, this would change all those figures - but the same principle aplpies.
 
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A

Alun

Well-Known Member
Oct 5, 2001
763
83
118
here's the formula again....

d = ( (d_1 + 10)(d_2 + 10) ) - 10
( --------------------------- )
( 10 )

hey, my theoretical physics phd wasn't a waste of time after all! :)
 
Bill

Bill

Baron of Breathold
Oct 17, 2001
1,805
332
188
82
Jim
I'll leave formulae to those more learned than I but let me remind you of one 'real world' fact. If you clear too much it will cost you one or two percent of your dive time. If you fall behind on clearing, you will have to stop or slow to catch up. Got a very good lesson a few hours ago when I not only had to stop at 190, I had to turn head up to get cleared enough for 200 (bragging rites for the day). How much do you think that affected the dive?
Aloha
Bill
 
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