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Bruce, this is good for the limits of altitude diving, generally up to
10,000 feet.  The actual equation is:

P = Po * e^-[(M*g*h)/(kB*T)]

Where M is the molecular mass of air
g is the acceleration due to gravity
h is the altitude
kB is the Boltzman constant
T is the absolute temperature

this will reduce to:

P = Po * e^ -0.0383*(h/1000)  (for imperial units-feet)


Brad

Bruce Stewart wrote:
> 
> Brad,
> 
> if the below is correct theres zero pressure at the summit
> of Everest!
> 
> B
> 
> At 11:01 AM 3/30/01 -0800, bgilmore@at*.ne* wrote:
> >For a back of envelope calculation
> >
> >Atmospheric pressure drops approximately 1 inHg per 1,000 ft. of rise in
> >elevation. Sea level atmospheric pressure is 29.92 inHg
> >
> >For example:
> >Diving at a lake with an elevation of 3150’ MSL
> >3150 ft  = 3.15 inHg
> >Therefore, the correction factor will be:
> >(29.92 ­ 3.15)/ 29.92 = .895
> >And
> >...895 * 14.7 psia = 13.1 psia
> >...895 * 1013 mBar = 907 mBar
> >
> >This is good enough for Decoplanner.
> >
> >Brad
> >
> >
> >--
> >Send mail for the `techdiver' mailing list to `techdiver@aquanaut.com'.
> >Send subscribe/unsubscribe requests to `techdiver-request@aquanaut.com'.
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