Re: pinhole Sierra questions

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From: Paul Doherty (pauld@exploratorium.edu)
Date: Thu Jun 27 2002 - 15:32:01 PDT 

Message-Id: <l0311070fb9414065f535@[192.174.3.125]>
Date: Thu, 27 Jun 2002 15:32:01 -0700
From: Paul Doherty <pauld@exploratorium.edu>
Subject: Re: pinhole Sierra questions

Note that to 20% 5.6 km is 15,000 feet

I was giving the easy to remember rule that the atmospheric pressure
decreases 1 inch of mercury per thousand feet (which is precise.)

Using this linear approximation you reach the half height pressure at
15,000 feet.

Paul Hewitt was giving the half-height for the exponential model of the
atmosphere which is more correct. However the expopnential model depends on
temperature and latitude and so can vary by over 1000 feet.

For a "standard" atmosphere at mid latitudes Paul H's 5.6 km or 18,000 ft.
half-height is more precise.

My approximation is easy to remember.

So we both are right within the models we were using.

Paul D

>Paul,
>
>I was just watching the Paul Hewitt video on gases, and he said that
>at an altitude of 5.6 km in height, one-half of the atmosphere will
>be underneath you. In a previous pinhole e-mail, you said that at
>15,000 feet, you're above half the atmosphere.
>
>Paul vs Paul . . . the battle to the final answer?
>
>- Geoff
>
>
>>The simple rule of thumb for lower elevations (below 15,000 feet (Which is
>>great for the continental US for which the high point, Mt. Whitney is below
>>15,000 feet.) is to subtract an inch of mercury for every 1000 feet of
>>elevation gain.
>>
>>So at 15,000 feet you subtract 15 inches of mercury from sea level 30
>>inches of mercury and find you are above 1/2 of the atmosphere.
>
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