Re: [Phys-l] Atmospheric pressure calculus deduction

[John Denker <jsd@av8n.com>]

On 06/23/2008 09:51 AM, Robert Cohen wrote:

> This seems needless complicated.

Agreed.
 
> If you want to "show how atmospheric pressure can be computed from
> the weight of a column air", that seems to imply that you already
> know the weight of the air column.  If so, the air pressure is
> straightforward as long as you assume hydrostatic balance: 

Forsooth, if you already know the weight of the air column, you 
don't need calculus and you don't even need to assume hydrostatic 
balance.  All you need to assume is mechanical equilibrium, in which
case
     Pressure = (weight of air column above you) / (area of air column)

This is the ultra-simple case.

> the air pressure is
> straightforward as long as you assume hydrostatic balance: dP/dz = -
> density*g (i.e., gravitational force is balanced by pressure gradient
> force).
>
> In that case, integrate from surface to infinity (where pressure is
> zero) to get that P = Weight/Area.

Again:  If you already know the weight of the air column, you can go
directly to P = Weight/Area ... without any derivatives or integrals,
and without knowing the density everywhere or anywhere.

=================

As for the not-so-simple case:

In the real world, in aviation in particular, you often want to know
the weight of the air column above height H (i.e. the pressure-versus-H
profile) and you don't have reliable _a priori_ information about that, 
since it changes with the weather ... hence the need for parameterized 
models of the atmosphere.