"jim rosinski" wrote in message
oups.com...
Sriram Narayan wrote:

That still wouldn't help since the pressure change for a 1000ft
change in
altitude at 18k would be smaller than at sea level. It would have to
have
some non-linear spring compensation as a function of absolute
pressure.

Yes but the non-linearity is very weak. I coded up the formula for the
US standard atmosphere (described below) and plotted it. See
http://www.burningserver.net/rosinsk...atmosphere.jpg
Perhaps the non-linearity is so weak that altimeters neglect it? I
don't know.

The formula for US standard atmosphere can be derived from the ideal
gas approximation (p = rho*R*T) and the hydrostatic approximation
(dp/dz = -rho*g). The final equation is:

z = T0/gamma * (1 - (p/p0)**(R*gamma/g)))

where R=287, T0 = 288K, p0 = 1013.25 mb = 29.92 in, gamma = 6.5 deg/km,
g = 9.8.

The formula assumes that temperature decreases linearly with altitude,
an assumption which becomes invalid above the tropopause. The equation
and its derivation can be found in Wallace & Hobbs, "Atmospheric
Science", pg. 60-61.

Jim Rosinski


It doesn't look that linear to me. I found a website with a similar graph
and it appears that at sea level and at 10000ft the slope of the curve is at
least 2x different. Your curve is quite a bit more linear (maybe 20%
increase in slope at 10k). There must some sort of mechanical compensation
involved otherwise altimeters would be off quite a bit even at 10k (even
with your curve). Isn't it something like 75ft accuracy requirement for
altimeters?

http://www.atmosphere.mpg.de/enid/16h.html